Disha NEET Physics Guide for classes 11 and 12.pdf

NEET questions bank for physics

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Typeset by Disha DTP Team
(ii)
By
Sanjeev Kumar Jha

3. 1. Physical World 1-7
2. Units and Measurements 8-34
3. Motion in a Straight Line 35-66
4. Motion in a Plane 67-98
5. Laws of Motion 99-138
6. Work, Energy and Power 139-172
7. System of Particles and Rotational Motion 173-210
8. Gravitation 211-240
9. Mechanical Properties of Solids 241-259
10. Mechanical Properties of Fluids 260-279
11. Thermal Properties of Matter 280-302
12. Thermodynamics 303-325
13. Kinetic Theory 326-345
14. Oscillations 346-372
15. Waves 373-400
Contents
(iii)

4. 16. Electric Charges and Fields 401-429
17. Electrostatic Potential and Capacitance 430-463
18. Current Electricity 464-511
19. Moving Charges and Magnetism 512-542
20. Magnetism and Matter 543-566
21. Electromagnetic Induction 567-587
22. Alternating Current 588-609
23. Electromagnetic Waves 610-623
24. Ray Optics and Optical Instruments 624-660
25. Wave Optics 661-684
26. Dual Nature of Radiation and Matter 685-710
27. Atoms 711-727
28. Nuclei 728-748
29. Semiconductor Electronics :
Materials, Devices and Simple Circuits 749-784
(iv)

5. WHAT IS SCIENCE ?
Science refers to a system of acquiring knowledge. This system
uses observation and experimentation to describe and explain
natural phenomena. The term science also refers to the organized
body of knowledge that people have gained using that system.
Less formally, the word science often describes any systematic
field of study or the knowledge gained from it.
The purpose of science is to produce useful models of reality.
Most scientific investigations use some form of the scientific
method.
The scientific method is a logical and rational order of steps by
which scientists come to conclusions about the world around
them. The Scientific method helps to organize thoughts and
procedures so that scientists can be confident in the answers
they find. Scientists use observations, hypotheses, and deductions
to make these conclusions.
Science as defined above, is sometimes called pure science to
differentiate it from applied science, which is the application of
research to human needs. Fields of science are commonly divided
into two major categories.
1.Natural science : The science in which we study about the
natural world. Natural science includes physics, chemistry,
biology, etc.
2.Social science : It is the systematic study of human behavior
and society.
WHAT IS PHYSICS ?
The word physics originates from a Greek word which means
nature. Physics is the branch of science that deals with the study
of basic laws of nature and their manifestation of various natural
phenomena.
There are two main thrusts in physics :
1.Unification : In physics, attempt is made to explain various
physical phenomena in terms of just few concepts and laws.
Attempts are being made to unify fundamental forces of
nature in the persuit of unification.
2.Reductionism. Another attempt made in physics is to explain
a macroscopic system in terms of its microsocopic
constituents. This persuit is called reductionism.
Keep in Memory
1.I
nformation received through the senses is called observation.
2.An idea that may explain a number of observations is called
hypothesis.
3.A hypothesis that has been tested many times is called
scientific theory.
4.A scientific theory that has been tested and has always
proved true is called scientific law.
SCOPE AND EXCITEMENT OF PHYSICS
The scope of physics is very vast. It covers a tremendous range
of magnitude of physical quantities like length, mass, time, energy,
etc. Basically, there are two domains of interest : macroscopic and
microscopic. The macroscopic domain includes phenomena at
the laboratory, terrestrial and astronomical scales. The microscopic
domain includes atomic, molecular and nuclear phenomena.
Classical physics deals mainly with macroscopic phenomena
consisting of the study of heat, light, electricity, magnetism, optics,
acoustics, and mechanics. Since the turn of the 20th century, the
study of quantum mechanics and relativity has become more
important. Today, physics is often divided into branches such as
nuclear physics, particle physics, quantum physics, theoretical
physics, and solid-state physics. The study of the planets, stars,
and their interactions is known as astrophysics, the physics of
the Earth is called geophysics, and the study of the physical laws
relating to living organisms is called biophysics.
Physics is exciting in many ways. Application and exploitation of
physical laws to make useful devices is the most interesting and
exciting part and requires great ingenuity and persistence of effort.
PHYSICS, TECHNOLOGY AND SOCIETY
Physics generates new technologies and these technologies have
made our lives comfortable and materially prosperous. We hear
terms like science, engineering , technology all in same context
though they are not exactly same.
1
Physical World

6. 2 PHYSICS
Scientists
Investigate the
natura
l world
Technology
The products and
processes created
by engineers
Scientific
knowled
ge
What scientists
have learned about
the natural world
Engineers
Create the
designed world

Values
Needs
Environment
Economy
Society
Science:
•A body of knowl
edge
•Seeks to describe and understand the natural world and its
physical properties
•Scientific knowledge can be used to make predictions
•Science uses a process--the scientific method--to generate
knowledge
Engineering:
•Design under constraint
•Seeks solutions for societal problems and needs
•Aims to produce the best solution from given resources and
constraints
•Engineering uses a process–the engineering design
process–to produce solutions and technologies
Technology:
•The body of knowledge, processes and artifacts that result
from engineering
•Almost everything made by humans to solve a need is a
technology
•Examples of technology include pencils, shoes, cell phones,
and processes to treat water
In the real world, these disciplines are closely connected.
Scientists often use technologies created by engineers to
conduct their research. In turn, engineers often use
knowledge developed by scientists to inform the design of
the technologies they create.
Link between technology and physics
Technology Scientific principle(s)
Aeroplane Bernoulli’s principle in fluid dynamics
Bose-Einstein condensate Trapping and cooling of atoms by laser beams and magnetic fields
Computers Digital logic
Electric generator Faraday’s laws of electromagnetic induction
Electron microscope Wave nature of electrons
Fusion test reactor (Tokamak) Magnetic confinement of plasma
Giant Metrewave Radio Telescope (GMRT) Detection of cosmic radio waves
Hydroelectric power Conversion of gravitational potential energy into electrical energy
Lasers Light amplification by stimulated emission of radiation
Non-reflecting coatings Thin film optical interference
Nuclear reactor Controlled nuclear fission
Optical fibres Total internal reflection of light
Particle accelerators Motion of charged particles in electromagnetic fields
Photocell Photoelectric effect
Production of ultra high magnetic fieldsSuperconductivity
Radio and television Generation, propagation and detection of electromagnetic waves
Rocket propulsion Newton’s laws of motion
Sonar Reflection of ultrasonic waves
Steam engine Laws of thermodynamics

7. 3Physical World
Som
e physicists from different countries of the world and their major contributions
Name of physicists Major contribution/discovery Country of origin
Abdus Salam Unification of weak and electromagnetic interactions Pakistan
Albert Einstein Explanation of photoelectric effect; Theory of relativityGermany
Archimedes Principle of buoyancy; Principle of the lever Greece
C.H. Townes Maser; Laser U.S.A.
Christiaan Huygens Wave theory of light Holland
C.V. Raman Inelastic scattering of light by molecules India
Edwin Hubble Expanding universe U.S.A.
Enrico Fermi Controlled nuclear fission Italy
Ernest Orlando Lawrence Cyclotron U.S.A.
Ernest Rutherford Nuclear model of atom New Zealand
Galileo Galilei Law of inertia Italy
Heinrich Rudolf Hertz Generation of electromagnetic waves Germany
Hideki Yukawa Theory of nuclear forces Japan
Homi Jehangir Bhabha Cascade process of cosmic radiation India
Issac Newton Universal law of gravitation; Laws of motion; Reflecting telescope U.K.
James Clark Maxwell Electromagnetic theory; Light - an electromagnetic wave U.K.
James Chadwick Neutron U.K.
J.C. Bose Ultra short radio waves India
J.J. Thomson Electron U.K.
John Bardeen Transistors; Theory of super conductivity U.S.A.
Lev Davidovich Landau Theory of condensed matter; Liquid helium Russia
Louis Victor de Broglie Wave nature of matter France
Marie Sklodowska Curie Discovery of radium and polonium; Studies on natural radioactivity Poland
Michael Faraday Laws of electromagnetic induction U.K.
M.N. Saha Thermal ionisation India
Niels Bohr Quantum model of hydrogen atom Denmark
Paul Dirac Relativistic theory of electron; Quantum statistics U.K.
R.A. Millikan Measurement of electronic charge U.S.A.
S. Chandrashekhar Chandrashekhar limit, structure and evolution of stars India
S.N. Bose Quantum statistics India
Victor Francis Hess Cosmic radiation Austria
Werner Heisenberg Quantum mechanics; Uncertainty principle Germany
W.K. Roentgen X-rays Germany
Wolfgang Pauli Exclusion principle Austria Example 1. Does imagination play any role in physics?
Solution : Yes, imagination has clayed an important role in the
development of physics. Huygen’s principle, Bohr’s theory,
Maxwell equations, Heisenberg’s uncertainty principle, etc. were
the imaginations of the scientists which successfully explained
the various natural phenomena.
Example 2. How is science different from technology?
Solution : Science is the study of natural laws while technology is
the practical application of these laws to the daily life problems.
FUNDAMENTAL FORCES IN NATURE
We come across several forces in our day-to-day lives eg.,
frictional force, muscular force, forces exerted by springs and
strings etc. These forces actually arise from four fundamental forces
of nature.
Following are the four fundamental forces in nature.
1.Gravitational force
2.Weak nuclear force
3.Electromagnetic force
4.Strong nuclear force
Among these forces gravitational force is the weakest and strong
nuclear force is the strongest force in nature.

8. 4 PHYSICS
Fundament
al forces of nature
Name Relative
strength
Range Exchange
particles
Major role Important properties
Gravitational force (Force of
attraction between any two boies
by virtue of their masses)
10
–39
InfiniteGravitons
Large-scale
structure
Universal attractive, weakest,
long range, central, conservative
force.
Weak nuclear force (Changes
quark types as in Beta-decay of
nucleus
10
–13
Very short,
sub-nuclear
size (on
10
–16
m)
Weak
bosons
Nuclear
reactions
Govern process involving
neutrino and antineutrino,
operates only through a range of
nuclear size.
Electromagnetic force (Force
between particles with
charge/magnetism)
10
–2
InfinitePhotons
Chemistry
and Biology
Either attractive or repulsive, long
range, central, conservative force.
Strong nuclear force (Strong
attractive force which binds
together the protons and neutrons
in nucleus)
1
Very Short,
nuclear size
(on 10
–15
m)
Gluons
Holding
nuclei
together
Basically an attractive becomes
repulsive (when distance between
nucleons < 0.5 fermi) strongest,
short range, non-central, non-
conservative force.
Physicists are trying to derive a unified theory that would describe all the forces in nature as a single fundamental law. So far, they have
succeeded in producing a unified description of the weak and electromagnetic forces, but a deeper understanding of the strong and
gravitational forces has not yet been achieved. Theories that postulate the unification of the strong, weak, and electromagnetic forces
are called Grand Unified Theories (often known by the acronym GUTs). Theories that add gravity to the mix and try to unify all four
fundamental forces into a single force are called Superunified Theories. The theory that describes the unified electromagnetic and
weak interactions is called the Standard Electroweak Theory, or sometimes just the Standard Model.
Progress in unification of different forces/domains in nature
Name of the physicist Year Achievement in unification
Issac Newton 1687
Unified celestial and terrestrial mechanics; showed that the same laws of
motion and the law of gravitation apply to both the domains.
Hans Christian Oersted 1820
Michael Faraday 1830
James Clark Maxwell 1873
Unified electricity, magnetism and optics; showed that light is an
electromagnetic wave.
Sheldon Glashow, Abdus Salam,
Steven Weinberg
1979
Showed that the 'weak' nuclear force and the electromagnetic force could
be viewed as different aspects of a single electro-weak force.
Carlo Rubia, Simon Vander Meer1984
Verified experimentally the predictions of the theory of electro-weak
force.
Showed that electric and magnetic phenomena are inseparable aspects of
a unified domain : electromagnetism.
Example 3 . What is electromagnetic force?
Solution : It is the force due to interaction between two moving
charges. It is caused by exchange of photons (g) between two
charged particles.
NATURE OF PHYSICAL LAWS
A physical law, scientific law, or a law of nature is a scientific
generalization based on empirical observations of physical
behavior. Empirical laws are typically conclusions based on
repeated scientific experiments and simple observations over many
years, and which have become accepted universally within the
scientific community. The production of a summary description of
nature in the form of such laws is a fundamental aim of science.
Physical laws are distinguished from scientific theories by their
simplicity. Scientific theories are generally more complex than laws.
They have many component parts, and are more likely to be
changed as the body of available experimental data and analysis
develops. This is because a physical law is a summary observation
of strictly empirical matters, whereas a theory is a model that
accounts for the observation, explains it, relates it to other
observations, and makes testable predictions based upon it. Simply
stated, while a law notes that something happens, a theory explains
why and how something happens.

9. 5Physical World

10. 6 PHYSICS
1.The man who
has won Nobel Prize twice in physics is
(a) Einstein (b) Bardeen
(c) Heisenberg (d) Faraday
2.Prof. Albert Einstein got nobel prize in physics for his work
on
(a) special theory of relativity
(b) general theory of relativity
(c) photoelectric effect
(d) theory of specific heats
3.Which of the following is wrongly matched ?
(a) Barometer-Pressure
(b) Lactometer-Milk
(c) Coulomb’s law-charges
(d) Humidity-Calorimeter
4.C.V. Raman got Nobel Prize for his experiment on
(a) dispersion of light(b) reflection of light
(c) deflection of light (d) scattering of light
5.A scientific way of doing things involve
(a) identifying the problem
(b) collecting data
(c) hypothesising a possible theory
(d) All of the above
6.The scientific principle involves in production of ultra high
magnetic fields is
(a) super conductivity(b) digital logic
(c) photoelectric effect (d) laws of thermodynamics
7.Which of the following has infinite range?
(a) Gravitational force(b) Electromagnetic force
(c) Strong nuclear force (d) Both (a) and (b)
8.Which of the following is the correct decreasing order of the
strengths of four fundamental forces of nature ?
(a) Electromagnetic force > weak nuclear force >
gravitational force > strong nuclear force
(b) Strong nuclear force > weak nuclear force >
electromagnetic force > gravitational force
(c) Gravitational force > electromagnetic force > strong
nuclear force > weak nuclear force
(d) Strong nuclear force > electromagnetic force > weak
nuclear force > gravitational force
9.The exchange particles for the electromagnetic force are
(a) gravitons (b) gluons
(c) photons (d) mesons
10.Louis de-Broglie is credited for his work on
(a) theory of relativity
(b) electromagnetic theory
(c) matter waves
(d) law of distribution of velocities
11.Madam Marie Curie won Nobel Prize twice which were in the
field of
(a) Physics and chemistry(b) Chemistry only
(c) Physics only (d) Biology only
12.The man who is known as the Father of Experimental Physics
is
(a) Newton (b) Albert Einstein
(c) Galileo (d) Rutherford
13.The person who has been awarded the title of the Father of
Physics of 20th century is
(a) Madame Curie (b) Sir C.V. Raman
(c) Neils Bohar (d) Albert Einstein
14.Which of the following is true regarding the physical science?
(a) They deal with non-living things
(b) The study of matter are conducted at atomic or ionic
levels
(c) Both (a) and (b)
(d) None of these
15.The branch of science which deals with nature and natural
phenomena is called
(a) Sociology (b) Biology
(c) Civics (d) Physics
16.Who gave general theory of relativity?
(a) Einstein (b) Marconi
(c) Ampere (d) Newton
17.Who discovered X-rays?
(a) Chadwick (b) Roentgen
(c) Thomson (d) Madam Curie
18.Which of the following is the weakest force?
(a) Nuclear force (b) Gravitational force
(c) Electromagnetic force (d) None of these
19.The field of work of S. Chandrashekar is
(a) theory of black hole (b) Cosmic rays
(c) theory of relativity(d) X-rays
20.Two Indian born physicists who have been awarded Nobel
Prize in Physics are
(a) H. J. Bhabha and APJ Kalam
(b) C.V. Raman and S. Chandrasekhar
(c) J.C. Bose and M.N. Saha
(d) S. N. Bose and H. J. Bhabha
21.Science is exploring, ...x... and ...y... from what we see
around us. Here, x and y refer to
(a) qualitative, modify(b) experiment, predict
(c) verification, predict (d) reasoning, quantitative
22.Macroscopic domain includes
(a) phenomena at the laboratory
(b) terrestrial scales
(c) astronomical scales
(d) All of the above

11. 7Physical World
EXERCISE - 1
1. (b) 2. (c) 3. (d) 4. (d) 5. (d)
6. (a) 7. (d) 8. (d) 9. (c)10. (c)
11. (a) 12. (c) 13. (d)14. (c)15. (d)
16. (a)17. (b) 18. (b) 19. (a)20. (b)
21. (b) Science is exploring, experimenting and predicting
from what we see around us.
22. (d) The macroscopic domain includes phenomena at the
laboratory, terrestrial and astronomical scales.
23. (b) The alpha particle scattering experiment of Rutherford
gave the nuclear model of the atom as shown in figure
q
Flash of
light
a
Lead
block
Fluorescent
screen
Scatte
ring
angleB
A
Gold foil
Polonium
sam
ple
23.In Rutherford, alpha particle scattering experiment as shown
in given figure, A and B refer to
q
Flash of
light
a
Lead
block
Fluorescent
screen
Scatt
ering
angle
B
A
Microscope
(a) polonium sample and aluminium foil
(b) polonium sample and gold foil
(c) uranium sample and gold foil
(d) uranium sample and aluminium foil
DIRECTIONS for Qs. (24-25) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following
(a) Statement -1 is false, Statement-2 is true
(b) Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c) Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d) Statement -1 is true, Statement-2 is false
24. Statement-1 : The concept of energy is central to Physics
and expression for energy can be written for every
physical system.
Statement-2 : Law of conservation of energy is not valid
for all forces and for any kind of transformation between
different forms of energy.
25. Statement-1 : Electromagnetic force is much stronger than
the gravitational farce.
Statement-2 : Electromagnetic force dominates all
phenomena at atomic and molecular scales.
24. (d) The concept of energy is central to Physics and the
expressions for energy can be written for every physical
system. When all forms of energy e.g., Heat, mechanical
energy, electrical energy etc., are counted, it turns out
that energy is conserved. The general law of
conservation of energy is true for all forces and for any
kind of transformation between different forms of
energy.
25. (b) It is mainly the electromagnetic force that governs the
structure of atoms and molecules, the dynamics of
chemical reactions and the mechanical, thermal and
other properties of materials.
Hints & Solutions

12. 8 PHYSICS
PHYSICAL QUANTI
TY
All the quantities in terms of which laws of physics are described
and which can be measured directly or indirectly are called
physical quantities. For example mass, length, time, speed, force
etc.
Types of Physical Quantity
1.Fundamental quantities : The physical quantities which do
not depend upon other physical quantities are called
fundamental or base physical quantities. e.g. mass, length,
time temperature electric current, luminous intensity and
amount of substance.
2.Derived quantities :The physical quantities which depend
on fundamental quantities are called derived quantities
e.g. speed, acceleration, force, etc.
UNIT
The process of measurement is a comparison process.
Unit is the standard quantity used for comparision.
The chosen standard for measurement of a physical quantity,
which has the same nature as that of the quantity is called the
unit of that quantity.
Choice of a unit (Characteristics of a unit):
(1) It should be suitable in size (suitable to use)
(2) It should be accurately defined (so that everybody
understands the unit in same way)
(3) It should be easily reproducible.
(4) It should not change with time.
(5) It should not change with change in physical conditions
i.e., temperature, pressure, moisture etc.
(6) It should be universally acceptable.
Every measured quantity (its magnitude) comprises of a number
and a unit. Ex: In the measurement of time, say
15 s
Number (n) Unit (u)
If Q is the magnitude o
f the quantity (which does not depend on
the selection of unit) then
Q = n u = n
1
u
1
= n
2
u
2
Þ
1
µn
u
Where u
1
and u
2
are the units and n
1
and n
2
are the numerical
values in two different system of units.
Fundamental (or Base) and Derived Units
Fundamental units are those, which are independent of unit of
other physical quantity and cannot be further resolved into any
other units or the units of fundamental physical quantities are
called fundamental or base units. e.g., kilogram, metre, second
etc,
All units other than fundamental are derived units (which are
dependent on fundamental units) e.g., unit of speed (ms
–1
) which
depends on unit of length (metre) and unit of time (second), unit
of momentum (Kgms
–1
) depends on unit of mass, length and time
etc.
SYSTEM OF UNITS
A system of units is a complete set of fundamental and derived
units for all physical quantities.
Different types of system of units
F.P.S. (Foot - Pound - Second) system. (British engineering system
of units.): In this system the unit of length is foot, mass is pound
and time is second.
C.G.S. (Centimetre - Gram - Second) system. (Gaussian system
of units): In this system the unit of length is centimetre, mass is
gram and time is second.
M.K.S (Metre - Kilogram - Second) system. This system is related
to mechanics only. In this system the unit of length is metre, mass
is kilogram and time is second.
S.I. (International system) units: (Introduced in 1971) Different
countries use different set of units. To avoid complexity, by
international agreement, seven physical quantities have been
chosen as fundamental or base physical quantities and two as
supplementary. These quantities are
S.No Base physical quantityFundamental unit Symbol
1 Mass kilogram kg
2 Length metre m
3 Time second s
4 Temperature kelvin K
5 Electric current ampere A
6 Luminous intensity candela cd
7 Amount of subs tance mole mol
2
Units and
Measurements

13. 9Units and Measurements
S.No Supplementary physical Supplementary unit Symbol
quantity
1 Pl
ane angle radian rad2 Solid angle steradian sr
Merits of S.I. Units :
(1) SI is a coherent system of units: This means that all derived
units are obtained by multiplication and division without
introducing any numerical factor.
(2) SI is a rational system of units: This is because it assigns
only one unit to a particular physical quantity.
(3) SI is an absolute system of units: There is no gravitational
unit in this system.
(4) SI system is applicable to all branches of science.
Conventions of writing of Units and their Symbols
·Unit is never written with capital initial letter.
·For a unit named after scientist the symbol is a capital
letter otherwise not.
·The unit or symbol is never written in plural form.
·Punctuations marks are not written after the symbol.
Definitions of Fundamental Units
(i)Metre : One metre is equal to 1650763.73 wavelength in
vacuum of the radiation corresponding to transition between
the levels 2p
10
and 5d
5
of the krypton – 86 atom
Or
The distance travelled by light in vacuum in

458,792,299
1
second is cal
led 1 metre.
(ii)Kilogram : The mass of cylinder (of height and diameter
39 cm) made of Platinum-iridium alloy kept at International
Bureau of weights and measures in paris is defined as 1kg.
(iii)Second : It is the duration of 9,192,631,770 periods of
radiation corresponding to the transition between the two
hyperfine levels of the ground state of Caesium (133) atom.
(iv)Ampere : It is the current which when flows through two
infinitely long straight conductors of negligible cross-
section placed at a distance of one metre in air or vacuum
produces a force of 2 × 10
–7
N/m between them.
(v)Candela : It is the luminous intensity in a perpendicular
direction, of a surface of 1/600,000 square metre of a black
body at the temperature of freezing platinum under a pressure
of 1.013 × 10
5
N/m
2
.
(vi)Kelvin : It is the 1/273.16 part of thermodynamic temperature
of triple point of water.
(vii)Mole : It is the amount of substance which contains as many
elementary entities as there are in 0.012 kg of Carbon-12.
S.I. Prefixes :
The magnitudes of physical quantities vary over a wide range.
For example, the atomic radius, is equal to 10
–10
m, radius of earth
is 6.4×10
6
m and the mass of electron is 9.1×10
–31
kg. The
internationally recommended standard prefixes for certain powers
of 10 are given in the table:
Prefix Power of 10 Sym bol
exa 18 E
peta 15 P
tera 12 T
giga 9 G
mega 6 M
kilo 3 k
hecto 2 h
deca 1 da
deci –1 d
centi –2 c
milli –3 m
micro –6 m
nano –9 n
pico –12 p
femto –15 f
atto –18 a
Some Important
Practical Units :
(1) For large distance (macro-cosm)
(a) Astronomical unit: It is the average distance of the
centre of the sun from the centre of the earth.
1 A.U. = 1.496 × 10
11
m
(b) Light year: It is the distance travelled by the light in
vacuum in one year. 1 ly = 9.46 × 10
15
m
(c) Parsec: One parsec is the distance at which an arc
1A.U. long subtends an angle of one second.
1 parsec = 3.1 × 10
16
m
(2) For small distance (micro-cosm)
1 micron = 10
–6
m 1nanometre = 10
–9
m
1angstorm = 10
–10
m 1fermi = 10
–15
m
(3) For small area 1 barn = 10
–28
m
2
(4) For heavy mass 1 ton = 1000kg
1quintal = 100kg 1slug = 14.57kg
1 C.S.L (chandrasekhar limit) = 1.4 times the mass of the sun
(5) For small mass 1 amu = 1.67 x 10
–27
kg
1 pound = 453.6g = 0.4536 kg
(6) For small time 1 shake = 10
–8
s
(7) For large time
Lunar month: It is the time taken by the earth to complete
one rotation about its axis with respect to sun.
1L.M. = 27.3 days.
Solar day: It is the time taken by the earth to complete one
rotation about its axis with respect to sun.
Sedrial day: It is the time taken by earth to complete one
rotation on its axis with respect to distant star.
(8) For measuring pressure
1 bar = 1atm pressure = 10
5
N/m
2
= 760mmHg
1torr = 1 mmHg
1 poiseuille = 10 Poise.
DIMENSIONS
The powers to which the fundamental units of mass, length and
time must be raised to represent the physical quantity are called
the dimensions of that physical quantity.

14. 10 PHYSICS
For examp
le : Force = mass × acceleration
= mass ×
1
v–u [LT]
[M]
t [T]
-
= = [MLTT
–2
]
Henc
e the dimensions of force are 1 in mass 1 in length and (– 2)
in time.
Dimensional Formula :
Unit of a physical quantity expressed in terms of M, L and T is
called dimensional formula. It shows how and which of the
fundamental quantities represent the dimensions.
For example, the dimensional formula of work is [ML
2
T
–2
]
Dimensional Equation :
When we equate the dimensional formula with the physical
quantity, we get the dimensional equation.
For example Work = [ML
2
T
–2
]
Classification of Physical Quantities (On the basis of dimensions) :
Physical
Quantity
Dime
nsional
physical quantity
Dimensional
constant
For e.g. c (velocity of light in vaccum)
For e.g.
distance, displacement, force, mass, time etc.
For e.g. 0, 1, 2, ....., e, , sin , cos , tan , etc. qp qq
For e.g. plane angle, solid angle, strain,
refractive index, dielectric constant, relative density,
specific gravity, poisson's ratio etc.
R(universal gas constant),
stefan's constant), h (Planck's constant),
k (Boltzmann constant), G (universal gravitational constant) etc.
s(
Dimensionless
constant
Dimensional
variable
Dimensionless
variable
Dimensionless
physical quantity
Dimensional Formula of Some Important Physical Quantities :
S.No. Physical quantity Relation w ith other quantities Dimensional formula
1.Velocity (v)
Length
Time
[M
0
LT
–1
]
2.Acceleration (a)
Velocity
Time
[M
0
LT
–2
]
3.Momentum (p) Mass × ve locity [MLT
–1
]
4.Force (F) Mass × acce leration [MLT
–2
]
5.Work Force × displacemen t [ML
2
T
–2
]
6.Power (P)
Work
Time
[ML
2
T
–3
]
7.Universal gravitational constant
21
2
mm
Fr
G= [M
–1
L
3
T
–2
]
8.Torque Fr´=t [ML
2
T
–2
]
9.Surface tension
F
S=
l
[MT
–2
]
10.Gravitational potential G
W
V
m
= [M
0
L
2
T
–2
]
11.Coefficient of viscosity
dx
dv
A
F
=h [ML
–1
T
–1
]
12.Impulse Force×time (F×t) [MLT
–1
]

15. 11Units and Measurements
13.Strain
Changein length L
Originallength L
Dæö
ç÷
èø
[M
0
L
0
T
0
]
14.Pressure gr adient
Pressure P
Distance
æö
ç÷
èøl
[ML
–2
T
–2
]
15.Plane a ngle
Arcs
Radius of circler
æö
ç÷
èø
[M
0
L
0
T
0
]
16.Angular v elocity
Angle
Timet
qæö
ç÷
èø
[M
0
L
0
T
–1
]
17.Radius of gyration
i
Moment of inertia of body I
Total mass of the body m
æö
ç÷
ç÷
å
èø
[M
0
L
1
T
0
]
18.Moment of for ce, moment of couple Force × distance (F × s) [ML
2
T
–2
]
19.Angular fr equency 2p×frequency (2pn) [M
0
L
0
T
–1
]
20.Pressure
Force
Area
[ML
–1
T
–2
]
21.Efficien cy
Output work or energyW
Input work or energy Q
æö
ç÷
èø
[M
0
L
0
T
0
]
22.Angular impul se forque×time (t × t) [ML
2
T
–1
]
23.Planck ’s constant
EnergyE
Frequency
æö
ç÷
nèø
[ML
2
T
–1
]
24.Heat capacity, Entropy
Heat energyQ
Temperature T
æö
ç÷
èø
[ML
2
T
–2
K
–1
]
25.Specific heat capacity
Heat energy Q
Mass temperature m T
æö
ç÷
´ ´Dèø
[M
0
L
2
T
–2
K
–1
]
26.Thermal conduc tivity, K
Heat energythickness Q T
KA
Area temperature time t x
´D æö
=-
ç÷
´´D èø
[MLT
–3
K
–1
]
27.Thermal Re sistance, R
Length
Thermal conductivity × area
[M
–1
L
–2
T
3
K]
28.Bulk modulus (B ) or (compressibility)
–1
Volume (changein pressure) P
V
Change in volume V
´D æö
-
ç÷
Dèø
[ML
–1
T
–2
]
29.Stefan’ s constant (s)
4
4 4
Q AtT(Energy/area)
Time (temperature) EQ/A.tT
æö=s
ç÷
ç÷
´ = =sèø
[ML
0
T
–3
K
–4
]
30.Univer sal gas constant R
Pressure volume PV
Mole temperature nT
´ æö
ç÷
´ èø
]molKTML[
1
122---
31.Voltage, electric potential (V) or
WorkW
Charge q
æö
ç÷
èø
[ 132
ATML
--]
electromotive force (e)

16. 12 PHYSICS
32.Capacitance (C)
Charge q
Potential differenceV
æö
ç÷
èø
[ 2421
ATLM
-- ]
33.Electric field ()E
r Electric force F
Charge q
æö
ç÷
èø
[ 13
AMLT
--]
34.Magnetic fie
ld
(B)
r
, magnetic induction,
Force
[F IBsin]
Current length
=q
´
l
[ML
0
T
–2
A
–1
]
magnetic flux d
ensity
35.Magnetic flux (f
m
) f = BA cosq [ML
2
T
–2
A
–1
]
36.Inductance
mMagneticflux
Current I
fæö
ç÷
èø
[ 222
ATML
--]
coefficient of self i
nductance (L) or
coefficient of mutual inductance (M)
37.Magnetic field strength or
magnetic moment density (I)
MagneticmomentM
Volume V
æö
ç÷
èø
[M
0
L
–1
T
0
A]
38.Permittivity constant in free space e
o
22
22
(Charge) q
4πelectrostaticforce(distance) 4 F r
æö
ç÷
ç÷
´ p´´èø
1 342
M L TA
--éù
ëû
39.Faraday constant (F), charge Avagadro constant × elementry charge ]molTALM[
100 -
40.Mass defect, (Dm) (Sum of masses of nucleons
00
[MLT]
– mass of nucleus)(M
P
+M
N
–M
n
ucleus
)
41.Resonant frequen cy (f
r
)
r
1
T
]TLM[
100-
42.Power of lens
-11
(Focal length)
f
æö
ç÷
èø
0 10
MLT
-éù
ëû
43.Refractive index
Speed of light invacuum
Speed of lightin medium
000
[MLT]
44.Wave number
2
Wavelength
p 0 10
MLT
-éù
ëû
45.Binding energy of nucleus Mass defect × (speed of light in vacuum)
2
22
MLT
-éù
ëû
46.Conductance ( c)
1
Resistance
1 2 23
M L AT
--éù
ëû
47.Fluid flow rate
4
(Pressure) (radius)
8 (Viscosity coefficient) (length)
p´æö
ç÷
èø ´
0 3 –1
M LTéù
ëû
48.Inductive reactance (Angular frequency × inductance)
232
MLTA
--éù
ëû
49.Capacitive reactance (Angular frequency × capacitance)
–1
232
MLTA
--éù
ëû
50.Magnetic dipole moment
Torque
or Current area
Magnetic field
´
020
MLTAéù
ëû

17. 13Units and Measurements
Short
cuts / Time saving techniques
1. To find dimensions of a typical physical quantity which is
involved in a number of formulae, try to use that formula
which is easiest for you. For example if you want to find the
dimensional formula of magnetic induction then you can use
the following formulae
q=t=m=
q
p
m
= sinMB,qvBF,nIB,
r
sinIdl
4
dB
0
2
0
Out of these the easiest is probably the third one.
2. If you have to find the dimensional formula of a combination
of physical quantities, then instead of finding thedimensional formula of each, try to correlate the combinationof physical quantities with a standard formula. For example,
if you have to find the dimension of CV
2
, then try to use
formula
2
CV
2
1
E= where E is energy of a capacitor..
3.==
em
c
1
00
velocity of light in vacuum
• Dimensions of the following are same
2
22
2
LIt
R
V
c
q
CVqVnRTPVWork =´=
=====
[ML
2
T
–2
]
• Dimensions of the following are same
Force = Impulse / time
= q v B = q E
= Thrust= weight = energy gradient [MLT
–2
]
• The dimension of RC =
R
L
is same as th
at of time
• Dimensions of the following are same
velocity = 00
T1
f= = ´l
m me
[M°LT
T
–1
]
• Dimensions of the following are same
Frequency
RkMB1
LmI LC
==== [M°L°T
–1
]
• Dime
nsions of the following are same
(E) Modulus of elasticity = Y (Young's modulus)
= B (Bulk modulus)
= h (Modulus of rigidity)
= Stress
= Pressure =
ilityCompressib
1
[ML
–1
T
–2
]
• D
imensions of the following are same
Acceleration, retardation, centripetal acceleration, centrifugal
acceleration, gravitational intensity/strength. [M°LT
–2
]
• Dimensions of the following are same
Water equivalent, thermal capacity, entropy, Boltzmann's
constant. [ML
2
T
–2
K
–1
]
Keep in Memory
Th
e dimensional formula of
•all trigonometric ratio is [M
0
L
0
T
0
]
•x in e
x
is [M
0
L
0
T
0
]
• e
x
is [M
0
L
0
T
0
]
•x in log x is [M
0
L
0
T
0
]
•log x is [M
0
L
0
T
0
]
Ex
ample 1.
Find out the unit and dimensions of permittivity of free
space.
Solution :
According to Coulomb’s law22
2
1 342
0
2 22
AT
[q]
[M L TA]
4 [F][r ][MLT ][L ]
--
-
éù
ëû
e===
p
Its unit =
22
2
coulomb (coulom b) coulomb
joule metre volt metrenewton metre
==
´´´
CV
–1
m
–1
Example 2.
Find out the unit and dimensions of coefficient of self or
mutual inductance of.
Solution :
e = ÷
ø
ö
ç
è
æ
dt
d
L
I
or ÷
ø
ö
ç
è
æ
dt
d
M
I
, where e is induced electromotive
force (e.m.f.)
dt Wt
Le
dq
æ ö æö
\==
ç ÷ ç÷
è ø èøII
or [L] = ]ATML[
]A][AT[
]T][TML[ 222
22
--
-
=
Its unit is volt × se
c/amp or ohm × sec or henry.
Example 3.
Find out the unit and dimensions of magnetic field intensity .
Solution :
As B = mH, hence
2
r
sind
4
1B
H
q
p
=
m
=
lI
;
1
2
[A][L]
H [M L T A]
[L]
-
\= =°°
Its unit is ampere /metre in SI system. In c.g.s. system, the
unit is oersted.
Example 4.
Find out the unit and dimensions of magnetic permeability
of free space or medium.
Solution :
According to Biot-Savart’s law
2
0
r
sind
4
B
q
p
m
=
lI
and
F = BI l sinq
or
0
2
d sinF
B
sin 4r
mq
==
q p
l
l
I
I
;

18. 14 PHYSICS
2
0
22
Fr
or\m m=
lI
(dimensionally)
Hence [m] or ]AMLT[
]L][A[
]L][MLT[
][
22
22
22
0
--
-
==m
Its units are
2
2222
N m newton joule/ metre volt coulomb
amp amp metreamp m amp amp
-´
===
´´-

ohm sec henry tesla metre
metre metre amp.
´´
= ==
Example 5.
The dimensions of physical quantity X in the equation
Force =
X
Density
is given by
(a) [ML
4
T
–2
] (b) [M
2
L
–2
T
–1
]
(c) [M
2
L
–2
T
–2
] (d) [ML
–2
T
–1
]
Solution :
(c)Q Force =
X
denisty
\X = Force × density
Hence X has dimensions
[MLT
–2
] 3
[M]
[L]
= [M
2
L
–2
T
–2
]
Example 6.
If force, acceleration and time are taken as fundamental
quantities, then the dimensions of length will be-
(a) [FT
2
] (b) [F
–1
A
2
T
–1
]
(c) [FA
2
T] (d) [AT
2
]
Solution :
(d)
x yz
L FAT=
0 1 0 2x 2y z
MLT [MLT][LT]T
--
=
=
x xy 2x2yz
MLT
+ --+
x = 0, x + y = 1, – 2x – 2y + z = 0
x = 0, y = 1, z = 2
Hence,
2
L AT=
DIMENSIONAL ANALYSIS AND ITS APPLICATIONS
Principle of Homogeneity :
Only those physical quantities can be added /subtracted/equated
/compared which have the same dimensions.
Uses of Dimensions :
(1) Conversion of one system of unit into another
Example : Convert a pressure of 10
6
dyne/cm
2
in S.I units.
Sol.We know that 1N = 10
5
dyne Þ 1 dyne = 10
–5
N
Also 1m = 100 cmÞ 1cm = 10
–2
m
Now, the pressure 10
6
dyne/cm
2
in SI unit is
10
6
25
22–
5
6
m/N10
m10m10
N10
10
cmcm
dyne
=
´
´=
´
-
-
(2) Checking the accuracy of various formulae
Example : Check the correctness of the following equation dimensionally
dv
F=ηA sinθ
dx
where F = force,
h = coefficient of viscosity,,
A = area, dv
= velocity
dx
gradient w.r.t distance,
q = angle of
contact
Sol.L.H.S = force = [MLT
–2
]
R.H.S =
1
11122dv LT
A (sin ) M L T L MLT
dxL
-
--- éùh q= ´=
ëû
The equation is dimensionally correct.
CAUTION : Please note that the above equation is not correct
numerically. The conclusion is that an equation, if correct
dimensionally, may or may not be numerically correct. Also
remember that if an equation is dimensionally incorrect, we can
conclude with surety that the equation is incorrect.
(3) Derivation of formula
Example : The air bubble formed by explosion inside water
performed oscillation with time period T which is directly
proportional to P
a
d
b
E
c
where P is pressure, d is density and E
is the energy due to explosion. Find the values of a, b and c.
Sol.Let us assume that the required expression for time period is
T = K P
a
d
b
E
c
where K is a dimensionless constant.
Writing dimensions on both sides,
c22b3a21100
]TML[]ML[]TML[]TLM[
----
=
]T[]T[]L[]M[
1c2a2c2b3acba
==
--+--++
Equating the powers,
a + b + c = 0....(1)
0c2b3a=+-- ....(2)
– 2a – 2c = 1 ....(3)
Solving these equations, we get,
a =
6
5
-, b =
2
1
, c =
3
1
.
Limitations of Dimensional Analysis :
(1) No information about the dimensionless constant is obtained
during dimensional analysis
(2) Formula cannot be found if a physical quantity is dependent
on more than three physical quantities.
(3) Formula containing trigonometrical /exponential function
cannot be found.
(4) If an equation is dimensionally correct it may or may not be
absolutely correct.
Example 7.
Find the dimensions of a and b in the Van der waal's
equation
æö
ç÷
èø
2
a
P + (V - b) = RT
V
where P is pressure and V is volume of gas.

19. 15Units and Measurements
Solut
ion :
Dimensionally
2
a
P [By principleof homogeneity]
V
=
]T[MLa
V
a
TML
25
2
21 ---
=Þ=Þ
Also dimen
sionally V=b [By principle of homogeneity]
\b = [L
3
]
Example 8.
The formula
2
2
3mg(1cos)
v
M(1sin)
-q
=
+q
l
is obtained as the
soluti
on of a problem. Use dimensions to find whether
this is a reasonable solution (v is a velocity, m and M are
masses,
l is a length and g is gravitational acceleration).
Solution :
Dimensions of L.H.S. = [LT
–1
]
2
= [L
2
T
–2
]
Dimensions of R.H.S. =
2
22[M][LT ][L]
[LT]
[M]
-
-
=
Hence fo
rmula is reasonable.
Example 9.
In the formula;
21
21
nn
ND
xx
éù-
=-êú
-
ëû
,
D = diff
usion coefficient, n
1
and n
2
is number of molecules
in unit volume along x
1
and x
2
which represents distances
where N is number of molecules passing through per unitarea per unit time calculate the dimensions of D.
Solution :
By homogeneity theory of dimension
Dimension of (N)
= Dimension of D ×
21
21
dimension of(n n )
dimension of (xx )
-
-

2
1
LT
= Dimension of D ×
3
L
L
-
Þ Dimensions of 'D' =
32
L
L LT
-
´
=
2
L
T
= [L
2
T
–1
]
Example 10.
Let
us consider an equation
21
mv mgh
2
= where m is the
mass of the body, v its velocity, g is the acceleration due to
gravity and h is the height. Check whether this equation is
dimensionally correct.
Solution :
The dimensions of LHS are
[M] [L T
–1
]
2
= [M] [ L
2
T
–2
] = [M L
2
T
–2
]
The dimensions of RHS are
[M][L T
–2
] [L] = [M][L
2
T
–2
] = [M L
2
T
–2
]
The dimensions of LHS and RHS are the same and hence
the equation is dimensionally correct.
Example 11.
A calorie is a unit of heat or energy and it equals about 4.2
J where 1J = 1 kg m
2
s
–2
. Suppose we employ a system of
units in which the unit of mass equals
a kg, the unit of
length equals
b m, the unit of time is g s. Show that a calorie
has a magnitude 4.2
a
–1

b
–2
g
2
in terms of the new units.
Solution :
1 cal = 4.2 kg m
2
s
–2
.
SI system New system
n
1
= 4.2 n
2
= ?
M
1
= 1 kg M
2
= a kg
L
1
= 1m L
2
= b metre
T
1
= 1s T
2
= g second
Dimensional formula of energy is [ML
2
T
–2
]
Comparing with [M
a
L
b
T
c
], we find that
a = 1, b = 2, c = – 2
Now,
a bc
1 11
21
2 22
MLT
nn
MLT
é ùé ùéù
=ê úê úêú
ë ûë ûëû
=
1 22
1 kg 1m 1s
4.2
kgms
-
éù é ùéù
êú ê úêú
a bgë ûëûëû
= 4.2 a
–1
b
–2
g
2
SIGNIFIC
ANT FIGURES
The number of digits, which are known reliably in our
measurement, and one digit that is uncertain are termed as
significant figures.
Rules to determine the numbers of significant figures:
1.All non-zero digits are significant. 235.75 has five significant
figures.
2.All zeroes between two non-zero digits are significant.
2016.008 has seven significant figures.
3.All zeroes occurring between the decimal point and the non-
zero digits are not significant. provided there is only a zero
to left of the decimal point. 0.00652 has three significant
figures.
4.All zeroes written to the right of a non-zero digit in a number
written without a decimal point are not significant. This rule
does not work if zero is a result of measurement. 54000 has
two significant figures whereas 54000m has five significant
figures.
5.All zeroes occurring to the right of a non-zero digit in a
number written with a decimal point are significant. 32.2000
has six significant figures.
6.When a number is written in the exponential form, the
exponential term does not contribute towards the significant
figures. 2.465 × 10
5
has four significant figures.
Keep in Memory
1.The
significant figures depend upon the least count of the
instrument.
2.The number of significant figure does not depend on the
units chosen.

20. 16 PHYSICS
ROUNDING OFF
1. If digi
t to be dropped is less than 5 then preceding digit
should be left unchanged.
2. If digit to be dropped is more than 5 then one should raise
preceding digit by one.
3. If the digit to be dropped is 5 followed by a digit other than
zero then the preceding digit is increased by one.
4. If the digit to be dropped is 5 then the preceding digit is not
changed if it is even.
5. If digit to be dropped is 5 then the preceding digit is increased
by one if it is odd.
Arithmetical Operations with Significant Figures and
Rounding off :
(1) For addition or subtraction, write the numbers one below
the other with all the decimal points in one line. Now locate
the first column from the left that has a doubtful digit. All
digits right to this column are dropped from all the numbers
and rounding is done to this column. Addition subtraction
is then done.
Example : Find the sum of 23.623 and 8.7 to correct
significant figures.
Sol.Step-1 :- 23.623 + 8.7Step-2 :- 23.6 + 8.7=32.3
(2) In multiplication and division of two or more quantities, the
number of significant digits in the answer is equal to the
number of significant digits in the quantity, which has
minimum number of significant digits.
The insignificant digits are dropped from the result if they
appear after the decimal point. They are replaced by zeroes
if they appear to the left of the decimal point. The least
significant digit is rounded off.
Example : 107.88 (5. S. F.)
× 0.610 (3 S. F.)
= 65.8068 @ 65.8
ACCURACY, PRECISION OF INSTRUMENTS AND ERRORS
IN MEASUREMENTS :
Accuracy and Precision are two terms that have very different
meanings in experimental physics. We need to be able to
distinguish between an accurate measurement and a precise
measurement. An accurate measurement is one in which the
results of the experiment are in agreement with the ‘accepted’
value. This only applies to experiments where this is the goal like
measuring the speed of light.
A precise measurement is one that we can make to a large number
of decimal places.
The following diagrams illustrate the meaning of terms accuracy
and precision :
In the above figure : The centre of the target represents the
accepted value. The closer to the centre, the more accurate the
experiment. The extent of the scatter of the data is a measure of
the precision.
A- Precise and accurate, B- Accurate but imprecise, C- Precise but
not accurate, D- Not accurate nor precise
When successive measurements of the same quantity are repeated
there are different values obtained. In experimental physics it is
vital to be able to measure and quantify this uncertainty. (The
words "error" and "uncertainty" are often used interchangeably
by physicists - this is not ideal - but get used to it!)
Error in measurements is the difference of actual or true value
and measured value.
Error = True value – Measured value
Keep in Memory
1.Accuracy d
epends on the least count of the instrument used
for measurement.
2.In the addition and subtraction operation, the result contains
the minimum number of decimal places of the figures being
used
3.In the multiplication and division operation, the result
contains the minimum number of significant figures.
4.Least count (L.C.) of vernier callipers = one MSD – one VSD
where MSD = mains scale division
VSD = vernier scale division
5.Least count of screw gauge (or spherometer)
pitch
=
no of divisions on circular scale
where pitch is the ratio of number of divisions moved on
linear scale and number of rotations given to circular scale.
6.Pure number or unmeasured value do not have significant
numbers
7.Change in the position of decimal does not change the
number of significant figures.
Similarly the change in the units of measured value does not
change the significant figures.

21. 17Units and Measurements
Metho
ds of Expressing Error:
Absolute error: It is the difference between the mean value and
the measured value of the physical quantity.
|DX
1
| = |X
mean
–X
1
|
..................................
..................................
|DX
n
| = |X
mean
–X
n
|
Mean absolute error:
DX
mean
or

DX=
12
| | | | . ........ | |D +D + +D
n
XXX
n
Relative error: It is the rati
o of the mean absolute error and the
value of the quantity being measured.
()
D
d=
mean
mean
X
Relative error a
X
Percen
tage error: It is the relative error expressed in percent
Percentage error
100%
D
=´
X
X
To find
the maximum error in compound quantities we proceed as :
(i)Sum and difference : We have to find the sum or difference
of two values given as (a ± Da) and (b ± Db), we do it as
follows
X ± DX = (a ± Da) + (b ± Db) = (a + b) ± (Da + Db)
Þ X = a + b and DX = Da + Db in case of sum
And X = (a – b) and DX = Da + Db in case of difference.
(ii)Product and quotient : We add the fractional or percentage
errors in case of finding product or quotient.
If P = ab then
P ab
P ab
D DDæö
=±+ç÷
èø
If
b
a
Q= then
Q ab
Q ab
D DDæö
=±+ç÷
èø
(iii)Power of a quantity : If x = a
n
then
Xa
n
Xa
DD æö
=ç÷
èø
Example :
For
y
x
Q
2
=, If %3100
x
x
=´
D
and
%4100
y
y
=´
D
then
Q
QD
× 100 = (2 × 3 + 4)% = 1
0%
Similarly :
If
pq
r
mn A m nc
A then p qr
A m ncc
D D DDéù
= =± ++
êú
ëû
Keep in Memory
1.
More the accuracy, smaller is the error.
2. Absolute error |DX| is always positive.
3.|DX| has the same dimensions as that of X.
4. If the least count of measuring instrument is not given and
the measured value is given the least error in the measurement
can be found by taking the last digit to be 1 and rest digit to
be zero. For e.g. if the measured value of mass m = 2.03 kg
then 01.0m
±=D kg.
5. If a numb
er of physical quantities are involved in an
expression then the one with higher power contributes more
in errors and therefore should be measured more accurately.
6. Relative error is a dimensionless quantity.
Example 12.
Each side of a cube is measured to be 7.203 m. What are the
total surface area and the volume of the cube to appropriate
significant figures?
Solution :
The number of significant figures in the measured length
7.203 m is 4. The calculated area and the volume should
therefore be rounded off to 4 significant figures.
Surface area of the cube = 6(7.203)
2
m
2
= 311.299254 m
2
= 311.3 m
2
Volume of the cube = (7.203)
3
m
3
= 373.714754 m
3
= 373.7 m
3
Example 13.
5.74 g of a substance occupies volume 1.2 cm
3
. Express its
density by keeping the significant figures in view.
Solution :
There are 3 significant figures in the measured mass whereas
there are only 2 significant figures in the measured volume.
Hence the density should be expressed to only 2 significant
figures.
Density = mass
volume
=
35.74
gcm
1.2
-
= 4.8 g cm
–3
.
Example 14
.
The mass of a box measured by a grocer’s balance is 2.300kg. Two gold pieces of masses 20.15 g and 20.17 g are added
to the box. What is (a) the total mass of the box, (b) the
difference in the masses of the pieces to correct significant
figures ?
Solution :
(a) Total mass = 2.3403 kg = 2.3 kg (upto 2 S. F.)
(b) Difference = 20.17 g – 20.15 g (upto 4 S. F.)
COMMON ERRORS IN MEASUREMENTS
It is not possible to measure the 100% correct value of any physical
quantity, even after measuring it so many times. There always
exists some uncertainty, which is usually referred to as experimental
error.
Experimental errors :
(i)Random error : It is the error that has an equal chance of
being positive or negative.
It occurs irregularly and at random in magnitude and
direction. It can be caused
(a) by the lack of perfection of observer
(b) if the measuring instrument is not perfectly sensitive.
(ii)Systematic error : It tends to occur in one direction either
positive or negative. It occurs due to
(a) measuring instrument having a zero error.
(b) an instrument being incorrectly calibrated (such as
slow- running-stop clock)
(c) the observer persistently carrying out a mistimed action
(e.g., in starting and stopping a clock)
For measuring a particular physical quantity, we take a number of
readings. Let the readings be X
1
, X
2
............,X
n
. Then the mean
value is found as follows
.....
12
( )
+ ++
==
XXX
n
X or true value X
mean
n

22. 18 PHYSICS
7. We are al
ways interested in calculating the maximum possible
error.
Example 15.
In an
experiment, the refractive index of water was observed
as 1.29, 1.33, 1.34, 1.35, 1.32, 1.36, 1.30 and 1.33. Calculate the
mean value, mean absolute error and percentage error in the
measurement.
Solution :
Mean value of refractive index,
1.29 1.33 1.34 1.351.32 1.36 1.30 1.33
8
+++++++
m=
= 1.3
3
Absolute error in measurement,
1
1.33 1.29 0.04Dm = - =+ ,
2
1.33 1.33 0.00Dm=-= ,
3
1.33 1.34 0.01Dm = - =- ,
4
1.33 1.35 0.02Dm = - =- ,
5
1.33 1.32 0.01Dm = - =+ ,
6
1.33 1.36 0.03Dm = - =- ,
7
1.33 1.30 0.03Dm=-=+ ,
8
1.33 1.33 0.00Dm=-=
So, mean a
bsolute error, | 0.04 | | 0.01| |0.02 | | 0.01| | 0.03 | | 0.03 | | 0 |
()
8
++++++
Dm=
= 0.0175 0.02»
Relative error =
0.02
0.015 0.02
1.33
Dm
± =+ =± =±
m
Percent
age error =
0.02
100% 1.5%
1.33
± ´ =±
Example 16.
The length and b
readth of a rectangle are (5.7 ± 0.1) cm and
(3.4 ± 0.2) cm. Calculate area of the rectangle with error limits.
Solution :
Here, l = (5.7 ± 0.1) cm, b = (3.4 ± 0.2) cm
Area A = l × b
= 5.7 × 3.4 = 19.38 cm² = 19.0 cm² (Rounding off to two significant figures)
A
A
D
= ±
b
b
DDæö
+ç÷
èø
l
l
= ±
0.1 0.2
5.7 3.4
æö
+ç÷
èø
= ±
0.34 1.14
5.7 3.4
+æö
ç÷
èø´
A
A
D
= ±
1.48
19.38
Þ DA = ±
1.48
A
19.38
´
= ±
1.48
19.38 1.48
19.38
´ =±
DA = ± 1.5 (Rounding o
ff to two significant figures)
\ Area = (19.0 ± 1.5) sq.cm.
Example 17.
A body travels uniformly a distance of (13.8 ± 0.2) m in a
time (4.0 ± 0.3) s. Calculate its velocity with error limits.
What is percentage error in velocity?
Solution :
Here, s = (13.8 ± 0.2) m, t = (4.0 ± 0.3) s
velocity,
s 13.8
v
t 4.0
== = 3.45 ms
–1
= 3.5 ms
–1
(rounding off
to two significant figures)
v st
v st
D DDæö
=±+ç÷
èø
( )0.8 4.140.2 0.3
13.8 4.0 13.8 4 .0
+æö
=± + =±ç÷
èø ´
Þ
v 4.94
0.0895
v 13.8 4.0
D
=± =±
´
D v = ± 0.0895 × v = ± 0.0895 × 3.45 = ± 0.3087 = ± 0.31
(Rounding o
ff to two significant fig.)
Hence, v = (3.5 ± 0.31) ms
–1
% age error in velocity
=
v
100
v
D
´ = ± 0.0895 × 100 = ± 8.95 % = ± 9%
Example 18.
T
wo resistances are expressed as R
1
= (4 ± 0.5) W and
R
2
= (12 ± 0.5) W. What is the net resistance when they are
connected (i) in series and (ii) in parallel, with percentage
error ?
(a) 16W ± 23%, 3W ± 6.25% (b) 3W ± 2.3%, 3W ± 6.25%
(c)3W ± 23%, 16W ± 6.25% (d) 16W ± 6.25%, 3W ± 23%
Solution :
(d)
S 12
R R R 16= + =W ;
12 12
P
12S
RR RR
R3
RRR
= = =W
+
S 12
R R R1D=D+D=W
Þ
S
S
R 1
100 100%
R 16
D
´ =´
= 6.25%
Þ
S
R 16 6.25%= W±
Similarly,
12
P
S
RR
R
R
=
SP 12
P 12S
RR RR
R RRR
DD DD
=++
Þ
P
P
R 0.50.51
0.23
R 4 12 16
D
= + +=
Þ
P
P
R
100 23%
R
D
´=
Þ P
R 3 23%= W±

23. 19Units and Measurements

24. 20 PHYSICS
1.Tempera
ture can be expressed as derived quantity in terms
of
(a) length and mass
(b) mass and time
(c) length, mass and time
(d) None of these
2.What is the unit of “a” in Vander Waal’s gas equation?
(a) Atm litre
–2
mol
2
(b) Atm litre
2
per mol
(c) Atm litre
–1
mol
2
(d) Atm litre
2
mol
–2
3.Random error can be eliminated by
(a) careful observation
(b) eliminating the cause
(c) measuring the quantity with more than one instrument
(d) taking large number of observations and then their
mean.
4.If e is the charge, V the potential difference, T the temperature,
then the units of
eV
T
are the same as that of
(
a) Planck’s constant(b) Stefan’s constant
(c) Boltzmann constant (d) gravitational constant
5.If f = x
2
, then the relative error in f is
(a)
x
xD2
(b)
x
x
2
)(D
(c)
x
xD
(d) (Dx)
2
6.Two quantities
A and B have different dimensions which
mathematical operation given below is physically
meaningful?
(a) A/B (b) A + B
(c) A – B (d) A = B
7.Which of the following systems of units is not based on
units of mass, length and time alone
(a) SI (b) MKS
(b) CGS (d) FPS
8.Unit of latent heat is
(a) J Kg
–1
(b) J mol
–1
(c) N Kg
–1
(d) N mol
–1
9.Dyne-sec is the unit of
(a) momentum (b) force
(c) work (d) angular momentum
10.Illuminance of a surface is measured in
(a) Lumen (b) candela
(c) lux (d) lux m
–2
11.SI unit of electric polarisation is
(a) Cm
–2
(b) coulomb
(c) ampere (d) volt
12.The SI unit of coefficient of mutual inductance of a coil is
(a) henry (b) volt
(c) farad (c) weber
13.Light year is
(a) light emitted by the sun in one year.
(b) time taken by light to travel from sun to earth.
(c) the distance travelled by light in free space in one year.
(d) time taken by earth to go once around the sun.
14.The SI unit of pressure is
(a) atmosphere (b) bar
(c) pascal (d) mm of Hg
15.Electron volt is a unit of
(a) potential difference(b) charge
(c) energy (d) capacity
16.Dimensions of impulse are
(a) ]TML[
1
-
(b) ]MLT[
2
(c) ]MT[
2-
(d) ]TML[
31--
17.The S.I. unit of pole s
trength is
(a) Am
2
(b) Am
(c) A m
–1
(d) Am
–2
18.Which is dimensionless?
(a) Force/acceleration(b) Velocity/acceleration
(c) Volume/area (d) Energy/work
19.Potential is measured in
(a) joule/coulomb (b) watt/coulomb
(c) newton-second (d) None of these
20.Maxwell is the unit of
(a) magnetic susceptibility
(b) intensity of Magnetisation
(c) magnetic Flux
(d) magnetic Permeability
21.The mass of the liquid flowing per second per unit area of
cross-section of the tube is proportional to (pressure
difference across the ends)
n
and (average velocity of the
liquid)
m
. Which of the following relations between m and n
is correct?
(a) m = n (b) m = – n
(c)m
2
= n (d) m = – n
2
22.Which of the following is a derived physical quantity?
(a) Mass (b) Velocity
(c) Length (d) Time
23.N kg
–1
is the unit of
(a) velocity (b) force
(c) acceleration (d) None of these
24.Which physical quantities have same dimensions?
(a) Moment of couple and work
(b) Force and power
(c) Latent heat and specific heat
(d) Work and power
25.The expression [ML
–1
T
–2
] does not represent
(a) pressure (b) power
(c) stress (d) Young’s modulus

25. 21Units and Measurements
1.Wh
at are the units of magnetic permeability?
(a) Wb A
–1
m
–1
(b) Wb
–1
Am
(c) Wb A m
–1
(d) Wb A
–1
m
2.The dimensions of pressure gradient are
(a) [ML
–2
T
–2
] (b) [ML
–2
T
–1
]
(c) [ML
–1
T
–1
] (d) [ML
–1
T
–2
]
3.The dimensions of Rydberg’s constant are
(a) [M
0
L
–1
T] (b) [MLT
–1
]
(c) [M
0
L
–1
T
0
] (d) [ML
0
T
2
]
4.The dimensions of universal gas constant are
(a) [L
2
M
1
T
–2
K
–1
] (b) [L
1
M
2
T
–2
K
–1
]
(c) [L
1
M
1
T
–2
K
–1
] (d) [L
2
M
2
T
–2
K
–1
]
5.The dimensions of magnetic moment are
(a) [L
2
A
1
] (b) [L
2
A
–1
]
(c) [L
2
/ A
3
] (d) [LA
2
]
6.The dimensions of Wien’s constant are
(a) [ML
0
T K] (b) [M
0
LT
0
K]
(c) [M
0
L
0
T K] (d) [MLTK]
7.The unit and dimensions of impedance in terms of charge Q
are
(a) mho, [ML
2
T
–2
Q
–2
] (b) ohm, [ML
2
T
–1
Q
–2
]
(c) ohm, [ML
2
T
–2
Q
–1
] (d) ohm, [MLT
–1
Q
–1
]
8.If L denotes the inductance of an inductor through which a
current i is flowing, the dimensions of L i
2
are
(a) [ML
2
T
–2
]
(b) [MLT
–2
]
(c) [M
2
L
2
T
–2
]
(d) Not expressible in M, L, T
9.The dimensional formula of wave number is
(a) [M
0
L
0
T
–1
] (b) [M
0
L
–1
T
0
]
(c) [M
–1
L
–1
T
0
] (d) [M
0
L
0
T
0
]
10.The period of a body under S.H.M. is represented by: T = P
a
D
b
S
c
where P is pressure, D is density and S is surface
tension, then values of a, b and c are
(a)
1,
2
1
,
2
3
- (b) 3,2,1--
(c)
2
1
,
2
3
,
2
1
-- (d) 1, 2 1/3
11.The time o
f oscillation T of a small drop of liquid depends
on radius r, density r and surface tension S. The relation
between them is given by
(a)
3
r
S
T
r
µ (b)
S
r
T
3
r
µ
(c)
r
µ
32
rS
T (d)
S
r
T
3
r
µ
12.The poten
tial energy of a particle varies with distance x from
a fixed origin as
Bx
xA
V
+
= where
A and B are constants.
The dimensions of AB are
(a)
15/22
[ML T]
-
(b)
122
[MLT]
-
(c)
3/25/22
[M L T]
-
(d)
17/22
[ML T]
-
13.Distance travelled by a particle at any instant ‘t’ can be
represented as S = A (t + B) + Ct
2
. The dimensions of B are
(a)
011
[MLT]
-
(b)
001
[M LT]
(c)
012
[MLT]
--
(d)
022
[MLT]
-
14.The deBroglie wavelength associated with a particle of mass
m and energy E is h/ 2mE. The dimensional formula of
Planck’s constant h is
(a) ]TLM[
222-
(b) ]TLM[
12-
(c) ]LTM[
2-
(d) ]TLM[
22-
15.The veloci
ty of a body which falls under gravity varies as g
a
h
b
, where g is acc. due to gravity and h is the height. The
values of a and b are
(a) a = 1, b = 1/2 (b) a = b = 1
(c) a = 1/2, b = 1 (d) a = 1/2; b = 1/2
16.The velocity v of a particle at time t is given by
b
v at
tc
=+
+
The dimensions of a, b c are respectively
(a) [LT
–2]
, [L], [T](b) [L
2
], [T] and [LT
2
]
(c) [LT
2
], [LT] and [L](d) [L], [LT] and [T
2
]
17.The dimensions of Hubble’s constant are
(a) ]T[
1-
(b) ]TLM[
200-
(c) ]MLT[
4
(d)
1
[MT]
-
18.Error in the measurement of radius of a sphere is 1%. Then
error in the measurement of volume is
(a) 1% (b) 5%
(c) 3% (d) 8%
19.Subtract 0.2 J from 7.26 J and express the result with correct
number of significant figures.
(a) 7.1 J (b) 7.06 J
(c) 7.0 J (d) 7 J
20.Multiply 107.88 by 0.610 and express the result with correct
number of significant figures.
(a) 65.8068 (b) 65.807
(c) 65.81 (d) 65.8

26. 22 PHYSICS
21.When 97.52 is divide
d by 2.54, the correct result is
(a) 38.3937 (b) 38.394
(c) 38.39 (d) 38.4
22.Relative density of a metal may be found with the help of
spring balance. In air the spring balance reads (5.00 ± 0.05)
N and in water it reads (4.00 ± 0.05) N. Relative density
would be
(a) (5.00 ± 0.05)N (b) (5.00 ± 11%)
(c) (5.00 ± 0.10) (d) (5.00 ± 6%)
23.Area of a square is (100 ± 2) m
2
. Its side is
(a) (10 ± 1) m (b) (10 ± 0.1) m
(c)
m)210(± (d) %210±
24.Let Q denote the ch
arge on the plate of a capacitor of
capacitance C. The dimensional formula for
C
Q
2
is
(a) ]TML[
22
(b) ]LMT[
2
(
c) ]MTL[
22-
(d) ]TML[
222
25.If L and R denote inductance and resistance then dimension
of L/R is
(a) ]TLM[
000
(b) ]TLM[
00
(c) ]TLM[
202
(d) ]MLT[
2
26. The dimensional formula of current density is
(a)
021
[M L T Q]
--
(b) ]QTLM[
1120-
(c) ]QMLT[
1-
(d) ]QTML[
21
2--
27.The least count of a stop watch is 0.2 second. The time of 20
oscillations of a pendulum is measured to be 25 second.
The percentage error in the measurement of time will be
(a) 8% (b) 1.8%
(c) 0.8% (d) 0.1%
28.The dimensional formula for relative density is
(a) [M L
–3
] (b) [M
0
L
–3
]
(c) [M
0
L
0
T
–1
] (d) [M
0
L
0
T
0
]
29.The solid angle sustended by the total surface area of a
sphere, at the centre is
(a)4p (b) 2p
(c)p (d) 3p
30.If C and L denote the capacitance and inductance, the
dimensions of LC are
(a) [M
0
L
0
T
–1
] (b) [M
0
L
–1
T
0
]
(c) [M
–1
L
–1
T
0
] (d) [M
0
L
0
T
2
]
31.The dimensions of solar constant is
(a) [M
0
L
0
T
0
] (b) [MLT
–2
]
(c) [ML
2
T
–2
] (d) MT
–3
32.The dimensions of
hc
e1
2
o
Î
are
(a)M
–1
L
–3
T
4
A
2
(b) ML
3
T
–4
A
–2
(c)M
0
L
0
T
0
A
0
(d)
M
–1
L
–3
T
2
A
33.The dimensional formula for entropy is
(a) [MLT
–2
K
1
] (b) [ML
2
T
–2
]
(c) [ML
2
T
–2
K
–1
] (d) [ML
2
T
–2
K]
34.Dimensions of specific heat are
(a) [ML
2
T
–2
K] (b) [ML
2
T
–2
K
–1
]
(c) [ML
2
T
2
K
–1
] (d) [L
2
T
–2
K
–1
]
35.L, C, R represent physical quantities inductance, capacitance
and resistance respectively. The combinations which have
the dimensions of frequency are
(a) 1/RC (b) R/L
(c)
LC/1 (d) C/L
36.The physic
al quantity which has the dimensional formula
[M
1
T
–3
] is
(a) surface tension(b) solar constant
(c) density (d) compressibility
37.Which of the following is the most accurate?(a) 200.0 m (b) 20 × 10
1
m
(c) 2 × 10
2
m (d) data is inadequate
38.The velocity of water waves (v) may depend on their
wavelength l, the density of water r and the acceleration
due to gravity, g. The method of dimensions gives the
relation between these quantities is
(a)v (b)
2
vgµl
(c)
22
vgµl (d)
2 12
vg
-
µl
39.The time dependence of a physical quantity p is given by
p = p
0
exp. (– a t
2
), where a is a constant and t is the time.
The constant a
(a) is dimensionless(b) has dimensions T
–2
(c) has dimensions T
2
(d) has dimensions of p.
40.In the eqn.
2
a
P (V b) constant ,
V
æö
+ -=
ç÷
èø
the unit of a is
(a) d
yne × cm
5
(b) dyne × cm
4
(c) dyne/cm
3
(d) dyne × cm
2
41.Dimensions of ‘ohm’ are same as (where h is Planck’s
constant and e is charge)
(a)
e
h
(b)
e
h
2
(c)
2
e
h
(d)
2
2
e
h
42.The Richards
on equation is given by I = AT
2
e
–B/kT
. The
dimensional formula for AB
2
is same as that for
(a)I T
2
(b) kT
(c)I k
2
(d)I k
2
/T
43.The unit of current in C.G..S. system is
(a) 10 A (b) 1/10 A
(c) 1/100 A (d) 1/1000 A

27. 23Units and Measurements
44.Wh
ich of the following do not have the same dimensional
formula as the velocity?
Given that m
0
= permeability of free space, e
0
= permittivity
of free space, n = frequency, l = wavelength, P = pressure, r
= density, w = angular frequency, k = wave number,
(a)
0o
1me (b)n l
(c)P/r (d)wk
45.A cube has numer
ically equal volume and surface area. The
volume of such a cube is
(a) 1000 unit (b) 200 unit
(c) 216 unit (d) 300 unit
46.A spherical body of mass m and radius r is allowed to fall in
a medium of viscosity h. The time in which the velocity of
the body increases from zero to 0.63 times the terminal
velocity (v) is called time constant t. Dimensionally t can be
represented by
(a)
hp6
mr
2
(b)
÷
÷
ø
ö
ç
ç
è
æ hp
2
g
rm6
(c)
m
6 rvph
(d) None of these
47.A quantity is represented by X = M
a
L
b
T
c
. The % error in
measurement of M, L and T are a%, b% and g%
respectively. The % error in X would be
(a) %)cba(g+b+a (b) %)cba(g+b-a
(c) %100)cba(´g-b-a(d) None of these
48.In a Vernier calliper, N divisions of vernier scale coincide
with (N–1) divisions of main scale (in which one division
represents 1 mm). the least count of the instrument in cm.
should be
(a)N (b) N – 1
(c)
N10
1
(d)
1N
1
-
49.What is the
fractional error in g calculated fromg/2Tlp= ? Given fraction errors in T and l are ± x and
± y respectively..
(a) x + y (b) x – y
(c) 2x + y (d) 2x – y
50.Conversion of 1 MW power in a New system of units having
basic units of mass, length and time as 10 kg, 1 dm and 1
minute respectively is
(a) 2.16 × 10
10
unit (b) 2 × 10
4
unit
(c) 2.16 × 10
12
unit (d) 1.26 × 10
12
unit
51.A resistor of 10 k W having tolerance 10% is connected in
series with another resistor of 20 k W having tolerance 20%.
The tolerance of the combination will be
(a) 10% (b) 13%
(c) 30% (d) 20%
52.Using mass (M), length(L), time (T) and electric current (A)
as fundamental quantities the dimensions of permittivity
will be
(a) [MLT
–1
A
–1
] (b) [MLT
–2
A
–2
]
(c) [M
–1
L
–3
T
+4
A
2
] (d) [M
2
L
–2
T
–2
A
2
]
53.The percentage errors in the measurement of mass and speed
are 2% and 3% respectively. The error, in kinetic energy
obtained by measuring mass and speed, will be
(a) 12 % (b) 10 %
(c) 8 % (d) 2 %
54.The density of a cube is measured by measuring its mass
and length of its sides. If the maximum error in the
measurement of mass and length are 4% and 3% respectively,
the maximum error in the measurement of density will be
(a) 7% (b) 9%
(c) 12% (d) 13%
55. The speed of sound in a gas is given by
RT
v
M
g
=
R = univer
sal gas constant,
T = temperatureM = molar mass of gasThe dimensional formula of
g is
(a) ]TLM
[
000
(b) ]LTM[
10-
(c) ]MLT[
2-
(d) ]TLM[
100-
56. Specific gravity has ............ dimensions in mass, ............
dimensions in length and ............ dimensions in time.(a) 0, 0, 0 (b) 0, 1, 0
(c) 1, 0, 0 (d) 1, 1, 3
57. If I is the moment of inertia and
w the angular velocity,,
what is the dimensional formula of rotational kinetic energy
2
I
2
1
w?
(a) ]TML[
12-
(b) ]
TLM[
212--
(c) ]TML[
22-
(d) ]TLM[
212--
58. Given that r = m
2
sin pt , where t represents time. If the unit
of m is N, then the unit of r is
(a)N (b)
2
N
(c) Ns (d)sN
2
59. The dimensional formula of farad is
(a) ]TQLM[
21--
(b)
1 222
[M L TQ]
--
(c) ]TQLM[
2
21
--
(d) ]QTLM[
221--
60. If time T, acceleration A and force F are regarded as base
units, then the dimensional formula of work is
(a) [FA] (b) ]FAT[
(c) ]FAT[
2
(d) ]TFA[
2

28. 24 PHYSICS
61. Turpenti
ne oil is flowing through a capillary tube of length
l and radius r. The pressure difference between the two
ends of the tube is p. The viscosity of oil is given by :
22
p(r x)
4v
-
h=
l
. Here v is velocity of oil at a distance x from
the axis of the tube. From this relation, the dimensional
formula of h is
(a) ]TML[
11--
(b) ]MLT[
1-
(c) ]TML[
22-
(d) ]TLM[
000
62. The dimensional formula of velocity gradient is
(a) ]TLM[
100-
(b) ]MLT[
1
(c)
01
[MLT]
-
(d) ]LTM[
20-
63. The thrust developed by a rocket-motor is given by
)PP(AmvF
21-+= where m is the mass of the gas ejected
per unit time, v is velocity of the gas, A is area of cross-
section of the nozzle,
1P and
2Pare the pressures of the
exhaust gas and surrounding atmosphere. The formula is
dimensionally
(a) correct
(b) wrong
(c) sometimes wrong, sometimes correct
(d) Data is not adequate
64. If E, m, J and G represent energy, mass, angular momentum
and gravitational constant respectively, then the
dimensional formula of
252
Gm/EJ is
(a) angle (b) le
ngth
(c) mass (d) time
65. In a vernier callipers, ten smallest divisions of the vernier
scale are equal to nine smallest division on the main scale. If
the smallest division on the main scale is half millimeter,
then the vernier constant is
(a) 0.5 mm (b) 0.1 mm
(c) 0.05 mm (d) 0.005 mm
66. A vernier calliper has 20 divisions on the vernier scale, which
coincide with 19 on the main scale. The least count of the
instrument is 0.1 mm. The main scale divisions are of
(a) 0.5 mm (b) 1 mm
(c) 2 mm (d)
4
1
mm
67. The pitch of the sc
rew gauge is 0.5 mm. Its circular scale
contains 50 divisions. The least count of the screw gauge is
(a) 0.001 mm (b) 0.01 mm
(c) 0.02 mm (d) 0.025 mm
68.If x = a – b, then the maximum percentage error in the
measurement of x will be
(a)
%100
b
b
a
a
´÷
ø
ö
ç
è
æ D
+
D
(b) %100
b
b
a
a
´÷
ø
ö
ç
è
æ D
-
D
(c) %100
ba
b
ba
a
´÷
ø
ö
ç
è
æ
-
D
+
-
D
(d) %100
ba
b
ba
a
´÷
ø
ö
ç
è
æ
-
D
-
-
D
69.The heat generated i
n a circuit is given by
2
IQ=Rt, where
I is current, R is resistance and t is time. If the percentage
errors in measuring I, R and t are 2%, 1% and 1% respectively,
then the maximum error in measuring heat will be
(a) 2% (b) 3%
(c) 4% (d) 6%
70. The pressure on a square plate is measured by measuring
the force on the plate and length of the sides of the plate by
using the formula
2
F
P=
l
. If the maximum errors in the
measurement of force and length are 4% and 2%
respectively, then the maximum error in the measurement of
pressure is
(a) 1% (b) 2%
(c) 8% (d) 10%
71.In a simple pendulum experiment for the determination of
acceleration due to gravity, time period is measured with an
accuracy of 0.2% while length was measured with an
accuracy of 0.5%. The percentage accuracy in the value of
g so obtained is
(a) 0.25% (b) 0.7%
(c) 0.9% (d) 1.0%
72.The dimensions of a rectangular block measured with
callipers having least count of 0.01 cm are 5 mm × 10 mm × 5
mm. The maximum percentage error in the measurement of
the volume of the block is
(a) 5% (b) 10%
(c) 15% (d) 20%
73.Consider the following pairs of quantities
1. Young's modulus; pressure
2. Torque; energy
3. Linear momentum; work
4. Solar day; light year.
In which cases are the dimensions, within a pair, same?
(a) 1 and 3 (b) 1 and 4
(c) 1 and 2 (d) 2 and 4
74.Which one of the following has the same dimension as that
of time, if R is resistance, L inductance and C is capacitance?
(a ) RC (b)
LC
(c)
R
L (d) All of the above
75.T
he equation of a wave is given by
x
yasink
v
æö
= w-
ç÷
èø
, where w is angular velo city and v is
linear velocity. The dimensions of K will be(a) [T
2
] (b) [T
–1
]
(c) [T] (d) [L T]

29. 25Units and Measurements
76.In the e
quation X = 3YZ
2
, X and Z are dimensions of
capacitance and magnetic induction respectively. In MKSQ
system, the dimensional formula for Y is
(a) [M
–3
L
–2
T
–2
Q
–4
](b) [M L
–2
]
(c) [M
–3
L
–2
Q
4
T
8
] (d) [M
–3
L
–2
Q
4
T
4
]
77.A force is given by F = at + bt
2
, where t is time, the
dimensions of a and b are
(a) [M L T
–4
] and [M L T
–1
]
(b) [M L T
–1
] and [M L T
0
]
(c) [M L T
–3
] and [M L T
–4
]
(d) [M L T
–3
] and [M L T
0
]
78.The frequency of vibration of a string is given by f =
L2
n
T
m
, where T is tension in the string, L is the length, n is
number of harmonics. The dimensional formula for m is
(a) [M
0
L T] (b) [M
1
L
–1
T
–1
]
(c) [M
1
L
–1
T
0
] (d) [M
0
L T
–1
]
79.The dimensions of voltage in terms of mass (M), length (L)
and time (T) and ampere (A) are
(a) [ML
2
T
–2
A
–2
] (b) [ML
2
T
3
A
–1
]
(c) [ML
2
T
–3
A
1
] (d) [ML
2
T
–3
A
–1
]
80.Suppose the kinetic energy of a body oscillating with
amplitude A and at a distance x is given by
22
Bx
K
xA
=
+
The dimensio
ns of B are the same as that of
(a) work/time (b) work × distance
(c) work/distance (d) work × time
81.The dimensions of magnetic field in M, L, T and C (coulomb)are given as
(a) [MLT
–1
C
–1
] (b) [MT
2
C
–2
]
(c) [MT
–1
C
–1
] (d) [MT
–2
C
–1
]
82.Two full turns of the circular scale of a screw gauge cover a
distance of 1mm on its main scale. The total number of
divisions on the circular scale is 50. Further, it is found that
the screw gauge has a zero error of – 0.03 mm. While
measuring the diameter of a thin wire, a student notes the
main scale reading of 3 mm and the number of circulr scale
divisions in line with the main scale as 35. The diameter of
the wire is
(a) 3.32 mm (b) 3.73 mm
(c) 3.67 mm (d) 3.38 mm
83.If the dimensions of a physical quantity are given by
M
a
L
b
T
c
, then the physical quantity will be
(a) velocity if a = 1, b = 0, c = – 1
(b) acceleration if a = 1, b = 1, c = – 2
(c) force if a = 0, b = – 1, c = – 2
(d) pressure if a = 1, b = – 1, c = – 2
84.The density of a material in CGS system of units is 4g/cm
3
.
In a system of units in which unit of length is 10 cm and unit
of mass is 100 g, the value of density of material will be
(a) 0.4 (b) 40
(c) 400 (d) 0.04
85.In an experiment four quantities a, b, c and d are measured
with percentage error 1%, 2%, 3% and 4% respectively.
Quantity P is calculated as follows
P =
32
ab
cd
% error in P is
(a) 10% (b) 7%
(c) 4% (d) 14%
86.What is the fractional error in g calculated from
g/2Tlp= ? Given fractional errors in T and l are ± x
and ± y respectively..
(a) x + y (b) x – y
(c) 2x + y (d) 2x – y
87.The resistance R of a wire is given by the relation R =
2
.
r
r
p
l
Percentage error in the measurement of r, l and r is 1%, 2%
and 3% respectively. Then the percentage error in the
measurement of R is :
(a)6 (b) 9
(c)8 (d) 10
88.What are the dimensions of permeability ?
(a) [M
1
L
1
T
1
A
–2
] (b) [M
1
L
1
T
–2
A
–2
]
(c) [M
2
L
2
T
1
A
0
] (d) [M
1
L
2
T
2
A
–2
]
89.The physical quantity having the dimensions
[M
–1
L
–3
T
3
A
2
] is
(a) resistance (b) resistivity
(c) electrical conductivity (d) electromotive force
90.The time of reverberation of a room A is one second. What
will be the time (in seconds) of reverberation of a room,
having all the dimensions double of those of room A?
(a)2 (b) 4
(c)
1
2
(d) 1
91.Which of
the following is the unit of molar gas constant?
(a) JK
–1
mol
–1
(b) Joule
(c) JK
–1
(d) J mol
–1
92. Density of liquid is 16.8 g cm
–3
. Its value in the International
System of Units is
(a)
3
16.8 kgm
-
(b)
3
168 kgm
-
(c)
3
1680 kgm
-
(d)
3
16800 kgm
-
93. The dimensional formula of couple is
(a) ]TML[
22-
(b) ]MLT[
2
(c)
13
[MLT]
--
(d) ]TML[
22--

30. 26 PHYSICS
94.The ref
ractive index of water measured by the relation
m =
real depth
apparent depth
is found to
have values of 1.34, 1.38,
1.32 and 1.36; the mean value of refractive index with
percentage error is
(a) 1.35 ± 1.48 % (b) 1.35 ± 0 %
(c) 1.36 ± 6 % (d) 1.36 ± 0 %
95.Diameter of a steel ball is measured using a Vernier callipers
which has divisions of 0.1 cm on its main scale (MS) and 10
divisions of its vernier scale (VS) match 9 divisions on the
main scale. Three such measurements for a ball are given
below:
S.No. MS(cm) VSdivisions
1. 0.5 8
2.0.54
3.
0.5 6
If the zero error is – 0.03 cm, then mean corrected diameter is(a) 0.52 cm (b) 0.59 cm
(c) 0.56 cm (d) 0.53 cm
96.The period of oscillation of a simple pendulum is T =
L
2
g
p.
Measured value
of L is 20.0 cm known to 1 mm accuracy and
time for 100 oscillations of the pendulum is found to be 90 susing a wrist watch of 1s resolution. The accuracy in thedetermination of g is(a) 1% (b) 5%
(c) 2% (d) 3%
Exemplar Questions
1.The number of significant figures in 0.06900 is
(a)5 (b) 4
(c)2 (d) 3
2.The sum of the numbers 436.32, 227.2 and 0.301 in
appropriate significant figures is
(a) 663.821 (b) 664
(c) 663.8 (d) 663.82
3.The mass and volume of a body are 4.237 g and 2.5 cm
3
,
respectively. The density of the material of the body in
correct significant figures is
(a) 1.6048 g cm
–3
(b) 1.69 g cm
–3
(c) 1.7 g cm
–3
(d) 1.695 g cm
–3
4.The numbers 2.745 and 2.735 on rounding off to 3
significant figures will give
(a) 2.75 and 2.74 (b) 2.74 and 2.73
(c) 2.75 and 2.73 (d) 2.74 and 2.74
5.The length and breadth of a rectangular sheet are 16.2 ± 0.1
cm and 10.1 ± 0.1 cm, respectively. The area of the sheet in
appropriate significant figures and error is
(a) 164 ± 3 cm
2
(b) 163.62 ± 2.6 cm
2
(c) 163.6 ± 2.6 cm
2
(d) 163.62 ± 3 cm
2
6.Which of the following pairs of physical quantities does
not have same dimensional forrmula?
(a) Work and torque
(b) Angular momentum and Planck's constant
(c) Tension and surface tension
(d) Impulse and linear momentum
7.Measure of two quantities along with the precision of
respective measuring instrument is
A = 2.5 ms
–1
± 0.5 ms
–1
,
B = 0.10 s ± 0.01 s. The value of AB will be
(a) (0.25 ± 0.08) m(b) (0.25 ± 0.5) m
(c) (0.25 ± 0.05) m(d) (0.25 ± 0.135) m
8.You measure two quantities as A = 1.0 m ± 0.2 m, B = 2.0 m ±
0.2 m. We should report correct value for
AB as
(a) 1.4 m ± 0.4 m (b) 1
.41 m ± 0.15 m
(c) 1.4 m ± 0.3 m (d) 1.4 m ± 0.2 m
9.Which of the following measurement is most precise?(a) 5.00 mm (b) 5.00 cm
(c) 5.00 m (d) 5.00 km
DIRECTIONS for Qs. (97 to 100) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following
(a) Statement -1 is false, Statement-2 is true
(b) Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c) Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d) Statement -1 is true, Statement-2 is false
97. Statement 1 : The number of significant figures depends on
the least count of measuring instrument.
Statement 2 : Significant figures define the accuracy of
measuring instrument.
98. Statement 1 : Absolute error may be negative or positive.
Statement 2 : Absolute error is the difference between the
real value and the measured value of a physical quantity.
99. Statement 1 : In the measurement of physical quantities
direct and indirect methods are used.
Statement 2 : The accuracy and precision of measuring
instruments along with errors in measurements should be
taken into account, while expressing the result.
100. Statement 1 : Energy cannot be divided by volume.
Statement 2 : Because dimensions for energy and volume
are different.

31. 27Units and Measurements
10.The
mean length of an object is 5 cm. Which is the following
measurement is most accurate?
(a) 4.9 cm (b) 4.805 cm
(c) 5.25 cm (d) 5.4 cm
11.Young's modulus of steel is 1.9 × 10
11
N/m
2
. When expressed
in CGS units of dyne/cm
2
, it will be equal to (1N = 10
5
dyne,
1 m
2
= 10
4
cm
2
)
(a) 1.9 × 10
10
(b) 1.9 × 10
11
(c) 1.9 × 10
12
(d) 1.9 × 10
13
12.If momentum (p), area (A) and time (T) are taken to be
fundamental quantities, then energy has the dimensional
formula
(a)[pA
–1
T
1
] (b) [p
2
AT]
(c)[pA
–1/2
T] (d) [pA
1/2
T] NEET/AIPMT (2013-2017) Questions
13.In an experiment four quantities a, b, c and d are measured
with percentage error 1%, 2%, 3% and 4% respectively.
Quantity P is calculated as follows P =
32
ab
cd
% error in P is:
(a) 10% (b) 7% [2013]
(c) 4% (d) 14%
14.The pair of quantities having same dimensions is
(a) Young’s modulus and energy [NEET Kar. 2013]
(b) impulse and surface tension
(c) angular momentum and work
(d) work and torque
15.If force (F), velocity (V) and time (T) are taken as fundamental
units, then the dimensions of mass are : [2014]
(a) [F V T
– 1
] (b) [F V T
– 2
]
(c) [F V
– 1
T
– 1
] (d) [F V
– 1
T]
16.If energy (E), velocity (V) and time (T) are chosen as the
fundamental quantities, the dimensional formula of surface
tension will be : [2015]
(a) [EV
–1
T
–2
] (b) [EV
–2
T
–2
]
(c) [E
–2
V
–1
T
–3
] (d) [EV
–2
T
–1
]
17.If dimensions of critical velocity u
c
of a liquid flowing
through a tube are expressed as
x yx
[ r]hr , where h, r a nd
r are the coefficient of viscosity of liquid, density of liquidand radius of the tube respectively, then the values of x, yand z are given by : [2015 RS]
(a) –1, –1, 1 (b) –1, –1, –1
(c) 1, 1, 1 (d) 1, –1, –1
18.A physical quantity of the dimensions of length that can be
formed out of c, G and
2
0
e
4pe
is [c is velocity of light, G is
universal constant of gravitation and e is charge] [2017]
(a)
1/2
2
2
0
e
cG
4
éù
êú
peêúëû
(b)
1/2
2
2
0
1e
G4c
éù
êú
peêúëû
(c)
2
0
1e
G
c4pe
(d)
1/2
2
2
0
1e
G
4c
éù
êú
peêúëû

32. 28 PHYSICS
EXERCISE - 1
1. (d) Temperature is one of the basic physical quantities.
2. (b) The vander Waal’s gas equation is
RT)bV)(
V
a
P(
2
=-+ for one mole.
& =m-
÷
÷
ø
ö
ç
ç
è
æ m
+ )bV(
V
a
P
2
mRT for m mole
Dime
nsionally in first bracket on L.H.S
m
=Þ
ú
û
ù
ê
ë
ém
=
2
2
V]P[
]a[
V
a
]P[
Dimension of
ú
ú
û
ù
ê
ê
ë
é
=
mole
litre.atm
]a[
2
3. (d)
4
. (c)
R
T
PV
T
W
T
eV
===
and
N
R
= Boltzmann con
stant.
5. (a) 6.(a)
7. (a) SI is based on seven fundamental units.
8. (a)
1
kgJ
kg
J
m
Q
L
-
===
9. (a) As force =
change in momentum/time.
\ force × time = change in momentum
10. (c) Illuminance is intensity of illumination measured in lux.
11. (a)
2
2
i
Cm
m
C
A
Q
P
-
===
12. (a)
13. (c
) 1 light year = speed of light in vacuum × no. of seconds
in one year = (3 × 10
8
) × (365 × 24 × 60 × 60) = 9.467 ×
10
15
m.
14. (c) 1 pascal = 1 N / m
2
.
15. (c) Electron volt is a unit of energy &
1eV = 1.6×10
–19
joule
16. (a) Impulse = force × time ]LTM[TMLT
112 --
=´= .
17. (b) Pole strength,
.mA
m
mA
2
M
m
2
===
l
18. (d) Both energy and work have same unit.
\ energy/work is a pure number.
19. (a) Potential is work done per unit charge.
20. (c) Maxwell is the unit of magnetic flux in C.G.S system.
1 Wb(S.I unit) = 10
8
maxwell
21. (b) Let M = p
n
v
m
m1n2112
)LT()TML(TML
-----
=

mn2mnn
TLM
--+-
=
2mn;1n-=+-=\
112n2m-=+-=+-=\ nm-=\
22. (b)
23. (c) N
kg
–1
= force/mass = acceleration
24. (a) Moment of couple = force × distance ]TLM[
221-
=
work = force × distance ]TLM[
221-
= .
25. (b)
]TML[
T
TML
time
energy
Power
32
22
-
-
=== .
EXERCISE - 2
1. (a)
From Biot Savart’s law 2
0
r
sindli
4
B
q
p
m
=
2 22
11
0
4Br Wbmm
WbAm
idl sin Am
-
--p
m===
q
2. (a) Pre
ssure gradient
Pressure difference
distance
= .
[Pressur
e gradient]
12
MLT
L
--
=
22
MLT
--éù=
ëû
3. (c) From ,
n
1
n
1
R
1
2
2
2
1
÷
÷
ø
ö
ç
ç
è
æ
-=
l
dimensions of ]TLM[L
L
1
R
0101--
===
4. (a)
22
PV W MLT
R
T T molK
-
===
mm
where m is num
ber of mole of the gas
= [M
1
L
2
T
–2
K
–1
mol
–1
]
5. (a) M = current × area = AL
2
= [L
2
A
1
]
6. (b)
0101
m
b T LK [M LT K]=l==
7. (b)
22
1
V W MLT
Impedance
Q Q QT
-
-
===
II
]QTML[
212--
= .
Hints & Solutions

33. 29Units and Measurements
8.
(a) Energy stored in an inductor
]TML[iL
2
1 222-
==
9. (b) Wa
ve number
]TLM[
L
11 010-
==
l
=n .
10. (a) T
= P
a
D
b
S
c
c2b3a21100
)MT()ML()TML(TLM
----
=
c2a2b3acba
TLM
----++
=
Applying principle of homogeneity
a + b + c = 0; – a – 3b = 0; – 2a – 2c = 1
on solving, we get a = – 3/2, b = 1/2, c = 1
11. (b) T
MT
LML
S
r
2
333
==
r
-
-
12. (d) xVA;VxxA];L[xB ====

22/52/122
TLMLTLM
--
==
]TLM[)L()TML(AB
22/7122/5 --
==
13. (b
) In S = A (t + B) + Ct
2
; B is added to time t. Therefore,
dimensions of B are those of time.
14. (b)
)TML(MLEm2h
22-
=l= ]TLM[
12-
=
15. (d)
ab
v gh;=
a2baba210
TLL)LT(]LTM[
-+--
==
2/1a1a2;1ba=-=-=+\
2/1b=\
16. (a) A
s c is added to t, \ c = [T]
1
2v LT
a [LT]
tT
-
-
=== ,
1
b v(t c)LT T [L]
-
= + = ´=
17
. (a) Hubble’s constant,
]L[
]LT[
cetandis
velocity
H
1-
==
]T[
1-
= .m/N1070
3-
´=
18. (c) ;r
3
4
V
3
p=
%3%13100
r
r
3100
V
V
=´=´÷
ø
ö
ç
è
æD
=´
D
19. (a
) Subtraction is correct upto one place of decimal,
corresponding to the least number of decimal places.7.26 – 0.2 = 7.06 = 7.1 J.
20. (d) Number of significant figures in multiplication is three,
corresponding to the minimum number107.88 × 0.610 = 65.8068 = 65.8
21. (d)
4.38393.38
54.2
52.97
== (with least nu mber of
significant figures, 3).
22. (d) Relative density =
Weight of body in air
Loss of weight in water

5.00 5.00
5.00 – 4.00 1.00
==

0.05 0.05
100 100
5.00 1.00
Dr æö
´=
+´
ç÷
èør
= (0.01 + 0.05) × 100
= 0.06 × 100 = 6%
\ Relative density = ±
5.00 6%
23. (a) Area = (Length)
2
or length
12
(Area)=
12
(100 2)=±
121
(100) 2
2
= ±´
(10 1)m=±
24. (c)
We know that
C2
Q
2
is energ
y of capacitor so it represent
the dimension of energy ]TML[
22-
= .
25. (b)
Volt/amp.
sec/amp
.Volt
L/R
´
= = sec. ]TLM[
00
=
26. (a)
Current Q
Current densit
y area area t
==
´
27. (c) 8.0100
25
2.0
=´
28. (d) 29. (
a)
30. (d) From
1
2 LC
n=
p
2 002
2 12
11
LC T [ML T]
(2 ) (T)
-
=
= ==
pn
31. (d) Solar constant = energy/sec/area

]MT[
TL
TLM 3
2
22
-
-
==
32. (c) F
rom
2
2
or
e
4
1
F
ep
=
2
o
2
rF4
e
p=
e
\ (dimensionally)
2 2 22
211
o
e 4 Fr (M LT )L
hc hc MLT[LT]
-
--
p
==
e
]ATLM[
0000
= ,
hc
e
o
2
e
is called f
ine structure constant & has value
137
1
.

34. 30 PHYSICS
33. (c)
K
TML
T
Q
Entropy
22-
== ]KTML[
1
22--
=
34. (d)
22
221Q MLT
s [LT K]
m MK
-
--
===
q
35. (c)
2 2 2 1 242
11
LC
(ML T A ) (M L T A )
- - --
=
´
=
1
2
1
T
T
-
=
36. (b) So
lar constant = energy/area/time

]TM[
TL
TLM 31
2
22
-
-
== .
37. (a)
38. (b)
abc
vkg= lr
c2b3a10
)LT
()ML(L]LTM[
---
=

c2cb3ab
TLM
-+-
=
1cb3a;0b=+-=\
1c2
-=- Þ 2/1c=
2
1
a=\
1/2 0 1/2 2
v gorvgµl r µl
39. (b) In p = p
0
exp. (– a t
2
) is dimensionless
]T[
T
1
t
1 2
22
-
===a\
40. (b) As P
V
a
2
=
2
a PV\=
324
2
dyne
(cm ) dyne cm
cm
= =´
41. (c)
232
2
12
2
ATML
)AT(
TML
e
h --
-
==
= Resistance
(ohm)
42. (c) I = AT
2
e
–B/kT
Dimensions of A = I /T
2
; Dimensions of B = kT
(Q power of exponential is dimensionless)
2 22
2
AB (kT)k
T
==
I
I
43. (a) The C.G.S unit of current is called biot (Bi) i.e.,
1C (1/10)e.m.u of charge 1
1A Bi
1sec sec 10
= ==
or 1Bi = 10A
44.
(d)
]TL[
L
1
T
1
k
11--
=´=w
The dimensions
of the quantities in a, b, c are of velocity
[LT
–1
]
45. (c) Volume (L
3
) = surface area (6L
2
)
\ L = 6, volume = 6
3
= 216
46. (d) None of the expressions has the dimensions of time.
47. (a) ;TLMX
cba
=
X aMbLcT
100 100
X MLT
D D DDæö
´= ++´ç÷
èø
)%
cba(g+b+a=
48. (c) N VD = (N – 1) MD
MD
N
1N
VD1 ÷
ø
ö
ç
è
æ-
=
L.C. = Least coun
t = 1MD – 1VDMD
N
1N
1.C.L÷
ø
ö
ç
è
æ -
-=
cm
N10
1
cm
N
1.0
.D.M
N
1
===

scale on vernier parts ofnumber
scalemain on part 1 of value
=
where V.D. = vernier division, M.D. Main scale division.
49. (c) From
2
2
T
4g;
g
2T
ll
p=p=
)x2y(
T
T2
g
g
+=
D
+
D
=
D
l
l
50. (c) We have, n
1
u
1
= n
2
u
2
n
2

1
1
2
æö
=
ç÷
èø
u
n
u
23
6 111
222
10
-
æ öæöæö
=´
ç ÷ç÷ç÷
è øèøèø
MLT
MLT
32
6
1
111
10
10 601 10
kgms
kgs m
-
-
æ ö æöæö
=´
ç÷ç ÷ ç÷
èøè ø èø´
()()
2361
10 10 60
10
æö
=´ ç÷
èø
()
37
10 60=´ = 2.16 × 10
12
units.
51. (c)
Effective resistance
S
R (10k 10%) (20k 20%)= W± + W±
\ Tole
rance of the combination =
30 30W±( k %)
52. (c) Force, F
2
21
0r
qq
4
1
pe
=
2
21
0
Fr4
q.q
p
=eÞ

35. 31Units and Measurements
So d
imension of e
0 ]ATLM[
]L][MLT[
]AT[ 2431
22
2
--
-
==
53. (c
) Percentage error in mass
%2100
m
m
=÷
ø
ö
ç
è
æ
´
D
and
perce
ntage error in speed
%3100
v
v
=÷
ø
ö
ç
è
æ
´
D
.
21
E mv
2
=
EmV
100 100 2 100
EmV
DDD
\´=
´+´
= 2% + 2 × 3% = 8%.
54. (d) Density = Mass
Volume
3
L
M
=r,
L
L
3
M
M D
+
D
=
r
rD
% erro
r in density = % error in Mass + 3 (% error
in length)
= 4 + 3(3) = 13%
55. (a) Ratio of specific heat,
p
v
C
C
g=
56. (a
) Specific gravity is the ratio of density of substance
and density of water at 4°C. The ratio of like quantities
is dimensionless.
57. (c) Dimensionally K.E = Work
58. (b) Trigonometric ratio are a number and hence
demensionless
59. (b)
1 222QQ
[C] [M L TQ]
VW
2
--
éù
éù
=== êú
êú
ëûêúëû
60. (c) ]LT
[]A[
2-
= or ]AT[]L[
2
=
[Work] = [Force ×Distance] = ]FAT[]FL[
2
=
61. (a)
h is the coefficient of viscosity..
62. (a) Velocity gradient is velocity per unit distance.
63. (a) Use principle of homogeneity.
64. (a) ]TLM[
]TLM][M[
]TML][TML[ 000
22315
21222
=
--
--
= angle.
65
. (c) 10 VD = 9MD, 1VD =
9
MD
10
Verni
er constant = 1 MD – 1 VD
=
9 1 11
1 MD MD
10 10 10 2
æö
- = =´ç÷
èø
= 0.05 mm
66. (c
)
0.1 19
1 MSD
10 20
æö
=-ç÷ èø
Þ 11
1MSD
100 20
=´
Þ 1 MSD
=
1
102
5
´=
67. (b) Least count
0.5
0.01mm
50
==
68. (c
) Maximum absolute error is baD+D. Therefore the
percentage error =
absoluteerror
100
actual value
´
6
9. (d)
100
t
t
100
R
R
100
I
I2
100
Q
Q
´
D
+´
D
+´
D
=´
D
2 2% 1% 1%=´++ = 6%.
70. (c)
PF
100 100 2 10 0
PF
DDD
´=´+ ´=
l
l
4% + 2 × 2%
= 8%
71. (c
)
2
T
g,
g
2T
ll
µp=
\ 100
g
g
´
D
= 0.5% + 2 × 0.2% = 0.9%
72.
(a) % error =
100
5.0
01.0
100
0.1
01.0
100
5.0
01.0
´+´+´
= 2 + 1 +
2 = 4 + 1 = 5
73. (c)
74. (b)
2 2 2 1 2 42
[LC]M LTA.MLTAT
- - --
==
75. (c) The quantity (
v
wx
– wk) has dimension of angle and
hence wk is dimensionless being angle.
76. (d)
1 242
3 2 44
2 242
[X]
M L TA
[Y] M L QT
[Z] MTA
--
--
--
=
==
Q
A
T
æö
=ç÷
èø
Q
77. (c
) [at] = [F] amd [bt
2
] = [F]
Þ [a] = MLT
–3
and [b] = MLT
–4
78. (c) Clearly,
2
22
nT
m
4fL
= ;
2
22
MLT
[m]
T .L
-
-
=
79. (d)
22
2 13W MLT
[V] MLAT
Q AT
-
--éù
===
êú
ëû
80. (b)
From
222
Bx Bx B
K
xxAx
= ==
+
\ B = K × x
= K.E. × distance = work × distance

36. 32 PHYSICS
81. (c) W
e know that F = q v B
2
11
1
F MLT
B MTC
qvC LT
-
--
-
\===
´
82. (d) Lea
st count of screw gauge =
0.5
mm 0.01mm
50
=
\ Reading = [Main scale reading
+ circular scale reading × L.C] – (zero error)
= [3 + 35 × 0.01] – (–0.03) = 3.38 mm
83. (d) Pressure =
[ ]
–2
–1 –2
2
MLT
MLT
L
=
Þ a = 1, b = – 1, c = – 2.
84. (b) In CGS system,
3
g
4
cm
d=
The unit of
mass is 100g and unit of length is 10 cm, so
density =
3
100g
4
100
10
cm
10
æö
ç÷
èø
æö
ç÷
èø
=
33
4
(100g)100
(10cm)1
10
æö
ç÷
èø
æö
ç÷
èø
3
3
4 100g
(10) ·
100 (10cm)
=´
= 40 unit
85. (d)
P =
32
ab
cd
,
P
P
D
× 100% =
a
3
a
D
× 100% +
b
2
b
D
×
100% +
c
c
D
× 100% +
d
d
D
× 100%.
= 3 × 1% + 2 × 2% + 3% + 4% = 14%
8
6. (c) From
2
2
T
4g;
g
2T
ll
p=p=

)x2y(
T
T2
g
g
+=
D
+
D
=
D
l
l
87. (b) Given =
2
R,
r
r
=
p
l
then
R
100
R
D
´
=
r
100 100 2 100
r
DrDD
´+ ´+ ´
r
l
l
= 1% + 2% + 2 × 3% = 9%
88. (b) The magnetic field at a point near a long straight
conductor is given by
B =
02
.
4
I
r
m
p
Þ
0
m =
4
2
rB
I
p
\[m
0
] =
[][]
[]
rB
I
=
-2 -1
L.MT A
I

F
B
qv
æö
=
ç÷
èø
Q
= M LT
–2
A
–2
89. (c)
Resistivity,
m
ne
2t
r=
3 –3 –2
[MLTA]r=
So, electrica
l conductivity
1
s=
r
–1–332
[M L TA]s=
90. (a) Rever
beration time,
t =
forward backward
dd
VV
æö æö
+ç÷ ç÷
èø èø
when dim
ensions double then, d¢ = 2d
\
dd 2d2d dd
t2
VV V V VV
¢¢ æö
¢=+= + = + ç÷
èø
Þt¢ = 2t
dd
t
VV
æö
+=ç÷
èø
Q
91. (a)
11
molKJ
Kmol
J
nT
PV
R
--
===
.
92. (d) 16.8 gcm
–3
= 16800 Kgm
–3
.
93.
(a) Dimensionally couple = Torque = Work
94. (a) The mean value of refractive index,
1.34 1.38 1.32 1.36
1.35
4
+++
m==
and
| (1.35 1.34) | |(1.35 1.38) | | (1.35 1.32) | | (1.35 1.36) |
4
-+-+-+-
Dm=
= 0.02
Thus
100
Dm
´
m
=
0.02
100 1.48
1.35
´=
95. (b) L
east count =
0.1
10
= 0.01 cm
d
1
= 0.5 + 8 × 0.01 + 0.03 = 0.61 cm
d
2
= 0.5 + 4 ×
0.01 + 0.03 = 0.57 cm
d
3
= 0.5 + 6 × 0.01 + 0.03 = 0.59 cm
Mean diameter =
0.61 0.57 0.59
3
++
= 0.59 cm
96
. (d) As,
2
2
g4
T
=p
l
So,
gT
100 100 2 100
gT
DDD
´=´+´
l
l
0.11
100 2 100
20 90
= ´ +´´ = 2.72 ; 3%
97. (c) 98. (b)
99
. (b) 100. (a)

37. 33Units and Measurements
EXERCISE - 3
Exemplar Questions
1. (b) In 0.06900, the two zeroes before six are not siginificant
and two zeroes on right side of 9 are significant figures.
Hence, number of significant figures are four (6900).
2. (b) In addition the result will be in least number of places
after decimal and minimum number of significant figure.
The sum of the given numbers can be calculated as
663.821. The number with least decimal places is 227.2
is correct to only one decimal place but in addition of
numbers, the final result should be rounded off to one
decimal place i.e., 664.
3. (c) As we know that in multiplication or division, the final
result should retain as many significant figures as are
there in the original number with the least significant
figures.
The significant figure in given numbers 4.237 g and 2.5
cm
3
are four and two respectivey so, density should
be reported to two significant figures.
3
3
4.237 g
Density 1.6948 1.7 g.cm
2.5 cm
-
= ==
\ As r
ounding off the number upto 2 significant
figures, we get density = 1.7.
4. (d) Rounding off 2.745 upto 3 significant figures here IVth
digit is 5 and its preceding is even, so no change in 4.Thus answer would be 2.74. Rounding off 2.735 upto 3significant figures, here IV digit is 5 and its precedingdigit is 3 (odd). So 3 is increased by 1 answer becomewould be 2.74.
5. (a) If Dx is error in a physical quantity, then relative error is
calculated as
Dx
x
.
Given th
at,
Length l = (16.2 ± 0.1) cm
Breadth b = (10.1 ± 0.1) cm
Q Dl = 0.1 cm, Db = 0.1 cm
Area (A) = l × b = 16.2 × 10.1 = 163.62 cm
2
In significant figure rounding off to three significantdigits, area A = 164 cm
2
0.1 0.1
16.2 10.1
D DD
=+=+
Alb
Alb

1.01 1.62 2.63
16.2 10.1 163.62
+
==
´
So,
2.63 2.63
164
163.62 163.62
D=´ =´AA
= 2
.636 cm
2
Now rounding off up to one significant figureDA = 3 cm
2
.
So, Area A = A ± DA = (164 ± 3) cm
2
.
6. (c)a. Work = force × distance
= [MLT
–2
][L] = [ML
2
T
–2
]
Torque = F × d = [ML
2
T
–2
]
LHS and RHS has same dimensions.
b.Angular momentum (L)
= mvr = [M][LT
–1
][L]
= [ML
2
T
–1
]
Planck's constant h = n
E

22
21
1
[MLT]
[MLT]
[T]
-
-
-
==
1
and
æö
= n n=ç÷
èø
Eh
T
Q
So, dim
ensions of h and L are equal.
c.Tension = force = [MLT
–2
]
Surface tension
2
force[MLT]
length [L]
-
==
= [ML
0
T
–2
]
d.Impulse =
force × time = [MLT
–2
][T]
= [MLT
–1
]
Momentum = mass × velocity
= [M][LT
–1
] = [MLT
–1
]
LHS and RHS has same dimensions.
Hence only (c) options of Physical quantities does not
have same dimensions.
7. (a) By applying the Rule of significant figure in
multiplication and addition.
As given that,
A = 2.5 ms
–1
± 0.5 ms
–1
, B = 0.10 s ± 0.01 s
x = AB = (2.5)(0.10) = 0.25 m (consider only significant
figure value)
DDD
=+
x AB
x AB
0.5 0.01 0.05 0.025 0.075
2.5 0.10 0.25 0.25
D+
=+==
x
x
Dx = 0.
075 = 0.08 m
(rounding off to two significant figures.)AB = (0.25 ± 0.08) m
8. (d) As given that,
A = 1.0 m ± 0.2 m, B = 2.0 m ± 0.2 m
So, X =
(1.0)(2.0)=AB = 1.414 m
Roundin
g off upto two significant digit
X = 1.4 m = (r) (Let)
1 1 0.2 0.2
2 2 1.0 2.0
D DDéùéù
=+=+
êúêú
ëûëû
x AB
x AB

0.6
2 2.0
=
´
0.6 0.6 1.4
0.212
2 2.0 2 2.0
´
ÞD
===
´´
x
x
Rounding off upto one significant digitDx = 0.2 m = Dr (Let)
So, correct value of
= +DABrr = (1.4 ± 0.2) m

38. 34 PHYSICS
9. (a)
For the most precise measurement, the unit must be
least and number of digits including zeroes after
decimal must be zero.
Now, take first option,
As here 5.00 mm has the smallest unit and the error in
5.00 mm is least (commonly taken as 0.01 mm if not
specified), hence, 5.00 mm is most precise.
10. (a) Now, checking the errors with each options one by
one,
|Dl
1
| = |5 – 4.9| = 0.1 cm
|Dl
2
| = |5 – 4.805| = 0.195 cm
|Dl
3
| = |5.25 – 5| = 0.25 cm
|Dl
4
| = |5.4 – 5| = 0.4 cm
Error Dl
1
is least or minimum.
So, 4.9 cm is most precise.
11. (c) It is given that Young's modulus (Y) is,
Y = 1.9 × 10
11
N/m
2
1N = 10
5
dyne
So, Y = 1.9 × 10
11
× 10
5
dyne/m
2
Convert meter to centimeter
Q 1 m = 100 cm
Y = 1.9 × 10
11
× 10
5
dyne/(100)
2
cm
2
= 1.9 × 10
16 – 4
dyne/cm
2
Y = 1.9 × 10
12
dyne/cm
2
12. (d) Given that fundamental quantities are momentum (p),
area (A) and time (T).
Let us consider the dimensional formula for
[]
abc
E p ATµ
]=
a bc
E kp AT
where k is dimensionless constant of proportionality.
Dimensions of energy [E] = [ML
2
T
–2
] and Dimension
of momentum p = mv = [MLT
–1
]
Dimension of Area [A] = [L
2
]
Dimension of Time [T] = [T]
Dimension of energy [E] = [K] [p]
a
[A]
b
[T]
c
Putting all the dimensions, valueML
2
T
–2
= [MLT
–1
]
a
[L
2
]
b
[T]
c
=
M
a
L
2b + a
T
–a + c
By principle of homogeneity of dimensions,
a = 1, 2b +a = 2 Þ 2b + 1 = 2 Þ
1
2
=b
2- + =-ac
c = –2 + a = –2 + 1
= –1
So, Dimensional formula (of energy)
E = [pA
1/2
T
–1
]1/21
[]
-
=E pAT
NEET/AIPMT (2013-2017) Questions
13. (d) P =
32
ab
cd
,
P
P
D
× 100% =
a
3
a
D
× 100% +
b
2
b
D
×
100% +
c
c
D
× 100% +
d
d
D
× 100%.
= 3 × 1% + 2 × 2% + 3% + 4% = 1
4%
14. (d) Work = Force × displacement
Torque = Force × force arm
= mass × acceleration × length
= [M] × [LT
–2
] × [L] = [M L
2
T
–2
]
15. (d) Force = mass × acceleration
Þ[Mass]
=
force
acceleration
éù
êú
ëû
=
force
velocity / time
éù
êú
ëû
= [F V
– 1
T]
16. (b
) Let surface tension
s = E
a
V
b
T
c
b–2
2–2aCMLTL
(ML T ) (T)
LT
æö
= ç÷
èø
Equating the dimensio
n of LHS and RHS
ML
0
T
–2
= M
a
L
2a + b
T
–2a – b + c
Þa = 1, 2a + b = 0, –2a – b + c = –2
Þa = 1, b = – 2, c = – 2
Hence, the dimensions of surface tension are
[E V
–2
T
–2
]
17. (d) Applying dimensional method :
v
c
= h
x
r
y
r
z
[M
0
LT
–1
] = [ML
–1
T
–1
]
x
[ML
–3
T
0
]
y
[M
0
LT
0
]
z
Equating powers both sides
x + y = 0; –x = –1 \ x = 1
1 + y = 0 \ y = –1
–x – 3y + z = 1
–1 – 3(–1) + z = 1
–1 + 3 + z = 1
\ z = –1
18. (d) Let dimensions of length is related as,
L
z
2
xy
0
e
[c] [G]
4
éù
= êú
peêúëû
2
0
e
4pe
= ML
3
T
–2
L = [LT
–1
]
x
[M
–1
L
3
T
–2
]
y
[ML
3
T
–2
]
z
[L] = [L
x

+ 3y
+ 3z
M
–y + z
T
–x – 2y – 2z
]
Comparing both sides–y + z = 0 Þ y = z ...(i)
x + 3y + 3z = 1 ...(ii)
–x – 4z = 0 (Q y = z) ...(iii)
From (i), (ii) & (iii)
z = y =
1
,
2
x = –2
Hence, L =
1/2
2
2
0
e
cG
4
-
éù
×êú
peêúëû

39. BASIC DEFINITIONS
Mechan
ics : Branch of physics, which deals with the study of
objects in rest and in motion.
Statics : Study of objects at rest or in equilibrium.
Kinematics : Study of motion of objects without considering the
cause of motion.
Dynamics : Study of motion of objects considering the cause of
motion.
Rest : An object is said to be at rest if it does not change its
position with time, with respect to its surrounding (a reference
point which is generally taken as origin in numerical problems)
Motion : An object is said to be in motion if it changes its position
with time, with respect to its surroundings.
Rest and motion are relative terms.
Point mass/Point object : An object is said to be a point mass if
during its motion it covers distance much greater than its own
size.
One dimensional motion : An object travels in a straight line. It is
also called rectilinear or linear motion. The position change of
the object with time in one dimension can be described by only
one co-ordinate.
Ex. A stone falling freely under gravity.
Two dimensional motion or motion in a plane : For an object
travelling in a plane two coordinates say X and Y are required to
describe its motion.
Ex. An insect crawling over the floor.
Three dimensional motion : An object travels in space.To describe
motion of objects in three dimension require all three coordinates
x, y and z.
Ex. A kite flying in the sky.
DISTANCE AND DISPLACEMENT
Distance or Path length : The length of the actual path travelled
by an object during motion in a given interval of time is called
the distance travelled by that object or path length. It is a scalar
quantity.
Displacement : It is the shortest distance between the initial and
final position of an object and is directed from the initial position
to the final position. It is a vector quantity.
Keep in Memory
1.Displ
acement may be positive, negative or zero but distance
is always positive.
2.Displacement is not affected by the shift of the coordinate
axes.
3.Displacement of an object is independent of the path
followed by the object but distance depends upon path.
4.Displacement and distance both have same unit as that of
length i.e. metre.
5.
Distance
1
|Displacement|
³
6.For a moving body distance always increases with time
7.For a body undergoing one dimensional motion, in the samedirection distance = | displacement |. For all other motion
distance > | displacement |.
SPEED
It is the distance travelled per unit time by an object. It is a scalar
quantity. It cannot be negative.
Uniform speed : An object is said to be moving with a uniform
speed, if it covers equal distances in equal intervals of time,
howsoever small the time intervals may be.
Non-uniform speed : If an object covers unequal distances in
equal interval of time or equal distances in unequal interval of
time.
Instantaneous speed : The speed of an object at a particular instant
of time is called the instantaneous speed.
Instantaneous speed, V
inst
=
0
lim
D®
D
=
Dt
s ds
t dt
.
Average sp
eed : It is ratio of the total distance travelled by the
object to the total time taken.
Average speed V
av
x
t
D
=
D
Dimensions : [M
0
LT
-1
]; Unit: In SI systems.
VELOCITY
It is the displacement of an object per unit time. It is a vector
quantity. It can be positive negative or zero.
Uniform velocity : An object is said to be moving with a uniform
velocity, if it covers equal displacements in equal intervals of
time, howsoever small the time intervals may be.
Non-uniform velocity : If an object covers unequal displacements
in equal interval of time or equal displacements in unequal interval
of time.
Instantaneous velocity : The velocity of an object at a particular
instant of time is called the instantaneous velocity.
3
Motion in a
Straight Line

40. 36 PHYSICS
Variable or n
on-uniform acceleration : If the velocity of body
changes in different amounts during same time interval, then the
acceleration of the body is known as variable acceleration.
Acceleration is variable if either its direction or magnitude or
both changes with respect to time. A good example of variable
acceleration is the acceleration in uniform circular motion.
EQUATIONS FOR UNIFORMLY ACCELERATED MOTION
When the motion is uniformly accelerated i.e., when acceleration
is constant in magnitude and direction :
(i)v = u + at
(ii) 2
at
2
1
uts+=
(iii) v
2
– u
2
= 2as
where u = in
itial velocity; v = final velocity;
a = uniform acceleration ands = distance travelled in time t,
(iv)
uv
st
2
+æö
=ç÷
èø
(v) Distance travelled in n
th
second s
n
= 1)(2n
2
a
u -+;
s
n
= distance cov
ered in n
th
second
Above equations in vector form ,ta
2
1
tus,tauv
2
+=+=
)s.a2u.uv.vor(s.a2uv
22 rrrrrrrrrr
=+=-
t)vu(
2
1
s
rrr
+=
When displacement (s)
is given as a function of time t [s = f(t)]
then
0
tsµ Body at rest v = 0 a = 0
1
tsµ Uniform velocity;
acceleration zero
0
tvµ a = 0
2
tsµ Uniform acceler ation
1
tvµ
0
taµ
moreor3
tsµ Non uniform
acceleration
moreor
2
tvµ
moreor1
taµ
We use calculus method (integration and differentiation)
for displacement, velocity, acceleration as a function of time.
We know that
ò
=Þ=dtvs
dt
ds
v ; ò
=Þ=dtav
dt
dv
a ;
when a = f(s
)
òò
=Þ=dvvdsa
ds
dv
va ,
where s = displace
ment, v = instantaneous velocity,
a = instantaneous acceleration
Instantaneous velocity,
0
lim
D®
D
==
D
rr
r
inst
t
r dr
V
t dt
.
Average Velocity
: It is ratio of the total displacement to the total
time taken.
Average velocity,
D
=
D
r
av
r
v
t
Dimensions : [M
0
LT
–1
] ; Unit: In SI system, m/s
Keep in Memory
1.| Average veloci
ty | can be zero but average speed cannot be
zero for a moving object.
2.
|Average velocity|
1
Average speed
£
3.| Instantaneous velocity | = Instantaneous speed.
4.A particle may have constant speed but variable velocity. It
happens when particle travels in curvilinear path.
5.If the body covers first half distance with speed v
1
and
next half with speed v
2
then
Average speed
12
12
2
=
+
vv
v
vv
6.If a body covers first one-third distance at a speed v
1
, next
one-third at speed v
2
and last one-third at speed v
3
, then
Average speed
123
12 23 13
3
=
++
vvv
v
vv v v vv
7.If a body travels with uniform speed v
1
for time t
1
and with
uniform speed v
2
for time t
2
, then
Average speed
11 22
12
+
=
+
vt vt
v
tt
ACCELERATION
The rate of ch
ange of velocity with respect to time is called
acceleration. It is a vector quantity.
Let velocity changes by
v
r
D during some interval of time tD.
Average acceleration
av
a
r
is given by
D
=
D
r
r
av
v
a
t
Instantaneous acceleration a
r
is given by

0
lim
D®
D
==
D
rr
r
t
v dv
a
t dt
Its SI unit is meter/se c
2
(ms
–2
).
A body moving with uniform velocity has zero acceleration. It
means that neither its speed nor its direction of motion is changing
with time.
Uniform acceleration : If the velocity of the body changes in
equal amount during same time interval, then the acceleration of
the body is said to be uniform. Acceleration is uniform when
neither its direction nor magnitude changes with respect to time.

41. 37Motion in a Straight Line
VERTICAL MOTION UNDER GRAVITY
(i) For a body thrown downward with initial velocity u from a
height h, the equations of motion are
v = u +gt ;
22
v u 2gh=+
h = ut +
21
gt
2
; th
n
h

=
g
u (2n 1)
2
+-
(ii) If initial velocity is zero, then the equations are
v = gt ; v 2gh=
h =
21
gt
2
; th
n
g
h (2n 1)
2
=-
(iii)When a body is thrown upwards with initial velocity u, the
equations of motion are
v = u – gt
h = ut –
21
gt
2
v
2
= u
2
–

2gh.
To summa
rise :
Motion of a body
Displacement V
elocity Acceleration
Velocity constant
Uniformly accelerated
motion
Non-uniformly
accelerated moti on
Accelerated motion
Velocity =
D
isplacement
Time
t)vu(
1
s
)1n2(
2
a
us
as2uv
at
2
1
uts
atuv
nth
22
2
´+=
-
+=
=-
+=
+=
dt
ds
v=
dt
dv
a=
Differentiation
Integration
Differentiation
Integration
òò
=dtv
ds
òò
=dtadv
dv
v
ds
=
2
N OTE: Calculus method as shown in non-uniformly accelerated
motion may also be used for uniformly accelerated motion.
Uniformly Accelerated Motion : A Discussion
While using equations of motion we can have two approaches.
Approach 1 :Take a = +ve when velocity increases and
a = –ve when velocity decreases.
Take rest of physical quantities such as u, v, t and s as positive.
Approach 2 : (Vector method)
Assume one direction to be positive and other negative. Assign
sign to all the vectors (u, v, a, s), +ve sign is given to a vector
which is directed to the positive direction and vice-versa
–
–
–
+
+
+
Normally the direction taken is as drawn above. But it is important
to note that you can take any direction of your choice to be positiveand the opposite direction to be negative.
N OTE: The second method (or approach) is useful only when
there is reversal of motion during the activity concerned.
Keep in Memory
1.Th
e direction of average acceleration vector is the direction
of the change in velocity vector.
t
vv
a
if
rr
r -
=
a
r
has a direction of )v(–vvv
ifif
rrrr
+=-
i.e., the resultant of
f
v
r
and
i
v–
r
2.There is no definite relationship between velocity vector
and acceleration vector.
3.For a body starting from rest and moving with uniform
acceleration, the ratio of distances covered in t
1
sec.,
t
2
sec, t
3
sec, etc. are in the ratio t
1
2
: t
2
2
: t
3
2
etc.
4.A body moving with a velocity v is stopped by application
of brakes after covering a distance s. If the same body moves
with a velocity nv, it stops after covering a distance n
2
s by
the application of same retardation.
Example 1.
The displacement of particle is zero at t = 0 and
displacement is ‘x’ at t = t. It starts moving in the positive
x-direction with a velocity which varies as v=kx where
k is cons
tant. Show that the velocity varies with time.

42. 38 PHYSICS
Solution :
dtk
x
dx
orxk
dt
dx
orxkv ===
Given that
when t = 0, x = 0 and when t = t, x = x,
Hence
òò
=
t
0
x
0
dtk
x
dx
;
òò
=\
-
t
0
x
0
½
dtkdxx or []
x
1
1
2
t
0
0
x
kt
1
1
2
-+
éù
êú
=êú
êú-+
êúëû
or
2
tk
xortkx2==
Now,
2
tk
2
tk
kv
2
=
ú
û
ù
ê
ë
é
´=
Thus the velocity var
ies with time.
Example 2.
A particle covers each 1/3 of the total distance with speed
v
l
, v
2
and v
3
respectively. Find the average speed of the
particle.
Solution :
Average speed
123
Total distance travelled s
v
sssTotal time taken
3v 3v 3v
==
++
Þ
123
1 2 2 3 31
3vvv
v
vv vv vv
=
++
Example 3.
A cheetah can accelerate from
0 to 96 km/h in 2 sec., whereas
a cat requires 6 sec. Compute the average accelerations forthe cheetah and cat.
Solution :
For cheetah |
r
aav
| =
fi
|v v|
t
-
D
rr
=
96km/h0
2sec
-
=
1000m
96
3600sec
2sec
´
= 15 m/s
2
For cat |
r
aav
| =
10
96
36
6
´
= 5 m/s
2
.
Example 4.
A particle is moving i
n east direction with speed 5 m/s.
After 10 sec it starts moving in north direction with same
speed. Find average acceleration.
Solution :
|
r
vf
| = |
r
vi
| = 5 m/s
Accelerati
on ¹ 0
E
S
N
W
(due to change in directio
n
of velocity Av. acceleration,
D
r
v =
r
vf
–
r
vi
=
r
vf
+ (–
r
vi
)
r
a=
fi
vv
t
-
D
rr
=
ˆˆ
5j 5i
10
-
Þ
r
a = –
1
2
$
i
+
1
2
$
j
|
r
a| =
1
2
1
2
2 2
F
H
G
I
K J+
F
H G
I
K J =
1
2
m/s
2
.
Keep in Memory
1.An obje
ct moving under the influence of earth's gravity in
which air resistance and small changes in g are neglected is
called a freely falling body.
2.In the absence of air resistance, the velocity of projection
is equal to the velocity with which the body strikes the
ground.
3.Distance travelled by a freely falling body in 1st second is
always half of the numerical value of g or 4.9 m, irrespective
of height h.
4.For a freely falling body with initial velocity zero
(i) Velocity µ time (v = gt)
(ii) Velocity
Distance fallenµ (v
2
= 2gs)
(iii)
Distance fallen a (time)
2
÷
ø
ö
ç
è
æ
=
2
gt
2
1
s , where g is
the
acceleration due to gravity.
5.If maximum height attained by a body projected vertically
upwards is equal to the magnitude of velocity of projection,
then velocity of projection is 2g ms
–1
and time of flight is 4
sec.
6.If maximum height attained by a body projected upward is
equal to magnitude of acceleration due to gravity i.e., 'g',
the time of ascent is
2 sec. and velocity of projection is
2g.
7.Ratio of maxi
mum heights reached by different bodies
projected with velocities u
1
, u
2
, u
3
etc. are in the ratio of
3
3
2
2
2
1
u:u:u etc. and ratio of times of ascent are in ratio of
u
1
: u
2
: u
3
etc.
8.During free fall velocity increases by equal amount every
decend and distance covered during 1st, 2nd, 3rd seconds
of fall, are 4.9m, 14.7m, 24.5m.
t = 3
4.9 m
14.7 m
24.5 m
t = 2
t = 1
t = 0u = 0
9.8 m/s
19.6 m/s
29.4 m/s
9.If a body is projected horizontally from top of a tower, the
time taken by it to reach the ground does not depend on
the velocity of projection, but depends on the height of
tower and is equal to t =
g
h2
.

43. 39Motion in a Straight Line
10If velocity v of a body changes its direction by q without
change in magnitude then the change in velocity will be
2
sinv2
q
.
11.From the top of a tower a body is projected upward with a
certain speed, 2nd body is thrown downward with same
speed and 3rd is let to fall freely from same point then
213ttt=
where t
1
= time taken by the body projected upward,
t
2
= time taken by the body thrown downward and
t
3
= time taken by the body falling freely.
12.If a body falls freely from a height h on a sandy surface and
it buries into sand upto a depth of x, then the retardation
produced by sand is given by
hx
ag
x
+æö
=ç÷
èø
.
13.In case of air resistance, the time of ascent is less than time of descent of a body projected vertically upward i.e. t
a
< t
d
.
14.When atmosphere is effective, then buoyancy force always acts in upward direction whether body is moving in upward or downward direction and it depends on volume of the body. The viscous drag force acts against the motion.
15.If bodies have same volume but different densities, the buoyant force remains the same.
CAUTION : Please note that dropping body gets the velocity
of the object but if the object is in acceleration, the body dropped
will not acquire the acceleration of the object.
COMMON DEFAULT
Incorrect. In the question, if it is given that a body is
dropped, taking its initial velocity zero.
Correct. The initial velocity is zero if the object dropping
the body is also at rest (zero velocity). But if the object
dropping the body is having a velocity, then the body being
dropped will also have initial velocity which will be same as
that of the object.
For example :
(a) When an aeroplane flying horizontally drops a bomb.
(b) An ascending helicopter dropping a food packet.
(c) A stone dropped from a moving train etc.
Incorrect. Applying equations of motion in case of non-
uniform acceleration of the body.
Correct. The equations of motion are for uniformly
accelerated motion of the body.
Please note that when the case is of non-uniform acceleration
we use calculus (differentiation and integration).
dt
ds
v;
ds
vdv
dt
dv
a === .
In fact calculus method is a universal method which can be used both in case of uniform as well as non-uniform acceleration.
Incorrect. Taking average velocity same as that of
instantaneous velocity.
Correct. Average velocity
if
if
av
t–t
rr
t
r
V
-
=
D
D
==
(where
i
r is position vector at time t
i
and
f
r is position
vector at time t
f
).
Whereas instantaneous velocity
inst.
t0
r dr
V lim
t dtD®
D
==
D
ur ur
u ur
It is important to note that average velocity is equal to
instantaneous velocity only when the case is of uniform
velocity.
Incorrect. Taking acceleration as negative (– a) even when
acceleration is an unknown.
Correct. Take acceleration as (a) when it is unknown even if
we know that the motion is a case of deceleration or retardation. On solving, we will find the value of (a) to be
negative .
Incorrect. Magnitude of instantaneous velocity is different
from instantaneous speed .
Correct. Magnitude of instantaneous velocity is equal to
the instantaneous speed in any case.
Example 5.
The numerical ratio of average velocity to average speed is
(a) always less than one (b) always equal to one
(c) always more than one (d) equal to or less than one
Solution : (d)
It is equal to or less than one as average velocity depends
upon displacement whereas average speed depends upon
path length.
Example 6.
The distance travelled by a body is directly proportional
to the time taken. Its speed
(a) increases (b) decreases
(c) becomes zero (d) remains constant
Solution : (d)
When s µ t, so
s
v constant.
t
==
Example 7.
A particle is projected vertically upwards. Prove that it will be
at 3/4 of its greatest height at time which are in the ratio 1 : 3.
Solution :
If u is the initial velocity of a particle while going vertically
upwards, then the maximum height attained is
g2
u
h
2
=.
If t is the time when particle reaches at a height
4
3
h, then
using the relation
2
at
2
1
uts+= ; we have
2
t)g(
2
1
uth
4
3
-+=
or,
2
2
gt
2
1
ut
g2
u
4
3
-=
÷
÷
ø
ö
ç
ç
è
æ
or
0
g
u
4
3
t
g
u2
t
2
2
2
=+-

44. 40 PHYSICS
Solving
it for t, we have
g2
u
g
u
2
g
u
4
3
14
g
u4
g
u2
t
2
2
2
2
±=
´´-±
=
t = u/g
t = 3u/2g
2
3h/4
t
1
h
Taking negative sign,
g2
u
g2
u
g
u
t
1 =-=;
Taking positive sign, .g2/u3
g2
u
g u
t
2 =+=
3
1
g2/u3
g2/u
t
t
2 1
=
÷
÷
ø
ö
ç
ç
è
æ
=
.
Example 8.
A police party in a j
eep is chasing a dacoit on a straight
road. The jeep is moving with a maximum uniform speed v.
The dacoit rides on a motorcycle of his waiting friend when
the jeep is at a distance d from him and the motorcycle
starts with constant acceleration a. Show that the dacoit
will be caught if
³v 2ad.
Solution :
Suppose the
dacoit is caught at a time t sec after the
motorcycle starts. The distance travelled by the motorcycle
during this interval is
2
at
2
1
s= ...(1)
During the interval, the jee
p travels a distance,
s + d = vt ...(2)
By (1) and (2),
0dvtat
2
12
=+-or
a
ad2vv
t
2
-±
=
The dacoit will be
caught if t is real and positive. This will be
possible if
ad2vor0ad2v
2
>>-.
Example 9.
Two trains, each
of length 100 m, are running on parallel
tracks. One overtakes the other in 20 second and onecrosses the other in 10 second. Calculate the velocities ofeach train.
Solution :
Let u and v be the velocities of the trains.The relative velocity of overtaking is u – v while the relativevelocity of crossing is u + v.
Total distance = 100 + 100 = 200 m.
10vuor
vu
200
20 =-
-
=\
and 20vuor
vu
200
10 =+
+
=
By solving, w
e get, u = 15 m/sec and v = 5 m/sec.
Example 10.
A body starts from rest and moves with a uniformacceleration. The ratio of the distance covered in the nthsec to the distance covered in n sec is
(a)
2
21
–
nn
(b)
2
11
–
nn
(c)
22
21
–
nn
(d)
2
21
+
nn
Solution : (a)
The d
istance covered in nth second
)1n2(
2
a
us
n -+= or )1n2(
2
a
0s
n -+=...(1)
Further di
stance covered in n second
22
na
2
1
0ta
2
1
tus +=+= ...(2)
22
n
n
1
n
2
)2/an(
)1n2(
2
a
s
s
-=
-
=\
Example 11.
The water drop falls at regular i
ntervals from a tap 5 m
above the ground. The third drop is leaving the tap at
instant the first drop touches the ground. How far above
the ground is the second drop at that instant?
(a) 1.25 m (b) 2.50 m
(c) 3.75 m (d) 4.00 m
Solution : (c)
See fig. Let t be the time interval between any two drops.
5 m
x
h
Ist drop
3rd drop
ground
2nd drop
Tap
For third drop 2/5tgor)t2(g
2
1
5
22
==
For second drop
2
tg
2
1
x= or m25.1
4
5
2
5
2
1
x ==´=
Therefor
e h = 5 – x = 5 – 1.25 = 3.75 m
Hence option (c)

45. 41Motion in a Straight Line
Example 12.
The height of a tower is h metre. A body is thrown from the
top of tower vertically upward with some speed, it takes
t
1
, second to reach the ground. Another body thrown from
the top of tower with same speed downwards and takes t
2
seconds to reach the ground. If third body, released from
same place takes ‘t’ second to reach the ground, then
(a)
12
t +t
t=
2
(b)
1
2
t
t=
t
(c)
12
211
=+
ttt
(d)
12
t= (tt)
Solution : (d)
Let u be the initial velocity of the body. Then at time t
1
2
11tg
2
1
tuh+-= ... (1)
and at time
t
2
, h =
2
22
tg
2
1
tu+ ...(2)
From eqs. (1) an
d (2), we get
2
22
2
11
tg
2
1
tutg
2
1
tu +=+-
or )tt(u)tt(g
2
1
21
2
2
2
1
+=-
or )tt(u)tt()tt(g
2
1
212121+=-+
or )tt(g
2
1
u
21
-= ...(3)
Substituting th
e value of u in eqn. (2) we get
2
2221tg
2
1
t)tt(g
2
1
h +-= or 21ttg
2
1
h= ...(4)
For th
ird body,
2
tg
2
1
h= ...(5)
From eqs
. (4) and (5) we get,
12
t tt=
Example 13.
A bullet m
oving with a speed 10 m/s hits the wooden plank
and is stopped when it penetrates the plank 20 cm deep.
Calculate retardation of the bullet.
Solution :
v
0
= 10 m/s, v = 0 and s = 20 cm. =
2
0.02m
100
=
Using v
2
– v
0
2
=
2ax
2 100
0 (10) 2a (0.02) a
2 0.02
-
-= Þ=
´
or a = – 2500 m/s
2
Ret
ardation = 2500 m/s
2
Example 14.
A body covers a distance of 20m in the 7th second and 24m
in the 9th second. How much distance shall it cover in
15th sec?
Solution :
From formula,
th
n
S u a(2n –1 )
2
1
=+
7
a
S th u (2 7 1)
2
= + ´- but
7
S th 20m=
\
a 13a
20 u 13 20 u
22
=+´ Þ =+ ... (1)
a
lso
9
s th 24m= \
17a
24u
2
=+ ... (2)
From eq
n
. (1
)
13a
u 20
2
=- ... (3)
Substitute
this value of in eq
n
. (2)
13a 17a
24 20
22
=-+
17a 13a
24 20
22
-=-
4a
4
2
= Þ 4 = 2a Þ
24
a 2m/s
2
==
Substituting this value
of a in eqn.

(3)
13a
u 20
2
=-
Þ
132
u 20 u 20 13
2
´
=- Þ=-
\ u = 7 m/s
Now, s
15th
=
a
u (2 15 1)
2
+ ´-
=
2
7 (29) 7 29 36m
2
+ =+=
Example 15.
A tra
in starts from rest and for the first kilometer moves
with constant acceleration, for the next 3 kilometers it has
constant velocity and for another 2 kilometers it moves
with constant retardation to come to rest after 10 min.
Find the maximum velocity and the three time intervals in
the three types of motion.
Solution :
Let the three time intervals be t
1
min, t
2
min, and t
3
min.
respectively.
Let the maximum velocity attained be v m/min.
A B C D
1000m 3000m 2000m
t
1 v–t
2 t
3
For A to
B
1
0v
1000 t
2
+æö
=ç÷
èø
Þ
1
vt2000= ........... (1)
For
B to C
2
vt3000= ........... (2)
[Using in both equations disp. = mean vel. × time]
and for C to D
3
v0
2000 t
2
+æö
=ç÷
èø
Þ
3
vt4000= ...... (3)
Addin
g eqs. (1), (2) and (3), we get
)ttt(v9000
321
++=

46. 42 PHYSICS
\
3
9000 900 10
v 900 m / min. km 54 km / hr.
10 1/ 60 hr.
-
´
====
No
w, from eqs. (1), (2) and (3) we get
1
2000 20 2
t 2 min.
90099
= == ,
2
3000 10 1
t 3 min.
900 9 3
= ==
and 3
4000 40 4
t 4 min.
900 9 9
= ==
Example 16.
A falling
stone takes 0.2 seconds to fall past a window
which is 1m high. From how far above the top of the window
was the stone dropped ?
Solution :
21
h gt
2
= ;
21
h 1 g(t 0.2)
2
+=+
2221111
gt 1 gt g(0.2) g 2 0.2t
2222
+= + + ´´
u=0
1m
h
1
1 0.2gt
5
=+ ;
42
2tt
55
= Þ=
144
hgm
2255
==
Example 17.
From the top of a multi-storeye
d building 40m tall, a boy
projects a stone vertically upwards with an initial velocityof 10 ms
–1
such that it eventually falls to the ground. (i)
After how long will the stone strike the ground ? (ii) Afterhow long will it pass through the point from where it wasprojected ? (iii) What will be its velocity when it strikes theground ? Take g = 10 ms
–2
.
Solution :
Using,
21
S ut at
2
r rr
=+
(i)
21
40 10t gt
2
-=- or
2
40 10t 5t-=-
[Q taking g = 10m/s
2
]
or
2
5t 10t 40 0- -=
h
v
u
u=10m/s
40m
or
2
10 10 4 5( 40) 10 100 800
t
2 5 10
+ -´- ++
==
´

10 30
4 sec.
10
+
==
(ii)It will pass from whe
re it was projected after
2 10
t 2sec.
g
´
==
(iii)Velocity with
which stone strikes the ground
V = 10 + g × 2 = 30 m/s
VARIOUS GRAPHS RELATED TO MOTION
(a)Displacement-time graph - In this graph time is plotted on
x-axis and displacement on y-axis.
(i) For a stationary body (v = 0) the time-displacement
graph is a straight line parallel to time axis.
Displ
Velocity = 0
Time
Body is at rest
(ii) When the velocity of a body is constant then time-
displacement graph will be an oblique straight line.
Greater the slope ÷
ø
ö
ç
è
æ
q=tan
dt
dx
of the straight line,
higher will be the velocity.
(iii) If the velocity of a body is not constant then the time-
displacement curve is a zig-zag curve.
(iv) For an accelerated motion the slope of time-
displacement curve increases with time while for
decelerated motion it decreases with time.

47. 43Motion in a Straight Line
a > 0
t
Displ.
a < 0
t
Displ.
(v) W
hen the particle returns towards the point of
reference then the time-displacement line makes an
angle q > 90° with the time axis.
t
Displ
q > 90º
(b)Velocity-time graph - In this curve time is plotted along
x-axis and velocity is plotted along y-axis.
(i) When the velocity of the particle is constant or
acceleration is zero.
t
VelocityVel . const, a = 0
(ii) When the particle is moving with a constant
acceleration and its initial velocity is zero.
(iii)When the particle is moving with constant retardation.
t
V Retardation
(iv) When the particle moves with non-uniform acceleration
and its initial velocity is zero.
(v) When the acceleration decreases and increases.
t
V
a d
e c r e a s i n
g
a in creasin g
(vi) The total area enclosed by the time - velocity curve
represents the distance travelled by a body.
N OTE: While finding displacement through v – t graph,
keeping sign under consideration.
(c)Acceleration-time graph - In this curve the time is plotted
along X-axis and acceleration is plotted along Y-axis.
(i) When the acceleration of the particle is zero.
t
a Acc = 0
(ii) When acceleration is constant
t
a = const
a

48. 44 PHYSICS
(iii)
When acceleration is increasing and is positive.
t
a a increasing
(iv) When acceleration is decreasing and is negative
t
a d
e
c
r
e
a
s
i
n
g
a
(v) When initial acceleration is zero and rate of change of
acceleration is non-uniform
t
a
(vi) The change in velocity of the particle = area enclosed
by the time-acceleration curve.

t
a
Example 18.
The velocity-ti
me graph of a body moving in a straight line
is shown in fig. Find the displacement and distance
travelled by the body in 10 seconds.
20
10
0
–10
–20
A
42 6 8 10
C
D
t (s)
u (m/s)
Solution :
The area e
nclosed by velocity-time graph with time axis
measures the distance travelled in a given time. Displacement
covered from 0 to 6 seconds is positive; from 6 to 8 seconds
is negative and from 8 to 10 seconds is positive; whereas
distance covered is always positive.
Total distance covered in 10 s
102
2
1
202
2
1
206
2
1
´´+´´+´´== 90 m
Total disp
lacement in 10s m50102
2
1
202
2
1
206
2
1
=´´+´´-´´=
Example 19.
Two trains, w
hich are moving along different tracks in
opposite directions, are put on the same track due to a
mistake. Their drivers, on noticing the mistake, start slowing
down the trains when the trains are 300 m apart. Graphs
given below show their velocities as function of time as the
trains slow down, The separation, between the trains when
both have stopped, is
10
20
40
V(m/s)
t (s)
Train I
8
–20
V(m/s)
t (s)
Train II
(a) 120 m (b) 28 0 m
(c) 60 m (d) 20 m.
Solution : (d)
Initial distance between trains is 300m. Displacement of 1st
train is calculated by area under v – t.
Curve of train
1
I displacement 10 40 200m
2
= =´´=
Displa
cement of train
1
II 8 ( 20) 80m
2
= ´ ´ - =-
\ Dis
tance between the two trains is 20m.
Example 20.
Figure given below shows the variation of velocity of a
particle with time.

1234567
v
e
l
o
c
i
t
y
(m/s) 2
4
time(sec)
Y
X
6
8
Find the following :
(i) D
isplacement during the time intervals
(a) 0 to 2 sec, (b) 2 to 4 sec. and (c) 4 to 7 sec.
(ii) Accelerations at –
(a) t = 1 sec, (b) t = 3 sec. and (c) t = 7 sec.
(iii) Average acceleration –
(a) between t = 0 to t = 4 sec.
(b) between t = 0 to t = 7 sec.
(iv) Average velocity during the motion.

49. 45Motion in a Straight Line
Solution :
(i) (a)Displacement between t = 0 sec. to t = 2 sec.
Þ m8s/m8sec2
2
1
=´´
(b)Between t = 2 sec.
to t = 4 sec.Þ m16s/m8sec2 =´
(c)Between t = 4 sec.
to t = 7 sec.
Þ m12s/m8sec3
2
1
=´´
(ii)Acceler
ation = slope of v – t curve
(a)At t = 1 sec,
slope =
2
s/m4sec/m
sec2
sec/m8
=
(b)At t =
3 sec, slope = 0
(c)At t = 7 sec,
slope =
2
s/m
3
2
2
3
8
-=-
(iii)Averag
e acceleration =
Total change in velocity
Total change in time
(a)Between t = 0 to t = 4 sec.,
Average acceleration =
2
s/m2
4
8m/s
=
(b)Betwe
en t = 0 to t = 7 sec.,
Average acceleration =
0
7
0
=
(iv)Aver
age velocity =
Total displacement
Total time
= s/m
7
1
5
7
36
7
12168
==
++
Example 21.
The v
elocity-time graph of a particle moving along a
straight line is shown below.
The acceleration and deceleration are same and it is equal
to 4 m/s². If the average velocity during the motion is 15 m/
s and total time of motion is 20 second then find
t
t 2020–t
20 – 2t
v
(a) the value of t (b)
the maximum velocity of the particle
during the journey. (c) the distance travelled with uniform
velocity.
Solution : v = 0 + at
Total displacement =
1
(20 2t 20) 4t 2t (40 2t)
2
-+´=-
Av
erage velocity =
Total displacement
Total time

2t (402t)
15
20
-
=
Solving quadrat
ic equation, 150 = 40t – 2t
2
Þ t = 5 sec.
(another solution not acceptable think why!)
Maximum velocity = 4t = 4 × 5 = 20 m/s
Distance travelled with uniform velocity
= (20 – 2t) V = (20 – 2 × 5) × 20 = 200 m
RELATIVE VELOCITY (In one dimension)
The velocity of A relative to B is the velocity with which A appears
to be moving w.r.t.an observer who is moving with the velocity of B
Relative velocity of A w.r.t. B
AB AB
v vv=-
r rr
Similarly, relative velocity of B w.r.t. A
BA BA
v vv=-
r rr
Case 1: Bodies moving in same direction :
v
A
v
B
AB AB
v vv=-
r rr
Þ v
AB
= v
A
– v
B
Case 2: Bo
dies moving in opposite direction :
– +v
A
v
B
()= --
rrr
ABAB
vvv Þ v
AB
= v
A
+ v
B
Example 22.
T
wo trains, one travelling at 54kmph and the other at
72kmph, are headed towards each other on a level track.
When they are two kilometers apart, both drivers
simultaneously apply their brakes. If their brakes produces
equal retardation in both the trains at a rate of 0.15 m/s
2
,
determine whether there is a collision or not.
Solution :
Speed of first train = 54 kmph = 15m/s.
Speed of second train = 72kmph = 20 m/s
As both the trains are headed towards each other, relative
velocity of one train with respect to other is given as
v
r
=15 + 20 = 35 m/s
Both trains are retarded by acceleration of 0.15 m/s
2
. Relative
retardation a
r
= 0.15 + 0.15 = 0.3 m/s
2
.
Now, we assume one train is at rest and other is coming at
35m/s retarded by 0.3 m/s
2
at a distance of two kilometer.
The maximum distance travelled by the moving train while
retarding is
s
max
=
v
a
r
r
2
2
=
()
.
35
2 03
2
´
= 2041.6
6m
It is more than 2km, which shows that it will hit the secondtrain.
Example 23.
Two cars started simultaneously towards each other from
towns A and B which are 480 km apart. It took first car
travelling from A to B 8 hours to cover the distance and
second car travelling from B to A 12 hours. Determine
when the car meet after starting and at what distance from
town A. Assuming that both the cars travelled with constant
speed.
Solution : Velocity of car from A =
hour/km60
8
480
=
velocity of ca
r from B =
hour/km40
12
480
=
Let the two c
ars meet at t hour
total distance 480
t 4.8
relative velocity of meeting 60 40
\= ==
+
hour
The distance s = v
A
× t = 60 × 4.8 = 288 km.

50. 46 PHYSICS

51. 47Motion in a Straight Line
1.The study of motion, without consideration of its cause is
studied in
(a) statistics (b) kinematics
(c) mechanics (d) modern physics
2.The ratio of the numerical values of the average velocity
and average speed of a body is always:
(a) unity (b) unity or less
(c) unity or more (d) less than unity
3.A particle has moved from one position to another position
(a) its distance is zero
(b) its displacement is zero
(c) neither distance nor displacement is zero
(d) average velocity is zero
4.The displacement of a body is zero. The distance covered
(a) is zero
(b) is not zero
(c) may or may not be zero
(d) depends upon the acceleration
5.Which of the following changes when a particle is moving
with uniform velocity?
(a) Speed (b) Velocity
(c) Acceleration (d) Position vector
6.The slope of the velocity time graph for retarded motion is
(a) positive (b) negative
(c) zero (d) can be +ve, –ve or zero
7.The area of the acceleration-displacement curve of a body
gives
(a) impulse
(b) change in momentum per unit mass
(c) change in KE per unit mass
(d) total change in energy
8.If the displacement of a particle varies with time as
7tx+=, the
(a) ve
locity of the particle is inversely proportional to t
(b) velocity of the particle is proportional to t
(c) velocity of the particle is proportional to
t
(d) the
particle moves with a constant acceleration
9.The initial velocity of a particle is u (at t = 0) and the
acceleration a is given by f t.
Which of the following relation is valid?
(a) v = u + f t
2
(b) v = u + f t
2
/2
(c)v = u + f t (d) v = u
10.The displacement x of a particle moving along a straight line
at time t is given by
x = a
0
+ a
1
t + a
2
t
2
What is the acceleration of the particle
(a)a
1
(b) a
2
(c) 2 a
2
(d) 3 a
2
11.The displacement-time graphs of two particles A and B are
straight lines making angles of respectively 30º and 60º with
the time axis. If the velocity of A is v
A
and that of B is v
B
, the
value of v
A
/v
B
is
(a) 1/2 (b)
3/1
(c)3 (d) 1/3
12.A pe
rson travels along a straight road for the first half time
with a velocity v
1
and the second half time with a velocity
v
2
. Then the mean velocity
v is given by
(a)
2
vv
v
21+
= (b)
21v
1
v
1
v
2
+=
(c)
21
vvv= (d)
1
2
v
v
v=
13.A particle cov
ers half of the circle of radius r. Then the
displacement and distance of the particle are respectively
(a)2pr, 0 (b) 2r, pr
(c)
r2,
2
rp
(d)pr, r
14.The d
istance through which a body falls in the nth second
is h. The distance through which it falls in the next second is
(a)h (b)
2
g
h+
(c) h – g (d) h
+ g
15.It is given that t = px
2
+ qx, where x is displacement and t is
time. The acceleration of particle at origin is
(a)
3
q
p2
- (b)
3
p
q2
- (c)
3
q
p2
(d)
3
p
q2
16.Figure shows the v-t graph for two particles P and Q. Which
of the following statements regarding their relative motion
is true ?
Their relative velocity is
O
V
Q
T
P
(a) is zero
(b) is non-zero but constant
(c) continuously decreases
(d) continuously increases
17.A stone is dropped into a well in which the level of water is
h below the top of the well. If v is velocity of sound, the time
T after which the splash is heard is given by
(a) T = 2h/v (b)
2hh
T
gv
æö
=+
ç÷
èø
(c)
g
h
v
h2
T +÷
ø
ö
ç
è
æ
= (d)
v
h2
g2
h
T +
÷
÷
ø
ö
ç
ç
è
æ
=

52. 48 PHYSICS
18.A point trav
ersed half of the distance with a velocity v
0
.
The half of remaining part of the distance was covered with
velocity v
1
& second half of remaining part by v
2
velocity.
The mean velocity of the point, averaged over the whole
time of motion is
(a)
3
vvv
210++
(b)
3
vvv2
210++
(c)
3
v2v2.v
210++
(d)
)vvv2(
)vv(v2
210
210
++
+
19.The acceleration of a particle is increasing linearly with time
t as bt. The particle starts from the origin with an initial
velocity v
0
. The distance travelled by the particle in time t
will be
(a)
2
0 bt
3
1
tv+ (b)
3
0 bt
3
1
tv+
(c)
3
0 bt
6
1
tv+ (d)
2
0
bt
2
1
tv+
20.The decelerati
on experienced by a moving motorboat after
its engine is cut off, is given by dv/dt = – kv
3
where k is
constant. If v
0
is the magnitude of the velocity at cut-off,
the magnitude of the velocity at a time t after the cut-off is
(a)
)1ktv2(
v
2
0
0
+
(c)
kt
0
ev
-
(c) 2/v
0
(d)
0
v
21.The displacement x o
f a particle varies with time according
to the relation
).e1(
b
a
x
bt-
-= Then select the false
alternatives.
(a) At
b
1
t=, the displacement of the particle is nearly
÷
ø
ö
ç
è
æ
b
a
3
2
(b) the velocity and acceleration of the particle at t = 0 are
a and –ab respectively
(c) the particle cannot go beyond
b
a
x=
(d) the particle will
not come back to its starting point at
t ® ¥
22.The displacement of a particle is given by
x t1=+. Which
of the f
ollowing statements about its velocity is true ?
(a) It is zero
(b) It is constant but not zero
(c) It increases with time
(d) It decreases with time
23.Two bodies of masses m
1
and m
2
fall from heights h
1
and

h
2
respectively. The ratio of their velocities, when they hit the
ground is
(a)
2
1
h
h
(b)
2
1
h
h
(c)
11
12
mh
mh
(d)
2
2
2
1
h
h
24.Two cars A an
d B are travelling in the same direction with
velocities v
A
and v
B
(v
A
>v
B
). When the car A is at a distance
d behind the car B the driver of the car A applies brakes
producing a uniform retardation a. There will be no collision
when
(a)
a2
)vv(
d
2
BA-
< (b)
a2
vv
d
2
B
2
A-
<
(c)
a2
)vv(
d
2
BA-
> (d)
a2
vv
d
2
B
2
A-
>
25.A body is thrown upw
ards and reaches its maximum height.
At that position
(a) its acceleration is minimum
(b) its velocity is zero and its acceleration is also zero
(c) its velocity is zero but its acceleration is maximum
(d) its velocity is zero and its acceleration is the acceleration
due to gravity.
1.The position x of a particle varies with time (t) as
x = A t
2
– B t
3
. The acceleration at time t of the particle will be
equal to zero. What is the value of t?
(a)
B3
A2
(b)
B
A
(c)
B3
A
(d) zero
2.The accele
ration of a particle, starting from rest, varies with
time according to the relation
tsinsa
2
ww-=
The displacement of this particle at a time t will be
(a) s sin w t (b) s w cos w t
(c) s w sin w t (d)
22
t)tsins(
2
1
ww-
3.The displacement o
f a particle is given by
y = a + b t + c t
2
– d t
4
The initial velocity and acceleration are respectively
(a) b, – 4 d (b) – b, 2 c
(c) b, 2 c (d) 2 c, – 4 d
4.A passenger travels along the straight road for half the
distance with velocity v
1
and the remaining half distance
with velocity v
2
. Then average velocity is given by
(a)v
1
v
2
(b) v
2
2
/ v
1
2
(c) (v
1
+ v
2
)/2 (d) 2v
1
v
2
/ (v
1
+ v
2
)
5.A point moves with uniform acceleration and v
1
, v
2
and v
3
denote the average velocities in t
1
, t
2
and t
3
sec. Which of
the following relation is correct?
(a)
)tt(:)tt()vv(:)vv(
32213221
+-=--
(b) )tt(:)tt()vv(:)vv(
32213221 ++=--
(c) )tt(:)tt()vv(:)vv(
31213221 --=--
(d) )tt(:)tt()vv(:)vv(
32213221
--=--

53. 49Motion in a Straight Line
6.A bus starts moving with acceleration 2 m/s
2
. A cyclist 96 m
behind the bus starts simultaneously towards the bus at
20 m/s. After what time will he be able to overtake the bus?
(a) 4 sec (b) 8 sec
(c) 12 sec (d) 16 sec
7.When the speed of a car is v, the minimum distance over
which it can be stopped is s. If the speed becomes n v, what
will be the minimum distance over which it can be stopped
during same retardation
(a) s/n (b) n s
(c) s/n
2
(d) n
2
s
8.The two ends of a train moving with constant acceleration
pass a certain point with velocities u and v. The velocity with
which the middle point of the train passes the same point is
(a) (u + v)/2 (b) (u
2
+ v
2
)/2
(c)
22
(u v )/2+u (d) 22
uv+
9.A particle accelerates from rest at a constant rate for some
time and attains a velocity of 8 m/sec. Afterwards it
decelerates with the constant rate and comes to rest. If the
total time taken is 4 sec, the distance travelled is
(a) 32 m (b) 16 m
(c) 4 m (d) None of the above
10.The velocity of a particle at an instant is 10 m/s. After 5 sec,
the velocity of the particle is 20 m/s. Find the velocity at 3
seconds before from the instant when velocity of a particle
is 10m/s.
(a) 8 m/s (b) 4 m/s
(c) 6 m/s (d) 7 m/s
11.A particle experiences constant acceleration for 20 seconds
after starting from rest. If it travels a distance s
1
in the first
10 seconds and distance s
2
in the next 10 seconds, then
(a)s
2
= s
1
(b) s
2
= 2 s
1
(c)s
2
= 3 s
1
(d) s
2
= 4 s
1
12.A train of 150 m length is going towards north direction at a
speed of 10 ms
–1
. A parrot flies at a speed of 5 ms
–1
towards
south direction parallel to the railway track. The time taken
by the parrot to cross the train is equal to
(a) 12 s (b) 8 s
(c) 15 s (d) 10 s
13.A particle is moving along a straight line path according to
the relation
s
2
= at
2
+ 2bt

+ c
s represents the distance travelled in t seconds and a, b, c
are constants. Then the acceleration of the particle varies
as
(a)s
–3
(b) s
3/2
(c)s
–2/3
(d) s
2
14.A stone thrown vertically upwards with a speed of 5 m/sec
attains a height H
1
. Another stone thrown upwards from
the same point with a speed of 10 m/sec attains a height H
2
.
The correct relation between H
1
and H
2
is
(a)H
2
= 4H
1
(b) H
2
= 3H
1
(c)H
1
=2H
2
(d) H
1
= H
2
15.A body covers 26, 28, 30, 32 meters in 10
th
, 11
th
, 12
th
and
13
th
seconds respectively. The body starts
(a) from rest and moves with uniform velocity
(b) from rest and moves with uniform acceleration
(c) with an initial velocity and moves with uniform
acceleration
(d) with an initial velocity and moves with uniform velocity
16.A stone thrown upward with a speed u from the top of the
tower reaches the ground with a velocity 3u. The height of
the tower is
(a) 3u
2
/g (b) 4u
2
/g
(c) 6u
2
/g (d) 9u
2
/g
17.A smooth inclined plane is inclined at an angle q with
horizontal. A body starts from rest and slides down the
inclined surface.
h
q
Then the time taken by it to reach the bottom is
(a) ÷
÷
ø
ö
ç
ç
è
æ
g
h2
(b) ÷
÷
ø
ö
ç
ç
è
æ
g
2l
(c)
1 2h
singq
(d)
g
)h2(
sinq
18.A ball is droppe
d downwards, after 1 sec another ball is
dropped downwards from the same point. What is the
distance between them after 3 sec?
(a) 25 m (b) 20 m
(c) 50 m (d) 9.8 m
19.Two trains are each 50 m long moving parallel towards each
other at speeds 10 m/s and 15 m/s respectively. After what
time will they pass each other?
(a)
sec
3
2
5 (b) 4 sec
(c)
2 sec (d) 6 sec
20.A ball is projected vertically upwards with kinetic energy E.
The kinetic energy of the ball at the highest point of its
flight will be
(a)E (b)
2/E
(c) E/2 (d) zer
o
21.A particle is moving in a straight line with initial velocity
and uniform acceleration a. If the sum of the distance travelled
in t
th
and (t + 1)
th
seconds is 100 cm, then its velocity after
t seconds, in cm/s, is
(a) 80 (b) 50
(c) 20 (d) 30
22.Similar balls are thrown vertically each with a velocity
20 ms
–1
, one on the surface of earth and the other on the
surface of moon. What will be ratio of the maximum heights
attained by them? (Acceleration on moon = 1.7 ms
–2
approx)
(a)6 (b)
6
1
(c)
5
1
(d) None of these

54. 50 PHYSICS
23.The relativ
e velocity V
AB
or V
BA
of two bodies A & B may
be
(1) greater than velocity of body A
(2) greater than velocity of body B
(3) less than the velocity of body A
(4) less than the velocity of body B
(a) (1) and (2) only (b) (3) and (4) only
(c) (1), (2) and (3) only(d) (1), (2), (3) and (4)
24.A stone is thrown vertically upwards. When the particle is
at a height half of its maximum height, its speed is 10m/sec,
then maximum height attained by particle is (g = 10m/sec
2
)
(a) 8 m (b) 10 m
(c) 15 m (d) 20 m
25.From a 10m high building a stone ‘A’ is dropped &
simultaneously another stone ‘B’ is thrown horizontally with
an initial speed of 5 m/sec
–1
. Which one of the following
statements is true?
(a) It is not possible to calculate which one of two stones
will reach ground first
(b) Both stones ‘A’ & ‘B’ will reach the ground
simultaneously.
(c) ‘A’ stones reach the ground earlier than ‘B’
(d) ‘B’ stones reach the ground earlier than ‘A’
26.An automobile travelling with a speed of 60 km/h, can apply
brake to stop within a distance of 20m. If the car is going
twice as fast i.e., 120 km/h, the stopping distance will be
(a) 60 m (b) 40 m
(c) 20 m (d) 80 m
27.The motion of a particle is described by the equation u = at.
The distance travelled by particle in first 4 sec is
(a) 4a (b) 12a
(c) 6a (d) 8a
28.If you were to throw a ball vertically upward with an initial
velocity of 50 m/s, approximately how long would it take for
the ball to return to your hand? Assume air resistance is
negligible.
(a) 2.5 s (b) 5.0 s
(c) 7.5 s (d) 10 s
29.A body travels 2 m in the first two second and 2.20 m in the
next 4 second with uniform deceleration. The velocity of the
body at the end of 9 second is
(a) – 10
1
sm
-
(b) – 0.20
1
sm
-
(c) – 0.40
1
sm
-
(d) – 0.80
1
sm
-
30.From a 200 m high towe
r, one ball is thrown upwards with
speed of 10
1
sm
-
and another is thrown vertically
downwards at the same speeds simultaneously. The time
difference of their reaching the ground will be nearest to
(a) 12 s (b) 6 s
(c) 2 s (d) 1 s
31.A rocket is fired upward from the earth’s surface such that it
creates an acceleration of 19.6
2
sm
-
. If after 5 s, its engine
is switched off, the maximum height of the rocket from earth’s
surface would be
(a) 980 m (b) 735 m
(c) 490 m (d) 245 m
32.A particle travels half the distance with a velocity of 6
1
sm
-
.
The remaining h
alf distance is covered with a velocity of 4
1
sm
-
for half the time and with a velocity of 8
1
sm
-
for the
res
t of the half time. What is the velocity of the particle
averaged over the whole time of motion ?
(a) 9 ms
–1
(b) 6 ms
–1
(c) 5.35 ms
–1
(d) 5 ms
–1
33.A ball released from a height falls 5 m in one second. In 4
seconds it falls through
(a) 20 m (b) 1.25 m
(c) 40 m (d) 80 m
34.A food packet is released from a helicopter rising steadily at
the speed of 2 m/sec. After 2 seconds the velocity of the
packet is
(g = 10 m/sec
2
)
(a) 22 m/sec (b) 20 m/sec
(c) 18 m/sec (d) none of these
35.The displacement x of a particle along a straight line at time
t is given by : x = a
0
+
2
ta
1
+
3
a
2
t
2
. The accelera
tion of the
particle is
(a)
3
a
2
(b)
3
a2
2
(c)
2
a
1
(d) a
0
+
3
a
2
36.A rubber ball
is dropped from a height of 5 metre on a plane
where the acceleration due to gravity is same as that onto
the surface of the earth. On bouncing, it rises to a height of
1.8 m. On bouncing, the ball loses its velocity by a factor of
(a)
5
3
(b)
25
9
(c)
5
2
(d)
25
16
37.A boy moving
with a velocity of 20 km h
–1
along a straight
line joining two stationary objects. According to him both
objects
(a) move in the same direction with the same speed of
20 km h
–1
(b) move in different direction with the same speed of
20 km h
–1
(c) move towards him
(d) remain stationary
38.A man leaves his house for a cycle ride. He comes back to
his house after half-an-hour after covering a distance of
one km. What is his average velocity for the ride ?
(a) zero (b) 2 km h
–1
(c) 10 km s
–1
(d)
-1
1
kms
2
39.A car travels from A to B at a speed of 20 km h
–1
and returns
at a speed of 30 km h
–1
. The average speed of the car for the
whole journey is(a) 5 km h
–1
(b) 24 km h
–1
(c) 25 km h
–1
(d) 50 km h
–1

55. 51Motion in a Straight Line
40.A body dropped from a height ‘h’ with an initial speed zero,
strikes the ground with a velocity 3 km/hour. Another body
of same mass dropped from the same height ‘h’ with an
initial speed u' = 4 km/hour. Find the final velocity of second
mass, with which it strikes the ground
(a) 3 km/hr (b) 4 km/hr
(c) 5 km/hr (d) 6 km/hr
41.An electron starting from rest has a velocity that increases
linearly with time i.e. v = kt where
2
sm2k
-
= . The distance
covered in the first 3 second is
(a) 9 m (b) 16 m
(c) 27 m (d) 36 m
42.A body released from the top of a tower falls through half
the height of the tower in 2 s. In what time shall the body fall
through the height of the tower ?
(a) 4 s (b) 3.26 s
(c) 3.48 s (d) 2.828 s
43.The displacement x of a particle at the instant when its
velocity is v is given by
16x3v+=. Its acceleration and
initial velocity are
(a) 1.5 units, 4 units(b) 3 units, 4 units
(b) 16 units, 1.6 units(d) 16 units, 3 units
44.Let A, B, C, D be points on a vertical line such that
AB = BC = CD. If a body is released from position A, the
times of descent through AB, BC and CD are in the ratio.
(a)
23:23:1+-(b) 23:12:1--
(c) 3:12:1- (d) 13:2:1-
45.A body mo
ves in a straight line along Y-axis. Its distance y
(in metre) from the origin is given by y = 8t – 3t
2
. The average
speed in the time interval from t = 0 second to t = 1 second
is
(a) – 4 ms
–1
(b) zero
(c) 5 ms
–1
(d) 6 ms
–1
46.The acceleration due to gravity on planet A is nine times the
acceleration due to gravity on planet B. A man jumps to a
height 2m on the surface of A. What is height of jump by
same person on planet B?
(a) 2/3 m (b) 2/9 m
(c) 18 m (d) 6 m
47.In the given figure the distance PQ is constant. SQ is a
vertical line passing through point R. A particle is kept at R
and the plane PR is such that angle
q can be varied such
that R lies on line SQ. The time taken by particle to comedown varies, as the
q increases
q
S
P Q
R
(a) decreases continuously
(b) increases
(c) increases then decreases
(d) decreases then increases
48.The displacement x of a particle varies with time t as
x = ae
-at
+ be
bt
, where a, b, a and b are positive constants.
The velocity of the particle will
(a) be independent of a and b
(b) drop to zero when a = b
(c) go on decreasing with time
(d) go on increasing with time
49.Which one of the following equations represents the motion
of a body with finite constant acceleration ? In these
equations, y denotes the displacement of the body at time t
and a, b and c are constants of motion.
(a) y = at (b)
2
btaty+=
(c)
32
ctbtaty++= (d) bt
t
a
y+=
50.The de
pendence of velocity of a body with time is given by
the equation
2
v 20 0.1t.=+ The body is in
(a) uniform retardation
(b) uniform acceleration
(c) non-uniform acceleration
(d) zero acceleration.
51.Two stones are thrown from the top of a tower, one straight
down with an initial speed u and the second straight up
with the same speed u. When the two stones hit the ground,
they will have speeds in the ratio
(a) 2 : 3 (b) 2 : 1
(c) 1 : 2 (d) 1 : 1
52.A graph of acceleration versus time of a particle starting
from rest at t = 0 is as shown in Fig. The speed of the particle
at t = 14 second is
.)secin(t
) msin(a
2
-0
2
2
4
46 810
12
14
2-
4-
(a) 2 ms
–1
(b ) 34 ms
–1
(c) 20 ms
–1
(d) 42 ms
–1
53.In the displacement d versus time t graph given below,
the value of average velocity in the time interval 0 to 20 s is
(in m/s)
0
0
10
10
30
30
40
50
20
20
s/t
m
/
d
(a) 1.5 (b) 4
(c
)1 (d) 2

56. 52 PHYSICS
54.Figure s
hows the position of a particle moving along the
X-axis as a function of time.
2 4 6
10
20
)m(x
)s(t
Which of the
following is correct?
(a) The particle has come to the rest 6 times
(b) The maximum speed is at t = 6 s.
(c) The velocity remains positive for t = 0 to t = 6 s.
(d) The average velocity for the total period shown is
negative.
55.A steel ball is bouncing up and down on a steel plate with a
period of oscillation of 1 second. If g = 10 ms
–2
, then it
bounces up to a height of
(a) 5 m (b) 10 m
(c) 2.5 m (d) 1.25 m
56.A body starts from rest and travels a distance x with uniform
acceleration, then it travels a distance 2x with uniform speed,
finally it travels a distance 3x with uniform retardation and
comes to rest. If the complete motion of the particle is along
a straight line, then the ratio of its average velocity to
maximum velocity is
(a) 2/5 (b) 3/5
(c) 4/5 (d) 6/7
57.When two bodies move uniformly towards each other, the
distance decreases by 6 ms
–1
. If both bodies move in the
same directions with the same speeds (as above), the
distance between them increases by 4 ms
–1
. Then the speeds
of the two bodies are
(a) 3 ms
–1
and 3 ms
–1
(b) 4 ms
–1
and 2 ms
–1
(c) 5 ms
–1
and 1 ms
–1
(d) 7 ms
–1
and 3 ms
–1
58.A ball is thrown vertically upward with a velocity ‘u’ from
the balloon descending with velocity v. The ball will pass
by the balloon after time
(a)
g2
vu-
(b)
g2
vu+
(c)
2(u v)
g
-
(d)
g
)vu(2+
59.Two bodies be
gin a free fall from the same height at a time
interval of N s. If vertical separation between the two bodies
is 1 after n second from the start of the first body, then n is
equal to
(a)
nN (b)
gN
1
(c)
2
N
gN
1
+ (d)
4
N
gN
1
-
60.A particle starting f
rom rest falls from a certain height.
Assuming that the acceleration due to gravity remain the
same throughout the motion, its displacements in three
successive half second intervals are S
1
, S
2
and S
3
then
(a)S
1
: S
2
: S
3
= 1 : 5 : 9 (b) S
1
: S
2
: S
3
= 1 : 3 : 5
(c)S
1
: S
2
: S
3
= 9 : 2 : 3 (d) S
1
: S
2
: S
3
= 1 : 1 : 1
61.For the velocity time graph shown in the figure below the
distance covered by the body in the last two seconds of its
motion is what fraction of the total distance travelled by it in
all the seven seconds?
10
8
6
4
2
012345678
B C
DA
­
ms
velocity
–1
(a)
2
1
(b)
4
1
(c)
3
2
(d)
3 1
62.Which of the f
ollowing graph cannot possibly represent
one dimensional motion of a particle?
(a)
t
x
(b)
x
t
(c)
t
speed
(d) All of the above
63.Th
e distance time graph of a particle at time t makes angles
45° with the time axis. After one second, it makes angle 60°
with the time axis. What is the acceleration of the particle?
(a)
31-(b)31+(c)3 (d) 1
64.Two particl
es A and B are connected by a rigid rod AB. The
rod slides along perpendicular rails as shown here. The
velocity of A to the left is 10 m/s. What is the velocity of B
when angle a = 60°?
B
Aa
(a) 9.8 m/s (b) 10 m/s (c) 5.8 m/s (d) 17.3 m/s
65.A balloon starts rising from the ground with an accelerationof 1.25 ms
–2
. After 8 s, a stone is released from the balloon.
The stone will (Taking g = 10 m s
–2
)
(a) begin to move down after being released
(b) reach the ground in 4 s
(c) cover a distance of 40 m in reaching the ground
(d) will have a displacement of 50 m.
66.A point initially at rest moves along x-axis. Its acceleration
varies with time as
2
a (6t 5)m /s=+ . If it starts from origin,
the distance covered in 2 s is
(a) 20 m(b) 18 m(c) 16 m(d) 25 m

57. 53Motion in a Straight Line
67.The relation between time t and distance x is
2
txx=a +b
where a an d b are constants. The retardation is
(a)
3
2va (b)
3
2vb (c)
3
2vab (d)
23
2vb
68.A stone
is just released from the window of a train moving
along a horizontal straight track. The stone will hit the ground
following a
(a) straight line path(b) circular path
(c) parabolic path (d) hyperbolic path
69.A parachutist after bailing out falls 50 m without friction.
When parachute opens, it decelerates at 2 m/s
2
. He reaches
the ground with a speed of 3 m/s. At what height, did he bail
out ?
(a) 182 m (b) 91 m
(c) 111 m (d) 293 m
70.A car accelerates from rest at a constant rate a for some time
after which it decelerates at a constant rate b to come to
rest. If the total time elapsed is t, the maximum velocity
acquired by the car is given by
(a)
t
22
÷
÷
ø
ö
ç
ç
è
æ
ba
b+a
(b)
t
22
÷
÷
ø
ö
ç
ç
è
æ
ba
b-a
(c)
t
÷
÷
ø
ö
ç
ç
è
æ
ba
b+a
(d)
t
÷
÷
ø
ö
ç
ç
è
æ
b+a
ab
71.A car mov
ing with a speed of 40 km/hour can be stopped by
applying brakes after at least 2m. If the same car is moving
with a speed of 80km/hour, what is the minimum stopping
distance.
(a) 8 m (b) 6 m
(c) 4 m (d) 2 m
72.A man throws balls with same speed vertically upwards one
after the other at an interval of 2 sec. What should be the
speed of throw so that more than two balls are in air at any
time
(a) only with speed 19.6 m/s
(b) more than 19.6 m/s
(c) at least 9.8 m/s
(d) any speed less then 19.6 m/s.
73.A ball is dropped from a high rise platform at t = 0 starting
from rest. After 6 seconds another ball is thrown downwards
from the same platform with a speed v. The two balls meet at
t = 18s. What is the value of v?
(take g = 10 m/s
2
)
(a) 75 m/s (b) 55 m/s
(c) 40 m/s (d) 60 m/s
74.A particle moves a distance x in time t according to equation
x = (t + 5)
–1
. The acceleration of particle is proportional to
(a) (velocity)
3/2
(b) (distance)
2
(c) (distance)
–2
(d) (velocity)
2/3
75.A particle has initial velocity
(2 3)ij+
rr
and acceleration
(0.3 0.2)ij+
rr
. The magnitude of velocity after 10 seconds
will be
(a)92 units (b)52 units
(c) 5 un
its (d) 9 units
76.A stone falls freely under gravity. It covers distances h
1
, h
2
and h
3
in the first 5 seconds, the next 5 seconds and the next
5 seconds respectively. The relation between h
1
, h
2
and h
3
is
(a)h
1
=
2h
3
=
3
h
5
(b) h
2
= 3h
1
and h
3
= 3h
2
(c)h
1
= h
2
= h
3
(d) h
1
= 2h
2
= 3h
3
77.A ball is released from the top of a tower of height h meters.
It takes T seconds to reach the ground. What is the position
of the ball at
3
T
second
(a)
9
h8
meters from the ground
(b)
9
h7
meters from the ground
(c)
9
h
meters from the ground
(d)
18
h17
meters from the ground
78.The motion of particle is described by the equation x = a +
bt
2
, where a = 15 cm and b = 3 cm/sec
2
. Its instant velocity at
time 3 sec will be
(a) 36 cm/sec (b) 9 cm/sec
(c) 4.5 cm/sec (d) 18 cm/sec
79.Which one of the following equation represents the motion
of a body moving with constant finite acceleration? In these
equation, y denotes the displacement in time t and p, q and
r are constant:
(a) y = (p + qt )(t + pt)
(b) y = p + t/r
(c) y = (p + t) (q + t ) (r + t)
(d)
(p qt)
y
rt
+
=
80.A ball is thr
own up with velocity 19.6 m/s. The maximum
height attained by the ball is
(a) 29.2 m (b) 9.8 m
(c) 19.6 m (d) 15.8 m
81.A train A which is 120 m long is running with velocity 20 m/
s while train B which is 130 m long is running in opposite
direction with velocity 30 m/s. What is the time taken by
train B to cross the train A ?
(a) 5 sec (b) 25 sec
(c) 10 sec (d) 100 sec
82.A car travelling at a speed of 30 km h
–1
is brought to a halt in
8 m by applying brakes. If the same car is travelling at 60 km
h
–1
, it can be brought to a halt with the same braking power
in
(a) 32 m (b) 24 m
(c) 16 m (d) 8 cm
83.Velocity time curve for a body projected vertically upwards
is
(a) parabola (b) ellipse
(c) hyperbola (d) straight line
84.A train is moving towards east and a car is along north, both
with same speed. The observed direction of car to the
passenger in the train is
(a) east-north direction (b) west-north direction
(c) south-east direction (d) None of the above

58. 54 PHYSICS
85.A metro train
starts from rest and in 5 s achieves 108 km/h.
After that it moves with constant velocity and comes to rest
after travelling 45 m with uniform retardation. If total distance
travelled is 395 m, find total time of travelling.
(a) 12.2 s (b) 15.3 s
(c) 9 s (d) 17.2 s
86.A particle moves along a straight line OX. At a time t (in
second) the distance x (in metre) of the particle from O is
given by x = 40 + 12t – t
3
. How long would the particle travel
before coming to rest?
(a) 24 m (b) 40 m
(c) 56 m (d) 16 m
87.The displacement of particle is given by
x =
2
12
0
.
23
at at
a+- What is its accele ration?
(a)
2
2
3
a
(b)
22
3
a
-
(c)a
2
(d) zero
88.Figure her
e gives the speed-time graph for a body. The
displacement travelled between t = 1.0 second and t = 7.0
second is nearest to
t
.)secin(
)
m
s
in(v
1
-
4-
4
4
6
8
2
0
(a) 1.5 m (b) 2 m
(c
) 3 m (d) 4 m
89.A boat takes 2 hours to travel 8 km and back in still waterlake. With water velocity of 4 km h
–1
, the time taken for
going upstream of 8 km and coming back is(a) 160 minutes (b) 80 minutes
(c) 100 minutes (d) 120 minutes
90.A lift in which a man is standing, is moving upwards with aspeed of 10 ms
–1
. The man drops a coin from a height of 4.9
metre and if g = 9.8 ms
–2
, then the coin reaches the floor of
the lift after a time
(a)
2 s (b) 1 s
(c)
2
1
(d)
2
1
91.If a ball is thrown vertically upwards with a velocity of
40m/s, then velocity of the ball after two seconds is :
(g = 10 m/sec
2
)
(a) 15 m/s (b) 20 m/s
(c) 25 m/s (d) 28 m/s
92.If a car at rest accelerates uniformly to a speed of 144 km/h
in 20 sec., it covers a distance of
(a) 20 cm (b) 400 m
(c) 1440 cm (d) 2980 cm
93.The water drops fall at regular intervals from a tap 5 m above
the ground. The third drop is leaving the tap at an instant
when the first drop touches the ground. How far above the
ground is the second drop at that instant ?
(Take g = 10 m/s
2
)
(a) 1.25 m (b) 2.50 m
(c) 3.75 m (d) 5.00 m
94.The displacement time graph of a moving particle is shown
below
C
D
E F
D
I
S
P
L
A
C
E
M
E
N
T
S
Time
The instantaneous velocity of the particle is negative at the
point
(a)D (b) F
(c)C (d) E
95.A particle moves along a straight line such that its
displacement at any time t is given by
s = (t
3
– 6t
2
+ 3t + 4) metres
The velocity when the acceleration is zero is
(a) 3 ms
–1
(b) – 12 ms
–1
(c) 42 ms
–2
(d) – 9 ms
–1
96.A body starts from rest, what is the ratio of the distance
travelled by the body during the 4th and 3rd seconds ?
(a)
5
7
(b)
7
5
(c)
3
7
(d)
7
3
97.Which of the following curve does not represent motion in
one dimension?
(a)
v
t
(b)
v
t
(c)
v
t
(d)
v
t
DIRECTIONS for Qs. (98 to 100) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following.
(a)Statement -1 is false, Statement-2 is true
(b)Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c)Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d)Statement -1 is true, Statement-2 is false
98. Statement 1 : Velocity-time graph for an object in
uniform motion along a straight path is a straight
line parallel to the time axis.
Statement 2 : In uniform motion of an object velocity
increases as the square of time elapsed.
99. Statement 1 : A positive acceleration can be associated with
a ‘slowing down’ of the body.
Statement 2 : The origin and the positive direction of an
axis are a matter of choice.
100. Statement 1 :In a free fall, weight of a body becomes
effectively zero.
Statement 2 : Acceleration due to gravity acting on a body
having free fall is zero.

59. 55Motion in a Straight Line
Exemplar Questions
1.Among the four graph shown in the figure there is only one
graph for which average velocity over the time interval (O,
T) can vanish for a suitably chosen T. Which one is it?
(a) t
x
(b)
t
x
(c)
x
t
(d)
t
x
2.A lift is coming from 8th floor and is just about to reach 4th
floor. Taking ground floor as origin and positive direction
upwards for all quantities, which one of the following is
correct?
(a)x < 0, v < 0, a > 0 (b)x > 0, v < 0, a < 0
(c)x > 0, v < 0, a > 0 (d)x > 0, v > 0, a < 0
3.In one dimensional motion, instantaneous speed v satisfies
0
0vv.£<
(a) The dis
placement in time T must always take non-
negative values
(b) The displacement x in time T satisfies
00
–<<vT x vT
(c) The acceleration is always a non-negative number
(d) The motion has no turning points
4.A vehicle travels half the distance l with speed v
1 and the
other half with speed v
2, then its average speed is
(a)
12
2
+vv
(b)
12
12
2+
+
vv
vv
(c)
12
12
2
+
vv
vv
(d)
12
12
()+Lvv
vv
5.The displac
ement of a particle is given by x = (t – 2)
2 where
x is in metre and t in second. The distance covered by the
particle in first 4 seconds is(a) 4 m (b) 8 m
(c) 12 m (d) 16 m
6.At a metro station, a girl walks up a stationary escalator in
time t
1. If she remains stationary on the escalator, then the
escalator take her up in time t
2. The time taken by her to
walk up on the moving escalator will be
(a)
12
2
+tt
(b)
12
21-
tt
tt
(c)
12
21+
tt
tt
(d)t
1 – t
2
NEET/AIPMT (2013-2017) Questions
7.A stone falls freely under gravity. It covers distances h
1
, h
2
and h
3
in the first 5 seconds, the next 5 seconds and the
next 5 seconds respectively. The relation between h
1
, h
2
and h
3
is [2013]
(a)h
1
=
2h
3
=
3h
5
(b) h
2
= 3h
1
and h
3
= 3h
2
(c)h
1
= h
2
= h
3
(d) h
1
= 2h
2
= 3h
3
8.The displacement ‘x’ (in meter) of a particle of mass ‘m’ (in
kg) moving in one dimension under the action of a force, is
related to time ‘t’ (in sec) by t = 3x+. The displacement
of the particle when its velocity is zero, will be
[NEET Kar. 2013]
(a) 2 m (b) 4 m
(c) zero (d) 6 m
9.A particle of unit mass undergoes one-dimensional motion
such that its velocity varies according to v(x) = bx
–2n
where b and n are constants and x is the position of the
particle. The acceleration of the particle as d function of x, is
given by: [2015]
(a) –2nb
2
x
–4n–1
(b) –2b
2
x
–2n+1
(c) –2nb
2
e
–4n+1
(d) –2nb
2
x
–2n–1
10.If the velocity of a particle is v = At + Bt
2
, where A and B are
constants, then the distance travelled by it between 1s and
2s is : [2016]
(a)
3
A 4B
2
+ (b) 3A + 7B
(
c)
37
AB
23
+ (d)
AB
23
+
11.Preeti reached the metro station and found that the escalator
was not working. She walked up the stationary escalator intime t
1
. On other days, if she remains stationary on the
moving escalator, then the escalator takes her up in time t
2
.
The time taken by her to walk up on the moving escalatorwill be: [2017]
(a)
12
21
tt
tt-
(b)
12
21
tt
tt+
(c) t
1
– t
2
(d)
12tt
2
+

60. 56 PHYSICS
EXERCISE - 1
1. (b) 2. (b) 3.(c)4.(c)
5. (d) 6. (b) 7.(c)
8. (b)
2
)7t(xor)7t(x+=+=
timevelocity),7t(2
dt
dx
µ\+=
9. (b) a = f t, tf
dt
dv
a == at t = 0, velocity = u
vt
u0
dv f t dt=òò
, v – u =
2
t
f
2
Þ v = u+
2
t
f
2
NOTE : Do not use v
= u + at directly because the
acceleration is not constant.
10. (c)
12
dx
v a 2at
dt
= =+ 2
dv
a 2a
dt
\==
11. (d)
AB
v tan 30º andv tan 60º==
A
B
vtan30º1/31
vtan60º3 3
\= ==
12. (a) Let for the first half time t, the person travels a distance
s
1
. Hence
1
1 11
s
v ors vt
t
==
For second
half time,
2
2 22
s
v ors vt
t
==
Now,
12
ssTotal displacement
v
Total time 2t
+
==

1 2 12
vtvtvv
2t2
++
==
13. (b) When
a particle cover half of circle of radius r, then
displacement is AB = 2r
& distance = half of circumference of circle = pr
A
B
r
r
pr
14. (d)
22
gn
2
1
)1n(g
2
1
y -+=
)1n2(
2
g
]n)1n[(
2
g 22
+=-+=......(i)
Also, )1n2(
2
g
h -= ......(ii)
From (i) and (ii)
y
= h + g
15. (a) Differentiate two times and put x = 0.
16. (d) The difference in velocities is increasing with time as
both of them have more constant but different
acceleration.
17. (b) Time taken by the stone to reach the water level
g
h2
t
1
=
Time taken by so
und to come to the mouth of the well,
v
h
t
2=
\ Total time
v
h
g
h2
tt
21
+=+
18. (d) Let the
total distance be d. Then for first half distance,
time =
0v2
d
, next distance. = v
1
t and la
st half distance
= v
2
t
12
d
vt vt
2
\ += ;
12
d
t
2(v v)
=
+
Now average speed
)vv(2
d
)vv(2
d
v2
d
d
t
21210
+
+
+
+
=
012
120
2v(v v)
(v v ) 2v
+
=
++
19. (c) a = bt or bt
dt
dv
=. Integrating, we get
c
2
bt
v
2
+=, where c is a co
nstant of integration.
At t = 0, v = v
0
. Thus v
0
= c.
Now,
o
2
v
2
bt
dt
ds
v +== dtv
2
bt
ds
o
2
÷
÷
ø
ö
ç
ç
è
æ
+=\
Integrating we get, tv
6
bt
s
o
3
+=
20. (a)
3
kv
dt
dv
-=or dtk
v
dv
3
-=
Integrating we get, ckt
v2
1
2
+-=- ...(1)
Hints & Solutions

61. 57Motion in a Straight Line
At t = 0, v = v
0

2
o
1
c
2v
\-=
Putting in (1)
2 2 22
00
1 1 11
kt or kt
2v 2v 2v 2v
- =- - - =-
or
22
0 v2
1
kt
v2
1
=
ú
ú
û
ù
ê
ê
ë
é
+ or
[]
2
2
02
0
v
v
ktv21=+
or
ktv21
v
vor
ktv21
v
v
2
0
0
2
0
2
02
+
=
+
=
21. (d)
)
e
1
1(
b
a
)e1(
b
a
)e1(
b
a
x
1b
1
b
-=-=-=
-
´-

a (e 1)
be
-
= b/a
3
2
~
b
a
637.0
718.2
)718.1(
b
a
718.2
)1718.2(
b
a
-==
-
=
velocity
bt
ae
dt
dx
v
-
== ,v
0
= a
accleratio
n
aba&abe
dt
dv
a
0
bt
-=-==
-
A
t t = 0,
0)11(
b
a
x =-=and
At
b
1
t=, b/a
3 2
)
e
1
1(
b a
)e1(
b a
x
1
=-=-=
-
At
a
t ,x
b
=¥=
It cannot go beyond this, so point
a
x
b
>is not reached
b
y the particle.
At t = 0, x = 0, at
a
t ,x
b
=¥= , therefore the particle
does not come back to its starting point at ¥=t.
22. (c)
2
x t 2t1=++
Hence
dx
v 2t2
dt
= =+ . It increases with time.
23. (b) When a body falls through a height h, it acquires a
velocity gh2.
24. (c
) Initial relative velocity v
A
– v
B
, is reduced to 0 in
distance d
¢ (<d) with retardation a.
da2)vv(0
2
BA
2
¢-=--\
22
AB AB
(v v) (v v)
dd
2a 2a
--
= \>¢
25.
(d)
EXERCISE -2
1. (c
) Given that x = A t
2
– B t
3
\ velocity
2dx
2At 3Bt
dt
==-
and a
cceleration
Bt6A2
dt
dx
dt
d
-=÷
ø
ö
ç
è
æ
=
For acce
leration to be zero 2A – 6Bt = 0.
B3
A
B6
A2
t ==\
2. (a) tsins
dt
xd
a
2
2
2
ww-==
On inte
grating,
tcoss
tcos
s
dt
dx 2
ww=
w
w
w=
Again on inte
grating, we get tsins
tsin
sx w=
w
w
w=
3. (c)
3dy
v b 2ct 4dt
dt
==+-
3
0
v b 2c(0) 4d(0) b=+-=
(Q for initial velocity, t = 0)
Now
2dv
a 2c 12dt
dt
= =-
2
0
a 2c 12d(0) 2c\=-= , (a t t = 0)
4. (d)
12
1221
12
12
xx
2vv122
xx vvvv
2v 2v 2vv
+
==
+æö+
+
ç÷
èø
5. (
b) Let u be the initial velocity
1 12 12
v u at,v u a(t t)¢¢\ =+ =+ +
and 3 123
v u a(t t t)¢=+ ++
Now
11
1
uv u(uat )
v
22
¢+ ++
== 1ta
2
1
u+=
12
2 12
vv 1
v u at at
22
¢¢+
= =++
23
3 123
vv 1
v u at at at
22
¢¢+
= =+++
So
,
1 2 12
1
v v a(t t)
2
-=-+
and 2 3 23
1
v v a(t t)
2
-=-+
1 2 2 3 1 2 23
(v v ):(vv ) (t t ):(t t )\- -=++

62. 58 PHYSICS
6. (b)
Let after a time t, the cyclist overtake the bus. Thent20t2
2
1
96
2
´=´´+ or t
2
– 20 t + 96 = 0
12
96440020
t
´
´-±
=\
.sec8
2
420
=
±
= .sec12and
7. (d) v
2
= u
2
+ 2 a s or
v
2
– u
2
= 2 a s
Maximum retardation, a = v
2
/2 s
When the initial velocity is n v, then the distance over
which it can be stopped is given by
2 2
20
n
2
u (n v)
s ns
2a2(v / 2s)
===
8.
(c) Let the length of train is s, then by third equation of
motion,
22
v u 2as= +´ ....(1)
Where v is fin
al velocity after travelling a distance s
with an acceleration a & u is initial velocity as per
question
Let velocity of middle point of train at same point is v',
then
)2/s(a2u)v(
22
´+=¢ ....(2)
By equations (1) and (2), we get
2
uv
v
22
+
=¢
9. (b) 8 = a t
1
a
nd 0 = 8 – a (4 – t
1
)
or
÷
ø
ö
ç
è
æ
-=\=
a
8
4a8
a
8
t
1
8 = 4 a – 8 or a
= 4 and t
1
= 8/4 = 2 sec
Now,
m8sor)2(4
2
1
20s
1
2
1
=´+´=
m8sor)2(4
2
1
28s
2
2
2
=´´-´=
m16ss
21
=+\
10. (b) u = 1
0 m/s, t = 5 sec, v = 20 m/s, a = ?
220 10
a 2 ms
5
--
==
From the formula v
1
=
u
1
+ a t, we have
10 = u
1
+ 2 × 3 or u
1
= 4 m/sec.
11. (c) Let a be the constant acceleration of the particle. Then
2
ta
2
1
tus+= or a50)10(a
2
1
0s
2
1 =´´+=
and
2
2
1
s 0 a(20) 50a 150a
2
éù
=+ -=
êú
ëû
21
s 3s\=
Alternatively :
Let a be consta
nt acceleration and
21
s ut at
2
=+ , then 1
1
s 0 a 100 50a
2
=+´´=
Velocity
after 10 sec. is v = 0 + 10a
So,
2 21
1
s 10a 10 a 100 150a s 3s
2
= ´+ ´ = Þ=
12.
(d) So by figure the velocity of parrot
w.r. t. train is = 5–(–10) = 15m/secso time taken to cross the train is
velocityrelative
trainoflength
= sec10
15
150
==
North
EastWest
South
train
10m/sec
parrot
5m/sec
13. (a) s
2
= at
2
+
2bt + c
b2at2
dt
ds
s2 +=\
or
s
bat
dt
ds +
= , again differe
ntiating
222
2
s
s
bat
)bat(as
dt
ds
.
s
)bat(s.a
dt
sd
÷
ø
ö
ç
è
æ+
+-
=
+-
=
\
3
22
2
2
s
)bat(as
dt
sd +-
=
2
3
2
ds
a s.
dt
-
\=¥
14. (a) F
rom third equation of motion
22
v u 2ah=+
In first case initial ve
locity u
1
= 5 m/sec
final velocity v
1
= 0, a = – g
and max. height obtained is H
1
, then,
g2
25
H
1=
In second case u
2
= 10 m/
sec, v
2
= 0, a = –g
and max. height is H
2
then,
g2
100
H
2= .
It implies that H
2
= 4H
1
1
5. (c) The distance covered in n
th
second is
a)1n2(
2
1
uS
n -+=
where u is initial v
elocity & a is acceleration
then
2
a19
u26+= ....(1)
2
a21
u28+= ....(2)
2
a23
u30+= ....(3)
2
a25
u32+= ....(4)
From eqs. (1) and (2) we
get u = 7m/sec, a=2m/sec
2
\ The body starts with initial velocity u =7m/sec
and moves with uniform acceleration a = 2m/sec
2

63. 59Motion in a Straight Line
16. (b) The stone rises up till its vertical velocity is zero and
again reached the top of the tower with a speed u
(downward). The speed of the stone at the base is 3u.
–
+
v, g, h
u
Hence
2
22 4u
(3u) ( u) 2gh or h
g
=-+=
17.
(c) So by second equation of motion, we get
21
S ut at
2
=+
here S = l ,
u = 0, a = g sinq
2
2 2h 1 2h
t
a singgsin
===
qq
l h
sin
æö
q=ç÷
èø
Q
l
h
g

c
o
s
qg
s in
q
g
q
18. (a) S
= ut + ½at
2
here a = g
For first body u
1
=0
ÞS
1
=½g × 9
For secon
d body u
2
=0
ÞS
2
= ½g × 4
So diff
erence between them after 3 sec. = S
1
– S
2
= ½ g×5
If g = 10m/sec
2
then S
1
–S
2
= 25 m.
19. (b) Relative speed of each train with respect to each other
be, n = 10 + 15 = 25 m/s
Here distance covered by each train = sum of their
lengths = 50 + 50 = 100 m
\ Required time
.sec4
25
100
==
20. (d
) At highest point of the trajectory velocity becomes
zero and all kinetic energy changes to potential energy
so at highest point, K.E. = 0
21. (b) The distance travel in n
th
second is
S
n
= u + ½ (2n–1)a....(1)
so distance travel in t
th
& (t+1)
th
second are
S
t
= u +½ (2t–1)a ....(2)
S
t+1
= u+½ (2t+1)a ....(3)
As per question,
S
t
+S
t+1
= 100 = 2(u + at) ....(4)
Now from first equation of motion the velocity, of
particle after time t, if it moves with an accleration a is
v = u + a t ....(5)
where u is initial velocity
So from eq(4) and (5), we get v = 50cm./sec.
22. (b) Since v
2
= u
2
+ 2as
For first case u
1
=20m/sec, v
1
= 0, a
1
= g = 10, s
1
= ?
So
m20
102
400
s
1
=
´
=
For sec
ond case (at moon) u
2
=20m/sec, v
2
=0,?s,sec/m7.1
6
g
a
2
2
2 ===
6/102
400
7.12
400
s
2
´
=
´
= so
6
1
s
s
2
1
=
23. (d
) All options are correct :
(i) When two bodies A & B move in opposite
directions then relative velocity between A & B
either V
AB
or V
BA
both are greater than V
A
& V
B
.
(ii) When two bodies A & B move in parallel direction
then
AABBAAB
VVVVV<Þ-=
BBAABBA
VVVVV<Þ-=
24. (b) From third equation of motion
v
2
= u
2
– 2gh
( a g)=-Q
Given, v = 10 m/sec at h/2. But v = 0, when particle
attained maximum height h. Therefore (10)
2
= u
2
– 2gh/2
or 100 = 2gh –2gh/2 (
Q0 = u
2
– 2gh)
Þ h = 10 m
2
5. (b) Since in both case the height of building and down
ward acceleration ‘g’ is same. So both stones reach
simultaneously i.e.,
22
t10
2
1
10gt
2
1
S ´=Þ=
or t 2 sec,= for both st one.
26. (d) Speed 1
5
v 60
18
=´ m/s
50
3
=m/s
= = ´=
11
5 100
d 20m, v' 120
183
m/s
Let dcelera
tion be a
2
11
0 v 2ad\=- ....(1)
or
2
11
v 2ad=
2
12
(2v ) 2ad= ...(2)
(2) divided by (1)gives,
m80204d
d
d
4
2
1
2
=´=Þ=
27. (d) E
quation of motion isatu
=
we know that
ds
u
dt
= at
dt
ds
=Þ atdtdsor=
integr
ating it we get, ò=ò
4
0
s
0
tdtads
a8]t[
2
a
s
4
0
2
==
28. (d) T
he only force acting on the ball is the force of gravity.
The ball will ascend until gravity reduces its velocity
to zero and then it will descend. Find the time it takes
for the ball to reach its maximum height and then double
the time to cover the round trip.

64. 60 PHYSICS
Using v
a
t maximum height
= v
0
+ at = v
0
– gt, we get:
0 m/s = 50 m/s – (9.8 m/s
2
) t
Therefore,
t = (50 m/s)/(9.8 m/s
2
) ~ (50 m/s)/ (10 m/s
2
) ~ 5s
This is the time it takes the ball to reach its maximum
height. The total round trip time is 2t ~ 10s.
29. (b)
A B C
A to B
1
2 u 2 a 2 2 1 u a,
2
=´+´´´Þ=+
A to C
66a
2
1
6u20.4 ´´+´= Þ a3u7.0+= ,
2
sm15.0aor3.0a2
-
-=-= ,
11
sm15.1sm)15.01(a1u
--
=+=-=
Velocity at t = 9 s
ec.
1
sm2.035.115.1915.015.1v
-
-=-=´-=
30. (c) The ball thrown upward will lose velocity in 1s. It return
back to thrown point in another 1 s with the samevelocity as second. Thus the difference will be 2 s.
31. (b) Velocity when the engine is switched off
1
v 19.6 5 98ms
-
= ´=
21max
hhh+= wh
ere
2
2
12
1v
h at &h
2 2a
==
max
1 98 98
h 19.655
2 2 9.8
´
= ´ ´´+
´
= 245 + 4
90 = 735 m
32. (b) Average velocity for the second half of the distance is
=
112
vv 48
6ms
22
-++
==
Given that firs
t half distance is covered with a velocity
of
1
sm6
-
. Therefore, the average velocity for the
whole time of motion is
1
sm6
-
33. (d) Since S = ut + ½ gt
2
where u is initial velocity & a is acceleration.
In this case u = 0 & a = g
so distance travelled in 4 sec is,
S = ½ × 10 × 16 = 80m
34. (c) The food packet has an initial velocity of 2 m/sec in
upward direction, therefore
v = – u + gt or v = –2 + 10 × 2 = 18 m /sec.
35. (b) Differentiated twice.
36. (c) Downward motion
2
v 0 29.85
2
-=´´
Þ v 98 9.9==
Also for upwar
d motion
22
0 u 2 ( 9.8) 1.8-=´-´
Þ 94.53528u==
Fractional los
s
4.0
9.9
94.59.9
=
-
=
37. (a) Use ABABv vv=-
r rr
.
38. (a) Since d
isplacement is zero.
39. (b) Average velocity
1
hkm24
3020
30202 -
=
+
´´
= .
40. (c) Fr
om third equation of motion, v
2
= u
2
+ 2as
where v & u are final & initial velocity, a is acceleration,
s is distance.
For first case v
1
= 3km/hour, u
1
= 0, a
1
=g & s
1
=?
203636
1009
s
1
´´
´
= metre
For second case v
2
=?
, u
2
=4km/hour ,
a
2
= g = 10m/sec
&
203636
1009
ss
21
´´
´
==
so
363620
1009102
36003600
1000100016
v
2
2
´´
´´´
+
´
´´
=
or v
2
= 5 km/hou
r
41. (a)
kt
dt
ds
= Þ
211
s kt 233
22
= = ´´´ = 9 m.
42. (d
) For constant acceleration and zero initial velocity
2
thµ
2
11
2
2 2
ht
ht
=

2
211
1
h
t t2t22s
h
Þ= =´=´
43. (a) 16x3v+= Þ 16x3v
2
+=
Þ x316v
2
=-
Comparing with ,aS
2uv
22
=- we get, u = 4 units, 2a
= 3 or a = 1.5 units
44. (b)
2
1
tg
2
1
ABS== Þ
2
12
1
2S AC g(t t)
2
==+
and
2
321
)ttt(g
2
1
ADS3 ++==
g
S2
t
1
=
g
S2
g
S4
t,
g
S4
tt
221
-==+
2S
A
S
B
S
C
S
3S
D
g
S6
ttt
321
=++
g
S4
g
S6
t
3
-=
)23(:)12(:1::t:t:t
321
--

65. 61Motion in a Straight Line
45. (c)
1(8131 1)0
v 5ms
1
-´- ´´ -
==
46.
(c) Since the initial velocity of jump is same on both planets
So 0 = u
2
– 2g
A
h
A
0 = u
2
–2g
B
h
B
or
m182
1
9
hh
g
hg
BB
B
AA
=´=Þ=
´
47. (d)
q
S
R
P Q
mg
q
s in
m
g
q
cos
m
g
Let distance (PR) is covered by the particle in time ‘t’.
Þ PR = q=q+singt
2
1
t.sing
2
1
0
22
Further PR =
qcos
PQ
(Given P
Q = constant)
Þ PQ = qqcossingt
2
12
= q2singt
4
12
Þ t = 2
q2sing
PQ
\
q
µ
2sin
1
t
So as q increases, q2sin first increases and then
decreases. Hence ‘t’ first decreases and then increases.
48. (d) Given x = ae
–at
+ be
bt
Velocity, v =
dx
dt
= –aae
–at
+ bbe
bt
= - +
a
e
be
t
ta
b
a
b
i.e., go o
n increasing with time.
49. (b)
2
y t ;v- t';a t°µ µµ
50. (c) On differentiating, acceleration = 0.2t )t(fa=Þ
51. (d) Use v
2
– u
2
= 2aS. In both the cases, (u positive or
negative) u
2
is positive.
52. (b) Area under a-t graph is change in velocity.
Area
1 11
(44)64 24 22
2 22
= ´ +´+ ´´ - ´´

1
36 2 34 ms
-
= -=
As initial v
elocity is zero therefore, the velocity at 14
second is
1
sm34
-
.
53. (c) At t = 20s, d = 20 m
54. (a) At six points in the graph the tangents have zero slope
i.e. velocity is zero.
55. (d) Time fall is
1
second.
2
2
1 1 10
hg
228
æö
==ç÷
èø
= 1.25 m
56. (b)
321
av
ttt
x3x2x
v
++
++
=
1
max
2x
t
v
= , 2
max
2x
t
v
= ,3
max
6x
t
v
=
max
av
6xv
v
10x
=
av
max
v 3
v5
=
57. (c)Let
A
vand B
v are the velocities of two bodies.
In first case,
A
v+ B
v= 6m/s ........(1)
I
n second case,
A
v– B
v= 4m/s .........(2)
From
(1) & (2) we get,
A
v= 5 m/s and B
v=1 m/s.
58.
(d)
BBv
r
= Relative velocity of ball w.r.t. balloon = vu+
gt)vu(0++-= of
g
vu
t
+
=
Total time
g
)vu(2+
=
59. (c)
2
2
2
1 )Nn(g
2
1
y,gn
2
1
y -==
\ ])Nn(n[g
2
1
yy
22
21 --=-
Þ
g
1 (2n N)N
2
=- ]1yy[
21
=-Q
Þ
2
N
gN
1
n +=
60. (b) S
1

=
2
1
g
2
2
1
÷
ø
ö
ç
è
æ
, S
1
+ S
2
=
1
2
g(1)
2
2
1
g
2
2
3
÷
ø
ö
ç
è
æ
= S
1
+ S
2
+ S
3
By
solving we get
S
1
: S
2
: S
3
= 1 : 3 : 5
61. (b) Distance in last two second
=
2
1
× 10 × 2
= 10 m.
Total distance = 2
1
× 10 ×
(6 + 2) = 40 m.
62. (d) In (a), at the same time particle has two positions which
is not possible. In (b), particle has two velocities at the
same time. In (c), speed is negative which is not
possible.
63. (a) Velocity at time t is tan 45° = 1. Velocity at time (t = 1) is
tan603°=. Acceleration is change in velocity in
one second = 31-.
64. (c)
Here, the particle B moves upwards. Let the upward
velocity of B be v then
v
tan 60
10
=°

66. 62 PHYSICS
65. (b)
v = 1.25 × 8 ms
–1
= 10 ms
–1
1
s 1.25 8 8m 40m
2
=´ ´´=
Now ,
21
40 10t 10 t
2
=- +´´
or
2
5t 10t 40 0- -=
or
2
t 2t80- -= or t = 4 s.
66.
(b) Given acceleration
a 6t5=+
\
dv
a 6t5
dt
= =+ , dv (6t 5)dt=+
Integrating it, we have
vt
00
dv (6t 5)dt=+
òò
2
v 3t 5t C,= ++ where C is constant of integration.
When t = 0 , v = 0 so C = 0
\
2ds
v 3t 5t
dt
==+ or
2
ds (3t 5t)dt=+
Integrating it within the co
nditions of motion, i.e., as t
changes from 0 to 2 s, s changes from 0 to s, we have
s2
2
00
ds (3t 5t)dt=+
òò
\
2
32
0
5
s t t 8 10 18m
2
=+ =+=
67. (a)
2
txx=a +b
Differentiating w.r.t. time on both sides, we get
dx dx
1 2 .x
dt dt
= a +b
\
3
2
dx 1 dv 2v
v ; 2v
dt 2 x dt ( 2 x)
-a
= = = =-a
b+a b+a
Negative sign show
s retardation.
68. (c) The horizontal velocity of the stone will be the same as
that of the train. In this way, the horizontal motion will
be uniform motion. The vertical motion will be controlled
by the force of gravity, i. e., vertical motion is accelerated
motion. Thus the resultant motion will be along a
parabolic trajectory.
69. (d) Initial velocity of parachute
after bailing out,
u =
gh2
u = 508.92´´ = 514
s/m2a-=
s/m3
v
m50
2
The velocity at ground,
v = 3m/s
S =
22
uv
22
´
-
=
4
9803
2
-
» 243 m
Initially he has
fallen 50 m.
\ Total height from where he bailed out
= 243 + 50 = 293 m
70. (d) As per question,
Let max. velocity is v
then v = a t
1
& v – b t
2
= 0, where t = t
1
+ t
2
Now t
1
+ t
2
= t or
vv
t+=
ab
t
vt
11
æöab
\== ç÷
a+bæö èø
+
ç÷
abèø
and
12
sss=+
222
v v v 11
222
æö
=+=+
ç÷
a b ab èø
71. (a) From third equation of motion :
22
v u 2as=+
for first case sec/m
36
1040
u
´
= ,
v =0, a = ?, s =
2 m
so,
2
2
sec/m
4
1
36
1040
a ÷
ø
ö
ç
è
æ´
=
for second case sec/m
36
1080
u
´
= , v = 0,
So
22
2
80 10 1 40 10
s / 2 8meter
36 4 36
´´æö æö
= ´´=
ç÷ ç÷
èø èø
72. (b) Height attained by balls in 2 sec is
m6.1948.9
2
1
=´´=
the same distance w
ill be covered in 2 second (for
descent)
Time interval of throwing balls, remaining same. So, for
two balls remaining in air, the time of ascent or descent
must be greater than 2 seconds. Hence speed of balls
must be greater than 19.6 m/sec.
73. (a) Clearly distance moved by 1
st
ball in 18s = distance
moved by 2
nd
ball in 12s.
Now, distance moved in 18 s by 1
st
ball
=
1
2
× 10 × 18
2
= 90 × 18 = 1620 m
Distance moved in 12 s by 2
nd
ball
= ut +
1
2
gt
2
\ 1620 = 12 v + 5 × 144
Þ v = 135 – 60 = 75 ms
–1
74. (a
)
1
5
x
t
=
+
\ v =
dx
dt
2
1
( 5)t
-
=
+
\ a =
2
2
dx
dt
= 3
2
( 5)t+
= 2x
3
Now
1
2
1
( 5)
v
t
µ
+
\
3
2
3
1
( 5)
va
t
µµ
+

67. 63Motion in a Straight Line
75. (b) v u at=+
rrr
ˆˆ ˆˆ(2 3 ) (0.30.2 ) 10vij ij=+++´ = ˆˆ55ij+
||v
r
= 22
55+
; | |52v=
r
76. (a)Qh =
1
2
gt
2
\h
1
=
1
2
g(5)
2
= 125
h
1
+ h
2
=
1
2
g(10)
2
= 500
Þh
2

= 375
h
1
+ h
2
+ h
3
=
1
2
g(15)
2
= 1125
Þh
3
= 625
h
2
=
3h
1
, h
3
= 5h
1
orh
1
=
2
h
3
=
3
h
5
77. (a) h =
2
gT
2
1
now for t = T/3 second
vertical distance moved is given by
h¢
2
1T
g
23
æö
=
ç÷
èø
Þ h¢
2
1gTh
299
=´=
\ position of ball from ground
9
h
h-=
9
h8
=
78. (d) x = a + b
t
2
= 15 + 3t
2
v =
dx
dt
= 3 × 2t = 6t
Þ
t 3s
v
=
= 6 × 3 = 18cm/s
79.
(a) Motion with constant acceleration is represented by a
quadratic equation of t
Y = (p + qt) (r + pt) = pr + qrt + p
2
t + pqt
2
80. (c) Let the maximum height attained by the ball be h.
At maximum height , velocity of ball, v = 0
Given, initial velocity, u = 19.6 m/s
Using the equation of motion,
v
2
= u
2
+ 2gh
We get 0 = (19.6)
2
+ 2 (– 9.8) × h
Þ h=
2
(19.6)
2 9.8´
= 19.6 m
8
1. (a) Here, length of train A, L
A
= 120 m
length of train B, L
B
= 130 m
velocity of train A, v
A
= 20 m/s
velocity of train B, v
B
= 30 m/s
Train B is running in opposite direction to train B,
\ velocity of train B relative to train A,
v
BA
= v
B
+ v
A
= (30 + 20) m/s
= 50 m/s
Total distance to be covered by train B
= L
A
+ L
B
= (120 + 130) m
= 250 m
Hence, time required by train B to cross train A
t=

250
50
sec = 5sec
82. (
a) Retardation,
222
0 (25 / 3)
2 28
vu
a
s
--
==
´
2
125
163
a
æö
=-´ç÷
èø
For u = 60
km h
–1
=
50
3
ms
–1
2
2
0 (50 / 3)
2 [ (25/3) 1/16]
s
-
=
´-´
= 32 m
83
. (d) Velocity time curve will be a straight line as shown:
o
t
v
At the highest point v = 0.
84. (b) Let O be the origin, then
O
Q
N
Car
E
TrainP
N
W
S
E
passenger in the train at P observes the car at Q along
the direction PQ; i.e. west north direction.
85. (d) Given : u = 0, t = 5 sec, v = 108 km/hr = 30m/s
By eq
n
of motion
v = u + at
or
2v 30
a 6 m / s [ u 0]
t5
==== Q
2
1
1
S at
2
=

21
6 5 75m
2
=´´=
Distanc
e travelled in first 5 sec is 75m.
Distance travelled with uniform speed of 30 m/s is S
2
395 = S
1
+ S
2
+ S
3
395 = 75 + S
2
+ 45
\ S
2
= 395 – 120 = 275 m
Time take n to travel 275 m =
275
9.2sec
30
=
For retarding
motion, we have
0
2
– 30
2
= 2 (– a) × 45

68. 64 PHYSICS
We get, a = 10 m/s
2
N
ow by, S = ut +
21
at
2
45 = 30t +
21
(–10)t
2
45 = 30t – 5t
2
Pn solving we g
et, t = 3 sec
Total time taken = 5 + 9.2 + 3 = 17.2 sec.
86. (c) When particle comes to rest,
V = 0 =
dxd
dt dt
=(40 + 12t – t
3
)
Þ12 – 3t
2
= 0
Þt
2
=
12
4
3
= \ t = 2 sec
The
refore distance travelled by particle before coming
to rest,
x = 40 + 12t – t
3
= 40 + 12 × 2 – (2)
3
= 56m
87. (b) We get acceleration by double differentiation of
displacement.
V =
12
0
atadxd
at
dt dt 2 3
2æö
= +-
ç÷
èø
=
1
2
a2
at
23
-
a =
1
2
2
a2
d at
dv2 23
a
dt dt 3
æö
-ç÷
èø -
==
88. (b)
1 11
(2 4) 2 1 4 3 4 2m
2 22
+ ´+´´-´´=
8
9. (a) Velocity of boat
1
hkm8
2
88 -
=
+
=
Velocity of w
ater
1
hkm4
-
=
160h
3
8
48
8
48
8
t ==
+
+
-
= minute
90. (b
) Using relative terms
s/m0u
.rel
=
?t,m9.4S,sm8.9a
2
===
-
2
t8.9
2
1
t09.4´´+´=
Þ 9.4t9.4
2
= Þ t = 1 s
91. (b
) From first equation of motion v = u + a t
here u = 40, a = g = – 10, t = 2
so v = 40 – 10 × 2 = 20 m/sec
92. (b) v = [144 × 1000/(60 × 60)] m/sec.
v = u + at
or (144 × 1000)/(60 × 60) = 0 + a × 20
2144 1000
a 2 m / sec
60 60 20
´
\==
´´
Now
21
s u t at
2
=+ m400)20(2
2
1
0
2
=´´+=
93. (c) He
ight of tap = 5m and (g) = 10 m/sec
2
.
For the first drop, 5 = ut
21
2
+gt
21
(0 ) 10
2
= ´+´tt = 5t
2
or t
2
= 1 or t = 1 s ec.
It means that the third drop leaves after one second ofthe first drop. Or, each drop leaves after every 0.5 sec.
Distance covered by the second drop in 0.5 sec
211
(0 0.5) 10
22
= + = ´ +´ut gt × (0.5)
2
= 1.25
m.
Therefore, distance of the second drop above the
ground = 5 – 1.25 = 3.75 m.
94. (d) At E, the slope of the curve is negative.
95. (d) Velocity,
2
3–123==+
ds
v tt
dt
Acceleration, 6 –12;==
dv
at
dt
For a = 0, we have,
0 = 6 t – 12 or t = 2s. Hence, at t = 2 s the velocity will be
2 –1
3 2 –12 2 3 9 ms= ´ ´ + =-v
96. (a)
4
3
0 (2 4 1)
72
5
0 (2 3 1)
2
+ ´-
==
+ ´-
a
D
aD
97. (b) In on
e dimensional motion, the body can have at a time
one velocity but not two values of velocities.
98. (d) In uniform motion the object moves with uniform velocity,
the magnitude of its velocity at different instane i.e., at
t = 0, t =1, sec, t = 2sec ..... will always be constant. Thus
velocity-time graph for an object in uniform motion along
a straight path is a straight line parallel to time axis.
99. (b) 100. (d)
EXERCISE -3
Exemplar Questions
1. (b) If we draw a line parallel to time axis from the point (A)
on graph at t = 0 sec. This line can intersect graph at B.
In graph (b) for one value of displacement there are
two different points of time. so, for one time, the average
velocity is positive and for other time is equivalent
negative.
As there are opposite velocities in the inteval 0 to T
hence average velocity can vanish in (b). This can be
seen in the figure given below.
t
x
O
A B
T
Here, OA = BT (same displac ement) for two different
points of time.

69. 65Motion in a Straight Line
2. (a) As the lift is moving downward directions so
displacement is negative (zero). We have to see whether
the motion is accelerating or retarding.
Due to downward motion displacement is negative the
lift reaches 4th floor is about to stop hence, motion is
retarding (–a) downward in nature hence, x < 0; a > 0.
8th floor
x < 0
4th floor
Ground floor
x < 0O
6th floor
As dis
placement is in negative direction,
x < 0 velocity will also be negative i.e., v < 0 but net
acceleration is +ve a > 0, that can be shown in the
graph.
3. (b) In one dimensional motion, for the maximum and
minimum displacement we must have the magnitudeand direction of maximum velocity.
As maximum velocity in positive direction is v
0, hence
maximum velocity in opposite direction is also –v
0.
Maximum displacement in one direction = v
0T
Maximum displacement in opposite directions = –v
0T.
Hence,
00
- <<vT x vT
4. (c)Time taken to travel first half distance,
1
11
/2
2
==
ll
t
vv
1
( / 2)=QLl
Time taken to travel second half distance,
2
22
=
l
t
v
2
( / 2)=QLl
So, total
time taken to travel full distance
= t
1 + t
2
1 2 12
11
222
éù
=+=+
êú
ëû
l ll
v v vv
Total time
21
12
2
éù+
=
êú
ëû
vvl
vv
So, averag
e speed,
v
av. =
Total distance
Total time
=
12
av.
12
12
2
11
2
Þ==
+éù
+
êú
ëû
vvl
v
vvl
vv
12
av.
12
2
\=
+
vv
v
vv
5. (b)As given that, x = (t – 2)
2
Now, velocity
2
( 2)==-
dxd
vt
dt dt
=
2 (t – 2) m/s
Acceleration,
[2( 2)]==-
dvd
at
dt dt
= 2
[1 – 0] = 2 m/s
2 = 2 ms
–2
at t = 0; v
0 = 2 (0 – 2) = –4 m/s
t = 2 s; v
2 = 2 (2 – 2) = 0 m/s
t = 4 s; v
4 = 2 (4 – 2) = 4 m/s
B
DOA
V
t
42
C
–4 m/s
4 m/s
(Time)
v-t graph is shown in diagram.
Distance travelled
= area between time axis of the graph
= area OAC + are ABD
11
22
=´+´OA OC AD BD
421
2 4 8m
22
´
= +´´=
If di
splacement occurs
11
22
=´´+´´OA OC AD BD
11
2( 4) 2 4 0
22
=´-+´´=
6. (c)Let us consider, displacement is L, then
velocity of girl with respect to ground,
1
=
g
L
v
t
Velocity of escalator with respect to ground,
2
=
e
L
v
t
Net velocity of the girl on moving escalator with respect
to ground

12
= +=+
ge
LL
vv
tt
Þv
ge
12
12
éù+
=êú
ëû
tt
L
tt

70. 66 PHYSICS
Now, if t is t
otal time taken by girl on moving escalator
in covering distance L, then
distance
speed
=t

12
1212
12
==
+æö+
ç÷
èø
ttL
tttt
L
tt
NEET/AIPMT (2013-2017) Questions
7. (a)Q h =
1
2
gt
2
\h
1
=
1
2
g(5)
2
= 125
h
1
+ h
2
=
1
2
g(10)
2
= 500
Þh
2
= 375
h
1
+
h
2
+ h
3
=
1
2
g(15)
2
= 1125
Þh
3
= 625
h
2
= 3h
1
,
h
3
= 5h
1
orh
1
=
2
h
3
=
3h
5
8. (c)Q t = 3x+
Þ x= t – 3 Þ x = (t – 3)
2
v =
dx
dt
= 2(t – 3) = 0
Þt = 3
\x = (3 – 3)
2
Þx = 0.
9
. (a) According to question,
V (x) = bx
–2n
So,
dv
dx
= – 2 nb x
– 2n – 1
A
cceleration of the particle as function of x,
a = v
dv
dx
= bx
–2n
{ }
–2n –1
b (–2n) x
= – 2nb
2
x
–4n–1
10.
(c)Given : Velocity
V = At + Bt
2
Þ
dx
dt
= At + Bt
2
By integratin
g we get distance travelled
Þ ( )
x2
2
01
dx At Bt dt=+
òò
Distance travell
ed by the particle between 1s and 2s
x =
( ) ( )
22 33A B 3A 7B
21 21
2 3 23
-+ -=+
11. (b) V
elocity of preeti w.r.t. elevator v
1
=
1
d
t
Velocity of elevator w.r.t. ground
2
2
d
v
t
= then
velocity of pr
eeti w.r.t. ground
v = v
1
+ v
2
12
ddd
ttt
=+
12
111
ttt
=+
\ t =
12
12
tt
(t t)+
(time taken by preeti to w
alk up on
the moving escalator)

71. SCALARS AND VECTORS
Scalars : The physical quantities which have only magnitude
but no direction, are called scalar quantities.
For example - distance, speed, work, temperature, mass, etc.
·Scalars are added, subtracted, multiplied and divided by
ordinary laws of algebra.
Vectors: For any quantity to be a vector,
(i) it must have magnitude.
(ii) it must have direction.
(iii) it must satisfy parallelogram law of vector addition.
For example – displacement, velocity, force, etc.
Electric current has magnitude as well as direction but
still it is not treated as a vector quantity because it is added by
ordinary law of algebra.
Types of Vectors
(i) Like vectors : Vectors having same direction are called like
vectors. The magnitude may or may not be equal.
A
B
A
r
and B
r
are like vectors. These are also called parallel
vectors or collinear vectors.
(ii) Equal vectors : Vectors having same magnitude and same
direction are called equal vectors.
A
B
Here A
r
and B
r
are equal vectors AB=
rr
Thus, equal vector is a special case of like vector.
(iii) Unlike vectors : Vectors having exactly opposite directions
are called unlike vectors. The magnitude may or may not be
equal.
A
B
A
r
and B
r
are unlike vectors.
(iv) Negative vectors : Vectors having exactly opposite direction
and equal magnitudes are called negative vectors.
A
B
Here A
r
and B
r
are negative vectors, AB=-
rr
Thus negative vectors is a special case of unlike vectors.
(v) Unit vector : Vector which has unit magnitude. It represents
direction only. For example take a vector B
r
. Unit vector in
the direction of B
r
is
|B|
B
, which
is denoted as
B
ˆ. B
ˆ, is
read as “B
cap” or "B caret".
(vi) Orthogonal unit vector : A set of unit vectors, having the
directions of the positive x, y and z axes of three dimensional
rectangular coordinate system are denoted by
ˆˆˆ
i, j and k.
They are
called orthogonal unit vectors because angle
between any of the two unit vectors is 90º.
o
X
Y
Z
j
ˆ
k
ˆ
i
ˆ
The coordinate system which has shown in fig. is called
right handed coordinate system. Such a system derives its
name from the fact that right threaded screw rotated through
90º from OX to OY will advance in positive Z direction as
shown in the figure.
(vii) Null vector (zero vector) : A vector of zero magnitude is
called a zero or null vector. Its direction is not defined. It is
denoted by 0.
Properties of Null or Zero Vector :
(a) The sum of a finite vector
A
ur
and the zero vector is
equal to the finite vector
i.e., AA+=0
ur ur
(b) The multiplication of a zero vector by a finite number n
is equal to the zero vector
i.e., 0 n = 0
(c) The multiplication of a finite A
ur
by a zero is equal to
zero vector
i.e., A0=0
ur
4
Motion in a
Plane

72. 68 PHYSICS
(viii) A
xial vector : Vector associated with rotation about an axis
i.e., produce rotation effect is called axial vector. Examples
are angular velocity, angular momentum, torque etc.
(ix) Coplanar vectors : Vectors in the same plane are called
coplanar vectors.
(x) Position vectors and displacement vectors : The vector
drawn from the origin of the co-ordinate axes to the position
of a particle is called position vector of the particle. If A (x
1
,
y
1
, z
1
) and B (x
2
, y
2
, z
2
) be the positions of the particle at
two different times of its motion w.r.t. the origin O, then
position vector of A and B are
X
Y
O
r
A
r
B
A
B
Path o
f

p
a
r
t
i
c
l
e
Displacem
e
n
t
vec
t
o
r
A 1 11
ˆˆˆ
r OA xi yj zk= = ++
uur uuur
B 2 22
ˆˆˆ
r OB xi yj zk==++
uur uuur
.
The displacement
vector is
AB = OAOB-.

2 2 2 1 21
ˆˆˆ
(x x )i (y y )j (z z )k= - +- +-
Laws of Vector
Algebra
1.
ABBA
rrrr
+=+ (Commutative law of addition)
2. C)BA()CB(A
rrrrrr
++=++ (Associative law of addition)
3. mAAm
rr
=
4. A)mn()An(m
rr
=
5. AnAmA)nm(
rrr
+=+
6. BmAm)BA(m
rrrr
+=+
ADDITION OF VECTORS
Trian
gle Law of Vector Addition
It states that if two vectors acting on a particle at the same time
are represented in magnitude and direction by the two sides of a
triangle taken in one order, their resultant vector is represented
in magnitude and direction by the third side of the triangle taken
in opposite order.
O
P
B
Q
q
A
R
N
b
Magnitude of R
ur
is given by
22
2 cos=++qR A B AB
whereq is the angle between A
ur
and B
ur
.
Direction of R
ur
: Let the resultant R
ur
makes an angle b with the
direct
ion of
A
ur
. Then from right angle triangle QNO,
tan
sin
cos
QN QN B
ON OP PN A B
q
b===
+ +q
(i)|R
ur
| is maximum, if cosq = 1 , q = 0° (parallel vector)
AB
¾ ¾® ¾ ¾®
u r ur
P
R
max

22
A B 2AB= ++ = A + B
(ii)|R
ur
| is minimum, if cosq = –1, q = 180° (o pposite vector)
BA
¬ ¾¾ ¾ ¾®
uuuruur
R
min

22
A B 2AB AB= + - =-
(iii) If the vectors A and B are orthogonal,
i.e.,
o 22
90,R ABq= =+
Parallelogr
am Law of Vector Addition
It states that if two vectors are represented in magnitude and
direction by the two adjacent sides of a parallelogram then their
resultant is represented in magnitude and direction by the
diagonal of the parallelogram.
Let the two vectors
A
ur
and B
ur
, inclined at angle q are represented
by
sides
OPand OS
uuur uuur
of parallelogram OPQS, then resultant vector
R
ur
is represented by diagonal OQ
uuur
of the parallelogram.
S Q
P
O
q
b
R
B
A
22 sin
2 cos ; tan
cos
q
= + + q b=
+q
B
R A B AB
AB
If q < 90° , (acute angle) R
ur
=A
ur
+B
ur
,R
ur
is called main
(major) di
agonal of parallelogram
If
q > 90° , (obtuse angle) R
ur
=A
ur
+B
ur
,R
ur
is called minor
diago
nal.
Polygon Law of Vector Addition
If a number of non zero vectors are represented by the (n–1)
sides of an n-sided polygon then the resultant is given by the
closing side or the n
th
side of the polygon taken in opposite
order.
B
D
E
C
O
E
B
A
CD
A
R
So,R ABCDE=++++
ur ururururur
or, OA AB BC CD DE OE++++=
uuur uuur uuur uuuruuur uuur

73. 69Motion in a Plane
15.Magnitude of a vector is independent of co-ordinate axes
system.
16.Component of a vector perpendicular to itself is zero.
17.(a) Resultant of two vectors is maximum when angle
between the vectors is zero i.e., q = 0°
R
max
= A + B
(b) Resultant of two vectors is minimum when
q = 180°
R
min
= A – B
(c) The magnitude of resultant of A and B can vary
between (A + B) and (A – B)
SUBTRACTION OF VECTORS
We convert vector subtraction into vector addition.
q
180° – q
A
– B
B
)B(ABA-+=-
If the angl
e between
A and B is q then the angle between
1
8
0
°

–

q
A
R
– B
a
A and B- is (180° – q).
q-+=-cosAB2BA|BA|
22
o
o
sin(180 )
tan
A Bcos (180 )
B -q
a=
+ -q

sin
A Bcos
Bq
=
-q
RESOLUTION OF A VECTOR
Rectangular Components of a Vector in Plane
X
Y
q
A
Aj
ˆ
yj
ˆ
A
y
i
ˆ
A
x
The vector A may be written as
ˆˆ
xy
A Ai Aj=+
u ur
where i
ˆ
A
x
is t he component of vector A in X-direction and
j
ˆ
A
y is the component of vector
A in the Y-direction.
1. Resultant of two unequal vectors cannot be zero.
2. Resultant of three co-planar vectors may or may not be
zero.
3. Minimum no. of coplanar vectors for zero resultant is 2 (for
equal magnitude) and 3 (for unequal magnitude).
4. Resultant of three non coplanar vectors cannot be zero.
Minimum number of non coplanar vectors whose sum can
be zero is four.
5. Polygon law should be used only for diagram purpose for
calculation of resultant vector (For addition of more than 2
vectors) we use components of vector.
Keep in Memory
1.If BA
rr
=, then 0AB-=
rr
is a null vector..
2.Null vector or zero vector is defined as a vector whose
magnitude is zero and direction indeterminate. Null vector
differs from ordinary zero in the sense that ordinary zero is
not associated with direction.
3.
|A|
A
A
ˆ
r
r
= is called a unit vector. It is unitless and
dimensionless vector. Its magnitude is 1. It represents
direction only.
4.If BA
rr
=, then |B| |A|
rr
= and B
ˆ
A
ˆ
=, where
ˆ ˆ
AandB are
unit vector
s of A and B respectively.
5.A vector can be divided or multiplied by a scalar.
6.Vectors of the same kind can only be added or subtracted.It is not possible to add or subtract the vectors of differentkind. This rule is also valid for scalars.
7.Vectors of same as well as different kinds can be multiplied.
8.A vector can have any number of components. But it canhave only three rectangular components in space and two
rectangular components in a plane. Rectangular
components are mutually perpendicular.
9.The minimum number of unequal non-coplanar whose
vector sum is zero is 4.
10.When
xyz
ˆˆˆ
A Ai A j Ak=++
ur
222
xyz
|A|AAA= ++
r
, where | A|
r
is modulus or
magn
itude of vector
A
r
.
11.ˆˆ
ij+ makes 45° with both X and Y-axes. It makes angle 90°
with Z-axis.
12. k
ˆ
j
ˆ
i
ˆ
++ makes angle 54.74° with each of the X, Y and
Z-axes.
13. ABBA -¹-
14.If |BA||BA|-=+then angle between
A and B is
2
p
.

74. 70 PHYSICS
Also A
x
= A cos q and A
y
= A sin q
j
ˆ
)sinA(i
ˆ
)cosA(A q+q=\
Þ A cos q and A sin q are the magnitudes of the components of
A in X and Y-direction respectively..
Also
22
||
xy
A AA=+
u ur
Rectangular components of a vector in 3D : Three rectangular
components along X, Y and Z direction are given by
xyz
ˆˆˆ
A i,A j,A k. Therefore,
vector A
r
may be written as
x yz
ˆˆˆ
A Ai A j Ak=++
ur
and
222
x yz
A AAA= ++
If a, b and g are the angles subtended by the rectangular
components of vector then
cos a =
x
A
,
A
cos b=
y
A
A
and cos g=
zA
A
Also, cos
2
a+ cos
2
b + cos
2g = 1
CAUTION : Do not resolve the vector at its head. The vector
is always resolved at its tail.
B cos q
B sin q
Wrong
q
B

B cos q
B sin q
Correct
q
B
Example 1.
X and Y component of vector A are 4 and 6 m respectively.
The X and Y component of A +B
uur uur
are 10 m and 9 m
respectively. Calculate the length of vector B and its angle
with respect to X-axis
Solution :
j
ˆ
6i
ˆ
4A+= and j
ˆ
9i
ˆ
10BA+=+
j
ˆ
3i
ˆ
6)j
ˆ
6i
ˆ
4(–)j
ˆ
9i
ˆ
10(A–)BA(B +=++=+=\
\ length of Bis m5336|B|
22
=+=
Also ÷
ø
ö
ç
è
æ
=qÞ===q
-
2
1
tan
2
1
6
3
B
B
tan
1
x
y
where q is the angle which (A B)+
rr
is making with
X-axis.
Example 2.
Find the resultant of vectors given in figure
X
O
Y
P4 cm
M
Q
3 cm
60°
6cm
Solution :
Here j
ˆ
2.5i
ˆ
3j
ˆ
60sin6i
ˆ
60cos6OP +=°+°=`
j
ˆ
3QMandi
ˆ
4PQ ==
j
ˆ
2.8i
ˆ
7QMPQOPOM +=++=\
Example 3.
The resultant of two forces F
1
and F
2
is P. If F
2
is reversed,
the resultant is Q. Show that
22 22
12
2+=+P Q (F F) .
a
q
q
2
F
2
F
P
1
F
1
F
Q
Solution :
Suppose q be the angle between the forces
1
F
r
and
2
F
r
,
then q++=cosFF2FFP
21
2
2
2
1
2
......(i)
2
F 2
F
P
1
F
1
F
q
Q
When the force F
2
is reversed, then the magnitude of their
resultant is
2 22
1 2 12
2 cos(180)Q F F FF= + + °-q
= q-+ cosFF2FF
21
2
2
2
1
......(ii)
Adding equations (i) and (ii),
)FF(2F2F2QP
2
2
2
1
2
2
2
1
22
+=+=+
Example 4.
Find the components of vector ˆˆ
A = 2i + 3j
r
along the
directions of ˆˆ
i+j and ˆˆ
i–j.
Solution :
Here j
ˆ
3i
ˆ
2A+=
r
In order to find the component of A
r
along the direction of
j
ˆ
i
ˆ
+, let us find out the unit vector along j
ˆ
i
ˆ
+. If aˆ is the
unit vector along j
ˆ
i
ˆ
+, then
2
j
ˆ
i
ˆ
|j
ˆ
i
ˆ
|
j
ˆ
i
ˆ
aˆ
+
=
+
+
=
Hence, the magnitude of the component vector of A
r
along
j
ˆ
i
ˆ
+
=
2
5
)32(
2
1
2
j
ˆ
i
ˆ
).j
ˆ
3i
ˆ
2(aˆ.A =+=
+
+=
r

75. 71Motion in a Plane
Solution :
F
2
F
1

2
ˆ
F 250j=
r
,
1
ˆ
F 250i=
r
21
ˆˆ
F F 250 j250i-=-
rr
,
22
21
| F F | 250 250 250 2 N-= +=
rr
(N-W direction)
135°
W
N
E
S
250
tan1
250
q= =-
-
°=qÞ135
Example 9.
If a,b
rr
and c
r
are unit vectors such that a+b+c=0
r
rr
, then
find the angle bet
ween
a
r
and b
r
.
Solution :
Gi
ven :
abc0++=
rrr
c (a b)Þ=-+
rrr
Also, |a| |b| |c|1===
rrr
Let angle between a
r
and b
r
= q
\
22
1 1 1 211c os= ++´´ ´q
\cos 1/2q=- Þ 120 2 / 3q= °=p
PRODUCT OF TWO VECTORS
Scalar or Dot Product
The scalar or dot product of two vectors A and B is a scalar,
which is equal to the product of the magnitudes of A
r
and B
r
and
cosine of the smaller angle be
tween them.
i.e.,
A.B
rr
= A B cosq
q
X
B cosq
B
A
e.g. W Fs;P Fv=× =×
rrr r
Properties of Scalar or Dot Product:
1. ˆˆˆˆ ˆˆ. . .1ii j j kk===
ˆˆˆˆˆˆ. . .0i j jk ki===
2. .AB
rr
= A (B cosq) = B (A cosq)
The dot product of two vect
ors can be interpreted as the
product of the magnitude of one vector and the magnitude
of the component of the other vector along the direction of
the first vector.
Therefore, component vector of A along
ˆˆ
ij+
ˆˆ
5ij5
ˆˆˆˆ(A.a)a (i j)
222
æö+
= = =+
ç÷
èø
r
Similarly, if b
ˆ is the unit vector along the direction of j
ˆ
i
ˆ
-,
then magnitude of the component vector of A along j
ˆ
i
ˆ
-

=
2
1
2
)32(
2
)j
ˆ
i
ˆ
(
).j
ˆ
3i
ˆ
2(
|j
ˆ
i
ˆ
|
)j
ˆ
i
ˆ
(
).j
ˆ
3i
ˆ
2()b
ˆ
.A( -=
-
=
-
+=
-
-
+=
r
\ Component vector of A along ˆˆ
ij-
ˆˆ
1ij1
ˆˆ ˆˆ
(A.b)b (i j)
222
æö-
= =- =- -
ç÷
èø
r
Example 5.
If k
ˆ
c
j
ˆ
4.0i
ˆ
3.0 ++ is a unit vector, then find the value of c.
Solution :
Unit vector is a vector of unit magnitude.
\
1c4.03.0
222
=++
1c16.009.0
2
=++Þ
75.0c75.025.01c
2
=Þ=-=Þ
Example 6.
W
hat is the vector joining the points (3, 1, 14) and
(–2, –1, –6) ?
Solution :
If P and Q be the points represented by the coordinates
(3, 1, 14) and (–2, –1, –6) respectively then,
PQ = p.v. of Q – p.v. of P
= k
ˆ
20j
ˆ
2i
ˆ
5)k
ˆ
14j
ˆ
i
ˆ
3()k
ˆ
6j
ˆ
i
ˆ
2( ---=++----
and
ˆˆˆ
QP PQ 5i 2j 20k=- =++
uuur uuur
Example 7.
F
ind the angle between two vectors of magnitude 12 and
18 units, if their resultant is 24 units.
Solution :
Magnitude of first vector
(A)
r
= 12; Magnitude of second
vector (B)
r
= 18 and resultant of the given vectors (R)
r
= 24
\ 24 =
22
A B 2AB cos++q
24 =
22
(12) (18) 2 12 18 cos++´´q
or (24)
2
= 144 + 324 + 432 cos q or 432 cos q = 108
or cos q =
108
432
= 0.25 or q = 75°52¢
Example 8.
Two forces
1
F 250N=
r
due east and N250 F
2
=
r
due northth
have
their common initial point. Find the magnitude and
direction of
12
FF
rr
-

76. 72 PHYSICS
3. ..AB BA=
rrrr
Dot product of two vectors is commutative.
4.
2
.AAA=
rr
5. .()..A B C AB AC+=+
r r r rrrr
Dot product is dis tributive.
6.x y z xyz
ˆˆˆˆˆˆ
A.B (AiAjA k).(BiBjBk)= ++ ++
rr
= (A
x
B
x
+ A
y
B
y
+ A
z
B
z
)
Vector
or Cross Product
The vector product of two vectors is defined as a vector having
magnitude equal to the product of two vectors and sine of the
angle between them. Its direction is perpendicular to the plane
containing the two vectors (direction of the vector is given by
right hand screw rule or right hand thumb rule.
C AB=´
ur ur ur
= (AB sin q)ˆn
The direction of (A B)´
ur ur
perpendicular to the plane containing
vectors A
ur
and B
ur
in the sense of advance of a right handed
screw rotated from A
ur
to B
ur
is through the smaller angle between
them.
()ABC´=
ur ur ur q
B
A
e.g., ;;v r rFLrp=w´ t=´ =´
rrrr rrr rr
Properties of Vector or Cross Product
1. ˆˆˆˆ ˆˆ 0ii j j kk´=´=´=
2. ˆˆˆˆˆ ˆ ˆ ˆˆ;;ijkjkikij´= ´= ´=
i
j
k
3. 0AA´=
rr
4.AB BA´¹´
rrrr
(not commutative) []´ =-´
rrrr
QAB BA
5. ()()()A B C AB AC´+ =´+´
r r r rrrr
(follows dist ributive law)
6.
ˆˆˆˆ ˆˆ() ()
xyz xyz
A B Ai A j Ak Bi B j Bk´= ++ ´++
rr

xyz
xyz
ˆˆˆ
i jk
AAA
BBB
=
= (A
y
B
z
– A
z
B
y
) ˆ
i
+ (A
z
B
x
– AA
x
B
z
) ˆ
j
+ (A
x
B
y
– A
y
B
x
) ˆ
k
7.The cross product of two vectors represents the area of the
parallelogram formed by them.
k
q
A
B
P Q
R
S
| | (sin)A B AB´=q
rr
= area of parallelogram PQRS
= 2 (area of DPQR)
8.A unit vector which is perpendicular to A as well as B is
sin||
´´
=
q´
u ur u urrr
u ur u ur
A B AB
ABAB
Keep in Memory
1.
xx yy zz
222222
x y
z xyz
ABABAB
A.B
cos
|A||B|
AAABBB
++
q==
++ ++
ur ur
ur ur
2.tan q =
q
q
=
´
cosBA
sinBA
B.A
|BA|
rr
rr
3.
2 2 22.|A B| |AB| AB´+=
rrrr
4. |)A()B(|2|B–A||BA|
rrrrrr
´=´+
5.If 0CBA=++
rrr
, then ACCB BA
rrrrrr
´=´=´
6. q=´- 2cosBA|BA||B.A|
2222
rrrr
7.If B.A|BA|=´ then angle between A and B is
4
p
.
8.If B||Athen 0BA=´
9.Division
by a vector is not defined. Because, it is not
possible to divide by a direction.
10.The sum and product of vectors is independent of
co-ordinate axes system.
CONDITION OF ZERO RESULTANT VECTOR AND LAMI'S
THEOREM
If the three vectors acting on a point object at the same time are
represented in magnitude and direction by the three sides of a
triangle taken in order, then their resultant is zero and the three
vectors are said to be in equilibrium.
i.e.
123
FFF0++=
rrr

77. 73Motion in a Plane
P Q
R
F
3
F
2
F
1
a
b g
Lami's Theorem
It states
that if three forces acting at a point are in equilibrium,
then each force is proportional to the sine of the angle between
the other two forces.
312
FFF
sin sin sin
==
abg
rrr
F
3
F
1
F
2
a
b
g
or,
312
FFF
PQ QR PR
==
rrr
Example 10.
Calculate the
area of a parallelogram formed from the
vectors k
ˆ
3j
ˆ
2i
ˆ
A ++=
r
and k
ˆ
j3i
ˆ
2B +-=
r
, as adjacent
side
s.
Solution :
The area of a parallelogram is given by |BA|
rr
´
Here,
ˆˆˆ
i jk
AB 1 23
2 31
´=+
-
rr
=
ˆ
i [(2 × 1) – (–3 × 3)] +
ˆ
j [(3 × 2) – (1 × 1)]
+
ˆ
k [(1 × –3) – (2 × 2)]
=
11
ˆ
i + 5
ˆ
j – 7
ˆ
k
222
| A B | (11) (5) ( 7)\ ´ = + +-
ur ur
@ 14
Example 11.
A partic
le suffers three displacements 4m in the northward,
2 m in the south-east and 1 m in the south-west directions.
What is the displacement of the particle and the distance
covered by it?
Solution :
Taking a frame of reference with the x-axis in the eastward
and the y-axis in the northward direction
Y
North (N)
XEast (E)West (W)
South (S)
12
ˆ ˆˆ
s = 4j, s= 2cos 45º i 2 sin 45º j-
u r u ur
3
ˆˆ
s = 2 cos 45ºi 2 sin 45º j--
u ur
123
ˆˆˆˆˆ
s=s +s+s =4j+ 2i 2j 2i 2j---
r ur uuruur
\ Displacement ˆ
s (4 2 2)j=-
r
ˆ
= (1.17)j = (1.17)
(northward)
And total distance covered = 4 + 2 + 1 = 7m
Example 12.
Prove that vectors ˆˆˆA= i +2j+3k
r
and ˆˆB=2ij-
r
are
perpendicular to
each other.
Solution :
Here, k
ˆ
3j
ˆ
2i
ˆ
A ++=
r
and j
ˆ
i
ˆ
2B-=
r
Two vectors are perpendicular to each other if,
0B.A=
rr
Now B.A
rr
= )k
ˆ
3j
ˆ
2i
ˆ
( ++. )j
ˆ
i ˆ
2(- = 0
= 1 × 2
+ 2 × (–1) + 3 × (0) = 2 – 2 + 0 = 0
Since
0B.A=
rr
, therefore vectors A
r
and B
r
are
perpen
dicular to each other.
Example 13.
Find the angle between the vectors
ˆˆˆA i j 2k=+-
r
and
ˆˆˆB i2jk=-+-
r
.
Solution :
Her
e k
ˆ
2j
ˆ
iˆ
A -+=
r
, k
ˆ
j
ˆ
2i
ˆ
B -+-=
r
We know that q=cosABB.A
rr
or
AB
B.A
cos
rr
=q
Now
222
222
A 1 1
( 2) 6,
B(1)2(1)6
ˆˆ ˆ ˆ ˆˆA.B (ij2k).(i2jk) 123 3
3 31
cos or 60
6266
= +
+-=
=-++-=
= +- -+ - =-++=
q= = = q= °
´
rr
Exampl
e 14.
A particle is displaced from a point (3, – 4, 5) to another point
(–2, 6, – 4) under a force k
ˆ
j
ˆ
3i
ˆ
2 -+. Find the work done by
the force.
Solution :
k
ˆ
j
ˆ
3i
ˆ
2F-+=
r
The displacement of the particle is
s
r
= position vector of point (–2, 6, – 4) – position vector of
point (3, – 4, 5)
s
r
= k
ˆ
9j
ˆ
10i
ˆ
5) k
ˆ
5j
ˆ
4i
ˆ
3()k
ˆ
4j
ˆ
6i
ˆ
2( -+-=+---+-

78. 74 PHYSICS
\ work done by the force is
)k
ˆ
9j
ˆ
10i
ˆ
5).(k
ˆ
j
ˆ
3i
ˆ
2(s.FW -+--+==
r
r
W = (2) (–5) + (3) (10) + (–1) (–9) = 29 units.
Example 15.
A force ˆˆ
F 6i xj=+
r
acting on a particle displaces it from the
point A (3, 4) to the point B (1, 1). If the work done is 3 units,
then find value of x.
Solution :
ˆˆ
d 2i 3j=--
r
, ˆˆ
F 6i xj=+
r
\ W F.d=
rr
; 3 = – 12 – 3xÞ x = –5
MOTION IN A PLANE OR MOTION IN TWO
DIMENSIONS
The motion in which the movement of a body is restricted to a
plane is called motion in a plane.
Example : A ball is thrown with some initial velocity (u) and
making angle q with harizontal.
The general approach to solve problem on this topic is to resolve
the motion into two mutually perpendicular co-ordinates. One
along X-axis and other along Y-axis. These two motions are
independent of each other and can be treated as two separate
rectilinear motions.
The velocity v and acceleration a can be resolved into its x and y
components.
u sin q
u cos q
v
y
v
x
P(x,y)
v
b
u
O
q
X
Y
xy
xy
ˆˆ
v vi vj
a ai aj
=+
=+
x-component of motion y- compon
ent of motion
v
x
= u
x
+ a
x
t v
y
= u
y
+ a
y
t
x = u
x
t+
1
2
a
x
t
2
y = u
y
t +
1
2
a
y
t
2
v
x
2
– u
x
2

= 2 a
x
x v
y
2
– u
y
2
= 2 a
y
y
x =
xx
u +v
t
2
æö
ç÷
èø
y =
yy
uv
t
2
+æö
ç÷
ç÷
èø
Velocity
The ratio of th
e displacement and the corresponding time interval
is called the average velocity.
xy
ˆˆ
v vi vj=+
r
Average velocity ˆ ˆ ˆˆ
xy
rxy
v i j vi vj
ttt
DDD
== + =+
DDD
Instantaneous velocit
y
0
lim
inst
t
r
v
tD®
D
=
D

O
X
Y
v
v
y
j
^
v
xi
^
q
The magnitude of v =
22
+
xy
vv
The direction of the velocity,
y
x
v
tan
v
q= \
y1
x
v
tan
v
-
q=
Acceleration
The average acceleratio
n in a x–y plane in time interval Dt is the
change in velocity divided by the time interval.
22
ˆˆ
xy
xy
a ai aj
The magnitude of a aa
=+
=+
r
Average acceleration ˆˆ
yx
vvv
a ij
ttt
DDD
==+
DDD
Instantaneous acceleration
inst
000
ˆˆlim lim lim
D® D® D®
DDD
==+
DDD
yx
ttt
vvv
a ij
ttt
In two or three dimensions, the velocity and acceleration vectors
may have any angle between 0°and 180° between them.
RELATIVE VELOCITY IN TWO DIMENSIONS
If two objects A and B moving with velocities V
A
and V
B
with
respect to some common frame of reference, then :
(i) Relative velocity of A w.r.t B
AB AB
v vv=-
r rr
(ii) Relative velocity of B w.r.t. A
BABA
v vv=-
r rr
Therefore,
AB BA
vv=
rr
and
AB BA
vv
=
rr
PROJECTILE MOTION
Projectile is the name given to
a body thrown with some initial
velocity in any arbitrary direction and then allowed to move
under the influence of gravity alone.
Examples : A football kicked by the player, a stone thrown from
the top of building, a bomb released from a plane.
The path followed by the projectile is called a trajectory.
The projectile moves under the action of two velocities:
(1)A uniform velocity in the horizontal direction, which does
not change (if there is no air resistance)
(2)A uniformly changing velocity in the vertical direction due
to gravity.
The horizontal and vertical motions are independent of each other.
Types of Projectile:
1. Oblique projectile : In this, the body is given an initial
velocity making an angle
qwith the horizontal and it moves
under the infuence of gravity along a parabolic path.

79. 75Motion in a Plane
2. Horizontal projectile : In this, the body is given an initial
velocity directed along the horizontal and then it moves
under the influence of gravity along a parabolic path.
Motion along x-axis
u
x
= u, a
x
= 0
x = u
x
t +
1
2
a
x
t
2
ou= ux
P(x,y)
v = u = uxx
x
g
v
v =gty
y
x
y
b
u
=
0
y
x = ut + 0
\ t =
x
u
…… (1)
Motion along y-axi
s
u
y
= 0,a
y
= g
y = u
y
t +
1
2
a
y
t
2
Þ 0 +
1
2
gt
2
y =
1
2
gt
2
…… (2)
From e
quations (1) and (2) we get y =
2
2
g
x
2u
æö
ç÷
èø
which is the equation of a parabola.
Velocity at any instant :
xy
ˆˆ
v vi vj=+
r
v =
2 22
u gt+
If b is the ang
le made by
v
r
with the horizontal, then
tanb =
y
x
v
v
=
gt
u
Time of flight and horizontal range:
If h is the distance of the ground from the point of projection, T
is the time taken to strike the ground and R is the horizontal range
of the projectile then
T =
2h
g
and
2h
Ru
g
=
Case 1: If the p
rojectile is projected from the top of the tower of
height 'h', in horizontal direction, then the height of tower h,range x and time of flight t are related as :
21
h gt and x vt
2
==
h
u
x
vy
vx
v
1
b
Case 2: If a particle is projected at an angle (q) in upward direction
from the top of tower of height h with velocity u, then
u
y
= u sin q
a
y
= – g
u
x
= u cos q
a
x
= 0
2
gt
2
1
t.sinuh-q+= and x =
u cosq.t
x
h
A
B
q
u sinq
u cos
u
+
–
q
Case 3: If a body is projected at an angle (q) from the top of tower
in downward direction then
u
y
= – u sin q, u
x
= u cosq, a
x
= 0
a
y
=
21
g, h u sin.t gt
2
+ - =- q-
and x = u cosq.t
+
–
x
h
q
u
usinq
ucosq
Equat
ion of Trajectory
Let the point from which the projectile is thrown into space is
taken as the origin, horizontal direction in the plane of motion is
taken as the X-axis, the vertical direction is taken as the Y-axis,
Let the projectile be thrown with a velocity u making an angle q
with the X-axis.
u sin q
u cos q
v
y
v
x
P(x,y)
v
b
u
O
q
X
Y
The components of the initial velocity in the X-direction and Y-
direction are u cos q and u sin q respectively. Then at any instant
of time t,

80. 76 PHYSICS
Motion along
x – axis
u
x
= ucosq, a
x
= 0
2
xx
1
x ut at
2
=+
x = (u cosq) t ...(1)
M
otion along y–axis
u
y
= u sin q, a
y
= –g
y = u
y
t +
1
2
a
y
t
2
y = u sinq t +
1
2
gt
2
...(2)
From equations (1) an
d (2) we get
y = x tanq
2
22
g
x
2u cos
-
q
which is the equ ation of a parabola. Hence the path followed by
the projectile is parabolic.
Velocity at any Point
Let v
y
be the vertical velocity of projectile at time t. (at P)
And v
x
be the horizontal component of velocity at time t.
22
sin (1)
cos (2)
y
x
xy
v u gt
vu
v vv
q
q\ = - ¼¼
= ¼¼
=+

2 2 2 2 22
u cos u sin 2gt u sin g t= q+ q- q+
2 22
v u g t 2gt usin=+-q
and the instantaneous angle (b) with horizontal is given by
q
-q=
q
-q
==b
cosu
gt
tan
cosu
gtsinu
v
v
tan
x
y
Time of Fli
ght :
The time of flight of the projectile is given by
2u sin
T 2t
g
q
== ,
where 't' is the tim
e of ascent or descent.
Maximum Height :
Maximum height attained by the projectile is given by
2
2u
H sin
2g
=q .
In case of vertical motion
, q = 90º so maximum height attained
2
2
u
H
g
=
Horizontal Range : The h
orizontal range of the projectile is given by
2
u sin2
R
g
q
= and
2
max
u
R
g
= at q = 45º
(Q maximum value of sin2q = 1)
Keep in Memory
1.The hor
izontal range of the projectile is same at two angles
of projection for q and (90° – q).
2.The height attained by the projectile above the ground is
the largest when the angle of projection with the horizontal
is 90° (vertically upward projection). In such a case time of
flight is largest but the range is the smallest (zero).
3.If the velocity of projection is doubled. The maximum height
attained and the range become 4 times, but the time of flight
is doubled.
4.When the horizontal range of the projectile is maximum, (q =
45°), then the maximum height attained is ¼th of the range.
5.For a projectile fired from the ground, the maximum height is
attained after covering a horizontal distance equal to half of
the range.
The velocity of the projectile is minimum but not zero at the
highest point, and is equal to u cosq i.e. at the highest point
of the trajectory, the projectile has net velocity in the
horizontal direction (vertical component is zero). Horizontal
component of velocity also remains same as the component
of g in horizontal direction is zero i.e., no acceleration in
horizontal direction.
Example 16.
A boat takes 2 hours to travel 8 km and back in still water
lake. If the velocity of water is 4 km h
–1
, the time taken for
going upstream of 8 km and coming back is
(a) 2 hours
(b) 2 hours 40 minutes
(c) 1 hour and 20 minutes
(d) can't be estimated with given information
Solution : (b)
Total distance travelled by boat in 2 hours = 8 + 8 = 16 km.
Therefore speed of boat in still water, v
b
= 16/2 = 8 km h
–1
.
Effective velocity when boat moves upstream = v
b
– v
w
= 8 – 4 = 4 km h
–1
.
Therefore time taken to travel 8 km distance
= 8/4 = 2h.
Effective velocity when boat moves along the stream
= v
b
+ v
w
= 8 + 4 = 12 km h
–1
.
The time taken to travel 8 km distance = 8/12 = 2/3h = 40 min.
Total time taken = 2h + 40 min = 2hours 40 min.
Example 17.
A boat man can row a boat with a speed of 10 km/h in still
water. If the river flows at 5 km/h, the direction in which
the boat man should row to reach a point on the other
bank directly opposite to the point from where he started
(width of the river is 2 km).
(a) is in a direction inclined at 120? to the direction of river
flow
(b) is in a direction inclined at 90? to the direction of
river flow.
(c) is 60? in the north-west direction
(d) is should row directly along the river flow

81. 77Motion in a Plane
Solution : (a)
Refer to fig., the boat man should go along OC in order to
cross the river straight (i.e. along OB).
BC
O A
2km
q
bn
rn
º30sin
2
1
10
5
OC
CB
sin
b
r
===
n
n
==q ; q = 30º;
\ Boat man should go along in a direction inclined at
90º + 30º = 120º to the direction of river flow.
Example 18.
A man swims at an angle
q = 120? to the direction of water
flow with a speed v
mw
= 5km/hr relative to water. If the
speed of water v
w
= 3km/hr, find the speed of the man.
Solution :
r
vmw
=
r
vm
–
r
vw
r
vm
=
r
vmw
+
r
vw
Þv
m
= |
r
vmw
+
r
vw
| = v v v v
mw w mw w
2 2
2+ + . cosq
Þv
m
= 5 3 253 120
2 2
++ () ()cos ?
Þv
m
= 25 9 15+- = 19 m/sec.
Example 19.
A g
un throws a shell with a muzzle speed of 98 m/sec. When
the elevation is 45?, the range is found to be 900 m. How
much is the range decreased by air resistance?
Solution :
Without air resistance, the expected range
8.9
)98(
8.9
90sin)98(
g
2sin
u
R
222
=
´
=
q
= = 980 m
Decrease in range = 980 m – 900 m = 80 m
Example 20.
A particle is projected with velocity u at an angle
q with
the horizontal so that its horizontal range is twice the
greatest height attained. The horizontal range is
(a) u
2
/g (b) 2 u
2
/3 g
(c) 4 u
2
/5 g (d) None of these
Solution : (c)
Greatest height attained
g2
sinu
H
22
q
= …… (
1)
Horizontal range,
g
cossinu2
g
2s
inu
R
22
qq
=
q
= …… (2)
[Qsin 2q = 2 sin q cosq]
Given that R = 2 H
g2
sinu2
g
cossin
u2
222
q
=
qq
\
Solving we get tan q = 2 .....(3)
Hence
)5/1(cosand)5/2(sin =q=q
From eqn. (2)
g5
u4
5
1
5
2
g
u2
R
22
=´´=
Example 21.
A body is projec
ted downwards at an angle of 30? to the
horizontal with a velocity of 9.8 m/s from the top of a tower
29.4 m high. How long will it take before striking the
ground?
Solution :
The situation is shown in fig.
30°
u
BA
–
+
The time taken
by the body is equal to the time taken by the
freely falling body from the height 29.4 m with initial velocity
u sin q. This is given by
9.8
usin 4 .9 m/s
2
q==
Applyin
g the formula, s = u t +
2
1
g t
2
, we ha
ve
29.4 = 4.9 t + 2
1
(9.8) t
2
or 4.
9 t
2
+ 4.9 t – 29.4 = 0
(because s, u and g are all in downward direction)
t
2
+ t – 6 = 0or t = 2 or –3
\ Time taken to reach ground = 2 second
Example 22.
Two boys stationed at A and B fire bullets simultaneously
at a bird stationed at C. The bullets are fired from A and B
at angles of 53° and 37° with the vertical. Both the bullets
fire the bird simultaneously. What is the value of v
A
if
v
B
= 60 units? (Given : tan 37° = 3/4)
Solution :
The vertical components must be equal.

82. 78 PHYSICS
\v
A
cos 53° = v
B
c
os 37°
orv
A
= v
B

cos 37
cos (90 37 )
°
°-°
orv
A
= 60 cot 37° =
60
tan 37°
[Qv
B
= 60 units]
=
604
3
´
= 80 units
PR
OJECTILE ON AN INCLINED PLANE
Let a body is thrown from a plane OA inclined at an angle a with
the horizontal, with a constant velocity u in a direction making an
angle q with the horizontal.
The body returns back on the same plane OA. Hence the net
displacement of the particle in a direction normal to the plane OA
is zero.
a
q
(q-a)
g sin()a
g c
os()a
O
A
B
u
a
g
y
u
y
x
u
x
= u cos (q –a) along the incline, + x-axis)
u
y
= u sin (q –a) along the incline, + y-axis)
a
x
= g sin a along – x-axis, as retardation
a
y
= g cos a along – y-axis, as retardation
The time of flight of the projectile is given by
2
at
2
1
uts+=
or
21
0 u sin( )T g cos T
2
=q-a-a
usin()
T
gcos()
2 q-a
=
a
If maximum hei
ght above the inclined plane is H,
22
usin()
H
2g
q-a
=
The horizontal ran
ge R of the projectile is given by
OB = u cos q t =
2
u sin( )cos
R
g cos
2 q-aq
=
a
The range of the pro
jectile at the inclined plane is given by
OA =
2
2
OB u sin( )cos
R
cos gcos
2 q-aq
==
a a
Condition for horizontal r
ange R on the inclined plane to be
maximum :
Since
2
2
u
R [2sin( ) cos]
gcos
= q-aq
a

2
2
u
[sin(2 )sin]
gcos
= q-a-a
a
{2 sin A cos B =
sin(A+B)+sin(A–B)}
R is maximum when sin (2q –a) is maximum
i.e., sin (2q – a) = 1 or
42
pa
q=+
2
max
2
u
R [1sin]
g cos
Þ = -a
a
or R
max
(on incline
d plane)
max
R (on horizontal plane)
1 sin
=
+a
where R
max
(on horiz
ontal plane)
g2
u
2
=.
Condition for time of f
light T to be maximum :
2usin( )
T
gcos
q-a
=
a
so T is max when sin (q – a) is ma ximum
i.e., sin (q –a) = 1 or
2u
T
2 g cos
p
q=+aÞ=
a
It means that if q
1
is th e angle for projectile for which T is maximum
and q
2
is the angle for which R is maximum, then q
1
= 2q
2
.
Example 23.
The slopes of wind screen of two cars are b
1
= 30° and
b
2
= 15° respectively. At what ratio v
1
/v
2
of the velocities
of the cars will their drivers see the hailstorms bounced by
windscreen of their cars in the vertical direction? Assume
hailstorms falling vertically.
Solution :
From the fig tan
1
v
v
a= and
1
902a= °-b
where v is velocity of hail
11
1
v
tan(90 2 ) cot 2
v
°-b = b=
b
1
–v
1
b
1
b1
v
a
Similarly,
2
2
v
cot2
v
b=
12
21
v cot2 cot 30
3.
v cot 2 cot 60
b
= ==
b

83. 79Motion in a Plane
Example 24.
A particle is projected up an inclined plane of inclination
a to the horizontal. If the particle strikes the plane
horizontally then tan a = ... . Given angle of projection
with the horizontal is b.
(a) 1/2 tan
b (b) tanb
(c) 2 tanb (d) 3 tanb
Solution : (a)
If the projectile hits the plane horizontally then
b
a
A
BC
plane horizontalplane
1
TT
2
=
or
2u sin( ) u sin
gcosg
b-ab
=
a
2 sinb cosa –2 cosb sina = cosa sinb
o
r sinb cosa = 2 cosb sina or tan
tan
2
b
a=
Keep in Memory
1.Equ
ation of trajectory of an oblique projectile in terms of
range (R) is
y = x tan q
x
1
R
æö
--ç÷
èø
2.There are two unique times at which the projectile is at the
same height h(< H) and the sum of these two times.
Since, h = (u sin q)t
21
gt
2
- is a quadratic in time
, so it has
two unique roots t
1
and t
2
(say) such that sum of roots
(t
1
+ t
2
) is 2u sin
g
q
and product (t
1
t
2
) is
2h
g
.
The time lapse (t
1
– t
2
) is

22
2
4u sin 8h
.
gg
q
-
UNIFORM AND NON-UNIFORM CI
RCULAR MOTION
Uniform Circular Motion
An object moving in a circle with a constant speed is said to be
in uniform circular motion.Ex. Motion of the tip of the second
hand of a clock.
Angular displacement : Change in angular position is called
angular displacement (dq ). BDS
dq
O
q
1
q
2
A
Angular velocity : Rate of change of angular displacement is called
angular velocity w
i.e.,
d
dt
q
w=
Relation between linear
velocity (v) and angular velocity (w).
In magnitude,
vr
vr
=w´
=w
rrr
Angular acceleration : Rate of change of angular velocity is called
angular accelaeration.
i.e.,
2
2
dd
dtdt
wq
a==
Relation between lin
ear acceleration and angular accelaration.
In magnitude,
ar
ar
=a´
=a
r urr
Centripetal acceleration : Acceleration acting on a body moving
in uniform circular motion is called centripetal acceleration. It
arises due to the change in the direction of the velocity vector.
Magnitude of certipetal acceleration is

2
2
c
v
ar
r
= =w
21
2 frequency
TT
p æö
w= =pu u ==ç÷
èø
Q
22
4
c
ar\ = pu
This acceleration is always directed radially towards the centre of
the circle.
Centripetal force: The force required to keep a body moving in
uniform circular motion is called centripetal force.
2
2
c
mv
F mr
r
= =w
It is always directed radially inwards.
Centrifugal force : Centrifugal force is a fictitious force which
acts on a body in rotating (non-inertial frames) frame of reference.
Magnitude of the centrifugal force
2
mv
F
r
= .
This force
is always directed radially outwards and is also called
corolious force.

84. 80 PHYSICS
Non-unifo
rm Circular Motion :
An object moving in a circle with variable speed is said to be in
non-uniform circular motion.
If the angular velocity varies with time, the object has two
accelerations possessed by it, centripetal acceleration (a
c
) and
Tangential accelaration (a
T
) and both perpendicular to each other.
b
a
a
T
a
c
Net acceleration
22
24 22
42
()
and, tan
cT
c
T
aaa
arr
ar
a
a
=+
= w+a
= w +a
b=
Keep in Memory
1.Angular di
splacement behaves like vector, when its
magnitude is very very small. It follows laws of vector
addition.
2.Angular velocity and angular acceleration are axial vectors.
3.Centripetal acceleration always directed towards the centre
of the circular path and is always perpendicular to the
instantaneous velocity of the particle.
4.Circular motion is uniform if a
T
= ra = 0, that is angular
velocity remains constant and radial acceleration
2
2
c
v
ar
r
= =w is constant.
5.When a
T

or a is present, angular velocity varies with time
and net acceleration is
22
cT
aaa=+
6.If a
T
= 0 or a = 0, n o work is done in circular motion.
Example 25.
A sphere of mass 0.2 kg is attached to an inextensible string
of length 0.5 m whose upper end is fixed to the ceiling. The
sphere is made to describe a horizontal circle of radius
0.3 m. What will be the speed of the sphere?
Solution :
Centripetal force is provided by component T sin q, therefore
;
r
m
sinT
2
n
=q
and, T cos mg;q=
so,
2
22
22
1/2
2
2 2 1/2
1/2
1/2
vr
tan;
rg
r
r
tan
r
rg
v
( r)
0.09 10
(0.25 0.09)
= 1.5 m/s.
q
==
-
éù
q=êú
êú -ëû
éù
=êú
-ëû
éù ´
=êú
-ëû
l
Q
l
l
O
B
A
T cos q
T
T sin q
l
mg
q
r
q
Example 26.
A cyclist is ridi
ng with a speed of 27 km/h. As he approaches
a circular turn on the road of radius 80 m, he applies
brakes and reduces his speed at the constant rate of 0.50
m/s every second. What is the magnitude and direction of
the net acceleration of the cyclist on the circular turn ?
Solution :
Speed, v = 27 km/h =
115
27 ms 7.5ms
18
--
´=
centripetal acce
leration,
2
c
v
a
r
=
or
2
22
c
(7.5)
a ms 0.7ms
80
--
==
v
a
c
a
t
a
q
P
P is the point at which cyclist applies brakes. At this point,
tangential acceleration a
t
, being negative, will act opposite
to v
r
.
Total acceleration,
22
ct
a aa=+
2222
c
t
or, a (0.7) (0.5) ms 0 .86ms
a0.7
tan 1.4
a 0.5
54 28
--
=+=
q===
\ q= °¢

85. 81Motion in a Plane

86. 82 PHYSICS
1.It is found that .ABA+= This necessarily implies,
(a)B = 0
(b)A, B are antiparallel
(c)A, B are perpendicular
(d)A.B £ 0
2.Which one o
f the following statements is true?
(a) A scalar quantity is the one that is conserved in a
process.
(b) A scalar quantity is the one that can never take negative
values.
(c) A scalar quantity is the one that does not vary from
one point to another in space.
(d) A scalar quantity has the same value for observers
with different orientations of the axes.
3.Two balls are projected at an angle q and (90º – q) to the
horizontal with the same speed. The ratio of their maximum
vertical heights is
(a) 1 : 1 (b) tanq : 1
(c) 1 : tanq (d) tan
2
q : 1
4.Which of the following is not correct ?
(a)
AB BA´ =-´
rrr
(b)AB BA´¹´
rrr
(c) ()A BC ABAC´ + =´+´
r r r rrrr
(d) ()()ABC ABC´+ =´+
rrrrrr
5.The greatest height to which a man can through a ball is h.
What is the greatest horizontal distance to which he can
throw the ball?
(a) 2h (b)
4
h
(c)
2
h
(d) None of these
6.If A
and B are two vectors, then the correct statement is
(a) A + B = B + A (b) A – B = B – A
(c) A × B = B × A (d) None of these
7.Three vectors A, B and C satisfy the relation
A · B = 0 and A · C = 0. The vector A is parallel to
(a)B (b) C
(c) B · C (d) B × C
8.If
ˆn is a unit vector in the direction of the vector A, then
(a)
A
ˆn=
A
(b)ˆn=AA
(c)
A
ˆn=
A
(d)ˆˆn=n×A
9.A projectile thrown with a speed v at an angle q has a range
R on the surface of earth. For same v and q, its range on the
surface of moon will be
Earth
moon
g
g
6
éù
=
êú
ëû
(a) R/6 (b) R
(c) 6
R (d) 36 R
10.Given that A + B = R and A
2
+ B
2
= R
2
. The angle between
A and B is
(a)0 (b)p/4
(c)p/2 (d)p
11.Given that A + B = R and A = B = R. What should be the
angle between A and B ?
(a)0 (b)p/3
(c)2p/3 (d)p
12.Let A = iA cos q + jA sin q be any vector. Another vector B,
which is normal to A can be expressed as
(a) i B cos q – j B sin q(b) i B cos q + j B sin q
(c) i B sin q – j B cos q(d) i B sin q + j B cos q
13.Three particles A, B and C are projected from the same point
with same initial speeds making angles 30º, 45º and 60º
respectively with the horizontal. Which of the following
statements is correct?
(a) A, B and C have unequal ranges
(b) Ranges of A and C are equal and less than that of B
(c) Ranges of A and C are equal and greater than that of B
(d) A, B and C have equal ranges
14.Two particles of equal masses are revolving in circular paths
of radii r
1
and r
2
respectively with the same period. The
ratio of their centripetal force is
(a)r
1
/r
2
(b)
12
r/r
(c) (r
1
/r
2
)
2
(d) (r
2
/r
1
)
2
15.A bomb is
released from a horizontal flying aeroplane. The
trajectory of bomb is
(a) a parabola (b) a straight line
(c) a circle (d) a hyperbola
16.A projectile can have the same range for two angles of
projection. If h
1
and h
2
are maximum heights when the
range in the two cases is R, then the relation between R, h
1
and h
2
is
(a)
12
R 4 hh= (b)
12
R 2 hh=
(c)
12
R hh= (d) None of these
17.A b
omb is dropped from an aeroplane moving horizontally
at constant speed. If air resistance is taken intoconsideration, then the bomb
(a) falls on earth exactly below the aeroplane
(b) falls on the earth exactly behind the aeroplane
(c) falls on the earths ahead of the aeroplane
(d) flies with the aeroplane
18.Two vectors A and B lie in a plane, another vector C lies
outside this plane, then the resultant of these three vectors
i.e., A + B + C
(a) can be zero
(b) cannot be zero
(c) lies in the plane containing A + B
(d) lies in the plane containing A – B

87. 83Motion in a Plane
19.A cannon on a level plane is aimed at an angle q above the
horizontal and a shell is fired with a muzzle velocity n
0
towards
a vertical cliff a distance D away. Then the height from the
bottom at which the shell strikes the side walls of the cliff is
(a)
2
22
0
gD
D sin
2v sin
q-
q
(b)
2
22
0
gD
D cos
2 v cos
q-
q
(c)
2
22
0
gD
D tan
2 v cos
q-
q
(d)
2
22
0
gD
D tan
2 v sin
q-
q
20.A projectile thrown with velocity v making angle q with
vertical gains maximum height H in the time for which the
projectile remains in air, the time period is
(a) g/cosHq (b) g/cosH2q
(c) g/H4 (d) g/H8
21.Three
particles A, B and C are thrown from the top of a
tower with the same speed. A is thrown up, B is thrown
down and C is horizontally. They hit the ground with speeds
v
A
, v
B
and v
C
respectively then,
(a)v
A
= v
B
= v
C
(b) v
A
= v
B
> v
C
(c)v
B
> v
C
> v
A
(d) v
A
> v
B
= v
C
22.An aeroplane flying at a constant speed releases a bomb.
As the bomb moves away from the aeroplane, it will
(a) always be vertically below the aeroplane only if the
aeroplane was flying horizontally
(b) always be vertically below the aeroplane only if the
aeroplane was flying at an angle of 45° to the horizontal
(c) always be vertically below the aeroplane
(d) gradually fall behind the aeroplane if the aeroplane was
flying horizontally.
23.In uniform circular motion, the velocity vector and
acceleration vector are
(a) perpendicular to each other
(b) same direction
(c) opposite direction
(d) not related to each other
24.The time of flight of a projectile on an upward inclined plane
depends upon
(a) angle of inclination of the plane
(b) angle of projection
(c) the value of acceleration due to gravity
(d) all of the above.
25.At the highest point on the trajectory of a projectile, its
(a) potential energy is minimum
(b) kinetic energy is maximum
(c) total energy is maximum
(d) kinetic energy is minimum.
1.A projectile is projected with a kinetic energy E. Its range is R.
It will have the minimum kinetic energy, after covering a
horizontal distance equal to
(a) 0.25 R (b) 0.5 R
(c) 0.75 R (d) R
2.The range of a projectile when launched at an angle of 15º
with the horizontal is 1.5 km. What is the range of the
projectile when launched at an angle of 45º to the horizontal
(a) 1.5 km (b) 3.0 km
(c) 6.3 km (d) 0.75 km
3.A gun fires two bullets at 60º and 30º with horizontal. The
bullets strike at some horizontal distance. The ratio of
maximum height for the two bullets is in the ratio
(a) 2 : 1 (b) 3 : 1
(c) 4 : 1 (d) 1 : 1
4.The angular speed of a fly-wheel making 120 revolutions/
minute is
(a)p rad/sec (b) 4p rad/sec
(c)2p rad/sec (d) 4p
2
rad/sec
5.Consider two vectors
12
ˆˆ ˆˆ
F 2i 5k and F3j 4k=+ =+
rr
. The
magnitude of th
e scalar product of these vectors is
(a) 20 (b) 23
(c)
335 (d) 26
6.If ran
ge is double the maximum height of a projectile, then q is
(a) tan
–1
4 (b) tan
–1
1/4
(c) tan
–1
1 (d) tan
–1
2
7.A body is projected such that its KE at the top is 3/4th of its
initial KE. What is the angle of projectile with the horizontal?
(a) 30º(b) 60º(c) 45º(d) 120º
8.Consider a vector F = 4 i – 3 j. Another vector that is
perpendicular to F is
(a) 4 i + 3 j (b) 6 i
(c) 7 k (d) 3 i – 4 j
9.From a point on the ground at a distance 2 meters from the
foot of a vertical wall, a ball is thrown at an angle of 45º
which just clears the top of the wall and afterward strikes
the ground at a distance 4 m on the other side. The height of
the wall is
(a)
m
3
2
(b)m
4
3
(c)m
3 1
(d)
m
3 4

88. 84 PHYSICS
10.The velocit
y of projection of a body is increased by 2%.
Other factors remaining unchanged, what will be the
percentage change in the maximum height attained?
(a) 1% (b) 2 %
(c) 4 % (d) 8 %
11.A ball is thrown from the ground with a velocity of
320
m/s making an angle of
60º with the horizontal. The ball will
be at a height of 40 m from the ground after a time t equal to
(g = 10 ms
–2
)
(a)
2 sec (b)3 sec
(c) 2 sec (d) 3 sec
12.For
ces of 4 N and 5 N are applied at origin along X-axis and
Y-axis respectively. The resultant force will be
(a)
÷
ø
ö
ç
è
æ-
4
5
tan,N41
1 (b) ÷
ø
ö
ç
è
æ-
5
4
tan,N41
1
(c) ÷
ø
ö
ç
è
æ
-
-
4
5
tan,N41
1
(d) ÷
ø
ö
ç
è
æ
-
-
5
4
tan,N41
1
13.A particle mov
es in a circle of radius 25 cm at two revolutions
per second. The acceleration of the particle in meter per
second
2
is
(a)p
2
(b) 8 p
2
(c) 4 p
2
(d) 2 p
2
14.The vector sum of the forces of 10 N and 6 N can be
(a) 2 N (b) 8 N
(c) 18 N (d) 20 N
15.A bomb is dropped on an enemy post by an aeroplane flying
horizontally with a velocity of 60 km h
–1
and at a height of
490 m. At the time of dropping the bomb, how far the
aeroplane should be from the enemy post so that the bomb
may directly hit the target ?
(a) m
3
400
(b) m
3
500
(c) m
3
1700
(d) 498 m.
16.A proj
ectile is thrown horizontally with a speed of
20 ms
–1
. If g is 10 ms
–2
, then the speed of the projectile
after 5 second will be nearly
(a) 0.5 ms
–1
(b) 5 ms
–1
(c) 54 ms
–1
(d) 500 ms
–1
17.A ball is projected at such an angle that the horizontal range
is three times the maximum height. The angle of projectionof the ball is
(a)
÷
ø
ö
ç
è
æ-
4
3
sin
1
(b) ÷
ø
ö
ç
è
æ-
3
4
sin
1
(c) ÷
ø
ö
ç
è
æ-
3
4
cos
1
(d) ÷
ø
ö
ç
è
æ-
3
4
tan
1
18.A body is projected
horizontally from a point above the
ground and motion of the body is described by the equation
x = 2t, y = 5t
2
where x, and y are horizontal and vertical
coordinates in metre after time t. The initial velocity of the
body will be
(a)
29 m/s horizontal(b) 5 m/s horizontal
(c) 2 m/s vertical (d) 2 m/s horizontal
19.A vector
1
P
u ur
is along the positive x-axis. If its vector product
with another vector
2P is zero then
2P could be
(a)j
ˆ
4 (b)
i
ˆ
4-
(c) )k
ˆ
j
ˆ
(+ (d) )j
ˆ
i
ˆ
(+-
20.A whe
el rotates with constant acceleration of
2.0 rod/s
2
, if the wheel starts from rest the number of
revolutions it makes in the first ten seconds will be
approximately
(a) 32 (b) 24
(c) 16 (d) 8
21.A projectile of mass m is fired with velocity u making angle
q with the horizontal. Its angular momentum about the point
of projection when it hits the ground is given by
(a)
2
2mu sin cos
g
qq
(b)
32
2mu sin cos
g
qq
(c)
2
mu sin cos
2g
qq
(d)
32
mu sin cos
2g
qq
22.A bucket, full of water is revolved in a vertical circle of
radius 2 m. What should be the maximum time-period of
revolution so that the water doesn’t fall out of the bucket?
(a) 1 sec (b) 2 sec
(c) 3 sec (d) 4 sec
23.If |a b| |a b|+=-
rrrr
then angle between a&b
rr
is
(a) 45º (b) 30º
(c)
90º (d) 180º
24.If the sum of two unit vectors is a unit vector, then the
magnitude of their difference is
(a)
3 (b)2
(c)5 (d)
2
1
25.Out of the follo
wing sets of forces, the resultant of which
cannot be zero ?
(a) 10, 10, 10 (b) 10, 10, 20
(c) 10, 20, 20 (d) 10, 20, 40

89. 85Motion in a Plane
26.The magnitude of the vector product of two vectors is 3
times th
e scalar product. The angle between vectors is
(a)
4
p
(b)
6
p
(c)
3
p
(d)
2
p
27.If A = B
+ C and the magnitudes of A, B and C are 5, 4 and 3
units, the angle between A and C is
(a) cos
–1
(3/5) (b) cos
–1
(4/5)
(c)
2
p
(d) sin
–1
(3/
4)
28.A rectangular sheet of material has a width of 3 m and a
length of 4 m. Forces with magnitudes of 3 N and 4N.
respectively, are applied parallel to two edges of the sheet,
as shown in the figure below.
3m
4m4N
3N
F
q
A third force F, is applied to the centre of the sheet, along a
line in the plane of the sheet, at an angle q = tan 0.75 with
respect to the horizontal direction. The sheet will be intranslational equilibrium when F has what value?
(a) F = 3 N (b) F = 4N
(c) F = 5 N (d) F = 7N
29.The linear velocity of a rotating body is given by :
vr= w´
uur uur ur
If k
ˆ
3j
ˆ
4randk
ˆ
2j
ˆ
2i
ˆ
-=+-=w , then the magnitude of
v is
(a) units29 (b) units31
(c) units37 (d) units41
30.Two forc
es are such that the sum of their magnitudes is 18 N
and their resultant is 12 N which is perpendicular to the
smaller force. Then the magnitudes of the forces are
(a) 12 N, 6 N (b) 13 N, 5 N
(c) 10 N, 8 N (d) 16N, 2N.
31.For any two vectors
A
r
and B
r
, if A.B |AB|=´
rrrr
, the
magnitude
of
CAB=+
rr r
is
(a) A + B (b) AB2BA
22
++
(c) 22
BA+
(d)
2
AB
BA
2
++
2
32.The ang
les which the vector
ˆˆˆ
A 3i 6j2k=++
r
makes with
th
e co-ordinate axes are :
(a)
7
1
cos,
7
4
cos,
7
3
cos
111 ---
(b)
7
2
cos,
7
6
cos,
7
3
cos
111 ---
(c)
7
3
cos,
7
5
cos,
7
4
cos
111 ---
(d) None of th
ese
33.The resultant of two forces 3P and 2P is R. If the first force
is doubled then the resultant is also doubled. The angle
between the two forces is
(a) 60º (b) 120º
(c) 70º (d) 180º
34.Let
CAB=+
rr r
(a)|C|
r
is always greater than |A|
r
(b) It is possible to have |C| |A|<
rr
and |C| |B|<
rr
(c)C
r
is always equal toAB+
rr
(d)C
r
is never equal to AB+
rr
35.At what angle must the two forces (x + y) and (x – y) act so
that the resultant may be
22
(x y)+?
(a)
1 22 22
cos [(x y)/2(x y)]
-
-+-
(b)
1 22 22
cos [2(x y)/(x y)]
-
--+
(c)
1 22 22
cos [ (xy ) /(x y )]
-
-+-
(d)
1 22 22
cos [(x y)/(x y)]
-
--+
36.A force of k
ˆ
F– acts on O, the origin of the coordinate
system. The torque about the point (1, –1) is
X
Y
Z
O
(a) )j
ˆ
i
ˆ
(F- (b) )j
ˆ
i
ˆ
(
F+-
(c) )j
ˆ
i
ˆ
(F+ (d) )j
ˆ
i
ˆ
(F--
37.The component of vector
ˆˆ
a 2i 3j=+
r
along the vector
i + j
is
(a)
2
5
(b) 210
(c)25 (d) 5

90. 86 PHYSICS
38.A body is pro
jected at an angle of 30º to the horizontal with
speed 30 m/s. What is the angle with the horizontal after 1.5
seconds? Take g = 10 m/s
2
.
(a) 0º (b) 30º
(c) 60º (d) 90º
39.A projectile is moving at 60 m/s at its highest point, where it
breaks into two equal parts due to an internal explosion.
One part moves vertically up at 50 m/s with respect to the
ground. The other part will move at a speed of
(a) 110 m/s (b) 120 m/s
(c) 130 m/s (d)
s/m6110
40.A particle having a mass 0.5 kg is pro
jected under gravity
with a speed of 98 m/sec at an angle of 60º. The magnitudeof the change in momentum (in N-sec) of the particle after 10seconds is(a) 0.5 (b) 49
(c) 98 (d) 490
41.A large number of bullets are fired in all directions with thesame speed v. What is the maximum area on the ground onwhich these bullets will spread?
(a)
g
v
2
p
(b)
2
4
g
vp
(c)
2
2
2
g
v
p (d)
2
42
g
vp
42.A force of ˆˆ
(3i + 4j)N acts on a body and displaces it by
ˆˆ
(3i + 4j)m. The work done by the force is
(a) 5 J (b) 30 J
(c) 25 J (d) 10 J
43.A person aims a gun at a bird from a point at a horizontal
distance of 100 m. If the gun can impact a speed of 500 ms
–
1
to the bullet. At what height above the bird must he aim
his gun in order to hit it? (g = 10 ms
–2
)
(a) 10.4 cm (b) 20.35 cm
(c) 50 cm (d) 100 cms
44.If
ˆˆˆ ˆˆˆ
a i j kandb 2i j 3k=-+ = ++
rr
, then the unit vector along
ab
®®
+is
(a)
5
k4i3+
(b)
5
k4i3+-
(c)
5
k4i3--
(d) None of these
45.A bo
dy is thrown with a velocity of 9.8 ms
–1
making an
angle of 30º with the horizontal. It will hit the ground after atime
(a) 3.0 s (b) 2.0 s
(c) 1.5 s (d) 1 s
46.Two bullets are fired horizontally, simultaneously and with
different velocities from the same place. Which bullet will
hit the ground earlier ?
(a) It would depend upon the weights of the bullets.
(b) The slower one.
(c) The faster one.
(d) Both will reach simultaneously.
47.A cricket ball is hit with a velocity 25
1
sm
-
, 60° above the
horizontal. How far above the ground, ball passes over a
fielder 50 m from the bat (consider the ball is struck very
close to the ground)?
Take
7.13= and g = 10 ms
–2
(
a) 6.8 m (b) 7 m
(c) 5 m (d) 10 m
48.The equation of a projectile is
2
gx
x3y
2
-=
The angle of proje
ction is given by
(a)
3
1
tan=q (b) 3tan=q
(c)
2
p
(d) zero.
49.A body is thro
wn horizontally with a velocity
gh2 from
the top of a tower o
f height h. It strikes the level ground
through the foot of the tower at a distance x from the tower.The value of x is
(a) gh (b)
gh
2
(c) 2h (d)
2gh
3
50.A plane flying horizontally at a height of 1500 m with a
velocity of 200 ms
–1
passes directly overhead on antiaircraft
gun. Then the angle with the horizontal at which thegun should be fired from the shell with a muzzle velocity of400 ms
–1
to hit the plane, is
(a) 90° (b) 60°
(c) 30° (d) 45°
51.A projectile A is thrown at an angle of 30° to the horizontalfrom point P. At the same time, another projectile B is thrownwith velocity v
2
upwards from the point Q vertically below
the highest point. For B to collide with A,
1
2
v
v
should be
Highest
point
P Q
30°
A
B
v
1
v
2
(a)1 (b) 2
(c)
2
1
(d) 4

91. 87Motion in a Plane
52.The velocity of projection of oblique projectile is
1
sm)j
ˆ
8i
ˆ
6(
-
+ . The horizontal range of the projectile is
(a) 4.9 m (b) 9.6 m
(c) 19.6 m (d) 14 m
53.A gun is aimed at a horizontal target. It takes
2
1
s for the
bulle
t to reach the target. The bullet hits the target x metre
below the aim. Then, x is equal to
(a)
m
4
8.9
(b) m
8
8.9
(c) 9.8 m ( d) 19.6 m.
54.The equation of trajectory of projectile is given by
2
x gx
y
203
=- , where x and y are in metre.
The maximum range of the projectile is
(a)
8
m
3
(b)
4
m
3
(c)
3
m
4
(d)
3
m
8
55.A bullet is fired with a speed of 1500 m/s in order to hit a
target 100 m away. If g = 10 m/s
2
. The gun should be aimed
(a) 15 cm above the target
(b) 10 cm above the target
(c) 2.2 cm above the target
(d) directly towards the target
56.A projectile of mass m is thrown with a velocity v making an
angle 60° with the horizontal. Neglecting air resistance, the
change in momentum from the departure A to its arrival at B,
along the vertical direction is
60°
A
v
B
(a) 2 mv (b)3mv
(c) mv (d)
mv
3
57.A person sitting in the rear end of the compartment throws
a ball towards the front end. The ball follows a parabolic
path. The train is moving with velocity of 20 m/s. A person
standing outside on the ground also observes the ball. How
will the maximum heights (y
m
) attained and the ranges (R)
seen by the thrower and the outside observer compare with
each other?
(a) Same y
m
different R(b) Same y
m
and R
(c) Different y
m
same R (d) Different y
m
and R
58.A particle moves in a circle of radius 4 cm clockwise at
constant speed 2 cm/s. If
ˆx and ˆy are unit acceleration
vectors along X and Y-axis respectively (in cm/s
2
), the
acceleration of the particle at the instant half way between P
and Q is given by
(a) ˆˆ4(x y)-+
(b)ˆˆ4(x y)+
(c)ˆˆ(xy)/2-+ O
y
x
P
Q
(d)ˆˆ(x y)/4-
59.A projec
tile is thrown in the upward direction making an
angle of 60° with the horizontal direction with a velocity of
147 ms
–1
. Then the time after which its inclination with the
horizontal is 45°, is
(a) 15 s (b) 10.98 s
(c) 5.49 s (d) 2.745 s
60.A cyclist moving at a speed of 20 m/s takes a turn, if he
doubles his speed then chance of overturn
(a) is doubled (b) is halved
(c) becomes four times (d) becomes 1/4 times
61.A person swims in a river aiming to reach exactly on the
opposite point on the bank of a river. His speed of swimming
is 0.5 m/s at an angle of 120º with the direction of flow of
water. The speed of water is
(a) 1.0 m/s (b) 0.5 m/s
(c) 0.25 m/s (d) 0.43 m/s
62.The position vector of a particle is
.j
ˆ
)tsina(i
ˆ
)tcosa(r w+w=
r
The velocity of the particle is
(a) directed towards the origin
(b) directed away from the origin
(c) parallel to the position vector
(d) perpendicular to the position vector
63.A ball whose kinetic energy is E is thrown at an angle of 45º
with horizontal. Its kinetic energy at highest point of flight
will be
(a)E (b) E/2
(c)
2
E
(d) O
64.Two pro
jectiles are fired from the same point with the same
speed at angles of projection 60º and 30º respectively.
Which one of the following is true?
(a) Their maximum height will be same
(b) Their range will be same
(c) Their landing velocity will be same
(d) Their time of flight will be same
65.A body of 3kg. moves in X-Y plane under the action of force
given by j
ˆ
t4i
ˆ
t6+. Assuming that the body is at rest at time
t = 0, the velocity of body at t = 3 sec is
(a) j
ˆ
6i
ˆ
9+ (b) j
ˆ
6i
ˆ
18+
(c) j
ˆ
12i
ˆ
18+ (d) j
ˆ
68i
ˆ
12+

92. 88 PHYSICS
66.From a 10
m high building a stone A is dropped, and
simultaneously another identical stone B is thrown
horizontally with an initial speed of 5 ms
–1
. Which one of
the following statements is true?
(a) It is not possible to calculate which one of the two
stones will reach the ground first
(b) Both the stones (A and B) will reach the ground
simultaneously
(c) A stone reaches the ground earlier than B
(d) B stone reaches the ground earlier than A
67.The vector sum of two forces is perpendicular to their vector
differences. In that case, the forces
(a) cannot be predicted
(b) are equal to each other
(c) are equal to each other in magnitude
(d) are not equal to each other in magnitude
68.A particle moves along a circle of radius
m
20
÷
ø
ö
ç
è
æ
p
with
constant tange
ntial acceleration. It the velocity of particle
is 80 m/sec at end of second revolution after motion hasbegun, the tangential acceleration is(a) 40 p m/sec
2
(b) 40 m/sec
2
(c) 640 p m/sec
2
(d) 160 p m/sec
2
69.If
B.A3|BA|
rrrr
=´ then the va lue of |A B|+
rr
is
(a)
½22
)AB3BA(+ (b)
½22
)ABBA(++
(c)
½
22
3
AB
BA
÷
÷
ø
ö
ç
ç
è
æ
++ (d) A + B
70.If a vector238i
jk
ÙÙÙ
++ is perpen dicular to the vector
44jik
ÙÙÙ
- +a, then the va lue of a is
(a) 1/2 (b) –1/2
(c)1 (d) –1
Directions for Qs. (71 to 75) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following-
(a) Statement -1 is false, Statement-2 is true
(b) Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c) Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d) Statement -1 is true, Statement-2 is false
71. Statement -1 : Fr
r
rr
´=t and rF
r
r
r
´¹t
Statement -2 : Cross prod
uct of vectors is commutative.
72. Statement -1 : If dot product and cross product of
A
r
and
B
r
are zero, it implies that one of the vector A
r
and B
r
must
be a null
vector
Statement -2 : Null vector is a vector with zero magnitude.
73. Statement-1 Two stones are simultaneously projected from
level ground from same point with same speeds but different
angles with horizontal. Both stones move in same vertical
plane. Then the two stones may collide in mid air.
Statement-2 : For two stones projected simultaneously from
same point with same speed at different angles with
horizontal, their trajectories may intersect at some point.
74. Statement-1 : K.E. of a moving body given by as
2
where s is
the distance travelled in a circular path refers to a variable
acceleration.
Statement-2 : Acceleration varies with direction only in this
case of circular motion.
75. Statement-1 : Centripetal and centrifugal forces cancel each
other.
Statement-2 : This is because they are always equal and
opposite.
Exemplar Questions
1.The angle between
ˆˆAij=+ and ˆˆBij=- is
(a) 45° (b) 90°
(c)
– 45° (d) 180°
2.Which one of the following statements is true?
(a) A scalar quantity is the one that is conserved in a
process
(b) A scalar quantity is the one that can never take negative
values
(c) A scalar quantity is the one that does not vary from
one point to another in space
(d) A scalar quantity has the same value for observers
with different orientation of the axes
3.Figure shows the orientation of two vectors u and v in the
xy-plane.
X
Y
u
v
O
If ˆˆu ai bj=+ and ˆˆv pi qj=+
Which of the following is correct?
(a)a and p are positive while b and q are negative
(b)a, p and b are positive while q is negative
(c)a, q and b are positive while p is negative
(d)a, b, p and q are all positive

93. 89Motion in a Plane
4.The component of a vector r along X-axis will have maximum
value if
(a)r is along positive Y-axis
(b)r is along positive X-axis
(c)r makes an angle of 45° with the X-axis
(d)r is along negative Y-axis
5.The horizontal range of a projectile fired at an angle of 15° is
50 m. If it is fired with the same speed at an angle of 45°, its
range will be
(a) 60 m (b) 71 m
(c) 100 m (d) 141 m
6.Consider the quantities, pressure, power, energy, impulse,
gravitational potential, electrical charge, temperature, area.
Out of these, the only vector quantities are :
(a) impulse, pressure and area
(b) impulse and area
(c) area and gravitational potential
(d) impulse and pressure
7.In a two dimensional motion, instantaneous speed v
0 is a
positive constant. Then, which of the following are
necessarily true?
(a) The average velocity is not zero at any time
(b) Average acceleration must always vanish
(c) Displacements in equal time intervals are equal
(d) Equal path lengths are traversed in equal intervals
8.In a two dimensional motion, instantaneous speed v
0 is a
positive constant. Then, which of the following are
necessarily true?
(a) The acceleration of the particle is zero
(b) The acceleration of the particle is bounded
(c) The acceleration of the particle is necessarily in the
plane of motion
(d) The particle must be undergoing a uniform circular
motion
NEET/AIPMT (2013-2017) Questions
9.The velocity of a projectile at the initial point A is
$
()23ij+
$
m/s. It’s velocity (in m/s) at point B is [2013]
(a) $
23ij-+
$ (b) $
23ij-
$
(c) $
23ij+
$ (d) $
23ij--
$
10.Vectors ,AB
rr
and C
r
are such that 0AB×=
rr
and 0.AC×=
rr
Then the vector parallel to A
r
is [NEET Kar. 2013]
(a)andBC
rr
(b)AB´
rr
(c)BC+
rr
(d)BC´
rr
11.A particle is moving such that its position coordinate (x, y) are
(2m, 3m) at time t = 0 (6m, 7m) at time t = 2 s and (13m, 14m) at time t = 5s.
Average velocity vector
av
(V)
r
from t = 0 to t = 5s is : [2014]
(a)
1
ˆˆ
(13i + 14j)
5
(b)
7
ˆˆ
(i + j)
3
(c)ˆˆ
2(i + j) (d)
11
ˆˆ
(i + j)
5
12.A ship A is moving Westwards with a speed of 10 km h
–1
and a ship B 100 km South of A, is moving Northwards with a speed of 10 km h
–1
. The time after which the distance
between them becomes shortest, is : [2015]
(a) 5 h (b)
5 2h
(c)10 2h (d) 0 h
13.If vectors ˆˆ
A cos ti sin tj= w+w
r
and
tt
ˆˆ
Bcosisinj
22
ww
=+
r
are functions of time, then the value of t at which they are orthogonal to each other is : [2015 RS]
(a)
t
2
p
=
w
(b)t
p
=
w
(c) t = 0 (d)t
4
p
=
w
14.The position vector of a particle R
r
as a function of time is
given by:
ˆˆ
R 4sin(2 t)i 4cos(2 t)j=p+p
r
Where R is in meter, t in seconds and
ˆ
i and
ˆ
j denote unit
vectors along x-and y-directions, respectively.
Which one of the following statements is wrong for the
motion of particle? [2015 RS]
(a) Magnitude of acceleration vector is
2
v
R
, where v is
the velocity of particle
(b) Magnitude of the velocity of particle is 8 meter/second (c) path of the particle is a circle of radius 4 meter.
(d) Acceleration vector is along -
R
r
15.A particle moves so that its position vector is given by
ˆˆr cos tx sin ty= w+w
r
. Where w is a constant. Which of the
following is true ? [2016]
(a) Velocity and acceleration both are perpendicular to r
r
(b) Velocity and acceleration both are parallel to r
r
(c) Velocity is perpendicular to r
r
and acceleration is
directed towards the origin
(d) Velocity is perpendicular to r
r
and acceleration is
directed away from the origin
16.If the magnitude of sum of two vectors is equal to the
magnitude of difference of the two vectors, the angle
between these vectors is : [2016]
(a) 0° (b) 90°
(c) 45° (d) 180°
17.The x and y coordinates of the particle at any time are x = 5t
– 2t
2
and y = 10t respectively, where x and y are in meters
and t in seconds. The acceleration of the particle at t = 2s is
(a) 5 m/s
2
(b) –4 m/s
2
[2017]
(c) –8 m/s
2
(d) 0

94. 90 PHYSICS
EXERCISE - 1
1. (b) 2. (d)
3. (d) q=
q-
q
=
2
22
22
2
1
tan
g2/)º90(sinu
g2/sinu
H
H
4. (d) 5. (a)
6. (
a) In vector addition, the commutative law is obeyed
i.e.,
ABBA
rrrr
+=+
Vector subtraction does not follow commutative law.
7. (d) Since B.A0C.A
rrrr
==, it means that A
r
is perpendicular
to both B&C
rr
, hence A
r
is parallel to )C B(
rr
´or
)BC(
rr
´.
8. (a) The unit ve
ctor of any vector
A
r
is defined as
|A|
A
A
ˆ
r
r
=
9. (c) On ear
th, R = u
2
sin 2q/g.
On moon, g' = g/6R' = u
2
sin 2q/g' = 6u
2
sin2q/g = 6R.
10. (c)
BA2
BAR
cos
222
--
=q 0
BA2
RR
22
=
-
=
\ 2/p=q
11. (c) ]cosAB2BA[
R
222
q++=
q++=cosR2RRR
2222
2/1cosorcosR2R
22
-=qq=- or 3/2p=q
12. (c) The dot product should be zero.
13. (c)
)g/u(
2
3
g
º60sinu
R
2
2
º30 ==
g/u
g
90sinu
R
2
2
º45 ==
)2/3(
g
u
g
º30cosu
g
º120sinu
R
222
º60 ===
so
30º 60º 45º
RRR=> or
ACB
RRR=>
14. (a)
2
11
mrFw=;
2
22
mr
F w=
since period T is same, so w is same, because
w
p
=
2
T .
Hence ÷
÷
ø
ö
ç
ç
è
æ
=
2
1
2
1
r
r
F
F
15. (a) A par
abola
16. (a)
22
1
u sin
h
2g
q
=
22
2
usin(90)
h
2g
-q
= ,
2
u sin2
R
g
q
=
Range R is
same for angle q and (90° – q)
\
22 22
12
u sin u sin (90 )
hh
2g 2g
q -q
=´
422
2
u(sin)sin(90)
4g
q ´ -q
= [ sin(90 ) cos ]-q=qQ
422
2
u (sin ) cos
4g
q´q
= [ sin 2 2sin cos ]q= qqQ
42
2
u(sin cos)
4g
qq
=
42
2
u(sin2)
16g
q
=
2 22
2
(usin2)R
1616g
q
==
or, R
2
= 16 h
1
h
2
or
12
R 4 hh=
17. (b) If the
re is no resistance, bomb will drop at a place exactly
below the flying aeroplane. But when we take into
account air resistance, bomb will face deceleration in
its velocity. So, it will fall on the earth exactly behind
the aeroplane.
18. (b)
19. (c) From the resultant path of the particle, when it is
projected at angle q with its velocity u is
q
-q=
22
2
cosu
gx
2
1
tanxy
y
B
D
O
q
A
Ca
n
o n
Where y denotes the instantaneous height of particle
when it travels an instantaneous horizontal distance x.
here x = D, u = v
o
so
2
22
0
1 gD
y D tan
2v cos
= q-
q
20. (d) Max. height =
g2
)90(sinv
H
22
q-
= .....(i)
Time of fl
ight,
g
)90sin(v2
T
q-
= ...(ii)
Hints & Solutions

95. 91Motion in a Plane
Horizontal
V
e
r
t
i
c
a
l
v
q
From (i)
,
g
H2
g
cosv
=
q
,From (ii),
2H 8H
T = 2
gg
=
21. (a)For A: It go
es up with velocity u will it reaches its
maximum height (i.e. velocity becomes zero) and comes
back to O and attains velocity u.
Using as2uv
22
+=
Þ gh2uv
2
A +=
O v
x
u=
v
X
u=Av
Bv
Cv
u
u
h
2 2
c x y
v vv= +
For B, goin g down with velocity u
Þ gh2uv
2
B +=
For C, horizonta
l velocity remains same, i.e. u. Vertical
velocity =
gh20+ = gh2
The res
ultant
C
v =
2
y
2
xvv+ = gh2u
2
+.
Hence
CBA
v
vv==
22. (c) Since horizontal component of the velocity of the bomb
will be the same as the velocity of the aeroplane,
therefore horizontal displacements remain the same at
any instant of time.
23. (a) In uniform circular motion speed is constant. So, no
tangential acceleration.
It has only radial acceleration
R
v
a
2
R
=[directed t
owards
center]
and its velocity is always in tangential direction. So these
two are perpendicular to each other.
24. (d)2usin()
T
gcos
q-a
=
a
25 (d)
Velocity and kinetic energy is minimum at the highest
point. q=
22
cosvm
2
1
E.K
EXERCISE - 2
1.
(b) K.E. is minimum at the highest point. So, the horizontal
distance is half of the range R i.e., 0.5 R.
2. (b)
g
30sinu
g
2sinu
R
22
1 =
q
= or 3
g
u
or
g2
u
5.1
22
==
km3
g
u
g
90sinu
R
22
2 ===
3. (b)
The bullets are fired at the same initial speed
º30sin
º60sin
º30sinu
g
2
g2
º60sinu
H
H
2
2
22
22
=´=
¢
2
2
(3/2)3
1(1/ 2)
==
4. (b)
5. (
a) 20)k
ˆ
4j
ˆ
3).(k
ˆ
5i
ˆ
2(F.F
21
=++=
rr
6. (d)
g2
sinu
2
g
cossin2
u
222
q
´=
qq
or tan q = 2
7. (a)
221 31
m(ucos) mu
2 42
q=´
or
4
3
cos
2
=q or º30cos
2
3
cos ==q .
8. (c)
9.
(d)
g
u
6or
g
u
R
22
== ;
q
-q=
22
2
cosu2
xg
tanxy
or ÷
ø
ö
ç
è
æ
-=
6
1
º45cos2
4
º45tan2h
2
m
3
4
3
2
2=-=
10. (c
) We know that,
g2
)sinu(
Hy
2
m
q
==
g2
sinu
22
q
=
.
u
u2
H
H D
=
D
\ Give n %2
u
u
=
D
H
2 2 4%
H
D
\ =´=
11. (c
) As,
2
gt
2
1
tsinus-q=
so
2
t10
2
1
t)2/3(32040´´-´=
or 5t
2
– 30t +
40 = 0 or t
2
– 6t + 8 = 0
or t = 2 or 4.
The minimum time t = 2s.
12. (a)
22
54R += N41=
q
R5N
4N

96. 92 PHYSICS
The angle q will be given by tan
4
5
=q or
15
tan
4
-æö
q= ç÷
èø
13. (c) Here sec
2
1
T= the required
centripetal acceleration
for moving in a circle is
22
22
C
v (r)
a r r (2 / T)
rr
w
= = =w=´p
so
222
c
a 0.25 (2 / 0.5) 16 .25 4.0= ´p =p´=p
14. (b)
max
R (10 6) 16N= += ,
min
R (10 6) 4N= -=
ÞValues can be from 4N to 16N
15. (b) Time taken for vertical direction motion
2h 2 490
t 100 10s
g 9.8
´
== ==
The same t
ime is for horizontal direction.5 500
x vt 60 10 m
183
æö
\= = ´ ´=
ç÷
èø
16.
(c) Even after 5 second, the horizontal velocity
x
v will be
20
1
sm
-
. The vertical velocity y
v is given by
1
y sm505100v
-
=´+=
Now,
22221
xy
v v v 20 50 54 ms
-
= += +»
17. (d) Giv
en
2 22
u sin 2 u sin
3
g 2g
qq
=
Þ ÷
ø
ö
ç
è
æ
=q
-
3
4
tan
1
18. (d) The
horizontal velocity of the projectile remains
constant throughout the journey.
Since the body is projected horizontally, the initial
velocity will be same as the horizontal velocity at any
point.
Since, x = 2t ,
dx
2
dt
=
\ Horizontal velocity = 2 m/s
\ Initial velocity = 2 m/s
19. (b) Vector product of parallel vectors is zero.
20. (c) For circular angular motion, the formula for angular
displacement q and angular acceleration a is
21
tt
2
q=w+a where w = initial velocity
or
21
0t
2
q=+a or
21
(2)(10)
2
q=´
or q = 100 radian
2p ra
dian are covered in 1 revolution
\ 1 radian is covered in
1
2p
revolution
or 100
radian are covered in
100
2p
revolution
\ Number
of revolution
50
16
3.14
==
21. (b)L Rp=´
rr
Where R = range
2
u sin2
g
q
=
The angle bet
ween
R
r
and p
r
= q
Also, p = m
u.
Hence,
2
u sin2
L musin()
g
q
= ´q
32
2mu sin cos
g
qq
=
22. (c) Le
t its angular velocity be
w at all points (uniform
motion). At the highest point weight of the body is
balanced by centrifugal force, so
2 g
m r mg
r
w = Þw=
2 r 22
T 22
g 10 5
pp
= =p =p =
w
.sec3
1.2
14.32
=
´
=
23. (c)|a b| |a b|+=-Þ
rrrr

22
|a b| |a b|+ =-
rrrr
22 22
|a| |b| 2a.b |a| |b| 2a.bÞ++ =+-
rrrr
24. (a) 25. (d) R
2
=
P
2
+ Q
2
– PQ cos q
(40)
2
= (10)
2
+ (20)
2
– 2 × 10 × 20 × cosq
400 cos q = 500 – 1600 = – 1100
4
11
400
1100
cos -=-=q which is not poss
ible.
In this way, the set of forces given in option (d) can not
be represented both in magnitude and direction by the
sides of a triangle taken in the same order. Thus their
resultant can not be zero.
26. (c) Here,
|A B| 3|A.B|´=
rrrr
ABsin 3ABcosÞq=q tan3Þ q=
60
3
p
Þq= °=
27. (a) See fig. Clearly A is the resultant of B and C. Further B
is perpendicular to C
B
C=3
4
A
5
4
q
cos q = 3/5 or q = cos
–1
(3/5)

97. 93Motion in a Plane
28. (c) A body is in translational equilibrium when the
components of all external forces cancel.
For the sheet : F cos q = 4 N, F sinq = 3 N. The magnitude
of F is found by adding the squares of the components:
F
2
cos
2
q+ F
2
sin
2
q = F
2
= 4
2
+ 3
2
= 25 N
2
. Therefore F
= 5 N. The F vector points in the proper direction,
since tan q = 0.75 = 3/4.
29. (a)
ˆˆˆ
ijk
v1 22
043
=-
-
r
v
r
]04[k
ˆ
]30[j
ˆ
]86[i
ˆ
-+++-=
v
r
k
ˆ
4j
ˆ
3i
ˆ
2 ++-= | v | 29unitsÞ=
u ur
.
30. (b) Use
tan
q+
q
=a
cosPQ
sinP
Þ tan 90° ¥=
q+
q
=
cosPQ
sinP
\ Q + P cosq= 0 Þ P cosq= -Q.
R = q++cosPQ2QP
22
R =
222
Q2QP-+ or R =
22
QP- = 12
144
= (P + Q)(P - Q) or P - Q = 144/18 = 8.
\ P = 13 N and Q = 5 N.
31. (b)|A.B| |AB|=´
rrrr
AB cos ABsinÞ q=q
1
tanθ= 1 cosθ=
2
ÞÞ
Now,
22
| A B | A B 2ABcos+=++q
rr
22 1
A B 2AB.
2
= ++
22
A B 2AB= ++
32. (b)
ˆ ˆˆ
3ibj2 k 362
ˆˆˆˆ
A i jk
7779364
+++ æö
= = ++
ç÷
++ èø
. If a, b and g are
a
ngles made by
Awith coordinate axes, then
7
2
cosand
7
6
cos,
7
3
cos =g=b=a .
33. (b) S
olve two equations : R
2
= 9P
2
+ 4P
2
+ 12P
2
cosq and
4R
2
= 36P
2
+ 4P
2
+ 24P
2
cosq.
34. (b)
35. (a)
2
22 22
x y (x y) (x y) 2(x y)(x y) cos
æö
+ =++-++ - qç÷
èø
22 22 22
22
x y x y 2 xy x y 2xy 2(x y ) cosÞ+=++ ++- + - q
or
22 22
2(xy).cos (xy)- q=-+
22
1
22
(x y)
cos
2(x y)
-
éù-+
Þq= êú
-êúëû
36. (c) Torque
ˆˆˆ
r F (i j) ( Fk)t= ´ = - ´-
ur r ur
ˆˆˆˆ
F[ i k jk]= -´ +´ ˆˆ ˆˆ
F( j i )F( i j)=+=+
ˆˆˆ ˆˆˆ
Since k ij and j k iéù ´= ´=
ëû
37. (a)Component of a
r
along
a.b
b
|b|
=
rr
r
r
38. (a)
x
u 30cos30º 303 / 2,== ,u
y
= 30 sin 30°,
yv 30sin30 gt= °-
05.110º30sin30v
y
=´-=
A
s vertical velocity = 0,
Angle with horizontal a = 0º
It is a state, when a particle reach to a highest point of
its path.
39. (c) From conservation of linear momentum,
12
mm
mvvv
22
=+
rrr
q
120 m/s
y
E
xpl
o
d
e
= 50m/secv
1
v
250 m/s
(here we take particle & earth as a system so in this
case external force is zero & linear momentum is
conserved)
Where v
r
is velocity of particle before explosion &
12
v ,v
rr
are velocity of its equal pieces.
here ˆ
v 60i=
r
(in x direction),
1
ˆ
v 50j=
r
(in y direction)
so
2
ˆˆ
v 120i 50j=-
r
or
2
| v | 130m /sec=
r
&
[From c
onservation of linear momentum]
120
50
tan
-
=q
40. (b
) There is no change in horizontal velocity, hence no
change in momentum in horizontal direction. The
vertical velocity at t = 10sec is
10)8.9(º60sin98v ´-´= = –13.13 m/sec
so change in momentum in vertical direction is
=
(0.5 98 3 /2) [ (0.5 13.13)]´´ --´
= 42.4
34+6.56 = 48.997
»49
41. (b) M
aximum possible horizontal range = v
2
/g
Maximum possible area of the circle
2
24
2
vv
g g
æö p
=p=ç÷
ç÷
èø

2
v
Herer
g
éù
=êú
êúëû

98. 94 PHYSICS
42. (c)
Here ,
ˆˆ
F (3i 4j)N=+
r
ˆˆ
d (3i 4j)m=+
r
\ W F.d (9 16)J 25J==+=
rr
43. (b) Let the gun be fired with velocity u from point O on the
bird at B, making an angle q with the horizontal direction.
Therefore the height of the aims of the person is at height
BA (=h) above the bird.
O
B
A
h
u
q
100 m
Here, horizontal range = 100
g
2sinu
2
=
q
or
2
2
500 sin2
100 or s
in 2
10
100 10 1
sin14
250(500)
q
=q
´
= == ¢
or2q = 14' or q = 7' =
18060
7p
´ radian
As,
arc
angle
radius
= \
OB
AB
=q
or AB = q × OB
=
7
(100 100) cm 20.35 cm
60 180
p
´´´=
44. (a) ˆˆ
a b 3i 4k+=+
rr
\ Required unit vector
=
22
ˆˆ
a b 3i 4k
|a b|
34
++
=
+
+
rr
rr
ˆˆ
3i 4k
5
+
= .
45. (d) Time o
f flight =
g
sinu2 q
= .sec1
2
1
2
8.9
º30sin8.92
=´=
´
´
46. (d) The initial velocity in the vertically downward direction
is zero and same height has to be covered.
47. (c)
2
22
1 gx
y x tan
2u cos
= q-
q
°´´´
´´
-°=
60cos25252
505010
60tan50y
2
=5 m
48. (b)
Comparing the given equation with
2
22
gx
y x tan
2u cos
= q-
q
, we get
3tan=q
49. (c)
x
h
2gh
u
y
= 0, s
y
= –h, a
y
= –g, t
y
= ?
21
s ut at
2
=+
21
h gt
2
\- =-
2h
t
g
Þ=
velocity =
x
t
2h
x 2gh 2h
g
\= ´=
50. (b) Hori
zontal distance covered should be same for the
time of collision.
200cos400
=q or
2
1
cos=q or °=q60
51. (c) This
happen when vertical velocity of both are same.
\
21
v v sin30=° or
2
1
v1
v2
=
52. (b) j
ˆ
8i
ˆ
6v+=
810
6
q
Comparing with xy
ˆˆ
v vi vj=+
u ur
, we get
1
x
sm6v
-
= and
1
y
sm
8v
-
=
Also,
2
y
2
x
2
vvv+= = 36 + 64 = 100
or
1
sm10v
-
=
10
6
cosand
10
8
sin =q=q
g
cossinv2
g
2sinv
R
22
qq
=
q
=
861
R 2 10 10
10 10 10
=´´´´´ = 9.6 m
53. (b)
2
gt
2
1
x=

1 1 1 9.8
9.8m
2 228
=´ ´´=

99. 95Motion in a Plane
54. (b) Comparing the given equation with the equation of
trajectory of a projectile,
2
22
gx
y x tan
2u cos
= q-
q
we get,
1
tan 30
3
q= Þq=°
and
222
2
20
2u cos 20 u
2cos
q=Þ=
q
2
10
cos 30
=
°
2
10
3
2
=
æö
ç÷
ç÷
èø
40
3
=
Now,
2
max
u 40
R
g 3 10
==
´
4
m
3
=
55. (c)
The bullet performs a horizontal journey of 100 cm with
constant velocity of 1500 m/s. The bullet also performs
a vertical journey of h with zero initial velocity and
downward acceleration g.
\ For horizontal journey, time (t)
Distance
Velocity
=
\
100 1
t sec
1500 15
== … (1)
The
bullet performs vertical journey for this time.
For vertical journey,
21
h ut gt
2
=+
2
11
h 0 10
2 15
æö
=+´´
ç÷
èø
or,
10 10 100
h m cm
2 15 15 2 1 5 15
´
==
´´ ´´
or,
20
h cm 2.2cm
9
==
The gun should be aimed
20
9
æö
ç÷
èø
cm above the target.
56. (b)
60°
60°
A
B
v
P
=
m
v
s
i
n

6
0
°
A

P = mvsin 60°
B
As the figure drawn above shows that at points A and
B the vertical component of velocity is v sin 60° but
their directions are opposite.
Hence, change in momentum is given by :
p mvsin 60 ( mvsin 60 ) 2mvsin 60D=°-- °=°
3
2mv 3mv
2
==
57. (a
) The motion of the train will affect only the horizontal
component of the velocity of the ball. Since, verticalcomponent is same for both observers, the y
m
will be
same, but R will be different.
58. (c)
2
v
a
r
= = 1 cm/s. Centripetal acceleration is directed
towards the centre. Its magnitude = 1. Unit vector atthe mid point on the path between P and Q is
ˆˆ(xy)/2-+ .
59. (c)
60°
45°
v cos 45°
v
v sin 45°
u cos 60°
u
u sin 60°
O B
A
Velocity of projectile u = 147 ms
–1
angle of projection a = 60°
Let, the time take
n by the projectile from O to A be t
where direction
b = 45°. As horizontal component of
velocity remains constant during the projectile motion.
Þv cos 45° = u cos 60°
Þ v ×
11
147
22
=´ Þ
1147
v ms
2
-
=
For Ver
tical motion, v
y
= u
y
– gt
Þv sin 45° = 45sin 60° – 9.8 t
Þ
147 1 3
147 9.8 t
222
´ =´-
Þ9.8 t =
147
( 3 1)
2
- Þ t = 5.49 s
60. (d) When a cyclist moves on a circular path, it experiences
a centrifugal force which is equal to mv
2
/ r. It tries to
overturn the cyclist in outward direction. If speed
increases twice, the value of centrifugal force too
increases to 4 times its earlier value. Therefore the
chance of overturning is 1/4 times.
61. (c) Here v = 0.5 m/sec. u = ?
so
Þ=q
v
u
sin
2
1
5.
u
= or u = 0.25 ms
–1
u
v
120º
30º
CB
A
direction
of flow
ri
ver

100. 96 PHYSICS
62. (d) j
ˆ
)tsin
a(i
ˆ
)tcosa(rw+w=
r
dt
d
dt
)r(d
v ==
r
}j
ˆ
)tsina(i
ˆ
)tcosa{( w+
w
j
ˆ
)tcosa(i
ˆ
)tsina( ww+ww-=
]j
ˆ
)tcosa(i
ˆ
)tsina[( w+w-w=
r.v0=
rr
\ velocity is perpendicular to the displacement.
63. (b) Since
m
E2
vmv
2
1
E
2
=Þ=
Now at highest po
int of flight, the vertical component
of velocity is zero & only horizontal component is non
zero. So K.E. at highest point is
21
E ' m(v cos 45º)
2
=
= E / 2
64. (b) G
iven, u
1
= u
2
= u,q
1
= 60º, q
2
= 30º
In Ist case, we know that range
g
)3090sin(u
g
120sinu
g
)60(2sin
u
R
222
1
°+°
=
°
=
°
=
22
u (cos30 ) 3u
g 2g
°
==
In IInd case
when °=q30
2
, then
g2
3u
g
60sinu
R
22
2 =
°
= Þ R
1
= R
2
(we get same value of ran
ges).
65. (a)
ˆˆ
F = 6ti + 4tj or
xy
6t 4t
a ,a
33
==
so
2t
u = a dt=t
oxx
ò (u ) = 9m/sec
xt=3
Þ
and
2
2tt
u = a dt=
yyo
3
ò sec/m6)u(
3ty
=Þ
=
(because u
x
& u
y
= 0 at t =
0 sec)
66. (b)
2
gt
2
1
s= , s and g are same for bo
th the balls, so time
of fall ‘t’ will also be the same for both of them (s is
vertical height)
67. (c) BAsumvectorP
rrr
+==
Q vector differences A B= =-
rr r
Since P
r
and Q
r
are perpendicular
\ 0Q.P=
rr
Þ (A B).(A B) 0+ -=
rrrr
Þ AA
2
= B
2
Þ BA=
68. (b) Ci
rcumference of circle is 2pr = 40m
Total distance travelled in two revolution is 80m. Initialvelocity u =0, final veloctiy v = 80 m/sec
so from
v
2
=u
2
+2as
Þ (80)
2
= 0
2
+2×80×a
Þa = 40m/sec
2
69.
(b) q=´sinBA|BA|
rr
q=cosBAB.A
rr
B.A3|BA|
rrrr
=´
Þ AB sin q = Ö3 A
B cos q or tan q = Ö3
\ q = 60º
22
| A B| A B 2AB cos60º\+= ++
rr

ABBA
22
++=
70. (b) For two vec
tors to be perpendicular to each other
AB
®®
× = 0
(238i jk
ÙÙÙ
++ ) · (44jik
ÙÙÙ
- +a) = 0
–8 + 12 + 8a = 0 or
a=- =-
4
8
1
2
71. (d) 72. (c ) 73. (d) 74. (d) 75. (a)
EXERCISE - 3
Exemplar Questions
1. (b) Given, ˆˆAij=+
ˆˆBij=-
As we know that
||||cosAB AB×=q
rr
22 22ˆˆ ˆˆ( )( ) (1 1)(1 1)cosijij+×-=+ +q
()()22cosijij+-=´q
where q is the angle between A and B
1001
cos0
22
-+-
q==
\ q = 90°
2. (d) A
scalar quantity does not depend on direction so it
has the same value for observers with different
orientations of the axes.
3. (b) From the diagram,
ˆˆu ai bj=+
As u is in the first quadrant, so both components a
and b will be positive.
For ˆˆ,v pi qj=+ as it is in positive x-dir ection and
located downward so x-component p will be positive
and y-component q will be negative.
Hence, a, b and p are positive but q is negative.
4. (b) Let r makes an angle q with positive x-axis component
of r along x-axis.
| | cos
x
rr=q
maximum maximum
( ) | | (cos )
x
rr =q
||cos0 ||rr= °=
(cosqQ is maximum of q = 0°)
q = 0°.
H
ence the vector r has maximum value along positive
x-axis.

101. 97Motion in a Plane
5. (c) Consider, projectile is fired at an angle q.
According to question,
q = 15° and R = 50 m
R
X
Y
u
q
Range,
2
sin2u
R
g
q
=
2
sin(2 15 )
50m
u
R
g
´°
==
22 1
50 sin30
2
guu´= °=´
50 × g × 2 = u
2
2
50 9.8 2 100 9.8 980u=´´= ´=
980u= = 31.304 m/s 145=
2
( 9.8 m/s )=gQ
Now, 45;q=°
22
sin 2 45uu
R
gg
´°
==
2
(14 5) 14 145
100 m
9.8
R
g
´´
===
6. (b) As we
know that,
Impulse,
Dæö
= D= D=Dç÷
èøD
p
IFt tp
t
where F is force, Dt is time duration and Dp is change
in momentum. As Dp is a vector quantity, hence impulse
is also a vector quantity. Sometimes area can also be
treated as vector.
7. (d) As speed is a scalar quantity, hence it will be related
with path length (scalar quantity) only.
Hence, Speed
0
total distance travelled
time taken
v=
So, total distance tr
avelled = Path length
= (speed) × time taken
Hence, path length which is scalar and traversed in
equal intervals.
8. (c) As given that in two dimensional motion the
instanteous speed v
0 is positive constant and we know
that acceleration is rate of change of velocity or
instantaneous speed and hence it will also be in the
plane of motion.
NEET/AIPMT (2013-2017) Questions
9. (b) At point B the direction of velocity component of the
projectile along Y - axis reverses.
Hence,
B
V
®

= 2i 3j-
$$
10. (d) Vector triple product
( ) ( ) ( )0A B C BAC CAB´´= ×- ×=
r r rr rrrrr
Þ ||()A BC´
rrr
[ 0 and 0]×= ×=
r rrr
QAB AC 1.
(a)
22
( ) ()ABC+=
rrr
22
2.A B ABC
2
Þ++=
rr
22
342.5AB
2
Þ++=
rr
2.0ABÞ=
rr
or .0ABÞ=
rr
AB\^
rr
Here A
2
+ B
2
= C
2
. H ence, AB^
uur uur
11. (d)
av
v
r
=
r (displacement)
t (time taken)
D
D
r
=
ˆˆ
(13 2)i (14 3) j
50
-+-
-
=
11
ˆˆ
(i j)
5
+
12. (a) ()AV 10 –i=
ur
$
()BV 10j=
ur
$
BA
ˆ
V 10 j 10 i 10 2 km / h=+=
ur
$
Distance OB = 100 c
os 45° =
50 2 km
O
S
100 km
w 10 km/h A
45°
100 km
N(
j)
$
BAV 102km/h=
B
Time taken to reach the shortest distance between
A and
BA
OB
B
V
=uuuur
502
5h
102
==
13. (b) T
wo vectors are
A
r
=
ˆˆ
cos ti sintjw+w
B
r
=
tt
ˆˆ
cosisinj
22
ww
+

102. 98 PHYSICS
For two vector
s
A
r
and B
r
to be orthogonal A.B
rr
= 0
A.B
rr
= 0 = cos wt.cos
t
2
w
+ sin wt.sin
t
2
w
= cos
tt
t cos
22
wwæ ö æö
w-=
ç ÷ ç÷
è ø èø
So,
t
22
wp
=\ t =
p
w
14. (b) Here,x = 4sin(2pt) ...(i)
y = 4cos(2pt) ...(ii)
Squaring and adding equation (i) and (ii)
x
2
+ y
2
= 4
2
Þ R = 4
Motion of the particle is circular motion, acceleration
vector is along –R
ur
and its magnitude =
2
V
R
Velocity of particle, V = wR = (2p) (4) = 8p
15. (c)Given: Position vector
r
r
= cos wt ˆx + sin wt ˆy
\Velocity, v
r
= – wsin wt ˆx + wcos wt ˆy
and acceleration,
a
r
= –w
2
cos wtˆx – w
2
sin wt ˆy = – w
2
r
r
r
r
. v
r
= 0 hence rv^
rr
and
a
r
is directed towards the origin.
16.(b)AB AB+=-
rrrr
Squaring on both sides
22
AB AB+ =-
rrrr
Þ A·A 2A·B B·B++
rrr r rr
= A·A–2A·B B·B+
rrr r rr
Þ 4A · B 0 4AB cos 0= Þ q=
rr
Þ cos q = 0Þq = 90°
1
7. (b) Given:
x = 5t – 2t
2
y = 10t
v
x
=
dx
dt
= 5 – 4t v
y
=
dy
10
dt
=
a
x
=
xdv
dt
= – 4 a
y
=
y
dv
0
dt
=
xy
a ai aj=+
r
2
a 4i m/s=-
r
Hence, acceleration of particle at (t = 2 s) = –4m/s
2

103. ARISTOTLE’S FALLACY
According to Aristotelian law an external force is required to keep
a body in motion. However an external force is required to
overcome the frictional forces in case of solids and viscous forces
in fluids which are always present in nature.
LINEAR MOMENTUM (p)
Linear momentum of a body is the quantity of motion contained
in the body. Momentum
p mv=
rr
It is a vector quantity having the same direction as the direction
of the velocity. Its SI unit is kg ms
–1
.
NEWTON’S LAWS OF MOTION
First law : A body continues to be in a state of rest or of uniform
motion, unless it is acted upon by some external force to change
its state.
Newton’s first law gives the qualitative definition of force according
to which force is that external cause which tends to change or
actually changes the state of rest or motion of a body.
Newton’s first law of motion is the same as law of inertia given by
Galileo.
Inertia is the inherent property of all bodies because of which
they cannot change their state of rest or of uniform motion unless
acted upon by an external force.
Second law : The rate of change of momentum of a body is directly
proportional to the external force applied on it and the change
takes place in the direction of force applied.
i.e.,
===
dp mdv
F ma
dt dt
rr
r r
This is the equation of motion of constant mass system. For
variable mass system such as rocket propulsion
()
=
d mv
F
dt
r
r
And,
()
=
m dv dm
F v
dt dt
+
r
r r
The SI unit of f orce is newton. (One newton force is that much
force which produces an acceleration of 1ms
–2
in a body of mass
1 kg.The CGS unit of force is dyne. (1N = 10
5
dyne)
The gravitational unit of force is kg-wt (kg-f) or g-wt (g-f)
1 kg-wt (kg-f) = 9.8 N, 1 g-wt (g-f) = 980dyne
Third law : To every action there is an equal and opposite
reaction. For example – walking , swimming , a horse pulling a
cart etc.
= –
AB BA
FF
rr
Action and reaction act on different bodies and hence cannot
balance each other. Action and reaction occur simultaneously.
Forces always occur in pairs.
EQUILIBRIUM OF A PARTICLE
A body is said to be in equilibrium when no net force acts on the
body.
i.e., = 0FS
r
Then
xy
F0,F0S= S= and z
F0S=
Stable equilibrium : If a body is slightly displaced from equilbrium
position, it has the tendency to regain its original position, it issaid to be in stable equilibrium.
In this case, P.E. is minimum. 2
2
du
ve
dr
æö
=+ç÷
ç÷
èø
So, the centre of gravity is lowest.
Unstable equilibrium : If a body, after being displaced from the
equilibrium position, moves in the direction of displacement, it is
said to be in unstable equilibrium.
In this case, P.E. is maximum.
2
2
du
ve
dr
æö
=-ç÷
ç÷
èø
So, the centre of gravity is highest.
Neutral equilibrium : If a body, after being slightly displaced
from the equilibrium position has no tendency to come back or to
move in the direction of displacement the equilibrium is known to
be neutral.
In this case, P.E. is constant
2
2
du
constant
dr
æö
=ç÷
ç÷
èø
The centre of gravity remains at constant height.
COMMON FORCES IN MECHANICS
1. Weight : It is the force with which the earth attracts a body
and is called force of gravity, For a body of mass m, where
acceleration due to gravity is g, the weight
W = mg
5
Laws of Motion

104. 100 PHYSICS
2. Tension :
The force exerted by the ends of a loaded/stretched
string (or chain) is called tension. The tension has a sense
of pull at its ends.
Case 1
T
T
Case 2
2T
2T
T
T
m
1
m
2
m
1
gm
2
g
T
T
Massless
pulley
Case 3
T
TTT'T'
T
m
a
T
T
1
T – T = ma
1
If m = 0, T = T
1
i.e tension is same
The tension in a string remains the same throughout the string if
(a) string is massless,
(b) pulley is massless or pulley is frictionless
Case 4 : String having mass
Let the total mass of the string be M and length be L. Then mass
per unit length is
L
M
Let x be the distan
ce of the string from the mass m. Then the mass
of the shaded portion of string is
÷
ø
ö
ç
è
æ
´x
L
M
If the string is at res
t then the tension T has to balance the wt of
shaded portion of string and weight of mass m.gx
L
M
mT ÷
ø
ö
ç
è
æ
+=\
Þ as x increase
s, the tension increases. Thus tension is non-
uniform in a string having mass.
3. Normal force : It measures how strongly one body presses
the other body in contact. It acts normal to the surface of
contact.
Case 1
N
N = mg
mg
Case 2
N
a
m
N – mg = ma
Þ N = m(g + a) mg
Case 3
q
q
N = mg cos q
N
mg cos q
mg
mg sinq
4. Sp
ring force : If an object is connected by spring and spring
is stretched or compressed by a distance x, then restoring
force on the object F = – kx
where k is a spring contact on force constant.
5. Frictional force : It is a force which opposes relative motion
between the surfaces in contact. f = mN
This will be discussed in detail in later section.
6. Pseudo force : If a body of mass m is placed in a non-inertial
frame having aceleration
a
r
, then it experiences a Pseudo
force acting in a direction opposite to the direction of a
r
.
pseudo
F – ma=
r r
Negative sign shows that the pseudo force is always directed
in a direction opposite to the direction of the acceleration of
the frame.
x
y
z
a
Fpseudo
m
CONSTRAI
NT MOTION :
When the motion of one body is dependent on the other body, therelationship of displacements, velocities and accelerations of
the two bodies are called constraint relationships.
Case 1 Pulley string system :
X
x
F
Block
Step 1 : Find the distance of the two bodies from fixed points.
Step 2 : The length of the string remain constant. (We use of
this condition)
Therefore X + (X – x) = constant Þ 2X – x = constant
dt
dx
dt
dX
20
dt
dx
–
dt
dX
2 =Þ=Þ
pBp
dX
2V v V velocity o f pulley
dt
é
Þ = ==
ê
ë
Q

105. 101Laws of Motion
B
dx
v velocity of block
dt
ù
==
ú
û
Again differentiating we get, 2a
p
= a
B
B
pB
dvdVp
a anda
dt dt
éù
==
êú
ëû
a
p
= acceleration of pulley, a
B
= acceleration of block
Case 2 Here =++yxh
22
constt. On differentiating w.r.t ‘t’
F
x
y
h
q
1 2
[Negative sign with dy/dt shows that with increase in time, y
decreases]
22
1 2x dx dy
0
dt dt
2hx
´
-=
+
Þ cos q (v
1
– v
2
) = 0
22
x
cos
hx
éù
q=êú
êú +ëû
Q
Case 3 Wedge block system : Thin lines represents the condition
of wedge block at t = 0 and dotted lines at t = t
ax
c
Ax
Ax
A
q
q
B
ay
ax
ay
A
x
= acceleration of wedge towards left
a
x
, a
y
= acceleration of block as shown
From D ABC ,
y
xx
a
tan
aA
q=
+
Frame of Reference :
Reference frames are co-ordinate systems in which an event is
described.
There are two types of reference frames
(a) Inertial frame of reference: These are frames of reference
in which Newton’s laws hold good. These frames are at rest
with each other or which are moving with uniform speed
with respect to each other.
All reference frames present on surface of Earth are
supposed to be inertial frame of reference.
(b) Non – inertial frame of reference: Newton’s law do not
hold good in non-inertial reference frame.
All accelerated and rotatory reference frames are non –
inertial frame of reference. Earth is a non-intertial frame.
When the observer is in non-inertial reference frame a
pseudo force is applied on the body under observation.
Free Body Diagram (FBD) :
Free body diagram of a mass is a separate diagram of that mass.
All forces acting on the mass are sketched. A FBD is drawn to
visualise the direct forces acting on a body.
Case 1 : Masses M
1
and M
2
are tied to a string, which goes over
a frictionless pulley
(a)If M
2
> M
1
and they move with acceleration a
1M
M
2
Mg
2
Mg
1
T
T
a
a
FBD of M
1
, FBD of M
2
T
M1
1M g
a
T
M
2
Mg
2
a
11
T Mg Ma-=
22
Mg T Ma-=
where T is the tension in the string. It gives
21
12
MM
ag
MM
-
=
+
and
12
12
2MM
Tg
MM
=
+
(b)If the pulley begins to move with acceleration f,
downwards

21
12
()
MM
a gf
MM
-
=-
+
u ur u ur u ur
and
12
12
2
()
MM
T gf
MM
=-
+
u r u ur u ur
Case 2 : Three masses M
1
, M
2
and M
3
are connected with strings
as shown in the figure and lie on a frictionless surface. They are
pulled with a force F attached to M
1
.
M
3 M
2
M
1
F
T
1
T
1
T
2
T
2
The forces on M
2
and M
3
are as follows
23
1
123
MM
TF
MMM
+
=
++
and
3
2
123
M
TF
MMM
=
++
;
Acceleration of the system is
123
F
a
MMM
=
++
Case 3 : Two blocks of masses M
1
and M
2
are suspended
vertically from a rigid support with the help of strings as shown in the figure. The mass M
2
is pulled down with a force F.

106. 102 PHYSICS
M
1
T
1
T
1
T
2
T
2
M
2
M
2g
M
1g
F
The tension between the masses M
1
and M
2
will be
T
2
= F + M
2
g
Tension between the support and the mass M
1
will be
T
1
= F + (M
1
+ M
2
)g
Case 4 : Two masses M
1
and M
2
are attached to a string which
passes over a pulley attached to the edge of a horizontal table.
The mass M
1
lies on the frictionless surface of the table.
1M
2
M
T
a
T
Mg2
Let the tensio
n in the string be T and the acceleration of the
system be a. Then
T = M
1
a ...(1)
M
2
g – T = M
2
a ...(2)
Adding eqns. (1) and (2), we get 2
12
M
ag
MM
éù
=êú
+
ëû
and
12
12
MM
Tg
MM
éù
=
êú
+
ëû
Case 5 : Two mass es M
1
and M
2
are attached to the ends of a
string, which passes over a frictionless pulley at the top of theinclined plane of inclination q. Let the tension in the string be T.
q
M
1
Mg
1
Mg cos
1
qM g sin
1
qq
N
M
2
M
2
g
(i)When the mass M
1
moves u pwards with acceleration a.
From the FBD of M
1
and M
2
,
T – M
1
g sin q = M
1
a ...(1)
M
2
g – T = M
2
a ...(2)
Solving eqns. (1) and (2) we get,
21
12
sinMM
ag
MM
éù-q
=
êú
+
ëû
FBD of mass M
1
R=N
T
M g cos
1 q
M g sin
1 qMg
1
x
y
21
12
(1+sin)
MM g
T
MM
éù
=
êú
+q
ëû
T
Mg
2
a
FBD of M
2
(ii)When the mass M
1
mov
es downwards with
acceleration a.
Equation of motion for M
1
and M
2
,
M
1
g sin q – T = M
1
a ...(1)
T – M
2
g = M
2
a ...(2)
Solving eqns. (1) and (2) we get,
12
12
sin
;
MM
ag
MM
éù q-
=
êú
+
ëû
21
12
(1sin)
MM g
T
MM
éù
=
êú
+ +q
ëû
(a) If (M
2
/M
1
= sinq) then th
e system does not accelerate.
(b) Changing position of masses, does not affect the
tension. Also, the acceleration of the system remains
unchanged.
(c) If M
1
= M
2
= M (say), then2
cossin;
2 22
g
a
qqæ öæö
=-
ç ÷ç÷
è øèø
2
cos sin
222
Mg
T
qqæ öæö
=+
ç ÷ç÷
è øèø
Case 6 : Two masses M
1
and M
2
are attached to the ends of a
string over a pulley attached to the top of a double inclinedplane of angle of inclination a and b.
Let M
2
move downwards with acceleration a and the tension in
the string be T then
a b
2
M
1
M
FBD of M
1 a
T
Mg
1
M gcos
1a
a
M
gsin
1a
M
1
Equation of motion for M
1
T – M
1
g sin a = M
1
a
or T = M
1
g sin a + M
1
a ...(1)

107. 103Laws of Motion
FBD
of M
2 aT
Mg
2
M gcos
2b
Mgs
i
n
2
b
M
2
b
Equat
ion of motion for M
2
M
2
g sinb – T = M
2
a
or T = M
2
g sin b – M
2
a ...(2)
Using eqn. (1) and (2) we get,
M
1
g sin a + M
1
a = M
2
g sin b – M
2
a
Solving we get, ( )21
12
sin sinM Mg
a
MM
b-a
=
+
and
12
12
[sin sin]
MMg
T
MM
= b+a
+
Case 7 : A pe
rson/monkey climbing a rope T
Mg
a
a
(a) A per
son of mass M climbs up a rope with acceleration a.
The tension in the rope will be M(g+a).
T – Mg = Ma Þ T = M(g + a)
(b) If the person climbs down along the rope with acceleration
a, the tension in the rope will be M(g–a).
a
a
Mg
T
Mg – T = Ma Þ T = M(g – a)
(
c) When the person climbs up or down with uniform speed,
tension in the string will be Mg.
Case 8 : A body starting from rest moves along a smooth inclined
plane of length l, height h and having angle of inclination q.
h
l
q
FBD of body

N=R
mg
mg sinq mg cosq
q
(where N=R is normal reaction applied by plane on the body
of mass m)
For downward motion, along the inclined plane,
q=Þ=qsingamasinmg
By work-energy theorem loss in P.E. = gain in K.E.
gh2vmv
2
1
mgh
2
=Þ=Þ
Also, from t
he figure, h = l sin q.
q==\sing2gh2vl
(a) Acceler
ation down the plane is g sin q.
(b) Its velocity at the bottom of the inclined plane will be
2 2 singhg=ql
(c) Time taken to reach the bottom will be
1/21/2
2 1/2
2 2 1 12
sin sinsin
sin
2
hh
t
gg g g
h
æöæö
==== ç÷ç÷
ç÷qq qèø æöèø
q
ç÷
èø
l
(d) If
angles of inclination are q
1
and q
2
for two inclined planes
Keeping the length constant then
½
12
21
sin
sin
t
t
æöq
=ç÷
qèø
Case 9 : Weight of a man in a lift :
(i)When lift is accelerated upward : In this case the man also
moves in upward direction with an acceleration a
r
.
N
a a
mg
Then from Newton’ second law
N – mg = ma or N = m(g + a)
or W
app
= m(g + a) (1 /)
o
W ag=+ (as W = mg)
Where W
app
i
s apparent weight of the man in the lift, W
o
is
the real weight, N is the reaction of lift on the man. It is clear
that N = W
app
When the lift moves upward and if we measure the weight
of the man by any means (such as spring balance) then we
observe more weight (i.e., W
app
) than the real weight (W
o
)
W
app
>W
o
(ii)When lift is accelerated downward : In this case from
Newton’s second law
N
a
mg
mg – N = ma
orN = m(g – a) = W
o
(1– a/g)
or W'
app
= W
o
(1– a/g){ }mgW
o
=Q
If we measure the weight of man by spring balance, we
observe deficiency because W
app
< W
o
.

108. 104 PHYSICS
(iii)When lift is a
t rest or moving with constant velocity : From
Newton’s second law N –mg = 0 or N = mg
In this case spring balance gives the true weight of the man.
Case 10 : Three masses M
1
, M
2
and M
3
are placed on a smooth
surface in contact with each other as shown in the figure.
A force F pushes them as shown in the figure and the three
masses move with acceleration a,
M
3
M
2
F
2
F
1
F
2 F
1
M
1
F
a
M
1
FF
1
Þ F – F
1
= m
1
a ...(i)
M
2
F
2
F
1
Þ F
1
– F
2
= m
2
a ...(ii)
F
2
M
3
Þ F
2
= M
3
a ...(iii)
Adding eqns. (i)
, (ii) and (iii) we get,
123
F
a
MMM
=
++
Þ
3
2
123
MF
F
MMM
=
++
and
23
1
123
()M MF
F
MMM
+
=
++
Keep in Memory
1. When a
man jumps with load on his head, the apparent
weight of the load and the man is zero.
2. (i) If a person sitting in a train moving with uniform
velocity throws a coin vertically up, then coin will fall
back in his hand.
(ii) If the train is uniformly accelerated, the coin will fall
behind him.
(iii) If the train is retarded uniformly, then the coin will fall
in front of him.
Example 1.
A chain of length
l is placed on a smooth spherical surface
of radius R with one of its ends fixed at the top of the sphere. What will be the acceleration a of each element of
the chain when its upper end is released? It is assumed
that the length of chain
R
π
2
æö
<
ç÷
èø
l .
Solution :
Let m be
the mass of the chain of length l. Consider an
element of length dl of the chain at an angle q with vertical,
R
q
dl
dq
From figure
, dl = R d q ;
Mass of the element,
dm =
l
l
d
m
; or dm = qdR.
m
l
Force responsible for acceleration, dF = (dm)g sinq ;
dF = qq=q÷
ø
ö
ç
è
æ
q dsin
mgR
)sing(dR
m
ll
Net force on the chain can be obtained by integrating the
above relation between 0 to a, we have
ò
a
a
a-=q-=qq=
0
0
]cos1[
Rmg
)cos(
Rmg
dsin
Rmg
F
lll
ú
û
ù
ê
ë
é
-=
R
cos1
Rmg l
l
;
\ Acceleration,
F gR
a 1 cos
mR
æö
==-
ç÷
èø
l
l
.
Example 2.
A block slides d
own a smooth inclined plane to the ground
when released at the top, in time t second. Another block
is dropped vertically from the same point, in the absence
of the inclined plane and reaches the ground in t/2 second.
Then find the angle of inclination of the plane with the
vertical.
Solution :
If q is the angle which the inclined plane makes with the
vertical direction, then the acceleration of the block sliding
down the plane of length l will be g cosq.
A
C B
q
l h
Using the formula,
2
at
2
1
uts+= , we have s = l, u = 0, t
= t
and a = g cos q .
so
22
t)cosg(
2
1
tcosg
2
1
t0 q=q+´=l ...(i)
Taking vertical down
ward motion of the block, we get
4/gt
2
1
)2/t(g
2
1
0h
22
=+= ...(ii)
Dividing eq
n
. (ii
) by (i), we get
q
=
cos4
1h
l
]/hcos[ lQ =q
or
q
=q
cos4
1
cos ; or
4
1
cos
2
=q ; or
2
1
cos=q
or q = 60º

109. 105Laws of Motion
Exam
ple 3.
A large mass M and a small mass
m hang at the two ends of a string
that passes through a smooth
tube as shown in fig. The mass m
moves around a circular path in
a horizontal plane. The length of
the string from mass m to the top
of the tube is l, and
q is the angle
the string makes with the
vertical. What should be the
frequency (
n) of rotation of mass
m so that mass
M remains stationary?
M
m
l
r
q
T
Solution :
Ten
sion in the string T = Mg.
Centripetal force on the body = mrw
2
=mr ( 2p
n)
2
. This is
provided by
the component of tension acting horizontally
i.e. T sinq ( = Mg sinq).
\ mr ( 2pn)
2
= Mg sinq = Mgr/l. or
1 Mg
2m
n=
pl
Example 4.
A string
of negligible mass going over a clamped pulley of
mass m supports a block of mass M as shown in fig. The
force on the pulley by the clamp is given by
(a)
2Mg
(b)2 mg
M
m
(c)
22
[ (M m) m]++ g
(d)
22
[(M m) M]++ g
Solution : (c)
Forc
e on the pulley by the clamp = resultant of
T = (M + m)g and mg acting along horizontal and vertical
respectively
\ F
22
[(M m)g] (mg)=++
22
[ (M m) m ]g= ++
Example 5.
The
masses of 10 kg and 20 kg respectively are connected
by a massless spring in fig. A force of 200 newton acts on the
20 kg mass. At the instant shown, the 10 kg mass has
acceleration 12 m/sec
2
. What is the acceleration of 20 kg
mass? 10 kg
20 kg
200 newton
Solution :
For
ce on 10 kg mass = 10 × 12 = 120 N
The mass of 10 kg will pull the mass of 20 kg in the backward
direction with a force of 120 N.
\ Net force on mass 20 kg = 200 – 120 = 80 N
Its acceleration
force
a
mass
=
2
s/m4
kg20
N80
==
Example 6.
Tw
o masses each equal to m are lying on X-axis at (–a, 0)
and (+ a, 0) respectively as shown in fig. They are connectedby a light string. A force F is applied at the origin and along
the Y-axis. As a result, the masses move towards each other.
What is the acceleration of each mass? Assume the
instantaneous position of the masses as (– x, 0) and (x, 0)
respectively
m O m
(–a, 0) (a, 0)
–X X
F
Solution :
O(–x, 0) (x, 0)
F
T T
q
B C
A
From fi
gure F = 2 T cos q or T = F/(2 cos q)
The force responsible for motion of masses on X-axis is T
sin q
q´
q
=q=\ sin
cos2
F
sinTam

)xa(
x
2
F
OA
OB
2
F
tan
2
F
22
-
´=´=q=
so,
)xa(
x
m2
F
a
22
-
´=
Example 7.
A block
of mass M is pulled along horizontal frictionless
surface by a rope of mass m. Force P is applied at one end
of rope. Find the force which the rope exerts on the block.
Solution :
The situation is shown in fig
M
TO
m
P
Let a be the common acceleration of the system. Here
T = M a for block
P – T = m a for rope
\ P – M a = m a or P = a (M + m) or
)mM(
P
a
+
=
\
)mM(
PM
T
+
=

110. 106 PHYSICS
Example 8.
In the system show
n below, friction and mass of the pulley
are negligible. Find the acceleration of m
2
if
m
1
= 300 g, m
2
= 500 g and F = 1.50 N
Solution :
When the pu
lley moves a distance d, m
1
will move a distance
2d. Hence m
1
will have twice as large an acceleration as m
2
has.
For mass m
1
, T
1
= m
1
(2a) ...(1)
For mass m
2
, F – T
2
= m
2
(a) ...(2)
Putting
2
1
T
T
2
=in eq
n
. (1) gives T
2
= 4 m
1
a
Substituting value of T
2
in e quation (2),
F = 4m
1
a + m
2
a = (4m
1
+ m
2
)a
Hence
2
12
F 1.50
a 0.88m/s
4m m 4(0.3) 0.5
===
++
LAW OF CO
NSERVATION OF LINEAR MOMENTUM
A system is said to be isolated, when no external force acts on it.
For such isolated system, the linear momentum
()P mv=
rr
is
constant i.e., conserved.
The line
ar momentum is defined as
vmP
rr
= .....(1)
where v
r
is the velocity of the body, whose mass is m. The direction
of P
r
is same as the direction of the velocity of the body. It is a
vector quantity. From Newton’s second law,
P
dt
d
)vm(
dt
d
F
.ext
rrr
== .....(2)
i.e., time ra
te of change in momentum of the body is equal to total
external force applied on the body.
If
0)P(
dt
d
0F
.ext
=Þ=
rr
or P
r
= constant .....(3)
Th
is is called law of conservation of momentum.
Now let us consider a rigid body consisting of a large number of
particles moving with different velocities, then total linear
momentum of the rigid body is equal to the summation of individual
linear momentum of all particles
i.e.,
n
i123n
i1
p p p p ..........p
=
å=+++
rrrrr
or 123
1
..........
n
total i n
i
Pppppp
=
=å=++++
r rrrrr
where
n21
p...............p,p
rrr
are individual linear momentum of first,
second and n
th
particle respectively.
If this rigid body is isolated i.e., no external force is applied on it,
then =
totalP
r
constant (from Newton’s second law).
Further we know that internal forces (such as intermolecular forces
etc.) also act inside the body, but these can only change individual
linear momentum of the particles (i.e., p
1
,

p
2
.........), but their total
momentum
totalP
r
remains constant.
Gu
n Firing a Bullet
If a gun of mass M fires a bullet of mass m with velocity v. Thenfrom law of conservation of momentum, as initially bullet & gunare at rest position i.e., initial momentum is zero, so final momentum(gun + bullet) must also be zero.
Since on firing, the bullet moves with velocity
bv
r
in forward
direc
tion, then from Newton’s third law, the gun moves in backward
direction g
v
r
. So,
Initial momentum = final
momentum
bg
Momentum
Momentum
of bulletof gun
0 mv MV=+
rr

b
g
mv
V
M
-
\=
u u ur
uuur
(–ve sign shows th
at the vel. of gun will have the opposite
direction to that of bullet)
IMPULSE
According to Newton’s second law the rate of change of
momentum of a particle is equal to the total external force applied
on it (particle) i.e.,
extF
dt
Pdr
r
= ...(i)
or dt.FPd
ext
rr
= or
f
i
t
f i ext
t
P P P F .dtD= -= ò
rrrr
...(ii)
Where
iP
r
is momentum of the particle at initial time t
i
and when
we apply some external force
extF
r
its final momentum is
fP
r
at
time t
f
. The quantity
ext
F dt×
r
on R.H.S in equation (ii) is called the
impulse.We can write equation (ii) as
f
i
t
ext
t
I F.dtP= =Dò
rr
...(iii)
So, the impulse
of the force
extF
r
is equal to the change in
mom
entum of the particle. It is known as impulse momentum
theorem.
Area=
impulse t
t
f
t
i
F
ext.
(a)

111. 107Laws of Motion
Example 10
.
A hammer of mass M strikes a nail of mass m with velocity
of u m/s and drives it ‘s’ meters in to fixed block of wood.
Find the average resistance of wood to the penetration of
nail.
Solution :
Applying the law of conservation of momentum,
m u = (M + m) v
0
Þ
u
Mm
M
v
0 ÷
ø
ö
ç
è
æ
+
=
There accel
eration a can be obtained using the formula
(v
2
= u
2
+ 2as).
Here we have 0 – v
0
2
= 2as or a = v
0
2
/2s
s2
u
Mm
M
a
22
÷
ø
ö
ç
è
æ
+
=\
Resist
ance = (M + m) a
22
Mu
m M 2s
æö
=ç÷
ç÷+
èø
Example 11.
A ba
ll of mass 0.5 kg is thrown towards a wall so that it
strikes the wall normally with a speed of 10 ms
–1
. If the ball
bounces at right angles away from the wall with a speed of
8ms
–1
, what impulse does the wall exert on the ball ?
Solution :
f
10
8
Approaching wall
u = –10 ms
–1
Leaving wall
v = +8 ms
–1
Taking the direction of the impulse J as positive and using
J = mv – mu
we have
11
J 8 (10 )9
22
=´--= N-s
Therefo
re the wall exerts an impulse of 9 N-s on the ball.
Example 12.
Two particles, each of mass m, collide head on when theirspeeds are 2u and u. If they stick together on impact, findtheir combined speed in terms of u.
Solution :
2u
Before impa
ct
After impact
m
u
v
2m
m
Using conservation of linear momentum (in the direction of
the velocity 2u) we have
(m) (2u) – mu = 2m × V Þ
1
Vu
2
=
The comb
ined mass will travel at speed u/2.
(Note that the momentum of the second particle before impactis negative because its sense is opposite to that specifiedas positive.)
Force vary with time and impulse is area under force versus
time curve
Area=F
ext.
Dt
t
t
f
t
i
F
ext.
Fav
F
ext.
(b)
.
Force constant w
ith time i.e.,
.extF
r
constant with time (shown
by horizontal line) and it would give same impulse to particlein time Dt = t
f
–t
i
as time varying force described.
It is a vector quantity having a magnitude equal to the area underthe force-time curve as shown in fig. (a). In this figure, it is assumedthat force varies with time and is non-zero in time interval Dt = t
f
–
t
i
. Fig.(b) shows the time averaged force
.extF
r
i.e., it is constant
in time interval Dt, then equation (iii) can be written as
f
i
t
ext.
t
I F dt=ò
r
)tt(F
if.ext-=
r
tFI
.extD=
r
...(iv)
The directio
n of impulsive vector I is same as the direction of
change in momentum. Impulse I has same dimensions as that ofmomentum i.e, [MLT
–1
]
Rocket propulsion (A case of system of variable mass ) : It is
based on principle of conservation of linear momentum.
In rocket, the fuel burns and produces gases at high temperature.These gases are ejected out of the rocket from nozzle at the
backside of rocket and the ejecting gas exerts a forward force on
the rocket which accelerates it.
Let the gas ejects at a rate
dt
dM
r-= and at co
nstant velocity u
w.r.t. rocket then from the conservation of linear momentum
rtM
ru
M
ru
dt
dv
0-
== where M
= M
0
- rt and M
0
is mass of rocket
with fuel and solving this equation, we get
÷
÷
ø
ö
ç
ç
è
æ
-
=
rtM
M
loguv
0
0
e
whe
re v = velocity of rocket w.r.t. ground.
Example 9.
Two skaters A and B approach each other at right angles.
Skater A has a mass 30 kg and velocity 1 m/s and skater
B has a mass 20 kg and velocity 2 m/s. They meet and
cling together. Find the final velocity of the couple.
Solution :
Applying principle of conservation of linear momentum,
2
22
2
1121
2
2
2
1
)vm()vm(v)mm(;ppp+=++=
( ) ( )( )
22
30 20 v 30 1 20 2 50+ = ´+´=
50
v 1 m/s
50
==

112. 108 PHYSICS
FRICTION
When a body i
s in motion on a rough surface, or when an object
moves through water (i.e., viscous medium), then velocity of the
body decreases constantly even if no external force is applied on
the body. This is due to friction.
So “an opposing force which comes into existence, when two
surfaces are in contact with each other and try to move relative
to one another, is called friction”.
Frictional force acts along the common surface between the two
bodies in such a direction so as to oppose the relative movement
of the two bodies.
(a)The force of static friction f
s
between book and rough
surface is opposite to the applied external force F
ext.
The
force of static friction f
s
=
ext
F
r
.
Book
R=N
W
(a)
f
s
F
ext.
(b) When
ext
F
r
. exceeds the certain maximum value of static
friction, the book starts accelerating and during motion
Kinetic frictional force is present.
Book
R=N
W
(b)
f
k
F
ext.
Body just starts moving
(c)A graph
ext
F
r
. versus | f | shown in f igure. It is clear that
f
s, ,max
> f
k
=mN
s
f
k
=mN
k
Body starts with
acceleration
Body is
at rest
static
region
kinetic
region
O
|f|
(c)
(f)
s max
Fig.(
a) shows a book on a horizontal rough surface. Now if
we apply external force
.extF
r
, on the book, then the book
will remain stationary if
ext.
F
r
is not too large. If we increase
.extF
r
then frictional force f also increase up to
s max
(f)
(called maximum force
of static friction or limiting friction)
and
s max
(f)= m
s
N. At any in stant when
.extF
r
is slightly
greater tha
n
s max
(f)then the book moves and accelerates to
the right.
Fig.(b) when the book is in motion, the retarding frictional
force become less than,
s max
(f)
Fig.(c)
s max
(f)is equal to m
k
N. When th e book is in motion,
we call the retarding frictional force as the force of kinetic
friction f
k
.
Since f
k
<
s max
(f), so it is clear that, we require more force to
start motion than to maintain it against friction.
By experiment one can find that
s max
(f)and f
k
are
proporti
onal to normal force N acting on the book (by rough
surface) and depends on the roughness of the two surfaces
in contact.
Note :
(i) The force of static friction between any two surfaces
in contact is opposite to
.extF
r
and given by Nf
ss
m£
and
smaxs
(f)N=m (when the body just moves in the
right direction).where N = W = weight of book and m
s
is called
coefficient of static friction, f
s
is called force of static
friction and
s max
(f) is called limiti ng friction or
maximum value of static friction.
(ii) The force of kinetic friction is opposite to the direction
of motion and is given by f
k
= m
k
N
where m
k
is coefficient of kinetic friction.
(iii) The value of m
k
and m
s
depends on the nature of
surfaces and m
k
is always less then m
s
.
Friction on an inclined plane : Now we consider a book on an
inclined plane & it just moves or slips, then by definition
R=N
mg cos q
mg=W
qm
gsinq
q
B
ook
(fs
max)
max
() =m
ss
fR
Now from figure, q=sinmgf
max,s and R = mg cosq
Þm
s
= tanq or q = tan
–1
(m
s
)
where a
ngle q is called the angle of friction or angle of repose
Some facts about friction :
(1) The force of kinetic friction is less than the force of static
friction and the force of rolling friction is less than force of
kinetic friction i.e.,
f
r
< f
k
< f
s
or m
rolling
< m
kinetic
< m
static
hence it is easy to roll the drum in comparison to sliding it.
(2) Frictional force does not oppose the motion in all cases,
infact in some cases the body moves due to it.
A
B
F
ext
In the figure, book B moves to the right due to friction
between A and B. If book A is totally smooth (i.e., frictionless)then book B does not move to the right. This is because ofno force applies on the book B in the right direction.

113. 109Laws of Motion
Laws of
limiting friction :
(i) The force of friction is independent of area of surfaces
in contact and relative velocity between them (if it is
not too high).
(ii) The force of friction depends on the nature of material
of surfaces in contact (i.e., force of adhesion).
m depends upon nature of the surface. It is
independent of the normal reaction.
(iii) The force of friction is directly proportional to normal
reaction i.e., F µ N or F = mn.
While solving a problem having friction involved, follow
the given methodology
Check
(a) F
app
(b) L
imiting
friction (f )
l
If F < f
app
app
l
Body does not move and
F = frictional force
If F = f
Body is on the verge of movement
if the body is initially at rest
Body moves with constant velocity
app
l
Rolling Friction :
The name rolling friction is a misnomer. Rolling friction has nothing
to do with rolling. Rolling friction occurs during rolling as well as
sliding operation.
Cause of rolling friction : When a body is kept on a surface of
another body it causes a depression (an exaggerated view shown
in the figure). When the body moves, it has to overcome the
depression. This is the cause of rolling friction.
Rolling friction will be zero only when both the bodies
incontact are rigid. Rolling friction is very small as compared to
sliding friction. Work done by rolling friction is zero
CONSERVATIVE AND NON-CONSERVATIVE FORCES
If work done on a particle is zero in complete round trip, the
force is said to be conservative. The gravitational force,
electrostatics force, elastic force etc., are conservative forces.
On the other hand if the work done on a body is not zero during
a complete round trip, the force is said to be non-conservative.
The frictional force, viscous force etc. are non-conservative
forces.
A
B
C
f
i
Initial
pos
ition
Final
position
Figure shows three processes A, B and C by which we can reach
from an initial position to final position. If force is conservative,
then work done is same in all the three processes i.e., independent
of the path followed between initial and final position.
If force is non conservative then work done from i to f is different
in all three paths A,B and C.
Hence it is clear that work done in conservative force depends
only on initial & final position irrespective of the path followed
between initial & final position. In case of non-conservative
forces the work done depends on the path followed between
initial and final position.
We can say also that there is no change in kinetic energy of the
body in complete round trip in case of conservative force. While
in case of non conservative forces, when a body return to its
initial position after completing the round trip, the kinetic energy
of the body may be more or less than the kinetic energy with
which it starts.
Example 13.
Pushing force making an angle
q to the horizontal is
applied on a block of weight W placed on a horizontal
table. If the angle of friction is
f, then determine the
magnitude of force required to move the body.
Solution :
The various forces acting on the block are shown in fig.
N
F
f
mg
f cosq
q
F sinq
Here,
f
tan
N
m= f= ; or f = N t anf...(i)
The condition for the block just to move is
Fcosq = f = N tanf ...(ii)
and F sinq + W = N ...(iii)
From (ii) and (iii),
F cosq = (W + F sinq ) tan f = W tanf + F sin q tanf ;
or F cos q – F sinq sinf/cosf = W sinf/cosf
or F (cosq cosf – sinq sinf) = W sinf ;
or F cos (q + f) = W sinfor F = W sinf / cos (q + f)
Example 14.
An object of weight W is resting on an inclined plane at an
angle
q to the horizontal. The coefficient of static friction
is
m. Find the horizontal force needed to just push the object
up the plane.

114. 110 PHYSICS
Solution :
The
situation is shown in fig.

F sinq
F co
sq
W cosq
W sinq
F
R
q
W
q
q
f=Rm
Let F be the horiz
ontal force needed to just push the object
up the plane. From figure R = W cos q + F sin q
Now f = mR = m [W cos q + F sin q]...(1)
Further, F cos q = W sin q + f ...(2)
F cos q = W sin q + m [W cos q + F sin q]
F cos q – m F sin q = W sin q + m W cos q
)sin(cos
)cos(sinW
F
qm-q
qm+q
=\
CASES OF CIRCULAR MOTIONS
Motion in a Vertical Circle :
Let us consider a particle of mass m attached to a string of length
R let the particle be rotated about its centre O.
At t = 0 the particle start with velocity u from the point A (lowest
point of vertical circle) and at time t its position is P. Then the
tension at point P is given by
T
P
v
P
q
q
O
B
R
Au
mg
mg
mg cos q
sin q
R
mv
cosmgT
2
P
P
=q- or
R
mv
cosmgT
2 P
P
+q= ...(1)
So te
nsion at point A (lowest point of vertical circle) is
R
mv
mgT
2 A
A
=- (Q q = 0º) ...(2)
and tension at poin
t B (highest point of vertical circle) is
R
mv
mgT
2 B
B
=+ (Q q =180º) ...(3)
Where
2
mv
r
is centripetal force req uired for the particle to move
in a vertical circle.
Now from law of conservation of energy
mgR2mv
2
1
mv
2
1 2
B
2
A =-
or,
22
AB
v v 4gR-= ...(4)
(change in kin
etic energy of particle)
= (change in potential energy of particle)
or
(loss in kinetic energy of the particle) = (gain in potential energy)
In conservative force system (such as gravity force) the
mechanical energy (i.e., kinetic energy + potential energy) must
be constant.
Total energy will be constant
Now from eq
ns
.(2) and (3), we get

mg
R
T
B
v
BB
A
D

mg
R
T
A
v = v
A C
B
A
D
)VV(
R
m
mg2TT
2
B
2
ABA -+=- )gR4(
R
m
mg2+=
Þ mg6TT
BA
=- ...(5)
orT
A
= T
B
+ 6mg ...(6
)
So it is clear from eq
n
. (6) that tension in string at lowest point
of vertical circle is greater then the tension at highest point of
vertical circle by 6mg.
Condition to complete a vertical circle :
If we reduce the velocity v
A
in equation (2), then T
A
will be reduce
and at some critical velocity v
c
, T
B
will be zero, then put T
B
= 0
and v
B
= v
C
in equation (3) and we obtain
CB
v v gR== ...(7)
In this condition the nece
ssary centripetal force at point B is
provided by the weight of the particle [see again equation (3)]
then from equation (4), we get2
45
AA
v gR gR v gR-= Þ= ...(8)
then the te
nsion at the point A will be
mg6
R
)gR5(m
mgT
A
=+= ...(9)
Hence if w
e rotate a particle in a vertical circle and tension in
string at highest point is zero, then the tension at lowest point of
vertical circle is 6 times of the weight of the particle.
Some Facts of Vertical Motion :
(i) The body will complete the vertical circle if its velocity at
lowest point is equal to or greater then
gR5
(ii) The bo
dy will oscillate about the lowest point if its velocity
at lowest point is less then
gR2. This will happen when
the velocity at the halfway mark, i.e.
ú
û
ù
ê
ë
é
== mgRmv
2
1
0v
2
AD
Q
(iii) Th
e string become slack and fails to describe the circle
when its velocity at lowest point lies betweengR5togR2

115. 111Laws of Motion
Example 15
.
A mass m is revolving in a vertical circle at the end of a
string of length 20 cm. By how much does the tension of the
string at the lowest point exceed the tension at the topmost
point?
Solution :
The tension T
1
at the topmost point is given by,
gm
20
vm
T
2
1
1 -=
Centrifugal f
orce acting outward while weight acting
downward
The tension T
2
at the lowest point,
gm
20
vm
T
2
2
2 +=
Centrifugal f
orce and weight (both) acting downward
gm2
20
vmvm
TT
2
1
2
2
12 +
-
=- ; hg2vv
2
2
2
1
-= or
g80)40(g2vv
2
1
2
2
==-
gm6gm2
20
gm80
TT
12 =+=-\
Example 16.
A stone of mass 1 kg tied to a light
inextensible string of
length L = (10/3) m is whirling in a circular path of radius
L in a vertical plane. If the ratio of the maximum to the minimum
tension in the string is 4 and g = 10 m/s
2
, then find the speed
of the stone at the highest point of the circle.
Solution :
O
T
P
V
P
V
O
Q
q
q
mg cos q
mg
L
The tensi
on T in the string is given by
ú
ú
û
ù
ê
ê
ë
é
+=
L
v
gmT
2
Q
max and
ú
ú
û
ù
ê
ê
ë
é
+-=
L
v
gmT
2
P
min
Accordin
g to the given problem
4
)L/v(g
)L/v(g
2
P
2
Q
=
+-
+
or
L
v
4g4
L
v
g
2
P
2
Q
+-=+
or
L
v
4g4
L
Lg4v
g
2
P
2
P
+-=
+
+
L = (10
/3) m and g = 10 m/s
2
(given)
Solving we get v
P
= 10 m/s.
Negotiating a Curve :
Case of cyclist
To safely negotiate a curve of radius r, a cyclist should bend at an
angle q with the vertical.
q
Nsinq
N Ncosq
Which is given by tan q =
rg
v
2
. Angle q is
also called as angle of
banking.
r
mv
sinN
2
=q an
d mgcosN
=q
Case of car on a levelled road
A vehicle can safely negotiate a curve of radius r on a rough levelroad when coefficient of sliding friction is related to the velocity
as 2
s
v
rg
m³ .
Now co
nsider a case when a vehicle is moving in a circle, the
centrifugal force is
r
mv
2
wherea
s m is mass of vehicle, r = radius
of circle and v is its velocity.
f
s
mv
r
2
The frictional force is static since wheels are in rolling motion
because point of contact with the surface is at rest
r
mv
f
2
s=\ mgff
smaxs
m=£
mg
r
mv
s
2
m£ or
rg
v
2
s
³m
Case of banking of road (frictionless)
A
vehicle can safely negotiate a curve of radius r on a smooth
(frictionless) road, when the angle q of banking of the road is
given by
2
tan
v
rg
q=.
N
mg
Horizontal
q
q
V
e
r
t
i
c
a
l

116. 112 PHYSICS
When th
e banked surface is smooth, the force acting will be gravity
and normal force only.N
q
mv
r
2
mg
Balancing forces
cosN mgq= ...(1)
2
sin
mv
N
r
q= ...(2)
2
tan
v
rg
=q ...(3)
Case of banking of road (with friction)
The m
aximum velocity with which a vehicle can safely negotiate
a curve of radius r on a rough inclined road is given by
v
2
=
( tan)
1 tan
rgm+q
-mq
; where m is the coefficient of friction of the
rough surface on which the vehicle is moving, and q is the angle
of inclined road with the horizontal.
Suppose a vehicle is moving in a circle of radius r on a rough
inclined road whose coefficient of friction is
μ
M
and angle of
banking is q.
f
s f
s
q
N
mv
r
2 mv
r
2
mg mg
N
Let velocity of object (vehicle) be V.
If we apply pseudo force on body, centrifugal force is
r
mv
2
when v is max. and fric
tion force will be acting down the slope.
Balancing the force horizontally,
q+q=sinNcosf
r
mv
s
2
...(1)
Balancing the
force vertically,
mgsinfcosN
s
+q=q ...(2)
when v = maximum, f = f
max
= f
s
= mN ...(3)
From eq
n
. (2),
mgsinNcosN +qm=q mg)sin(cosN =qm-qÞ
or
qm-q
=
sincos
mg
N
From eq
ns
.(1) and (3),
qm-q
q+qm
=
sincos
sinmgcosmg
r
mv
2
Þ
qm-
q+m
=
tan1
)tan(mg
r
mv
2

2
max
( tan)
1 tan
v rg
m+q
Þ=
-mq
Now in the case of minimum velocity with which body could move
in a circular motion, the direction of friction will be opposite to
that one in maximum velocity case.
f
s
q
N
mv
r
2
mg
and
2
min
tan
1 tan
v rg
æöm-q
=
ç÷
+mqèø
Keep in Memory
1.Whenever a
particle is moving on the circular path then
there must be some external force which will provide the
necessary centripetal acceleration to the particle.
For examples :
(i) Motion of satellite around a planet : Here the centripetal
force is provided by the gravitational force.
i.e.
r
mv
r
GMm
2
2
=
V
(M) Planet
Satellite
(
m)
(ii) Motion of electron around the nucleus : Here the
required centripetal force is provided by the
Coulombian force
i.e.
r
mv
r
)e)(ze(
4
1
2
2
o
=
pe

Nucleus
(Ze)
r
Electron
(e)
(iii) Mo
tion of a body in horizontal and vertical circle:
Here the centripetal force is provided by the tension.
Horizontal circle
r
mv
T
2
=
T
V
(m)

117. 113Laws of Motion
(d)Th
e vertical depth h of P below A is independent of the
length of the string since from eq
n
. (1) and (4)
h mg
T mgT
h
= Þ=
l
l
but
2
Tm=wl
Therefore
2
2
mgg
mh
h
w= Þ=
w
l
l
which is
independent of l.
Example 17.
A particle of mass m is moving in a circular path of constant
radius r such that its centripetal acceleration a
c
is varying
with time t as a
c
= k
2
rt
2
, where k is a constant. Determine
the power delivered to the particle by the forces acting on
it.
Solution :
Here tangential acceleration also exists which requires power.
Given that centripetal acceleration
a
c
= k
2
rt
2
also, a
c
= v
2
/r ;
\ v
2
/r = k
2
rt
2
or v
2
= k
2
r
2
t
2
or v = k r t ;
Tangential acceleration,
rk
dt
dv
a ==
Now, force
F = ma = m k r ;
So, power, P = F v = m k r × k r t = m k
2
r
2
t.
Example 18.
The string of a pendulum is horizontal. The mass of the bob
is m. Now the string is released. What is the tension in the
string in the lowest position?
Solution :
mg
v
T
O
Let v be the velocity of the bob at the lowest position. In
this position, The P.E. of bob is converted into K.E. Hence,
2
vm
2
1
gm=l or lg2v
2
= ...(1)
If T be
the tension in the string, then
÷
÷
ø
ö
ç
ç
è
æ
=-
l
2
vm
gmT
...(2)
From eq
ns
. (1
) and (2).
T – m g = 2 m g or T = 3 m g
Vertical circle
At point A,
r
mv
T
2
A
A
= ;
At point
B,
r
mv
mgT
2
B
B
=+
T
V
BT
A
B
mg
mg
B
V
C
T
C
V
mg
C
And at point C,
r
mv
mgT
2
C
C
=-
CONICAL PENDULUM
Con
sider an inextensible string of length l which is fixed at
one end, A. At the other end is attached a particle P of mass
m describing a circle with constant angular velocity w in a
horizontal plane.
Horizontal Plane
mg
A
P
r
O
h
P
Tsin
O
2r
Vertical section
As P rotates, the string AP traces out the surface of a cone.
Consequently the system is known as a conical pendulum.
Vertically, T cos mgq= ... (1)
Hori
zontally,
2
T sin mrq=w ... (2)
In trian
gle AOP,
r sin=ql ... (3)
and h cos=ql ... (4)
Seve
ral interesting facts can be deduced from these
equations :
(a)It is impossible for the string to be horizontal.
This is seen from eq
n
. (1) in which
mg
cos
T
q= cannot be
z
ero. Hence q cannot be 90°.
(b)The tension is always greater than mg.
This also follows from eq
n
. (1) as cos q < 1 (q is acute but
not zero). Hence, T > mg
(c)The tension can be calculated without knowing the
inclination of the string since, from eq
n
. (2) and (3)
2
T sin m sinq= qwl Þ
2
Tm=wl

118. 114 PHYSICS

119. 115Laws of Motion
1.A rec
tangular block is placed on a rough horizontal surface
in two different ways as shown, then
F
F
(a) (b)
(a) friction will be more in case (a)
(b) friction will be more in case (b)
(c) friction will be equal in both the cases
(d) friction depends on the relations among its dimensions.
2.Centripetal force :
(a) can change speed of the body.
(b) is always perpendicular to direction of motion
(c) is constant for uniform circular motion.
(d) all of these
3.When a horse pulls a cart, the horse moves down to
(a) horse on the cart.
(b) cart on the horse.
(c) horse on the earth.
(d) earth on the horse.
4.The force of action and reaction
(a) must be of same nature
(b) must be of different nature
(c) may be of different nature
(d) may not have equal magnitude
5.A body is moving with uniform velocity, then
(a) no force must be acting on the body.
(b) exactly two forces must be acting on the body
(c) body is not acted upon by a single force.
(d) the number of forces acting on the body must be even.
6.The direction of impulse is
(a) same as that of the net force
(b) opposite to that of the net force
(c) same as that of the final velocity
(d) same as that of the initial velocity
7.A monkey is climbing up a rope, then the tension in the rope
(a) must be equal to the force applied by the monkey on
the rope
(b) must be less than the force applied by the monkey on
the rope.
(c) must be greater than the force applied by the monkey
on the rope.
(d) may be equal to, less than or greater the force applied
by the monkey on the rope.
8.A uniform rope of length L resting on a frictionless horizontal
surface is pulled at one end by a force F. What is the tension
in the rope at a distance l from the end where the force is
applied.
(a)F (b) F (1 + l/L)
(c) F/2 (d) F (1 – l/L)
9.A particle of mass m is moving with velocity v
1
, it is given
an impulse such that the velocity becomes v
2
. Then
magnitude of impulse is equal to
(a) )vv(m
12
rr
- (b) )vv(m
21
rr
-
(c) )vv(
m
12
rr
-´ (d) )vv(m5.0
12
rr
-
10.A const
ant force F = m
2
g/2 is applied on the block of mass
m
1
as shown in fig. The string and the pulley are light and
the surface of the table is smooth. The acceleration of m
1
is
F
m
1
m
2
(a)
righttowards
)
mm(2
gm
21
2
+
(b)
lefttowards
)mm
(2
gm
21
2
-
(c)
righttowards
)m
m(2
gm
12
2
-
(d)
lefttowards
)m
m(2
gm
12
2
-
11.A mass is hanging on a spring balance which is kept in a lift.
The lift ascends. The spring balance will show in its readings
(a) an increase
(b) a decrease
(c) no change
(d) a change depending on its velocity
12.A cart of mass M has a block of mass m attached to it as
shown in fig. The coefficient of friction between the block
and the cart is m. What is the minimum acceleration of the
cart so that the block m does not fall?
(a)mg
(b) g/m
M
m
(c)m/g
(d) M mg/m

120. 116 PHYSICS
13.A particle of mas
s m moving eastward with a speed v collides
with another particle of the same mass moving northward
with the same speed v. The two particles coalesce on
collision. The new particle of mass 2m will move in the north-
external direction with a velocity :
(a) v/2 (b) 2v
(c)
2/v (d) None of these
14.A s
pring is compressed between two toy carts of mass m
1
and m
2
. When the toy carts are released, the springs exert
equal and opposite average forces for the same time on
each toy cart. If v
1
and v
2
are the velocities of the toy carts
and there is no friction between the toy carts and the ground,
then :
(a)v
1
/v
2
= m
1
/m
2
(b) v
1
/v
2
= m
2
/m
1
(c)v
1
/v
2
= –m
2
/m
1
(d) v
1
/v
2
= –m
1
/m
2
15.Two mass m and 2m are attached with each other by a rope
passing over a frictionless and massless pulley. If the pulley
is accelerated upwards with an acceleration ‘a’, what is the
value of T?
(a)
3
ag+
(b)
3
ag-
(c)
3
)ag(m4 +
(d)
3
)ag(m-
16.A rider on a horse back falls forward when the horse
suddenly stops. This is due to
(a) inertia of horse
(b) inertia of rider
(c) large weight of the horse
(d) losing of the balance
17.A ball of mass m is thrown vertically upwards. What is the
rate at which the momentum of the ball changes?
(a) Zero (b) mg
(c) Infinity (d) Data is not sufficient.
18.A small block is shot into each of the four tracks as shown
below. Each of the tracks rises to the same height. The
speed with which the block enters the track is the same in all
cases. At the highest point of the track, the normal reaction
is maximum in
(a)
v
(b)
v
(c)
v
(d)
v
19.A weight W res
ts on a rough horizontal plane. If the angle
of friction be
q, the least force that will move the body
along the plane will be
(a) qcosW (b) qcotW
(c) qtanW (d) qsinW
20.A particle starts sliding do
wn a frictionless inclined plane.
If
n
S is the distance traveled by it from time t = n – 1 sec to
t = n sec, the ratio S
n
/S
n+1
is
(a)
1n2
1n2
+
-
(b)
n2
1n2+
(c)
1n2
n2
+
(d)
1n2
1n2
-
+
21.A block is kept
on a inclined plane of inclination q of length
l.
The velocity of
particle at the bottom of inclined is (the
coefficient of friction is
m)
(a)
2/1
)]sincos(g2[q
-qm
l (b) )cos(sing2 qm-ql
(
c)
)cos(sing2 qm+ql (d) )sin(cosg2 qm+ql
22.A b
ird is in a wire cage which is hanging from a spring
balance . In the first case, the bird sits in the cage and in the
second case, the bird flies about inside the cage. The reading
in the spring balance is
(a) more in the first case
(b) less in first case
(c) unchanged
(d) zero in second case.
23.In an explosion, a body breaks up into two pieces of unequal
masses. In this
(a) both parts will have numerically equal momentum
(b) lighter part will have more momentum
(c) heavier part will have more momentum
(d) both parts will have equal kinetic energy
24.A block of mass m on a rough horizontal surface is acted
upon by two forces as shown in figure. For equilibrium of
block the coefficient of friction between block and surface is
m
q
F
1
F
2
(a)
q+
q+
cosFmg
sinFF
2
21
(b)
q-
+q
sinFmg
FcosF
2
21
(c)
q+
q+
sinFmg
cosFF
2
21
(d)
12
2
FsinF
mg F cos
q-
-q
25.A plate of mass M is placed on a horizontal of frictionless
surface (see figure), and a body of mass m is placed on thisplate. The coefficient of dynamic friction between this body
and the plate is
m . If a force 2m mg is applied to the body
of mass m along the horizontal, the acceleration of the plate
will be
M
m
2 mgm
(a) g
M
mm
(b) g
)mM(
m
+
m
(c) g
M
m2m
(d) g
)mM(
m2
+
m
.

121. 117Laws of Motion
1.An
object of mass 10 kg moves at a constant speed of
10 ms
–1
. A constant force, that acts for 4 sec on the object,
gives it a speed of 2 ms
–1
in opposite direction. The force
acting on the object is
(a) –3 N (b) –30 N
(c) 3 N (d) 30 N
2.A solid sphere of 2 kg is suspended from a horizontal beam
by two supporting wires as shown in fig. Tension in each
wire is approximately (g = 10 ms
–2
)
(a) 30 N
(b) 20 N
(c) 10 N
mg
30? 30?
T T
(d) 5 N
3.A toy g
un consists of a spring and a rubber dart of mass 16
g. When compressed by 4 cm and released, it projects the
dart to a height of 2 m. If compressed by 6 cm, the height
achieved is
(a) 3 m (b) 4 m
(c) 4.5 m (d) 6 m
4.A player stops a football weighting 0.5 kg which comes
flying towards him with a velocity of 10m/s. If the impact
lasts for 1/50th sec. and the ball bounces back with a velocity
of 15 m/s, then the average force involved is
(a) 250 N (b) 1250 N
(c) 500 N (d) 625 N
5.A car travelling at a speed of 30 km/h is brought to a halt in
4 m by applying brakes. If the same car is travelling at 60 km/h,
it can be brought to halt with the same braking power in
(a) 8 m (b) 16 m
(c) 24 m (d) 32 m
6.A body of mass 4 kg moving on a horizontal surface with an
initial velocity of 6 ms
–1
comes to rest after 3 seconds. If
one wants to keep the body moving on the same surface
with the velocity of 6 ms
–1
, the force required is
(a) Zero (b) 4 N
(c) 8 N (d) 16 N
7.A machine gun has a mass 5 kg. It fires 50 gram bullets at the
rate of 30 bullets per minute at a speed of 400 ms
–1
. What
force is required to keep the gun in position?
(a) 10 N (b) 5 N
(c) 15 N (d) 30 N
8.A force time graph for the motion of a body is shown in Fig.
Change in linear momentum between 0 and 8s is
2
0
1
2 4 678
x
t (s)
F (N)
(a) zero ( b) 4 N-s
(c) 8 Ns (d) None of these
9.Fig. shows a uniform rod of length 30 cm having a mass of
3.0 kg. The strings shown in the figure are pulled by constant
forces of 20 N and 32 N. All the surfaces are smooth and the
strings and pulleys are light. The force exerted by 20 cm part
of the rod on the 10 cm part is
10
cm
20 cm
20 N 32 N
(a) 20 N (b) 2 4 N
(c) 32 N (d) 52 N
10.A force of 10 N acts on a body of mass 20 kg for 10 seconds.Change in its momentum is
(a) 5 kg m/s (b) 100 kg m/s
(c) 200 kg m/s (d) 1000 kg m/s
11.Consider the system shown in fig. The pulley and the string
are light and all the surfaces are frictionless. The tension in
the string is (take g = 10 m/s
2
)
1 kg
1 kg
(a) 0 N (b) 1 N
(
c) 2 N (d) 5 N
12.The elevator shown in fig. is descending with an accelerationof 2 m/s
2
. The mass of the block A = 0.5 kg. The force exerted
by the block A on block B is
(a) 2 N
(b) 4 N
(c) 6 N
A
2 m/s
2
B
(d) 8 N

122. 118 PHYSICS
13.Two blo
cks of masses 2 kg and 1 kg are placed on a smooth
horizontal table in contact with each other. A horizontal force
of 3 newton is applied on the first so that the block moves
with a constant acceleration. The force between the blocks
would be
(a) 3 newton (b) 2 newton
(c) 1 newton (d) zero
14.A 4000 kg lift is accelerating upwards. The tension in the
supporting cable is 48000 N. If
2
sm10g
-
= then the
acceleratio
n of the lift is
(a) 1
2
sm
-
(b) 2
2
sm
-
(c) 4
2
sm
-
(d) 6
2
sm
-
15.A rocket has a mass of 100 kg. Ninety percent of this is fuel. It
ejects fuel vapors at the rate of 1 kg/sec with a velocity of 500
m/sec relative to the rocket. It is supposed that the rocket is
outside the gravitational field. The initial upthrust on the
rocket when it just starts moving upwards is
(a) zero (b) 500 newton
(c) 1000 newton (d) 2000 newton
16.A 0.1 kg block suspended from a massless string is moved
first vertically up with an acceleration of
2
ms5
-
and then
moved vertically down with an acceleration of
2
ms5
-
. If
1
T and
2
T are the respective tensions in the two cases,
then
(a)
12
TT>
(b)
21
TT-= 1 N, if
2
ms10g
-
=
(c)
12
T T 1kgf-=
(d)
12
T T 9.8N,-= if
2
g 9.8 ms
-
=
17.The coefficient
of friction between two surfaces is 0.2. The
angle of friction is
(a) )2.0(sin
1-
(b) )2.0(cos
1-
(c) )1.0(tan
1-
(d) )5(cot
1-
18.A man weighing 80 kg is standing on a trolley weighing 320
kg. The trolley is resting on frictionless horizontal rails. If
the man starts walking on the trolley along the rails at a
speed of one metre per second, then after 4 seconds, his
displacement relative to the ground will be :
(a) 5 metres (b) 4.8 metres
(c) 3.2 metres (d) 3.0 metres
19.Starting from rest, a body slides down a 45º inclined plane in
twice the time it takes to slide down the same distance in the
absence of friction. The coefficient of friction between the
body and the inclined plane is:
(a) 0.33 (b) 0.25
(c) 0.75 (d) 0.80
20.A ball of mass 0.5 kg moving with a velocity of 2 m/sec
strikes a wall normally and bounces back with the same
speed. If the time of contact between the ball and the wall is
one millisecond, the average force exerted by the wall on
the ball is :
(a) 2000 newton (b) 1000 newton
(c) 5000 newton (d) 125 newton
21.The mass of the lift is 100 kg which is hanging on the string.
The tension in the string, when the lift is moving with
constant velocity, is(g = 9.8 m/sec
2
)
(a) 100 newton (b) 980 newton
(c) 1000 newton (d) None of these
22.In the question , the tension in the strings, when the lift is
accelerating up with an acceleration 1 m/sec
2
, is
(a) 100 newton (b) 980 newton
(c) 1080 newton (d) 880 newton
23.A block of mass 5 kg resting on a horizontal surface is
connected by a cord, passing over a light frictionless pulley
to a hanging block of mass 5 kg. The coefficient of kinetic
friction between the block and the surface is 0.5. Tension in
the cord is : (g = 9.8 m/sec
2
)
5 kg
5 kg
A
B
(a) 49 N (b) Zero
(c)
36.75 N (d) 2.45 N
24.A 40 kg slab rests on frictionless floor as shown in fig. A 10kg block rests on the top of the slab. The static coefficientof friction between the block and slab is 0.60 while the kineticfriction is 0.40. The 10 kg block is acted upon by a horizontalforce of 100 N. If g = 9.8 m/s
2
, the resulting acceleration of
the slab will be:

40 kg
100 N
No friction
(a) 0.98 m/s
2
(b) 1.47 m/s
2
(c) 1.52 m
/s
2
(d) 6.1 m/s
2
25.Two blocks are connected over a massless pulley as shown infig. The mass of block A is 10 kg and the coefficient of kineticfriction is 0.2. Block A slides down the incline at constant speed.The mass of block B in kg is:
A
B30?
(a) 3.5 (b) 3.3
(c)
3.0 (d) 2.5

123. 119Laws of Motion
26.Tw
o trolleys of mass m and 3m are connected by a spring.
They were compressed and released at once, they move off
in opposite direction and come to rest after covering a
distance S
1
, S
2
respectively. Assuming the coefficient of
friction to be uniform, ratio of distances S
1
: S
2
is :
(a) 1 : 9 (b) 1 : 3
(c) 3 : 1 (d) 9 : 1
27.A particle of mass 10 kg is moving in a straight line. If its
displacement, x with time t is given by x = (t
3
– 2t – 10) m,
then the force acting on it at the end of 4 seconds is
(a) 24 N (b) 240 N
(c) 300 N (d) 1200 N
28.When forces F
1
, F
2
, F
3
are acting on a particle of mass m
such that F
2
and F
3
are mutually perpendicular, then the
particle remains stationary. If the force F
1
is now removed
then the acceleration of the particle is
(a)F
1
/m (b) F
2
F
3
/mF
1
(c) (F
2
– F
3
)/m (d) F
2
/m
29.One end of massless rope, which passes over a massless
and frictionless pulley P is tied to a hook C while the
other end is free. Maximum tension that the rope can
bear is 360 N. With what value of maximum safe
acceleration (in ms
–2
) can a man of 60 kg moves
downwards on the rope? [Take g = 10 ms
–2
]
P
C
(a) 16 (b) 6
(
c)4 (d) 8
30.A force k
ˆ
10j
ˆ
6i
ˆ
8F--=
r
newton produces an acceleration
of 1 ms
–2
in a body. The mass of the body is
(a) 10 kg (b) 210 kg
(c) 310kg (d) 200 kg
31.A unifo
rm chain of length 2 m is kept on a table such that a
length of 60 cm hangs freely from the edge of the table. Thetotal mass of the chain is 4 kg. What is the work done in
pulling the entire chain on the table ?
(a) 12 J (b) 3.6 J
(c) 7.2 J (d) 1200 J
32.A body of mass 1 kg moving with a uniform velocity of
1
ms1
-
. If the value of g is
2
ms5
-
, then the force acting on
the frictionless horizontal surface on which the body is
moving is
(a) 5 N (b) 1 N
(c) 0 N (d) 10N
33.A trailer of mass 1000 kg is towed by means of a rope
attached to a car moving at a steady speed along a level
road. The tension in the rope is 400 N. The car starts to
accelerate steadily. If the tension in the rope is now 1650 N,
with what acceleration is the trailer moving ?
(a) 1.75 ms
–2
(b) 0.75 ms
–2
(c) 2.5 ms
–2
(d) 1.25 ms
–2
34.A rocket of mass 5000 kg is to be projected vertically upward.
The gases are exhausted vertically downwards with velocity
1000 ms
–2
with respect to the rocket. What is the minimum
rate of burning the fuel so as to just lift the rocket upwards
against gravitational attraction ?
(a) 49 kg s
–1
(b) 147 kg s
–1
(c) 98 kg s
–1
(d) 196 kg s
–1
35.Blocks A and B of masses 15 kg and 10 kg, respectively, are
connected by a light cable passing over a frictionless pulley
as shown below. Approximately what is the acceleration
experienced by the system?
(a) 2.0 m/s
2
(b) 3.3 m/s
2
(c) 4.9 m/s
2
A
B
(d) 9.8 m/s
2
36.A 50 kg
ice skater, initially at rest, throws a 0.15 kg snowball
with a speed of 35 m/s. What is the approximate recoil speed
of the skater?
(a) 0.10 m/s (b) 0.20 m/s
(c) 0.70 m/s (d) 1.4 m/s
37.Block A is moving with acceleration A along a frictionless
horizontal surface. When a second block, B is placed on top
of Block A the acceleration of the combined blocks drops to
1/5 the original value. What is the ratio of the mass of A to
the mass of B?
(a) 5 : 1 (b) 1 : 4
(c) 3 : 1 (d) 2 : 1
38.A force F is used to raise a 4-kg mass M from the ground to
a height of 5 m.
60°
F
M
What is the work
done by the force F? (Note : sin 60° = 0.87;
cos 60° = 0.50. Ignore friction and the weights of the pulleys)
(a) 50 J (b) 100 J
(c) 174 J (d) 200 J
39.A 5000 kg rocket is set for vertical firing. The exhaust speed
is 800 m/s. To give an initial upward acceleration of 20 m/s
2
,
the amount of gas ejected per second to supply the needed
thrust will be (Take g = 10 m/s
2
)
(a) 127.5 kg/s (b) 137.5 kg/s
(c) 155.5 kg/s (d) 187.5 kg/s

124. 120 PHYSICS
40.A bullet is f
ired from a gun. The force on the bullet is given
by F = 600 – 2 × 10
5
t
Where, F is in newtons and t in seconds. The force on the
bullet becomes zero as soon as it leaves the barrel. What is
the average impulse imparted to the bullet?
(a) 1.8 N-s (b) Zero
(c) 9 N-s (d) 0.9 N-s
41.A rifle man, who together with his rifle has a mass of 100 kg,
stands on a smooth surface and fires 10 shots horizontally.
Each bullet has a mass 10 g and a muzzle velocity of 800 ms
–
1
. The velocity which the rifle man attains after firing 10
shots is
(a) 8
1
ms
-
(b) 0.8
1
ms
-
(c) 0.08
1
ms
-
(d) – 0.8
1
ms
-
42.A block of mass 4 kg rests on an inclined plane. The
inclination to the plane is gradually increased. It is found
that when the inclination is 3 in 5, the block just begins to
slidedown the plane. The coefficient of friction between the
block and the plane is
(a) 0.4 (b) 0.6
(c) 0.8 (d) 0.75.
43.The minimum velocity (in ms
-1
) with which a car driver must
traverse a flat curve of radius 150 m and coefficient of friction
0.6 to avoid skidding is
(a) 60 (b) 30
(c) 15 (d) 25
44.A body of mass 2 kg is placed on a horizontal surface
having kinetic friction 0.4 and static friction 0.5. If the force
applied on the body is 2.5 N, then the frictional force acting
on the body will be [g = 10 ms
–2
]
(a) 8 N (b) 10 N
(c) 20 N (d) 2.5 N
45.A bag of sand of mass m is suspended by a rope. A bullet of
mass
20
m
is fired at
it with a velocity v and gets embedded
into it. The velocity of the bag finally is
(a)
v
21
20
´ (b)
20v
21
(c)
v
20
(d)
v
21
46.For the arrangement shown in the Figure the tension in the
string is [Given :
1
tan (0.8) 39
-
=°]
39°
m = 1 kg
m = 0.8
(a) 6 N (b) 6.4 N
(c) 0.4 N (d)
zero.
47.A 1 kg block and a 0.5 kg block move together on a horizontal
frictionless surface . Each block exerts a force of 6 N on the
other. The block move with a uniform acceleration of
a
1 kg 0.5 kgF
(a)3
2
ms
-
(b) 6
2
ms
-
(c)
9
2
ms
-
(d) 12
2
ms
-
48.A body of mass 32 kg is suspended by a spring balancefrom the roof of a vertically operating lift and goingdownward from rest. At the instant the lift has covered 20 mand 50 m, the spring balance showed 30 kg and 36 kg
respectively. Then the velocity of the lift is
(a) decreasing at 20 m, and increasing at 50 m
(b) increasing at 20m and decreasing at 50 m
(c) continuously decreasing at a steady rate throughout
the journey
(d) constantly increasing at constant rate throughout the
jour n ey.
49.An object at rest in space suddenly explodes into three
parts of same mass. The momentum of the two parts are
i
ˆ
p2 and j
ˆ
p. The momentum of the third part
(a) will have a magnitude
3p
(b) will hav
e a magnitude
5p
(c) will have a magn
itude p
(d) will have a magnitude 2p.
50.A triangular block of mass M with angles 30°, 60°, and 90°
rests with its 30°–90° side on a horizontal table. A cubical
block of mass m rests on the 60°–30° side. The acceleration
which M must have relative to the table to keep m stationary
relative to the triangular block assuming frictionless contact is
(a)g (b)
2
g
(c)
3
g
(d)
5
g
51. A body of mass 1.0 kg is falling with an acceleration of 10 m/
sec
2
. Its apparent weight will be (g = 10 m/sec
2
)
(a) 1.0 kg wt (b) 2.0 kg wt
(c) 0.5 kg wt (d) zero
52.In the figure a smooth pulley of negligible weight is
suspended by a spring balance. Weight of 1 kg f and
5 kg f are attached to the opposite ends of a string passing
over the pulley and move with acceleration because of
gravity, During their motion, the spring balance reads a
weight of
(a) 6 kg f
(b) less then 6 kg f
(c) more than 6 kg f
1 kg
5 kg
(d) may be more or less then 6 kg f

125. 121Laws of Motion
53.A par
ticle moves so that its acceleration is always twice its
velocity. If its initial velocity is 0.1 ms
–1
, its velocity after it
has gone 0.1 m is
(a) 0.3 ms
–1
(b) 0.7 ms
–1
(c) 1.2 ms
–1
(d) 3.6 ms
–1
54.An object is resting at the bottom of two strings which are
inclined at an angle of 120° with each other. Each string can
withstand a tension of 20N. The maximum weight of the
object that can be supported without breaking the string is
(a) 5 N (b) 10 N
(c) 20 N (d) 40 N
55.On a smooth plane surface (figure) two block A and B are
accelerated up by applying a force 15 N on A. If mass of B is
twice that of A, the force on B is
(a) 30 N (b) 15 N
(c) 10 N (d) 5 N
BA
15 N
56.A 10 kg ston
e is suspended with a rope of breaking strength
30 kg-wt. The minimum time in which the stone can be raised
through a height 10 m starting from rest is (Take
kg/N10g= )
(a) 0.5 s (b) 1.
0 s
(c)
3/2s (d) 2 s
57.A ba
ll of mass 0.4 kg thrown up in air with velocity 30
1
ms
-
reaches the highest point in 2.5 second . The air resistance
encountered by the ball during upward motion is
(a) 0.88 N (b) 8800N
(c) 300 dyne (d) 300 N.
58.A player caught a cricket ball of mass 150 g moving at a rate
of 20 m/s. If the catching process is completed in 0.1s, the
force of the blow exerted by the ball on the hand of the
player is equal to
(a) 150 N (b) 3 N
(c) 30 N (d) 300 N
59.In the system shown in figure, the pulley is smooth and
massless, the string has a total mass 5g, and the two
suspended blocks have masses 25 g and 15 g. The system
is released from state 0
=l and is studied upto stage 0'=l
During the process, the acceleration of block A will be
(a) constant at
g
9
(b) constant at
g
4
A
25 g
B
15 g
l l'
(c) incr
easing by factor of 3
(d) increasing by factor of 2
60.A horizontal force F is applied on back of mass m placed ona rough inclined plane of inclination
q. The normal reaction
N is
F
(a) qcosmg (b) qsinmg
(c)mg cos Fcosq-q (d) q+qsinFcosmg
61.The coeffici
ent of friction between the rubber tyres and the
road way is 0.25. The maximum speed with which a car can
be driven round a curve of radius 20 m without skidding is
(g = 9.8 m/s
2
)
(a) 5 m/s (b) 7 m/s
(c) 10 m/s (d) 14 m/s
62.A bucket tied at the end of a 1.6 m long string is whirled in a
vertical circle with constant speed. What should be the
minimum speed so that the water from the bucket does not
spill when the bucket is at the highest position?
(a) 4 m/sec (b) 6.25 m/sec
(c) 16 m/sec (d) None of the above
63.A cane filled with water is revolved in a vertical circle of
radius 4 meter and the water just does not fall down. The
time period of revolution will be
(a) 1 sec (b) 10 sec
(c) 8 sec (d) 4 sec
64.A circular road of radius r in which maximum velocity is v,
has angle of banking
(a)
÷
÷
ø
ö
ç
ç
è
æ
-
rg
v
tan
2
1
(b)
÷
÷
ø
ö
ç
ç
è
æ
-
2
1
v
rg
tan
(c)
÷
÷
ø
ö
ç
ç
è
æ
-
rg
v
tan
1
(d) ÷
ø
ö
ç
è
æ-
v
rg
tan
1
65.A smal
l sphere is attached to a cord and rotates in a vertical
circle about a point O. If the average speed of the sphere is
increased, the cord is most likely to break at the orientation
when the mass is at
C
O
D
m
l
A
B
(a) bottom point B (b) the point C
(c) the point D (d) top point A

126. 122 PHYSICS
66.A perso
n with his hand in his pocket is skating on ice at the
rate of 10m/s and describes a circle of radius 50 m. What is
his inclination to vertical : (g = 10 m/sec
2
)
(a) tan
–1
(½) (b) tan
–1
(1/5)
(c) tan
–1
(3/5) (d) tan
–1
(1/10)
67.When the road is dry and the coefficient of the friction is m,
the maximum speed of a car in a circular path is 10 ms
–1
. If
the road becomes wet and
2
'
m
=m, what is the maximum
speed pe
rmitted?
(a) 5 ms
–1
(b) 10 ms
–1
(c)
1
ms210
-
(d)
1
ms25
-
68.Two pulley arr
angements of figure given are identical. The
mass of the rope is negligible. In fig (a), the mass m is lifted
by attaching a mass 2m to the other end of the rope. In fig
(b), m is lifted up by pulling the other end of the rope with a
constant downward force F = 2mg. The acceleration of m in
the two cases are respectively
m2m
(a)

m
(b)
F = 2 mg
(a) 3g, g (b)g/3, g
(c)g/3, 2g (d) g, g/3
69.The linear momentum p of a body moving in one dimension
varies with time according to the equating P = a + bt
2
where
a and b are positive constants. The net force acting on thebody is
(a) proportional to t
2
(b) a constant
(c) proportional to t
(d) inversely proportional to t
70.Three blocks of masses m
1
, m
2
and m
3
are connected by
massless strings, as shown, on a frictionless table. They are
pulled with a force T
3
= 40 N. If m
1
= 10 kg, m
2
= 6 kg and m
3
= 4kg, the tension T
2
will be
M
1
M
2
M
3
T
1
T
2
T
3
(a) 20 N (b) 40 N
(c) 1
0 N (d) 32 N
71.A ball of mass 400 gm is dropped from a height of 5 m. A boyon the ground hits the ball vertically upwards with a bat
with an average force of 100 newton so that it attains a
vertical height of 20 m. The time for which the ball remains in
contact with the bat is (g = 10 m/s
2
)
(a) 0.12 s (b) 0.08 s
(c) 0.04 s (d) 12 s
72.Block A of weight 100 kg rests on a block B and is tied with
horizontal string to the wall at C. Block B is of
200 kg. The coefficient of friction between A and B is 0.25
and that between B and surface is
1
3
. The horizontal force F
necessary to move the block B should be (g = 10 m/s
2
)
A
C
B
F
(a) 1050 N (b) 1450 N
(
c) 1050 N (d) 1250 N
73.An open topped rail road car of mass M has an initial velocity
v
0
along a straight horizontal frictionless track. It suddenly
starts raising at time t = 0. The rain drops fall vertically with
velocity u and add a mass m kg/sec of water. The velocity of
car after t second will be (assuming that it is not completely
filled with water)
(a)
0
u
vm
M
+ (b)
0
mv
M mt+
(c)
0
Mv ut
M ut
+
+
(d)0
mut
v
M ut
+
+
74.A ball mass m falls vertically to the ground from a height h
1
and rebounds to a height h
2
. The change in momentum of
the ball of striking the ground is
(a)
12
m2g(h h)+ (b)
12
n 2g(m m )+
(c)
12
mg(h h)- (d)
12
m( 2gh 2gh )-
75.In the given figu
re, the pulley is assumed massless and
frictionless. If the friction force on the object of mass m is f,
then its acceleration in terms of the force F will be equal to
m
F
(a)(F f)/m- (b)
F
f /m
2
æö
-
ç÷
èø
(c) F/m (d) None of th ese

127. 123Laws of Motion
76.A s
mooth block is released at rest on a 45° incline and then
slides a distance ‘d’. The time taken to slide is ‘n’ times as
much to slide on rough incline than on a smooth incline.
The coefficient of friction is
(a)
k
m =
2
n
1
1- (b)
k
m =
2
n
1
1-
(c)
s
m =
2
n
1
1- (d)
s
m =
2
n
1
1-
77.The upper h
alf of an inclined plane with inclination
f is
perfec
tly smooth while the lower half is rough. A body
starting from rest at the top will again come to rest at the
bottom if the coefficient of friction for the lower half is given
by
(a) 2 cos
f (b) 2 sin f
(c) tan f (d) 2 tan f
78.A particle of mass 0.3 kg subject to a force F = – kx with k =
15 N/m . What will be its initial acceleration if it is released
from a point 20 cm away from the origin ?
(a) 15 m/s
2
(b) 3 m/s
2
(c) 10 m/s
2
(d) 5 m/s
2
79.A block is kept on a frictionless inclined surface with angle
of inclination ‘a’ . The incline is given an acceleration ‘a’ to
keep the block stationary. Then a is equal to
a
(a) g coseca (b) g / tana
(c) g tana (d) g
80.Conside
r a car moving on a straight road with a speed of 100
m/s . The distance at which car can be stopped is [µ
k
= 0.5]
(a) 1000 m (b) 800 m
(c) 400 m (d) 100 m
81.A round uniform body of radius R, mass M and moment ofinertia I rolls down (without slipping) an inclined planemaking an angle q with the horizontal. Then its acceleration
is
(a)
2
gsin
1 MR /I
q
,
(b)
2
g sin
1 I / MR
q
∗
(c)
2
g sin
1 MR /I
q
∗
(d)
2
gsin
1 I / MR
q
,
82.A block of m ass m is placed on a smooth wedge of
inclination q. The whole system is accelerated horizontally
so that the block does not slip on the wedge. The force
exerted by the wedge on the block (g is acceleration due to
gravity) will be
(a) mg/cos q (b) mg cos q
(c) mg sin q (d) mg
83.The coefficient of static friction, m
s
, between block A of
mass 2 kg and the table as shown in the figure is 0.2. What
would be the maximum mass value of block B so that the
two blocks do not move? The string and the pulley are
assumed to be smooth and massless. (g = 10 m/s
2
)
A
B
2 kg
(a) 0.4 kg (b) 2.0 kg
(c)
4.0 kg (d) 0.2 kg
84.A body under the action of a force
ˆˆˆ
F = 6 i –8 j+10k,
r
acquires an acceleration of 1 m/s
2
. The
mass of this body must be
(a) 10 kg (b) 20 kg
(c) 102kg (d) 210 kg
85.A convey
or belt is moving at a constant speed of 2m/s. A box
is gently dropped on it. The coefficient of friction between
them is µ = 0.5. The distance that the box will move relative to
belt before coming to rest on it taking g = 10 ms
–2
, is
(a) 1.2 m (b) 0.6 m
(c) zero (d) 0.4 m
86.A person of mass 60 kg is inside a lift of mass 940 kg and
presses the button on control panel. The lift starts moving
upwards with an acceleration 1.0 m/s
2
. If g = 10 ms
–2
, the
tension in the supporting cable is
(a) 8600 N (b) 9680 N
(c) 11000 N (d) 1200 N
87.The upper half of an inclined plane of inclination q is
perfectly smooth while lower half is rough. A block
starting from rest at the top of the plane will again come to
rest at the bottom, if the coefficient of friction between the
block and lower half of the plane is given by
(a)m =
2
tanq
(b)m = 2 tan q
(c)m =
tan q (d)m =
1
tanq
88.A bridge is in the from of a semi-circle of radius 40m. The
greatest speed with which a motor cycle can cross the bridge
without leaving the ground at the highest point is
(g = 10 m s
–2
) (frictional force is negligibly small)
(a) 40 m s
–1
(b) 20 m s
–1
(c) 30 m s
–1
(d) 15 m s
–1

128. 124 PHYSICS
89.An explosio
n breaks a rock into three parts in a horizontal
plane. Two of them go off at right angles to each other. The
first part of mass 1 kg moves with a speed of 12 ms
–1
and the
second part of mass 2 kg moves with speed 8 ms
–1
. If the
third part flies off with speed 4 ms
–1
then its mass is
(a) 5 kg (b) 7 kg
(c) 17 kg (d) 3 kg
90.Two particles of masses m and M (M > m ) are connected by
a cord that passes over a massless, frictionless pulley. The
tension T in the string and the acceleration a of the particles
is
(a)
2mM Mm
Tg;ag
(M m) (M m)
==
-+
(b)
2mM Mm
Tg ;ag
(M m) (M m)
æö-
==
ç÷
+ è +ø
(c)
m M Mm
Tg ;ag
(Mm) (Mm)
æöæö-
==
ç÷ç÷
è +ø è +ø
(d)
mM 2Mm
Tg ;ag
(Mm) (Mm)
æöæö
==
ç÷ç÷
è +ø è +ø
91.A bullet of mass m is fired from a gun of mass M. The
recoiling gun compresses a spring of force constant k by a
distance d. Then the velocity of the bullet is
(a)kd M/m (b)
d
km
M
(c)
d
kM
m
(d)
kM
d
m
92.A spring of force constant k is cut into two pieces whose
lengths are in the ratio 1 : 2. What is the force constant of
the longer piece ?
(a)
2
k
(b)
3
2
k
(c) 2 k (d) 3k
93.A motor cyc
le is going on an overbridge of
radius R. The driver maintains a constant speed. As the
motor cycle is ascending on the overbridge, the normal force
on it
(a) increases (b) decreases
(c) remains the same(d) fluctuates erratically
94.A body of mass M hits normally a rigid wall with velocity V
and bounces back with the same velocity. The impulse
experienced by the body is
(a) MV (b) 1.5 MV
(c) 2 MV (d) zero
95.A heavy uniform chain lies on horizontal table top. If the
coefficient of friction between the chain and the table surface
is 0.25, then the maximum fraction of the length of the chain
that can hang over one edge of the table is
(a) 20% (b) 25%
(c) 35% (d) 15%
96.A body of mass 5 kg explodes at rest into three fragments
with masses in the ratio 1 : 1 : 3. The fragments with equal
masses fly in mutually perpendicular directions with speeds
of 21 m/s. The velocity of heaviest fragment in m/s will be
(a)
27 (b)25
(c)23 (d)2
97.Two bodies of
masses m and 4m are moving with equal
kinetic energies. The ratio of their linear momenta will be
(a) 1 : 4 (b) 4 : 1
(c) 1 : 2 (d) 2 : 1
Directions for Qs. (98 to 100) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following-
(a) Statement -1 is false, Statement-2 is true
(b) Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c) Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d) Statement -1 is true, Statement-2 is false
98. Statement -1 : The work done in bringing a body down from
the top to the base along a frictionless incline plane is the
same as the work done in bringing it down the vartical side.
Statement -2 : The gravitational force on the body along
the inclined plane is the same as that along the vertical side.
99. Statement -1 : On a rainy day, it is difficult to drive a car or
bus at high speed.
Statement -2 : The value of coefficient of friction is lowered
due to wetting of the surface.
100. Statement -1 : The two bodies of masses M and m (M > m)
are allowed to fall from the same height if the air resistance
for each be the same then both the bodies will reach the
earth simultaneously.
Statement -2 : For same air resistance, acceleration of both
the bodies will be same.

129. 125Laws of Motion
Exemplar Questions
1.A ball is travelling with uniform translatory motion. This
means that
(a) it is at rest
(b) the path can be a straight line or circular and the ball
travels with uniform speed
(c) all parts of the ball have the same velocity (magnitude
and direction) and the velocity is constant
(d) the centre of the ball moves with constant velocity and
the ball spins about its centre uniformly
2.A metre scale is moving with uniform velocity. This implies
(a) the force acting on the scale is zero, but a torque about
the centre of mass can act on the scale
(b) the force acting on the scale is zero and the torque
acting about centre of mass of the scale is also zero
(c) the total force acting on it need not be zero but the
torque on it is zero
(d) neither the force nor the torque need to be zero
3.A cricket ball of mass 150 g has an initial velocity
1ˆˆ(3 4 )msu ij
-
=+
r
and a final velocity
1ˆˆ(3 4 )msv ij
-
=-+
r
,
after being hit. The change in momentum (final momentum-
initial momentum) is (in kgms
–1)
(a) zero (b) ˆˆ(0.45 0.6 )ij-+
(c) ˆˆ(0.9 1.2 )jj-+ (d)ˆ ˆˆ5()i ji-+
4.In the previous problem (3), the magnitude of the momentum transferred during the hit is
(a) zero (b) 0.75 kg-m s
–1
(c) 1.5 kg-m s
–1 (d) 1.4 kg-m s
–1
5.Conservation of momentum in a collision between particles
can be understood from
(a) Conservation of energy
(b) Newton's first law only
(c) Newton's second law only
(d) both Newton's second and third law
6.A hockey player is moving northward and suddenly turns
westward with the same speed to avoid an opponent. The
force that acts on the player is
(a) frictional force along westward
(b) muscle force along southward
(c) frictional force along sotuh-west
(d) muscle force along south-west
7.A body of mass 2 kg travels according to the law
23
()x t pt qt rt=++ where, q = 4 ms
–2, p = 3 ms
–1 and r = 5
ms
–3. The force acting on the body at t = 2s is
(a) 136 N (b) 134 N
(c) 158 N (d) 68 N
8.A body with mass 5 kg is acted upon by a force
ˆˆ( 3 4 )NF ij=-+
r
. If its initial velocity at t = 0 is
1ˆˆ(6 12 ) msvij
-
=- , the time at which it will just have a
velocity along the y-axis is
(a) never (b) 10 s
(c) 2 s (d) 15 s
9.A car of mass m starts from rest and acquires a velocity
along east, ˆ( 0)v viv=>
r
in two seconds. Assuming the
car moves with uniform acceleration, the force exerted on
the car is
(a)
2
mv
eastward and is exerted by the car engine
(b)
2
mv
eastward and is due to the friction on the tyres
exerted by the road
(c) more than
2
mv
eastward exerted due to the engine and
overcomes the friction of the road
(d)
2
mv
exerted by the engine
NEET/AIPMT (2013-2017) Questions
10.Three blocks with masses m, 2 m and 3 m are connected by
strings as shown in the figure. After an upward force F is
applied on block m, the masses move upward at constant
speed v. What is the net force on the block of mass 2m?
(g is the acceleration due to gravity) [2013]
(a) 2 mg
(b) 3 mg
(c) 6 mg (d) zero
11.A car is moving in a circular horizontal track of radius 10 m
with a constant speed of 10 m/s. A bob is suspended from
the roof of the car by a light wire of length 1.0 m. The angle
made by the wire with the vertical is [NEET Kar. 2013]
(a) 0° (b)
3
p
(c)
6
p
(d)
4
p
12.A balloon with mass ‘m’ is descending down with an
acceleration ‘a’ (where a < g). How much mass should be
removed from it so that it starts moving up with an
acceleration ‘a’? [2014]
(a)
2ma
ga+
(b)
2ma
ga-
(c)
ma
ga+
(d)
ma
ga-

130. 126 PHYSICS
13.The force ‘F’ acting on a particle of mass ‘m’ is indicated by
the force-time graph shown below. The change in momentum
of the particle over the time interval from zero to 8 s is :
[2014]
6
3
0
–3
46 8
F
(
N
)
t(s)
2
(a) 24 Ns (b) 20 Ns
(c) 12 Ns (d) 6 Ns
14.A system consists of three masses m
1
, m
2
and m
3
connected
by a string passing over a pulley P. The mass m
1
hangs
freely and m
2
and m
3
are on a rough horizontal table (the
coefficient of friction = m). The pulley is frictionless and of
negligible mass. The downward acceleration of mass m
1
is :
(Assume m
1
= m
2
= m
3
= m) [2014]
(a)
g(1–g)
g
m
(b)
2g
3
m

P
m
1
m
2
m
3
(c)
g(1–2)
3
m
(d)
g(1 – 2 )
2
m
15.Three blocks A, B and C of masses 4 kg, 2 kg and 1 kg
respectively, are in contact on a frictionless surface, as
shown. If a force of 14 N is applied on the 4 kg block then
the contact force between A and B is [2015]
A
B
C
(a) 6 N (b) 8 N
(c) 18 N (d) 2 N
16.A block A of mass m
1
rests on a horizontal table. A light
string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m
2
is suspended. The coefficient of kinetic friction
between the block and the table is µ
k
. When the block A is
sliding on the table, the tension in the string is [2015]
(a)
21
12
(m – km )g
(m m)
m
+
(b)
12k
12
m m (1 )g
(m m)
+m
+
(c)
12k
12
m m (1– )g
(m m)
m
+
(d)
2 k1
12
(m m )g
(m m)
+m
+
17.Two stones of masses m and 2 m are whirled in horizontal
circles, the heavier one in radius
r
2
and the lighter one in
radius r. The tangential speed of lighter stone is n times that
of the value of heavier stone when they experience same
centripetal forces. The value of n is : [2015 RS]
(a) 3 (b) 4
(c) 1 (d) 2
18.A plank with a box on it at one end is gradually raised about
the other end. As the angle of inclination with the horizontal
reaches 30º the box starts to slip and slides 4.0 m down the
plank in 4.0s. The coefficients of static and kinetic friction
between the box and the plank will be, respectively :
[2015 RS]
mg
q
(a) 0.6 and 0.5 (b) 0.5 and 0.6
(c) 0.4 and 0.3 (d) 0.6 and 0.6
19.What is the minimum velocity with which a body of mass m
must enter a vertical loop of radius R so that it can complete
the loop ? [2016]
(a)
gR (b)2gR
(c)3gR (d)5gR
20.One end of string of length l is connected to a particle of mass 'm' and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed 'v' the net force on the particle (directed towards centre) will be (T represents the tension in the string) : [2017]
(a)
2
mv
T
l
+ (b) T –
2
mv
l
(c) Zero (d) T
21.Two blocks A and B of masses 3 m and m respectively are connected by a massless and inextensible string. The whole system is suspended by a massless spring as shown in figure. The magnitudes of acceleration of A and B immediately after the string is cut, are respectively : [2017]
(a)
g
,g
3
A
B
3m
m
(b) g, g
(c)
gg
,
33
(d)
g
g,
3

131. 127Laws of Motion
EXERCISE - 1
1. (c) 2. (b) 3. (d) 4. (a) 5. (c)
6. (a) 7. (a)
8. (d) Let n be the mass per unit length of rope. Therefore,
mass of rope = nL.
Acceleration in the rope due to force F will be
a = F/nL.
Mass of rope of length (L – l) will be n (L – l).
Therefore, tension in the rope of length (L – l), is equal
to pulling force on it
= n (L – l) a = n (L – l) × F/nL = F (1 – l/L)
9. (a) Impulse = change in momentum =
12vmvm
rr
-
10. (a)
Let a be the acceleration of mass m
2
in the downward
direction. Then
T – m
2
(g/2) = m
1
a ....(i)
and m
2
g – T = m
2
a ....(ii)
Adding eqs. (1) and (2), we get
(m
1
+ m
2
) a = m
2
g – m
2
(g/2) = m
2
g/2
\
)mm(2
gm
a
21
2
+
=
11. (a)
Let acceleration of alift=and
let reaction at spring balance = R
R
mg
Applying Newton’s law
R – mg = ma ()agmR+=Þ
thus n
et weight increases,
So reading of spring balance increases.
12. (b) See fig.
ma
mN
N
mg
If a = acceleration of the cart, then N = ma
\ mN = mg or m ma = mg or a = g/m
13. (c) p
1
= mv northwards, p
2
= mv eastwards
N
E
S
W
m
m
v
v
Let p = momentum after collision. Then,
21ppp
rrr
+= or
22
)mv()mv(p+=
r
2mvvm2=¢ or
2
v
v=¢ m/sec
14.
(c) Applying law of conservation of linear momentum
m
1
v
1
+ m
2
v
2
= 0,
1
2
2
1
v
v
m
m
-=or
1 2
2
1
m
m
v
v
-=
15. (c)
The equations of motion are
2 mg – T = 2ma
T– mg = ma
ÞT = 4ma & a = g/3 so T = 4mg/3
If pulley is accelerated upwards with an accleration a,
then tension in string is
)ag(
3
m4
T +=
16. (b)
Inertia is resistance to change.
17. (b) The time rate of change of momentum is force.
18. (a) At the highest point of the track,
2
mv'
N mg
r
+=

mg
N
where r is the radius of curvature at that point and v¢ is
the speed of the block at that point.Now
2
mv'
N mg
r
=-
N will be
maximum when r is minimum (v¢ is the same
for all cases). Of the given tracks, (a) has the smallest
radius of curvature at the highest point.
19. (c)
fW=m
f W tan=q [\ q=mtan]
20. (a) )1n2(
2
a
S
n
-=
)1n2(
2
a
S
1n +=
+
1n2
1n2
S
S
1n
n
+
-
=
+
21. (b)
From the F.B.D.
N = mg cos
q
F = ma = mg sinq – Nm
)cos(singa qm-q=Þ
x
qsing
qg
Nm
q
N
m m
mg
cos
Hints & Solutions

132. 128 PHYSICS
Now using, as2uv
22
=-
or, l)cos(sing2v
2
qm-q´=
(l = length of incline)
or, v = )cos(sing2 qm-ql
22
. (a) Based on Newton’s third law of motion.
23. (a) If m
1
, m
2
are masses and u
1
, u
2
are velocity then by
conservation of momentum m
1
u
1
+ m
2
u
2
= 0 or11 22
|mu| |mu|=
24. (a) Here, on resolving force F
2
and applying the concept
of equilibrium
F
1
F
2
sin q
F
2
cos q
mg
N
f
m
q+=cosFmgN
2
, and f = µN
\ ]cosFmg[f
2
q+m=… (i)
Also q+=sinF
Ff
21
… (ii)
From (i) and (ii)
]cosFmg[
2
q+m = F
1
+ F
2
sin q
Þ
q+
q+
=m
cosFmg
sinFF
2
21
25. (
a) The frictional force acting on M is µmg
\ Acceleration =
M
mgm
EXERCISE - 2
1. (b) Here u = 10 ms
–1
, v = –2 ms
–1
,
t = 4 s, a = ?
Using
2
s/m3
4
102
t
uv
a -=
--
=
-
=
\ Force, F =
ma = 10×(–3) = –30 N
2. (b) 2 T cos 60º = mg
or T = mg = 2×10 = 20 N.
3. (c) If k is the spring factor, then P.E. of the spring
compressed by distance x
÷
ø
ö
ç
è
æ
=
2
kx
2
1
will equal to gain
in P
.E. of the dart ( = mgh) i.e.
mghkx
2
1 2
=
\ 200g16)4(k
2
1
2
´´= ....(i)
and hg16)6(k
2
1
2
´´= ...(ii)
On so
lving, (i) and (ii), we get h = 450 cm = 4.5 m.
4. (d) Here m = 0.5 kg ; u = – 10 m/s ;
t = 1/50 s ; v = + 15 ms
–1
Force = m (v– u)/t = 0.5 (10 + 15) × 50 = 625 N
5. (b) As, (1/2)m v
2
= Fs
So
4F)30(m
2
1
2
´= and sF)60(m
2
1
2
´=
\ s/4 = (60)
2
/ (30)
2
= 4 or
s = 4 × 4 = 16 m.
6. (c) Acceleration,
2vu 06
a 2 ms
t3
---
= = =-
Force =
m×a = 4×2 = 8 N
7. (a)
takentime
momentuminchange
requiredFor
ce =
=
3
(50 10 30) 400 (5 0)
10N
60
-
´ ´ ´ -´
=
8. (a) Ch
ange in momentum = Force × time = Area which the
force-time curve encloses with time axis.
9. (b)
20cm10cm
32N(F)
2
20N(F)
1
l
1
l
2
L
F F
It is clear F
2
> F
1
, so rod mo ves in right direction with
an acceleration a, whereas a is given by
(F
2
–F
1
)= mL×a................(i)
where m is mass of rod per unit length.
Now consider the motion of length l
1
from first end,
then
F– F
1
= ml
1
a..................(ii)
Dividing eq (ii) by (i), we get
L
l
FF
FF
1
12
1
=
-
-
or
1
1
12
F
L
l
)FF(F+´-=
here l
1
= 10 cm
., L = 30 cm., F
1
= 20 N, F
2
= 32N
so F = 24 N
10. (b) Change in momentum = F × t
= 10 × 10 = 100 Ns or 100 kg. m/s
11. (d) See fig.
1 kg
1 kg
T
T
T
From figu
re, 1 g – T = 1 a ...(i)
and T = 1 a ....(ii)
From eqs. (i) and (ii), we get1g – 1a = 1a or 2a = g
\ a = (g/2) = (10/2) = 5 m/s
2
So, T = ma = 1 × 5 = 5 N
12. (b) R = mg – ma = 0.5 × 10 – 0.5 × 2 = 5 – 1 = 4 N
13. (c) See fig. Let F be the force between the blocks and a their
common acceleration. Then for 2 kg block,

133. 129Laws of Motion
3 N 2 kg
1 kg
F F
3 – F = 2
a ...(1)
for 1 kg block, F = 1 × a = a....(2)
\ 3 – F = 2 F or 3 F = 3 or F = 1 newton
14. (b) )ag(mT+=
)a10(400048000 +=
Þ a = 2
2
sm
-
15. (b) Initial thrust on the rocket =
rel
v
t
m
D
D
= 500 × 1 = 500 N
where
m
t
D
D
= rate of ejection of fuel.
16. (b) )ag(mT
1
+= = 0.1(10 5)1.5N+=
N5.0)510(1.0)ag(mT
2
=-=-=
Þ N1N)5.05.1(TT
21
=-=-
17. (d) A
ngle of friction =
1
tanµ
-
18. (c)
Displacement of the man on the trolley = 1 × 4 = 4m
Now applying conservation of linear momentum
80 × 1 + 400 v = 0 or
5
1
v-= m/sec.
The d
istance travelled by the trolley
= – 0.2 × 4 = –0.8 m.
(In opposite direction to the man.)
Thus, the relative displacement of the man with the
ground = (4 – 0.8) = 3.2 m.
19. (c) In presence of friction a = (g sinq – mg cos q)
\ Time taken to slide down the plane
)cos(sing
s2
a
s2
t
1
qm-q
==
In absence of friction
q
=
sing
s2
t
2
Given
:
21t2t=
\
22
12
t = 4t or
q
´
=
qm-qsing
4s2
)cos(sing
s2
sin q = 4 sinq – 4m cos q
75.0
4
3
tan
4
3
==q=m (since q
= 45°)
20. (a)
()
3
10
25.02
t
mv2
t
mvmv
F
-
´´
==
--
= = 2 × 10
3
N
2
1. (b) T = m (g + a) = 100 (9.8 + 0) = 980 N
22. (c) T = m(g+a) = 100(9.8+1) = 1080N
23. (c) For block A, T – mN = 5a and N = 5g
5 kg
5 kgB
A
mN
T
a
T
N
a
5 kg
for bl
ock B, 5g – T = 5a
Þ T = 36.75N, a = 2.45 m/sec
2
24. (a) Force on the slab (m = 40 kg) = reaction of frictional
force on the upper block

40 kg
10 kg
100 N
m
k ? 10 ? g
\ 40a = m
k
× 10 × g or a = 0.98 m/sec
2
25. (b) Considering the equilibrium of B
–m
B
g + T = m
B
a
Since the block A slides down with constant speed.
a = 0.
Therefore T = m
B
g
Considering the equilibrium of A, we get
10a = 10g sin 30º – T – mN
where N = 10g cos 30°
B
A
T
T
10g cos30?
mg
B
mN
10g
10g sin 30?
N
a
a
\ º30cosg10Tg
2
10
a10 ´m--=
but a =
0, T = m
B
g
2
32.0
gmg50
B--= × 10 ×g
Þ m
B
= 3.268 » 3.3 kg
26.
(d) mv
1
+ 3mv
2
= 0 or
3
v
v
2
1
-=
Now 11
2
1 S.mg.S.Fmv
2
1
m==
22
2
2 S.mg3.S.Fv)m3(
2
1
m==
or
1
9
v
v
S
S
2
2
2
1
2
1
==

134. 130 PHYSICS
27. (b) m =
10 kg, x = (t
3
– 2t – 10) m
2t3
dt
dx 2
-=n= t6a
dt
xd
2
2
==
At the end
of 4 seconds, a = 6 × 4 = 24 m/s
2
F = ma = 10 × 24 = 240 N
28. (a)
m
F
1
F
2
F
3
The formula for
force is given by F
1
= ma
Acceleration of the particle
m
F
a
1
=,
because F
1
is equal to the ve
ctor sum of F
2
& F
3
.
29. (c)
P
C
mg
T
a
mg T ma-=
60 10 360
a
60
´-
=
2
a 4 ms
-
=
30. (b)
222
8 ( 6) ( 10)
m
1
+- +-
= = 10 2kg
31. (b) Mass of over hanging chain m’ kg)6.0(
2
4
´=
Let at the
surface PE = 0
C.M. of hanging part = 0.3 m below the table 30.0106.0
2
4
gxmU
i
´´´-=¢-=
U m'gx 3.6JD== = Work done in putting the entire
chain on the table
32. (a) Weight of body = m g = 5 N
33. (d) Here, the force of friction is 400N.
F
net
= (1650 400) 1250N-=
\ a =
2
ms25.1
1000
1250 -
=
34. (a)
r
dm mg
dtv
= =
1
skg49
1000
8.95000 -
=
´
35. (a) Two external forces, F
A
and F
B
, act on the system and
move in opposite direction. Let’s arbitrarily assume that
the downward direction is positive and that F
A
provides
downward motion while F
B
provides upward motion.
F
A
= (+15 kg) (9.8 m/s
2
) = 147 N
and F
B
= (–10 kg)(9.8 m/s
2
) = – 98 N
F
total
= F
A
+ F
B
= 147 N + (–98 N) = 49 N
The total mass that must be set in motion is
15 kg + 10 kg = 25kg
Since ,amF
totaltotal
= a = F
total
/ m
total
= 49 N / 25 kg
@ 2 m/s
2
36. (a) Moment
um is always conserved. Since the skater and
snowball are initially at rest, the initial momentum is
zero. Therefore, the final momentum after the toss must
also be zero.
0PP
snowballskater
=+
or 0vmvm
snowballsnowballskaterskater
=+
skatersnowballsnowballskater
m/vmv-=
(0.15kg)(35m /s)
0.10m/s
(50kg)
=-=
-
The negativ
e sign indicates that the momenta of the
skater and the snowball are in opposite directions.
37. (b) Apply Newton’s second law
F
A
= F
AB
, therefore :
m
A
a
A
= (m
A
+ m
B
)a
AB
and a
AB
= a
A
/ 5
Therefore : m
A
a
A
= (m
A
+ m
B
)a
A
/5 which reduces to
4 m
A
= m
B
or 1 : 4
38. (d) Work is the product of force and distance. The easiest
way to calculate the work in this pulley problem is to
multiply the net force or the weight mg by the distance
it is raised: 4 kg x 10 m/s
2
x 5 m = 200 J.
39. (d)Given : Mass of rocket (m) = 5000 Kg
Exhaust speed (v) = 800 m/s
Acceleration of rocket (a) = 20 m/s
2
Gravitational acceleration (g) = 10 m/s
2
We know that upward force
F = m (g + a) = 5000 (10 +20)
= 5000 × 30 = 150000 N.
We also know that amount of gas ejected
s/kg5.187
800
150000
v
F
dt
dm
===÷
ø
ö
ç
è
æ
40. (d) Giv
en F = t102–600
5
´
The force is zero at time t, given by
t102–6000
5
´=
3–
5
103
102
600
t ´=
´
=Þ seconds
\ Impulse
–3
t 3
10
5
00
Fdt (600 – 2 10 t ) dt
´
==´òò
–3
0
3 10
52
2 10t
600t –
2
´
éù ´
=êú
êúëû
23–53–
)103(10
–103600´´´=
1.8 – 0.9 0.9Ns==
41. (b) According to law of conservation of momentum,
10
100v 10 800
1000
=- ´´
ie, v = 0.
8 ms
–1
.

135. 131Laws of Motion
42
. (d)
5
3
sin=q
q
5
3
4
\ tan
4
3
=q Þ
3
tan 0.75
4
m= q==
43
. (b) The condition to avoid skidding, v =
rgm
= 101506.0´´ = 30 m/s.
44. (d) L
imiting friction =
0.5 2 10 10N´´=
The applied force is less than force of friction, therefore
the force of friction is equal to the applied force.
45. (d) Applying law of conservation of momentum
Momentum of bullet = Momentum of sand-bullet
system
m m 21
v m V mV
20 20 20
æö
=+=
ç÷
èø
46
. (d) Here 8.0tan
=q
where q is angle of repose
°==q
-
39)8.0(tan
1
The given angle of inclination is equal to the angle of
repose. So the 1 kg block has no tendency to move.
\ =qsinmgforce of fr iction
Þ T = 0
47.
(d) For 0.5 kg block, 6 = 0.5 a
48. (b) While moving down, when the lift is accelerating the
weight will be less and when the lift is decelerating the
weight will be more.
49. (b) Total momentum = j
ˆ
pi
ˆ
p2+
Magnitude
of total momentum
=
p5p5p)p2(
222
==+
This must
be equal to the momentum of the third part.
50. (c)
90°
30°
M
mg sin 30° ma cos 3
0°
30°
60°
ma (pseudo force)
a
°=°30sinmg30cosma
\
3
g
a=
51. (d
) Apparent weight when mass is falling down is given
by
W' m(g a)=-
\W ' 1 (1010) 0=´-=
52. (b) Reading of spring balance
2T =
12
12
4mm 4 5 1 10
mm 63
´´
==
+
kgf
53. (a
) a = 2v (given)
Þ
dv
v 2v
ds
=
or dv 2ds=
v
0.1
0
0.1
dv 2[s] 0.2==
ò
v 0.1 0.2-=
Þ
1
v 0.3ms
-
=
54. (c
) If W is the maximum weight, then
W = 2T cos 60°
or W = T = 20N
55. (c) The acceleration of both the blocks =
155
3xx
=
\ Force on
5
B 2x 10N
x
=´=
56. (
b) The maximum acceleration that can be given is a
\ a10g10g30
+=
Þ
2
ms20g2a
-
==
We know th
at
2
at
2
1
uts+=
\
20
102
a
s2
t
´
== = 1 s
57.
(a) Let the air resistance be F. Thenmg F ma+= Þ ]ga[mF-=
Here
2
ms12
5.2
30
a
-
==
58. (c) N30
1.0
)200(15.0
t
)uv(m
F =
-
=
-
=
59. (c
) Considering the two masses and the rope a system,
then
Initial net force =
[ ]25 (15 5) g5g-+=
Final net force = ()25 5 15 g 15g+-=éù
ëû
Þ (acceleration)
final
= 3 (acceleration)
initial
60. (d)
m
g sin q
mg cosq
F co
s q
qF sin
F
mg
q
N
From figure q+q=sinFcosmgN

136. 132 PHYSICS
61. (b) r/
vmmg
2
=m or
rgvm=
or s/m7)208.925.0(v =´´=
62. (a) Since wat
er does not fall down, therefore the velocity
of revolution should be just sufficient to provide
centripetal acceleration at the top of vertical circle. So,)16()}6.1(10{)rg(v=´== = 4 m/sec.
63. (
d) The speed at the highest point must be
rgv³
Now ()T/2rrvp=w=
\ rg)T/2(r>p or
÷
÷
ø
ö
ç
ç
è
æ
p<
p
<
g
r
2
gr
r2
T
\ sec4
8.9
4
2T =÷
ø
ö
ç
è
æ
p=
64. (a) From fi
gure,
qsinN =
r
mv
2
....... (i)
qcosN = mg...... (ii)
Dividing, we get
qtan =
rg
v
2
or q=
rg
v
tan
2
1-
65. (a) In t
he case of a body describing a vertical circle,
O
D
B
T
A
C
q
Mg
Mg sin q
Mg cos q
q
l
2
m
cosmgT
n
=q-
l
2
m
cosmgT
n
+q=
Tension is
maximum when cos q = +1 and velocity is
maximum
Both conditions are satisfied at q = 0º (i.e. at lowest
point B)
66. (b) Since surface (ice) is frictionless, so the centripetal
force required for skating will be provided by inclination
of boy with the vertical and that angle is given as
2
v
tanθ
rg
= where v is speed of skating & r is radius
of circle in which he moves.
67. (d) grv
max
m=
68. (b)
Let a and a' be the accelerations in both cases
respectively. Then for fig (a),
(a)
a
a
mg 2mg
TT
T – mg = ma … (1)
and 2mg – T 2ma= … (2)
Adding (1) and (2), we get
mg 3ma=
\
g
a
3
=
For fig (b),
(b)
a¢a¢
mg
F = 2mg
T¢T¢
T ' mg ma '-= … (3)
and2mgT'0-= … (4)
Solving (3) and (4)
a'g=
\
g
a
3
= and a'g=
69. (c) Linear momentum,
2
P a bt=+
dP
2bt
dt
= (on differe
ntiation)
\ Rate of change of momentum, dP
t
dt
µ
By 2nd law of motion,
dP
F
dt
µ
\ Ftµ
70. (d) For equilibrium of all 3 masses,
T
3
= (m
1
+ m
2
+ m
3
)a or
3
123
T
a
mmm
=
++

137. 133Laws of Motion
For
equilibrium of m
1
& m
2
2 12
T (m m ) .a=+
or,
1 23
2
123
(m m )T
T
mmm
+
=
++
Given m
1
= 10
kg, m
2
= 6 kg, m
3
= 4 kg, T
3
= 40 N
\
2
(10 6).40
T 32N
1064
+
==
++
71.
(a) Velocity of ball after dropping it from a height of
5 m
10 m/sec 20 m/sec
(using v
2
= u
2
+ 2gh)
v
2
= 0
+ 2 × 10 × 5 Þ v = 10 m/s
Velocity gained by ball by force exerted by bat
0 = u
2
– 2gh
u
2
= 2 × 10 × 20 or u = 20 m/s
Change in momentum = m(u + v)
= 0.4 (20 + 10) = 12 kg m/s
P
F
t
D
=
D
or
P
t
F
D
D=
12
t 0.12s ec
100
D==
72. (d)
F
1
= Force of friction between B and A
11
mg=m
= 0.25 × 100 × g
= 25 g newton
F
2
= Force of friction between (A + B) and surface
222
mg=m =m (mass of A and B) g
1 300
(100 200)g g 100g
33
=+== newton
\
12
FFF=+
= 25 g + 100 g = 25g = 125 × 10 N
\ F = 1250 N
73. (b) The rain drops falling vertically with velocity u do not
affect the momentum along the horizontal track. A
vector has no component in a perpendicular direction
Rain drops add to the mass of the car
Mass added in t sec = (mt) kg
Momentum is conserved along horizontal track.
Initial mass of car = M
Initial velocity of car = v
0
Final velocity of (car + water) = v
Mass of (car + water) after time t = (M + mt)
\ final momentum = initial momentum
0
(M mt)vMv+=
\
0
Mv
v
(M mt)
=
+
74. (d)
Let v
1
= velocity when height of free fall is h
1
v
2
= velocity when height of free rise is h
2
\
22
11
v u 2gh=+ for free fall
or
F
or free rise after impact on ground
2
22
0 v 2gh=- or
2
22
v 2gh=
Initial momentum = mv
1
F
inal momentum = mv
2
\ Change in momentum = m(v
1
– v
2
)
22
m(2gh gh)=-
75. (b
) T = tension is the string
\ Applied force F = 2T
T = F/2 … (i)
m
F
TT
T
f
For block
of mass m, force of friction due to surface f.
For sliding the blockT – f = force on the block = mass × acceleration
or acceleration of block
Tf
m
-
= . Put T from (i)
\Acceleration
F
f
2
m
-
=
76. (b)
smooth
d
rough
qm-q cosgsing
q
sin
g
°45°45
d
When surface is When surface is
smooth rough
d = q
2
1
1
(g sin )t
2
, d =
2
2
1
(gsingcos )t
2
q-mq
q
=
sing
d2
t
1
,
qm-q
=
cosgsing
d2
t
2
Ac
cording to question,
12
ntt=
qsing
d2
n =
qm-qcosgsing
d2
m, applicable here, is coefficient of kinetic friction as
the block moves over the inclined plane.
n =
k1
1
m-

÷
÷
ø
ö
ç
ç
è
æ
=°=°
2
1
45sin45cosQ
k
2
1
1
n
m-
= or
k
1m- =
2
n
1
or
2
k
n
1
1-=m

138. 134 PHYSICS
77. (d) Acceleration of block while sliding down upper half =
fsing ;
retardation of block while sliding down lower half = –
)cosgsing( fm-f
For the block to come to rest at the bottom, acceleration
in I half = retardation in II half.
)cosgsing(singfm-f-=f
Þ f=mtan2
Alternative method : According to work-energy
theorem, W = DK = 0
(Since initial and final speeds are zero)
\ Work done by friction + Work done by gravity
= 0
i.e., (µmgcos)mgsin0
2
- f + f=
l
l
µ
or cos sin
2
f=f or µ 2 tan=f
78. (c) Mass (m) = 0.3 kg Þ F = m.a = – 15 x
a = – x50x
3
150
x
3.0
15
-=
-
=
a = – 50 × 0.2 =
2
s/m10
79. (c) From free body diagram,
asinmg
cosg
a
a
ma
mg
mg cos
+ ma sin
a
a
N
a
a
For block to remain stationary,
mg sin ma cosa=a Þ a = atang
80. (a) as2uv
22
=- or s)g(2u0
k
22
m-=-
s10
2
1
2100
2
´´-´=-
s = 1000 m
81. (b) This is a standard formula and should be memorized.
2
g sin
a
I
1
MR
q
=
+
82. (a) N = m a sin q + mg cos q ......(1)
also m g sin q = m a cos q ......(2)
from (2) a = g tan q
q+
q
q
=\ cosmg
cos
sin
mgN
2
,
or
q
=
cos
mg
N
m
acosq
m
g
s
i
n
q
mgcosq
ma cosq
q
N
ma
mg
83. (a) m
B
g = m
s
m
A
g{Q m
A
g = m
s
m
A
g}
Þ m
B
= m
s
m
A
or m
B
= 0.2 × 2 = 0.4 kg
84. (c) ˆˆˆ
F = 6 i –8 j+10k,
r
| F | 36 64 100 10 2 N= ++= ( )
222
xyz
F FFF= ++Q
a = 1 ms
–2
Q F = ma
\
02
m 102
1
1
== kg
85. (d) Frictional force on the box f = mmg
\ Acceleration in the box
a = mg = 5 ms
–2
v
2
= u
2
+ 2as
Þ 0 = 2
2
+ 2 × (5) s
Þ s = –
2
5
w.r.t. belt
Þ distance = 0.4 m
86. (c) a = 1
m = 1000 kg
Total mass = (60 + 940) kg = 1000 kg
Let T be the tension in the supporting cable, then
T – 1000g = 1000 × 1
Þ T = 1000 × 11 = 11000 N
87. (b)
S/2 sin q
S/2 sin q
Smooth
Rough
S/2
S/2
q
q
For upper half of inclined plane v
2
= u
2
+ 2a S/2 = 2 (g sin q) S/2 = gS sin q
For lower half of inclined plane
0 = u
2
+ 2 g (sin q – m cos q) S/2
Þ – gS sin q = gS ( sinq – m cos q)
Þ 2 sin q = m cos q
Þ m= 2sin
cos
q
q
= 2 tan q
88. (b)
1
v gr 10 40 20 m s
-
= = ´=
89. (a)
4m / s ec
m
3
2 kg m
2
P
resultant
1 kg
x
y
8 m/sec
12 m/sec
m
1

139. 135Laws of Motion
P
resu
ltant
=
22
12 16+
= 144 256+ = 20
m
3
v
3
= 20 (moment
um of third part)
or, m
3
=
20
4
= 5 kg
90. (b) Mg
– T = Ma
T – mg = ma
m
mg mg
TT
M
On solving, we get
a =
(M m)g
Mm
-
+
and T =
2Mmg
Mm+
91. (c)Let velocity of bullet be v. If velocity of gun is V, then
mv + MV = 0
ÞV =
mv
M
-
As spring com
presses by d, so
2211
kd MV
22
=
or
2
21 1 mv
kdM
2 2M
æö
=
ç÷
èø
Þv =
d
kM
m
92. (b) Here, l
2
= 2l
1
As for a spring, force constant
1
k
l
µ
\12
1 2 12
111
,,kkk
l l ll
µ µµ
+
1 12
1
k ll
kl
+
=
and
2 12
2
k ll
kl
+
=
or
2
1
1
1
l
kk
l
éù
+=
êú
ëû
and
1
2
2
1
l
kk
l
éù
+=
êú
ëû
\
1
[12]3kkk= += [using (i)]
2
31
1
22
kkk
éù
==+
êú
ëû
[using (i)]
9
3. (a) R = mg cos q –
2
mV
r
V
car
R
O
mg
qmg cos q
Over bridge
q
When q decreases, cosq increases
i.e. R increases
94. (c) Impulse experienced by the body
= change in momentum = MV – (–MV) = 2MV.
95. (a) The force of friction on the chain lying on the table
should be equal to the weight of the hanging chain.
Let
r = mass per unit length of the chain
µ = coefficient of friction
l = length of the total chain
x = length of hanging chain
Now, µ(l – x) rg = xrg or µ(l – x) = x
or µl = (µ + 1)x or x = µl/(µ + 1)
0.25 0.25
0.2
(0.25 1) 1.25
ll
xl\= ==
+
0.2 20%
x
l
==
96. (a)
Masses of the pieces are 1, 1, 3 kg. Hence
222
(1 21) (1 21) (3 )V´ +´ =´
That
is,
72V= m/s
97. (c)
2
112
2
21 2
K pm
Km p
=´
[Q p = mv Þ
2
=
2
p
K
m
]
Hence,
11
22
11
42
pM
pM
= ==
98. (d) Work done in moving an object against gravitational
force depends only on the initial and final position of
the object, not upon the path taken. But gravitational
force on the body along the inclined plane is not same
as that along the vertical and it varies with angle of
inclination.
99. (b) On a rainy day, the roads are wet. Wetting of roads
lowers the coefficient of friction between the types
and the road. Therefore, grip on a road of car reduces
and thus chances of skidding increases.
100. (a) The force acting on the body of mass M are its weight
Mg acting vertically downward and air resistance F
acting vertically upward.
\ Acceration of the body ,
M
F
ga-=
Now M > m, th
erefore, the body with larger mass will
have great acceleration and it will reach the ground
first.
EXERCISE - 3
Exemplar Questions
1. (c) In a uniform translatory motion if all the parts of the
body moves with (same velocity in same straight line,so the velocity is constant.
A
v
v
v
The situation is shown in (figure) where a body A is in
unfirom translatory motion.

140. 136 PHYSICS
2. (b) Ac
cording to question we have to apply Newton's
second law of motion, in terms of force and change in
momentum.
We know that
dp
F
dt
=
As given that the me
ter scale is moving with uniform
velocity, hence
Force (F) = m × 0 = 0
No change in its velocity i.e., acceleration of it zero by
Newton's second law.
Hence, net or resultant force must act on body is zero.,rFt=´
rrr
so,
As all part
of the scale is moving with uniform velocity
and total force is zero, hence, torque will also be zero.
3. (c) As given that,
Mass of the ball = 150 g = 0.15 kg
ˆˆ
(3 4 ) m/su ij=+
r
ˆˆ(3 4 ) m/sv ij=-+
r
(Dp) Change in momentum
= Final momentum – Initial momentum
mv mu=-
rr
ˆˆ ˆˆ( ) (0.15)[ (3 4 ) (3 4 )]mvu ij ij= -= -+ -+
rr
ˆˆ(0.15)[ 6 8 ]ij= --
ˆˆ[0.15 6 0.15 8 ]ij=-´+´
ˆˆ[0.9 1.20 ]ij=-+
ˆˆ[0.9 1.2]p ijD=-+
Hence verifies o
ption (c).
4. (c) From previous solutionˆˆ ˆˆ(0.9 1.2 ) 0.9 1.2p ij ijD=- + =--
Magnitude
of
22
| | ( 0.9) ( 1.2)pD = - +-
0.81 1.44=+
2.25 1.5== kg-m s
–1
Verifies th
e option (c).
5. (d) By Newton's second law :
ext
dp
F
dt
=
r
As
ext
F
r
in law of conservation of momentum is zero.
If 0,0
extFdpp= =Þ = constant
He
nce, momentum of a system will remain conserve if
external force on the system is zero.
In case of collision between particle equal and opposite
forces will act on individual particle by Newton's third
la w.
So,
12 21
FF=-
rr
( 0)
ext
F=
r
Q
Total force on the system will be zero.
12 21dp dp
dt dt
=-
rr
or
12 21
dp dp=-
rr
12 21
( 0)dp dpÞ +=
rr
So prove the law of conservation of momentum and
verifies the option (d).
6. (c) Consider the adjacent diagram.
AB
B
O
E
S
N
P
2
P
2
P
1
–( )
W
P
1
The force on player is due to rate of change of
momentum. The direction of force acting on player will
be the same as the direction of change in momentum.
Let OA = P
1 i.e., Initial momentum of player northward
AB = P
2 i.e., Final momentum of player towards west.
Clearly, OB = OA + OB
Change in momentum
= P
2 – P
1
= AB – OA = AB + (–OA)
= Clearly resultant AR will be along south-west.
So, it will be also the direction of force on player.
7. (a) As given that, mass = 2 kg
p = 3 m/s, q = 4 m/s
2, r = 5 m/s
3
As given equation is
23
()x t pt qt rt=++
2()
23
dst
v p qt rt
dt
= =++
2
2
()
026
dv d xt
a q rt
dtdt
= = =++

2
2
( 2)
()
2 12
t
d xt
qr
dt
=
éù
=+êú
ëû
2 12qr=+
= 2 × 4 + 12 × 5
= 8 + 60 = 68 m/s
2
Force acting on body ()=F ma
r
= 2 × 68 = 136 N.
8. (b
) As given that mass = m = 5 kg
Acting force = ˆˆ
(3 4)NF ij=-+
r

141. 137Laws of Motion
Initial
velocity at t = 0,
ˆˆ(6 12 ) m/suij=-
r
Retardation,
2
ˆˆ34
ˆ m/s
55
F ij
a
m
æö
==-+ç÷
èø
r
As the final velocity along Y-component only. So its x-
component must be zero.
From ,
x xx
v u at=+ for X-co mponent only,,
ˆ3
ˆ06
5
i
it
-
=+
ˆ3
ˆ()6
5
=
i
ti
56
10s
3
t
´
==
t = 10 sec. Hence ver
ifies the option (b).
9. (b) As given that mass of the car = m
As car starts from rest, u = 0
Velocity acquired along east
ˆ()v vi=
r
Time (t) = 2 s.
We know that, v u at=+
ˆ02viaÞ =+´
ˆ
2
Þ=
v
ai
r
(Force, by engine is internal force)
ˆ
2
mv
Fmai==
rr
Hence, force acting on the car is
2
mv
towards east due
to force of friction is
ˆ
2
mv
i which moves the car in
eastward direction. Hence, force by engine is internal
force.
NEET/AIPMT (2013-2017) Questions
10. (d)
2m
3m
2mg
mg
mg
m v
m 2m 3m
T T' T "
F T T'
mg 2mg 3mg
6 mg
From fi
gure
F = 6 mg,
As speed is constant, acceleration a = 0
\ 6 mg = 6ma = 0, F = 6 mg
\ T = 5 mg , T
¢
= 3 mg
T² = 0
F
net
on block of mass 2 m
= T – T' – 2 mg = 0
ALTERNATE :
Q v = constant
s
o, a = 0, Hence, F
net
= ma = 0
11. (d) Given; speed = 10 m/s; radius r = 10 m
Angle made by the wire with the vertical
22
10
tan1
10 10
v
rg
q===
´
Þ 45
4
p
q= °=
12. (a)Let upthrust of air be F
a
then
For downward motion of balloonF
a
= mg – ma
mg – F
a
= ma
For upward motionF
a
– (m – Dm)g = (m – Dm)a
Therefore Dm =
2ma
ga+
13. (c)Change in momentum,
Dp = Fdtò
= Area of F-t graph
= ar of D – ar of + ar of
=
1
263243
2
´´-´+´ = 12 N-s
14.
(c) Acceleration
=
Net force inthe direction of motion
Total mass ofsystem
=
1 23
123
m g (m m )g
mmm
-m+
++
=
g
(1 2)
3
-m
(Qm
1
= m
2
= m
3
= m given)
15. (a) Acceleration of system
net
total
F
a
M
=
=
214 14
2m/s
4217
==
++
A
4kg
B
2kg
C
1kg
14N
The conta
ct force between A and B
= (m
B
+ m
C
) × a = (2 + 1) × 2 = 6N
16. (b) For the motion of both the blocks
m
1
a = T – m
k
m
1
g
m
2
g – T = m
2
a

142. 138 PHYSICS
m
1
m
2
m
k
mg
2
a
mg
1m
k
T
a
a =
2 k1
12
mg– mg
mm
m
+
m
2
g – T = (m
2
)
2 k1
12
m g– mg
mm
æö m
ç÷
+
èø
solving we get tension in the string
T =
12k
12
m m g(1 )g
mm
+m
+
17. (d) According to question, two stones experience same
centripetal force
i.e.
12CC
FF=
or,
22
12
mv 2mv
r (r / 2)
= or,
22
12
V 4V=
So, V
1
= 2V
2
i.e., n = 2
18. (a) Coefficient of static friction,
m
s
= tan 30° =
1
3
= 0.577 @ 0.6
S = ut +
21
at
2
4 =
1
2
a(4)
2
Þ a =
1
2
= 0.5
[Qs = 4m and t = 4s given]
a = gsinq – m
k
(g) cosq
Þ m
k
=
0.9
3
= 0.5
19. (d) To complete the loop a body must enter a vertical loop
of radius R with the minimum velocity v = 5gR.
20. (d) Net force on particle in uniform circular motion is cen-
tripetal force
2
mv
l
æö
ç÷
ç÷
èø
which is provided by tension in
string so the net force will be equal to tension i.e., T.
21. (a) m
mg
Before cutting the string
kx = T + 3 mg ...(i)
T = mg ...(ii)
Þ kx = 4mg
After cutting the string T = 0
A
4mg 3mg
a
3m
-
=
4mg
A
g
a
3
=
and B
mg
a g
m
==

143. WORK
Work Done by a Constant Force
Work done (W) by a force F
ur
in displacing a body through a
displacement x is given by
W = F.x
rr
= Fx cos q
F
x
Body
F cos q
q
Fsinq
Where q is the angle between the applied force F
r
and displacement
x
r
.
The S.I. unit of work is joule, CGS unit is erg and its dimensions
are [ML
2
T
–2
].
1 joule = 10
7
erg
(a) When q = 0° then W = Fx
(b) When q is between 0 and p/2 then
W = Fx cos q = positive
(c) When q = p/2 then W = Fx cos 90° = 0 (zero)
Work done by centripetal force is zero as in this case angle
q = 90°
(d)\ When q is between / 2 andpp then
W = Fx cos q = negative
Work Done by a Variable Force
When the force is an arbitrary function of position, we need the
techniques of calculus to evaluate the work done by it. The figure
shows F
x
as function of the position x. We begin by replacing
the actual variation of the force by a series of small steps.
The area under each segment of the curve is approximately equal
to the area of a rectangle. The height of the rectangle is a constant
value of force, and its width is a small displacement Dx. Thus, the
step involves an amount of work DW
n
= F
n
Dx
n
. The total work
done is approximately given by the sum of the areas of the
rectangles.
i.e., W »
n
FåDx
n
.
As the size of the steps is reduced, the tops of the rectangle more
closely trace the actual curve shown in figure. If the limit Dx ® 0,
which is equivalent to letting the number of steps tend to infinity,
the discrete sum is replaced by a continuous integral.
0
lim
n
nnx
x
W F x F dx
D®
= D=å ò
Thus, the work done by a force F
x
from an initial point A to final
point B is
n
A
x
ABx
x
W F dx
®
=ò
The work done by a variable force in displacing a particle from x
1
to x
2
ò
=
2
1
x
x
FdxW = area under force displacement graph
CAUTION : When we find work, we should be cautious about
the question, work done by which force? Let us take an example
to understand this point. Suppose you are moving a body up
without acceleration.
Work done by applied force
app appapp
W F x Fx·==
u ur u ur
Work done by gravitational force
x
mg
F
applied
ggrav
W F x mgx= =-
uuur uur uur
6
Work, Energy
and Power

144. 140 PHYSICS
ENERGY
It is the capaci
ty of doing work. Its units and dimensions are
same as that of work.
Potential Energy
The energy possessed by a body by virtue of its position or
configuration is called potential energy. Potential energy is
defined only for conservative forces. It does not exist for non
conservative forces.
(a) Elastic potential energy (Potential energy of a spring) :
Let us consider a spring, its one end is attached to a rigid
wall and other is fixed to a mass m. We apply an external
force
.ext
F
r
on mass m in the left direction, so that the spring
is compressed by a distance x.
l
m
F
ext
x
If spring constant is k, then energy stored in spring is given
by
P.E. of compressed spring = ½kx
2
Now if the external force is removed, the mass m is free to
move then due to the stored energy in the spring, it starts
oscillating
(b) Gravitational potential energy : When a body is raised to
some height, above the ground, it acquires some potential
energy, due to its position. The potential energy due to
height is called gravitational potential energy. Let us
consider a ball B, which is raised by a height h from the
ground.
h
Ground
mg
F
app
B
In doing so, we do work against gravity and this work is
stored in the ball B in the form of gravitational potential
energy and is given by
W = F
app
. h = mgh = gravitational potential energy...(i)
Further if ball B has gravitational P.E. (potential energy) U
o
at ground and at height h, U
h
, then
U
h
–U
o
=mgh ...(ii)
If we choose U
o
= 0 at ground (called reference point) then
absolute gravitational P.E of ball at height h is
U
h
= mgh ...(iii)
In general, if two bodies of masses m
1
and m
2
are separated
by a distance r, then the gravitational potential energy is
12
mm
UG
r
=-
Kinetic Energy
The energy possess
ed by a body by virtue of its motion is called
kinetic energy.
The kinetic energy E
k
is given by
E
k
= ½ mv
2
...(i)
Where m is mass of body, which is moving with velocity v. We
know that linear momentum (p) of a body which is moving with a
velocity v is given by
p = mv ...(ii)
So from eqs. (i) and (ii), we have
2
2
k
p
E
m
= ...(iii)
This is the relation bet
ween momentum and kinetic energy.
The graph between
k
E and p is a straight line
k
p 2mE=Q
p
E
k
The graph between
k
E and
p
1
is a rectangular hyperb
ola
E
k
1
p
The graph between E
k
and
1
m
is a rectangular hyperbola
E
k
1
m
p is constant
Keep in Memory
1. Work
done by the conservative force in moving a body in
a closed loop is zero.
Work done by the non-conservative force in moving a
body in a closed loop is non-zero.
2. If the momenta of two bodies are equal then the kinetic
energy of lighter body will be more.Q
221121
Em2Em2orpp ==

145. 141Work, Energy and Power
\
1
2
2
1
m
m
E
E
=
3. If t
he kinetic energies of two bodies are same then the
momentum of heavier body will be more.
21
EE=Q
2
1
2
1
m
m
p
p
=\
WORK-ENERG
Y THEOREM
Let a number of forces acting on a body of mass m have a resultant
force .extF
r
And by acting over a displacement x
(in the direction of .extF
r
), .extF
r
does work on the body, and there
by changing its velocity from u (initial velocity) to v (final velocity).
Kinetic energy of the body changes.
So, work done by force on the body is equal to the change in
kinetic energy of the body.
22
W ½ mv ½mu=-
This e
xpression is called Work energy (W.E.) theorem.
LAW OF CONSERVATION OF MECHANICAL ENERGY
The sum of the potential energy and the kinetic energy is called
the total mechanical energy.
The total mechanical energy of a system remains constant if
only conservative forces are acting on a system of particles and
the work done by all other forces is zero.
i.e., DK + DU =0
orK
f
– K
i
+ U
f
– U
i
=0
or K
f
+ U
f
=K
i
+ U
i
= constant
VARIOUS FORMS OF ENERGY : THE LAW OF
CONSERVATION OF ENERGY
Energy is of many types – mechanical energy, sound energy,
heat energy, light energy, chemical energy, atomic energy, nuclear
energy etc.
In many processes that occur in nature energy may be transformed
from one form to other. Mass can also be transformed into energy
and vice-versa. This is according to Einstein’s mass-energy
equivalence relation, E = mc
2.
In dynamics, we are mainly concerned with purely mechanical
energy.
Law of Conservation of Energy :
The study of the various forms of energy and of transformation
of one kind of energy into another has led to the statement of a
very important principle, known as the law of conservation of
energy.
"Energy cannot be created or destroyed, it may only be
transformed from one form into another. As such the total amount
of energy never changes".
Keep in Memory
1.Wo
rk done against friction on horizontal surface
= m mgx and work done against force of friction on inclined
plane = (mmg cosq) x where m = coefficient of friction.
2.If a body moving with velocity v comes to rest after
covering a distance ‘x’ on a rough surface having
coefficient of friction m, then (from work energy theorem),
2m gx = v
2
. Here retardation is
ag= -m
ur ur
3.Work done by a centripetal force is always zero.
4.Potential energy of a system decreases when a
conservative force does work on it.
5.If the speed of a vehicle is increased by n times, then its
stopping distance becomes n
2
times and if momentum is
increased by n times then its kinetic energy increases by
n
2
times.
6.Stopping distance of the vehicle
forceStopping
energ
yKinetic
=
7.Two vehicles of masses M
1
and M
2
are moving with
velocities u
1
and u
2
respectively. When they are stopped
by the same force, their stopping distance are in the ratio
as follows :
Since the retarding force F is same in stopping both the
vehicles. Let x
1
and x
2
are the stopping distances of vehicles
of masses M
1
& M
2
respectively, then
)M mass thestoppingin donework (F.x
)M mass thestoppingin donework (x.F
22
11
2
k
111
2
k22 2
E½Mu
E½Mu
==
....(i)
where u
1
a
nd u
2
are initial velocity of mass M
1
& M
2
respectively & final velocity of both mass is zero.
2
1
k
k
2
1
E
E
x
x
=Þ ....(ii)
Let us app
ly a retarding force F on M
1
& M
2
, a
1
& a
2
are
the decelerations of M
1
& M
2
respectively. Then from third
equation of motion
( )
22
v u 2ax=+ :
1
2
1
111
2
1
x2
u
axa2u0=Þ-= ....(iii a)
and
2
2
2
222
2
2
x2
u
axa2u0=Þ-= ....(iii b )
I
f t
1
& t
2
are the stopping time of vehicles of masses
M
1
& M
2
respectively, then from first equation of motion
(v = u+at)
1
1
1111
a
u
ttau0=Þ-= ....(iv a)
and
2
2
2222
a
u
ttau0=Þ-= ....(iv b)

146. 142 PHYSICS
Then by re
arranging equation (i), (iii) & (iv), we get
÷
÷
ø
ö
ç
ç
è
æ
´
÷
÷
ø
ö
ç
ç
è
æ
==
1
2
2
1
22
11
2
1
u
u
x
x
a/u
a/u
t
t
÷
÷
ø
ö
ç
ç
è
æ
=´=Þ
22
11
1
2
2
22
2
11
2
1
uM
uM
u
u
)uM(½
)uM(½
t
t
(a) If
2 1
2
1
21
M
M
t
t
uu =Þ=
(b) If
2 1
2
1
21
u
u
t
t
MM =Þ=
(c)If M
1
u
1
= M
2
u
2
Þ t
1
= t
2
and
2
1 11 212
2
2 12122
x (Mu)M xM
x M xM(Mu)
= ´ Þ=
(d) Co
nsider two vehicles of masses M
1
& M
2
respectively.
If they are moving with same velocities, then the ratio
of their stopping distances by the application of same
retarding force is given by
2
1
2
1
M
M
x
x
=
and let M
2
>
M
1
then x
1
< x
2
Þlighter mass will cover less distance then the
heavier mass
And the ratio of their retarding times are as follows :
2
1
2
1
M
M
t
t
=
i.e
2
1
2 1
t
t
x
x
=
8.If kinetic energy o
f a body is doubled, then its momentum
becomes
2 times,
2
kk
p
E pE
2m
= Þµ
9.If two bodies
of masses m
1
and m
2
have equal kinetic
energies, then their velocities are inversely proportional
to the square root of the respective masses. i.e.
22
11 22
11
m v m v then
22
=
1
2
2
1
m
m
v
v
=
10.(a) The spring
constant of a spring is inversely
proportional to the no. of turns i.e.
1
K
n
µ
or kn = const.
(
b) Greater the no. of turns in a spring, greater will be the
work done i.e. W µ n
(c) The greater is the elasticity of the spring, the greater
is the spring constant.
11. Spring constant : The spring constant of a spring is
inversely proportional to length i.e.,
l
1
Kµ or
Kl = constant.
(a)
If a spring is divided into n equal parts, the spring
constant of each part = nK.
(b) If spring of spring constant K
1
, K
2
, K
3
.......... are
connected in series, then effective force constant
......
K
1
K
1
K
1
K
1
321eff
+++=
(c) If sprin
g of spring constant K
1
, K
2
, K
3
........... are
connected in parallel, then effective spring constant
K
eff
= K
1
+ K
2
+ K
3
+.... .........
POWER
Power of the body is defined as the time rate of doing work by
the body.
The average power P
av
over the time interval Dt is defined by
av
W
P
t
D
=
D
...(i)
And the insta
ntaneous power P is defined by
0t
W dW
P Lim
t dtd®
D
==
D
...(ii)
Power is a s
calar quantity
The S.I. unit of power is joule per second
1 joule/sec = 1watt
The dimensions of power are [ML
2
T
–3
]
(.)..
dW d dS
P FS F Fv
dt dt dt
== ==
r
rrrr r
(force is constant over a small time interval)
So instantaneous power (or instantaneous rate of working) of aman depends not only on the force applied to body, but also onthe instantaneous velocity of the body.
Example 1.
A metre stick of mass 600 mg, is pivoted at one end and
displaced through an angle of 60º. The increase in its
potential energy is (g = 10 ms
–2
)
(a) 1.5 J (b) 15 J
(c) 150 J (d) 0.15 J
Solution : (a)
G
M
O
h
60º
x
l/2
l/2
G´
A´
A
The C.G. of stick rises from G to G´.
\ Increase in P.E. = mgh
= mg (l/2 – x) = (1 cos 60º )
2
mg
-
l
J5.1
2
1
1
2
1x10x6.0
=
ú
û
ù
ê
ë
é
-=

147. 143Work, Energy and Power
Example 2.
A block of mass M slides along the sides of a flat bottomed
bowl. The sides of the bowl are frictionless and the base
has a coefficient of friction 0.2. If the block is released
from the top of the side which is 1.5 m high, where will the
block come to rest if, the length of the base is 15 m ?
(a) 1 m from P (b) Mid point of flat part PQ
(c) 2 m from P (d) At Q
Solution : (b)
P.E. of the block at top of side = 1.5 mg.
This is wasted away in doing work on the rough flat part,
15 m
1.5 m
P Q
\ 1.5 mg = m mg.x or .m5.7
5.1
x =
m
=
i.e, the block
comes to rest at mid-point of PQ.
Example 3.
Fig. given below shows a smooth curved track terminating
in a smooth horizontal part. A spring of spring constant
400 N/m is attached at one end to the wedge fixed rigidly
with the horizontal part. A 40 g mass is released from rest
at a height of 4.9 m on the curved track. Find the maximum
compression of the spring.
4.9 m
Solution :
Acco
rding to the law of conservation of energy,
2
xk
2
1
hgm=
where x is
maximum compression.
2mgh
x
k
æö
\= ç÷
èø
or .cm8.9
)m/N400(
)m9.4(0)s/m8.9()kg4.0(2
x
2
=
ï
þ
ï
ý
ü
ï
î
ï
í
ì ´´´
=
Example 4.
The K.
E. of a body decreases by 19%. What is the
percentage decrease in momentum?
Solution :
As
p 2m K.E=
fifi
i i
K.E K.Epp
100 100
p K.E
81 100
100
100
10%
--
´=´
æö-
=´
ç÷
èø
=-
Example
5.
A particle of mass m is moving along a circular path of
constant radius R. The centripetal acceleration varies as
a = K
2
Rt
2
, where K is a constant and t is the time elapsed.
What is the power delivered to the particle by the force
acting on it?
Sol.For circular motion, a
c
= v
2
/ R here K
2
Rt
2
=
R
v
2
or v
2
=
K
2
R
2
t
2
Now,
2222
tRK.m
2
1
mv
2
1
KE ==
Work do
ne in time t = W = DK
(from work energy theorem)
222222
tRK
2
m
0)tRK(
2
m
K =-=D\
2 2 22dWm
Power K.R.2t mKRt
dt2
===
Exam
ple 6.
An electron of mass 9.0 × 10
–28
g is moving at a speed of
1000 m/sec. Calculate its kinetic energy if the electron
takes up this speed after moving a distance of 10 cm from
rest. Also calculate the force in kg weight acting on it.
Solution :
K.E. =
2 31 3211
m 9 10 (10 )
22
-
n= ´´ = 4.5
× 10
–25
J ;
From
2
n– u
2
=2as,
2
n= 2asQ u = 0
\
2 32
1
(10)
a
2s2 10
-
n
==
´
F = ma =
9 × 10
–31
(0.5 × 10
7
) N
24
4.5 10
9.8
-
´
= kg wt.

= 0.46 × 10
–24
kg wt.
Example 7.
The bob of a simple pendulum of length 1 m is drawn aside
until the string becomes horizontal. Find the velocity of
the bob, after it is released, at the equilibrium position.
Solution :
Wh
en the bob is raised from A to B the height through
which it is raised is the length of the pendulum.h = 1m
Taking A as the standard level.
P.E. at B = mgh = m × 9.8 × 1 = (9.8) m joule, where m is the
mass of the bob.
Since it is at rest at B it has no K.E.

148. 144 PHYSICS
When the bob
reaches A after it is released from B, its
energy at A is kinetic one. P.E. at A is zero.
If v be the velocity at A, from the law of conservation of
energy
K.E. at A = P.E. at B
1
2
mv
2
= mgh or v
2
= 2gh
Þ v =2gh 2 9.8 1 196 4.427 m/s=´´==
The bob
has a velocity 4.427 m/s at A.
COLLISION
Collision between two bodies is said to take place if either of
two bodies come in physical contact with each other or even
when path of one body is affected by the force exerted due to
the other.
Collision
Elastic collision (e = 1)
• Total ener
gy conserved
• Total momentum conserved
• K.E is conserved
Head on collision
(Impact parameter b = 0)
Inelastic collision
Oblique collision
(Impact parameter b 0)¹
Perfectly inelastic collision
(e = 0)
Inelastic collision (0 < e < 1)
• Total energy conserved
• Total m
omentum conserved
• K.E is not conserved
Head on
collision
Oblique
collision
(1) Elastic collision : The collision in which both the
momentum and kinetic energy of the system remains
conserved is called elastic collision.
Forces involved in the interaction of elastic collision are
conservative in nature.
(2) Inelastic collision : The collision in which only the
momentum of the system is conserved but kinetic energy
is not conserved is called inelastic collision.
Perfectly inelastic collision is one in which the two bodies
stick together after the collision.
Forces involved in the interaction of inelastic collision are
non-conservative in nature.
Coefficient of Restitution (or coefficient of resilience) :
It is the ratio of velocity of separation after collision to the
velocity of approach before collision.
i.e., e = | v
1
– v
2
|/ | u
1
– u
2
|
Here u
1
and u
2
are the velocities of two bodies before collision
and v
1
and v
2
are the velocities of two bodies after collision.
1. 0 < e < 1 (Inelastic collision)
Collision between two ivory balls, steel balls or quartz ball
is nearly elastic collision.
2. For perfectly elastic collision, e = 1
3. For a perfectly inelastic collision, e = 0
Oblique Elastic Collision :
When a body of mass m collides obliquely against a stationary
body of same mass then after the collision the angle between
these two bodies is always 90°.
Elastic Collision in One Dimension (Head on)
Let two bodies of masses M
1
and M
2
moving with velocities u
1
and u
2
along the same straight line, collide with each other. Let
u
1
>u
2
. Suppose v
1
and v
2
respectively are the velocities after
the elastic collision, then:
According to law of conservation of momentum
22112211
vMvMuMuM+=+ ...(1)
M
1
M
1u
1
v
1
u
2
v
2
M
2
M
2
Before collision After collision
Fr
om law of conservation of energy

22 22
11 22 11 22
11 11
Mu Mu Mv Mv
22 22
+=+ ...(2)

12 12
uu –(vv)-=- ...(3)
Relative ve
locity of a Relative velocity of a
body before collision body after collision
Solving eqs. (1) and (2) we get,
)MM(
uM2
)MM(
u)M–M(
v
21
22
21
121
1
+
+
+
=
...(4)
)MM(
uM2
)MM(
u)M–M(
v
21
11
21
212
2
+
+
+
=
...(5)
From eqns. (4) an
d (5), it is clear that :
(i) If M
1
= M
2
and u
2
= 0 then v
1
= 0 and v
2
= u
1
. Under this
condition the first particle comes to rest and the second
particle moves with the velocity of first particle after
collision. In this state there occurs maximum transfer of
energy.
(ii) If M
1
>> M
2
and (u
2
=0) then, v
1
= u
1
, v
2
= 2u
1
under this
condition the velocity of first particle remains unchanged
and velocity of second particle becomes double that of
first.
(iii) If M
1
<< M
2
and (u
2
= 0) then v
1
= –u
1
and v
2
=
»
1
2
1
u
M
M2
0
under this c
ondition the second particle remains at rest
while the first particle moves with the same velocity in the
opposite direction.
(iv) If M
1
= M
2
= M but u
2

¹ 0 then v
1
= u
2
i. e., the particles
mutually exchange their velocities.
(v) If second body is at rest i.e., u
2
= 0, then fractional decrease
in kinetic energy of mass M
1
, is given by
2
21
21
2
1
2
1
k
kk
)MM(
MM4
u
v
1
E
'EE
1
11
+
=-=
-

149. 145Work, Energy and Power
Inelastic Collision :
Let two bodies A and B collide inelastically. Then from law of
conservation of linear momentum
M
1
u
1
+ M
2
u
2
= M
1
v
1
+M
2
v
2
...(i)
velocity of separation
Coefficient of restitution
velocity ofapproach
æö
=-
ç÷
èø



e
)uu(
)vv(
21
21
-
-
-= ...(ii)
From eqns
.(i) and (ii), we have,
122
1 12
12 12
M eM M
(1 e)
v uu
MM MM
æ öæö-+
=+
ç ÷ç÷
++è øèø
...(iii)
2
21
12
1
21
1
2
u
MM
eMM
u
MM
M)e1(
v
÷
÷
ø
ö
ç
ç
è
æ
+
-
+
÷
÷
ø
ö
ç
ç
è
æ
+
+
= ...(iv)
Loss in
kinetic energy (–DE
k
) = initial K.E. – final K.E
Þ
ú
û
ù
ê
ë
é
+-
ú
û
ù
ê
ë
é
+=D-
2
22
2
11
2
22
2
11k
vM
2
1
vM
2
1
uM
2
1
uM
2
1
E
Þ
( )( )
2
21
2
21
21
k uu1e
MM
MM
2
1
E --
÷
÷
ø
ö
ç
ç
è
æ
+
=D- ...(v)
Negativ
e sign indicates that the final kinetic energy is less than
initial kinetic energy.
Perfectly Inelastic Collision
In this collision, the individual bodies A and B move with velocities
u
1
and u
2
but after collision move as a one single body with
velocity v.
So from law of conservation of linear momentum, we have
M
1
u
1
+M
2
u
2
=(M
1
+M
2
)V ...(i)
or
÷
÷
ø
ö
ç
ç
è
æ
+
+
=
21
2211
MM
uMuM
V ...(ii)
BA
u
1
M
1
u
2
M
2
(u > u
12
)
AB
V
M + M
12
Before collisi
on After
collision
And loss in kinetic energy, –DE
k
= total initial K.E
– total final K.E
2
21
2211
21
2
22
2
11
MM
uMuM
)MM(
2
1
uM
2
1
uM
2
1
÷
÷
ø
ö
ç
ç
è
æ
+
+
+-+=
or,
2
21
21
21
k )uu(
)MM(
MM
2
1
E -
+
=D- ...(iii)
Obliqu
e Collision :
This is the case of collision in two dimensions. After the collision,
the particles move at different angle.
A
B
Before collision
1u
0u
2= A
B
1v
2v
q
f
After collision
m
1
m
1
m
2
m
2
x axis
y axis
We will apply the principle of conservation of momentum in the
mutually perpendicular direction.
Along x-axis, m
1
u
1
= m
1
v
1
cosq + m
2
v
2
cosf
Along y-axis,0 = m
1
v
1
sinq - m
2
v
2
sinf
Keep in Memory
1.Suppo
se, a body is dropped from a height h
0
and it strikes
the ground with velocity v
0
. After the (inelastic) collision
let it rise to a height h
1
. If v
1
be the velocity with which the
body rebounds, then the coefficient of restitution.
1
2
11 00
v 2gh
e
v 2ghæö
==
ç÷
èø
=
2
1
0
h
h
÷
÷
ø
ö
ç
ç
è
æ

h
0
v
0 v
1
h
1
2.If after n collisions with the ground, the velocity is v
n
and
the height to which it rises be h
n
, then
2
1
o
n
o
nn
h
h
v
v
e
÷
÷
ø
ö
ç
ç
è
æ
==
3.When a b
all is dropped from a height h on the ground,
then after striking the ground n times , it rises to a height
h
n
= e
2n
h
o
where e = coefficient of restitution.
4.If a body of mass m moving with velocity v, collides
elastically with a rigid ball, then the change in the
momentum of the body is 2 m v.
(i) If the collision is elastic then we can conserve the
energy as
2
22
2
11
2
11 vm
2
1
vm
2
1
um
2
1
+=
(ii) If
two particles having same mass and moving at right
angles to each other collide elastically then after the
collision they also move at right angles to each other.
(iii) If a body A collides elastically with another body of
same mass at rest obliquely, then after the collision
the two bodies move at right angles to each other, i. e.
(q + f) =
2
p
5.In an elastic co
llision of two equal masses, their kinetic
energies are exchanged.

150. 146 PHYSICS
6.When two bo
dies collide obliquely, their relative velocity
resolved along their common normal after impact is in
constant ratio to their relative velocity before impact
(resolved along common normal), and is in the opposite
direction.
1
m
2m
1
q
2
q
b
2m
1
m
1
u
2
u
v
1
v
2
After collision
Before collision
1122
12
v sin v sin
e
u sin u sin
q-q
=-
a-b
Example 8.
A body of mass m m
oving with velocity v collides head on
with another body of mass 2m which is initially at rest.
What will be the ratio of K.E. of colliding body before and
after collision?
Solution :
mv + 2m × 0 = mv
1
+2mv
2
;
v = v
1
+ 2v
2
or
2
vv
v
1
2
-
= ;
2
2
2
1
2
mv2
2
1
mv
2
1
mv
2
1
+=
2
12
1
2
2
2
1
2
2
vv
2vv2vv÷
ø
ö
ç
è
æ-
+=+=
2
vv2vv
vv
1
22
12
1
2 -+
+= or
1
22
1
2
1
2
vv2vvv2v2-++=
or 0vvv2v3
2
1
2
1
=--;
3
v
v
1
-=
collisionafterbodycollidingof.E.K
collisionbeforebodycollidingof.E.K

2
2
3
v
m
2
1
mv
2
1
÷
ø
ö
ç
è
æ
= = 9 : 1
Example 9.
A bu
llet of mass m moving horizontally with velocity v hits
a block of wood of mass M, resting on a smooth horizontal
plane. Find the fraction of energy of the bullet dissipated
in the collision itself (assume collision to be inelastic).
Solution :
Applying the law of conservation of momentum, we have,
m v = (m + M) v
1
÷
ø
ö
ç
è
æ
+
=
Mm
mv
v
1
Loss of K.E.
22
1
11
mv (m M)v
22
= -+
2
2
mM
vm
)Mm(
2
1
vm
2
1
÷
ø
ö
ç
è
æ
+
+-=
ú
û
ù
ê
ë
é
+
=
ú
û
ù
ê
ë
é
+
-=
Mm
M
vm
2
1
Mm
m
1vm
2
1
22
Fraction of K.E. dissi
pated
.E.KInitial
.E.KofLoss
= ÷
ø
ö
ç
è
æ
+
=
Mm
M
Example 10.
A smooth sphere of
mass 0.5 kg moving with horizontal
speed 3 m/s strikes at right angles a vertical wall and bounces
off the wall with horizontal speed 2 m/s. Find the coefficient
of restitution between the sphere and the wall and the
impulses exerted on the wall at impact.
Solution :
At impact
Just after impact
Just before impact
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/2m/s
3m/s
J J
e = separation speed : approach speed = 2 : 3
Therefore the coefficient of restitution is 2/3.
Using impulse = change in momentum for the sphere we
have : = 0.5 × 2 – 0.5 (–3) = 2.5
The equal and opposite impulse acting on the wall is
therefore 2.5 N s.

151. 147Work, Energy and Power

152. 148 PHYSICS
1.The mag
nitude of work done by a force :
(a) depends on frame of reference
(b) does not depend on frame of reference
(c) cannot be calculated in non-inertial frames.
(d) both (a) and (b)
2.Work done by a conservative force is positive if
(a) P.E. of the body increases
(b) P.E. of the body decreases
(c) K.E. of the body increases
(d) K.E. of the body decreases
3.A vehicle is moving with a uniform velocity on a smooth
horizontal road, then power delivered by its engine must be
(a) uniform (b) increasing
(c) decreasing (d) zero
4.Which of the following force(s) is/are non-conservative?
(a) Frictional force(b) Spring force
(c) Elastic force (d) All of these
5.A ball of mass m and a ball B of mass 2m are projected with
equal kinetic energies. Then at the highest point of their
respective trajectories.
(a) P.E. of A will be more than that of B
(b) P.E of B will be more than that of B
(c) P.E of A will be equal to that of B
(d) can’t be predicted.
6.In case of elastic collision, at the time of impact.
(a) total K.E. of colliding bodies is conserved.
(b) total K.E. of colliding bodies increases
(c) total K.E. of colliding bodies decreases
(d) total momentum of colliding bodies decreases.
7.The engine of a vehicle delivers constant power. If the
vehicle is moving up the inclined plane then, its velocity,
(a) must remain constant
(b) must increase
(c) must decrease
(d) may increase, decrease or remain same.
8.A ball projected from ground at a certain angle collides a
smooth inclined plane at the highest point of its trajectory.
If the collision is perfectly inelastic then after the collision,
ball will
(a) come to rest
(b) move along the incline
(c) retrace its path.
9.The vessels A and B of equal volume and weight are
immersed in water to depth h. The vessel A has an opening
at the bottom through which water can enter. If the work
done in immersing A and B are W
A
and W
B
respectively,
then
(a)W
A
= W
B
(b) W
A
< W
B
(c)W
A
> W
B
(d) W
A

>
< WW
B
10.A metallic wire of length L metre extends by l metre when
stretched by suspending a weight Mg from it. The mechanical
energy stored in the wire is
(a) 2 Mg l (b) Mg l
(c)
Mg
2
l
(b)
Mg
4
l
11.A body of mass m accelerates uniformly from rest to v
1
in
time t
1
. As a function of t, the instantaneous power delivered
to the body is
(a)
1
2
mt
t
n
(b)
2
1
1
mt
t
n
(c)
2
1
1
mt
t
n
(d)
2
1
2
1
mt
t
n
12.A block is acted upon by a force, which is inversely
proportional to the distance covered (x). The work done
will be proportional to
(a)x (b) x
1/2
(c)x
2
(d) None of these
13.A small body is projected in a direction inclined at 45º to the
horizontal with kinetic energy K. At the top of its flight, its
kinetic energy will be
(a) Zero (b) K/2
(c) K/4 (d) K/
2
14.A motor cycle is mov
ing along a straight horizontal road
with a speed v
0
. If the coefficient of friction between the
tyres and the road is m, the shortest distance in which the
car can be stopped is
(a)
2
0
v
2gm
(b)
2
v
m
(c)
2
0v
g
æö
ç÷
mèø
(d)
0
v
gm
15.Consider the following two statement:
I. Linear momentum of a system of particles is zero.
II. Kinetic energy of a system of particles is zero.
Then
(a) I implies II but II does not imply I.
(b) I does not imply II but II implies I.
(c) I implies II and II implies I.
(d) I does not imply II and II does not imply I.
16.Which of the following must be known in order to determine
the power output of an automobile?
(a) Final velocity and height
(b) Mass and amount of work performed
(c) Force exerted and distance of motion
(d) Work performed and elapsed time of work

153. 149Work, Energy and Power
17.The work done in stretching a spring of force constant k
from length
1
l and
2
l is
(a) )(k
2
1
2
2
ll- (b) )(k
2
1 2
1
2
2
ll-
(c) )(k
12
ll- (d) )(
2
k
12
ll+
18.If the for
ce acting on a body is inversely proportional to its
velocity, then the kinetic energy acquired by the body in
time t is proportional to
(a)t
0
(b) t
1
(c)t
2
(d) t
4
19.The engine of a truck moving along a straight road delivers
constant power. The distance travelled by the truck in time
t is proportional to
(a)t (b) t
2
(c)
t (d) t
3/2
20.A bullet o
f mass ‘a’ and velocity ‘b’ is fired into a large
block of wood of mass ‘c’. The bullet gets embedded intothe block of wood. The final velocity of the system is
(a)
b
c
ab
´
+
(b)
ab
a
c
+
´
(c)
a
b
ac
´
+
(d)
ac
b
a
+
´
21.A ball is dropped from a height h. If the coefficient of
restitution be e, then to what height will it rise after jumping
twice from the ground?
(a) e h/2 (b) 2 e h
(c) e h (d) e
4
h
22.A ball of mass m moving with a constant velocity strikes
against a ball of same mass at rest. If e = coefficient of
restitution, then what will be the ratio of velocity of two
balls after collision?
(a)
1e
1e
-
+
(b)
e1
e1
-
+
(c)
1e
1e
+
-
(d)
2e
e1
+
-
23.Which o ne of the following physical quantities is
represented by the shaded area in the given graph?
F
o
r
c
e
Distance
(a) Torque (b) Impulse
(c) Power (d) Work done
24.A particle of mass m
1
moving with velocity v collides with
a mass m
2
at rest, then they get embedded. Just after
collision, velocity of the system
(a) increases (b) decreases
(c) remains constant(d) becomes zero
25.A mass m
1
moves with a great velocity. It strikes another
mass m
2
at rest in a head on collision. It comes back along
its path with low speed, after collision. Then
(a)m
1
> m
2
(b) m
1
< m
2
(c)m
1
= m
2
(d) cannot say
1.A particle describe a horizontal circle of radius 0.5 m with
uniform speed. The centripetal force acting is 10 N. The
work done in describing a semicircle is
(a) zero (b) 5 J
(c) 5 p J (d) 10 p J
2.A cord is used to lower vertically a block of mass M,
a distance d at a constant downward acceleration of g/4.
The work done by the cord on the block is
(a)
d
Mg
4
(b)
d
3Mg
4
(c)
d
3Mg
4
- (d) Mg d
3.A boy pus
hes a toy box 2.0 m along the floor by means of a
force of 10 N directed downward at an angle of 60º to the
horizontal. The work done by the boy is
(a) 6 J (b) 8 J
(c) 10 J (d) 12 J
4.A particle moving in the xy plane undergoes a displacement
of )j
ˆ
3i
ˆ
2(s+=
r
while a constant force )j
ˆ
2i
ˆ
5(F+=
r
N
acts on th
e particle. The work done by the force F is
(a) 17 joule (b) 18 joule
(c) 16 joule (d) 15 joule
5.A simple pendulum 1 metre long has a bob of 10 kg. If the
pendulum swings from a horizontal position, the K.E. of the
bob, at the instant it passes through the lowest position of
its path is
(a) 89 joule (b) 95 joule
(c) 98 joule (d) 85 joule
6.A particle moves under the effect of a force F = cx from
x = 0 to x = x
1
, the work done in the process is
(a)
2
1
cx (b)
2
1
1
cx
2
(c)
2
1
2 cx (d) zero

154. 150 PHYSICS
7.A motor of 100 H.P.
moves a load with a uniform speed of 72
km/hr. The forward thrust applied by the engine on the car
is
(a) 1111 N (b) 3550 N
(c) 2222 N (d) 3730 N
8.Two bodies A and B having masses in the ratio of 3 : 1
possess the same kinetic energy. The ratio of linear
momentum of B to A is
(a) 1 : 3 (b) 3 : 1
(c)
1:3 (d)3:1
9.When a U
238
nucleus, o riginally at rest, decays by emitting
an a-particle, say with speed of v m/sec, the recoil speed of
the residual nucleus is (in m/sec.)
(a) – 4 v/234 (b) – 4 v/238
(c) 4 v/238 (d) – v/4
10.Calculate the K.E and P.E. of the ball half way up, when a
ball of mass 0.1 kg is thrown vertically upwards with an
initial speed of 20 ms
–1
.
(a) 10 J, 20 J (b) 10 J, 10 J
(c) 15 J, 8 J (d) 8 J, 16 J
11.A spring of force constant 800 N/m has an extension of 5
cm. The work done in extending it from 5 cm to 15 cm is
(a) 16 J (b) 8 J
(c) 32 J (d) 24 J.
12.If the linear momentum is increased by 5%, the kinetic energy
will increase by
(a) 50% (b) 100%
(c) 125% (d) 10%
13.A cord is used to lower vertically a block of mass M, through
a distance d at a constant downward acceleration of g/8.
Then the work done by the cord on the block is
(a) Mg d/8 (b) 3 Mg d/8
(c) Mg d (d) – 7 mg d/8
14.A force
F (5i 3j 2k)N= ++
rrrr
is applied over a particle which
displaces it from its origin to the point r (2i j)m.=-
rrr
The
work done o
n the particle in joule is
(a) +10 (b) +7
(c) –7 (d) +13
15.A spring of spring constant 5 × 10
3
N/m is stretched initially
by 5 cm from the unstretched position. Then the workrequired to stretch it further by another 5 cm is(a) 18.75 J (b) 25.00 J
(c) 6.25 J (d) 12.50 J
16.Two solid rubber balls A and B having masses 200 &400 gm respectively are moving in opposite direction withvelocity of A equal to 0.3 m/sec. After collision the twoballs come to rest when the velocity of B is(a) 0.15 m/sec (b) 1.5 m/sec
(c) –0.15 m/sec (d) None of these
17.A bomb of mass 9 kg explodes into the pieces of masses3 kg and 6 kg. The velocity of mass 3 kg is 16 m/s. Thekinetic energy of mass 6 kg in joule is(a) 96 (b) 384
(c) 192 (d) 768
18.A long string is stretched by 2 cm and the potential energyis V. If the spring is stretched by 10 cm, its potential energywill be(a) V / 25 (b) V/5
(c) 5 V (d) 25 V
19.When the kinetic energy of a body is increased to threetimes, then the momentum increases(a) 6 times (b) 1.732 times
(c)
2 times (d) 2 times
20.Two
bodies of masses 2 m and m have their KE in the ratio
8 : 1. What is the ratio of their momenta ?(a) 8 : 1 (b) 4 : 1
(c) 2 : 1 (d) 1 : 1
21.A body of mass 5 kg initially at rest explodes into 3fragments with mass ratio 3 : 1 : 1. Two of fragments each ofmass ‘m’ are found to move with a speed 60 m/s in mutuallyperpendicular directions. The velocity of third fragment is
(a)
602 (b)203
(c)102 (d)202
22.A machine, which is 75% efficient, uses 12 J of energy in
lifting up a 1 kg mass through a certain distance. The mass
is then allowed to fall through that distance. The velocity
at the end of its fall is (in m/s)
(a)
24 (b)12
(c)18 (d)9
23.A body accelerates uniformly from rest to a velocity of 1
ms
–1
in 15 seconds. The kinetic energy of the body will be
9
2
J when 't' is equal to [Take
mass of body as 1 kg]
(a) 4s (b) 8s
(c) 10s (d) 12s
24.A machine gun fires a bullet of mass 40 g with a velocity
1200 ms
–1
. The man holding it can exert a maximum force of
144 N on the gun. How many bullets can he fire per second
at the most?
(a)2 (b) 4
(c)1 (d) 3
25.A crane is used to lift 1000 kg of coal from a mine 100 m
deep. The time taken by the crane is 1 hour. The efficiency
of the crane is 80%. If g = 10 ms
–2
, then the power of the
crane is
(a) 10
4
W (b) 10
5
W
(c)
W
836
10
4
´
(d) W
836
10
5
´
26.In figure, a carriage P is
pulled up from A to B. The relevant
coefficient of friction is 0.40. The work done will be(a) 10 kJ
(b) 23 kJ
(c) 25 kJ
P 50
kg
50
30 m
B
A
q
C
m
(d) 28 kJ

155. 151Work, Energy and Power
27.A neutron with velocity v strikes a stationary deuterium
atom, its K.E. changes by a factor of
(a)
16
15
(b)
2
1
(c )
1
2
(d) None of th
ese
28.A body moves a distance of 10 m along a straight line under
the action of a force of 5 newtons. If the work done is 25
joules, the angle which the force makes with the direction
of motion of body is
(a) 0º (b) 30º
(c) 60º (d) 90º
29.A sphere of mass 8m collides elastically (in one dimension)
with a block of mass 2m. If the initial energy of sphere is E.
What is the final energy of sphere?
(a) 0.8 E (b) 0.36 E
(c) 0.08 E (d) 0.64 E
30.Johnny and his sister Jane race up a hill. Johnny weighs
twice as much as jane and takes twice as long as jane to
reach the top . Compared to Jane
(a) Johnny did more work and delivered more power.
(b) Johnny did more work and delivered the same amount
of power.
(c) Johnny did more work and delivered less power
(d) Johnny did less work and johnny delivered less power.
31.A body of mass m moving with velocity v makes a head on
elastic collision with another body of mass 2m which in
initially at rest. The loss of kinetic energy of the colliding
body (mass m ) is
(a)
2
1
of its initial kinetic energy
(b)
9
1
of its initial kinetic energy
(c)
9
8
of its initial kinetic energy
(d)
1
4
of its initial kinetic energy
32.In the non-relativistic regime, if the momentum, is increase
by 100% , the percentage increase in kinetic energy is
(a) 100 (b) 200
(c) 300 (d) 400
33.A body is dropped from a height of 20m and rebounds to a
height 10m. The loss of energy is
(a) 10% (b) 45%
(c) 50% (d) 75%
34.A moving body with a mass m
1
and velocity u strikes a
stationary body of mass m
2
. The masses m
1
and m
2
should
be in the ratio m
1
/m
2
so as to decrease the velocity of the
first body to 2u/3 and giving a velocity of u to m
2
assuming
a perfectly elastic impact. Then the ratio m
1
/m
2
is
(a)5 (b)5/1
(c)25/1 (d) 25.
35.A body of mass m is suspended from a massless spring of
natural length l. It stretches the spring through a vertical
distance y. The potential energy of the stretched spring is
(a) )y(mg+l (b) )y(mg
2
1
+l
(c)mgy
2
1
(d)mgy
36.Figure here shows the frictional force versus displacement
for a particle in motion. The loss of kinetic energy in
travelling over s = 0 to 20 m will be
15
10
5
0
0 5 10 20
x(m)
f(N)
(a) 250 J (b) 2 00 J
(c) 150 J (d) 10 J
37.Ten litre of water per second is lifted from a well through 10
m and delivered with a velocity of 10 ms
–1
. If
g = 10 ms
–2
, then the power of the motor is
(a) 1 kW (b) 1.5 kW
(c) 2 kW (d) 2.5 kW
38.A nucleus ruptures into two nuclear parts which have their
velocity ratio equal to 1:2. The ratio of their respective
nuclear sizes (nuclear radii )is
(a) 1 : 2 (b)2:1
(c) 1 : 2
1/3
(d) 1
: 8
39.The rest energy of an electron is 0.511 MeV. The electron is
accelerated from rest to a velocity 0.5 c. The change in its
energy will be
(a) 0.026 MeV (b) 0.051 MeV
(c) 0.08 MeV (d) 0.105 MeV
40.A one-ton car moves with a constant velocity of
15 ms
–1
on a rough horizontal road. The total resistance to
the motion of the car is 12% of the weight of the car. The
power required to keep the car moving with the same
constant velocity of 15ms
–1
is
[Take g = 10 ms
–2
]
(a) 9 kW (b) 18 kW
(c) 24 kW (d) 36 kW
41.Hail storms are observed to strike the surface of the frozen
lake at 30
0
with the vertical and rebound at 60
0
with the
vertical. Assume contact to be smooth, the coefficient of
restitution is
(a)
1
e
3
= (b)
1
e
3
=
(c)e3= (d) e = 3

156. 152 PHYSICS
42.A body s
tarts from rest and acquires a velocity V in time T.
The work done on the body in time t will be proportional to
(a)
V
t
T
(b)
2
2V
t
T
(c)
2
2
V
t
T
(d)
2
2
2
V
t
T
43.A ball is allowed to
fall from a height of 10 m. If there is 40%
loss of energy due to impact, then after one impact ball will
go up to
(a) 10 m (b) 8 m
(c) 4 m (d) 6 m
44.The potential energy of a conservative system is given by
U = ay
2
– by, where y represents the position of the particle
and a as well as b are constants. What is the force acting on
the system ?
(a) – ay (b) – by
(c) 2ay – b (d) b – 2ay
45.An automobile engine of mass M accelerates and a constant
power p is applied by the engine. The instantaneous speed
of the engine will be
(a)
1/2
[Pt/M] (b)
1/2
[2Pt/M]
(c)
1/2
[Pt /2M] (d)
1/2
[Pt / 4M]
46.A bullet fired into a fixed target loses half of its velocity
after penetrating 3 cm. How much further it will penetrate
before coming to rest assuming that it faces constant
resistance to motion ?
(a) 2.0 cm (b) 3.0 cm
(c) 1.0 cm (d) 1.5 cm
47.A bomb of mass 16kg at rest explodes into two pieces of
masses 4 kg and 12 kg. The velolcity of the 12 kg mass is
4ms
–1
. The kinetic energy of the other mass is
(a) 144 J (b) 288 J
(c) 192 J (d) 96 J
48.Given that a force
ˆ
F acts on a body for time t, and displaces
the body by
ˆ
d. In which of the following cases, the speed
of the body must not increase?(a) F > d (b) F < d
(c)
ˆˆ
Fd= (d)
ˆˆ
Fd^
49.A body is attached to the lower end of a vertical helical
spring and it is gradually lowered to its equilibrium position.This stretches the spring by a length x. If the same bodyattached to the same spring is allowed to fall suddenly,what would be the maximum stretching in this case?(a)x (b) 2 x
(c) 3 x (d) x/2
50.A bag of mass M hangs by a long thread and a bullet (massm) comes horizontally with velocity V and gets caught inthe bag. Then for the combined (bag + bullet) system
(a)
mvM
momentum
Mm
=
+
(b)
2
mV
kinetic energy
2
=
(c)
mV(M m)
momentum
M
+
=
(d)
22
mv
kinetic energy
2(M m)
=
+
51.A particle, initially at res
t on a frictionless horizontal surface,
is acted upon by a horizontal force which is constant in
magnitude and direction. A graph is plotted of the work
done on the particle W, against the speed of the particle v.
If there are no other horizontal forces acting on the particle,
the graph would look like
(a)
W
v
(b)
W
v
(c)
W
v
(d)
W
v
52.A particle of mass m moving eastward with a speed v
collides with another particle of the same mass moving
northwards with the same speed. If two particles coalesce
on collision, the new particle of mass 2 m will move in the
north-east direction with a velocity
(a) v / 2 (b)
v2
(c)v/2 (d) None of these
53.A sm
all block of mass m is kept on a rough inclined surface
of inclination q fixed in an elevator. The elevator goes up
with a uniform velocity v and the block does not slide on
the wedge. The work done by the force of friction on the
block in time t as seen by the observer on the inclined p
lane will be
(a) zero (b) mgvt cos
2
q
(c) mgvt sin
2
q (d) mgvt sin 2q
54.A nucleus moving with a velocity
v
r
emits an a-particle.
Let the vel
ocities of the a-particle and the remaining nucleus
be
1
v
r
and
2
v
r
and their masses be m
1
and m
2
.
(a)v
r
,
1
v
r
and
2
v
r
must be parallel to each other
(b) None of the two of v
r
,
1
v
r
and
2
v
r
should be parallel to
each other
(c)
11 22
mv mv+
rr
must be parallel to
12
(m m )v+
r
.
(d) None of these
55.A s
hell is fired from a cannon with a velocity V at an angle
q with the horizontal direction. At the highest point in its
path, it explodes into two pieces of equal masses. One ofthe pieces retraces its path to the cannon. The speed of theother piece immediately after the explosion is(a) 3 V cos q (b) 2 V cos q
(c)
3
V cos
2
q (d) V cos q

157. 153Work, Energy and Power
56.The bob A of a simple pendulum is released when the string
makes an angle of 45º with the vertical. It hits another bob
B of the same material and same mass kept at rest on the
table. If the collision is elastic
45º
A
B
(a) both A and B rise to the same height
(b) both A and B come to rest at B
(c) both A and B move with the velocity of A
(d) A comes to rest and B moves with the velocity of A
57.Two masses m
a
and m
b
moving with velocities v
a
and v
b
in
opposite direction collide elastically and after the collision
m
a
and m
b
move with velocities V
b
and V
a
respectively.
Then the ratio m
a
/m
b
is
(a)
ab
ab
VV
VV
-
+
(b)
ab
a
mm
m
+
(c)1 (d)
1
2
58.A particle moves in a straight line with retardation
proportional to its displacement. Its loss of kinetic energy
for any displacement x is proportional to
(a)x (b) e
x
(c)x
2
(d) log
e
x
59.A particle of mass
1
mmoving with velocity v strikes with a
mass m
2
at rest, then the condition for maximum transfer of
kinetic energy is
(a)m
1
>> m
2
(b) m
2
>> m
2
(c)m
1
= m
2
(d) m
1
= 2m
2
60.Two blocks of masses 10 kg and 4 kg are connected by a
spring of negligible mass and placed on a frictionless
horizontal surface. An impulse gives a velocity of 14 m/s to
the heavier block in the direction of the lighter block. The
velocity of the centre of mass is
(a) 30 m/s (b) 20 m/s
(c) 10 m/s (d) 5 m/s
61.A mass m is moving with velocity v collides inelastically
with a bob of simple pendulum of mass m and gets embedded
into it. The total height to which the masses will rise after
collision is
(a)
g8
v
2
(b)
g4
v
2
(c)
g2
v
2
(d)
g
v2
2
62.A motor
drives a body along a straight line with a constant
force. The power P developed by the motor must vary with
time t according to
(a)
P
t
(b)
P
t
(c)
P
t
(d)
P
t
63.A glass marble dropped from a certain height above the
horizontal surface reaches the surface in time t and thencontinues to bounce up and down. The time in which themarble finally comes to rest is
(a)te
n (b)te
2
(c)
e
t
1e
1+éù
êú
-ëû
(d) ú
û
ù
ê
ë
é
+
-1
e1
e
t
64.A weight sus
pended from the free end of a vertically
hanging spring produces an extension of 3 cm. The spring
is cut into two parts so that the length of the longer part is
3
2
of the o
riginal length, If the same weight is now
suspended from the longer part of the spring, the extension
produced will be
(a) 0.1 cm (b) 0.5 cm
(c) 1 cm (d) 2 cm
65.A 10 m long iron chain of linear mass density 0.8 kg m
–1
is
hanging freely from a rigid support. If g = 10 ms
–2
, then the
power required to left the chain upto the point of support in
10 second
(a) 10 W (b) 20W
(c) 30 W (d) 40 W
66.A smooth sphere of mass M moving with velocity u directly
collides elastically with another sphere of mass m at rest.
After collision, their final velocities are V and v respectively.
The value of v is
(a)
m
uM2
(b)
M
um2
(c)
M
m
1
u2
+
(d)
2u
M
1
m
+
67.The kinetic ene
rgy of particle moving along a circle of radius
R depends upon the distance covered S and is given by K
= aS where a is a constant. Then the force acting on the
particle is
(a)
R
aS
(b)
R
)aS(2
2
(c)
2
2
R
aS
(d)
R
aS2

158. 154 PHYSICS
68.A ramp is con
structed in parabolic shape such that the
height y of any point on its surface is given in terms of the
point's horizontal distance x from the bottom of the ramp be
y = x
2
/2L. A block of granite is to be set on the ramp; the
coefficient of static friction is 0.80. What is the maximum x
coordinate at which the block can be placed on the ramp
and remain at rest, if L = 10 m?
y
x
(a) 8 m (b) 8. 4 m
(c) 9 m (d) 9.4 m
69.The force F acting on a body moving in a circle of radius ris always perpendicular to the instantaneous velocity v.The work done by the force on the body in half rotation is(a) Fv (b) F·2pr
(c) Fr (d) 0
70.The negative of the distance rate of change of potentialenergy is equal to(a) force acting on the particle in the direction of
displacement
(b) acceleration of the particle, perpendicular to
displacement
(c) power(d) impulse.
71.n small balls each of mass m impinge elastically each secondon a surface with velocity v. The force experienced by thesurface will be
(a)
mnv
2
1
(b) 2 mnv
(c) mnv (d) 2 mnv
72.A
horse drinks water from a cubical container of side 1 m.
The level of the stomach of horse is at 2 m from the ground.
Assume that all the water drunk by the horse is at a level of
2m from the ground. Then minimum work done by the horse
in drinking the entire water of the container is
(Take r
water
= 1000 kg/m
3
and g = 10 m/s
2
) –
(a) 10 kJ
(b) 15 kJ
(c) 20 kJ
(d) zero
73.Th
e ball rolls down without slipping (which is at rest at a)
along ab having friction. It rolls to a maximum height h
c
where bc has no friction. K
a
, K
b
and K
c
are kinetic energies
at a, b and c.
Which of the following is correct ?
a
c
hc ha
b
(a)K
a
= K
c
, h
a
= h
c
(b) K
b
> K
c
,
h
a
= h
c
(c)K
b
> K
c
, h
a
< h
c
(d) K
b
> K
c
, h
a
> h
c
74.A body falls freely under gravity. Its velocity is v when ithas lost potential energy equal to U. What is the mass ofthe body ?(a)U
2
/v
2
(b) 2U
2
/v
2
(c) 2U/v
2
(d) U /v
2
75.If v be the instantaneous velocity of the body droppedfrom the top of a tower, when it is located at height h, thenwhich of the following remains constant ?
(a)
2
ghv+ (b)
2
v
gh
2
+
(c)
2
v
gh
2
- (d) gh – v
2
76.The c
oefficient of friction between the tyres and the road is
m. A car is moving with momentum p. What will be the
stopping distance due to friction alone ? The mass of the
car is m.
(a)p
2
/2mg (b) p
2
/2mmg
(c)p
2
/2m
2
mg (d) p
2
/2mg
77.A particle moves in the X–Y plane under the influence of a
force
F
r
such that its instantaneous momentum is
ˆˆ
p i2cost j2sint=+
r
.
What is the ang
le between the force and instantaneous
momentum ?
(a) 0° (b) 45°
(c) 90° (d) 180°
78.A particle of mass 10 kg moving eastwards with a speed 5
ms
–1
collides with another particle of the same mass moving
north-wards with the same speed 5 ms
–1
. The two particles
coalesce on collision. The new particle of mass 20 kg will
move in the north-east direction with velocity
(a) 10 ms
–1
(b) 5 ms
–1
(c)
1
(5 / 2)ms
-
(d) none of these
79.A metal
ball of mass 2 kg moving with a velocity of
36 km/h has a head on collision with a stationary ball ofmass 3 kg. If after the collision, the two balls move together,the loss in kinetic energy due to collision is(a) 140 J (b) 100 J
(c) 60 J (d) 40 J
80.A force acts on a 30 gm particle in such a way that the positionof the particle as a function of time is given by x = 3t – 4t
2
+ t
3
, wh er e x i s i n m et r es a n d t i s i n s econ d s. T h e wo r k d on e
during the first 4 seconds is(a) 576 mJ (b) 450 mJ
(c) 490 mJ (d) 530 mJ
81.A rubber ball is dropped from a height of 5m on a plane, wherethe acceleration due to gravity is not shown. On bouncingit rises to 1.8 m. The ball loses its velocity on bouncing bya factor of
(a)
25
16
(b)
5
2
(c)
5
3
(d)
25
9

159. 155Work, Energy and Power
82.A 3 kg ball strikes a heavy rigid wall with a speed of 10 m/
s at an angle of 60º. It gets reflected with the same speed and
angle as shown here. If the ball is in contact with the wall for
0 . 2 0 s , w h a t i s t h e a v e r a g e f o r c e e x e r t e d o n t h e b a l l b y t h e w a l l ?
(a) 150 N
60º
60º
(b) Zero
(c) N3150
(d) 300 N
83.A b
o m b o f m a s s 1 k g i s t h r o w n v e r t i c a l l y u p w a r d s w i t h a s p e e d
of 100 m/s. After 5 seconds it explodes into two fragments.
One fragment of mass 400 gm is found to go down with a
speed of 25 m/s. What will happen to the second fragment
just after the explosion? (g = 10 m/s
2
)
(a) It will go upward with speed 40 m/s
(b) It will go upward with speed 100 m/s
(c) It will go upward with speed 60 m/s
(d) It will also go downward with speed 40m/s
84.In a simple pendulum of length l the bob is pulled aside
from its equilibrium position through an angle q and then
released. The bob passes through the equilibrium position
with speed
(a)
)cos1(g2q+l (b) qsing2l
(c) lg2 (d) )cos1(g2q-l
85.A stationary particle e
xplodes into two particles of masses
m
1
a n d m
2
which move in opposite directions with velocities
v
1
and v
2
. The ratio of their kinetic energies E
1
/E
2
is
(a)m
1
v
2
/m
2
v
1
(b) m
2
/m
1
(c)m
1
/m
2
(d) 1
86.A mass of 0.5 kg moving with a speed of 1.5 m/s on a
horizontal smooth surface, collides with a nearly weightless
spring of force constant k = 50 N/m. The maximum
compression of the spring would be
(a) 0.5 m ( b) 0.15 m
(c) 0.12 m (d) 1.5 m
87.A ball moving with velocity 2 m/s collides head on withanother stationary ball of double the mass. If the coefficientof restitution is 0.5, then their velocities (in m/s) aftercollision will be
(a) 0, 1 (b) 1, 1
(c) 1, 0.5 (d) 0, 2
88.A body projected vertically from the earth reaches a height
equal to earth's radius before returning to the earth. The
power exerted by the gravitational force is greatest
(a) at the highest position of the body
(b) at the instant just before the body hits the earth
(c) it remains constant all through
(d) at the instant just after the body is projected
89.A mass m moving horizontally (along the x-axis) with
velocity v collides and sticks to mass of 3m moving vertically
upward (along the y-axis) with velocity 2v. The final velocity
of the combination is
(a)
13
ˆˆ
vi vj
42
+ (b)
12
ˆˆ
vi vj
33
+
(c)
21
ˆˆ
vi vj
33
+ (d)
31
ˆˆ
vi vj
24
+
90.The potent
ial energy of particle in a force field is
2
AB
U
rr
=- , where A and B are positive constants and r is
the distance of particle from the centre of the field. For
stable equilibrium, the distance of the particle is
(a)B / 2A (b) 2A / B
(c)A / B (d)B / A
91.A uniform force of
ˆˆ(3)ij+ newton acts on a particle of
mass 2 kg. The particle is displaced from position $
(2)ik
$
+
meter to position $$
(43)i jk
$
+- meter. The work done by
the force on the particle is(a) 6 J (b) 13 J
(c) 15 J (d) 9 J
92.An explosion breaks a rock into three parts in a horizontalplane. Two of them go off at right angles to each other. Thefirst part of mass 1 kg moves with a speed of 12 ms
–1
and
the second part of mass 2 kg moves with speed 8 ms
–1
. If the
third part flies off with speed 4 ms
–1
then its mass is
(a) 5 kg (b) 7 kg
(c) 17 kg (d) 3 kg
93.If the kinetic energy of a body is increased by 300%, themomentum of the body is increased by(a) 300% (b) 200%
(c) 100% (d) 50%
94.If the mass of the body is halved and velocity gets doubledthen final kinetic energy would be .........of initial.(a) same (b) four times
(c) double (d) eight times
95.A train of weight 10
7
N is running on a level track with
uniform speed of 36 km h
–1
. The frictional force is 0.5 kg per
quintal. If g = 10 m/s
2
, then power of engine is
(a) 500 kW (b) 50 kW
(c) 5 kW (d) 0.5 kW
DIRECTIONS (Qs. 96 to 100): Each question contains Statement-1 and Statement-2. Choose the correct answer (ONLY ONE optionis correct ) from the following.(a)Statement -1 is false, Statement-2 is true
(b)Statement -1 is true, Statement-2 is true; Statement -2 is acorrect explanation for Statement-1
(c)Statement -1 is true, Statement-2 is true; Statement -2 is nota correct explanation for Statement-1
(d)Statement -1 is true, Statement-2 is false
96. Statement-1 : A quick collision between two bodies is
more violent than slow collision, even when initial andfinal velocities are identical.
Statement -2 : The rate of change of momentum
determines that the force is small or large.

160. 156 PHYSICS
Exemplar Questions
1.An electron and a proton are moving under the influence of
mutual forces. In calculating the change in the kinetic energy
of the system during motion, one ignores the magnetic force
of one on another. This is, because
(a) the two magnetic forces are equal and opposite, so
they produce no net effect
(b) the magnetic forces do not work on each particle
(c) the magnetic forces do equal and opposite (but non-
zero) work on each particle
(d) the magnetic forces are necessarily negligible
2.A proton is kept at rest. A positively charged particle is
released from rest at a distance d in its field. Consider two
experiments; one in which the charged particle is also a
proton and in another, a positron. In the same time t, the
work done on the two moving charged particles is
(a) same as the same force law is involved in the two
experiments
(b) less for the case of a positron, as the positron moves
away more rapidly and the force on it weakens
(c) more for the case of a positron, as the positron moves
away a larger distance
(d) same as the work done by charged particle on the
stationary proton.
3.A man squatting on the ground gets straight up and stand.
The force of reaction of ground on the man during the
process is
(a) constant and equal to mg in magnitude
(b) constant and greater than mg in magnitude
(c) variable but always greater than mg
(d) at first greater than mg and later becomes equal to mg
4.A bicyclist comes to a skidding stop in 10 m. During this
process, the force on the bicycle due to the road is 200N
and is directly opposed to the motion. The work done by
the cycle on the road is
(a) + 2000 J (b) – 200 J
(c) zero (d) – 20,000 J
5.A body is falling freely under the action of gravity alone in
vaccum. Which of the following quantities remain constant
during the fall?
(a) Kinetic energy
(b) Potential energy
(c) Total mechanical energy
(d) Total linear momentum
6.During inelastic collision between two bodies, which of the
following quantities always remain conserved?
(a) Total kinetic energy
(b) Total mechanical energy
(c) Total linear momentum
(d) Speed of each body
7.Two inclined frictionless tracks, one gradual and the other
steep meet at A from where two stones are allowed to slide
down from rest, one on each track as shown in figure.
Which of the following statement is correct?
h
I
A
B
q
1
q
2
II
(a) Both the stones reach the bottom at the same time but
not with the same speed
(b) Both the stones reach the bottom with the same speed
and stone I reaches the bottom earlier than stone II
(c) Both the stones reach the bottom with the same speed
and stone II reaches the bottom earlier than stone I
(d) Both the stones reach the bottom at different times
and with different speeds
8.The potential energy function for a particle executing linear
SHM is given by
21
()
2
V x kx= where k is the force
constant of the oscillator (Fig.). For k = 0.5 N/m, the graph
of V(x) versus x is shown in the figure. A particle of total
energy E turns back when it reaches x = ±x
m
. If V and K
indicate the PE and KE, respectively of the particle at x =
+x
m
, then which of the following is correct?
x
Vx()
–x
m
x
m
(a)V = O, K = E(b)V = E , K = O
(c)V < E, K = O(d)V = O, K < E
97. Statement -1 : If co llision occurs between two elastic
bodies their kinetic energy decreases during the time of
collision.
Statement -2 : During collision intermolecular space
decreases and hence elastic potential energy increases.
98. Statement -1 :A work done by friction is always negative.
Statement -2 : If frictional force acts on a body its K.E.
may decrease.
99. Statement -1 : An object of mass m is initially at rest. A
constant force F acts on it. Then the velocity gained by
the object during a fixed displacement is proportional to
1/m.
Statement -2 : For
a given force and displacement velocity
is always inversely proportional to root of mass.
100. Statement -1 : Mechanical energy is the sum of
macroscopic kinetic & potential energies.Statement - 2 : Mechanical energy is that part of total
energy which always remain conserved.

161. 157Work, Energy and Power
9.Two identical ball bearings in contact with each other and
resting on a frictionless table are hit head-on by another ball
bearing of the same mass moving initially with a speed v as
shown in figure.
1 23
v
If the collision is elastic, which of the following (figure) is a
possible result after collision?
(a)
1
v = 0 v/2
(b)
12 3
vv = 0
(c)
123
v/3
(d)
1 2 3
v/1v/2v/3
10.A body of mass 0.5 kg travels in a straight line with velocity
v = a x
3/2
where a = 5 m
–1/2
s
–1
. The work done by the net
force during its displacement from x = 0 to x = 2 m is
(a) 15 J (b) 50 J
(c) 10 J (d) 100 J
11.A body is moving unidirectionally under the influence of a
source of constant power supplying energy. Which of thediagrams shown in figure correctly shown the displacement-time curve for its motion?
(a)
t
d
(b)
t
d
(c)
t
d
(d)
t
d
12.Which of the diagrams shown in figure most closely shows
the variation in kinetic energy of the earth as it moves oncearound the sun in its elliptical orbit?
(a)
KE
t
(b)
KE
(c)
KE
(d)
KE
13.Which of the diagrams shown in figure represents variation
of total mechanical energy of a pendulum oscillating in airas function of time?
(a)
E
t
(b)
t
E
(c)
t
E
(d)
t
E
14.A mass of 5 kg is moving along a circular path of radius 1 m.
If the mass moves with 300 rev/min, its kinetic energy wouldbe
(a) 250 p
2
(b) 100 p
2
(c) 5 p
2
(d) 0
15.A raindrop falling from a height h above ground, attains a
near terminal velocity when it has fallen through a height(3/4)h. Which of the diagrams shown in figure correctly
shows the change in kinetic and potential energy of thedrop during its fall up to the ground?
(a)
PE
KE
t
h
(b)
PE
KE
t
h/4
h
(c)
PE
KE
t
h
x
O
(d) PE
KE
t
h
O
16.In a shotput event an athlete throws the shotput of mass 10
kg with an initial speed of 1 m s
–1
at 45° from a height 1.5 m
above ground. Assuming air resistance to be negligibleand acceleration due to gravity to be 10 m s
–2
, the kinetic
energy of the shotput when it just reaches the ground willbe
(a) 2.5 J (b) 5.0 J
(c) 52.5 J (d) 155.0 J

162. 158 PHYSICS
17.Which of t
he diagrams in figure correctly shows the change
in kinetic energy of an iron sphere falling freely in a lake
having sufficient depth to impart it a terminal velocity?
(a)
KE
Depth
tO
(b)
KE
Depth
tO
(c)
KE
Depth
tO
(d)
KE
Depth
tO
18.A cricket ball of mass 150 g moving with a speed of 126 km/
h hits at the middle of the bat, held firmly at its position by
the batsman. The ball moves straight back to the bowler
after hitting the bat. Assuming that collision between ball
and bat is completely elastic and the two remain in contact
for 0.001s, the force that the batsman had to apply to hold
the bat firmly at its place would be
(a) 10.5 N (b) 21 N
(c) 1.05 × 10
4
N (d) 2.1 × 10
4
N
NEET/AIPMT (2013-2017) Questions
19.A uniform force of ˆˆ(3)ij+ newton acts on a particle of
mass 2 kg. The particle is displaced from position $
(2)ik
$
+
meter to position $$
(43)i jk
$
+- meter. The work done by
the force on the particle is [2013]
(a) 6 J (b) 13 J
(c) 15 J (d) 9 J
20.An explosion breaks a rock into three parts in a horizontal
plane. Two of them go off at right angles to each other. The
first part of mass 1 kg moves with a speed of 12 ms
–1
and
the second part of mass 2 kg moves with speed 8 ms
–1
. If
the third part flies off with speed 4 ms
–1
then its mass is
(a) 5 kg (b) 7 kg [2013]
(c) 17 kg (d) 3 kg
21.A person holding a rifle (mass of person and rifle together
is 100 kg) stands on a smooth surface and fires 10 shots
horizontally, in 5 s. Each bullet has a mass of 10 g with a
muzzle velocity of 800 ms
–1
. The final velocity acquired by
the person and the average force exerted on the person are
[NEET Kar. 2013]
(a) –1.6 ms
–1
; 8 N(b) –0.08 ms
–1
; 16 N
(c) – 0.8 ms
–1
; 8 N(d) –1.6 ms
–1
; 16 N
22.A particle with total energy E is moving in a potential energy
region U(x). Motion of the particle is restricted to the region
when [NEET Kar. 2013]
(a)U(x) > E (b)U(x) < E
(c)U(x) = O (d)U(x) £ E
23.One coolie takes 1 minute to raise a suitcase through a
height of 2 m but the second coolie takes 30 s to raise the
same suitcase to the same height. The powers of two coolies
are in the ratio of [NEET Kar. 2013]
(a) 1 : 2 (b) 1 : 3
(c) 2 : 1 (d) 3 : 1
24.A body of mass (4m) is lying in x-y plane at rest. It suddenly
explodes into three pieces. Two pieces, each of mass (m)
move perpendicular to each other with equal speeds (v).
The total kinetic energy generated due to explosion is :
(a) mv
2
(b)
23
mv
2
[2014]
(c) 2 mv
2
(d) 4 mv
2
25.A particle of mas
s m is driven by a machine that delivers a
constant power of k watts. If the particle starts from rest the
force on the particle at time t is [2015]
(a)
–1/2
mkt (b)
–1/2
2mkt
(c)
–1/21
mkt
2
(d)
–1/2mk
t
2
26.Two simila
r springs P and Q have spring constants K
P
and
K
Q
, such that K
P
> K
Q
. They are stretched, first by the
same amount (case a,) then by the same force (case b). Thework done by the springs W
P
and W
Q
are related as, in
case (a) and case (b), respectively [2015]
(a)W
P
= W
Q
; W
P
= W
Q
(b) W
P
> W
Q
; W
Q
> W
P
(c)W
P
< W
Q
; W
Q
< W
P
(d) W
P
= W
Q
; W
P
> W
Q
27.A block of mass 10 kg, moving in x direction with a constantspeed of 10 ms
–1
, is subject to a retarding force F = 0.1 × J/m
during its travel from x = 20 m to 30 m. Its final KE will be:
(a) 450 J (b) 275 J [2015]
(c) 250 J (d) 475 J
28.The heart of man pumps 5 litres of blood through the arteries
per minute at a pressure of 150 mm of mercury. If the density
of mercury be 13.6 ×10
3
kg/m
3
and g = 10m/s
2
then the
power of heart in watt is : [2015 RS]
(a) 2.35 (b) 3.0
(c) 1.50 (d) 1.70

163. 159Work, Energy and Power
29.A ball is thrown vertically downwards from a height of 20 m
with an initial velocity v
0
. It collides with the ground loses
50 percent of its energy in collision and rebounds to the
same height. The initial velocity v
0
is : [2015 RS]
(Take g = 10 ms
–2
)
(a) 20 ms
–1
(b) 28 ms
–1
(c) 10 ms
–1
(d) 14 ms
–1
30.On a frictionless surface a block of mass M moving at speed
v collides elastically with another block of same mass M
which is initially at rest. After collision the first block moves
at an angle q to its initial direction and has a speed
v
3
. The
seco
nd block's speed after the collision is : [2015 RS]
(a)
3
v
4
(b)
3
v
2
(c)
3
v
2
(d)
22
v
3
31.Two particles
A and B, move with constant velocities
1v
r
and
2v
r
. At the initial moment their position vectors are
1r
r
and
2
r
r
respectively. The condition for particles A and B for
their collision is: [2015 RS]
(a)
11 22r .v r .v=
rrrr
(b)
1122
rv rv´=´
rrrr
(c)
12 12
rr vv-=-
rr rr
(d)
12 21
12 21
rr vv
|r r| |v v|
--
=
--
rr rr
rr rr
32.A body of mass 1 kg begins to move under the action of a
time dependent force
2ˆˆ
F (2ti3t j)=+
r
N, where
ˆ
i and
ˆ
j are
unit
vectors alogn x and y axis. What power will be
developed by the force at the time t? [2016]
(a) (2t
2
+ 3t
3
)W (b) (2t
2
+ 4t
4
)W
(c) (2t
3
+ 3t
4
) W (d) (2t
3
+ 3t
5
)W
33.A particle of mass 10 g moves along a circle of radius 6.4 cm
with a constant tangential acceleration. What is the
magnitude of this acceleration if the kinetic energy of the
particle becomes equal to 8 × 10
–4
J by the end of the
second revolution after the beginning of the motion ? [2016]
(a) 0.1 m/s
2
(b) 0.15 m/s
2
(c) 0.18 m/s
2
(d) 0.2 m/s
2
34.Consider a drop of rain water having mass 1 g falling from a
height of 1 km. It hits the ground with a speed of 50 m/s.
Take 'g' constant with a value 10 m/s
2
.
The work done by the (i) gravitational force and the (ii)
resistive force of air is [2017]
(a) (i) 1.25 J (ii) –8.25 J
(b) (i) 100 J (ii) 8.75 J
(c) (i) 10 J (ii) – 8.75 J
(d) (i) – 10 J (ii) –8.25 J

164. 160 PHYSICS
EXERCISE - 1
1. (a) 2. (b)
3. (d) 4. (a) 5. (c)
6. (c) 7. (a) 8. (b)
9. (b) Because water enters into the vessel A, it becomes
heavier. Gravity helps it sink. External work required
for immersing A is obviously less than that for
immersing B.
10. (c) Weight Mg moves the centre of gravity of the spring
through a distance
2/
2
)0(
l
l
=
+
\ Mechanical energy stored = Work done = Mg l/2.
11. (d) From
11
v u at, v 0 at=+ =+
1
1
v
a
t
\=
11
F ma mv /t==
Velocity acquir
ed in t sec = at =
1
1
v
t
t
Power
2
111
2
11 1
mv v t mv t
Fv
tt t
=´= ´=
12. (d)
W = F × s
0
xW)x(
x
1
W µ\µ
13. (b) At
the top of flight, horizontal component of
velocity
u cos 45º u / 2==
2
1u
K.E.m
2 2
æö
\=
ç÷
èø

.K
2
1
2
um
2
1
2
=
÷
÷
ø
ö
ç
ç
è
æ
=
14. (a) From
22
v u 2as-=
s)g(2v0
2
0
m=-
g2
v
s
2
0
m
=\
15. (b) If K0LÞ=
r
.E may or may not be zero.
If K.E = 0 Þ 0L=
r
.
16. (d) Power
is defined as the rate of doing work. For the
automobile, the power output is the amount of work
done (overcoming friction) divided by the length of
time in which the work was done.
17. (b)
)(k
2
1
k
2
1
k
2
1
W
2
1
2
2
2
1
2
2
llll-=-=
18. (b)
1
F (given)
v
µ
Then s.FEW
k
==
Beco
me =
k
s
E
v
µ Þk
s
E
s/t
µ
Þ tE
k
µ
19. (d)
22
3
mv ms
t t
=
since both P
and m are constants
\
2
3
s
t
= constant
20. (
c)
11 22
12
mv mv a b 0 a (b)
v.
m m ac ac
+ ´+
= ==
+ ++
21. (d) As
2/1
0
nn
h
h
e
÷
÷
ø
ö
ç
ç
è
æ
=
.heheheh
422
0
n2
n
===\
´
22. (a) As u
2
= 0 and m
1
= m
2
, therefore from
m
1
u
1
+ m
2
u
2
= m
1
v
1
+ m
2
v
2
we get u
1
= v
1
+ v
2
Also,
2 1 2 1 12
1 2 1 12
v v v v 1 v /v
e
u v v 1 v /v
- --
===
++
,
which gives
1
2
v1e
v 1e
-
=
+
23. (d) Work done Fdx=
ò
24. (b)
1
m
2
m 21
mm+
vu 0u=
1vu=
applying conse
rvation of momentum
()
12121
v)mm(0mvm+=+
v
)mm(
m
v
21
1
1
+
=Þ
as )mm(m
211
+
< so velocity decreases.
25. (b)
1 21
1
12
(m m )u
v
mm
-
=
+
As
1
v is negative and less than
u
1
, t
herefore, m
1
< m
2
. EXERCISE - 2
1. (a) W = F s c
os 90º = zero
2. (c) As the cord is trying to hold the motion of the block,
work done by the cord is negative.
W = – M (g – a) d
4
dgM3
d
4
g
gM
-
=÷
ø
ö
ç
è
æ
--=
3. (c) W = F s cos q =
10 × 2 cos 60º = 10 J.
4. (c) )j
ˆ
3i
ˆ
2).(j
ˆ
2i
ˆ
5(s.FW++==
rr
= 10 + 6 = 16 J.
Hints & Solutions

165. 161Work, Energy and Power
5. (c) From horizontal position to lowest position, height
through which the bob falls = l
\ At lowest position, v 2g=l
K.E. at lowest point
211
mv m(2 g) mg
22
=== ll = 10 × 1 × 9.8 = 98 J.
6
. (b)
òò ú
û
ù
ê
ë
é
===
1
1
1 x
0
x
0
2
x
0
xc
2
1
dxxcdxFW
2
1
2
1 xc
2
1
)0x(c
2
1
=-=
7. (d)
Forward thrust,
P 100 746
F 3730 N.
v 20
´
===
8. (
c) As
22
AA BB
11
mv mv
22
=
AB
BA
mv
vm
= ;
B BB
A
AA
P mv
P mv
=

A
AA
BB
B
mmm 1
mmm 3
= ==
9. (a)
From
11 22
m v m v 0,+=
11
2
2
mv 4
v
m 234
-
= =-
10. (
b) Total energy at the time of projection
2211
m v 0.1(2 0) 20J
22
==´=
Half wa
y up, P.E. becomes half the P.E. at the top i.e.J10
2
20
.E.P==\ K.E. =
20 – 10 = 10J.
11. (b) Workdone, W
()
22
21
1
kxx
2
=-
()
2 21
k 0.15 (0.05)
2
éù
=-
ëû
1
800 0.02 8J
2
=´´=
12. (d) As
m2
p
E
2
=
dE dp
2 2 5% 10%
Ep
æö
\ = =´=
ç÷
èø
13
. (d)
Mg
8
7
)8/gg(MT=-=
Work don
e by the cord = T × d cos 180º.8/dgM7)1(dgM
8
7
-=-=
14. (b) x.
FW
rr
= )j
ˆ
i
ˆ
2).(k
ˆ
2j
ˆ
3i
ˆ
5( -++= joules7310=- =
15. (a)
23
1 )05.0(105
2
1
W ´´=
23
2 )10.0(105
2
1
W ´´=Þ
.J75.1805.015.0105
2
1
W
3
=´´´´=D\
16. (a)
12
12
m 0.2 kg, m 0.4 kg, v 0.3m / s, v ?====
Applying law of conservation of momentum
m
1
v
1
– m
2
v
2

0.2 0.3
0.15 m / s.
0.4
´
==
17. (c) 11
2
2
mv 3
v 16 8m /s
m6
- -
= = ´ =-
22
2 22
11
E m v 6( 8 ) 192 J.
22
= =´-=
18
. (d)
22
)2(k
2
1
)x(k
2
1
V == or
2
V
4
V2
k ==
V25)10(
2
V
2
1
)10(k
2 1
V
22
=÷
ø
ö
ç
è
æ
´==¢
19. (b
) If P = momentum, K = kinetic energy, then
P
1
2
= 2 mK
1
,
2
22
P 2mK=
\
2
2 21
1 11
P K 3K
3
P KK
æö
===ç÷
èø
\
2
1
P 3
1.732
P1
==
20. (b)
1 11
2 22
p 2mK
p 2mK
=
21. (d
) Applying the principle of conservation of linear
momentum, we get
22
3m v (m 60) (m 60)´= ´ +´ 260m´=
v 20 2m/s=
22. (c
) Energy stored (E)
J9)12(
100
75
=´=
As
2
mv
2
1
E=
sec/m18
1
92
m
E2
v =
´
==
23. (c)
The uniform acceleration is
2101
a ms
15 15
--
==
Let v be the
velocity at kinetic energy
J
9
2
therefor
e
12
ms
3
2
vor
9
2
v1
2
1 -
==´´
Using v =
u + at
21
0 t t 10s
3 15
=+ ´Þ=

166. 162 PHYSICS
24. (d
) Let n be the number of bullets that the man can fire in
one second.
\ change in momentum per second
mvn´= = F
[ m= mass of bullet, v = v
elocity] (
Q F is the force)
120040
1000144
mv
F
n
´
´
==\ = 3
25. (d) Po
wer supplied =
t
mgh
Power used by crane =
80
100
t
mgh
´
W
836
10
80
100
3600
100101000
5
´
=´
´´
=
26. (b) Work d
one against gravity
W
g
= 50 × 10 × 30 = 15 kJ
Work done against friction
W
f

mgcoss=m q´=
4
0.4 50 10 50 8 kJ
5
´´´´=
Total wor
k done kJ23kJ8kJ15WW
fg
=+=+=
27. (d) Let mass of neutron = m
then mass of deuterium = 2m
[
Q it has double nuclides thus has neutron].
Let initial velocity of neutron = v and final velocities
of neutron and deuterium are v
1
and v
2
respectively.
m
2m
v=0
m 2m
v
1
v® 2
v®
neutron
deuterium
Applying conservation of momentum
21
mv2mv)0(m2mv+=+
)i.(..........v2vv
21
+=Þ
applying conservation of energy
222
12
111
mv mv 2m.v
222
=+
222
12
v v 2v ......(ii)Þ=+
from (i) and (ii),
2
2
2
2
2
v
2)v2v(v+-=
2
22
2
2
22
v2vv4v4vv+-+=Þ
0vv4v6
2
2
2
=-
3
v
v&
3
v2
v
12 -==Þ
now fractional
change in kinetic energy
if
i
KK
K
-
=

2
2
1
2
mv
2
1
mv
2
1
mv
2
1
-
=
2
2
2
v
v
89
9v
-
==
28. (c) W = F s cos q, ,
2
1
105
25
sF
W
cos =
´
==q q = 60º.
29. (b) Fo
r elastic collision in one dimension
1
v =
21
22
mm
um2
+
+
)mm(
u)mm(
21
121
+
-
As mass 2m, is a
t rest, So
2
u = 0
Þ 1
v=
m2m8
u)m2m8(
+
-
= u
5
3
Final energy
of sphere =
f
.)E.K(
=
2
5
u3
)m8(
2
1
÷
ø
ö
ç
è
æ
=
2
2
5
3
u)m8(
2
1
÷
ø
ö
ç
è
æ
´
= E
25
9
= 0.36 E
30.
(b) The work is done against gravity so it is equal to the
change in potential energy. W = E
p
= mgh
For a fixed height, work is proportional to weight lifted.
Since Johnny weighs twice as much as Jane he works
twice as hard to get up the hill.
Power is work done per unit time. For Johnny this is
W/Dt. Jane did half the work in half the time, (1/2 W)/
(1/2 Dt) = W/Dt which is the same power delivered by
Johnny.
31. (c) Fraction of energy transferred =
9
8
)21(
24
2
=
+
´
32. (c)
m2
p
E
2
=
Þ
2
2
2
1
2
1
p
p
E
E
= Þ 4EE
12
´=
\
112
E3EE=-
33. (c) Since th
e new height gained is half , therefore there is
50% loss of energy.
34. (a)
um
3
u2
mum
211+= (By condition of linear
momentum)
Þ
12
1
mu mv
3
= ...... (i)
Also
|uu|
|vv|
e
12
21
-
-
=
Þ
2u
vu
3
-= Þ
5
vu
3
= ...... (ii)
From (i)
and (ii),
um
3
5
um
3
1
21= Þ 5
m
m
2
1
=
35. (c) At
Equilibrium , ky = mg
Þ
y
mg
k=
mgy
2
1
y
y
mg
2
1
U
2
=
÷
÷
ø
ö
ç
ç
è
æ
=
36. (a) Loss in K.
E = Area under the curve

167. 163Work, Energy and Power
37. (b) In this case, P =
t
mv
2
1
mgh
2
+
Þ
ú
ú
û
ù
ê
ê
ë
é
+=
2
v
gh
t
m
P
2
Þ W
2
1010
1010
1
10
P
ú
û
ù
ê
ë
é ´
+´= = 1500W
38. (c
) By conservation of linear momentum,
12
21
mv
mv
=
or
2
1
r
3
4
r
3
4
3
2x
3
x
1
=
pr
pr
Þ
3/1
2
1
2
1
r
r
=
39. (c)
222 00
22
22
mm
mccc
v (0.5c)
11
cc
==
--
2
0
2020
cm15.1c
75.0
m
c
25.01
m
==
-
=
Change in
energy = 1.15 m
0
c
2
– m
0
c
2
= 0.15 × 9.1 × 10
–31
× (3 × 10
8
)
2
= 12.285 × 10
–15
J MeV
106.1
10285.12
13
15
-
-
´
´
=
= 0.07678 Me
V
40. (b)
12
F 1000 10 N 1200 N
100
=´´=
P =
Fv = 1200 N × 15 ms
–1
= 18 kW.
41. (b) Components of velocity before and after collision
parallel to the plane are equal, So
v sin 60° = u sin 30°.......(1)
Components of velocity normal to the plane are related
to each other
v cos 60° = e u (cos 30°) ........(2)
Þ cot 60° = e cot 30° Þ
cos60
e
cot 30
°
=
°
Þ
3
3
1
e= Þ .
3
1
e=
42. (d
) Work done on the body is gain in the kinetic energy.
Acceleration of the body is a = V/T.
Velocity acquired in time t is v = at
V
t
T
=
K.E. acquired µ v
2
. That is work done
22
2
Vt
T
µ
43. (d
) Kinetic energy of ball when reaching the ground
= mgh = mg × 10
Kinetic energy after the impact
60
mg 10 6mg
100
= ´ ´=
If th
e ball rises to a height h, then mgh = 6 mg.
Hence, h = 6 m.
44. (d)
dU
F b 2ay
dy
=- =-
45.
(b) Fv = P . or
dv
M vP
dt
=
That is
P
vdv dt
M
=
òò
Hence
1/2
v [2Pt/M]=
46. (c)
cm3
u
2
u 0v=
x
Case I :
3.a.2u
2
u 2
2
=-÷
ø
ö
ç
è
æ
or –
4
u3
2
= 2. a
. 3
Þ a = –
8
u
2
Case II :
2
2
u
0 ÷
ø
ö
ç
è
æ
- = 2. a.
x or –
4
u
2
= 2.
÷
÷
ø
ö
ç
ç
è
æ
-
8
u
2
× x
Þ x = 1 cm
A
lternative method : Let K be the initial energy and F
be the resistive force. Then according to work-energy
theorem,
W = DK
i.e.,
2
21 1v
3Fmvm
2 22
æö
=-
ç÷
èø
211
3F mv1
24
æö
=-
ç÷
èø
231
3F mv
42
æö
=
ç÷
èø
...(1)
and
2
21v1
Fx m m(0)
222
æö
=-
ç÷
èø
i.e.,
211
mv Fx
42
æö
=
ç÷ èø
...(2)
Comparing e
qns. (1) and (2)
F = Fx
or x = 1 cm

168. 164 PHYSICS
47. (b)
Let the velocity and mass of 4 kg piece be v
1
and m
1
and that of 12 kg piece be v
2
and m
2
.
Applying conservation of linear momentum
m
2
v
2
= m
1
v
1
1
1 ms12
4
412
v
-
=
´
=Þ
J2881444
2
1
vm
2
1
.E.K
2
111
=´´==\
48. (d) Veloci
ty will increase when force is along the direction
of displacement i.e. d
ˆ
F
ˆ
=.
49. (b) When body is lowered gradually, its weight acts at
C.G. of the spring. When same body is allowed to fall
freely, the same weight acts at lower end of the spring.
In the latter case, original length (L) of spring is double.
As DL
µL, therefore, DL become s twice in second
case i.e. 2x.
50. (d) If V is velocity of combination (bag + bullet), then
from principle of conservation of linear momentum
mv
(m M)V mvo rV
(m M)
+==
+
22
21 mv
K.E. (m M
)V
2 2(m M)
=+=
+
51. (c) W
orkdone W = [ML
2
T
–2
]
It shows that
122
W (LT ) i.e.W v.
-
µµ
\ graph betwee
n W & v is a parabola.
Alternatively :
According to work energy theorem 222
mv
2
1
mu
2
1
mv
2
1
W =-= (Q u = 0)
Þ W µ v
2
52. (c)
Applying principle of conservation of linear momentum
22
(2m)V (mv) ( mv) mv 2= +=
v2v
V
2 2
\==
53. (a) Sin
ce block does not slide on wedge so displacement
is zero & hence work done by force is zero.
f
m
gsinq
q
u
mg
54. (c) If ma
ss of nucleus is m, mass of a particle is m
1
&
mass of remaining nucleus is m
2
, then from the law of
conservation of momentum.
nucleusremaining
ofmomentum
22
particle
ofmomentum
11
nucleus of
momentum initial
vmvmvm
rrr
+=
a
55. (a) Le
t M be the mass of shell. Applying law of
conservation of linear momentum
MV cos
q =
MVM
cosv
22
æö
- q+ç÷
èø
ie, MV cos q +
M
2
Vcosq=
M
2
v
or v = 3 Vcosq.
56. (d) As
bob B is of same material and same mass as the
bob A, therefore, on elastic collision, their velocities
are exchanged. Bob A comes to rest and B moves with
the velocity of A.
57. (c) As velocities are exchanged on perfectly elastic
collision, therefore masses of two objects must be
equal.
.mmor1
m
m
ba
b
a
==\
58. (c)a kx=-
dv
kx
dt
Þ =-
Also
dv
dx
dtorv
dt
dx
==
v x2
v0
1
vdv
kx v dv kxdx
dx
\ =- Þ =-òò
( )
2
22
21
kx
vv
2
- =- ( )
2
22 211 1 kx
mvvm
2 22
æö-
Þ -= ç÷
èø
2
Kx\Da
59. (c)
2
i 11
1
K mu,
2
=
1
21
21
1
2
11f u
mm
mm
v,vm
2
1
K
+
-
==
Fractional loss
2
11
2
11
2
11
i
fi
um
2
1
vm
2
1
um
2
1
K
KK
-
=
-

( )
( )
2
21
2
21
2
1
2
1
mm
mm
1
u
v
1
+
-
-=-=
( )
2
21
21
mm
mm4
+
=
mm
2
=; nmm
1
=
( )
2
i
f
n1
n4
K
K
1
+
=-
Energy tran
sfer is maximum when 0K
f
=
( )
2
4n
1
1n
=
+
2
4n 1 n 2nÞ=++ Þ 0n21n
2
=- +
()01n
2
=- 1n=ie. mm
2
=, mm
1
=
Trans
fer will be maximum when both masses are equal
and one is at rest.

169. 165Work, Energy and Power
60. (c) s/m10
410
041410
V
c =
+
´+´
= ; since spring force is
internal force.
61. (a) For inelastic collision, linear momentum is conserved
Þ
1
mv =
2
mv2 Þ
2
v
v
1
2=
Loss i
n K.E. = Gain in P.E.
=
mgh2v)m2(
2
1
mv
2
1 2
2
2
1 =-
Þ4 mgh =
2
1
mv –
2
mv
2
1
=
2
mv
2
1
=
2
mv
2
Þh =
g8
v
2
62. (d)P Fv=´ Þ P = F a t\ Ptµ
63. (c)
g
h2
t
AB=
1
BC CB
2
2h
tt2
g
2e h 2h
2 2e
gg
+=
==
h
h
A
B
h
1
2
D
C
g
h2
e2tt
2
DBBD =+
\ Total time taken by the body in coming to rest

.........
g
h2
e2
g
h2
e2
g
h2 2
+++=
22h 2h
2e [1 e e .........]
gg
= + +++
=
e1
1
g
h2
e2
g
h2
-
´+
2h1e 1e
t
g 1e 1e
++éùæö
==
ç÷êú
--ëûèø
64. (d) Initially, mgk3= or
3
mg
k=
New force cons
tant of longer part
2
mg
3
mg
2
3
k
2
3
'k =´==
Finally,
mgy'k=
cm22
mg
mg
'k
mg
y =´==
65. (d) kg8
kg8.010m=´=
height of iron chain = 5m
mgh 8 10 5
P W 40W
t 10
´´
===
66.
(c) By law of conservation of momentum,
Mu = MV + mv ....(i)
Also e =
MVMvMu
|uu|
|vv|
21
21
-=Þ
- -
....(ii)
From (i) an
d (ii), 2Mu = (M + m)v
Þ
2uM 2u
vv
mMm
1
M
= Þ=
+
+
67. (
d) Centripetal force =
R
mv
2

R
aS2
R
K2
R
2
mv
2
12
==÷
ø
ö
ç
è
æ
=
68. (a
) As the block is at rest at P.
qm=qcosmgsinmg
dx
dy
slopetan ==q
=m
L
x
L2
x
dx
d
2
=
÷
÷
ø
ö
ç
ç
è
æ
=
mg
mg cosq
mg sinq
q
N
P
x
y
8.0
L
x
=\
Þ x = 0.8
× 10 = 8 m
69. (d) Work done by centripetal force is zero.
70. (a)
dU
F
dx
=-
71. (c)
The change in momentum in the ball after the collision
with surface is m(0–v) = –mv
Since n balls impinge elastically each second on the
surface, then rate of change of momentum of ball per
second is
mvn (consider magnitude only)
Now According to Newton’s second law
rate of change of momentum per second of ball = force
experienced by surface.
72. (d)
h=1.5m
The ma
ss of water is m = 1 × 10
3
kg
\ The increase in potential energy of water is
= mgh = (1 × 10
3
) (10) (1.5) = 15 kJ

170. 166 PHYSICS
73. (d)
74. (
c)
2
U (1/ 2)Mv=
75. (b) P. E + K.E
= constant, mass being constant
gh + v
2
/2 = constant
76. (c)
2
K p / 2m mgx= =m
Hence,
22
x p / 2m g.=m
77. (c)
dp
ˆˆ
F i2sint j2c ott
dt
==-+
r
r
Hence F.p0=
rr
, hence angle between F
r
and p
r
is 90°
78. (c) Here ˆˆ
imv jmv 2mV+=
r
That is
v
ˆˆ
V (i j)
2
=+
r
Hence
vv
V2
2 2
=´= . Hence v = 5 ms
–1
79. (c) Apply conservation of momentum,
m
1
v
1
= (m
1
+ m
2
)v
v =
)mm(
vm
21
11
+
Here v
1
= 36 km/hr = 10 m
/s,
m
1
= 2 kg, m
2
= 3 kg
s/m4
5
210
v =
´
=
K.E. (initial) = J100)10(2
2
1 2
=´´
K.E. (Final) = J40)4()23(
2
1 2
=´+´
Loss in K.E
. = 100 – 40 = 60 J
Alternatively use the formula
( )
( )
212
k 12
12
mm1
E uu
2mm
-D=-
+
80. (a)
x = 3t –4t
2
+ t
3
2
t3t83
dt
dx
+-=
Acceleratio
n =
t68
dt
xd
2
2
+-=
Acceleration af
ter 4 sec = –8 + 6 × 4 = 16
Displacement in 4 sec = 3 ×4 – 4 × 4
2
+ 4
3
= 12 m
\Work = Force × displacement
= Mass × acc. × disp. = 3 × 10
–3
× 16 × 12 = 576 mJ
81. (b) According to principle of conservation of energy
Potential energy = kinetic energy
Þ
gh2vmv
2
1
mgh
2
=Þ=
If h
1
and h
2
are initial and final he
ights, then
Þ
2211
gh2v,gh2v==
Loss in velocity,
2121
gh2–gh2v–vv==D
\ fractional lo
ss in velocity
1
v
vD
= =
1
2
1
21
h
h
–1
gh2
gh2–gh2
=
6.0–136.0–1
5
8.1
–1 ===
5
2
4.0==
82. (c) Chan
ge in momentum along the wall
= mv cos60º – mv cos 60º = 0
Change in momentum perpendicular to the wall
= mv sin60º – (– mv sin60º) = 2mv sin60º
\ Applied force =
Time
momentuminChange
=
20.0
º60sinmv2
=
23103
2 0.20
´´´
´
= 50 33´ = 3150 newton
83. (b) S
peed of bomb after 5 second,
v = u – gt = 100 –10×5 = 50m/s
Momentum of 400 g fragment
=
)25(
1000
400
-´(downward)
Moment
um of 600g fragment =
v
1000
600
Momentum of bomb just before explosion
= 1 × 50 = 50
From conservation of momentum
Total momentum just before collision = Total
momentum just after collision
Þ v
1000
600
25
1000
400
50 +´-=
Þ v = 100 m/s (upwar
d)
84. (d) If l is length of pendulum and q be angular amplitude
then height.
A
B
P
C
q
l
h
AC–ABh= = l – l cos q = l(1 – cos q)
At extreme po
sition, potential energy is maximum and
kinetic energy is zero; At mean (equilibrium) position
potential energy is zero and kinetic energy is
maximum, so from principle of conservation of energy.
Bat)PEKE(Pat)PEKE( +=+
0mv
2
1
mgh0
2
+=+
v 2gh 2g (1– cos )Þ==q l

171. 167Work, Energy and Power
85. (b) From law of conservation of momentum, before
collision and after collision linear momentum (p) will
be same.
or initial momentum = final momentum.
2
p
E
2m
=
Accordi
ng to question,
2
112
2
21 2
E p 2m
E 2m p
=´
1
2
2
1
m
m
E
E
=Þ
[p
1
= p
2
]
86. (b)
22
kx
2
1
mv
2
1
= Þ mv
2
= kx
2
or 0.5 × (
1.5)
2
= 50×x
2
\ x = 0.15 m
87. (a) Clearly v
1
= 2 ms
–1
, v
2
= 0
m
1
= m (say), m
2
= 2m
v
1
' = ?, v'
2
= ?
e =
12
21
v' v'
vv
-
-
....(i)
By con
servation of momentum,
2m = mv
1
' + 2mv
2
' ... (ii)
From (i), 0.5 =
21v' v'
2
-
\ v
2
' = 1 + v
1
'
F
rom (ii), 2 = v
1
'+ 2 + 2 v
1
'Þv
1
= 0 and v
2
= 1 ms
–1
8
8. (b) Power exerted by a force is given by
P = F.v
When the body is just above the earth’s surface, its
velocity is greatest. At this instant, gravitational force
is also maximum. Hence, the power exerted by the
gravitational force is greatest at the instant just before
the body hits the earth.
89. (a) As the two masses stick together after collision, hence
it is inelastic collision. Therefore, only momentum is
conserved.
2v
3m
v
xm
ˆˆ
mvi 3m(2v)j (4m)v
r
\+=
v6
ˆˆ
v i vj
44
=+
r
=
v3
ˆˆ
i vj
42
+
90. (b) Fo
r equilibrium
0
dU
dr
= Þ
32
2
0
AB
rr
-
+=
r =
2A
B
for stable equilibrium
2
2
dU
dr
should be positive for the value of r.
here
2
2 43
62dU AB
dr rr
=- is +ve val ue for
2A
r
B
=So
91. (d) Gi
ven :
F
r
= 3ij+
$$
1
r
ur
=
$
()2ik+
$
,
2
r
u ur
= ()4i3jk+-
r
$$
r
r
=
21
rr-
uur ur
= ()4i3jk+-
r
$$
–
$
()2ik+
$
orr
r
= 2i 3j+
$$
–
$
2k
So work done by the given force w = f .r
rr
= ()
$
( )3i j . 2i 3j 2k+ +-
$$ $$
= 6 + 3 = 9J
92. (a)
4m / s ec
m
3
2 kg m
2
P
resultant
1 kg
x
y
8 m/sec
12 m/s
ec
m
1
P
resultant
=
22
12 16+
= 144 256+ = 20
m
3
v
3
= 20 (moment
um of third part)
or, m
3
=
20
4
= 5 kg
93. (c)p 2mK=
p' 2m[K 3K]=+ = 2p
p 100 2p p
100 100%
pp
D´-
= ´=
94. (c
) Let m and v be the mass and the velocity of the body.
Then initial K.E. ,
K
i
=
21
2
mv
Now, mass =
2
m
velocity = 2v

172. 168 PHYSICS
\ Final K.E.
=
21
(2)
22
m
v
æö
ç÷
èø
=
21
(2)
2
mv
=
21
2
2
mv
æö
ç÷
èø
= 2 K
i
95. (a) Power of
engine = Force × velocity
= Fv
Here, mass of engine =
7
10
10
kg = 10
6
kg
F = frictional fo
rce
= 0.5 kgf per quintal
= (0.5 × 10) N per quintal
= 5 N per quintal
=
6
5 10
100
N
æö´
ç÷
èø
= (5 × 10
4
) N
and v = 36 km h
–1

=
136 1000
60 60
kmh
-´
´
= 10 ms
–1
\Power of engine
= (5 × 10
4
× 10)W
= 5 × 10
5
W
= 500 kW
96. (b) 97. (b)
98. (a) When frictional force is opposite to velocity, kinetic
energy will decrease.
99. (c) 100. (d)
EXERCISE - 3
Exemplar Questions
1. (b)When electron and proton are moving under influence
of their mutual forces, then according to the flemings
left hand rule, the direction of force acting on a charge
particle is perpendicular to the direction of motion.
In magnetic field, work-done = F. s. cosq
= F . s. cos 90° = 0.
So magnetic forces do not work on moving charge
particle.
2. (c)Forces between two protons is same as that of between
proton and a positron.
As positron is much lighter than proton, it moves away
through much larger distance compared to proton.
Work done = Force × Distance
As forces are same in case of proton and positron but
distance moved by positron is larger, hence, work done
on positron will be more than proton.
3. (d)When the man squatting on the ground he is tilted
somewhat, hence he also has to apply frictional force
besides his weight.
R (reactional force) = friction force (f) + mg
i.e. R > mg
When the man does not squat and gets straight up in
that case friction (f) » 0
R (Reactional force) » mg
Hence, the reaction force (R) is larger when squatting
and become equal to mg when no squatting.
4. (c)According to the question, work done by the frictional
force on the cycle is :
= 200 × 10 = –2000 J
As the road is not moving, hence work done by the
cycle on the road is zero.
5. (c)As the body is falling freely under gravity and no
external force act on body in vaccum so law of
conservation, the potential energy decreases and
kinetic energy increases because total mechanical
energy (PE + KE) of the body and earth system will be
remain constant.
6. (c)According to the question, consider the two bodies
as system, the total external force on the system will
be zero.
Hence, in an inelastic collision KE does not conserved
but total linear momentum of the system remain
conserved.
7. (c)As the (inclined surface) are frictionless, hence,
mechanical energy will be conserved. As both the
tracks having common height, h (and no external force
acts on system).
KE & PE of stone I at top = KE + PE at bottom of I.
From conservation of mechanical energy,
2
1
1
00
2
mv mgh+ =+
1
2v ghÞ= similarly
2
2v gh=
Hence, speed is same for
both stones.
For stone I, acceleration along inclined plane a
1
= g
sin q
1
Similarly, for stone II a
2
= g sin q
2
sin q
1
< sin q
2
Thus, q
2
> q
1
hence a
2
> a
1
.
a
2
is greater than a
1
and both length for track II is also
less hence, stone II reaches earlier than stone I.
8. (b)Total Mechanical energy is E = PE + KE at any instant.
When particle is at x = x
m
i.e., at extreme position,
partical returns back and its velocity become zero foran instant. Hence, at x = x
m
; x = 0, K.E. = 0.
From Eq. (i),
E = PE + 0 = PE = V(x
m
) =
21
2
m
kx
but at mean position at origin V(x
m
) = 0.
9. (b)If two bodies of equal masses collides elastically, their
velocities are interchanged.
When ball 1 collides with ball-2, then velocity of ball-
1, v
1
becomes zero and velocity of ball-2, v
2
becomes
v, i.e., similarly then its own all momentum is mV.
So,
1 2120 , 0,v v v P P mV=Þ===
Now ball 2 collides to ball 3 and its transfer it's
momentum is mV to ball 3 and itself comes in rest.
So,
2 323
0 , 0,v v v P P mV=Þ===
So, ball 1 and ba
ll 2, become in rest and ball 3 move
with velocity v in forward direction.

173. 169Work, Energy and Power
10. (b)As we know that,
W.D.
22
11
0
xx
xx
F dx ma dx=×=×òò
u ur u urr r
As given
that,
m = 0.5 kg, a = 5 m
–1/2
s
–1
,
work done (W) = ?
3/2
v ax=
We also kno
w that,
Acceleration,
3/2 3/2
0
()
dv dv d
a v ax
ax
dt dx dx
==×=

3/ 2 1/2 2 233
22
ax a x ax= ´´´=
Now
, Force =
22
0
3
2
ma m ax=
From (i),
Wor
k done
2
0
x
x
Fdx
=
=
=ò

2
22
0
3
2
ma x dx
éù
=
êú
ëû
ò

2
3
2
0
3
23
x
ma
æö
=´ ç÷
ç÷
èø

21
8
2
ma=´

1
(0.5) (25) 850 J
2
= ´ ´ ´=
11.
(b)As given that power = constant
As we know that power (P)dW F dx F dx
P
dt dt dt
×
===
u urr
As the body is moving unidirectionally.
Hence, cos0F dx Fdx Fdx× = °=
constant
Fdx
P
dt
== (P=Q constant by question)
Now, by dimensional formula
0Fv×=
[F] [v] = constant
[MLT
–2
] [LT
–1
] = constant
[ML
2
T
–3
] = constant
3
2T
L
M
=
(As mas
s of body constant)
2 3 3/2
L T LTµ Þµ Þ Displacement
3/2
()dtµ
Verifies
the graph (b).
12. (d)The speed of earth around the sun can never be zero
or negative, so the kinetic energy of earth cannot be
zero and negative.
So, option (b) and (c) represents wrongly the variation
in kinetic energy of earth.
A BSUN
When the earth is closest to the sun, speed of the
earth is maximum, hence, KE is maximum. When theearth is farthest from the sun speed is minimum hence,
KE is minimum.
So, K.E. of earth increases (B to A) and then decrease,
variation is correctly represented by option (d).
13. (c)When a pendulum oscillates in air, due to air resistance
the force of friction acts between bob of pendulum
and air, so it will lose energy continuously in
overcoming. Therefore, total mechanical energy (KE
+ PE) of the pendulum decreases continuously with
time and finally becomes zero.
Sum of KE and PE can never be negative. So, option
(a) and (d) are incorrect. Hence option (c) is verifies.
14. (a)As given that, mass (m) = 5 kg,
n = 300 revolution
Radius (R) = 1 m
t = 60 sec
2
(300 2 ) rad / 60s
n
t
pæö
w= = ´ ´p
ç÷
èø

600
rad/s = 10 rad/s
60
´p
=p
linear sp
eed (v) = wR = (10p × 1)
v = 10p m/s
KE
21
2
mv=

21
5 (10)
2
=´´p

2 1
100 5
2
= p´´

2
250 J=p
So, verifies the option (a).
15. (b)P.E. is maximum when drop start falling at
t = 0 as it fall is P.E. decrease gradually to zero. So, it
rejects the graph (a), (c) and (d).
K.E. at t = 0 is zero as drop falls with zero velocity, its
velocity increases (gradually), hence, first KE also
increases. After sometime speed (velocity) is constant
this is called terminal velocity, so, KE also become
constant. It happens when it falls
3
4
æö
ç÷
èø
height or
remains at
4
4
æö
ç÷
èø
from ground, then PE decreases
continuously as the drop is falling continuously.
The variation in PE and KE is best represented by (b).

174. 170 PHYSICS
16. (d)As gi
ven that, h = 1.5 m, v = 1 m/s, m = 10 kg, g = 10 ms
–2
By the law of conservation of mechanical energy as
no force acts on shotput after thrown.
(PE)
i
+ (KE)
i
= (PE)
f
+ (KE)
f
21
0 (KE)
2
iif
mgh mv+ =+
(KE)
f
=
21
2
ii
mgh mv+
Total energy wh
en it reaches ground, so21
(KE) 10 10 1.5 10 (1)
2
f=´´ +´´
E = 150 + 5 = 155 J.
17. (
b)First velocity of the iron sphere
2V gh= after sometime its velo city becomes
constant, called terminal velocity. Hence, accordingfirst KE increases and then becomes constant due toresistance of sphere and water which is represented
by (b).
18. (c)As given that,
150
150 g kg 0.15 kg
1000
m===
Dt = time of
contact = 0.001 s
126×1000
126km/h = m/s
60 60
u=
´

5
126 35 m/s
18
= ´=
v = –126 km/h
5
126 35 m/s
18
=- ´ =-
So, final ve
locity is acc. to initial force applied by
batsman.
So, change in momentum of the ball
3
( ) ( 35 35) kg-m/ s
20
p mvuD= -= --

3 21
( 70)
202
= - =-
As we know that, force
421/2
N = –1.05 × 10 N
0.001
p
F
t
D-
==
D
Hence negative sign shown that direction of force will
be opposite to initial velocity which taken positivedirection. Hence verify the option (c).
NEET/AIPMT (2013-2017) Questions
19. (d) Given : F
r
= 3ij+
$$
1
r
ur
=
$
()2ik+
$
,
2
r
u ur
= ()4i3jk+-
r
$$
r
r
=
21
rr-
uur ur
= ()4i3jk+-
r
$$
–
$
()2ik+
$
orr
r
= 2i 3j+
$$
–
$
2k
So work done by the given force w = f .r
rr
=
()
$
( )3i j . 2i 3j2k+ +-
$$ $$ = 6 + 3 = 9J
20. (
a)
4m / s ec
m
3
2 kg m
2
P
resultant
1 kg
x
y
8 m/sec
12 m/sec
m
1
P
resulta
nt
=
22
12 16+
= 144 256+ = 20
m
3
v
3
= 20 (momentum of th
ird part)
or, m
3
=
20
4
= 5 kg
21. (c) A
ccording to law of conservation of momentum
MV + mnv = 0
Þ
0.01 kg 10 800 m/s
100
mNv
V
M
- - ´´
==
Þ – 0.8 m/s
According t
o work energy theorem,
Average work done = Change in average kinetic energy
i.e.,
2
av av rms
1
2
F S mV´=
Þ
2
av max rms1
2 22
FVtV
m=
Þ F
av
= 8 N
22. (d)
As the particle is moving in a potential energy region.
\ Kinetic energy ³ 0
And, total energy E = K.E. + P.E.
Þ U(x) £ E
23. (a)
Q Power
w
P
t
=
Þ
12
21
30 30 1
1 minute 60 2
Pt ss
Pts
== ==
(t
1
= 1 minute; t
2
= 30 secon d given)
24. (b) By conservation of linear momentum
2mv
1
= 1
v
2mvv
2
Þ=
m
v
v
m
2m
v
1
As two masses of each of mass m move perpendicular
to each other.
Total KE generated
=
222
1
11
mv mv (2m)v
222
1
++

175. 171Work, Energy and Power
=
2
22mv3
mv mv
22
+=
25. (d) A
s we know power P =
dw
dt
Þw = Pt =
1
2
mV
2
So, v =
2Pt
m
Hence, acceleration
dV 2P1
a.
dtm 2t
==
Therefo
re, force on the particle at time ‘t’
= ma =
2
–1/22Km 1 Km mK
.t
m 2t22t
==
26.
(b) As we know work done in stretching spring
21
w kx
2
=
where k =
spring constant
x = extension
Case (a) If extension (x) is same,
21
W Kx
2
=
So,W
P
> W
Q
(QK
P
> K
Q
)
Case (b) If spr
ing force (F) is same
2
F
W
2K
=
So,W
Q
> W
P
27. (
d) From, F = ma
a =
F
m
=
0.1x dV
0.01x V
10 dx
==
So,
2
1
v 30
v 20
x
vdV dx
100
=òò
–
30
222
20
1
V
V x 30 30 20 20
–
2 200 200 200
V
´´
==
= 4.5
– 2 = 2.5
( )
22
21
1
m V – V 10 2.5 J – 25J
2
=´=
Final
K.E.
=
22
21
111
mv mv – 25 10 10 10 – 25
222
= =´´´
= 500 – 25 =
475 J
28. (d) Power
F.V PAV ghAV= =r
rrr
F
P and P gh
A
éù
= =r
êú
ëû
Q
= 13.6 × 10
3
× 10 × 150 × 10
–3
× 0.5
× 10
–3
/60
=
102
60
= 1.70 watt
29. (a) When ball collides with the ground it loses its 50% of
energy
\
f
i
KE1
KE2
=
Þ
2
f
2
i
1
mV
12
1 2
mV
2
=
or
f
i
V 1
V 2
=
or, 2
0
2gh 1
2
V 2gh
=
+
or, 4gh =
2
0
V 2gh+
\ V
0
= 20ms
–1

M
M
30. (d) Here, M
1
= M
2
and u
2
= 0
u
1
= V, V
1
=
V
3
; V
2
= ?
u
1
=V
M
1
u
2
=0
M
2
M
1
M
2
q
f
V
1=V/3
V
2=?
From figure, along x-axis,
M
1
u
1
+ M
2
u
2
= M
1
V
1
cosq + M
2
V
2
cosf ...(i)
Along y-axis
0 = M
1
V
1
sinq – M
2
V
s
sinf ...(ii)
By law of conservation of kinetic energy
2222
11 22 1 1 22
1111
Mu Mu MV MV
2222
+=+ ...(iii)
Putting M
1
= M
2
and u
2
= 0
in equation (i), (ii) and
(iii) we get
q + f =
2
p
= 90°
and
2 22
112
u VV=+
V
2
=
2
2
2
V
V
3
æö
+
ç÷
èø
11
V
u VandV
3
éù
==
êú
ëû
Q
or, V
2
–
2
V
3
æö
ç÷
èø
=
2
2
V
2
2V
V
9
- =
2
2
V
or
2
2
V =
28
V
9
Þ V
2
=
22
3
V
31. (d)
For collision
B/AV
ur
should be along
()A/B
rBA®
r

176. 172 PHYSICS
So,
21 12
2 1 12
rrVV
V V rr
--
=
--
ur ur rr
A B
BA
1V 2V
32. (d) Given force
2ˆˆ
F 2ti 3tj=+
r
According to Newton's second law of motion,
2dv
ˆˆ
m 2ti 3tj
dt
=+
r
(m = 1 kg)
Þ
v
0
dv
ò
r
r
= ( )
t
2
0
ˆˆ
2ti 3t j dt+
ò
Þ
23ˆˆ
v ti tj=+
r
Power P =
2 23ˆˆˆˆ
F·v (2t i 3t j) · (t i t j)=++
rr
= (2t
3
+ 3t
5
)W
33. (a) Giv
en: Mass of particle, M = 10g =
10
kg
1000
radius of circle R = 6.4 cm
Kinetic energy E of particle = 8 × 10
–4
J
acceleration a
t
= ?
21
mv
2
= EÞ
21 10
v
2 1000
æö
ç÷
èø
= 8 × 10
–4
Þv
2
= 16 ×
10
–2
Þv = 4 × 10
–1
= 0.4 m/s
Now, using
v
2
= u
2
+ 2a
t
s(s = 4pR)
(0.4)
2
= 0
2
+ 2a
t

22 6.4
4
7 100
æö
´´ç÷
èø
Þa
t
= (0.4)
2
×
7 100
8 22 6.4
´
´´
= 0.1 m/s
2
34. (c
) From work-energy theorem,
W
g
+ W
a
= DK.E
or, mgh + W
a
=
21
mv0
2
-
3 3 32
a
1
10 10 10 W 10 (50)
2
--
´´+=´´
Þ W
a
= –8.
75 J
which is the work done due to air resistanceWork done due to gravity = mgh= 10
–3
× 10 × 10
3
= 10 J

177. CENTRE OF MASS
Centre of mass for a system of particles is defined as that point
where the entire mass of the system is imagined to be
concentrated.
If
321r,r,r......... be the position vectors of masses m
1
, m
2
, m
3
....... respectively from the origin O,
y
O
x
m
1
m
2
m
4
m
3
r
4
r
1
r
2
r
3
then the centre of mass of the system is
11 22 33
1
23
( .....)
( ......)
+++
=
+++
u r u ur uur
uuur
cm
mr mr mr
r
m
mm
1
1
n
ii
i
mr
M
=
=å
u ur
where M is the total mass of the system of particles. The product
of mass of the particles and its position vector w.r.t. the referencepoint is called moment of mass
i.e., moment of mass
mr=´
r
MOTION OF CENTRE OF MASS
The motion of the centre of mass is governed by the equation
cm ext
MAF=
r r
where
2
2
()
=
r
r
cm
d rcm
A
dt
Momentum c
onservation of a system of particles :
In the absence of external forces, the velocity of the centre ofmass remains constant.
We have, MA
cm
=F
ext
If F
ext
=0
( )cm
d
Mv
dt
=0
\ v
cm
= constant
.
Hence, momentum (Mv
cm
= constant) of the centre of mass
system is conserved.
Rigid Bodies
If a body does not undergo any change in shape by the action of
a force, it is said to be rigid.
If such body undergoes some displacement, every particle in it
undergoes the same displacement. No real body can be perfectly
rigid.
Rotatory Motion
A body rotating about a fixed axis then every particle of the
body moves in a circle and the centres of all these circles lie on
axis of rotation. The motion of the body is said to possess
rotational motion.
Keep in Memory
1.The c
entre of mass of a system of two identical particles
lies in between them on the line joining the particles.
2.If m
1
= m
2
11 22 12
cm
12
(mr mr) (r r)
r
(mm)2
+ +
==
+
u ur u uru r u ur
r
so, fo
r particles of equal masses the centre of mass is
located at the mean position vector of the particles.
3.The position of centre of mass remains unchanged in pure
rotatory motion. But it changes with time in translatory
motion or rolling motion.
4.The position of centre of mass of a body is independent
of the choice of co-ordinate system.
5.If we take the centre of mass at the origin, then the sum of
the moments of the masses at of the system about the
origin
ii
mrS
r
is zero.
6.In pure rotatory motion, the axis of rotation passes through
the centre of mass.
7.If external force is zero then the velocity of the centre of
mass of a body remains constant.
8.The centre of mass and centre of gravity of a body
coincide, if the value of g is same throughout the dimension
of the body.
9.In kinematics and dynamics, whole of the mass of a body
can be assumed to be concentrated at the centre of mass.
10.The location of the centre of mass depends on the shape
and nature of distribution of mass of the body.
(a) The position of centre of mass of continuous bodies
can be found using integration as
7
System of Particles
and Rotational Motion

178. 174 PHYSICS
Relation
ship between angular velocity and linear velocity:
v
r
w
P
x
y
z
q
Þ v = rw sin q
rrr
v= ×rw
where q is the angle between w & r.
MOMENT OF INERTIA AND RADIUS OF GYRATION
A rigid body having constituent particles of masses m
1
, m
2
,
....m
n
and r
1
, r
2
... r
n
be their respective distances from the axis of
rotation then moment of inertia is given by,
I = m
1
r
1
2
+ m
2
r
2
2
+ ... + m
n
r
n
2 2
1
n
ii
i
mr
=
=å
The moment of inertia
of continuous mass distribution is given
by

2
I r dm=ò
where r is the perpendicular distance of the small mass dm from
the axis of rotation.
Its SI unit is kgm
2
. It is a tensor.
Radius of gyration :
The radius of gyration of a body about its axis of rotation may
be defined as the distance from the axis of rotation at which, if
the entire mass of the body were concentrated, its moment of
inertia about the given axis would be same as with its actual
distribution of mass.
Radius of gyration k is given by,
I = MK
2
or
1
2
1 2
2I
n
ii
i
i
mr
K
Mm
éù
Sêú
éù
êú==
êú
Sêúëû ëû
where, M = Sm
i
r
1
m
1
r
2
m
2
r
3 m
3
k
Sm
r
4
m
4
X
Y
Also,
222 2
1 23
.....
n
rrr r
k
n
+++
=
Therefor
e, radius of gyration (k) equals the root mean square of
the distances of particles from the axis of rotation.
òòò
=== dmz
M
1
Z,dmy
M
1
Y,dmx
M
1
X
cmcmcm
where, x, y a
nd z are the co-ordinates of small mass
dm and M is the total mass of the system.
(b) The C.M. of a uniform rod of length L is at its middle
point.
(c) Centre of mass of a uniform semicircular wire is at
÷
ø
ö
ç
è
æ
p
R2
,0, where R is the radius of the semicircular wire.
It does not depend on mass.
O
Y
X
R
(d) For symmetrical bodies of uniform mass distribution,
the C. M. lies at the geometrical centre
ANGULAR VELOCITY AND ANGULAR ACCELERATION
The angular velocity is defined as the angle covered by the
radius vector per unit time. It is denoted by
w.
O
q
r
P
s
Q
X
Y
Average angular velocity
t
Dq
w=
D
The unit of angular velocity is rad/sec.
The instantaneous angular velocity w (similar to instantaneous
linear velocity) is defined as
o
lim
D®
Dqq
w==
Dt
d
t dt
The angular acceleration is the rate of change of angular
velocity. It is denoted by a.
The average angular acceleration a
avg.
of a rotating body is
21
.
21
w -w Dw
a==
-D
avg
ttt
In analogy to linear acceleration a
r
, the instantaneous angular
acceleration is defined as
lim
d®
Dww
a==
Dto
d
t dt
The unit of angular acceleration is rad/sec
2
.

179. 175System of Particles and Rotational Motion
GENERAL THEOREMS ON MOMENT OF INERTIA
Theorem of perpendicular axis :
According to this theorem “ the moment of inertia of a plane
lamina (a plane lamina is a 2-dimensional body. Its third
dimension is so small that it can be neglected.) about an axis,
perpendicular to the plane of lamina is equal to the sum of the
moment of inertia of the lamina about two axes perpendicular
to each other, in its own plane and intersecting each other at
the point, where the perpendicular axes passes through it.
If I
x
and I
y
be the moment of inertia of a plane lamina (or 2D rigid
body) about the perpendicular axis OX and OY respectively,
which lie in plane of lamina and intersect each other at O, then
moment of inertia (I
z
) about an axis passing through (OZ) and
perpendicular to its plane is given by
X
p(x, y)
x
Y
yo
Z
r
I
x
+ I
y
= I
z
Let u
s consider a particle of mass m at point P distance r from
origin O, where
22
yxr+=
soI
x
+ I
y
= Smy
2
+ Smx
2
= Smr
2
i.e., I
z
=
I
x
+ I
y
Theorem of parallel axes :
(Derived by Steiner) This theorem is true for both plane laminar
body and thin 3D body. It states that “the moment of inertia of a
body about any axis is equal to its moment of inertia about a
parallel axis through its centre of mass plus the product of the
mass of the body and the square of the distance between two
axes.
P A
Q B
M
c.m
O
r
Let AB
be the axis in plane of paper about which, the moment of
inertia (I) of plane lamina is to be determined and PQ an axis
parallel to AB, passing through centre of mass O of lamina is at a
distance ‘r’ from AB.
Consider a mass element m of lamina at point P distant x from PQ.
Now the moment of inertia of the element about AB = m (x + r)
2
so moment of inertia of whole lamina about AB is
I = Sm(x+r)
2
= Smx
2
+ Smr
2
+2 Smxr
Where first term on R.H.S is S mx
2
= I
c.m.
moment of inertia of
lamina about PQ through its centre of mass, second term on
R.H.S. is Smr
2
= r
2
Sm = Mr
2
, M is whole mass of lamina, third
term on R.H.S is (Smx) r = 0, because Smx is equal to moments of
all particles of lamina about an axis PQ, passing through its centre
of mass. Hence
I = I
c.m.
+ M.r
2
i.e., the moment of inertia of lamina about AB = its moment of
inertia about a parallel axis PQ passing through its centre of mass
+ mass of lamina×(distance between two axes)
2
Example 1.
Three rings each of mass P and radius Q are arranged as
shown in fig. What will be the moment of inertia of the
arrangement about YY’?
Y
Y’
3
Q
QQ
1 2
P P
P
Solution :
Mo
ment of inertia of each ring about its diameter
=
2
PQ
2
1
.
So total mome
nt of inertia of all three rings about Y Y' is
I
total
= I
1
+ I
2
+ I
3
Using theorem of parallel axes (for rings 1 and 2), we get
I
total
=
22222
PQ
2
1
PQPQ
2
1
PQPQ
2
1
+÷
ø
ö
ç
è
æ
++÷
ø
ö
ç
è
æ
+ =
27
PQ
2
Example 2.
F
our particles each of mass m are lying symmetrically on
the rim of a disc of mass M and radius R. Find MI of this
system about an axis passing through one of the particles
and perpendicular to plane of disc.
Solution :
According to the theorem of parallel axes, MI of disc about
an axis passing through K and perpendicular to plane of
disc, is
=
222
MR
2
3
MRMR
2
1
=+
R
RO
K
Particle 2
Particle 3
Particle 4
Particle 1
Total MI of th e system =
2222
)R2(m)R2(m)R2(mMR
2
3
+++
2
MR
19
2
=

180. 176 PHYSICS
Example 3.
Thr
ee identical rods, each of length
l, are joined to from a
rigid equilateral triangle. Find its radius of gyration about
an axis passing through a corner and perpendicular to
the plane of the triangle.
Solution :
B
A
60
o
D
l
C
22
2
ABACBCD
ml ml
IIII (Imh)
33
=++= +++
= ú
û
ù
ê
ë
é
++=++
4
3
12
1
3
2
m)60sin(m
12
m
m
3
2 22
2
2
ll
l
l
=
22
k)m3(m
2
3
=l or
2
k=
l
Example 4.
A uniform rod of mass m and lengt
h
l makes a constant
angle
q with the axis of rotation which passes through
one end of the rod. Determine its moment of inertia about
this axis.
Solution :
Mass of element of uniform rod =
dx
m
÷
ø
ö
ç
è
æ
l
x sinq
x
dx
q
Axis
Moment of inertia of the element about the axis
=
2
)sinx(dx
m
q÷
ø
ö
ç
è
æ
l
.
2
222
0
mm
I sinθ. xdx sin θ
3
== ò
l
l
l
Example 5.
A circular plate of uni
form thickness has a diameter of
56 cm. A circular portion of diameter 42 cm is removed
from one edge of the plate as shown. Find the position of
centre of mass of the remaining portion.
Solution :
O
m
2
C
1
m
1m
C
2
Area of whole plate = p (56/2)
2
= 784 p sq. cm.
Area of cut portion = p (42/2)
2
= 441 p sq. cm. ;
Area of remaining portion = 784p – 441p = 343 p cm
2
;
As mass µarea.
\
1
2
mMass of cut portion 441 9
Mass of remaining portion m 343 7
p
===
p
Let C
2
be centre of
mass of remaining portion and C
1
be
centre of mass of cut portion.
O is centre of mass of the whole disc.;
OC
1
= r
1
= 28 - 21 = 7 cm.
OC
2
= r
2
= ?;
Equating moments of masses about O,
we get m
2
× r
2
= m
1
× r
1
Þ
97
7
9
r
m
m
r
1
2
1
2 =´=´=
\ Centre of mas
s of remaining portion is at 9 cm to the left
of centre of disc.
Example 6.
Show that the centre of mass of a rod of mass M and length
L lies midway between its ends, assuming the rod has a
uniform mass per unit length.
x
O
x
dx
dm
Solution :
By symm
etry, we see that y
CM
= z
CM
= 0 if the rod is placed
along the x axis. Furthermore, if we call the mass per unitlength l (the linear mass density), then l = M/L for a uniform
rod.If we divide the rod into elements of length dx, then the
mass of each element is dm =
λdx. Since an arbitrary element
of each element is at a distance x from the origin, equation
gives
x
CM
=
LL 2
00
11L
x dm x dx
M M 2M
l
= l=òò
Because l = M/L,
this reduces to x
CM
=
2
LML
2ML2
æö
=
ç÷
èø
One can also argu
e that by symmetry, x
CM
= L/2.

181. 177System of Particles and Rotational Motion
MOMENT OF INERTIA AND RADIUS OF GYRATION OF DIFFERENT OBJECTS
Shape of bodyRotational axis Figure Moment of Radius of
inertia gyration
(1) Ring (a) Perpendicular to plane
M = mass passing through centre
R = radiusof mass
MR
2
R
(b) Diameter in the plane
1
2
MR
2
R
2
(c) Tangent perpendicular
to plane
2MR
2
2R
(d) Tangent in the plane
3
2
MR
2
3
R
2
(2) Disc (a) Perpendicular to plane
passing through centre
of mass
21
MR
2
R
2
(b) Diameter in the plane
2
MR
4
R
2
(c) Tangent in the plane
5
4
MR
2
5
2
R
M
R cm
C
I
M
R
cm
B
I
M
R
cm
CM
'
dI I
M
Rcm
D
I
I
c
R
M
cm
I
d
Y
Z
Ic
R
M
cm
I
d

182. 178 PHYSICS
(d) Tangent perpendicular
to plane
3
2
MR
2 3
2

R
(3) Thin walled(a) Geometrical axis MR
2
R
cylinder
(b) Perpendicular to length
22
RL
M
2 12
æö
+
ç÷
èø
22
RL
2 12
+
passing through centre of mass
(c) Perpendicular to length 22
RL
M
23
æö
+
ç÷
èø
22
RL
23
+
passing through one end
(4) Solid cylinder(a) Geometrical axis
2
MR
2
R
2
(b) Perpendicular to length
22
RL
M
4 12
æö
+
ç÷
èø
22
RL
4 12
+
passing through centre of mass
(c) Perpendicular to length
22
RL
M
43
éù
+êú
êúëû
22
RL
43
+passing through one end
I
cI
d
R
M
cm
Ic
M
cm
L
R
Ic
M
cm
L
R
Ic
M
cm
L
R
I
d
I
c
M
cm
R
L
M
L
Ic
cm
R
I
M
L
Ic
cm
I
d
R

183. 179System of Particles and Rotational Motion
(5) Annular disc(a) Perpendicular to plane
passing through centre
M
2
[
22
12
RR+]
22
12
RR
2
+
of mass
(b) Diameter in the plane
22
12M[R R]
4
+
22
12
RR
4
+
(6) Hollow cylinder(a) Geometrical axis
22
12
RR
M
2
éù+
êú
êúëû
22
12
RR
2
+
(b) Perpendicular to length
passing through centre
222
12
(R R)L
M
124
éù +
+êú
êúëû
222
12
RRL
124
+
+
of mass
(7) Solid sphere(a) Along the diameter
2
5
MR
2

2
5
R
(b) Along the tangent
7
5
MR
2

7
5
R
I
c
R
1
R2C
M
I
c
R
1
R
2
C
M
I
M
cm
L
R
1
R
2
M
cm
L
R
1
R
2
I
Ic
M
cm
R
Ic
M
cm
R
Id

184. 180 PHYSICS
(8) Thin sph
erical shell(a) Along the diameter
I
c
M
R
2
3
MR
2

2
3
R
(b) Along the
tangent
II
cd
M
R
5
3
MR
2

5
3
R
(9) Hollow sphereAlong th
e diameter
Cavity
Hollow sphere
R
r
I
2
5
M
55
33
Rr
Rr
éù-
êú
-êúëû
55
33
2(R r)
5(R r)
-
-
(10) Thin rod (a) Pe
rpendicular to length
2
ML
12
L
23
passing through centre
of mass
(b) Perpendicular to length
2
ML
3
L
3
passing through one end
M
cm
L
Ic
M
cm
L
Ic
Id

185. 181System of Particles and Rotational Motion
(11) Rectangular plate(a) Perpendicular to length in
2
Ma
12
a
23
the plane passing through
centre of mass
a
b O
AB
A
B
C
D
I
(b) Perpendicular to breadth in
2
Mb
12
b
23
the plane passing through
centre of mass
a
b
O
CD
A
B
C
D
I
(c) Perpendicular to plane
passing through centre of mass
a
b
O
CM
A
B
C
D
I
22
M(a b)
12
+
22
ab
23
+
(12) Square Plate (a) Pe
rpendicular to plane passing
I1
M
a
a
through centre of mass I
1
=
2
Ma
6
a
6
(b) Diagonal passing through
I2I3 M
a
a

2
23
Ma
II
12
=
a
23
centre of mass
(13) Cube (a) Perpendicular to plane passing
I
1
M
a
2
1
Ma
6
I=
6
a
through centre of mass
(b) Perpendicular to plane passing
M
a
I
2
2
2
2Ma
I=
3
2
3
a
through one end

186. 182 PHYSICS
TORQUE, ANGULAR MOMENTUM
AND ANGULAR
IMPULSE
Torque :
The moment of force is called torque. It is defined as the product
of force and the perpendicular distance of the force from the
axis of rotation.
O X
Y
q
qr
F
i.e.,rFt=´
rrr
or, sinrFt=q
where q is the angle between r
r
and F
r
.
Its S.I. unit is (N-m). The di
mensions of torque [ML
2
T
–2
] are
the same as that of energy but it is not energy.Note : If the line of action of a force passes through axis of
rotation then no torque will be formed.Angular Momentum :
The angular momentum of a particle about an arbitrry point 'O' is
the moment of linear momentum taken about that point.
r
p
O X
Y
q
q
It is given as L rp=´
rrr
or, sinL rp=q
where q is the angle between r
r
and F
r
.
Angular Impulse :
2
1
21
t
t
J dtLL=t=-
ò
(= Change
in angular momentum)
CONSERVATION OF ANGULAR MOMENTUM
From equation
dt
Ld
ext=t
r
.
If constantL0
ext =Þ=t
r
...(i)
This is called law
of conservation of angular momentum.
According to this “if resultant external torque
ext
t
r
acting on
the system is zero then total angular momentum of the system is
constant.
The
magnitude of angular momentum for a system is given by
iii
n
1i
v)rm(|L|
=
å= wå=
=
)rm(
2
ii
n
1i
[Q v = rw]
Þ w=w´å=
=
I)I(|L|
i
n
1i
...(ii)
Where I
i
is the mome
nt of inertia of the i
th
particle of that system
and I is total moment of inertia of the system
n321i
n
1i
I....................IIIII++++=å=
=
or
2
nn
2
22
2
11
rm....................rmrmI+++= ...(iii)
So if a system undergoes a redistribution of its mass, then itsmoment of inertia changes but since no external torque is appliedon the system so total angular momentum is constant before andafter the distribution of mass, even if moment of inertia of thesystem is changed.
Initial Final
LL =
u ur u ur
...(iv)
or,I
1
w
1
= I
2
w
2
...(v)
whe
re L initial denote the state previous to the redistribution of
mass and final denote the state after the redistribution of mass inthat system.
A comparison of useful relations in rotational and translational
or linear motion : Rotational motion about Linear motion
a fixed axis
Angular veloc
ity
d
dt
q
w= Linear velocity
dx
v
dt
=
Angular accelerat
ion
d
dt
w
a= Linear accelerat ion
dv
a
dt
=
Resultant torque t = Ia Resul
tant force F = ma
Equations of rotationalEquations of linear motion
motion
0
2
00
22
00
t
constant then t ½ t
2()
w=w +aì
ïï
a= q-q=w
+aí
ï
w = w + a q-qïî
2
22
v u at
1
a constant then s ut at
2
v u 2as
ì=+
ï
ï
= =+í
ï
ï
-=î
where (s = x – x
0
)
Wo
rk
o
Wd
q
q
= tqò
rr
work
o
x
x
W Fdx=ò
rr
Kinetic energy E
k
= ½ Iw
2
Kinetic energy E
k
=½ mv
2
Power P = tw Power P = Fv
rr
Angular momentum L = Iw
rr
Linear momentum p = mv
Torque
dL
dt
t= Force
dp
F
dt
=
Work Energy Theorem in Rotatio
nal Motion :
According to this theorem “the work done by external forces in
rotating a rigid body about a fixed axis is equal to the change
in rotational kinetic energy of the body.”

187. 183System of Particles and Rotational Motion
Since, we can express the torque as
t = Ia
dt
d
.
d
d
I
dt
d
I
q
q
w
=
w
=
q
w
w=t
d
d
I but tdq = dW
Þtdq = dW = Iwdw
B
y integrating the above expression, we get total work done by
all external force on the body, which is written as
2
1
2
2
I
2
1
I
2
1
dIdW
2
1
2
1
w-w=ww=qt=
òò
w
w
q
q
where the a
ngular velocity of the body changes from w
1
to w
2
as
the angular displacement

changes from q
1
to

q
2
due to external
force
.extF
r
on the body..
Rotational Kinetic Energy :
Let us consider a rigid body (collection of small particles) of high
symmetry which is purely rotating about z-axis with an angular
velocity w. Each particle has some energy, determined by m
i
and
v
i
. The kinetic energy of m
i
particle is
()
2
iiik
vm½E= ... (i)
x
y
v
i
m
ir
i
O
w
Fig. The total kinetic energy of the body is ½I
w
2
.
Now we know that in rigid body every particle moves with same
angular velocity, the individual linear velocities depends on the
distance r
i
from the axis of rotation according to the expression
(v
i
= r
i
w). Hence the total kinetic energy of rotating rigid body is
equal to the sum of kinetic energies of individual particles.
2
iiikk
vm½)E(ES=S=
2
)rm(
2
2
ii
w
S=
2
½
k
EI=w ... (ii)
where
2
i
i
rmIS= is moment of inertia of the rigid body..
Now consider a rigid body which is rolling without slipping. In
this case it possesses simultaneous translatory motion and
rotatory motion and the total kinetic energy of the rigid body
K.E
Total
= rotation K.E. + translational K.E. of C.M.
E
k
= naltranslatiokrotationalk
)E()E(+ ... (iii)
E
k
=
2
cm
2
Mv½I½+w ... (iv)
)K/r1(I½)r/K1(Mv½E
222222
cmk
+w=+= ... (v)
where,
v
cm
= linear velocity of the centre of mass of the rigid body
K = radius of gyration )rMMKI(
2
ii
2
S==
r = radius of moving rigid body
I = moment of inertia of the rigid body about centre of mass
½ Iw
2
= rotational kinetic energy about the centre of mass.
Hence it is clear from the expression that total kinetic energy of
rolling body is equal to the sum of rotational kinetic energy about
centre of mass (C.M.) and translational kinetic energy of the centre
of mass of body.
Body Rolling without Slipping on an Inclined Plane :
When a body performed translatory as well as rotatory motion
then we can say that the body is in rolling motion.
Acceleration for body rolling down an inclined plane without
slipping.
Let M is the mass of the body, R is its radius and I is the moment
of inertia about the centre of mass and K is the radius of gyration.
q
Mgsinq
Mg
Mgcosq
q
R
f
Force equa
tion, MafsinMg =-q ...(1)
Torque equation, a=IfR ...(2)
Also,
2
MKI= ...(3)
Þ
R
MK
f
2
a
= ...(4)
Adding, eqn
. (1) & (4)
÷
÷
ø
ö
ç
ç
è
æ
+
a
=q a
R
K
MsinMg
2
R
a
=a (Q Motion is p
ure rolling)
2
22
2
sin
sin
1
Kag
Mg M aa
RK
R
æö q
q= + Þ=ç÷
ç÷
èø
+
No
w assume in fig. that a body (which has high symmetry such
as cylinder, sphere etc.) is rolling down an incline plane without
slipping. This is possible only if friction is present between object
and incline plane, because it provides net torque to the body for
rotating about the centre of mass (since the line of action of the
other forces such as mg and R pass through the centre of mass of
rigid body, hence they do not produce torque in a body about
the centre of mass).
M
R
mgsinq
x
contact
pointmgh
q
w
f=Rm
s
v
c
c. m

188. 184 PHYSICS
A round obje
ct rolling down an incline, mechanical energy is
conserved if no slipping occurs.
But mechanical energy of the body remains constant despite of
friction because the contact point is at rest relative to the surface
at any instant.
For pure rolling motion v
c
= rw
so
2
c
2
ck
Mv½)r/v(I½E+=
2
c
2
c
k
vM
r
I
2
1
E
÷
÷
ø
ö
ç
ç
è
æ
+= ...(i)
As body rolls do
wn an incline, it loses potential energy Mgh (h
is the height of incline). Since body starts from rest at top, hence
its total kinetic energy at bottom given by eqn.(i) must be equal
to Mgh at top i.e.,
MghVM
r
I
2
1 2
c
2
c
=
÷
÷
ø
ö
ç
ç
è
æ
+
2/1
2
c
c
MR/I1
gh2
v
÷
÷
ø
ö
ç
ç
è
æ
+
=Þ
...(ii)
since h =x
sin q, where x is length of incline and I
c
=MK
2
(K is radius of gyration), then
½
22
c
R/K1
singx2
v
ú
û
ù
ê
ë
é
+
q
= ....(iii)
For roll
ing down an inclined plane without slipping, the
essential condition is :
q
+
³m tan
IMR
I
cm
2
cm
s ; m
s
is the coefficient of
static friction
where I
cm
is moment of inertia of the body about its C.M.
DIFFERENT TYPES OF MOTION
Pure translational motion : In this case the velocities at all three
points : (i) top most point P, (ii) C.M. and (iii) contact point O are
same.
(A)
P v
C.M (v)
v
O
Pure rotational motion : In this case the velocity of
(i) top most point P is Rw
(ii) C.M. is zero(iii) contact point O is – Rw
(B)
Rw
–Rw
w
O
C.M V
cm
= 0
Rolling Motin Co
mbination of translatory and rotatory
motion :
(i) In pure rolling motion, the contact point O remains at rest.
(ii) In pure rolling, the velocity of top most point is,
V = V
cm
+ wR = 2V
cm
VV
cm
V
cm
V+R
cm w
OwR
Contact
point
C.M
P
R
(iii) In ro
lling without slipping V
cm
= wR.
2 R = 2Vw
cm
V
cm
O
Surface at rest
Keep in Memory
1.The axis of
the rolling body is parallel to the plane on
which the body rolls in case of sphere, disc, ring.
2.Let (E
k
)
r
= rotational kinetic energy
(E
k
)
t
= translation kinetic energy
(a) For solid sphere, (E
k
)
r
= 40% of (E
k
)
t
(b) For shell (E
k
)
r
= 66% of (E
k
)
t
(c) For disc, (E
k
)
r
= 50% of (E
k
)
t
(d) For ring, (E
k
)
r
= (E
k
)
t
3.Translational kinetic energy is same for rolling bodies
having same M, R and w irrespective of their shape.
4.Total energy is minimum for solid sphere and maximum for
ring having same mass and radius.
5.Rotational kinetic energy is maximum for ring and minimum
for solid sphere of same mass and radius.
6.(i) The acceleration down the inclined plane for different
shapes of bodies of same mass and radius are as
follows :
Sphere > disc > shell > ring
(ii) The velocity down the plane is related as follows:
Sphere > disc > shell > ring
(iii) The time taken to reach the bottom of the inclined
plane is related as follows :
Ring > shell > disc > sphere.

189. 185System of Particles and Rotational Motion
This is because
(i) For ring 1
R
K
2
2
=
(ii) For dis
c
2
1
R
K
2
2
=
(iii)
For sphere
5
2
R
K
2
2
=
(iv) For s
hell
3
2
R
K
2
2
=
7.The angular spe
ed of all particles of a rotating/revolving
rigid body is same, although their linear velocities may be
different.
8.(i) Two particles moving with angular speed w
1
and w
2
on the same circular path and both in anti-clockwise
direction then their relative angular speed will be w
r
=
w
1
– w
2
.
1p
r
2
p
w
1
w
2
(ii) In the above case one particle will complete one
revolution more or less as compared to the other intime
1221
21
21r T
1
T
1
T
1
or
TT
TT22
T -=
-
=
w-w
p
=
w
p
=
9.Let two part
icles move on concentric circles having radius
r
1
and r
2
and their linear speeds v
1
, v
2
both along anti-
clockwise direction then their relative angular speed will
be
12
r
12
|v v|
ω
|r r|
-
=
-
ur ur
rr
1
r
2r
2
v
1v
10.As I µ K
2
and I = MK
2
, hence graph between I and K will be
a parabola. However graph between log I and log K will be
straight line.
Y
I
K
X
Y
log I
log K
X
11.Rotatio
nal KE,
2
rkI
2
1
)E(w= \ E
r
µ w
2
\ graph betwe
en
rk
)E( and w will be as below
Y
X
w
E
r
12.Angular
momentum L = Iw, hence L µ w.
\ Graph between L and w is a straight line.
Y
X
L
w
13.The moment of inertia is not a vector quantity because
clockwise or anti-clockwise, direction is not associated
with it. It is a tensor.
14.When a spherical/circular/cylindrical body is given a push,
it only slips when the friction is absent, It may roll with
slipping if friction is less than a particular value and it may
roll without slipping if the friction is sufficient. (i.e.
cm
2
cm
s
IMR(
tanI
+
q
³m )
15.When a body
rolls without slipping no work is done againstfriction.
Example 7.
A thin circular ring of mass M and radius R rotating about
its axis with a constant angular speed
w. Two blocks, each
of mass, m are attached gently to opposite ends of a
diameter of the ring. Find the angular speed of the ring.
Solution :
As L = Iw = constant. Therefore,

1122
wI=wI or
2
11
2
I
wI
=w =
m2M
M
K)m2M(
MK
2
2
+
w
=
+
w
Example
8
A solid sphere of mass 500 g and radius 10 cm rolls withoutslipping with a velocity of 20 cm/s. Find the total K.E. ofthe sphere.

190. 186 PHYSICS
Solution :
Tota
l K.E. =
2211
Iω mv
22
+ =
222121
mrω mv
252
´+
As, I
sphere
=
22
mr
5
=
222117
mv mv mv
5 2 10
+=
= J014.0
5
1
1000
500
10
7
2
=÷
ø
ö
ç
è
æ
´´ .
Example 9.
A body of radius R and mass M is rolling horizontally
w
ithout slipping with speed v, it then rolls up a hill to a
maximum height h. If h = 3v
2
/4 g, (a) what is the moment
of inertia of the body? (b) what might be the shape of the
body?
Solution :
(a)
22
total trans. r ot.
11
K K K MvI ω
22
=+=+
22
2
22
11vvI
MvIM
222 RR
æö éù
= + =+ ç÷
êúç÷
ëûèø
When it rolls up a hill to height h, the e
ntire kinetic
energy is converted into potential energy M g h
Thus
22
2
vIv
M Mgh Mg 3
2 4gR
éùéù
+== êúêú
ëû êúëû
or
2
2
I 3 MR
M MI
22R
éù
+ = \=
êú
ëû
(b) The
body may be a circular disc or a solid cylinder.
Example 10.
The angular velocity of a body changes from
w
1
to
w
2
without applying torque but by changing moment of
inertia. What will be the ratio of initial radius of gyration
to the final radius of gyration?
Solution :
Since I
1
w
1
= I
2
w
2
or
22
11 22
MK MKw=w
Þ
÷
÷
ø
ö
ç
ç
è
æ
w
w
=
1
2
2
1
K
K
Example 11.
A rod of mass M and length
l is suspended freely from its
end and it can oscillate in the vertical plane about the
point of suspension. It is pulled to one side and then
released. It passes through the equilibrium position with
angular speed
w. What is the kinetic energy while passing
through the mean position?
Solution :
K.E,
21
2
= Iw and I = (M l
2
/12) + M (l/2)
2
=
3
M
2
l
\
222
2
M
6
1
3
M
2
1
K w=w
ú
ú
û
ù
ê
ê
ë
é
= l
l
Example 12.
A cord is woun
d round the circumference of wheel of radius
r. The axis of the wheel is horizontal and moment of inertia
about it is I. A weight mg is attached to the end of the cord
and falls from rest. After falling through a distance h, what
will be the angular velocity of the wheel?
Solution :
2211
mgh Ιω mv
22
=+
222
rm
2
1
2
1
w+wI=
or
22
)rm(hg
m2 w+I=
1/2
2
2mgh
mr
éù
\ w=êú
I+êúëû
Example 13.
Let g be the acceleration due to gravity
at earth’s surface
and K be the rotational kinetic energy of the earth.Suppose the earth’s radius decrease by 2%, keeping all
other quantities same, then
(a) g decreases by 2% and K decreases by 4%
(b) g decreases by 4% and k increases by 2%
(c) g increases by 4% and K decreases by 4%
(d) g decreases by 4% and K increases by 4%
Solution : (c)
We know that,
2
R
MG
g=
Differentia
ting,
÷
ø
ö
ç
è
æ
-=
R
dR2
g
dg
....(1)
Further,
222
RM
5
3
2
1
I
2
1
K w
ú
û
ù
ê
ë
é
=w=
or
÷
ø
ö
ç
è
æ
´w=
R
dR2
M
10
3
K
dK 2
....(2)
When rad
ius decreases by 2%, then g increases by 4% and
K decreases by 4%.
Example 14.
A small solid ball rolls without slipping along the track
shown in fig. The radius of the circular part of track is R. If
the ball starts from rest at a height 8 R above the bottom,
what are the horizontal and vertical forces acting on it at
P?
8 R
R
R
O P
Ball
Solution :
Suppos
e m is the mass of the ball of radius r.
On reaching P, the net height through which the ball
descends is
8 R – R = 7R, (from the fig.)
\ decrease in P.E. of ball = mg × 7 R
This appears as total KE of ball at P.

191. 187System of Particles and Rotational Motion
Thus mg × 7R = KE of translation + KE of rotational
2211
mvI ω
22
=+
222
mr
5
2
2
1
mv
2
1
w÷
ø
ö
ç
è
æ
´+=

2
mv
10
7
=
\
2
v= 10 g R
(
where v = rw and r is radius of solid ball)
The horizontal force acting on the ball,
F
h
= centripetal force towards O
mg10
R
)Rg10(m
R
mv
2
===
Vertical fo
rce on the ball, F
v
= weight of the ball = mg.
Example 15.
A solid cylinder first rolls and then slides from rest down a
smooth inclined plane. Compare the velocities in the two
cases when the cylinder reaches the bottom of the incline.
Solution :
(i) We know that acc. of a body rolling down an inclined
plane is
22
1
R/K1
sing
a
+
q
=
For a s
olid cylinder
2
R
K
2
2
=;
q=
+
q
= sing
3
2
)2/11(
sing
a
1 ; From
2
v= u
2
+ 2 a s
2
1
2
v 0 2 gs
inθ
3
æö
=+´
ç÷
èø
l
4
gsin
3
= q´ l ...(i)
(i
i) Acc. of the sliding cylinder, a
2
=g sin q
From
2
v= u
2
+ 2 a s;
2
2
v 0 2gsinθ=+´ l ...(ii)
\
2
1
2
2
v 4g sinθ2
3 2gsinθ3v
´
==
´´
l
l
;
1
2
v
2/3
v
= .
Example 16.
A disc of mass
M and radius R is rolling with angular
speed
w on a horizontal plane (fig.). Determine the
magnitude of angular momentum of the disc about the
origin O.
Y
O
X
M
R
w
Solution :
The
angular momentum L is given by
cm
L
Ι ωMvR=+ R)R(MRM
2
1 2
w+w÷
ø
ö
ç
è
æ
= w=
2
RM
2
3

192. 188 PHYSICS

193. 189System of Particles and Rotational Motion
1.Centre of mass of the earth and the moon system lies
(a) closer to the earth
(b) closer to the moon
(c) at the mid-point of line joining the earth and the moon
(d) cannot be predicted
2.Four particles of masses m
1
,m
2
,m
3
and m
4
are placed at the
vertices A,B,C and D as respectively of a square shown.
The COM of the system will lie at diagonal AC if
(a)m
1
= m
3
(b)m
2
= m
4
A B
D
C
m
1 m
2
m
3
m
4
(c)m
1
= m
2
(d)m
3
= m
4
3.Tw
o spheres A and B of masses m and 2m and radii 2R and
R respectively are placed in contact as shown. The COM
of the system lies
A
B
2RR
(a) inside A (b) inside B
(c) at the point of contact (d) None of these
4.Moment of inertia does not depend upon(a) distribution of mass(b) axis of rotation
(c) point of application of force
(d) None of these
5.A disc is given a linear velocity on a rough horizontal surface
then its angular momentum is
(a) conserved about COM only
(b) conserved about the point of contact only
(c) conserved about all the points
(d) not conserved about any point.
6.A body cannot roll without slipping on a
(a) rough horizontal surface
(b) smooth horizontal surface
(c) rough inclined surface
(d) smooth inclined surface
7.A body is projected from ground with some angle to the
horizontal. The angular momentum about the initial position
will
(a) decrease
(b) increase
(c) remains same
(d) first increase then decrease
8.A ball tied to a string is swung in a vertical circle. Which of
the following remains constant?
(a) tension in the string
(b) speed of the ball
(c) centripetal force
(d) earth's pull on the ball
9.Angular momentum of a system of a particles changes,
when
(a) force acts on a body
(b) torque acts on a body
(c) direction of velocity changes
(d) None of these
10.Angular momentum is
(a) a polar vector (b) an axial vector
(c) a scalar (d) None of these
11.If a running boy jumps on a rotating table, which of the
following is conserved?
(a) Linear momentum
(b) K.E
(c) Angular momentum
(d) None of these
12.A gymnast takes turns with her arms & legs stretched. When
she pulls her arms & legs in
(a) the angular velocity decreases
(b) the moment of inertia decreases
(c) the angular velocity stays constant
(d) the angular momentum increases
13.Moment of inertia of a circular wire of mass M and radius R
about its diameter is
(a) MR
2
/2 (b) MR
2
(c) 2MR
2
(d) MR
2
/4.
14.A solid sphere is rotating in free space. If the radius of the
sphere is increased keeping mass same which one of the
following will not be affected ?
(a) Angular velocity
(b) Angular momentum
(c) Moment of inertia
(d) Rotational kinetic energy
15.One solid sphere A and another hollow sphere B are of
same mass and same outer radii. Their moment of inertia
about their diameters are respectively
A
Iand
B
I Such that
(a)
AB
II< (b)
AB
II>
(c)
ABII= (d)
AA
BB
Id
Id
=
where
A
d and
B
d are their densities.

194. 190 PHYSICS
16.Angular momentum
is
(a) moment of momentum
(b) product of mass and angular velocity
(c) product of M.I. and velocity
(d) moment of angular motion
17.The angular momentum of a system of particle is conserved
(a) when no external force acts upon the system
(b) when no external torque acts upon the system
(c) when no external impulse acts upon the system
(d) when axis of rotation remains same
18.Analogue of mass in rotational motion is
(a) moment of inertia(b) angular momentum
(c) gyration (d) None of these
19.Moment of inertia does not depend upon
(a) angular velocity of body
(b) shape and size
(c) mass
(d) position of axis of rotation
20.A hollow sphere is held suspended. Sand is now poured
into it in stages.
SAND
The centre of gravity of the sphere with the sand
(a) rises continuously
(b) remains unchanged in the process
(c) First rises and then falls to the original position
(d) First falls and then rises to the original position
21.A block Q of mass M is placed on a horizontal frictionless
surface AB and a body P of mass m is released on its
frictionless slope. As P slides by a length L on this slope of
inclination q, the block Q would slide by a distance
A B
M
Q
q
mP
(a) (m/M) L cos q (b) m L /(M + m)
(c) (M + m)/(m L cos q) (d) (m L cos q) / (m + M)
22.A solid sphere and a hollow sphere of the same materialand of same size can be distinguished without weighing
(a) by determining their moments of inertia about their
coaxial axes
(b) by rolling them simultaneously on an inclined plane
(c) by rotating them about a common axis of rotation
(d) by applying equal torque on them
23.A stick of length L and mass M lies on a frictionless
horizontal surface on which it is free to move in any way. A
ball of mass m moving with speed v collides elastically with
the stick as shown in fig.
L
M
m
v
If after the collision ball comes to rest, then what should be
the mass of the ball?
(a) m = 2 M (b) m = M
(c) m = M/2 (d) m = M/4
24.A flywheel rotates about an axis. Due to friction at the axis,
it experiences an angular retardation proportional to its
angular velocity. If its angular velocity falls to half while it
makes n rotations, how many more rotations will it make
before coming to rest?
(a) 2n (b) n
(c) n/2 (d) n/3
25.A raw egg and a hard boiled egg are made to spin on a table
with the same angular momentum about the same axis. The
ratio of the time taken by the two to stop is
(a) = 1 (b) < 1
(c) > 1 (d) None of these
1.A solid cylinder of mass 20 kg rotates about its axis with
angular speed 100 rad/s. The radius of the cylinder is 0.25
m. The K.E. associated with the rotation of the cylinder is
(a) 3025 J (b) 3225 J
(c) 3250 J (d) 3125 J
2.What is the moment of inertia of a solid sphere of density r
and radius R about its diameter?
(a)
r
5
R
176
105
(b) r
2
R
176
105
(c) r
5
R
105
176
(d) r
2
R
105
176
3.Two particles A and B, initially at rest, moves towards each
other under a mutual force of attraction. At the instant when
the speed of A is v and the speed of B is 2 v, the speed of
centre of mass is
(a) zero (b) v
(c) 1.5 v (d) 3 v
4.Point masses 1, 2, 3 and 4 kg are lying at the points (0, 0, 0),
(2, 0, 0), (0, 3, 0) and (–2, –2, 0) respectively. The moment of
inertia of this system about X-axis will be
(a) 43 kg –m
2
(b) 34 kg–m
2
(c) 27 kg – m
2
(d) 72 kg – m
2

195. 191System of Particles and Rotational Motion
5.A body having moment of inertia about its axis of rotation
equal to 3 kg-m
2
is rotating with angular velocity equal to 3
rad/s. Kinetic energy of this rotating body is the same as
that of a body of mass 27 kg moving with a speed of
(a) 1.0 m/s (b) 0.5 m/s
(c) 1.5 m/s (d) 2.0 m/s
6.A particle moves in a circle of radius 0.25 m at two
revolutions per second. The acceleration of the particle in
metre per second
2
is
(a)p
2
(b) 8p
2
(c)4p
2
(d) 2p
2
7.The radius of gyration of a body about an axis at a distance
6 cm from its centre of mass is 10 cm. Then, its radius of
gyration about a parallel axis through its centre of mass will
be
(a) 80 cm (b) 8 cm
(c) 0.8 cm (d) 80 m
8.A particle of mass m is moving in a plane along a circular
path of radius r. Its angular momentum about the axis of
rotation is L. The centripetal force acting on the particle is
(a)L
2
/mr (b) L
2
m/r
(c)L
2
/m r
3
(d) L
2
/m r
2
9.A small disc of radius 2 cm is cut from a disc of radius 6 cm.
If the distance between their centres is 3.2 cm, what is the
shift in the centre of mass of the disc?
(a) 0.4 cm (b) 2.4 cm
(c) 1.8 cm (d) 1.2 cm
10.A solid cylinder of mass m & radius R rolls down inclined
plane without slipping. The speed of its C.M. when it
reaches the bottom is
(a)
gh2
h
(b)4gh/3
(c) gh4/3
(d)
gh4
11.The ratio
of moment of inertia of circular ring & circular disc
having the same mass & radii about on axis passing the c.m
& perpendicular to plane is
(a) 1 : 1 (b) 2 : 1
(c) 1 : 2 (d) 4 : 1
12.A rod of length L is pivoted at one end and is rotated with
a uniform angular velocity in a horizontal plane. Let T
1
and
T
2
be the tensions at the points L/4 and
3L/4 away from the pivoted end. Then
(a)T
1
> T
2
(b) T
2
> T
1
(c)T
1
– T
2
(d) the relation between T
1
and T
2
depends on whether
the rod rotates clockwise and anticlockwise.
13.A particle of mass m is observed from an inertial frame of
reference and is found to move in a circle of radius r with a
uniform speed v. The centrifugal force on it is
(a)
r
mv
2
towards th
e centre
(b)r
mv
2
away fro
m the centre
(c)r
mv
2
along the tan
gent through the particle
(d) zero
14.A circular disc X of radius R is made from an iron plate of
thickness t, and another disc Y of radius 4R is made from an
iron plate of thickness
t
4
. Then the relation between the
moment of inertia
X
I and
Y
I is
(a)
YX
Ι32Ι= (b)
YX
Ι16Ι=
(c)
YX
ΙΙ= (d)YX
Ι64Ι=
15.A partic
les performing uniform circular motion. Its angular
frequency is doubled and its kinetic energy halved, then the
new angular momentum is
(a)
L
4
(b) 2 L
(c)
4 L (d)
L
2
16.A simple pendulum is vibrating with angular amplitude of
90° as shown in figure.
q
For what value of q is the acceleration directed
(i) vertically upwards
(ii) horizontally
(iii) vertically downwards
(a) °°
-
90,
3
1
cos,0
1
(b)
°°
-
90,0,
3
1
cos
1
(c) °°
-
0,
3
1
cos,90
1
(d)
°°
-
0,90,
3
1
cos
1
17.A mass is tie
d to a string and rotated in a vertical circle, the
minimum velocity of the body at the top is
(a)
gr (b) r/g
(c)
2/3
r
g
÷
ø
ö
ç
è
æ
(d) gr

196. 192 PHYSICS
18.A couple is ac
ting on a two particle systems. The resultant
motion will be
(a) purely rotational motion
(b) purely linear motion
(c) both a and b
(d) None of these
19.A mass m is moving with a constant velocity along a line
parallel to the x-axis, away from the origin. Its angular
momentum with respect to the origin
(a) is zero (b) remains constant
(c) goes on increasing(d) goes on decreasing.
20.A smooth sphere A is moving on a frictionless horizontal
plane with angular speed w and centre of mass velocity v. It
collides elastically and head on with an identical sphere B
at rest. Neglect friction everywhere. After the collision, their
angular speeds are w
A
and w
B
, respectively. Then
(a)w
A
< w
B
(b)w
A
= w
B
(c)w
A
= w (d)w
B
= w
21.A particle moves in a circle of radius 4 cm clockwise at
constant speed 2 cm s
–1
. If
xˆ and ˆy are unit acceleration
vec
tors along X and Y respectively (in cm s
–2
), the
acceleration of the particle at the instant half way between
P and Q is given by
X
Y
P
QO
(a) )y ˆxˆ(4+- (b) )yˆxˆ(4+
(
c)
2/)yˆxˆ(+- (d) 4/)yˆxˆ(-
22.A particle is confin
ed to rotate in a circular path decreasing
linear speed, then which of the following is correct?
(a)
L
r
(angular momentum) is conserved about the centre
(b) only direction of angular momentum L
r
is conserved
(c) It spirals to
wards the centre
(d) its acceleration is towards the centre.
23.Two rings of radius R and nR made up of same material
have the ratio of moment of inertia about an axis passing
through centre as 1 : 8. The value of n is
(a)2 (b)
22
(c)4 (d)
2
1
24.A cylinder rolls down an
inclined plane of inclination 30°,
the acceleration of cylinder is
(a)
3
g
(b) g
(c)
2
g
(d)
3
g2
25.A uniform bar o
f mass M and length L is horizontally
suspended from the ceiling by two vertical light cables as
shown. Cable A is connected 1/4th distance from the left
end of the bar. Cable B is attached at the far right end of the
bar. What is the tension in cable A?
Cable BCable A
1
4
L
L
(a) 1/4 Mg (b) 1/3 Mg
(c
) 2/3 Mg (d) 3/4 Mg
26.ABC is a triangular plate of uniform thickness. The sides are
in the ratio shown in the figure. I
AB
, I
BC
and I
CA
are the
moments of inertia of the plate about AB, BC and CA as axes
respectively. Which one of the following relations is correct?
(a)
BCAB
II>
(b)
ABBC
II>
B
A
C
3
90°
3
5
(c)
CABCAB
III=+
(d)
CA
Iis maximum
27.In carb
on monoxide molecule, the carbon and
the oxygen atoms are separated by a distance
1.12 × 10
–10
m. The distance of the centre of mass, from the
carbon atom is
(a) 0.64 × 10
–10
m (b) 0.56 × 10
–10
m
(c) 0.51 × 10
–10
m (d) 0.48 × 10
–10
m
28.A weightless ladder 20 ft long rests against a frictionless
wall at an angle of 60º from the horizontal. A 150 pound man
is 4 ft from the top of the ladder. A horizontal force is needed
to keep it from slipping. Choose the correct magnitude from
the following.
(a) 175 lb (b) 100 lb
(c) 120 lb (d) 17.3 lb
29.The moment of inertia of a disc of mass M and radius R
about an axis, which is tangential to the circumference of the
disc and parallel to its diameter, is
(a)
2
MR
2
3
(b)
2
MR
3
2
(c)
2
MR
4
5
(d)
2
MR
5
4
30.A tube one metre long is f
illed with liquid of mass 1 kg. The
tube is closed at both the ends and is revolved about one
end in a horizontal plane at 2 rev/s. The force experienced
by the lid at the other end is
(a)4p
2
N (b) 8p
2
N
(c) 16p
2
N (d) 9.8 N
31.If the linear density (mass per unit length) of a rod of length
3m is proportional to x, where x is the distance from one end
of the rod, the distance of the centre of gravity of the rod
from this end is
(a) 2.5 m (b) 1 m
(c) 1.5 m (d) 2 m

197. 193System of Particles and Rotational Motion
32.A composite disc is to be made using equal masses of
aluminium and iron so that it has as high a moment of inertia
as possible. This is possible when
(a) the surfaces of the disc are made of iron with aluminium
inside
(b) the whole of aluminium is kept in the core and the iron
at the outer rim of the disc
(c) the whole of the iron is kept in the core and the
aluminium at the outer rim of the disc
(d) the whole disc is made with thin alternate sheets of
iron and aluminium
33.A ball rolls without slipping. The radius of gyration of the
ball about an axis passing through its centre of mass is K. If
radius of the ball be R, then the fraction of total energy
associated with its rotational energy will be
(a)
22
2
RK
R
+
(b)
2
22
R
RK+
(c)
2
2
R
K
(d)
22
2
RK
K
+
34.When a buc
ket containing water is rotated fast in a vertical
circle of radius R, the water in the bucket doesn't spill
provided
(a) the bucket is whirled with a maximum speed of
2gR
(b) the bucket is whirled around with a minimum speed of
]gR)2/1[(
(c) the bu
cket is having a rpm of
)R/(g900
2
p
(d) the buck
et is having a rpm of
)R/(g3600
2
p
35.A coin p
laced on a gramophone record rotating at
33 rpm flies off the record, if it is placed at a distance of
more than 16 cm from the axis of rotation. If the record is
revolving at 66 rpm, the coin will fly off if it is placed at a
distance not less than
(a) 1 cm (b) 2 cm
(c) 3 cm (d) 4 cm
36.Two fly wheels A and B are mounted side by side with
frictionless bearings on a common shaft. Their moments of
inertia about the shaft are 5.0 kg m
2
and 20.0 kg m
2
respectively. Wheel A is made to rotate at 10 revolution per
second. Wheel B, initially stationary, is now coupled to A
with the help of a clutch. The rotation speed of the wheels
will become
(a)2
5 rps (b) 0.5 rps
(
c) 2 rps (d) None of these
37.A wheel is rolling straight on ground without slipping. If
the axis of the wheel has speed v, the instantenous velocity
of a point P on the rim, defined by angle q, relative to the
ground will be
(a)
÷
ø
ö
ç è
æ
q
2
1
cosv
(b) ÷
ø
ö
ç
è
æ
q
2
1
cosv2
q
P
(c) )sin1(v q+
(d) )cos1(v q+
38.Acertain bicycle can go up a gentle incline with constant
speed when the frictional force of ground pushing the rear
wheel is F
2
= 4 N. With what force F
1
must the chain pull on
the sprocket wheel if R
1
=5 cm and R
2
= 30 cm?
Road
ChainF
1
R
2
R
1
F=4N
2
Horizontal
(a) 4 N (b) 24 N
(c
) 140 N (d)
4
35
N
39.Auniform
rod of length l is free to rotate in a vertical plane
about a fixed horizontal axis through O. The rod beginsrotating from rest from its unstable equilibrium position.When it has turned through an angle q, its angular velocity
w is given as
q
O
(a) qsin
g6
l
(b)
2
sin
g6q
l
(c)
2
cos
g6q
l
(d) qcos
g6
l
40.A sphere o
f mass 2000 g and radius 5 cm is rotating at the
rate of 300 rpm .Then the torque required to stop it in 2p
revolutions, is
(a) 1.6 × 10
2
dyne cm(b) 1.6 × 10
3
dyne cm
(c) 2.5 × 10
4
dyne cm(d) 2.5 × 10
5
dyne cm
41.Five masses are placed in a plane as shown in figure. The
coordinates of the centre of mass are nearest to
y
2
3 kg
5 kg
2 kg1 kg
4 kg
1
0
0 2
x
1
(a) 1.2, 1.4 (b) 1.3, 1.1
(c) 1.1, 1.3 (d) 1.0, 1.0

198. 194 PHYSICS
42.A solid sphere
of mass 1 kg rolls on a table with linear
speed 1 ms
–1
. Its total kinetic energy is
(a) 1 J (b) 0.5 J
(c) 0.7 J (d) 1.4 J
43.A boy and a man carry a uniform rod of length L, horizontally
in such a way that the boy gets 1/4
th
load. If the boy is at
one end of the rod, the distance of the man from the other
end is
(a) L/3 (b) L/4
(c) 2 L/3 (d) 3 L/4
44.A solid sphere of mass M and radius R is pulled horizontally
on a sufficiently rough surface as shown in the figure.
Choose the correct alternative.
(a) The acceleration of the centre of mass is F/M
(b) The acceleration of the centre of mass is
2F
3M
(c) The friction force on the sphere acts forward
(d) The magnitude of the friction force is F/3
45.Three identical particles each of mass 1 kg are placed
touching one another with their centres on a straight line.
Their centres are marked A, B and C respectively. The
distance of centre of mass of the system from A is
(a)
AB AC BC
3
++
(b)
AB AC
3
+
(c)
AB BC
3
+
(d)
AC BC
3
+
46.M.I of a circular loop of radius R about the axis in figure is
Axis of rotation
O
R/2
x
y
(a) MR
2
(b) (3/4) MR
2
(c
) MR
2
/2 (d) 2MR
2
47.If the earth is treated as a sphere of radius R and mass M,
its angular momentum about the axis of its diurnal rotation
with period T is
(a) 2
4 MR
5T
p
(b)
2
2 MR
5T
p
(c)
2
MRT
T
(d)
3
MR
T
p
48.The wheel of a car
is rotating at the rate of 1200 revolutions
per minute. On pressing the accelerator for 10 seconds it
starts rotating at 4500 revolutions per minute. The angular
acceleration of the wheel is
(a) 30 radian / second
2
(b) 1880 degrees/ second
2
(c) 40 radian / second
2
(d) 1980 degree/second
2
49.Four masses are fixed on a massless rod as shown in the
adjoining figure. The moment of inertia about the dotted
axis is about
0.2 m 0.2 m
2kg 5kg 5kg 2kg
0.4 m 0.4 m
(a) 2 kg × m
2
(b) 1 kg × m
2
(c) 0.5 kg × m
2
(d) 0.3 kg × m
2
50.The moment of inertia of a body about a given axis is 1.2 kgm
2
. Initially, the body is at rest . In order to produce a
rotational kinetic energy of 1500 J, an angular accelerationof 25 rad s
–2
must be applied about that axis for a duration
of
(a) 4 s (b) 2 s
(c) 8 s (d) 10 s
51.Fig. shows a disc rolling on a horizontal plane with linear
velocity v. Its linear velocity is v and angular velocity is w.
Which of the following gives the velocity of the particle P
on the rim of the disc
v
O
q
M
NB
P
A
r
w
(a) v (1 + cos q) (b) v (1 – cos q)
(c) v (1 + sin q) (d) v (1 – sin q)
52.A toy car rolls down the inclined plane as shown in the fig.It loops at the bottom. What is the relation between H andh?
(a)
H
2
h
=
r
D
B
H
h
(b)
H
3
h
=
(c)
H
4
h
=
(d)
H
5
h
=

199. 195System of Particles and Rotational Motion
53.A sphere rolls down on an inclined plane of inclination q.
What is the acceleration as the sphere reaches bottom?
(a)
5
g sin
7
q (b)
3
g sin
5
q
(c)
2
g sin
7
q (d)
2
gsin
5
q
54.In a bicycle, the radius of rear wheel is twice the radius of
front wheel. If r
F
and r
F
are the radii, v
r
and v
r
are the speed
of top most points of wheel. Then
(a)
rFv 2v= (b)
Fr
v 2v=
(c)
Fr
vv= (d)
Frvv>
55.An annular ring with inner and outer radii
1
R and
2
R is
rolling without slipping with a uniform angular speed. The
ratio of the forces experienced by the two particles situated
on the inner and outer parts of the ring ,
2
1
F
F
is
(a)
2
2
1
R
R
÷
÷
ø
ö
ç
ç
è
æ
(b)
1
2
R
R
(c)
2R
1
R
(d) 1
56.A whee
l having moment of inertia 2 kg-m
2
about its vertical
axis, rotates at the rate of 60 rpm about this axis, The torque
which can stop the wheel’s rotation in one minute would
be
(a)
Nm
18
p
(b)
2
Nm
15
p
(c)Nm
12
p
(d)Nm
15
p
57.Consider a system of two particles having masses m
1
and
m
2
. If the particle of mass m
1
is pushed towards the centre
of mass particles through a distance d, by what distance
would the particle of mass m
2
move so as to keep the mass
centre of particles at the original position?
(a)d
m
m
1
2
(b) d
mm
m
21
1
+
(c) d
m
m
2
1
(d) d
58.The r
atio of the radii of gyration of a circular disc about a
tangential axis in the plane of the disc and of a circular ring
of the same radius about a tangential axis in the plane of
the ring is
(a) 1 : Ö2 (b) 1 : 3
(c) 2 : 1 (d)Ö5 : Ö6
59.A round disc of moment of inertia I
2
about its axis
perpendicular to its plane and passing through its centre
is placed over another disc of moment of inertia I
1
rotating
with an angular velocity w about the same axis. The final
angular velocity of the combination of discs is
(a)
1
21
I
)II(w+
(b)
21
2
II
I
+
w
(c)w (d)
21
1
II
I
+
w
60.Three pa
rticles, each of mass m gram, are situated at the
vertices of an equilateral triangle ABC of side/cm (as shown
in the figure). The moment of inertia of the system about a
line AX perpendicular to AB and in the plane of ABC, in
gram-cm
2
units will be
(a)
2
m
2
3
l
(b)
2
m
4
3
l
(c) 2 ml
2
m
l l
l mm
C
B
A
X
(d)
2
m
4
5
l
61.The momen
t of inertia of a uniform circular disc of radius ‘R’
and mass ‘M’ about an axis passing from the edge of the disc
and normal to the disc is
(a) MR
2
(b)
1
2
2
MR
(c)
3
2
2
MR (d)
7
2
2
MR
62.Two bodies
have their moments of inertia I and 2I
respectively about their axis of rotation. If their kinetic
energies of rotation are equal, their angular momenta will
be in the ratio
(a) 2 : 1 (b) 1 : 2
(c) 21: (d)12:
63.A drum of r
adius R and mass M, rolls down without
slipping along an inclined plane of angle q. The frictional
force(a) dissipates energy as heat.(b) decreases the rotational motion.
(c) decreases the rotational and translational motion.
(d) converts translational energy to rotational energy
64.A tube of length L is filled completely with an incompressible
liquid of mass M and closed at both ends. The tube is then
rotated in a horizontal plane about one of its ends with
uniform angular speed w. What is the force exerted by the
liquid at the other end?
(a)
2
ML
2
w
(b)
2
MLw
(c)
2
ML
4
w
(d)
2
ML
8
w

200. 196 PHYSICS
65.A circular disk o
f moment of inertia I
t
is rotating in a
horizontal plane, its symmetry axis, with a constant angular
speed
w
i
. Another disk of moment of inertia I
b
is dropped
coaxially onto the rotating disk. Initially the second disk
has zero angular speed. Eventually both the disks rotate
with a constant angular speed f
w. The energy lost by the
initially rotating disk to friction is
(a)
1
2
2
2
()
w
+
b
i
tb
I
II
(b)
2
2
()
w
+
t
i
tb
I
II
(c)
2
()
bt
i
tb
II
II
-
w
+
(d)
1
2

2
()
bt
i
tb
II
II
w
+
66.The moment of inertia o
f a thin uniform rod of mass M and
length L about an axis passing through its midpoint and
perpendicular to its length is I
0
. Its moment of inertia about
an axis passing through one of its ends and perpendicular
to its length is
(a)I
0
+ ML
2
/2 (b) I
0
+ ML
2
/4
(c)I
0
+ 2ML
2
(d) I
0
+ ML
2
67.The instantaneous angular position of a point on a rotating
wheel i s gi ven by t he equat i on q(t) = 2t
3
– 6t
2
. The torque
on the wheel becomes zero at
(a) t = 1s (b) t = 0.5 s
(c) t = 0.25 s (d) t = 2s
68.The moment of inertia of a uniform circular disc is
maximum about an axis perpendicular to the disc and
passing through
B
C
D
A
(a)B (b)C
(c)D (d)A
69.Three masse
s are placed on the x-axis : 300 g at origin, 500
g at x = 40 cm and 400 g at x = 70 cm. The distance of the
centre of mass from the origin is
(a) 40 cm (b) 45 cm
(c) 50 cm (d) 30 cm
70.If the angular velocity of a body rotating about an axis is
doubled and its moment of inertia halved, the rotational
kinetic energy will change by a factor of :
(a)4 (b) 2
(c)1 (d)
1
2
71.One quarter sector is cut from a uniform circular disc of
radius R. This sector has mass M. It is made to rotate about
a line perpendicular to its plane and passing through the
centre of the original disc. Its moment of inertia about the
axis of rotation is
(a)
21
mR
2
(b)
21
mR
4
(c)
21
mR
8
(d)
2
2mR
Directions for Qs. (72 to 75) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following-
(a) Statement -1 is false, Statement-2 is true
(b) Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c) Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d) Statement -1 is true, Statement-2 is false
72. Statement 1 : When you lean behind over the hind legs of
the chair, the chair falls back after a certain angle.
Statement 2 : Centre of mass lying outside the system
makes the system unstable.
73. Statement 1: A rigid disc rolls without slipping on a fixed
rough horizontal surface with uniform angular velocity. Then
the acceleration of lowest point on the disc is zero.
Statement 2 : For a rigid disc rolling without slipping on a
fixed rough horizontal surface, the velocity of the lowest
point on the disc is always zero.
74. Statement 1 : If no external force acts on a system of
particles, then the centre of mass will not move in any
direction.
Statement 2 : If net external force is zero, then the linear
momentum of the system remains constant.
75. Statement 1 : A wheel moving down a frictionless inclined
plane will slip and not roll on the plane.
Statement 2 : It is the frictional force which provides a
torque necessary for a body to roll on a surface.

201. 197System of Particles and Rotational Motion
Exemplar Questions
1.For which of the following does the centre of mass lie outside
the body?
(a) A pencil (b) A shotput
(c) A dice (d) A bangle
2.Which of the following points is the likely position of the
centre of mass of the system shown in figure?
A
B
C
D
Hollow sphere
Air
R/2
R/2
Sand
(a)A (b)B
(c)C (d)D
3.A particle of mass m is moving in yz-plane with a uniform
velocity v with its trajectory running parallel to +ve y-axis
and intersecting z-axis at z = a in figure. The change in its
angular momentum about the origin as it bounces elastically
from a wall at y = constant is
(a) ˆmva
x
e
v
z
a
y
(b) ˆ2mva
x
e
(c) ˆymv
x
e
(d) ˆ2ymv
x
e
4.When a disc rotates with uniform angular velocity, which of
the following is not true?
(a) The sense of rotation remains same
(b) The orientation of the axis of rotation remains same
(c) The speed of rotation is non-zero and remains same
(d) The angular acceleration is non-zero and remains same
5.A uniform square plate has a small piece Q of an irregular
shape removed and glued to the centre of the plate leaving
a hole behind in figure. The moment of inertia about the z-
axis is then,
Q
P
y
x
Q
P
y
x
hole
(a) increased
(b) decreased
(c) the same
(d) changed in unpredicted manner
6.In problem-5, the CM of the plate is now in the following
quadrant of x-y plane.
(a)I (b) II
(c) III (d) IV
7.The density of a non-uniform rod of length 1 m is given by
r(x) = a (1 + bx
2) where, a and b are constants and
0 1.x££ The centre of mass of the rod will be at
(a)
3(2)
4(3)
b
b
+
+
(b)
4(2)
3(3 )
b
b
+
+
(c)
3(3)
4(2)
b
b
+
+
(d)
4(3)
3(2)
b
b
+
+
8.A merry-go-round, made of a ring-like platform of radius R
and mass M, is revolving with angular speed w. A person of
mass M is standing on it. At one instant, the person jumps
off the round, radially away from the centre of the round (as
seen from the round). The speed of the round of afterwards
is
(a)2w (b)w
(c)
2
w
(d) 0
NEET/AIPMT (2013-2017) Questions
9.A small object of uniform density rolls up a curved surface
with an initial velocity ‘n’. It reaches upto a maximum height
of
2
3
4g
n
with respect to the initial position. The object is a
(a) solid sphere(b) hollow sphere [2013]
(c) disc (d) ring
10.A rod PQ of mass M and length L is hinged at end P. The
rod is kept horizontal by a massless string tied to point Q as
shown in figure. When string is cut, the initial angular
acceleration of the rod is [2013]
(a) g /L (b) 2g/L
(c)
2
3
g
L
(d)
3
2
g
L

202. 198 PHYSICS
11.Two dis
cs are rotating about their axes, normal to the discs
and passing through the centres of the discs. Disc D
1
has
2 kg mass and 0.2 m radius and initial angular velocity of 50
rad s
–1
. Disc D
2
has 4kg mass, 0.1 m radius and initial angular
velocity of 200 rad s
–1
. The two discs are brought in contact
face to face, with their axes of rotation coincident. The final
angular velocity (in rad s
–1
) of the system is
(a) 40 (b) 60 [NEET Kar. 2013]
(c) 100 (d) 120
12.The ratio of radii of gyration of a circular ring and a circular
disc, of the same mass and radius, about an axis passing
through their centres and perpendicular to their planes are
(a)
2 :1 (b)1:2 [NEET Kar. 2013]
(c) 3 : 2 (d)
2 : 1
13.The ratio of the accelerations for a solid sphere (mass ‘m’
and radius ‘R’) rolling down an incline of angle ‘q’ without
slipping and slipping down the incline without rolling is :
(a) 5 : 7 (b) 2 : 3 [2014]
(c) 2 : 5 (d) 7 : 5
14.A solid cylinder of mass 50 kg and radius 0.5 m
is free to rotate about the horizontal axis. A massless string
is wound round the cylinder with one end attached to it
and other hanging freely. Tension in the string required to
produce an angular acceleration of 2 revolutions s
– 2
is :
(a) 25 N (b) 50 N [2014]
(c) 78.5 N (d) 157 N
15.Three identical spherical shells, each of mass m and radius
r are placed as shown in figure. Consider an axis XX' which
is touching to two shells and passing through diameter of
third shell. Moment of inertia of the system consisting of
these three spherical shells about XX' axis is [2015]
(a) 3mr
2
(b)
216
mr
5
X
X¢
(c) 4mr
2
(d)
211
mr
5
16.A rod of weight W is supported by two parallel knife edges
A and B and is in equilibrium in a horizontal position. Theknives are at a distance d from each other. The centre ofmass of the rod is at distance x from A. The normal reaction
on A is [2015]
(a)
Wd
x
(b)
W(d – x)
x
(c)
W(d–x)
d
(d)
Wx
d
17.A mass m moves in a circle on a smooth horizontal plane
with velocity v
0
at a radius R
0
. The mass is attached to
string which passes through a smooth hole in the plane as
shown.
v
0
m
The tension in the string is increased gradually and finally
m moves in a circle of radius
0
R
2
. The final value of the
kinetic
energy is [2015]
(a)
2
0
1
mv
4
(b)
2
0
2mv
(c)
2
0
1
mv
2
(d)
2
0
mv
18.A force ˆˆˆ
F i 3j 6k=a++
r
is acting at a point
ˆˆˆ
r 2i 6j 12k= --
r
. The value of a for which angular
momentum about origin is conserved is : [2015 RS]
(a) 2 (b) zero
(c) 1 (d) –1
19.Point masses m
1
and m
2
are placed at the opposite ends of
a rigid rod of length L, and negligible mass. The rod is to be
set rotating about an axis perpendicular to it. The position
of point P on this rod through which the axis should pass
so that the work required to set the rod rotating with angular
velocity w
0
is minimum, is given by : [2015 RS]
m
1
m
2
w
0
P
x (L–x)
(a)
1
2
m
xL
m
=
(b)
2
1
m
xL
m
=
(c)
2
12
mL
x
mm
=
+
(d)
1
12
mL
x
mm
=
+
20.An automo
bile moves on a road with a speed of 54 km h
-1
.
The radius of its wheels is 0.45 m and the moment of inertia
of the wheel about its axis of rotation is 3 kg m
2
. If the
vehicle is brought to rest in 15s, the magnitude of average
torque transmitted by its brakes to the wheel is : [2015 RS]
(a) 8.58 kg m
2
s
-2
(b) 10.86 kg m
2
s
-2
(c) 2.86 kg m
2
s
-2
(d) 6.66 kg m
2
s
-2

203. 199System of Particles and Rotational Motion
21.A disk and a sphere of same radius but different masses roll
off on two inclined planes of the same altitude and length.
Which one of the two objects gets to the bottom of the
plane first ?
(a) Disk [2016]
(b) Sphere
(c) Both reach at the same time
(d) Depends on their masses
22.A uniform circular disc of radius 50 cm at rest is free to turn
about an axis which is perpendicular to its plane and passes
through its centre. It is subjected to a torque which produces
a constant angular acceleration of 2.0 rad s
–2
. Its net
acceleration in ms
–2
at the end of 2.0s is approximately :
(a) 8.0 (b) 7.0 [2016]
(c) 6.0 (d) 3.0
23.From a disc of radius R and mass M, a circular hole of diameter
R, whose rim passes through the centre is cut. What is the
moment of inertia of the remaining part of the disc about a
perpendicular axis, passing through the centre ? [2016]
(a) 15 MR
2
/32 (b) 13 MR
2
/32
(c) 11 MR
2
/32 (d) 9 MR
2
/32
24.Which of the following statements are correct ? [2017]
(A) Centre of mass of a body always coincides with the
centre of gravity of the body
(B) Centre of mass of a body is the point at which the total
gravitational torque on the body is zero
(C) A couple on a body produce both translational and
rotation motion in a body
(D) Mechanical advantage greater than one means that
small effort can be used to lift a large load
(a) (A) and (B) (b) (B) and (C)
(c) (C) and (D)(d) (B) and (D)
25.Two discs of same moment of inertia rotating about their
regular axis passing through centre and perpendicular to
the plane of disc with angular velocities w
1
and w
2
. They
are brought into contact face to face coinciding the axis of
rotation. The expression for loss of energy during this
process is:- [2017]
(a)
2
12
1
I()
4
w -w (b)
2
12
I()w -w
(c)
2
12
1
()
8
w -w (d)
2
12
1
I()
2
w +w
26.A rope
is wound around a hollow cylinder of mass 3 kg and
radius 40 cm. What is the angular acceleration of thecylinder if the rope is pulled with a force of 30 N ? [2017]
(a) 0.25 rad/s
2
(b) 25 rad/s
2
(c) 5 m/s
2
(d) 25 m/s
2

204. 200 PHYSICS
EXERCISE - 1
1. (a) 2. (b) 3.(
c) 4.(c)
5. (b) 6. (d) 7.(b)
8. (d) The pull of earth changes only when the body moves
so that g changes. It is an exceptional case and should
not be considered unless otherwise mentioned.
9. (b) If we apply a torque on a body, then angular momentum
of the body changes according to the relationdL
τ
dt
=
r
r
Þ if t
r
= 0 then, L = constant
10
. (b) Angular momentum
L
r
is defined as L r m(v)=´
rrr
so L
r
is, an axial vector..
11. (c) The boy does not exert a torque to rotating table by
jumping, so angular momentum is conserved i.e.,
dL
0L
dt
=Þ
r
r
= constant
12
. (b) Since no external torque act on gymnast, so angular
momentum (L=I
ω) is conserved. After pulling her arms
& legs, the angular velocity increases but moment ofinertia of gymnast, decreases in, such a way thatangular momentum remains constant.
13. (a)14. (b) Angular momentum will remain the same since external
torque is zero.
15. (a)
22
BA
1
I MR I MR
2
==
AB
II\<
16. (a) Angular momentum ( )r linear momentum=´
r
17. (b) We know that ext
dL
τ
dt
=
if angular momentu
m is conserved, it means change in
angular momentum = 0
0dL,or=,
ext
dL
0τ0
dt
=Þ=
Thus total external torque = 0.
18. (a) Analogue of mass in rotational motion is moment of
inertia. It plays the same role as mass plays intranslational motion.
19. (a) Basic equation of moment of inertia is given
by
2
ii n
1i
rmI
=
S=
i
m
ir
where
i
m is the m ass of
th
i particle at a distance of
i
r from axis of rotation.
Thus it does not depend on angular velocity.
20. (d) Initially centre of gravity is at the centre. When sand
is poured it will fall and again after a limit, centre ofgravity will rise.
21. (d) Here, the centre of mass of the system remains
unchanged. in the horizontal direction. When the mass
m moved forward by a distance L cos q, let the mass
(m + M) moves by a distance x in the backward
direction. hence
(M + m) x – m L cos q = 0
\ x = (m L cos q)/(m + M)
22. (b) Acceleration of solid sphere is more than that of
hollow sphere, it rolls faster, and reaches the bottom
of the inclined plane earlier.
Hence, solid sphere and hollow sphere can be
distinguished by rolling them simultaneously on an
inclined plane.
23. (d) Applying the law of conservation of momentum
m v = M V ...(1)
By conservation of angular momentum
2
ML
mv(L/2) ω
12
æö
=ç÷
ç÷
èø
...(2)
As the collision
is elastic, we have
2 22111
mv MV Ιω
222
=+ ...(3)
Substituting the values, w
e get m = M/4
24. (b)a is propotional to w
Let a = kw (
Qk is a constant)
dω
kω
dt
= [also
dθ dθ
ωdt
dt ω
=Þ= ]
ωdω
kω
dθ
\= Þ dw = kdq
Now
ω/2
ω
dωk dθ=òò
0 θ
1
ω/20
ωω
dωk dθ kθ kθ
22
= Þ-= Þ-=òò
(Qq
1
= 2pn)
\ θ =
1
θ or 2pn
1
= 2pn
n
1
= n
25.
(b) So raw egg is like a spherical shell & hard bioled egg is
like solid sphere. Let
1
I,
2
I be M. I. of raw egg and
boiled egg respectively.Given that angular momentum L, is same
\ I
1
w
1
= I
2
w
2
Þ w
2
>w
1 QI
1>
I
2
Now from first equati
on of angular motion (w
f
=w
i
+at)
here a is retarding decceleration for both cases &
w
f
= 0 for both case.
So
1 11
2 22
t (raw egg)ω/αt
1
t (hard egg)ω / αt
= Þ<
Hints & Solutions

205. 201System of Particles and Rotational Motion
EXERCISE - 2
1. (
d) K.E. of rotation =
21
2
Iw
=
22
11
22
mr
æö
´wç÷
èø
21
20 (0.25) 100100
4
=´´ ´´ = 3125 J
2
. (c) For solid sphere
232
RR
3
4
5
2
RM
5
2
÷
ø
ö
ç
è
æ
rp==I

r=r´=
55
R
105
176
R
7
22
15
8
3. (a) For
ce F
A
on particle A is given by A
A AA
mv
F ma
t
== ...(1)
Similarly
B
B BB
m 2v
F ma
t
´
== ...(2)
Now
AB
AB
m v m 2v
( F F)
tt
´
== Q
So m
A
= 2 m
B
For t
he centre of mass of the system
AA BB
AB
mv mv
v
mm
+
=
+
or
BB
BB
2 m v m 2v
v0
2mm
-´
==
+
Negati
ve sign is used because the particles are
travelling in opposite directions.
Alternatively, if we consider the two masses in a
system then no external force is acting on the system.
Mutual forces are internal forces. Since the centre of
mass is initially at rest, it well remain at rest
4. (a) Moment of inertia of the whole system about the axis
of rotation will be equal to the sum of the moments of
inertia of all the particles.
(–2,–2, 0)
4 kg
(0, 0, 9)
1 kg
(2, 0, 0)
2 kg
(0, 3, 0)
3 kg
X
Y
\ I = I
1
+ I
2
+ I
3
+ I
4
= 0 + 0 + 27 + 16 = 43 kg m
2
5. (a) J5.13)3(3
2
1
2
1
E
32
r
=´´=wI=
2211
K.E. m v 27 v 13.5
22
= =´´=
v 1m/s=
6. (c) C
entripetal acceleration22
c
a4r= pn
222
ms425.0224
-
p=´´´p=
7. (b) F
rom the theorem of parallel axes, the moment of inertia
I = I
CM
+ Ma
2
where I
CM
is moment of inertia about centre of mass
and a is the distance of axis from centre.mk
2
= m(k
1
)
2
+ m(6)
2
(Q I = mk
2
where k is radius of gyration)
or, k
2

= (k
1
)
2
+ 36
or,100 = (k
1
)
2
+ 36
Þ (k
1
)
2
= 64 cm
\ k
1
= 8 cm
8. (c) L = m v r or v = L /mr
Centripetal force
22
mv m(L/ m r)
rr
=
3
2
rm
L
=
9. (a)
The situation can be shown as : Let radius of complete
disc is a and that of small disc is b. Also let centre ofmass now shifts to O
2
at a distance x
2
from original
centre.
b
a
O
O
1
O
2 x-axis
x
2
x
1
The pos
ition of new centre of mass is given by
X
CM
=
2
1
22
. b .x
.a .b
-sp
sp -sp
Here, a = 6 cm, b = 2 cm, x
1
= 3.2 cm
Hence, X
CM
=
2
22
(2) 3.2
(6) (2)
-s´p´
s´p´ -
s´p´
=
12.8
32
p
p
= – 0.4 cm.
10. (b) By energy conservation
ffii
)E.P()E.K()E.P()E.K(+=+
0)E.P(,mgh)E.P(,0)E.K(
fii
===
22
f cm
(K.E) ½Iω½mv=+
Where
I (moment of inertia) = ½mR
2
(for solid cylinder)
so
2
22cm
cm
2
v
mgh ½(½mR ) ½mv
R
æö
=+ ç÷
èø
cm
v 4gh / 3Þ=

206. 202 PHYSICS
11. (b
) Moment of inertia (M.I) of circular ring I
1
= mr
2
moment of inertia (M.I) of circular disc I
2
= ½mr
2
1
2
I
I
2
1
=Þ
12. (a) Tensi
on provides the necessary centripetal force for
the rest of rod.
13. (d) Centrifugal force is observed when the observer is in
a frame undergoing circular motion.
14. (d)
21
mR
2
I=
2
MtRµµ
For disc X,
2
X
1
Ι (m)(R)
2
= =
221
(πr t).(R)
2
for disc Y,
22
Y
1
Ι [π(4R) .t /4 ][4R]
2
=
Þ
X
3
Y
Ι 1
Ι(4)
= Þ
YX
I 64I=
15. (a)
1
Angular momentum
Angular frequency
µ
Kinetic energyµ
Þ
K.E.
L
ω
=
r
1 12
212
L K.E ω
4
L ω K.E
æö
= ´=
ç÷
èø
Þ 2
L
L
4
=
16. (a) When q = 0° , the net force is directed vertically
upwards.
17. (a) Let velocity at
A
vA=and velocity at
B
vB=
A
T
r
B
Bv
Av
Applying conse
rvation of energy at A & B
2
B
2
A mv
2
1
gmr2mv
2
1
=+ [Q (P.E)
A
= mg (2r)]
)i(..........gr4v
v
2
A
2
B
+=
Now as it is moving in circular path it has centripetal
force.
At point A
r
mv
mgT
2
A
=+Þ
0Tvelocityminimumfor ³
2
2A
A
mv
or mg v gr
r
³Þ³ grv
A
³Þ
18. (a) A coup
le consists of two equal and opposite forces
whose lines of action are parallel and laterally
separated by same distance. Therefore, net force (or
resultant) of a couple is null vector, hence no
translatory motion will be produced and only
rotational motion will be produced.
19. (b) Angular momentum of mass m moving with a constant
velocity about origin is
line parallel
to x-axis
v
m
O
y
x
L = momentum × perpe
ndicular distance of line of
action of momentum from origin
L = mv × y
In the given condition mvy is a constant. Therefore
angular momentum is constant.
20. (c) Since the spheres are smooth, there will be no transfer
of angular momentum from the sphere A to sphere B.
The sphere A only transfers its linear velocity v to the
sphere B and will continue to rotate with the same
angular speed w.
21. (c)
2
)yˆxˆ(
nˆ
+-
= –a cos45 x
o
–a sin45 y
o
45
o
a
2
ˆˆv 4 (x y)
ˆan
r4 2
-+
= =´
ur
22. (b) Since v is changing (decreasing), L is not conserved
in magnitude. Since it is given that a particle is confinedto rotate in a circular path, it can not have spiral path.Si nce the par t i cl e has t wo accel er at i ons a
c
and a
t
therefore the net acceleration is not towards the centre.
L
a
c
a
t
v
The direction of L remains same even when the
speed decreases.
23. (a) The moment of inertia (I) of circular ring whose axis of
rotation is passing through its center,
2
11
RmI=
Also,
2
22
)nR(mI=
Since both rings have same density,
1
1
2
2
AR2
m
A)nR(2
m
´p
=
´p
Þ

207. 203System of Particles and Rotational Motion
Where A is cross-section of ring,
21
AA=(Given)
12
nmm=\
Given
8
1
I
I
2
1
= =
2
2
2
1
)nR(m
Rm
=
2
1
2
1
)nR(nm
Rm
3
n
1
8 1
=Þ or n = 2
24
. (a) Remember that acceleration of a cylinder down a
smooth inclined plane is
x
qsing
qg
Nm
q
N
m m
mg
cos
÷
ø
ö
ç
è
æ
+
q
=
2
mR
I
1
sing
a where
2
mR
I
2
= is the momen
t of
Inertia for cylinder
÷
÷
ø
ö
ç
ç
è
æ
´+
°
=
2
2
mR
1
2
mR
1
30sing
a
3
g
2
1
1
2
1
g
=
+
´
=
25. (c
) This is a torque problem. While the fulcrum can be
placed anywhere, placing it at the far right end of the
bar eliminated cable B from the calculation. There are
now only two forces acting on the bar ; the weight
that produces a counterclockwise rotation and the
tension in cable A that produces a clockwise rotation.
Since the bar is in equilibrium, these two torques must
sum to zero.
1
2
3
4
L
L
fulcrum
W = mg
Cable
0)
L2/1(Mg)L4/3(T
A
=-=tS
Therefore
3/Mg2)L3/4)(2/MgL()4/L3/()2/MgL(T
A ===
26. (b) The intersection of medians is the centre of mass of
the triangle. Since distances of centre of mass fromthe sides are related as : x
BC
< x
AB
< x
AC
therefore I
BC
> I
AB
> I
AC
or I
BC
> I
AB.
27. (a) C O
1.12×10
–10
(12 amu) (16 amu)
d
c.m.
1.12×10 – d
–10
Fr
om definition of centre of mass.
1216
0121012.116
d
10
+
´+´´
=
-

28
1012.116
10-
´´
=
= 0.64 × 10
–
10
m.
28. (d) AB is the ladder, let F be the horizontal force and W is
the weigth of man. Let N
1
and N
2
be normal reactions
of ground and wall, respectively. Then for verticalequilibrium
W = N
1
.....(1)
For horizontal equilibrium N
2
= F .....(2)
Taking moments about A
N
2
(AB sin60°) – W(B Cos 60°) = 0 ......(3)
Using (2) and AB = 20 ft, BC = 4 ft, we get
N
2
N
1 W
A
B
C
F
O
60°
4 ft
0
2
1
4w–
2
3
20F =÷
ø
ö
ç
è
æ
´
÷
÷
ø
ö
ç ç
è
æ
´
35
150
35
w
320
2w2
F ==
´
=Þ
pound (lb)
10 3 10 1.7317.3= =´= pound
2
9. (c) Moment of inertia of disc about its diameter is
2
d
MR
4
1
I=
I
Id
R
MI of disc ab
out a tangent passing through rim and in
the plane of disc is
2222
G MR
4
5
MRMR
4
1
MRII =+=+=
30. (b) F
= mrw
2
=
221
1 4 228N
2
´ ´p´´ =p
31. (d) Consider an element of length dx at a distance x from
end A.
Here : mass per unit length l of rod
)kxx(=lÞµl
\ dm = ldx = kx dx
dx BA
x

208. 204 PHYSICS
Positi
on of centre of gravity of rod from end A
ò
ò
=
L
0
L
0
CG
dm
dmx
x
3
3
3
3
00
CG
3 32
2
0
0
x
(3)
x(kx dx) 3
3
x 2m
(3)
kx dx x
2
2
éù
êú
êúëû
\ = = ==
éù
êú
êúëû
ò
ò
32.
(b) Density of iron > density of aluminium
moment of inertia = ò
dmr
2
.
r
dm
\ Since r
iron
> r
alum
inium
so whole of aluminium is kept in the core and the iron
at the outer rim of the disc.
33. (d) Rotational energy =
2
)(I
2
1
w =
22
)mK(
2
1
w
Linear kinetic en
ergy =
22
Rm
2
1
w
\ Required fraction
=
2222
22
Rm
2
1
)mK(
2
1
)mK(
2
1
w+w
w
=
22
2
RK
K
+
34. (c) At the h
ighest point
22
2
g
mg mR4
4R
= p n Þn=
p
Revolut
ion per minute
2
900 g
60
R
= n=
p
35. (d)
22
4mrmgnp
=m
22
22 11 11
22112
22
21
rr
rrr
4
nn
Þn=nÞ==
nn
cm4
4
r
r
1
2
==
36. (c) By
conservation of angular momentum,
50
5 10 5 20 2 rps
25
´=w+wÞw==
3
7. (b)
q
q
v
v
)cos1(v2cosv2vvv
2222
R
q+=q++=
2
cosv2
q
=
38. (b) For no a
ngular acceleration t
net
= 0
Þ F
1
× 5 = F
2
× 30 (given F
2
= 4N) N24F
1
=Þ
39. (c)
2
I
2
1
w = Loss of gravitatio
nal potental energy
2
21 m mg
(1 cos)
232
´ w= +q
ll
2
cos
g6
2
cos2
g3 22 q
=wÞ÷
ø
ö
ç
è
æ q
=wÞ
ll
40. (d) Use t = Ia
and aq=
w-w2
2
0
2
Here
2
MR
5
2
I=
Given M = 2000g = 2kg
R =
5cm = 5 × 10
–2
m
2 2 radiansq=p´p
n = 300 rpm = 5
rps
I\t=a =
22
2 02
MR
52
æöw -wæö
= ç÷ç÷
èø qèø
= ( )
()
222
2
2
2
4 50
2
2 5 10
5 24
-
-
p
´´´
´p
= 0.02
5 N – m.
= 2.5 × 10
5
dyne cm.
41. (c)
54321
1524032201
X
.M.C
++++
´+´+´+´+´
=
1.1
15
17
15
584
==
++
=
C.M
1020324251
Y
12345
´+´+´+´+´
=
++++
=
685
1.3
15
++
=
42. (c)
22
I
2
1
Mv
2
1
E w+=
J7.0Mv
10
7
R
v
MR
5
2
2
1
Mv
2
1 2
2
2
22
==´´+=

209. 205System of Particles and Rotational Motion
43. (a) So couple about the C.M must be zero for rotational
equilibrium, It means that
6
L
xx
4
W3
2
L
4
W
=Þ´=´
C.M
xL/2
L/2
W
4
3W
4
boy man
and the distance of man from other end is
2
L
– x
3
L
6
L
2
L
=-=
44. (b,c)
4
5. (b) Position of C.M w.r. to A
A B C
1 0 1 AB 1 AC AB AC
1113
´ +´ +´
+
==
++
46. (b) Use theorem of parallel axes.
47. (a) Angular momentum
222
I MR.
5T
p
= w=
2
4 MR
5T
p
=
48. (d)
21
21
tt
w -w
a=
-
1
2 1200
40
60
p´
w= =p ; 2
2 4500
150
60
p´
w= =p
2110
rad / sec
10
p
a=
Now, radian 180p =°
\
180
1 rad degree=
p
\
211 180
degree /sec
p´
a=
p
= 1980 degree/sec
2
4
9. (b)
2 22
I 2 5 (0.2) 2 2 (0.4) 1kg m=´´ +´ ´ =´
50. (b) E = 1500 = 1/2 × 1.2 w
2
23000
2500
1.2
w==
50rad / secw=
50
t 2sec
25
w
===
a
51. (c
) Velocity of
P (NP) (NM MP)= w= +w
r(r sin) v(1 sin)=+qw= +q
52. (d) Velocity at the bottom and top of the circle is 5gr
and gr. Therefore (1/2)M(5gr) = MgH and
(1/2) M (gr) = Mgh.
53. (a) Acceleration of a body rolling down an inclined plane
is given by ,
2
2
gsin
a
K
1
R
q
=
+
In case of a solid sp
here , we have
22
2222
K [(I/ M)] I (2 /5)MR 2
5R R MR MR
====
Substituting
2
2
K2
5R
= in equation (1) we get
5
a gsin
7
=q
54. (c) The velocity of the top point of the wheel is twice that
of centre of mass. And the speed of centre of mass is
same for both the wheels.
1a
2
a
11
Rvw=
1R 22
Rvw=
2R
55. (c) 1
2
1
2
1
2
1
2
1
1 R
R
R
R
v
a w=
w
==
2
2
2
2
2
2
R
R
v
a w==
Taking particle m
asses equal
2
1
2
1
2
1
2
1
R
R
a
a
ma
ma
F
F
===
Alternative method : T
he force experienced by any
particle is only along radial direction or the centripetal
force.
Force experienced by the particle, F = mw
2
R
11
22
FR
FR
\=
56. (d)t × D
t = L
0
{
Q since L
f
= 0}
Þ t × Dt = Iw
or t × 60 = 2 × 2 × 60p/60
(f 60rpm=Q
60
2f2
60
ö
\ w= p = p´
÷
ø
mN
15
-
p
=t
57. (c) m
1
d =
m
2
d
2

2
1
2
m
dm
d=Þ
58. (d)
y
1y¢
1
Circular disc( 1)
4
MR
I
2
1
y=

210. 206 PHYSICS
22
2
1
y MR
4
5
MR
4
MR
I =+=¢\
y
2y¢
2
Circular ring (2)
2
MR
I
2
2
y=
22
2
2
y MR
2
3
MR
2
MR
I =+=¢\
2
1
1
yMKI=¢ ,
2
2
2
yMKI=¢
2
y
1
y
2
2
2
1
I
I
K
K
¢
¢
=\ 6:5K:K
21
=Þ
59. (d) An
gular momentum will be conserved
I
1
w = I
1
w' + I
2
w'
21
1
II
I
+
w
=w¢Þ
60. (d) I
AX
= m(AB)
2
+ m(O
C)
2
= ml
2
+ m (l cos 60º)
2
= ml
2
+ ml
2
/4 = 5/4 ml
2
m
l l
l
mm
C
BA
O
X
60º
60º
61. (c) M.I
. of a uniform circular disc of radius ‘R’ and
mass ‘M’ about an axis passing through C.M.
and normal to the disc is
I MR
CM..
=
1
2
2
From parallel axis theorem
II MR MR MR MR
TCM
= + = + =
..
2 2 2 21
2
3
2
62. (d)K
L
I
=
2
2
Þ L
2
= 2KIÞ L KI=2
L
L
K
K
I
I
1
2
1
2
1
2
= ×

=×=
K
K
I
I2
1
2
L
1
: L
2
= 1 : 2
63. (d) Net work d
one by frictional force when drum rolls
down without slipping is zero.
R
M
f
q
W
net
= 0, W
trans.
+ W
rot.
= 0
DK
trans.
+ DK
rot.
= 0
DK
trans
= –DK
rot
.
i.e., converts translation energy to rotational energy.
64. (a) Tube may be treated as a particle of mass M at distance
L/2 from one end.
Centripetal force
22ML
Mr
2
=w=w
65. (d) By conservation of angular momentum, I
ti
w=(I
t
+I
b
) f
w
where f
w is the final angular velocity of disks
\ f
w =
t
i
tb
I
II
æö
wç÷
+
èø
Loss in K.E.,KD = Initial K.E. – Final K.E.
=
21
2
tiIw–
21
()
2
tbfII+w
Þ KD =
1
2
21
2
ti
Iw-
2
2
2
()
()
t
tbi
tb
I
II
II
+w
+
=
1
2
2
i
w
()
t
tbt
tb
I
III
II
+-
+
=
1
2
2
i
w
tb
tb
II
II+
66. (b) By theorem of parallel axes,
I = I
cm
+ Md
2
I = I
0
+ M (L/2)
2
= I
0
+ ML
2
/4
67. (a) When angular acceleration (a) is zero then torque on
the wheel becomes zero.
q(t) = 2t
3
– 6t
2
d
6t 12t
dt
2q
Þ=-
2
d
12t 12 0
dt
2
q
Þa= = -=
\ t = 1 sec.

211. 207System of Particles and Rotational Motion
68. (a) According to parallel axis theorem of the moment of
Inertia
I = I
cm
+ md
2
d is maximum for point B so I
max
about B.
69. (a)
11 22 33
cm
123
mx mx mx
X
mmm
++
=
++
cm
300 (0) 500(40)400 70
300 500 400
X
´+
+´
=
++
cm
500 40 400 70
1200
X
´+´
=
cm
50 70 120
33
X
+
== = 40 cm
7
0. (b)
21
KI
2
=w
211
K' (2 ) 2K
22
æö
= w=ç÷
èø
71. (
a) For complete disc with mass '4M', M.I. about given
axis = (4M)(R
2
/2) = 2 MR
2
Hence, by symmetry, for the given quarter of the disc
M.I. = 2 MR
2
/4
21
MR
2
=
72. (d)
73
. (a) For a disc rolling without slipping on a horizontal rough
surface with uniform angular velocity, the acceleration
of lowest point of disc is directed vertically upwards
and is not zero (Due to translation part of rolling,
acceleration of lowest point is zero. Due to rotational
part of rolling, the tangential acceleration of lowest
point is zero and centripetal acceleration is non-zero
and upwards). Hence statement 1 is false.
74. (a) 75.(b)
EXERCISE - 3
Exemplar Questions
1. (d) A bangle is in the form of a ring as shown in figure.
The centre of mass of the ring lies at its centre, whichis outside the ring or bangle.
C
Centre
2. (c)Centre of mass of a system lies towards the part of the
system, having bigger mass. In the above diagram,
lower part is heavier, hence CM of the system lies
below the horizontal diameter at C point.
3. (b) The initial velocity is ˆ
iy
v ve= and after reflection from
the wall, the final velocity is
ˆ
fy
v ve=-. The trajectory
is at constant distance a on z axis and as particle moves
along y axis, its y component changes .
So position vector (moving along y-axis),
ˆˆ
yz
r ye ae=+
r
Hence, the change in angular momentum is
ˆ( )2
fjx
r m v v mv ae´ -=
r
.
4. (d)
Angular acceleration
d
dt
w
a=
where w is an
gular velocity of the disc and is uniform
or constant. 0
d
dt
w
a==
Hence, angular acceleration is zero.
5. (b) According to the question, when the small piece Q
removed it is stick at P through axis of rotation passes,
but axis of rotation does not passes through Q and It
is glued to the centre of the plate, the mass comescloser to the z-axis, hence, moment of inertia decreases.
6. (c) Let us consider the diagram of problem 5, there is a
line shown in the figure drawn along the diagonal.First, centre of mass of the system was on the dottedline and was shifted towards Q from the centre (1st
quadrant).
Q
y
x
Q
x
y
x
hole
When the mass Q is decrease, it will be on the same
line but shifted towards other side of Q on the line
joining QPor away from the centre so, new CM will
lies in IIIrd quadrant in x-y plane.
7. (a) As given that density :
2
()(1)x a bxr=+
where a an
d b are constants and
01x££
At b = 0, ()xar == constant
In that case, centre of mass will be at
x = 0.5 m (i.e., mid-point of the rod)
Putting, b = 0 in options.
(a)
321
0.5m
432
´==

212. 208 PHYSICS
(b)
42
0.5m
33
´¹
(c)
33
0.5m
42
´¹
(d)
43
0.5m
32
´¹
So, only (a) option gives 0.5.
8. (a) As there is external torque acting on the system,
angular momentum should be conserved.
Hence Iw = constant...(i)
where, I is moment of inertia of the system and w is
angular velocity of the system.
From Eq. (i),
11 22
IIw=w
where w
1 and w
2 are angular velocities before and
after jumping.
2
()I mr=Q So, given that,
m
1 = 2M, m
2 = M, w
1 = w, w
2 = ?
r
1 = r
2 = R
22
111 222
mr mrw=w
22
2
2MR MRw=w
as mass reduced to half, hence, moment of inertia also
reduced to half.
2
2w=w
NEET/AIPMT (2013-2017) Questions
9. (c)
2
3V
4g
V
From law of conservation of mechanical energy
1
2
Iw
2
+ 0 +
1
2
mv
2
= mg ×
2
3v
4g
Þ
1
2
Iw
2
=
3
4
mv
2
–
1
2
mv
2
=
2
mv
2
3
1
2
æö
-
ç÷
èø
or,
1
2
I
2
2
V
R
=
2
mv
4
or, I =
1
2
mR
2
Hence, object is a disc.
10. (d)
Weight of the rod will produce the torque
t = mg
L
2
= I a =
2
mL
3
a
2
rod
ML
I
3
éù
=êú
êúëû
Q
Hence, angular acceleration a =
3g
2L
11. (c) Given:m
1
= 2 kg m
2
= 4 kg
r
1
= 0.2 m r
2
= 0.1 m
w
1
= 50 rad s
–1
w
2
= 200 rad s
–1
As, I
1
W
1
= I
2
W
2
= Constant
\
11 22
12
f
IW IW
W
II
+
=
+
=
22
111 222
22
11 22
11
22
11
22
mrw mrw
mr mr
+
+
By putting the value of m
1
, m
2
, r
1
, r
2
and solving we
get = 100 rad s
–1
12. (a)Q I = MK
2
\ K =
I
M
I
ring
= MR
2
and I
disc
=
21
2
MR
2
11
2
22
2 :1
2
KI MR
KI MR
===
æö
ç÷
èø
13. (a) For solid sphere rolling without slipping on inclined
plane, acceleration
a
1
=
2
2
g sin
K
1
R
q
+
For solid sphere slipping on inclined plane without
rolling, acceleration
a
2
= g sin q
Therefore required ratio =
1
2
a
a
=
2
2
1 15
27K
1
1
5
R
==
+
+

213. 209System of Particles and Rotational Motion
14. (d) Here a = 2 revolutions/s
2
= 4p rad/s
2
(given)
I
cylinder
=
2211
MR (50)(0.5)
22
=
=
25
4
Kg-m
2
As t = Iaso T
R = Ia
Þ T =
25
(4)
I 4
N
R (0.5)
æö
pç÷
èøa
= = 50 pN
= 157 N
15. (c) Moment of inertia of shell 1 along diameter
I
diameter
= 22
MR
3
Moment of inertia of shell 2 = m. i of shell 3
= I
tangential
=
22225
MR MR MR
33
+=
X
X¢
32
1
So, I of the system along x x
1
= I
diameter
+ (I
tangential
) × 2
or, I
total
=
2225
MR MR 2
33
æö
+´
ç÷
èø
=
2212
MR 4MR
3
=
16. (c)
By torque balancing about B
N
A
(d) = W (d – x)
A
W(d–x)
N
d
=
d
x d – x
B
N
A
N
B
W
A
17. (b
) Applying angular momentum conservation
V
0
m
mV
0
R
0
= (m) (V
1
)
0
R
2
æö
ç÷
èø
\ v
1
= 2V
0
Therefore
, new KE =
1
2
m (2V
0
)
2
=
2
0
2mv
18. (d) From Newton's second law for rotational motion,
t
r
=
dL
dt
r
, if L
r
= constant then t
r
= 0
So, t
r
= rF´
rr
= 0
( )
ˆˆ ˆ ˆˆˆ
2i 6j 12k( i 3j 6k) 0-- ´a+ +=
Solving we get a = –1
19. (c) Work required to set the rod rotating with angular
velocity w
0
K.E. =
21
I
2
w
Work is minimum when I is minimum.
I is minimum about the centre of mass
So, (m
1
) (x) = (m
2
) (L – x)
or, m
1
x = m
2
L – m
2
x
\ x =
2
12
mL
mm+
20. (d)Giv
en : Speed V = 54 kmh
–1
= 15 ms
–1
Moment of inertia, I = 3 kgm
2
Time t = 15s
w
i =
V
r
=
15 100
0.453
= w
f
= 0
w
f
= w
i
+ at
0 =
100
3
+ (–a) (15) Þ a =
100
45
Average torque transmitted by brakes to the wheel
t = (I) (a) = 3 ×
100
45
= 6.66 kgm
2
s
–2
21.
(b) Time of descent µ
2
2
K
R
Order of value of
2
2
K
R
for disc;
2
2
K
R
=
1
2
= 0.5
fo
r sphere;
2
2
K
R
=
2
5
= 0.4
(spher
e) < (disc)
\Sphere reaches first
22. (a)Given: Radius of disc, R = 50 cm
angular acceleration a = 2.0 rads
–2
; time t = 2s
Particle at periphery (assume) will have both radial
(one) and tangential acceleration
a
t
= Ra = 0.5 × 2 = 1 m/s
2

214. 210 PHYSICS
From equation
,
w = w
0
+ at
w = 0 + 2 × 2 = 4 rad/sec
a
c
= w
2
R = (4)
2
× 0.5 = 16 × 0.5 = 8m/s
2
Net acceleration,
a
total
=
22
tc
aa+ = 22
18+
» 8 m/s
2
23. (b) Mo
ment of inertia of complete disc about point 'O'.
I
Total disc
=
2
MR
2

M
R
R
O
Mass of removed disc
M
Removed
=
M
4
(Mass µ area)
Moment o
f inertia of removed disc about point 'O'.
I
Removed
(about same perpendicular axis)
= I
cm
+ mx
2
=
( )
2
M R/2
42
+
2 2
M R 3MR
4 2 32
æö
=ç÷
èø
Therefore the
moment of inertia of the remaining part
of the disc about a perpendicular axis passing through
the centre,
I
Remaing disc
= I
Total
– I
Removed
=
2
22MR 3 13
MR MR
2 32 32
-=
24. (d)
Centre of mass may or may not coincide with centre
of gravity. Net torque of gravitational pull is zero about
centre of mass.
0
g i i ig
rmt=St=S´=
Mechanical advantage , M. A.=
Load
Effort
If M.A. > 1 Þ Loa d > Effort
25. (a) Here,
12
I I 2Iw+w=w
Þ w =
12
2
w +w
(K.E.)
i
=
22
12
11
II
22
w+w
(K.E.)
f
=
21
2I
2
´w =
2
12
I
2
w +wæö
ç÷
èø
Loss in K.E.
fi
(K.E) (K.E)=- =
2
12
1
I()
4
w -w
26. (b) Given,
mass of cylinder m = 3kg
R = 40 cm = 0.4 m
40
cm
F = 30 N
F = 30 N ; a = ?
As w
e know, torque t = Ia
F × R = MR
2
a
2
FR
MR
´
a=
2
30 (0.4)
3 (0.4)
´
a=
´
or,
2
25rad / sa=

215. NEWTON'S UNIVER SAL LAW OF GRAVITATION
Gravitational force is an attractive force between any two point
masses M
1
and M
2
separated by any distance r.
It is given by F = G
12
2
æö
ç÷
èø
MM
r
where G is
the universal gravitational constant.
Its value, G = 6.67 × 10
–11
Nm
2
kg
–2
Dimensions of G are [M
–1
L
3
T
–2
]
(a) The gravitational force is the central force and follows
inverse square law. It acts along the line joining the particles.
(b) Since the work done by the gravitational force is
independent of the path followed and hence it is a
conservative force.
(c) It is the weakest force in nature. It is 10
38
times smaller
than nuclear force and 10
36
times smaller than electric force.
Strongest force being nuclear force (for small range)
followed by electric force.
(d) Gravitation is independent of the presence of other bodies
around it.
ACCELERATION DUE TO GRAVITY (g)
The force of attraction exerted by earth on a body of mass m is
the force of gravity.
So the force of gravity from Newton’s gravitational law is

2
=
e
GMm
F
r
……(i)
wher
eM
e
= mass of earth
r = distance of the body from the centre of earth.
The force of gravity can be written as F = mg……(ii)
where g is called the acceleration due to gravity.
From the expression (i) and (ii), we get2
=
e
GM
g
r
……(iii)
If b
ody is located at the surface of earth i.e., r = R
e
(radius of
earth) then
3
22
(4/3) 4 3
pr
= = =
pr
ee
e
ee
GMGR
g GR
RR
……(iv)
wh
ere r is the density of earth.
This is the relation between universal gravitational constant
'G' and acceleration due to gravity 'g'.
Change in the Value of Acceleration due to Gravity (g)
(i)Due to rotation or latitude of earth : Let us consider a
particle P at rest on the surface of earth at a latitude f.
()
22
cosf=-wf
e
g gR
f
r
Rg¢
w
At poles f = 90º, so g(f) = g ...(v)
At equator f = 0º so g(f) = g –w
2
R
e
...(vi)
(ii)Due to shape of earth : The shape of the earth is not
perfectly spherical. It is flattened at poles. The polar radius
is 21 km less than the equatorial radius. Hence acceleration
due to gravity at poles is greater than at the equator.
g
p
= 9.83 ms
–2
= value of g at poles
g
e
= 9.78 ms
–2
= values of g at equator
i.e. g
p
> g
e
(iii)At a depth 'd' below the earth surface : The acceleration
due to gravity at the surface of earth.
GR
3
4
R
GM
g
e
2
e
e
ú
û
ù
ê
ë
é
rp÷
ø
ö
ç
è
æ
== (from eq
n
. iv).. .(vii)

d
R-d
O R
P
Earth
Surface of earth
8
Gravitation

216. 212 PHYSICS
If a body is ta
ken at a depth d below the earth surface, then
the body is attracted by inner core of mass M' of earth
radius (R
e
– d). Then acceleration due to gravity at point P
is
3
e
22
ee
G (4/3) (R d)
GM
g
(
R d) (R d)
éùp -r
¢ ëû
¢==
--
()( )r-p= )dR(3/4G
e
...(viii)
From eq
ns
. (vii) an
d (viii) we have
1
æ öæö-
¢= =-ç ÷ç÷
è øèø
e
ee
Rd d
ggg
RR
...(ix)
The value of accelera
tion due to gravity at the centre of the
earth is zero.(iv)At a height 'h' above the earth surface : The acceleration
due to gravity at the surface of earth from expression (iv)is defined as
2
e
e
R
GM
g=
At a height h
above the earth surface, the acceleration due
to gravity is
2
e
2
e
e
r
GM
)hR(
GM
g =
+
=¢ ...(x)
Þ
2
e
e
R
hR
gg
-
÷
÷
ø
ö
ç
ç
è
æ+
=¢Þ [from eqs. (ix) & (x)]
If
h << R
e
e
2
1
R
æö
»-¢
ç÷
èø
h
gg
...(xi)
Keep in Memory
1.The val
ue of the acceleration due to gravity on the moon
is about one sixth of that on the earth and on the sun is
about 27 times that on the earth.
E
sEM
g
g 27g and g
6
æö
==
ç÷
èø
2.The value of g is minimum on the mercury, among all
planets.
3.For h<<R, the rate of decrease of the acceleration due togravity with height is twice as compared to that with depth.
4.The value of g increases with the increase in latitude. Itsvalue at latitude q is given by : g
q
= g – Rw
2
cos
2
q.
5.Rotation of the earth about its own axis is responsible fordecrease in the value of g with latitude.
6.The weight of the body varies along with the value of g (i.e.W = mg)
7.Inertial mass and gravitational mass-
(a) Inertial mass (M
i
) defined by Newton's law of motion
M
i
= F/a =
External force applied on the body
Acceleration produced in it
i12
i21
(M)a
(M)a
=
(ff applied fo
rce on (M
i
)
1
and (M
i
)
2
is same.)
(b) Gravitational mass M
g
defined by Newton's law of
gravitation
M
g
=
gravitytodueonAccelerati
bodyofWeig
ht
g
W
g
F
g
==
8.If two spheres of same material, mass and radius are put in
contact, the gravitational attraction between them is directly
proportional to the fourth power of the radius.
4
2
R
3
4
G
4
1
F ÷
ø
ö
ç
è
æ
ps=
A
R
R
B
m 2R ¬ ® m
This system may be considered to be made up of two mass
each separated by 2R distance.
9. If the earth stops rotating about its axis, the value of g at
the equator will increase by about 0.35%, but that at the
poles will remain unchanged.
If the earth starts rotating at the angular speed of about 17
times the present value, there will be weightlessness on
equator, but g at the poles will remain unchanged. In such a
case, the duration of the day will be about 84 min.
10.If the radius of planet decreases by n%, keeping the mass
unchanged, the acceleration due to gravity on its surface
increases by 2n%. i.e.,
R
R2
g
g D-
=
D
11.If the mass of
the planet increases by m% keeping the
radius constant, the acceleration due to gravity on itssurface increases by m% i.e.,
gM
gM
DD
= where R = constant.
12.If the density of planet decreases by r% keeping the radius
constant, the acceleration due to gravity decreases by r%.
13.If the radius of the planet decreases by r% keeping the
density constant, the acceleration due to gravity decreases
by r%.
Example 1.
W
hat will be the acceleration due to gravity on the surface
of the moon with its radius as 1/4th the radius of the earthand its mass as 1/80th the mass of earth?
Solution :
2
e
e
R
GM
g= and
2
m
m
R
GM
'g= ;
\
5
1
1
4
80
1
R
R
M
M
g
'g
22
m
e
e
m
=÷
ø
ö
ç
è
æ
´=
÷
÷
ø
ö
ç
ç
è
æ
= ; \ 5/g'g=

217. 213Gravitation
Exam
ple 2.
A mass M splits into two parts m and (M–m), which are
then separated by a certain distance. What will be the
ratio m/M which maximizes the gravitational force
between the two parts?
Solution :
2
x
)mM(Gm
F
-
= ; For maxima,
2
1
M
m
or0)m2M(
x
G
dm
dF
2
==-=
Example 3.
Ide
ntical point masses each equal to m are placed at
x = 0, x = 1, x = 2, x = 4, ... The total gravitational forceon mass m at x = 0 due to all other masses is
(a) infinite (b) (4/3) Gm
2
(c) (4/3 GM
2
) (d) zero
Solution : (c) o m m m
x=4x=2x=1x=0
....
4
Gm
2
Gm
1
Gm
F
2
2
2
2
2
2
0
+++=
ú
û
ù
ê
ë
é
+++= ...
4
1
2
1
1
1
GmF
222
2
0
It is a G.P. with
common ratio 1/4.
\
ú
û
ù
ê
ë
é
-
=
4/11
1
GmF
2
0
3
Gm4
F
2
0
=
Example 4.
F
our identical point masses each equal to m are placed
at the corners of a square of side a. The force on a point
mass m’ placed at the point of intersection of the two
diagonals is
(a) (4Gmm¢)/a
2
(b) (2Gmm¢)/a
2
(c) (Gmm¢)/a
2
(d) zero
Solution :(d)
The forces acting on the mass m¢ placed at the point of
intersection of diagonals are balanced by each other as
shown in figure. Therefore net force is zero.
m m
mm
m¢F
F
F
F
Example 5.
A t
unnel is dug along the diameter of the earth. Determine
the force on a particle of mass m placed in the tunnel at a
distance x from the centre.
Solution :
Force on a mass m placed at a distance x from the centre is
equal to the force of gravity due to a mass of spherical
volume of earth of radius x.
m
R
x
Center of earth
\
2
3
x
md
3
x4
G
F
÷
÷
ø
ö
ç
ç
è
æp
=
wher
e d = density of earth =
3
R4
M3
p
;
\ x.
R
GMm
F
3
= [M is total m
ass of earth]
Example 6.
If the radius of the earth were to shrink by two per cent,
its mass remaining the same, the acceleration due to
gravity on the earth’s surface would
(a) decrease by 2% (b) increase by 2%
(c) increase by 4% (d) decrease by 4%
Solution : (c)
As
2
R
GM
g= ; So, if R decrea
ses then g increases.
Taking logarithm of both the sides;log g = log G + log M – 2 log R
Differentiating it, we get
g
dg
= 0 + 0 –
R
dR2
;
\
g
dg

100
4
100
2
2 =÷
ø
ö
ç
è
æ-
-=
\ % incre
ase in
100
g
dg
g ´== %4100
100
4
=´
Example 7.
An appl
e of mass 0.25 kg falls from a tree. What is the
acceleration of the apple towards the earth? Also calculate
the acceleration of the earth towards the apple.
Given : Mass of earth = 5.983 × 10
24
kg,
Radius of earth = 6.378 × 10
6
metre,
G = 6.67 × 10
–11
Nm
2
kg
–2
.
Solution :
Mass of apple, m = 0.25 kg,
Mass of earth, M = 5.983 × 10
24
kg,
Radius of earth, R = 6.378 × 10
6
metre
Gravitational constant, G = 6.67 × 10
–11
Nm
2
kg
–2
Let F be the gravitational force of attraction between the
apple and earth.
Then,
2
GmM
F
R
=
Let a be th
e acceleration of apple towards the earth.
22
F GmM GM
a
mmRR
===
Þ
11 24
22
62
6.67 10 5.983 10
a ms 9.810m s
(6.378 10 )
-
--´´´
==
´

218. 214 PHYSICS
Let a¢ be the a
cceleration of the earth towards the apple.
2
11
2
62
252
F Gm
a
MR
6.67 10 0.25
ms
(6.378 10 )
4.099 10 ms .
-
-
--
==¢
´´
=
´
=´
Example 8.
The acc
eleration due to gravity at the moon’s surface is
1.67 ms
–2
. If the radius of the moon is 1.74 × 10
6
m, calculate
the mass of the moon. Use the known value of G.
Solution :
2
GM
g
R
= or
2
gR
M
G
=
This relation is true not only f
or earth but for any heavenly
body which is assumed to be spherical.
Here,
26
g 1.67ms , R 1.74 10 m
-
= =´ and

11 22
G 6.67 10 Nm kg
--
=´
\Mass of moon,
62
22
11
1.67 (1.74 10 )
M kg 7.58 10 kg
6.67 10
-
´´
= =´
´
GRAVITATIONAL FIELD
(or Grav
itational Field Intensity)
The region or space around a body in which its gravitational
influence is experienced by other bodies is the gravitational
field of that body.
The gravitational field strength E
g
, produced by a mass M at
any point P is defined as the force exerted on the unit mass
placed at that point P. Then
22
æö
===
ç÷
èø
g
F G Mm GM
Em
m rr
...(i)
where m = te
st mass
r = distance between M and m
(i) The direction of E
g
always points towards the mass
producing it.
(ii) The gravitational field can be represented by gravitational
lines of force.
(iii) The S.I unit of E
g
is newton/kg. Its dimensions are
[M
0
LT
–2
].
Gravitational intensity (E
g
) for spherical shell
r
|E |=0
g
Radius of
spherical shell
|E|
g
O
R
2
r
GM
|E
g|=
(i)|E
g
|= 0 at points inside the spherical shell (i.e. r < R)
(ii)
2
||
-
=
g
GM
E
R
at the surface of s hell (i.e. r = R)
(iii)
2
||
-
=
g
GM
E
r
for outside th e spherical shell (r >R).
It is clear that as r increases, E
g
decreases.
Gravitational intensity for solid sphere :
R
r
|E|
g
Solid
sphere
E
g
(i)
3
.
||=-
g
GMr
E
R
for points ins ide the solid sphere (r<R)
(ii) At the surface of solid sphere (r = R)
2
||=-
g
GM
E
R
(iii) Outside the solid sphere
2
||=-
g
GM
E
r
(r > R)
GRAVITATIONAL POTENTIAL
Gravitat
ional potential at any point is the work done by
gravitational force in carrying a body of unit mass from infinity
to that point in gravitational influence of source.
i.e.,
A
o
W GM
V
mr
-
==
where r is t
he distance of point from source mass M.
The S.I. unit of V is joule/kg. Its dimensions are [M
0
L
2
T
–2
]. It is
a scalar quantity.
Gravitational potential at the earth surface is
-
= =-
e
Ae
e
GM
V gR
R
Gravitational pot
ential due to a uniform ring at a point on its
axis
22
=-
+
g
GM
V
ar
a
O
r P
q
Gravitational potential due to a uniform thin spherical shell
O
V
g
r
R
GM
-

219. 215Gravitation
(i)
at a point outside the shell,
g
GM
V
r
-
= (r > R
)
(ii) at a point inside the shell,
g
GM
V
R
=- (r < R)
w
here R = radius of spherical shell.
Potential due to a uniform spheral shell is constant throughout
the cavity of the shell.
Gravitational potential due to a uniform solid sphere
(i) at an external point
=-
g
GM
V
r
(r³ R)
V
R
solid
state
(ii) at an internal point
22
3
(3)
2
=--
g
GM
V Rr
R
(r < R)
O
r
R
(iii)Potential at the centre of solid sphere is
3
2
-
=
GM
V
R
GRAVITATIONAL POTENTIAL ENERGY
Let V
g
be the gravitational potential at a point. If we place a mass
m at that point, then we say that the gravitational potential energy
possessed by the mass is
=´
gg
U Vm Þ =-´
g
GM
Um
r
g
GM
V
r
éù
\ =-
êú
ëû
We can also therefore say that the gravitational potential energy
of a system containing two masses m
1
and m
2
placed such that
the centre to centre distance between them is r then
12
=-
g
Gmm
U
r
This formula is valid taking into consideration the fact that we
have taken the gravitational potential of few at infinity. Pleaseremember that unless otherwise stated it is understood that thegravitational potential is taken to be zero at infinity.
Relation between gravitational potential energy andgravitational potential.
Gravitational potential energy = gravitational potential × mass
Keep in Memory
1.The gravitat
ional potential energy of a mass m at a point
above the surface of the earth at a height h is given by
g
GMm
U
Rh
-
=
+
. The –ve sign shows that if h increases,
the gravitational PE decreases and becomes zero at infinity.
2.(a) If we take reference level to be at the surface of earth
(not at infinity) i.e., we assume that the gravitational
P.E of a mass m is zero at the surface of earth, then the
gravitational potential energy at a height h above the
surface of earth is (mgh), where h << R
e
(radius of
earth)
(b) The gravitational P.E of mass m on the earth’s surface
is
e
g
e
GMm
U mgR
R
-
= =-
If we assume
that gravitational P.E of mass m is zero
at infinity i.e., we take reference level at infinity then
P.E.of any mass m at a height h above the earth surface
)hR(
mGM
U
e
e
h
+
-
=
So work d
one against the gravitational force, when
the particle is taken from surface of earth to a heighth above the earth surfaceWork done = change in potential energy
ÞU
h
–U
g
=
e
e
e
e
R
mGM
hR
mGM
+
+
-
)hR(R
hmGM
ee
e
+
´
@
÷
÷
ø
ö
ç
ç
è
æ
»
ee
e
R
h
R
mGM
(if h<<R)
e
2
e
GM
mh
R
æö
=ç÷
ç÷
èø
So, work done = mgh is stored in the particle as
particle potenital energy, when kept at height ‘h’.
Example 9.
Four particles each of mass 1 kg are at the four corners of
a square of side 1m. Find the work done to remove one of
the particles to infinity.
Solution :
The gravitational potential energy associated with any pair
of particle of mass m
1
and m
2
separated by a distance r is
r
mGm
U
21-
=
Initial P.E. of s
ystem, – U
1
=
2
11G2
1
11G
.4
´´
+
´´
or
ú
ú
û
ù
ê
ê
ë
é+
-=
2
224
GU
1

220. 216 PHYSICS
1 Kg 1 Kg
1 Kg1 Kg
1m
1m
1m 1m
2m
2m
When one mass is removed, then
P.E. = U
2
= ú
û
ù
ê
ë
é ´´
+
´´
-
2
11G
1
11G
.2
G
2
122
U
2
ú
ú
û
ù
ê
ê
ë
é +
-= ;
\ Work done =
ú
ú
û
ù
ê
ê
ë
é --+
2
122224
G =
ú
ú
û
ù
ê
ê
ë
é +
2
122
G
[Q Work done = cha
nge in P.E. = –U
2
–[–U
1
] = U
1
–U
2
]
Example 10.
A solid sphere of uniform density and mass M has a radius
of 4m. Its centre is at the origin of coordinate system. Two
spheres of radii 1m are taken out so that their centres are
at P(0, –2, 0) and Q(0, 2, 0) respectively. This leaves two
spherical cavities. What is the gravitational field at the
centre of each cavity?
O
P Q
R
z
y
Solution :
If the c
avities are not made, then the intensity at the point
P (or Q), due to whole sphere
32
GM
2
64
GM
x.
R
GM
I
3
=´== where
PQR
IIII=++
rrrr
\
QPR IIII
rrrr
--= …… (i)
Where I
R
is the intensity
of gravitational field at a point,
distance x from O due to remainder sphere (exclude cavity)
Mass of big sphere
3
d)4(4
M
3
´p
=
Mass of sma
ll sphere P or Q,
3
4π(1)
3
´
=
d
m \
M
m
64
=
At P, I
p
= 0, ÷
ø
ö
ç
è
æ
==
64
M
.
4
G
r
Gm
I
22Q
\ I
R
=
GM GM
32 1024
- [From eq (i)]
\ R
31GM
I
1024
=
Example 11.
Find the gravitational pote
ntial energy of a system of four
particles, each having mass m, placed at the vertices of a
square of side l. Also obtain the gravitational potential at
the centre of the square.
Solution :
The system has four pairs with distance l and two diagonal
pairs with distance
2l.
m
l
2l
m
m m
\
2 2 22
Gm Gm 2Gm 1 Gm
U 4 2 2 5.41
22
æö
=- - =- + =-
ç÷
èølll l
The gravitational potential at the centre of the square is
( )r 2 /2=l
V = Algebraic s
um of potential due to each particle
Þ
4 2Gm
V=-
l
Example 12.
The radius of earth is R. Find
the work done in raising a
body of mass m from the surface of earth to a height R/2.
Solution :
If a body of mass m is placed at a distance x from the centre
of earth, the gravitational force of attraction F between the
body and the earth is
2
x
mGM
F=
Small amount of
work done in raising a body through a
small distance dx is given by
dx
x
mGM
dxFdW
2
==
Total work done in
raising the body from the surface of
earth to a height R/2 is given by
W =
R R/2
2
R
GMm
dx
x
+
ò
=
3 R/2 3 R/2
2
RR
1
GMm x dx GMm
x
- æö
=- ç÷
èøò
=
21
GMm
3RR
éù
--
êú
ëû
=
2
Mm gRm
3R 3R
G
=
=
1
mgR
3
[\ GM = gR
2
]

221. 217Gravitation
SATELLITES
A satel
lite is any body revolving around a large body under
the gravitational influence of the latter.
The period of motion T of an artificial satellite of earth at a
distance h above the surface of the earth is given by,
p
3
2
( +)
=2
e
e
Rh
T
gR
where,
R
e
= radius of the earth
where, g = acceleration due to gravity on the surface of the earth.
If R
e
> > h, then
e
R
T2
g
=p ; R
e
= 6.4 × 10
6
m ;
g = 9.
8 ms
–2
i.e.
T 84.58 minutes5075sec»;
Assuming R
e
+ h = r, the distance of the satellite from the centre
of the earth,
3/2
T (r)µ
Orbital velocity (v
0
)
Le
t a satellite of mass m revolve around the earth in circular orbit
of radius r with speed v
0
. The gravitational pull between satellite
and earth provides the necessary centripetal force.
Centripetal force required for the motion =
2
0
mv
r
Gravitational force =
GMm
r²
2
0
2
mvGMm
r r
= or
2
0
GM
v
r
=
or 0
=
GM
v
r
.......... (1)
or 0
=
+
g
vR
Rh
[Q
2
GM
g
R
= and r = (R + h)]
Orbital velocity,
eR
==
e
e
GM
Vo gR
(i) Value of orbital velocity does not depend on the mass of the
satellite.
(ii) Around the earth the value of orbital velocity is
7.92 km/sec.
(iii) Greater is the height of the satellite, smaller is the orbital
velocity.
(iv) The direction of orbital velocity is along the tangent to the
path.
(v) The work done by the satellite in a complete orbit is zero.
Angular momentum (L) : For satellite motion, angular momentum
will be given by
GM
L mvr mr
r
== i.e., L = [m
2
GMr]
1/2
Angular momentum of a satellite depends on both, the mass of
orbiting and central body. It also depends on the radius of the
orbit.
Energy of a satellite
Total energy of satellite revolving in an orbit of radius r around
the earth can be calculated as follows :
(i) The gravitational potential energy of a satellite of mass m
is
–=
g
GMm
U
r
, where r is the radius of the orbit.
(ii)Kinetic energy of the satellite is
2
0
1
22
==
k
GMm
E mv
r
(iii) So, to tal energy of the satellite
2
= + =-
gk
GMm
EUE
r
The negative sign shows that satellite cannot leave the
orbit itself. It requires an energy equal to
2
GMm
R
, which is
ca
lled Binding Energy (B.E.) of satellite.
(iv) Total energy of a satellite at a height equal to the radius of
the earth
1
––
2()44
= ==
+
GMmGMm
mgR
RRR
GEO-STA
TIONARY SATELLITE
A satelite which appears to be stationary for a person on the
surface of the earth is called geostationary satellite.
It is also known as parking satellite or synchronous satellite.
(i) The orbit of the satellite must be circular and in the equatorial
plane of the earth.
(ii) The angular velocity of the satellite must be in the same
direction as the angular velocity of rotation of the earth i.e.,
from west to east.
(iii) The period of revolution of the satellite must be equal to the
period of rotation of earth about its axis.
i.e. 24 hours = 24 × 60 × 60 = 86400 sec.
24²
T r³
GM
p
= or
11
2 22
33
2 22
GMT GM R T
r
4
R4
æöæö
= =´ç÷ç÷
ppèøèø
=
1
362(86400)²
9.8 (6.38 10)
4²
éù
´´´
êú
pëû
= 42237 × 10
3
m = 42,237 km. » 42000 km.
h = r – R = 42000 – 6400 = 35600 km.
(a) Height of geostationary satellite from the surface of
the earth is nearly 35600 km.
(b) The orbital velocity of this satellite is nearly
3.08 km/sec.
(c) The relative velocity of geostationary satellite with
respect to earth is zero.
This type of satellite is used for communication
purposes. The orbit of a geostationary satellite is called
‘Parking Orbit’.
Polar Satellite :
Polar satellites travel around the earth in an orbit that travels
around the earth over the poles. The earth rotates on its axis as
the satellite goes around the earth. Thus over a period of many
orbits it looks down on every part of the earth.

222. 218 PHYSICS
Different orbital shapes correspon
ding to different velocities of
a satellite :
[1] When v < v
0
(i) The path is spiral. The satellite finally falls on the earth
(ii) Kinetic energy is less then potential energy
(iii) Total energy is negative
[2] When v = v
0
(i) The path is circular
(ii) Eccentricity is zero
(iii) Kinetic energy is less than potential energy
(iv) Total energy is negative
[3] When v
0
< v < v
e
(i) The path is elliptical
(ii) Eccentricity < 1
(iii) Kinetic energy is less than potential energy
(iv) Total energy is negative
[4] When v = v
e
(i) The path is a parabola
(ii) Eccentricity = 1
(iii) Kinetic energy is equal to potential energy
(iv) Total energy is zero
[5] When v > v
e
(i) The path is a hyperbola
(ii) Eccentricity > 1
(iii) Kinetic energy is greater than potential energy
(iv) Total energy is positive
If the orbit of a satellite is elliptical
(1)The energy
GMm
E
2a
=- = const. with ‘a’ as semi-major axis;
(2)KE will be maximum when the satellite is closest to the central
body (at perigee) and minimum when it is farthest from the
central body (at apogee) [as for a given orbit L = const.,
i.e., mvr = const., i.e., v µ 1/r]
(3)PE = (E – KE) will be minimum when KE = max, i.e., the
satellite is closest to the central body (at perigee) and
maximum when KE = min, i.e., the satellite is farthest from
the central body (at apogee).
ESCAPE SPEED (V
e
)
It is the minimum speed with which a body should be projected
from the surface of a planet so as to reach at infinity i.e., beyond
the gravitational field of the planet.
If a body of mass m is projected with speed v from the surface of
a planet of mass M and radius R, then
as
21
2
=K mv and
GMm
U
R
=- , E
S
=
21
2
mv–
GMm
R
Now if v' is the speed of body at ¥, then
2211
() 0 ()
22
¥= +=¢¢E mv mv [as U0
¥
=]
So by conservati
on of energy 2211
()
22
=-= ¢
GMm
mv mv
R
i.e.
221
()
22
1
=+ ¢
GMm
mv mv
R
so v will be minimum
when v¢ ® 0,
i.e.
min
2
2===
e
GM
v v gR
R
2
GM
asg
R
éù
=
êú
ëû
•The value of escape velocity does not depend upon the mass
of the projected body, instead it depends on the mass andradius of the planet from which it is being projected.
•The value of escape velocity does not depend on the angleand direction of projection.
•The value of escape velocity from the surface of the earth is11.2 km/sec.
•The minimum energy needed for escape is = GMm/R.
•If the velocity of a satellite orbiting near the surface of theearth is increased by 41.4%, then it will escape away fromthe gravitational field of the earth.
•If a body falls freely from infinite distance, then it will reachthe surface of earth with a velocity of 11.2 km/sec.
Relation between orbital velocity (V
0
) and escape speed (V
e
)
0
22==
e
V gRV
Keep in Memory
1.The esca
pe velocity on moon is low
6
æö
=ç÷
èø
E
m
g
asg hence
there is no
atmosphere on moon.
2.If the orbital radius of the earth around the sun be one
fourth of the present value, then the duration of the year
will be one eighth of the present value.
3.The satellites revolve around the earth in a plane that
coincides with the great circle around the earth.
KEPLER'S LAWS OF PLANETARY MOTION
1. The law of orbits : Each planet revolves about the sun in
an elliptical orbit with the sun at one of the foci of the
ellipse.
eaea
P
x
Perihelion AphelionSun
S¢
y
a a
S
a = Semi-major axis
A pl
anet of mass m moving in an elliptical orbit around the
sun. The sun of mass M, is at one focus S¢ of the ellipse.
The other focus is S, which is located in empty space. Each
focus is at distance ‘ea’ from the ellipse’s centre, with ‘e’
being the eccentricity of the ellipse and ‘a’ semimajor axis
of the ellipse, the perihelion (nearest to the sun) distance
r
min.
, and the aphelion (farthest from the sun) distance r
max.
are also shown.
r
max
= a + e a = (1 + e) a
r
min
= a – e a = (1 – e) a

223. 219Gravitation
The di
stance of each focus from the centre of ellipse is ea,
where e is the dimensionless number between 0 to 1 called
the eccentricity. If e = 0, the ellipse is a circle.
For earth e = 0.017.
2.Law of areas : An imaginary line that connects a planet to
the sun sweeps out equal areas in the plane of the planet's
orbit in equal times;
i.e., the rate dA/dt at which it sweeps out area A is constant.
S
P
4
P
3
P
2
P
1
A
1A
2
If12 34
=
PP PP
tt
then AA
1
= A
2
dA 1 (r)(v dt) 1
rv
dt 2 dt 2
== and as L = mvr
dA
r
vdt
so
2
=
dAL
dtm
....(1)
But a
s L = constt., (force is central, so torque = 0 and hence
angular momentum of the planet is conserved)
areal velocity (dA/dt) = constant which is Kepler's II
nd
law,
i.e., Kepler's II
nd
law or constancy of areal velocity is a
consequence of conservation of angular momentum.
3.Law of periods : The square of the period of revolution of
any planet is proportional to the cube of the semi-major
axis of the orbit,
i.e., T
2
µ r
3
.
or,
23
11
23
22
=
Tr
Tr
Focus Semi major
a
xis
a
Satellite
r
Perigee
KE = Max
PE
= Min
Apogee
KE = Min
PE = Max
r
min r
max
b
b = Semi
minor axis
Sun
WEIGHTL
ESSNESS
The “weightlessness” you may feel in an aircraft occurs any time
the aircraft is accelerating downward with acceleration g. It is
possible to experience weightlessness for a considerable length
of time by turning the nose of the craft upward and cutting power
so that it travels in a ballistic trajectory. A ballistic trajectory is
the common type of trajectory you get by throwing a rock or a
baseball, neglecting air friction. At every point on the trajectory,
the acceleration is equal to g downward since there is no support.
A considerable amount of experimentation has been done with
such ballistic trajectories to practice for orbital missions where
you experience weightlessness all the time.
The satellite is moving in a circular orbit, it has a radial acceleration
2
0
0
2
vGM GM
a asv
rrr
éù
æö
=== êú ç÷
èø
êúëû
i.e., it is falling towards earth's centre with acceleration a,
so apparent weight of the body in it W
ap
= m(g? – a)
where g? is the acceleration due to gravity of earth at the position
(height) of satellite, i.e. g´ = (GM/r
2
), so that
ap
22
GM GM
W m0
rr
éù
= -=
êú
ëû
i.e., the apparen
t weight of a body in a satellite is zero and is
independent of the radius of the orbit.Keep in Memory
1.Th
e moon takes 27.3 days to revolve around the earth. The
radius of its orbit is 3.85 × 10
5
km.
2.Kepler's second law is based on conservation of angular
momentum.
3.Perihelion distance is the shortest distance between the
sun and the planet.
4.Aphelion distance is the largest distance between the Sun
and the planet.
perihelion
aphelion
r
r
=
Vaphelion
Vper
ihelion
5.If e is the eccentricity of the orbit
then
aphelion
perihelion
r1e
1er
+
=
-
r
aphelion
+ r
per
ihelion
= 2r
6.If e > 1 and total energy (KE + PE) > 0, the path of the
satellite is hyperbolic and it escapes from its orbit.
7.If e < 1 and total energy is negative it moves in elliptical
path.
8.If e = 0 and total energy is negative it moves in circular
path.
9.If e = 0 and total energy is zero it will take parabolic path.
10.The path of the projectiles thrown to lower heights is
parabolic and thrown to greater heights is elliptical.
11.Kepler’s laws may be applied to natural and artificial
satellites as well.
12.Gravitational force does not depend upon medium so no
medium can shield it or block it.
13.The escape velocity and the orbital velocity are independent
of the mass of the body being escaped or put into the orbit.

224. 220 PHYSICS
Example 1
3.
If the earth is at 1/4 of its present distance from the sun,
what would be the duration of the year?
Solution :
We know that,
3
2
3
1
2
2
2
1
R
R
T
T
=
Substituting the giv
en values, we get
64)4(
T
1
or
4
R
R
T
)1( 3
2
2
3
3
2
2
2
==
÷
ø
ö
ç
è
æ
=
.year
8
1
Tor
64
1
T
2
2
2
==
Example 14.
T
wo satellites A and B go round a planet P in a circular
orbits having radii 4R and R respectively. If the speed of
satellite A is 3v, what will be the speed of satellite B?
Solution :
We know that
v (GM/r)=
Here
GM
3v
4R
æö
=ç÷
èø
and
GM
v'
R
æö
=
ç÷
èø
;
\
v' GM 4R
2
3v R GM
æöæö
=´= ç÷ç÷
èø èø
; v' 6v=
Example 15.
A sp
aceship is launched into a circular orbit close to earth’s
surface. What additional velocity has now to be imparted
to the spaceship in the orbit to overcome the gravitational
pull? Radius of earth = 6400 km and g = 9.8 m/s
2
.
Solution :
The orbital velocity of spaceship in circular orbit
0
GM GM
v
rR
==
(Q spaceship is very close to earth, r = R)
As
2
R
MG
g= hence G M =
g R
2
\ 6
0
v (R g) (6.4 10 ) (9.8)= = ´´ = 7.9195 km/s
Furt
her,
)9195.72(Rg2v
e
´== = 11.2 km/s
(wh
ere v
e
is escape velocity)
Additional velocity required = 11.2000 – 7.0195
= 3.2805 km/s
So the velocity 3.2805 km/s must be added to orbital velocityof spaceship.

225. 221Gravitation

226. 222 PHYSICS
1.A satel
lite is orbiting around the earth near its surface. If its
kinetic energy is doubled, then
(a) it will remain in the same orbit.
(b) it will fall on the earth.
(c) it will revolve with greater speed.
(d) it will escape out of the gravitational field of the earth.
2.There is no atmosphere on the moon because
(a) it is closer to the earth and also it has the inactive inert
gases in it.
(b) it is too for from the sun and has very low pressure in
its outer surface.
(c) escape velocity of gas molecules is greater than their
root mean square velocity.
(d) escape velocity of gas molecules is less than their
root mean square velocity.
3.The maximum kinetic energy of a planet moving around the
sun is at a position
B
A
Sun
D
C
(a)A (b) B
(c)C (d) D
4.A m
an waves his arms while walking. This is to
(a) keep constant velocity
(b) ease the tension
(c) increase the velocity
(d) balance the effect of earth’s gravity
5.A missile is launched with a velocity less than escape
velocity. The sum of its kinetic and potential energies is
(a) zero
(b) negative
(c) positive
(d) may be positive, negative or zero.
6.Which of the following graphs represents the motion of a
planet moving about the sun ?
(a)
T
2
R
3
(b)
T
2
R
3
(c)
T
2
R
3
(d)
T
2
R
3
7.Due to rotation o
f the earth the acceleration due to gravity
g is
(a) maximum at the equator and minimum at the poles
(b) minimum at the equator and maximum at the poles
(c) same at both places
(d) None of these
8.A planet moves around the sun. At a point P it is closest
from the sun at a distance d
1
and has a speed v
1
. At another
point Q, when it is farthest from the sun at a distnace d
2
its
speed will be
(a)
2
21
2
1
d/vd (b)
112
d/vd
(c)
211
d/vd (d)
2
11
2
2
d/vd
9.The weight of an object in the coal mine, sea level and at
the top of the mountain, are respectively W
1
, W
2
and W
3
then
(a)W
1
< W
2
> W
3
(b) W
1
= W
2
= W
3
(c)W
1
< W
2
< W
3
(d) W
1
> W
2
> W
3
10.Two planets of radii r
1
and r
2
are made from the same
material. The ratio of the acceleration due to gravity g
1
/g
2
at the surfaces of the two planets is
(a)r
1
/r
2
(b) r
2
/r
1
(c) (r
1
/r
2
)
2
(d) (r
2
/r
1
)
2
11.What would be the length of a sec. pendulum at a planet
(where acc. due to gravity is g/4) if it’s length on earth is l
(a)l/2 (b) 2 l
(c)l/4 (d) 4 l
12.Time period of a simple pendulum inside a satellite orbiting
earth is
(a) zero (b)
¥
(c)T (d) 2 T
13.The ratio o
f the radii of the planets R
1
and R
2
is k. The ratio
of the acceleration due to gravity is r. The ratio of the escape
velocities from them will be
(a) k r (b)
kr
(c)(k /r) (d)(r/k)
14.Ife
v and
0
v represent the escape velocity and orbital
velocity of a satellite corresponding to a circular orbit of
radius R, then
(a)
eo
vv= (b)
eo
v 2v=
(c)
eo
v (1/ 2)v= (d) v
e
and v
o
are not r elated
15.The kinetic energy needed to project a body of mass m from
the earth surface (radius R) to infinity is
(a) mgR/2 (b) 2mgR
(c) mgR (d) mgR/4.

227. 223Gravitation
16.The esca
pe velocity of a body depends upon mass as
(a)m
0
(b) m
1
(c)m
2
(d) m
3
.
17.The radius of a planet is 1/4
th
of R
e
and its acc. due to
gravity is 2g. What would be the value of escape velocity
on the planet, if escape velocity on earth is v
e
.
(a)
e
v
2
(b)
e
v2
(c) 2 v
e
(d)
ev
2
18.If the gravitational force had varied as r
–5/2
instead of r
–2
;
the potential energy of a particle at a distance ‘r’ from thecentre of the earth would be proportional to
(a)
1
r
-
(b)
2
r
-
(c)
3/2
r
-
(d)
5/2
r
-
19.Two sate
llites revolve round the earth with orbital radii 4R
and 16R, if the time period of first satellite is T then that of
the other is
(a) 4 T (b) 4
2/3
T
(c) 8 T (d) None of these
20.A planet revolves in an elliptical orbit around the sun. The
semi-major and semi-minor axes are a and b. Then the square
of time period, T is directly proportional to
(a)
3
a (b)
3
b
(c)
3
2
ba
÷
ø
ö
ç
è
æ+
(d)
3
2
ba
÷
ø
ö
ç
è
æ-
21.Which o
f the following quantities do not depend upon the
orbital radius of the satellite ?
(a)
R
T
(b)
R
T
2
(b)
2
2
R
T
(d)
3
2
R
T
22.The orbital
velocity of an artificial satellite in a circular orbit
very close to Earth is v. The velocity of a geosynchronous
satellite orbiting in a circular orbit at an altitude of 6R from
Earth's surface will be
(a)
7
v
(b)
6
v
(c)v (d)v6
23.Escape velo
city when a body of mass m is thrown vertically
from the surface of the earth is v, what will be the escape
velocity of another body of mass 4 m is thrown vertically
(a)v (b) 2v
(c) 4v (d) None of these
24.The potential energy of a satellite of mass m and revolving
at a height R
e
above the surface of earth where R
e
= radius
of earth, is
(a) – m g R
e
(b)
2
Rgm
e-
(c)
3
Rgm
e-
(d)
4
Rgm
e
-
25.Energ y required to move a body of mass m from an orbit of
radius 2R to 3R is
(a) GMm/12R
2
(b) GMm/3R
2
(c) GMm/8R (d) GMm/6R
1.The escape velocity from the earth's surface is 11 km/s. The
escape velocity from a planet having twice the radius and
same mean density as that of earth is
(a) 5.5 km/s (b) 11 km/s
(c) 22 km/s (d) None of these
2.Two point masses each equal to 1 kg attract one another
with a force of 10
–10
N. The distance between the two
point masses is (G = 6.6 × 10
–11
MKS units)
(a) 8 cm (b) 0.8 cm
(c) 80 cm (d) 0.08 cm
3.There are two bodies of masses 10
3
kg and 10
5
kg separated
by a distance of 1 km. At what distance from the smaller
body, the intensity of gravitational field will be zero
(a) 1/9 km (b) 1/10 km
(c) 1/11 km (d) 10/11 km
4.Taking the gravitational potential at a point infinte distance
away as zero, the gravitational potential at a point A is –5
unit. If the gravitational potential at point infinite distance
away is taken as + 10 units, the potential at point A is
(a) – 5 unit (b) + 5 unit
(c) + 10 unit (d) + 15 unit
5.A planet of mass 3 × 10
29
gm moves around a star with a
constant speed of 2 × 10
6
ms
–1
in a circle of radii 1.5 × 10
14
cm. The gravitational force exerted on the planet by the star is
(a) 6.67 × 10
22
dyne (d) 8 × 10
27
Newton
(c) 8 × 10
26
N (d) 6.67 × 10
19
dyne
6.The mass of the moon is 1/81 of earth’s mass and its radius
1/4 that of the earth. If the escape velocity from the earth’s
surface is 11.2 km/sec, its value from the surface of the
moon will be
(a) 0.14 kms
–1
(b) 0.5 kms
–1
(c) 2.5 kms
–1
(d) 5.0 kms
–1
7.If the mass of earth is eighty times the mass of a planet and
diameter of the planet is one fourth that of earth, then
acceleration due to gravity on the planet would be
(a) 7.8 m/s
2
(b) 9.8 m/s
2
(c) 6.8 m/s
2
(d) 2.0 m/s
2

228. 224 PHYSICS
8.Two bo
dies of masses 10 kg and 100 kg are separated by a
distance of 2m ( G = 6.67 × 10
–11
Nm
2
kg
–2
). The
gravitational potential at the mid point on the line joining
the two is
(a) 7.3 × 10
–7
J/kg (b) 7.3 × 10
–9
J/kg
(c) –7.3 × 10
–9
J/kg (d) 7.3 × 10
–6
J/kg
9.The time period of a satellite of earth is 5 hours. If the
separation between the earth and the satellite is increased
to 4 times the previous value, the new time period will become
(a) 10 hours (b) 80 hours
(c) 40 hours (d) 20 hours
10.At sea level, a body will have minimum weight at
(a) pole (b) equator
(c) 42° south latitude(d) 37° north latitude
11.The distance of neptune and saturn from the sun is nearly
10
13
and 10
12
meter respectively. Assuming that they move
in circular orbits, their periodic times will be in the ratio
(a) 10 (b) 100
(c)
10 10 (d) 1000
12.A geostatio
nary satellite is orbiting the earth at a height of
5R above that surface of the earth, R being the radius of the
earth. The time period of another satellite in hours at a
height of 2R from the surface of the earth is
(a)5 (b) 10
(c)
62 (d)
6
2
13.The escape velocity for a body projected vertically upwards
from the surface of earth is 11 km/s. If the body is projectedat an angle of 45º with the vertical, the escape velocity will be
(a) 22 km/s (b) 11 km/s
(c)
2
11
km/s (d) 211 km/s
14.If the length of a simple p
endulum is increased by 2%, then
the time period(a) increases by 2%(b) decreases by 2%
(c) increases by 1%(d) decreases by 1%
15.The kinetic energy of a satellite in its orbit around the earth
is E. What should be the kinetic energy of the satellite so as
to enable it to escape from the gravitational pull of the earth?
(a) 4 E (b) 2 E
(c)
E2 (d) E
16.The time period
of a satellite in a circular orbit of radius R is
T, the period of another satellite in a circular orbit of radius
4 R is
(a) 4 T (b) T/4
(c) 8 T (d) T/8
17.If the change in the value of g at the height h above the
surface of the earth is the same as at a depth ‘x’ below it,
then (both x and h being much smaller than the radius of the
earth)
(a) x = h (b) x = 2 h
(c) x = h/2 (d) x = h
2
18.A satellite of mass m revolves around the earth of radius R
at a height ‘x’ from its surface. If g is the acceleration due to
gravity on the surface of the earth, the orbital speed of the
satellite is
(a)
xR
gR
2
+
(b)
xR
gR
-
(c) gx (d)
2/1
xR
gR
2
÷
÷
ø
ö
ç
ç
è
æ
+
19.The time period of an
earth satellite in circular orbit is
independent of
(a) both the mass and radius of the orbit
(b) radius of its orbit
(c) the mass of the satellite
(d) neither the mass of the satellite nor the radius of its orbit.
20.Suppose the gravitational force varies inversely as the nth
power of distance. Then the time period of a planet in circular
orbit of radius ‘R’ around the sun will be proportional to
(a)R
n
(b)
÷
ø
ö
ç
è
æ-
2
1n
R
(c)
÷
ø
ö
ç
è
æ+
2
1n
R (d)
÷
ø
ö
ç
è
æ-
2
2n
R
21.An earth satellite of mass m revolves in a circular orbit at a
height h from the surface of the earth. R is the radius of the
earth and g is acceleration due to gravity at the surface of
the earth. The velocity of the satellite in the orbit is given
by
(a) g R
2
/(R + h)
(b) g R
(c) g R/(R – h)
(d)
2
gR / (R h)éù +
ëû
22.Mass of the Earth has been determined through
(a) use of Kepler's
3
2
R
T
constancy la
w and Moon's period
(b) sampling the density of Earth's crust and using Earth's
radius
(c) Cavendish's determination of G and using Earth radius
and g at its surface
(d) use of periods of satellites at different heights above
Earth's surface and known radius of Earth
23.Consider Earth to be a homogeneous sphere. Scientist A
goes deep down in a mine and scientist B goes high up in a
balloon. The gravitational field measured by
(a) A goes on decreasing and that by B goes on increasing
(b) B goes on decreasing and that by A goes on increasing
(c) each decreases at the same rate
(d) each decreases at different rates

229. 225Gravitation
24.Ther
e are _______ gravitational lines of force inside a
spherically symmetric shell.
(a) infinitely many
(b) zero
(c) varying number depending upon surface area
(d) varying number depending upon volume
25.A uniform ring of mass m and radius r is placed directly
above a uniform sphere of mass M and of equal radius. The
centre of the ring is directly above the centre of the sphere
at a distance
3r as shown in the figure.
The gravitational force exerted by the sphere on the ring
will be
(a)
2
r8
GMm
(b)
2
r4
GMm
r
3r 2r
(c)
2
r8
GMm
3
(d)
3r8
GMm
3
26.The gravitation
al potential difference between the surface
of a planet and a point 20 m above the surface is 2 joule/kg.
If the gravitational field is uniform, then the work done in
carrying a 5 kg body to a height of 4 m above the surface is
(a) 2 J (b) 20 J
(c) 40 J (d) 10 J
27.The ratio of the kinetic energy required to be given to a
satellite so that it escapes the gravitational field of Earth to
the kinetic energy required to put the satellite in a circular
orbit just above the free surface of Earth is
(a)1 (b) 2
(c)3 (d) 9
28.The radii of two planets are respectively R
1
and R
2
and
their densities are respectively r
1
and r
2
. The ratio of the
accelerations due to gravity at their surfaces is
(a)
2
2
2
2
1
1
21
R
:
R
g:g
rr
=
(b)
212121
:RRg:grr=
(c)
1221
21
R:Rg:grr=
(d)
221121
R:Rg:grr=
29.An artificial satellite moving in a circular orbit around the
earth has a total (K.E. + P.E.) is E
0
. Its potential energy is –
(a) –E
0
(b) 1.5 E
0
(c) 2E
0
(d) E
0
30.If value of acceleration due to gravity changes from one
place to another, which of the following forces will undergo
a change ?
(a) Viscous force (b) Buoyant force
(c) Magnetic force (d) All of the above
31.The amount of work done in lifting a mass ‘m’ from the
surface of the earth to a height 2R is
(a) 2mgR (b) 3mgR
(c)
2
3
mgR (d)
3 2
mgR
32.The radius o
f the earth is 4 times that of the moon and its
mass is 80 times that of the moon. If the acceleration due to
gravity on the surface of the earth is 10 m/s
2
, then on the
surface of the moon its value will be
(a) 1 ms
–2
(b) 2 ms
–2
(c) 3 ms
–2
(d) 4 ms
–2
33.A satellite of mass ‘m’, moving around the earth in a circular
orbit of radius R, has angular momentum L. The areal velocity
of satellite is (M
e
= mass of earth)
(a) L /2m (b) L /m
(c) 2L /m (d) 2L /M
e
34.A solid sphere of uniform density and radius R applies a
gravitational force of attraction equal to F
1
on a particle
placed at A, distance 2R from the centre of the sphere.
R R
A
A spherical cavity of radius R/2 is now made in the sphere
as shown in the figure. The sphere with cavity now appliesa gravitational force F
2
on the same particle placed at A.
The ratio F
2
/F
1
will be
(a) 1/2 (b) 3
(c)7 (d) 1/9
35.A body starts from rest from a point distance R
0
from the
centre of the earth. The velocity acquired by the body whenit reaches the surface of the earth will be (R representsradius of the earth).
(a)
÷
÷
ø
ö
ç
ç
è
æ
-
0
R
1
R
1
MG2 (b)
÷
÷
ø
ö
ç
ç
è
æ
-
R
1
R
1
MG2
0
(c)
÷
÷
ø
ö
ç
ç
è
æ
-
0R
1
R
1
MG (d)
÷
÷
ø
ö
ç
ç
è
æ
-
0
R
1
R
1
MG2
36.The larges
t and the shortest distance of the earth from the
sun are r
1
and r
2
. Its distance from the sun when it is at
perpendicular to the major-axis of the orbit drawn from the sun
(a)
12
(r r )/4+ (b)
12 12
(r r)/(r r)+-
(c)
1212
2rr /(r r)+ (d)
12
(r r )/3+

230. 226 PHYSICS
37.The orbital
velocity of an artificial satellite in a circular orbit
just above the centre’s surface is u. For a satellite orbiting
at an altitude of half of the earth’s radius, the orbital velocity is
(a)
0v
3
2
÷
÷
ø
ö
ç
ç
è
æ
÷
ø
ö
ç
è
æ
(b)
0
v
3
2
(c)
0
v
2
3
(d)
0
v
2 3
÷
ø
ö
ç
è
æ
38.A planet is revolvin
g around the sun in an elliptical orbit.
Its closests distance from the sun is r
min
. The farthest
distance from the sun is r
max
. If the orbital angular velocity
of the planet when it is nearest to the sun is w, then the
orbital angular velocity at the point when it is at the farthest
distance from the sun is
(a)
w)r/r(
maxmin
(b) w)r/r(
minmax
(c) w)r/r(
2
min
2
m
ax
(d) w)r/r(
2
max
2
min
39.Two spherical bodies of mass M and 5M and radii R and 2R
respectively are released in free space with initial separation
between their centres equal to 12 R. If they attract each
other due to gravitational force only, then the distance
covered by the smaller body just before collision is
(a) 2.5 R (b) 4.5 R
(c) 7.5 R (d) 1.5 R
40.If the radius of the earth were to shrink by one per cent, its
mass remaining the same, the acceleration due to gravity
on the earth’s surface would
(a) decrease (b) remain unchanged
(c) increase (d) None of these
41.If earth is supposed to be a sphere of radius R, if
30
g is
value of acceleration due to gravity at lattitude of 30° and g
at the equator, the value of g – g
30
is
(a)
R
4
12
w (b) R
4
32
w
(c)R
2
w (d) R
2
12
w
42.A ball is dropped
from a satellite revolving around the earth
at height of 120 km. The ball will
(a) continue to move with same speed along a straight
line tangentially to the satellite at that time
(b) continue to move with same speed along the original
orbit of satellite.
(c) fall down to earth gradually
(d) go far away in space
43.Two identical geostationary satellites are moving with equal
speeds in the same orbit but their sense of rotation brings
them on a collision course. The debris will
(a) fall down
(b) move up
(c) begin to move from east to west in the same orbit
(d) begin to move from west to east in the same orbit
44.If there were a small gravitational effect, then which of the
following forces will undergo a change?
(a) Viscous force (b) Electrostatic force
(c) Magnetic force (d) Archimedes' uplift
45.The gravitational force of attraction between a uniform
sphere of mass M and a uniform rod of length l and mass m
oriented as shown is
r l
m
M
(a)
)r(r
GMm
l+
(b)
2
r
GM
(c) l+
2
Mmr (d) mM)r(
2
l+
46.Explor
er 38, a radio-astronomy satellite of mass
200 kg, circles the Earth in an orbit of average radius
2
R3
where R is the
radius of the Earth. Assuming the gravitational
pull on a mass of 1 kg at the earth's surface to be 10 N,calculate the pull on the satellite(a) 889 N (b) 89 N
(c) 8889 N (d) 8.9 N
47.Suppose, the acceleration due to gravity at the Earth'ssurface is 10 m s
–2
and at the surface of Mars it is
4.0 m s
–2
. A 60 kg pasenger goes from the Earth to the Mars
in a spaceship moving with a constant velocity. Neglect allother objects in the sky. Which part of figure best representsthe weight (net gravitational force) of the passenger as afunction of time?
t
0
time
D
Weight
600 N
200 N
A
B
C
(a)A (b) B
(c)C (d) D
48.A proj
ectile is fired vertically from the Earth with a velocity
kv
e
where v
e
is the escape velocity and k is a constant less
than unity. The maximum height to which projectile rises,
as measured from the centre of Earth, is
(a)
k
R
(b)
1k
R
-
(c)
2
k1
R
-
(d)
2
k1
R
+

231. 227Gravitation
49.A diametr
ical tunnel is dug across the Earth. A ball is
dropped into the tunnel from one side. The velocity of the
ball when it reaches the centre of the Earth is .... (Given :
gravitational potential at the centre of Earth =
R
GM
2
3
– )
(a)
R (b)gR
(c) gR5.2 (d) gR1.7
50.A man of mass
m starts falling towards a planet of mass M
and radius R. As he reaches near to the surface, he realizes
that he will pass through a small hole in the planet. As he
enters the hole, he sees that the planet is really made of two
pieces a spherical shell of negligible thickness of mass 2M/
3 and a point mass M/3 at the centre. Change in the force of
gravity experienced by the man is
(a)
2
2 GMm
3R
(b) 0
(c)
2
1 GMm
3R
(d)
2
4 GMm
3R
51.In a region of only gravitational field of mass 'M' a particle
is shifted from A to B via three different paths in the figure.
The work done in different paths are W
1
, W
2
, W
3
respectively then
M
B
A
3
2
1
C
(a)W
1
= W
2
= W
3
(b) W
1
> W
2
> W
3
(c
)W
1
= W
2
> W
3
(d) W
1
< W
2
< W
3
52.Four similar particles of mass m are orbiting in a circle ofradius r in the same angular direction because of their mutualgravitational attractive force. Velocity of a particle is given by
m
m
m
m
r
(a)
1/2
GM1 22
r4
éùæö+
êúç÷
èøêúëû
(b)
3
GM
r
(c)()
GM
1 22
r
+ (d)
1/2
1GM12
2r2
éù æö+
êú ç÷
èøêúëû
53.The percentage change in the acceleration of the earth
towards the sun from a total eclipse of the sun to the point
where the moon is on a side of earth directly opposite to
the sun is
(a)
s2
m1
Mr
100
Mr
´
(b)
2
s 2
m1
Mr
100
Mr
æö
´
ç÷
èø
(c)
2
1m
2s
rM
2 100
rM
æö
´
ç÷ èø
(d)
2
s1
2m
Mr
100
rM
æö
´
ç÷ èø
54.A satellite i
s revolving round the earth in an orbit of radius
r with time period T. If the satellite is revolving round the
earth in an orbit of radius r + D r (Dr << r) with time period T
+ D T then,
(a)
T 3r
T 2r
DD
= (b)
T 2r
T 3r
DD
=
(c)
Tr
Tr
DD
= (d)
Tr
Tr
DD
=-
55.A cavi
ty of radius R/2 is made inside a solid sphere of
radius R. The centre of the cavity is located at a distance R/
2 from the centre of the sphere. The gravitational force on a
particle of mass ‘m’ at a distance R/2 from the centre of the
sphere on the line joining both the centres of sphere and
cavity is – (opposite to the centre of gravity)
[Here g = GM/R², where M is the mass of the sphere]
(a)
mg
2
(b)
3mg
8
(c)
mg
16
(d) None of these
56.A spherical uniform planet is rotating about its axis. The
velocity of a point on its equator is V. Due to the rotation of
planet about its axis the acceleration due to gravity g at
equator is 1/2 of g at poles. The escape velocity of a particle
on the pole of planet in terms of V is
(a)V
e
= 2V (b) V
e
= V
(c)V
e
= V/2 (d)
e
V 3V=
57.The esc
ape velocity from a planet is v
e
. A tunnel is dug
along a diameter of the planet and a small body is dropped
into it at the surface. When the body reaches the centre of
the planet, its speed will be
(a)v
e
(b)
e
v/2
(c)v
e
/2 (d) zero

232. 228 PHYSICS
58.A (nonro
tating) star collapses onto itself from an initial
radius R
i
with its mass remaining unchanged. Which curve
in figure best gives the gravitational acceleration a
g
on the
surface of the star as a function of the radius of the star
during the collapse
R
i
a
g
a
b
d
R
c
(a)a (b) b
(c)c (d) d
59.The ear
th is assumed to be sphere of radius R. A platform is
arranged at a height R from the surface of Earth. The escapevelocity of a body from this platform is kv, where v is itsescape velocity from the surface of the earth. The value ofk is
(a)
1
2
(b)
1
3
(c)
1
2
(d)2
60.Four equal masses (each of mass M) are placed at the corners
of a square of side a. The escape velocity of a body from
the centre O of the square is
(a)
a
GM2
4 (b)
a
GM28
(c)
a
GM4
(d)
a
GM24
61.A planet o
f mass m moves around the sun of mass M in an
elliptical orbit. The maximum and minimum distance of the
planet from the sun are r
1
and r
2
respectively. The time
period of planet is proportional to
(a)
5/2
1r (b)
2/3
21
rr÷
ø
ö
ç
è
æ
+
(c)
2/3
21
r–r÷
ø
ö
ç
è
æ
(d)
2/3
r
62.The change in potential energy, when a body of mass m is
raised to a height nR from the earth’s surface is (R = radius
of earth)
(a)
÷
÷
ø
ö
ç
ç
è
æ
1–n
n
mgR (b) nmgR
(c)
÷
÷
ø
ö
ç
ç
è
æ
+1n
n
mgR
2
2
(d) ÷
ø
ö
ç
è
æ
+1n
n
mgR
63.A satellite is launche
d into a circular orbit of radius R around
the earth. A second satellite is launched into an orbit of
radius 1.01 R. The period of second satellite is larger than
the first one by approximately
(a) 0.5% (b) 1.0%
(c) 1.5% (d) 3.0%
64.If g
E
and g
M
are the accelerations due to gravity on the
surfaces of the earth and the moon respectively and if
Millikan’s oil drop experiment could be performed on the
two surfaces, one will find the ratio
electronic chargeon the moon
to be
electronic charge onthe earth
(a)
ME
g /g (b) 1
(c)0 (d)
EM
g /g
65.The density of a newly discovered planet is twice that of
earth. The acceleration due to gravity at the surface of theplanet is equal to that at the surface of the earth. If theradius of the earth is R, the radius of the planet would be
(a) ½ R (b) 2 R
(c) 4 R (d) 1/4 R
66.Imagine a new planet having the same density as that of
earth but it is 3 times bigger than the earth in size. If the
acceleration due to gravity on the surface of earth is g and
that on the surface of the new planet is g’, then
(a)g’ = g/9 (b) g’ = 27g
(c)g’=9g (d) g’=3g
67.For a satellite moving in an orbit around the earth, the ratio
of kinetic energy to potential energy is
(a)
1
2
(b)
1
2
(c)2 (d)2
68.The figure shows elliptical orbit of a planet m about the
sun S. The shaded area SCD is twice the shaded area SAB.
If t
1
is the time for the planet to move from C to D and t
2
is
the time to move from A to B then :
A
B
m
v
S
C
D
(a)t
1
= 4t
2
(b) t
1
= 2t
2
(c)t
1
= t
2
(d) t
1
> t
2
69.The
radii of circular orbits of two satellites A and B of the
earth, are 4R and R, respectively. If the speed of satellite Ais 3 V, then the speed of satellite B will be
(a) 3 V/4 (b) 6 V
(c) 12 V (d) 3 V/2

233. 229Gravitation
70.A particle of
mass M is situated at the centre of a spherical
shell of same mass and radius a. The gravitational potential
at a point situated at
2
a
distance from the centre, will be
(a)
3GM
a
- (b)
2GM
a
-
(c)
GM
a
- (d)
4GM
a
-
71.A planet moving along an
elliptical orbit is closest to the
sun at a distance r
1
and farthest away at a distance of r
2
. If
v
1
and v
2
are the linear velocities at these points
respectively, then the ratio
1
2
v
v
is
(a) (r
1
/r
2
)
2
(b) r
2
/r
1
(c
) (r
2
/r
1
)
2
(d) r
1
/r
2
72.A particle of mass m is thrown upwards from the surface of
the earth, with a velocity u. The mass and the radius of the
earth are, respectively, M and R. G is gravitational constant
and g is acceleration due to gravity on the surface of the
earth. The minimum value of u so that the particle does not
return back to earth, is
(a)
2GM
R
(b)
2
2GM
R
(c)
2
2gR (d)
2
2GM
R
73.Which one of the following graphs represents correctly
the variation of the gravitational field intensity (I) with the
distance (r) from the centre of a spherical shell of mass M
and radius a ?
(a)
I
r = ar
(b)I
r = ar
(c)
I
r = ar
(d)I
r = ar
Directions for Qs. (74 to 75) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer(ONLY ONE option is correct ) from the following
(a)Statement -1 is false, Statement-2 is true
(b)Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c)Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d)Statement -1 is true, Statement-2 is false
74. Statement -1 : For the planets orbiting around the sun,
angular speed, linear speed and K.E. changes with time,
but angular momentum remains constant.
Statement -2 : No torque is acting on the rotating planet.
So its angular momentum is constant.
75. Statement -1 : Gravitational potential is maximum at infinity.
Statement -2 : Gravitational potential is the amount of work
done to shift a unit mass from infinity to a given point in
gravitational attraction force field.
Exemplar Questions
1.The earth is an approximate sphere. If the interior contained
matter which is not of the same density everywhere, thenon the surface of the earth, the acceleration due to gravity
(a) will be directed towards the centre but not the same
everywhere
(b) will have the same value everywhere but not directed
towards the centre
(c) will be same everywhere in magnitude directed towards
the centre
(d) cannot be zero at any point
2.As observed from the earth, the sun appears to move in an
approximate circular orbit. For the motion of another planetlike mercury as observed from the earth, this would
(a) be similarly true
(b) not be true because the force between the earth and
mercury is not inverse square law
(c) not be true because the major gravitational force on
mercury is due to the sun
(d) not be true because mercury is influenced by forces
other than gravitational force
3.Different points in the earth are at slightly different
distances from the sun and hence experience different
forces due to gravitation. For a rigid body, we know that if
various forces act at various points in it, the resultant motion
is as if a net force acts on the CM (centre of mass) causing
translation and a net torque at the CM causing rotation
around an axis through the CM. For the earth-sun system
(approximating the earth as a uniform density sphere).
(a) the torque is zero
(b) the torque causes the earth to spin
(c) the rigid body result is not applicable since the earth
is not even approximately a rigid body
(d) the torque causes the earth to move around the sun

234. 230 PHYSICS
4.Satellites o
rbitting the earth have finite life and sometimes
debris of satellites fall to the earth. This is because
(a) the solar cells and batteries in satellites run out
(b) the laws of gravitation predict a trajectory spiralling
inwards
(c) of viscous forces causing the speed of satellite and
hence height to gradually decrease
(d) of collisions with other satellites
5.Both the earth and the moon are subject to the gravitational
force of the sun. As observed from the sun, the orbit of the
moon
(a) will be elliptical
(b) will not be strictly elliptical because the total
gravitational force on it is not central
(c) is not elliptical but will necessarily be a closed curve
(d) deviates considerably from being elliptical due to
influence of planets other than the earth
6.In our solar system, the inter-planetary region has chunks
of matter (much smaller in size compared to planets) called
asteroids. They
(a) will not move around the sun, since they have very
small masses compared to the sun
(b) will move in an irregular way because of their small
masses and will drift away into outer space
(c) will move around the sun in closed orbits but not obey
Kepler's laws
(d) will move in orbits like planets and obey Kepler's laws
7.Choose the wrong option.
(a) Inertial mass is a measure of difficulty of accelerating
a body by an external force whereas the gravitational
mass is relevant in determining the gravitational force
on it by an external mass
(b) That the gravitational mass and inertial mass are equal
is an experimental result
(c) That the acceleration due to gravity on the earth is the
same for all bodies is due to the equality of gravitational
mass and inertial mass
(d) Gravitational mass of a particle like proton can depend
on the presence of neighbouring heavy objects but
the inertial mass cannot
8. Particles of masses 2M, m and M are respectively at points A,
B and C with
1
()
2
AB BC m=× is much-much smaller than
M and at time t = 0, they are all at rest as given in figure.
At subsequent times before any collision takes place.
rA B C
Mm2M
(a)m will remain at rest
(b)m w
ill move towards M
(c)m will move towards 2M
(d)m will have oscillatory motion
NEET/AIPMT (2013-2017) Questions
9.A body of mass ‘m’ is taken from the earth’s surface to the
height equal to twice the radius (R) of the earth. The change
in potential energy of body will be [2013]
(a)
2
3
mgR (b) 3 mgR
(c)
1
3
mgR (d) mg2R
10.Infinit
e number of bodies, each of mass 2 kg are situated on
x-axis at distances 1m, 2m, 4m, 8m, ..... respectively, from theorigin. The resulting gravitational potential due to thissystem at the origin will be [2013]
(a)
8
3
- G (b)
4
3
- G
(c) – 4 G (d) – G
1
1.The radius of a planet is twice the radius of earth. Bothhave almost equal average mass-densities. If V
P
and V
E
are
escape velocities of the planet and the earth, respectively,then [NEET Kar. 2013]
(a)V
E
= 1.5V
P
(b)V
P
= 1.5V
E
(c)V
P
= 2V
E
(d)V
E
= 3V
P
12.A particle of mass ‘m’ is kept at rest at a height 3R from the
surface of earth, where ‘R’ is radius of earth and ‘M’ is mass
of earth. The minimum speed with which it should beprojected, so that it does not return back, is (g is acceleration
due to gravity on the surface of earth) [NEET Kar. 2013]
(a)
1
2GM
R
æö
ç÷
èø
(b)
1
2
2
GM
R
æö
ç÷
èø
(c)
1
2
4
gRæö
ç÷
èø
(d)
1
22
4
gæö
ç÷
èø
13.Dependence of intensity of gravitational field (E) of earth
with distance (r) from centre of earth is correctly representedby: [2014]
(a)
E
O
R
r
(b)
E
O
Rr
(c)
E
O
R
r
(d)
E
O
Rr
14.A black hole is an object whose gravitational field is so
strong that even light cannot escape from it. To what
approximate radius would earth (mass = 5.98 × 10
24
kg)
have to be compressed to be a black hole? [2014]
(a) 10
– 9
m (b) 10
– 6
m
(c) 10
– 2
m (d) 100 m

235. 231Gravitation
15.Two sp
herical bodies of mass M and 5 M and radii R and 2
R released in free space with initial separation between their
centres equal to 12 R. If they attract each other due to
gravitational force only, then the distance covered by the
smaller body before collision is [2015]
(a) 4.5 R (b) 7.5 R
(c) 1.5 R (d) 2.5 R
16.Kepler's third law states that square of period of revolution
(T) of a planet around the sun, is proportional to third power
of average distance r between sun and planet i.e. T
2
= Kr
3
here K is constant. If the masses of sun and planet are M
and m respectively then as per Newton's law of gravitation
force of attraction between them is F =
2
GMm
,
r
here G is
gravitation
al constant. The relation between G and K is
described as [2015]
(a) GMK = 4p
2
(b) K = G
(c) K =
1
G
(d) GK = 4p
2
17.A re
mote - sensing satellite of earth revolves in a circular
orbit at a height of 0.25 × 10
6
m above the surface of earth.
If earth's radius is 6.38 × 10
6
m and g = 9.8 ms
–2
, then the
orbital speed of the satellite is: [2015 RS]
(a) 8.56 km s
–1
(b) 9.13 km s
–1
(c) 6.67 km s
–1
(d) 7.76 km s
–1
18.A satellite S is moving in an elliptical orbit around the earth.
The mass of the satellite is very small compared to the mass
of the earth. Then, [2015 RS]
(a) the total mechanical energy of S varies periodically
with time.
(b) the linear momentum of S remains constant in
magnitude.
(c) the acceleration of S is always directed towards the
centre of the earth.
(d) the angular momentum of S about the centre of the
earth changes in direction, but its magnitude remains
constant.
19.The ratio of escape velocity at earth (v
e
) to the escape
velocity at a planet (v
p
) whose radius and mean density are
twice as that of earth is :
(a) 1 : 2 (b) 1 : 2
2 [2016]
(c) 1 : 4 (d)
1 : 2
20.At what height from the surface of earth the gravitational
potential and the value of g are –5.4 × 10
7
J kg
–1
and 6.0 ms
–2
respectively ?
Take the radius of earth as 6400 km : [2016]
(a) 2600 km (b) 1600 km
(c) 1400 km (d) 2000 km
21.Two astronauts are floating in gravitation free space after
having lost contact with their spaceship. The two will
(a) move towards each other. [2017]
(b) move away from each other.
(c) become stationary
(d) keep floating at the same distance between them.
22.The acceleration due to gravity at a height 1 km above the
earth is the same as at a depth d below the surface of earth.
Then [2017]
(a) d = 1 km (b) d =
3
km
2
(c) d = 2 km (d) d =
1
km
2

236. 232 PHYSICS
EXERCISE - 1
1. (d) 2. (d) 3
. (a) 4. (d)
5. (b) 6. (c) 7. (b)
8. (c) In planetary motion
ext.
τ
r
= 0 Þ L
r
= constant
()=´
==
rrrr
L r p mv mrv (Q q = 90º)
So m
1
d
1
v
1
=

m
2
d
2
v
2
(here r = d)
Þ
11
2
2
vd
v
d
=
d
1 d
2 v
2v
1
sun
9. (a) At the surface of earth, the value of g = 9.8m/sec
2
. If
we go towards the centre of earth or we go above the
surface of earth, then in both the cases the value of g
decreases.
Hence W
1
=mg
mine
, W
2
=mg
sea level
, W
3
=mg
moun
So W
1
< W
2
> W
3
(g at the sea level = g at the
suface of earth)
10. (a) According to Gravitational Law
12
2
GMM
F
r
= …… (i)
Where M
1
is ma
ss of planet & m
2
is the mass of any
body. Now according to Newton’s second law, body ofmass m
2
feels gravitational acceleration g which is
F = m
2
g ……(ii)
So from (i) & (ii), we get
1
2
GM
g
r
=
So the ratio of
gravitational acceleration due to two
planets is
232
11212
2 23
22112
g M
r (4/3)πr ρ r
gMr r (4 / 3)
πrρ
´
=´=´
´
11
22
gr
gr
æö
=ç÷
èø
(both planet have same material, so
density is same)
11. (c) Since T(time period) =
l
2π
g
for second pendulum T=2sec
Now on the planet the value of acceleration due to gravity is
g/4.
So for the planet, the length of sec. pendulam l' is
2sec g/4
/4
2secg
l
l'l
l'
= ´ Þ=
12. (b) Si
nce
T 2π
g
l
=
but inside the satellite g = 0
S
o T = ¥
13. (b) We know that,
e
v (2gR)=
eP 111
eP 222
(v) (2gR)
(v) (2gR)
\=
11
22
gR
gR
æöæö
=´ç÷ç÷
èøèø
kr=
14. (b)
eo
v 2v= where v
e
and v
o
are the escape velocity
and orbital velocity respectively.
15. (c) KE = mgR)gR2(m
2
1
mv
2
1 2
esc
2
== .
16. (a) v
esc
= gR2, where R is radius of the planet.
Hence escape velocity is independent of m.
17. (a) The escape velocity on the earth is defined as
e ee
v 2gR=
Where R
e
& g
e
are the ra
dius & acceleration due to
gravity of earth.
Now for planet g
P
=2g
e
, R
P
=R
e
/4
So
e
P PP ee
v
v 2g R 2 2g R / 4
2
= =´´=
18. (c)
The potential energy for a conservative force is defined
as
dr
dU
F
-
= or
r
U F.dr
¥
=-ò
rr
…… (i)
or
r
12 12
r
2
GMM GMM
U dr
rr
¥
-
==ò
…… (ii)
(Q U
¥
= 0)
If we br
ing the mass from the infinity to the centre of
earth, then we obtain work, ‘so it has negative
(gravitational force do work on the object) sign &
potential energy decreases. But if we bring the mass
from the surface of earth to infinite, then we must do
work against gravitational force & potential energy of
the mass increases.
Now in equation (i) if
12
5/2
GMM
F
r
= instead o f
12
2
GMM
F
r
= then
r
12 12
r
5/2 3/2
GMM GMM 2
U dr
3rr
¥
-
==ò
r
3/2
1
U
r
+
Þµ
Hints & Solutions

237. 233Gravitation
19. (c
)
2/3
2
2/3
2
1
2
1
R16
R4
T
T
R
R
T
T
÷
ø
ö
ç
è
æ
=Þ
÷
÷
ø
ö
ç
ç
è
æ
=
T8T
2
=Þ
20. (a)2
Tµ
(major axis)
3
Þ T
2
a a
3
2
1. (d) According to 3
rd
law of Kepler
32
RTµ
32
KRT=Þ
where K i
s a constant
Thus
3
2
R
T
does not de
pends on radius.
22. (a)
12
11
v ,v
R 7R
µµ
R
7R
21
2
1
vv1v
v
v 7 77
=Þ==
23.
(a) Escape velocity does not depend upon the mass of
the body.
24. (b) At a height h above the surface of earth the
gravitational potential energy of the particle of mass
m is
e
h
e
GMm
U
Rh
=-
+
Where M
e
& R
e
are th e mass & radius of earth
respectively.
In this question, since h = R
e
So
22
=
-
=-=
e
ee
hR
e
GM m mgR
U
R
25. (d) E
nergy required to move a body of mass m from an
orbit of radius 2R to 3R is
GMm GMm G Mm
U
3R 2R 2R
-
D=--=
GMm GMm
3R 6R
-=
EXERCISE - 2
1. (
c)
2. (c) According to Newton’s Gravitation Law
12
g
2
GMM
F
r
= here F
g
= 10
–10
New ton,
m
1
= m
2
= 1kg,
G = 6.6 × 10
–11
So
11
2 12
10
g
GMM 6.61 0 11
r 0.66
F 10
-
-
´ ´´
===
or r = 0.81
25 metre = 81.25 cm
»80cm
3. (c)
2
5
2
3
)r1(
10G
)r(
10G
-
´
=
´
2
22
1 10
r (1 r)
=
-
r1r10
r1
10
r
1
-=Þ
-
=
\
1
11
=r km
4. (b)
The gravitational potential V at a point distant ‘r’ from
a body of mass m is equal to the amount of work done
in moving a unit mass from infinity to that point.
( )
r
r
V V E.d r GM 1/r 1/
µ
µ
- =- =- -µò
urr
GM
r
-
=
dV
AsE
dr
-æö
=ç÷
èø
r
(i) In the first case
when V
µ
= 0,
r
GM
V5
r
-
= =- unit
(ii) I
n the second case
V
µ
= + 10 unit
V
r
– 10 =
– 5
or V
r
= + 5 unit
5. (c) Gravitational force supplies centripetal force
\
2 29 82
14
mv 3
10 (2 10 )F dynes
r 1.5 10
´ ´´
==
´
= 8 × 1
0
31
dyne = 8 × 10
26
N (Q 1N = 10
5
dyne.)
6. (c)
e
e
eme
ee
M
2G
2GM 281
v; vv
RR9
4
= ==
= 2/9 × 11.2
kms
–1
= 2.5 kms
–1
7. (d) Since gravitational acceleration on earth is defined as
2
e
e
e
R
GM
g= …… (i)
mass o
f planet is
e
P
M
M
80
= & radius
e
P
R
R
4
=
So
P
P
2
P
GM
g
R
= …… (ii)
Fro
m (i) & (ii), we get
2
eeP
Pe
2
eP
RgM
gg
M5R
= ´= =2m/se
c
2
(as g=10m/sec
2
)
8. (c)
1
1001067.6
1
101067.6–
V
1111
g
´´
-
´´
=
--
= –6.67 × 10
–10
– 6. 67 × 10
–9
= – 6.67 × 10
–10
× 11 = –7.3 × 10
–9
J/kg

238. 234 PHYSICS
9. (c) 23
TRµ
Þ
3
2
1
2
2
1
R
R
T
T
÷
÷
ø
ö
ç
ç
è
æ
=
÷
÷
ø
ö
ç
ç
è
æ
Þ
2/32/3
2
1
2
1
4
1
R
R
T
T
÷
ø
ö
ç
è
æ
=
÷
÷
ø
ö
ç
ç
è
æ
=
÷
÷
ø
ö
ç
ç
è
æ
Þ 8)4(
T
T 2/3
1
2
==
Þ21
T 8 T 8 5 40=´ =´= hours
10. (b) At p
oles, the effect of rotation is zero and also the
distance from the centre of earth is least.
11. (c)
23
TRµ
(According to Kepler’s law)
2133 2123
12
T (10 ) and T (10 )µµ
2
311
2
22
TT
(10) or 10 10
TT
\==
.
12. (c) Accor
ding to Kepler’s law of period T
2
µ R
323 3
11
233
22
(6)
8
(3)
TR R
TRR
===
2
2
24 24
8
T
´
=
2
2
24 24
8
T
´
= = 72 = 36 × 2
2
62T=
13. (b) Since escap
e velocity
( )ee
v 2gR= is independe nt
of angle of projection, so it will not change.
14. (c) Given that l
2
= 1.02 l
1
We know that T 2π orT
g
æö
=µç÷
èø
l
l
221
111
T (1.02 )
101
T
æö
\===
ç÷
èø
ll
ll
Thus time period incr
eases by 1%.
15. (b) We know that
e0
v 2v= , where v
0
is orbi tal
velocity.
K.E. in the orbit,
2
0
1
E Mv
2
=
K.E. to escape
22
e0
11
E Mv M(2v)
22
==

2
0
1
Mv 2 2E
2
= ´=
16. (c)
1/2
3
2π(R h)
T
Rg
éù+
=êú
êúëû
Here (R + h) changes from R to 4 R. Hence period of
revolution changes from T to (4
3
)
1/2
T = 8 T.
17. (b)h
2h
g g1
R
æö
=-
ç÷
èø
x
x
g g1
R
æö
=-
ç÷
èø
Given that,
hx
gg=
\ x = 2 h
18. (d)
2
2
)xR(
GmM
)xR(
mv
+
=
+
also
2
R
GM
g=
22
22
mv GMR
m
(R x) R (R x)
æö
\=
ç÷
èø+ +
2
22
)xR(
R
mg
)xR(
mv
+
=
+
\
xR
gR
v
2
2
+
=\ Þ
2/1
2
xR
gR
v
÷
÷
ø
ö
ç
ç
è
æ
+
=
19. (c)
2
2
)xR(
GmM
xR
mv
+
=
+
x = height of satellite f
rom earth surface
m = mass of satellite
2GM GM
v orv
(Rx) Rx
Þ==
++
2(Rx) 2(Rx)
T
v GM
Rx
p+ p+
==
+
which is independen
t of mass of satellite
20. (c)
2n
MRKRF w==
-
)1n(2
KR
+-
=wÞ
or
2
)1n(
R'K
+-
=w [where K' = K
1/2
, a constant]
2
)1n(
R
T
2
+-
a
p
( n 1)
2
TR
+
\a
21. (d) Accordi
ng to Newton gravitational force
mg
R
mGM
2
e
e
= &
gm
)hR(
mGM
2
e
e
¢=
+
Þg'
2
e
2
e
2
e )hR(
gR
)R/h1(
1
g
+
=
+
=

239. 235Gravitation
wher
e M
e
is mass of earth, G is gravitational constant,
R
e
is radius & earth, h is height of satellite above the
surface of earth, g is value at the surface of earth, g' is
value at height h above the surface of earth.
so
gm
)hR(
mgR
)hR(
mv
2
e
2
e
e
2
¢=
+
=
+

hR
gR
v
e
2 e
+
=Þ
h
R
e
+R
e
O
Earth
earth
surface
satellite
orbit
v
22. (c)
23
. (d) Both decreases but variation are different.
24. (b) There is no gravitational field in the shell.
25. (c) The gravitational field due to the ring at a distance
r3
is given
by
22/322
r8
Gm3
E
])r3(r[
)r3(Gm
E =Þ
+
=
Attractive
force =
2
r8
GmM3
26. (a)
For uniform gravitational field
E
g
=
r
V
-
2
ms
10
1
20
2 -
=
-
-=
Now, W
= mgh =
1
5 4 2J
10
´ ´=
27. (b)
2
02
2
e1
mv
2
1
E,mv
2
1
E ==
0e
2
0
e
2
1
v2v2
v
v
E
E
==
÷
÷
ø
ö
ç
ç
è
æ
=\ Q
28. (d)gRµr
29. (c)
30
. (b) As buoyant force involves ‘g’ in it.
31. (d) Work done = U
f
– U
i
mgR
3
2
R
GmM
3
2
R
Gmm
RR
2
GMm
==÷
ø
ö
ç
è
æ
--
+
-=
32. (b) 33. (a)
34. (d) The gravitational force due to the whole sphere at A
point is
eo
1
2
GMm
F
(2R)
= , wher e m
0
is the assumed rest mass at
point A.
In the second case, when we made a cavity of radius
(R/2), then gravitational force at point A is
2
oe
2
)2/RR(
mGM
F
+
=
\ F
2
/F
1
= 1/9
35.
(b)
R
2R0
0
GMm 1 1
P.E. dr GMm
RRr
éù
= =-- êú
ëû
ò
The K.E. ac
quired by the body at the surface
21
mv
2
=
2
0
1 11
mv GMm
2 RR
éù
\=-- êú
ëû
0
11
v 2GM
RR
æö
=- ç÷
èø
36. (c) Applying the properties of ellipse, we have
12
1 2 12
rr211
R r r rr
+
=+=
R
Instant position
of satellite
major
axis
r
2
r
1
Sun
12
12
2rr
R
rr
=
+
37. (a)
GM
v
r
æö
=
ç÷
èø
where r is radius of the orbit of the
satellite.
here r = R
e
+ h =
e
ee
R 3
RR
22
+=
So, 0
e
2GM2
vv
3R3
==
,
where v
0
is the o
rbital velocity of the satellite, which
is moving in circular orbit of radius, r = R
e
38. (d)
22
11 22 1 122
vr vr o rrω rω==
(QL=mrv = constant)
or
22
min max
rωrω ¢=
22
min max
ω(r /r ) ω¢\=
39. (c)
Distance between the surface of the spherical bodies
= 12R – R – 2R = 9RForce Massµ
AccelerationMassµ
Distance Accelerationµ
Þ
1
2
a M1
a 5M5
== Þ
12
2
1
S5S
5
1
S
S
=Þ=

240. 236 PHYSICS
S
1
+ S
2
= 9
Þ 6S
1
= 9Þ 5.1
6
9
S
1
==
S
2
= 1.5 × 5 = 7
.5
Note: Maximum distance will be travelled by smaller
bodies due to the greater acceleration caused by the
same gravitational force
40. (c)
22
)R99.0(
GM
'g;
R
GM
g ==
2
2
g'R
g'g
g0.99R
æö
\= Þ>
ç÷
èø
41. (b) Ac
celeration due to gravity at lattitude’
l’ is given by
lw-=
l
2
2
ee
cosRgg
At equator,
l = 90°Þ lcos = cos 90° = 0
or
l
g =
e
g = g (as given in question)
At 30°,
222
30 R
4
3
g30cosRgg w-=w-=
or,
2
30R
4
3
gg w=-
42. (b) The orb
ital speed of satellite is independent of mass
of satellite, so ball will behave as a satellite & will
continue to move with same speed in original orbit.
43. (a) The total momentum will be zero and hence velocity
will be zero just after collisiion. The pull of earth will
make it fall down.
44. (d) Weight of liquid displaced = mg.
45. (a) The force of attraction between sphere and shaded
position
2
x
l
÷
ø
ö
ç
è
æ
=
dx
m
GMdF
x
r l
mM
r + l
22
21
2
1
1
21
1
()
++
+
+ -+
-
+
+
-
==
éù
==
êú
-+
ëû
éù
éù=- = -=
ëû êú
+ëû
òò
ò
rl rl
rr
rl
rl
r r
rl
rl
r
r
GMm GMm
F dx dx
lx lx
G
Mm GMm x
x dx
ll
GMm GMm GMm
x
l l x rrl
46. (a)
243
1 since
2
æöæö
= - = =+ç÷ç÷
èøèø
hgR
g g hR
h Rh
Force on th
e satellite =
mg
9
4
mg
h
=
N88910200
9
4
»´´=
47. (c)
2
R
1
gµ so we will not g
et a straight line.
Also F = 0 at a point where Force due to Earth = Force
due to mars
48. (c) Applying conservation of energy principle, we get
r
GMm
R
GMm
vmk
2
1 2
e
2
-=-
r
GMm
R
GMm
R
GM2
mk
2
12
-=-Þ
R
k
R
1
r
1
r
1
R
1
R
k
22
-=Þ-=-Þ
2
2
k1
R
r)k1(
R
1
r
1
-
=Þ-=Þ
49. (b) Lo
ss in potential energy = Gain in kinetic energy
2
mv
2
1
R
GMm
2
3
R
GMm
=÷
ø
ö
ç
è
æ
---
gR
R
GM
vmv
2
1
R2
GMm 2
==Þ=Þ
50. (a) Cha
nge in force of gravity
=
22
M
Gm
GMm 3
RR
- (only due
to mass M/3 due to shell
gravitational field is zero (inside the shell))
= 2
2GMm
3R
51. (a)
52. (a
) Centripetal force = net gravitational force
2
0
1
mv
2Fcos45F
r
= °+ =
22
22
2GM 1 Gm
2( 2r) 4r
+
r
V
0
F
F
F
1
2r
45°
2 2
0
2
mvGm
[2 2 1]
r 4r
=+
1/2
GM(221
4r
æö +
Þ
ç÷
èø

241. 237Gravitation
53.
(c) During total eclipse :
Total attraction due to sun and moon, se me
1
22
12
GM M GM M
F
rr
=+
When moon g
oes on the opposite side of earth.
Effective force of attraction,
se me
2
22
12
GM M GM M
F
rr
=-
Change in fo
rce,
me
12
2
2
2GMM
FFF
r
D=-=
Ch
ange in acceleration of earth
Da =
m
2
e 2
2GMF
M r
D
=
Average for
ce on earth, F
av
=
avs
2
e 1
F GM
M r
=
%age cha
nge in acceleration
=
2
m1
2
avs 2
2GMra
100 100
a GMr
D
´= ´´
2
1m
2s
rM
2 100
rM
æö
=´ç÷
èø
54. (a)Since,
32
krT=
Differentiating the above equation
Þ
Tr
23
Tr
DD
= Þ
T 3r
T 2r
DD
=
55. (b
) Gravitational field at mass m due to full solid sphere
10
00
rR1
E ........
3 6 4G
rr éù
= = e=
êú
eep ëû
r
r
Gravitational field at mass m due to cavity (–r)
33
2
22
00
()(R/2)ra
E ........
using E3R 3r
éù-rr
== êú
ee êúëû
r
r
=
3
2
00
()RR
2424R
-r -r
- =-
ee
R
m
R/2
CM
Net gravitational field
12
00
RR
EEE
6 24
rr
=+=-
ee
rrr

0
R
8
r
=
e
Net force on m
0
mR
F mE
8
r
®==
e
r
Here,
0
3
M1
&
4G(4/3)R
r
= e=
pp
then
3mg
F
8
=
56. (a) V
= wR
2
0
ggR= -w [g = at equato r, g
0
= at poles]
20
0
g
gR
2
= -w ;
2 0g
R
2
w= ;
2 0
gR
V
2
=
2
e0
V 2g R 4 V 2V= ==
57. (b)
2
e
1 GMm
mv
2R
= ;
e
GMm
v 2gR
R
==
In tunnel
body will perform SHM at centre
V
max
= Aw (see chapter on SHM)
=
e
vR2
gR
2R/g2
p
==
p
58. (b)
2
1
g
R
µ
R decreasin
g g increase hence, curve b represents
correct variation.
59. (a) Here,
2GM
v
R
= and
2GM
kv
RR
=
+
.
Solving
1
k
2
=
60. (b
) Potential energy of particle at the centre of square
GMm
4
a
2
æö
ç÷
=-ç÷
ç÷
èø
\
22GMm 1 8 2 GM
4 mv0v
a 2a
2
æö
ç÷
- + =Þ=
ç÷
èø
61. (b) T
2
a r
3
, where r = mean radius =
2
rr
21+
62. (d) ÷ ø
ö
ç
è
æ
--
+
-=-=D
R
GMm
hnR
GMm
UUU
if
mgR
1n
n
R
GMm
.
1n
n
+
=
+
=
63. (c)
GM
)hR(
2T
3
+
p=
GM
)R01.1(
2T,
GM
R
2T
3
2
3
1 p=p=
%5.1100
T
TT
1
12
=´
-
64. (b)Electronic charge does not depend on acceleration
due to gravity as it is a universal constant.
So, electronic charge on earth
= electronic charge on moon
\ Required ratio = 1.

242. 238 PHYSICS
65. (a)
2
R
MG
g= also
3
R
3
4
dMp´=
Rd
3
4
g p=\ at the surface o f planet
R)d2(
3
4
g
p
¢p=, R)d(
3
4
g
e p=
g
e
= g
p
Þ dR = 2d R'
Þ R' = R/2
6
6. (d) We know that
g =
GM
R
2 =
2
3
R
R
3
4
G r÷
ø
ö
ç
è
æ
p
=
4
3
prGR
g
g
R
R
R
R
''
= = =
3
3\=g g'3
67. (a) K.
E. of satellite moving in an orbit around the earth is
K =
r2
GMm
r
GM
m
2
1
mv
2
1
2
2
=
÷
÷
ø
ö
ç ç
è
æ
=
P.E. of satellite an
d earth system isU
GMm
r
= Þ
K
U
GMm
r
GMm
r
= =
2 1
2
68. (b) Ac
cording to Kepler’s law, the areal velocity of a
planet around the sun always remains constant.
SCD : A
1
– t
1
(areal velocity constant)
SAB : A
2
– t
2
12
12
AA
,
tt
=
t
1
= t
2
.
1
2
A
,
A
(given A
1
= 2A
2
)
= t
2
.
2
2
2A
A
\t
1
= 2t
2
69. (b) Orb
ital velocity of a satellite in a circular orbit of radius
a is given by
GM
v
a
= Þ va
1
a
Þ
2
1
v
v
=
1
2
a
a
\v
2
= v
1
4R
R
= 2 v
1
= 6V
70. (a)
Potential at the given point = Potential at the point
due to the shell + Potential due to the particle
=
2GM GM
aa
-- =
3GM
a
-
71. (b) Angu
lar momentum is conserved
\ L
1
= L
2
Þ mr
1
v
1
= mr
2
v
2
Þ r
1
v
1
= r
2
v
2
12
21
vr
vr
Þ=
72. (a) The velocity u should be equal to the escape velocity.
That is, u = 2gR
But g =
2
GM
R
2
GM
u 2· ·R
R
\=
2GM
R
Þ
73. (d) Intens
ity will be zero inside the spherical shell.
I = 0 upto r = a and
2
1
I
r
µ when r > a
74
. (b) 75. (b)
EXERCISE - 3
Exemplar Questions
1. (d) Let the density of earth as a sphere is uniform, then it
can be treated as point mass placed at its centre then
acceleration due to gravity g = 0, at the centre. But if
the density of earth is considered as a sphere of non-
uniform then value of 'g' will be different at different
points
4
3
g GR
æö
= pr
ç÷
èø
Q . So g cannot be zer o at any point.
2. (c) Force of attraction between any two objects obeys
the inverse square law.
As observed from the earth, the sun appears to move
in an approximate circular orbit. The gravitational force
of attraction between the earth and the sun always
follows inverse square law.
Due to relative motion between the earth and mercury,
the orbit of mercury, as observed from the earth will
not be approximately circular, since the major
gravitational force on mercury is due to the sun is
very large than due to earth and due to the relative
motion to sun and earth with mercury.
3. (a) As we know that, the torque on earth due to
gravitational attractive force on earth is zero.
As the earth is revolving around the sun in a circular
motion due to gravitational attraction. The force of
attraction will be of radial nature i.e., angle between
position vector r and force F is also, zero.
So, torque
sin00r F rF= t = ´ = °=
4. (c) As the total (P.E.) of the earth satellite orbiting in orbit
is negative
2
GM
r
-æö
ç÷
èø
, where r is radius of the satellite
and M is mass of the earth.
Due to the viscous force acting on satellite, energy
decreases continuously and radius of the orbit or
height decreases gradually.
5. (b) The major force acting on moon is due to gravitational
force of attraction by sun and earth and moon is not
always in line of joining sun and earth.
As observed from the sun, two types of forces are
acting on the moon one is due to gravitational
attraction between the sun and the moon and the other
is due to gravitational attraction between the earth
and the moon. So these two force have different lines
of action and it will not be strictly elliptical because
total force on the moon is not central.

243. 239Gravitation
6. (d) Asteroids are also being acted upon by central
gravitational forces, hence Asteroid will move in
circular orbits like planets and obey Kepler's laws.
7. (d) Gravitational mass of proton is equivalent to its inertial
mass and is independent of presence of neighbouring
heavy objects so verifies the option (d).
8. (c) Force of Gravitation,
2g
GMm
F
r
=
Let AB = r
So, force on B due to A
2
(2)
()
BA
G Mm
F
AB
==
towards BA.
2
(2)
2
g
G Mm
F
r
==
and force on B due to C
2
()
BC
GMm
F
BC
==
towards BC
As, (BC) = 2AB
2 22
4(2)4()4
g
BC
FGMm GMm GMm
F
ABABr
Þ= = ==
As F
BA > F
BC,
hence, m will move towards (BA) i.e., (2M).
NEET/AIPMT (2013-2017) Questions
9. (a)Ini tial P. E., U
i
=
GMm
R
-
,
Final P.E., U
f
=
GMm
3R
-
[Q R' = R + 2R = 3R]
\ Change in potential energy,
DU =
GMm
3R
-
+
GMm
R
=
GMm
R
1
1
3
æö
-
ç÷
èø
=
2
3
GMm
R
=
2
3
mgR
GMm
mgR
R
æö
=
ç÷
èø
Q
ALTERNATE : DU =
mgh
h
1
R
+
By placing the value of h = 2R we get
DU =
2
3
mgR.
10. (c)
m
Gravitational potential V =
Gm
r
-
V
0
= –
G2
1
´
–
G2
2
´
–
G2
4
´
–
G2
8
´
– 2G
111
1 ....
248
éù
++++¥
êú
ëû
= – 2G ×
1
1
1
2
-
= – 2G ×
1
1
2
= – 4 G..
11. (c) Escape velocity,
8
3
eV R GP=p
Þ V
e
µ R Þ 2
PP
EE
VR
VR
==
Þ V
P
= 2V
E
.
12. (b) As we know, the minimum speed with which a body is
projected so that it does not return back is called
escape speed.
222
4
===
+
e
GM GM GM
V
rRhR

1
2
( 3)
2
æö
==
ç÷
èø
Q
GM
hR
R
13. (b) First when (r < R) E µ r and then when r > R
2
1
E
r
µ.
Hence graph (b) correctly dipicts.
14. (c) From question,
Escape velocity
=
2GM
R
= c = speed of light
ÞR =
2
2GM
c
=
11 24
82
2 6.6 10 5.98 10
m
(3 10)
-
´´ ´´
´
= 10
– 2
m
15. (b)
12R–3R=9R
12R
M
R 2R
5M
Before collision
M
5M
R2R
At the time of collision
Let the distance moved by spherical body of mass M
is x
1
and by spherical body of mass 5m is x
2
As their C.M. will remain stationary
So, (M) (x
1
) = (5M) (x
2
) or, x
1
= 5x
2
and for tauching x
1
+ x
2
= 9R
So, x
1
= 7.5 R
16. (a) As we know, orbital speed,
orb
GM
V
r
=

244. 240 PHYSICS
Time period T
=
orb
2r 2r
r
v GM
pp
=
Squarring both sides,
T
2

=
2
2
32rr4
.r
GMGM
æöpp
=
ç÷
èø
Þ
22
3
T4
K
GMr
p
==
ÞGMK = 4p
2
.
17. (d)Giv
en: Height of the satellite from the earth's surface
h = 0.25 × 10
6
m
Radius of the earth R = 6.38 × 10
6
m
Acceleration due to gravity g = 9.8 m/s
2
Orbital velocity, V
0
= ?
V
0
=
2
2
GM GM R
.
(R h) (R h)R
=
++
=
6
9.8 6.38 6.38
6.63 10
´´
´
= 7.76 km/s
2
GM
g
R
éù
=
êú
ëû
Q
18. (c) The
gravitational force on the satellite will be aiming
towards the centre of the earth so acceleration of the
satellite will also be aiming towards the centre of the
earth.
19. (b) As we know, escape velocity,
V
e
=
32GM 2G4
·RR
R R3
æö
= prµrç÷
èø
\
e ee
p pp
VR
VR
r
=
r
Þ
e ee
p ee
VR
V 2R2
r
=
r
\Ratio
e
p
V
V
= 1 : 22
20. (a) As we know, gravitational potential (v) and
acceleration due to gravity (g) with height
V =
GM
Rh
-
+
= –5.4 × 10
7
…(1)
and g =
( )
2
GM
6
Rh
=
+
…(2)
Dividing (1) by (
2)
( )
7
2
GM
5.4 10Rh
GM 6
Rh
-
-´+
=
+
Þ
( )
7
5.4 10
6
Rh
´
=
+
ÞR + h = 9000 km so, h = 2600 km
21. (a) Both the astronauts are in the condition of
weightlessness. Gravitational force between them
pulls towards each other. Hence Astronauts move
towards each other under mutual gravitional force.
22. (c) Above earth surfaceBelow earth surface
g
h
=
e
2h
g1
R
æö
-ç÷
èø
g
d
=
e
d
g1
R
æö
-ç÷
èø
According to question, g
h
= g
d
ee
2hd
g1 g1
RR
æöæö
- =-ç÷ç÷
èøèø
Clearly,
d = 2h
= 2 km

245. INTERATOMIC AND INTERMOLECULAR FORCE
The force between atoms of an element is called interatomic
force. The force between molecules of a compound (or element)
is called intermolecular force. These forces are electrical in
nature. Depending on the distance between the atoms, this force
may be attractive or repulsive in nature. These forces are
responsible for the definite size or shape of a solid.
Detailed calculations as well as deductions from experiment show
that the interaction between any isolated pairs of atoms or
molecules may be represented by a curve that shows how the
potential energy varies with separation between them as shown
in the figure.
This curve describes the interatomic potential. The force between
the atoms can be found from the potential energy by using therelation
=-
dU
F
dr
The resulting interatomic force curve is shown in figure.
Force is along the line joining the atoms or molecules, and isshown negative for attraction and positive for repulsion.

O
F(R)
R
0
R
We see that as the distance decreases, the attractive force first
increases and then decreases to zero at a separation where thepotential energy is minimum.
For smaller distance force is repulsive, because at these distance
the negative charge distribution associated with one atom begins
to over lap with that associated with the neighboring atom.
The force of attraction between molecules may be written as
7
=-
a
a
F
r
The negative sign shows that F
a
is a force of attraction and ‘a’ is
a constant which depends on the kind of attractive force betweenthe molecules and the structure of the molecules. The force ofattraction results from the creation of induced dipole moment inone molecule by the neighbouring molecule.
When the molecules are brought closer there is a force of repulsion
between them. It can be shown that the repulsive force is
9
=+
r
b
F
r
,
where b is a
constant like a. The force of repulsion varies very
rapidly. F
r
is inversely proportional to the ninth power of distance
between the molecules. The resultant force acting on themolecules is
79
ab
F
rr
=-+
The d
istance between the molecules decides the sign of the
resultant force.
ELASTICITY
The property of the body by virtue of which it tends to regain its
original shape and size after removing the deforming force is
called elasticity. If the body regains its original shape and size
completely, after the removal of deforming forces, then the body
is said to be perfectly elastic.
·The property of the body by virtue of which it tends to
retain its deformed state after removing the deforming
force is called plasticity. If the body does not have any
tendency to recover its original shape and size, it is called
perfectly plastic.
STRESS AND STRAIN
Stress :
When a deforming force is applied to a body, an internal restoring
force comes into play.
The restoring force per unit area is called stress.
=
Restoring force
Stress
Area
9
Mechanical
Properties of Solids

246. 242 PHYSICS
Its S.I. Un
it is Nm
-2
Stress is a tensor as its value changes when direction changes.
Types of Stress :
Longitudinal stress or tensile stress, volumetric stress and
tangential stress are the types of stress.
Strain :
It is defined as the ratio of the change in shape or size to the
original shape or size of the body.
=
Change in dimension
Strain
Original dimension
Strain has no units or dimensions.
Types of Strain :Longitudinal strain : It is defined as the ratio of the change in
length to the original length.
Longitudinal strain=
Dl
l
Volume strain : It is th e ratio of the change in volume to the
original volume.
Volume strain=
DV
V
Shearing strain : It is the angular deformation produced in a
body.
Shearing strain (q)=
Dx
h
HOOKE’S LAW
It is the fundamental law of elasticity given by Robert Hooke in1679. It states that “the stress is directly proportional to strain
provided the strain is small “.
i.e. Stress µ strain
constantÞ==
stress
E
strain
This proportio
nality constant is called modulus of elasticity
(name given by Thomas Young) or coefficient of elasticity (E).
Since stress has same dimensions as that of pressure and strainis dimensionless. So the dimensions of E is same as, that of
stress or pressure i.e. [ML
–1
T
–2
]
The modulus of elasticity depends on the material and on thenature of deformation. There are three type of deformations andtherefore three types of modulus of elasticity.
(i)Young’s modulus (Y) : It measures the resistance of a solid
to elongation.
(ii)Shear modulus )(h or modulus of rigidity : It measures
the resistance to motion of the plane of a solid sliding parton each other.
(iii)Bulk modulus (B) : It measure the resistance that solid or
liquid offer to their volume change.
The stress under which the system breaks is called breakingstress.(i)Young's modulus (Y) : Let us consider a long bar (shown
in fig.) of cross-sectional area A and length l
o
, which is
clamped at one end. When we apply external force F
l
longitudinally along the bar, internal forces in bar resistdistortion, but bar attains equilibrium in which its length is
greater and in which external force is exactly balanced byinternal forces. The bar is said to be stressed in thiscondition.
l
0
F
l
Fixed
Dl
Now if the solid bar obey the Hooke’ law, the Young's
modulus, Y is defined as
Tensile stress
==
D
l
ll
/
/
FA
Y
Tensile strain
Where Dl is
change in the length of bar, when we
apply F
l
.
(ii)Shear modulus (h) : Shear modulus or modulus of
rigidity h ish==
D
/
/
t
FAShearing stress
Shearing strain x h
F
t
h
Dx
q
Fixed face
(a) As we see, there is no change in volume under
this deformation, but shape changes.
(b ) q»q=
D
tan
h
x
(see the figure), w
here q is shear
angle.
(iii)Bulk modulus (B) : The Bulk modulus B is defined
as
-D
= ==
DD
/
//
nFAVolume stress P
B
Volume stain VV VV
F
n
V
V – VD
Negative sign comes to make B positive, because with the
increase of pressure, the volume of body decreases or vice
versa.
The reciprocal of the Bulk modulus is called
compressibility of material
i.e., Compressibility = 1/B.

247. 243Mechanical Properties of Solids
Poisson's Ratio (s) :
Lateral strain/longitudinal strain = Poisson's ratio (s).
The theoretical value of s lies between –1 and 0.5 and practical
value of s lies between 0 and 0.5.
STRESS-STRAIN CURVE
Plastic
region
E (Fracture point)
B
A
C
DBreaking
strength
Proportiona
l limit
Elastic limit
Stress
StrainOO¢
(i)Propo rtional limit : The limit in which Hooke’s law is valid
i.e, stress is directly proportional to strain is called proportion
limit. Stress µ strain
(ii)Elastic limit : It is a maximum stress upto which the body
completely recovers its original state after the removal of
the deforming forces.
(iii)Yield point : The point beyond elastic limit, at which the
length of wire starts increasing without increasing stress,
is defined as the yield point.
(iv)Breaking point : The position when the strain becomes so
large that the wire breaks down at last, is called breaking
point. At this position the stress acting in that wire is called
breaking stress and strain is called breaking strain.
• Breaking stress is also known as the tensile strength.
• Metals with small plastic deformation are called brittle.
• Metals with large plastic deformation are called ductile.
Elastic fatigue : This is the phenomenon of a delay in recovering
the original configuration by a body, if it had been subjected to
stress for a longer time the body looses the property of elasticity
temporarily.
Elastic relaxation time : It is the time delay in regaining the
original shape after removal of deforming forces. Elastic relaxation
time for gold, silver and phosphor bronze is negligible.
Elastic Hysteresis
When the stress applied on a body, is decreased to zero, the
strain will not be reduced to zero immediately. For some
substances (e.g.-vulcanized rubber), the strain lags behind the
stress. This lagging of strain behind stress is called elastic
hysteresis.
Strain
S
t
r
e
s
s
The stress-strain graph for increasing and decreasing load
encloses a loop, as shown in figure. The area of the loop gives
the energy dissipated during its deformation.
Keep in Memory
1.Th
ermal stress =
,TY
A
F
aD= wher
e a is the coefficient of
linear expansion and DT is the change in temperature.
2.(i) The modulus of rigidity (h) for liquids is zero.
(ii) For a given tensile force, the increase in length is
inversely proportional to square of its diameter.
(iii) The pressure required to stop volume expansion of a
piece of metal is
P=BgdT,
where g = coefficient of volume expansion = 3a
(iv) To compare elasticities of different materials, their
identical small balls are made and they are dropped
from same height on a hard floor. The ball which rises
maximum after striking the floor, is most elastic. The
order of elasticity of different materials on this basis
is as follows :
Y
ivory
> Y
steel
> Y
rubber
> Y
clay
(v) For the construction of rails, bridges, girders and
machines, materials with high Young's modulus are
used so that they may not get permanently deformed.
For a spring, F =
.
YAL
kx
L
D
=
Hence
spring constant k =
L
YA
. Here DL = x
.
3.When equal force is applied on identical wires of different
materials then the wire in which minimum elongation is
produced is more elastic. For the same load, more elongation
is produced in rubber than in steel wire, hence steel is more
elastic than rubber.
Relation between Y, B, h and s
(i) Y = 3B(1 – 2s) (ii) Y = 2h (1 + s)
(iii)
3 –2
62
B
B
h
s=
+h
(iv)
9 31
YB
=+
h
Energy stored per u
nit volume in a strained body
Energy per unit volume =
2
1
stress × st
rain
=2
1
modulus o
f elasticity × (strain)
2
=2
1
(stress)
2
/mo
dulus of elasticity..
Work done in stretching a wire or work done per unit volume
=
´ ´ =´´
11
22
stress strain load extension KEEP I

248. 244 PHYSICS
Torsional
rigidity of a cylinder :
(i) The torsional rigidity C of a cylinder is given by,
4
2
R
C
ph
=
l
where h= modulus of rigidity of material of cylinder,,
R = radius of cylinder, l = length of cylinder
(ii) Restoring couple,
4
2
R
C
phf
t= f=
l
(iii) Work done in twisting the cylinder through an angle
f,
42
4
R
W
phf
=
l
joule =
21
2
Cf
Cantilever :
A beam fixed at one end and loaded at the other end is called a
cantilever.
(i) The depression y at a distance x from the fixed end is (when
the weight of cantilever is ineffective)
÷
÷
ø
ö
ç
ç
è
æ
-=
6
x
2
x
IY
mg
y
32
l
where mg = load
applied, l = length of cantilever,
I = geometrical moment of inertia of its cross section.
(ii) Maximum depression at free end of cantilever,
IY3
mg
y
3
max
l
=d=
(a) For rectang
ular beam,
12
bd
I
3
=
(b) For a circula
r cross section beam of radius r,
4
r
I
4
p
=
(iv) Depressi
on produced in a beam supported at two ends
and loaded at the middle,
IY48
mg
3
l
=d
For rectangular be
am,
3
3
4
mgl
Ybd
d=
and for circular
beam,
Yr12
mgl
4
3
p
=d
Keep in Memory
1.Wound spring p
ossess elastic potential energy.
2.A material which can be drawn into wires is called ductile
and a material which can be hammered into sheet is called
malleable.
3.Ductility, brittleness, malleability, etc., are not elastic
properties.
4.The substance which breaks just beyond the elastic limit
is called brittle.
5.(i) Breaking force of a wire
= breaking stress × area of cross section
(ii) Breaking stress does not depend on the length of
wire
(iii) Breaking stress depends on the material of wire.
Example 1.
A uniform rod of mass m, length L, area of cross section Aand Young’s modulus Y hangs from a ceiling. Itselongation under its own weight will be
(a)
2mgL
AY
(b)
mgL
AY
(c)
mgL
2 AY
(d) zero
Solution (c)
Mas
s of section BC of wire =
)xL(
L
m
- ;
Tension at B, g)xL(
L
m
T -=
(L–x)
A
B
C
x
Elongation of elemen
t dx at B,
dx
YAL
g)xL(m
Y
dx
A
T
d
-
==l
Total elongation
ò ò
=-==
L
0
AY2
Lmg
dx)xL(
YAL
mg
dl
Example 2.
If the poten
tial energy of the molecule is given by
U =
6 12
AB
rr
- . Then at equilibrium position its potential
energy is equal to
(a) –A
2
/4 B (b) A
2
/4 B
(c) 2 A/B (d) A/2 B
Solution : (b)
ú
û
ù
ê
ë
é
--=-=
126
r
B
r
A
dr
d
dr
dU
F
= ú
û
ù
ê
ë
é
+
-
137
r
B12
r
6xA
In equilibrium position F =
0.
so,
137
r
B12
r
A6
= or
A
B2
r
6
=
\ Potential ener
gy at equilibrium position
222
2
A B AAA
U
(2B/A) 2B 4B 4B(2B / A)
= - =-=
Example
3.
The normal density of gold is
r and its bulk modulus is K.
The increase in density of a lump of gold when a pressure
P is applied uniformly on all sides is
(a) K/r P (b) P/r K
(c)r P/K (d)r K/P
Solution : (a)
V/V
p
K
D
= or
K
p
V
V
=
D
;
Also
V
M
=r and
VV
M'
D-
=r ;
1'
V
V
1
)V/V1(
1
)VV(
V
-
÷
ø
ö
ç
è
æD
-=
D-
=
D-
=
r
r
\

249. 245Mechanical Properties of Solids
Now
l
l
ll
D
s-=
D
D
D
-=s
r
r
or
/
r/r
;
33
101)102(5.0
r
r --
´-=´´-=
D
\
Further, 0)101(2)102(
V
V 33
=´´-´=
D --
\ % incr
ease in volume is 0.
Example 6.
(a) A heavy machine is to be installed in a factory. To
absorb vibrations of the machine, a block of rubber is
placed between the machine and floor, which of the
two rubbers A and B would you prefer to use for the
purpose ? Why ?
StrainO StrainO
Stress Stress
A
B
(b)Which of the
two rubber materials would you choose
for a car tyre ?
Solution :
(a)Rubber B is preferred. The area of the hysteresis loop
measures the amount of heat energy dissipated by the
material. The area of loop B is more than A. So B can
absorb more vibrations.
(b)To avoid excessive heating of car tyre, rubber A would
be preferred over rubber B.

K
p
1
V
V
1 +=÷
ø
ö
ç
è
æD
+» or
K
p
1=-
r
r¢
or
K
pr
=r-r¢ (QD V << V)
Example 4.
A wire 3 m in length and 1 mm in diameter at 30ºC and
kept in a low temperature at –170ºC and is stretched
by hanging a weight of 10 kg at one end. Calculate the
change in the length of the wire. Given
a = 1.2 ×10
–5
/ºC
and Y = 2 × 10
11
N/m
2
. Take g = 10 m/s
2
.
Solution :
We know that,
611
10785.0102
31010
AY
LF
-
´´´
´´
==l = 1.91 × 10
–3
(wher
e A = p r
2
) = 0.785 × 10
–6
m
2
Contraction in length = –a L D T
= – (1.2 × 10
–5
) (3) (–170 – 30)
= 7.2× 10
–3
m
The resultant change in length
= 7.2 × 10
–3
– 1.91 × 10
–3
= 5.29 mm
Example 5.
A material has Poisson’s ratio 0.5. If a uniform rod of itsuffers a longitudinal strain of 2 × 10
–3
, what is the
percentage increase in volume?
Solution :
Here
l
ll
l
l
2
2
2
2
r
rr2r
r
)r(
V
V D+D
=
p
pD
=
D
or
r
r
2
V
V D
+
D
=
D
l
l

250. 246 PHYSICS

251. 247Mechanical Properties of Solids
1.Elastomers are the materials which
(a) are not elastic at all
(b) have very small elastic range
(c) do not obey Hooke’s law
(d) None of these
2.The load versus elongation graph for four wires is shown.
The thinnest wire is
S
R
Q
P
Elongation
Load
(a)P (b)Q
(c)R (d)S
3.Whi
ch of the following affects the elasticity of a substance?
(a) hammering and annealing
(b) change in temperature
(c) impurity in substance
(d) All of these
4.Which of the following has no dimensions ?
(a) strain (b) angular velocity
(c) momentum (d) angular momentum
5.Minimum and maximum values of Possion’s ratio for a metal
lies between
(a) – ¥ to + ¥ (b) 0 to 1
(c) – ¥ to 1 (d) 0 to 0.5
6.In solids interatomic forces are
(a) totally repulsive(b) totally attractive
(c) both (a) and (b)(d) None of these
7.Which one of the following is not a unit of Young’s
modulus ?
(a)Nm
–1
(b) Nm
–2
(d) dyne cm
–2
(d) mega pascal
8.Two rods A and B of the same material and length have
their radii r
1
and r
2
respectively. When they are rigidly fixed
at one end and twisted by the same couple applied at the
other end, the ratio
Angle of twist at the end of A
Angle of twist at the end of B
æö
ç÷
èø
is
(a)
2
2
2
1
r/r (b)
3
2
3
1
r/r
(c)
4
1
4
2
r/r (d)
4
2
4
1
r/
r
9.The value of tan (90 – q) in the graph gives
q
S
t
r
a
i
n
Stress
(a) Young's modulus of elasticity
(b) compressibility
(c) shear strain
(d) tensile strength
10.The length of a metal is l
1
when the tension in it is T
1
and
is l
2
when the tension is T
2
. The original length of the wire
is
(a)
2
21ll+
(b)
21
1221
TT
TT
+
+ll
(c)
12
1221
TT
TT
-
-ll
(d)
2121
TTll
11.Unifo
rm rod of mass m, length
l, area of cross-section AA
has Young’s modulus Y. If it is hanged vertically, elongation
under its own weight will be
(a)
AY2
mgl
(b)
AY
mg2l
(c)
AY
mgl
(d)
lA
mgY
12.Which on
e of the following affects the elasticity of a
substance ?
(a) Change in temperature
(b) Hammering and annealing
(c) Impurity in substance
(d) All of the above
13.According to Hooke’s law of elasticity, if stress is
increased, then the ratio of stress to strain
(a) becomes zero (b) remains constant
(c) decreases (d) increases
14.The length of an iron wire is L and area of corss-section is
A. The increase in length is l on applying the force F on its
two ends. Which of the statement is correct?
(a) Increase in length is inversely proportional to its length
(b) Increase in length is proportional to area of cross-
section
(c) Increase in length is inversely proportional to area of
cross-section
(d) Increase in length is proportional to Young's modulus

252. 248 PHYSICS
15.A and B are two
wires. The radius of A is twice that of B.
They are stretched by the same load. Then the stress on B
is
(a) equal to that on A(b) four times that on A
(c) two times that on A(d) half that on A
16.Hooke's law defines
(a) stress
(b) strain
(c) modulus of elasticity
(d) elastic limit
17.In case of steel wire (or a metal wire), the limit is reached
when
(a) the wire just break
(b) the load is more than the weight of wire
(c) elongation is inversely proportional to the tension
(d) None of these
18.A steel ring of radius r and cross sectional area A is fitted
onto a wooden disc of radius R (R > r). If the Young’s
modulus of steel is Y, then the force with which the steel
ring is expanded is
(a) A Y (R/r) (b) A Y (R – r)/r
(c) (Y/A)[(R – r)/r](d) Y r/A R
19.Which of the following relation is true ?
(a)
3 (1)YK= -s (b)
9Y
K
Y
h
=
+h
(c) (6)KYs= +h (d)
05.Y-h
s=
h
20.For a constant
hydraulic stress on an object, the fractional
change in the object volume
V
V
Dæö
ç÷
èø
and its bulk modulus
(B) are re
lated as
(a)
V
B
V
D
µ (b)
1V
VB
D
µ
(c)
2V
B
V
D
µ (d)
2V
B
V
-D
µ
21.The diagram s
hown below represents the applied forces
per unit area with the corresponding change X (per unit
length) produced in a thin wire of uniform cross section in
the curve shown. The region in which the wire behaves like a
liquid is
(a) ab
(b) bc
(c) cd
F
O X
a
b
c
d
(d) Oa
22.A steel wire
is suspended vertically from a rigid support.
When loaded with a weight in air, it extends by l
a
and when
the weight is immersed completely in water, the extension is
reduced to l
w
. Then the relative density of material of the
weight is
(a)
wa
/ll (b)
wa
a
ll
l
-
(c) )/(
waw
lll- (d)
aw
/ll
23.Whe
n an elastic material with Young’s modulus Y is
subjected to stretching stress S, elastic energy stored per
unit volume of the material is
(a) YS / 2 (b) S
2
Y / 2
(c)S
2
/ 2Y (d) S / 2Y
24.The ratio of shearing stress to the corresponding
shearing strain is called
(a) bulk modulus (b) Young's modulus
(c) modulus of rigidity (d) None of these
25.The Young’s modulus of a perfectly rigid body is
(a) unity
(b) zero
(c) infinity
(d) some finite non-zero constant
1.Two wires of same material and length but cross-sections
in the ratio 1 : 2 are used to suspend the same loads. The
extensions in them will be in the ratio
(a) 1 : 2 (b) 2 : 1
(c) 4 : 1 (d) 1 : 4
2.A body of mass 10 kg is attached to a wire of radius 3 cm. It’s
breaking stress is 4.8 × 10
7
Nm
–2
, the area of cross-section of
the wire is 10
–6
m
2
. What is the maximum angular velocity with
which it can be rotated in the horizontal circle ?
(a) 1 rad sec
–1
(b) 2 rad sec
–1
(c) 4 rad sec
–1
(d) 8 rad sec
–1
3.The Young’s modulus of brass and steel are respectively
10
10
N/m
2
. and 2 × 10
10
N/m
2
. A brass wire and a steel wire
of the same length are extended by 1 mm under the same
force, the radii of brass and steel wires are R
B
and R
S
respectively. Then
(a)
BS
R2R= (b) 2/RR
BS
=
(c)
BSR4R= (d) 4/RR
BS=
4.A steel wire of le
ngth 20 cm and uniform cross-section
1 mm
2
is tied rigidly at both the ends. The temperature of
the wire is altered from 40ºC to 20ºC. Coefficient of linear
expansion for steel a = 1.1 × 10
–5
/ºC and Y for steel is
2.0 × 10
11
N/m
2
. The change in tension of the wire is
(a) 2.2 × 10
6
newton (b) 16 newton
(c) 8 newton (d) 44 newton

253. 249Mechanical Properties of Solids
5.A cube is subjected to a uniform volume compression. If
the side of the cube decreases by 2% the bulk strain is
(a) 0.02 (b) 0.03
(c) 0.04 (d) 0.06
6.A wire suspended vertically from one of its ends is
stretched by attaching a weight of 200N to the lower end.
The weight stretches the wire by 1 mm. Then the elastic
energy stored in the wire is
(a) 0.2 J (b) 10 J
(c) 20 J (d) 0.1 J
7.A metal rod of Young's modulus 2 × 10
10
N m
–2
undergoes
an elastic strain of 0.06%. The energy per unit volume stored
in J m
–3
is
(a) 3600 (b) 7200
(c) 10800 (d) 14400
8.A force of 10
3
newton, stretches the length of a hanging
wire by 1 millimetre. The force required to stretch a wire of
same material and length but having four times the diameter
by 1 millimetre is
(a) 4 × 10
3
N (b) 16 × 10
3
N
(c)
31
10N
4
´ (d)
31
10N
16
´
9.A 2 m long rod of
radius 1 cm which is fixed from one end is
given a twist of 0.8 radian. The shear strain developed will
be
(a) 0.002 (b) 0.004
(c) 0.008 (d) 0.016
10.There are two wire of same material and same length while
the diameter of second wire is two times the diameter of
first wire, then the ratio of extension produced in the wires
by applying same load will be
(a) 1 : 1 (b) 2 : 1
(c) 1 : 2 (d) 4 : 1
11.For a given material, the Young's modulus is 2. 4 times that
of rigidity modulus. Its Poisson's ratio is
(a) 2.4 (b) 1.2
(c) 0.4 (d) 0.2
12.A cube at temperature 0ºC is compressed equally from all
sides by an external pressure P. By what amount should its
temperature be raised to bring it back to the size it had
before the external pressure was applied. The bulk modulus
of the material of the cube is B and the coefficient of linear
expansion is a.
(a) P/B a (b) P/3 B a
(c) 3 p a/B (d) 3 B/P
13.The compressibility of water is 4 × 10
–5
per unit atmospheric
pressure. The decrease in volume of 100 cm
3
of water under
a pressure of 100 atmosphere will be
(a) 0.4 cm
3
(b) 4 × 10
–5
cm
3
(c) 0.025 cm
3
(d) 0.004 cm
3
14.For the same cross-sectional area and for a given load, the
ratio of depressions for the beam of a square cross-section
and circular cross-section is
(a) 3 : p (b)p : 3
(c) 1 : p (d)p : 1
15.A massive stone pillar 20 m high and of uniform cross-
section rests on a rigid base and supports a vertical load of
5.0 × 10
5
N at its upper end. If the compressive stress in the
pillar is not to exceed 1.6 × 10
6
N m
–2
, what is the minimum
cross-sectional area of the pillar? Density of the stone
= 2.5 × 10
3
kg m
–3
. (Take g = 10 N kg
–1
)
(a) 0.15 m
2
(b) 0.25 m
2
(c) 0.35 m
2
(d) 0.45 m
2
16.A circular tube of mean radius 8 cm and thickness 0.04 cm is
melted up and recast into a solid rod of the same length.
The ratio of the torsional rigidities of the circular tube and
the solid rod is
(a)
4
44
)8.0(
)98.7()02.8(-
(b)
2
22
)8.0(
)98.7()02.8(-
(c)
44
2
)98.7()02.8(
)8.0(
-
(d)
23
2
)98.7()02.8(
)8.0(
-
17.From a ste
el wire of density r is suspended a brass block of
density r
b
. The extension of steel wire comes to e. If the
brass block is now fully immersed in a liquid of density r
l
,
the extension becomes e'. The ratio
'e
e
will be
(a)
lr-r
r
b
b
(b)
b
b
r
r-r
l
(c)
r-r
r-r
l
b
(d)
l
l
r-r
r
b
18.One en
d of a uniform wire of length L and of weight W is
attached rigidly to a point in the roof and
1
W weight is
suspended from looser end. If A is area of cross-section of
the wire, the stress in the wire at a height
4
L
from the up
per
end is
(a)a
WW
1+
(b)
a
4/W3W
1+
(c)
1
W W/4
a
+
(d)
a
W3W4
1+
19.A beam o
f metal supported at the two edges is loaded at the
centre. The depression at the centre is proportional to
(a) Y
2
(b) Y
(c) 1/Y (d) 1/Y
2
20.An iron rod of length 2m and cross-sectional area of 50
mm
2
stretched by 0.5 mm, when a mass of 250 kg is hung
from its lower end. Young’s modulus of iron rod is
(a)
220
m/N106.19´ (b)
218
m/N106.19´
(c)
210
m/N106.19´ (d)
215
m/N106.19´
21.A wire fixed at the upper end stretches by length l by
applying a force F. The work done in stretching is
(a) 2Fl (b) Fl
(c)
2l
F
(d)
2
lF

254. 250 PHYSICS
22.A metalic ro
d of length l and cross-sectional area A is made
of a material of Young modulus Y. If the rod is elongated by
an amount y, then the work done is proportional to
(a)y (b)
1
y
(c)y
2
(d)2
1
y
23.On stretching a wire, the elastic energy stored per unit
volume is
(a)Fl/2AL (b)FA/2L
(c)FL/2A (d)FL/2
24.When a pressure of 100 atmosphere is applied on a spherical
ball, then its volume reduces to 0.01%. The bulk modulus
of the material of the rubber in dyne/cm
2
is
(a) 10 × 10
12
(b) 100 × 10
12
(c) 1 × 10
12
(d) 10 × 10
12
25.What per cent of length of wire increases by applying a
stress of 1 kg weight/mm
2
on it?
(Y = 1 × 10
11
N/m
2
and 1 kg weight = 9.8 newton)
(a) 0.0067% (b) 0.0098%
(c) 0.0088% (d) 0.0078%
26.K is the force constant of a spring. The work done in
increasing its extension from l
1
to l
2
will be
(a)K(l
2
– l
1
) (b)
21
()
2
K
ll+
(c)
22
21
()Kll- (d)
22
21()
2
K
ll-
27.If a rubber ba
ll is taken at the depth of 200 m in a pool, its
volume decreases by 0.1%. If the density of the water is1 × 10
3
kg/m
3
and g = 10m/s
2
, then the volume elasticity in
N/m
2
will be
(a) 10
8
(b) 2 × 10
8
(c) 10
9
(d) 2 × 10
9
28.The diagram below shows the change in the length X of athin uniform wire caused by the application of stress F attwo different temperatures T
1
and T
2
. The variation shown
suggests that
(a)T
1
> T
2
X
T
2
T
1
F
(b) T
1
< T
2
(c)T
2
> T
1
(d) T
1
³ T
2
29.A mate
rial has poisson’s ratio 0.50. If a uniform rod of it
suffers a longitudinal strain of 2 × 10
–3
, then the percentage
change in volume is
(a) 0.6 (b) 0.4
(c) 0.2 (d) Zero
30.A 5 metre long wire is fixed to the ceiling. A weight of 10 kg
is hung at the lower end and is 1 metre above the floor. The
wire was elongated by 1 mm. The energy stored in the wire
due to stretching is
(a) zero (b) 0.05 joule
(c) 100 joule (d) 500 joule
31.Two wires A and B are of the same material. Their lengths
are in the ratio of 1 : 2 and the diameter are in the ratio 2 : 1.
If they are pulled by the same force, then increase in length
will be in the ratio of
(a) 2 : 1 (b) 1 : 4
(c) 1 : 8 (d) 8 : 1
32.Two wires are made of the same material and have the same
volume. However wire 1 has cross-sectional area A and
wire 2 has cross-sectional area 3A. If the length of wire 1
increases by Dx on applying force F, how much force is
needed to stretch wire 2 by the same amount?
(a) 4 F (b) 6 F
(c) 9 F (d)F
33.A rubber cord catapult has cross-sectional area 25 mm
2
and initial length of rubber cord is 10 cm. It is stretched to 5
cm and then released to project a missile of mass 5 gm.
Taking
rubber
Y = 5 × 10
8
N/m
2
. Velocity of projected missile
is
(a) 20 ms
–1
(b) 100 ms
–1
(c) 250 ms
–1
(d) 200 ms
–1
34.The potential energy U between two atoms in a diatomic
molecules as a function of the distance x between atoms
has been shown in the figure. The atoms are
A B Cx
U
O
(a) attracted when x lies between A and B and are repelled
when x lies between B and C
(b) attracted when x lies between B and C and are repelled
when x lies between A and B
(c) are attracted when they reach B from C(d) are repelled when they reach B from A
35.The diagram shows a force - extension graph for a rubberband. Consider the following statements :I. It will be easier to compress this rubber than expand itII. Rubber does not return to its original length after it is
stretched
III. The rubber band will get heated if it is stretched and
released
Extension
Force
Which of these can be deduced from the graph?
(a) III only (b) II and III
(c) I and III (d) I only

255. 251Mechanical Properties of Solids
36.A rod of length l and radius r is joined to a rod of length
l/2 and radius r/2 of same material. The free end of small rod
is fixed to a rigid base and the free end of larger rod is given
a twist of q°, the twist angle at the joint will be
(a)q/4 (b)q/2
(c)5q/6 (d) 8q/48
37.To break a wire, a force of 10
6
N/m
2
is required. If the density
of the material is 3 × 10
3
kg/m
3
, then the length of the wire
which will break by its own weight will be
(a) 34 m (b) 30 m
(c) 300 m (d) 3 m
38.The upper end of a wire of diameter 12mm and length 1m is
clamped and its other end is twisted through an angle of
30°. The angle of shear is
(a) 18° (b) 0.18°
(c) 36° (d) 0.36°
39.A steel wire of uniform cross-section of 1mm
2
is heated
upto 50°C and clamped rigidly at its ends. If temperature of
wire falls to 40°C, change in tension in the wire is (coefficient
of linear expansion of steel is 1.1 × 10
–5
/°C and Young's
modulus of elasticity of steel is 2 × 10
11
N/m
2
)
(a) 22 N (b) 44 N
(c) 88 N (d) 88 × 10
6
N
40.In Searle's experiment to find Young's modulus the diameter
of wire is measured as d = 0.05cm, length of wire is
l = 125cm and when a weight, m = 20.0 kg is put, extension
in wire was found to be 0.100 cm. Find maximum permissible
error in Young's modulus (Y). Use :
2
mg
Y
( /4)dx
=
p
l
.
(a) 6.3% (b) 5.3%
(
c) 2.3% (d) 1%
41.If the ratio of lengths, radii and Young’s modulus of steel
and brass wires shown in the figure are a, b, and c,
respectively. The ratio between the increase in lengths of
brass and steel wires would be
(a)
2
2
ba
c
4kg
2kg
Steel
Brass
(b)
2
2
bc
a
(c)
2
2
ba
c
(d)
2
2
a
bc
42.A uniform cube is subjected to volume compression. If each
side is decreased by 1%, then bulk strain is
(a) 0.01 (b) 0.06
(c) 0.02 (d) 0.03
43.When a 4 kg mass is hung vertically on a light spring that
obeys Hooke’s law, the spring stretches by 2 cms. The
work required to be done by an external agent in stretching
this spring by 5 cms will be (g = 9.8 m/sec
2
)
(a) 4.900 joule (b) 2.450 joule
(c) 0.495 joule (d) 0.245 joule
44.The following four wires are made of the same material.
Which of these will have the largest extension when the
same tension is applied ?
(a) Length = 100 cm, diameter = 1 mm
(b) Length = 200 cm, diameter = 2 mm
(c) Length = 300 cm, diameter = 3 mm
(d) Length = 50 cm, diameter = 0.5 mm
45.A steel rod of radius R = 10 mm and length
L= 100 cm is stretched along its length by a force F = 6.28 ×
10
4
N. If the Young’s modulus of steel is Y = 2 ×10
11
N/m
2
,
the percentage elongation in the length of the rod is :
(a) 0.100 (b) 0.314
(c) 2.015 (d) 1.549
DIRECTIONS for Qs. 46 to 50 : These are Assertion-Reason
type questions. Each of these question contains two statements:
Statement-1 (Assertion) and Statement-2 (Reason). Answer
these questions from the following four options.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a
correct explanation for Statement -1
(b) Statement-1 is True, Statement -2 is True; Statement-2 is
NOT a correct explanation for Statement - 1
(c) Statement-1 is True, Statement- 2 is False
(d) Statement-1 is False, Statement -2 is True
46. Statement-1 Identical springs of steel and copper are equally
stretched. More work will be done on the steel spring.
Statement-2 Steel is more elastic than copper.
47. Statement-1 Stress is the internal force per unit area of a
body.
Statement-2 Rubber is less elastic than steel.
48. Statement 1 : The stress-strain graphs are shown in the
figure for two materials A and B are shown in figure. Young's
modulus of A is greater than that of B.

A
BS
t
r
e
s
s
Strain
Statement 2 : The Young's modules for small strain is,
stress
Y
strain
= = slope of linear portion, of graph; and slope
of A is more than slope that of B.
49. Statement 1: Young’s modulus for a perfectly plastic body
is zero.
Statement 2: For a perfectly plastic body, restoring force is
zero.
50. Statement 1: Strain causes the stress in an elastic body.
Statement 2: An elastic rubber is more plastic in nature.

256. 252 PHYSICS
Exemplar Questions
1.Modulus of rigidity of ideal liquids is
(a) infinity
(b) zero
(c) unity
(d) some finite small non-zero constant value
2.The maximum load a wire can withstand without breaking,
when its length is reduced to half of its original length, will
(a) be double (b) be half
(c) be four times (d) remain same
3.The temperature of a wire is doubled. The Young's modulus
of elasticity
(a) will also double(b) will become four times
(c) will remain same(d) will decrease
4.A spring is stretched by applying a load to its free end. The
strain produced in the spring is
(a) volumetric
(b) shear
(c) longitudinal and shear
(d) longitudinal
5.A rigid bar of mass M is supported symmetrically by three
wires each of length l. Those at each end are of copper and
the middle one is of iron. The ratio of their diameters, if each
is to have the same tension, is equal to
(a)Ycopper / Yiron (b)
iron
copper
Y
Y
(c)
2
iron
2
copper
Y
Y
(d)
iron
copper
Y
Y
6.A mild ste
el wire of length 2L and cross-sectional area A is
stretched, well within elastic limit, horizontally between two
pillars (figure ). A mass m is suspended from the mid-point
of the wire. Strain in the wire is
m
x
2L
(a)
2
2
2
x
L
(b)
x
L
(c)
2
x
L
(d)
2
2
x
L
7.A rectangular frame is to be suspended symmetrically by
two strings of equal length on two supports (figure). It canbe done in one of the following three ways;
(a) (b) (c)
The tension in the strings will be
(a) the same in all cases (b) least in (a)
(c) least in (b) (d) least in (c)
8.Consider two cylindrical rods of identical dimensions, one
of rubber and the other of steel. Both the rods are fixed
rigidly at one end to the roof. A mass M is attached to each
of the free ends at the centre of the rods.
(a) Both the rods will elongate but there shall be no
perceptible change in shape
(b) The steel rod will elongate and change shape but the
rubber rod will only elongate
(c) The steel rod will elongate without any perceptible
change in shape, but the rubber rod will elongate and
the shape of the bottom edge will change to an ellipse
(d) The steel rod will elongate, without any perceptible
change in shape, but the rubber rod will elongate with
the shape of the bottom edge tapered to a tip at the
centre
NEET/AIPMT (2013-2017) Questions
9.The following four wires are made of the same material.
Which of these will have the largest extension when the
same tension is applied ? [2013]
(a) Length = 100 cm, diameter = 1 mm
(b) Length = 200 cm, diameter = 2 mm
(c) Length = 300 cm, diameter = 3 mm
(d) Length = 50 cm, diameter = 0.5 mm

257. 253Mechanical Properties of Solids
10.If the ratio of diameters, lengths and Young’s modulus of
steel and copper wires shown in the figure are p, q and s
respectively, then the corresponding ratio of increase in
their lengths would be [NEET Kar. 2013]
(a)
7
(5)
q
sp
(b) 2
5
(7)
q
sp
Steel
2m
Copper
5m
(c) 2
7
(5)
q
sp
(d)
2
(5)
q
sp
11.Copper of fixed volume ‘V; is drawn into wire of length ‘l’.
When this wire is subjected to a constant force ‘F’, theextension produced in the wire is ‘Dl’. Which of the
following graphs is a straight line? [2014]
(a)Dl versus
1
l
(b)Dl versus l
2
(c)Dl
versus
2
1
l
(d)Dl versus l
12.The ap
proximate depth of an ocean is 2700 m. The
compressibility of water is 45.4 × 10
–11
Pa
–1
and density of
water is 10
3
kg/m
3
.What fractional compression of water
will be obtained at the bottom of the ocean ? [2015]
(a) 1.0 × 10
–2
(b) 1.2 × 10
–2
(c) 1.4 × 10
–2
(d) 0.8 × 10
–2
13.The Young's modulus of steel is twice that of brass. Two
wires of same length and of same area of cross section, one
of steel and another of brass are suspended from the same
roof. If we want the lower ends of the wires to be at the
same level, then the weights added to the steel and brass
wires must be in the ratio of : [2015 RS]
(a) 2 : 1 (b) 4 : 1
(c) 1 : 1 (d) 1 : 2
14.The bulk modulus of a spherical object is 'B'. If it is subjected
to uniform pressure 'p', the fractional decrease in radius is
(a)
B
3p
(b)
3p
B
[2017]
(c)
p
3B
(d)
p
B

258. 254 PHYSICS
EXERCISE - 1
1. (c) 2. (b)
3. (d) 4. (a) 5. (d)
6. (c) 7. (a)
8. (c) Couple per unit angle of twist,
l2
r
C
4
hp
=
\Couple t =
l2
r
C
4
qhp
=q
Here h, l, C & t are sam
e. So, r
4
q = constant
\
÷
÷
ø
ö
ç ç
è
æ
=
q
q
4
1
4
2
2
1
r
r
9. (a)
strain
stress
)90tan(=q-
10. (c) If l is the original length of wire, then change in length
of first wire, )(
11
lll -=D
change in l
ength of second wire, )(
22
lll -=D
Now,
2
2
1
1
A
T
A
T
Y
l
l
l
l
D
´=
D
´=
or
2
2
1
1TT
llD
=
D
or
llll -
=
-
2
2
1
1 TT
orT
1l
2
– T
1l = T
2l
1
–lT
2
or
12
2112
TT
TT
-
-
=
ll
l
11. (c)
F F mg
Y
A YA YA
= ÞD==
D
l ll
l
l
12. (d) The elasticity of a material depends upon the
temperature of the material. Hammering & annealing
reduces elastic property of a substance.
13. (b) The ratio of stress to strain is always constant. If
stress is increased, strain will also increase so that
their ratio remains constant.
14. (c)
1FL
ll
YAA
= Þµ
15. (b) Stress =
2
Force1
Stress
Area πr
\µ
2
2
(2)4
BA
BA
AB
Sr
SS
Sr
æö
= = Þ=ç÷
èø
16. (c)
17. (d
) According to Hooke's Law, within the elastic limits
stress is directly proportional to strain.
18. (b) Let T be the tension in the ring, then
)rR(A
rT
)rR(2.A
r2.T
Y
-
=
-p
p
=
\
r
)rR(AY
T
-
=
19. (d)
0.5
2(1)
Y
Y
-h
= h +s Þs=
h
20.
(b)
1
/
pV
B
VV BV
DD
= Þµ
D
[Dp = constant]
21
. (b) The wire starts behaving like a liquid at point b. It
behaves like a viscous liquid in the region bc of the
graph.
22. (b) Let V be the volume of the load and r its relative
density
So,
aaA
LgV
A
LF
Y
ll
r
== .....(1)
When the load is imme
rsed in the liquid, then
ww A
L)g1VgV(
A
LF
Y
ll
´´-r
=
¢
=
....(2)
(Q Now net weight = weight – upthrust)
From eqs. (1) and (2), we get
wa
)1(
ll
-r
=
r
or
)(
wa
a
ll
l
-
=r
23. (c) Ene
rgy stored per unit volume
=
strainstress
2
1
´´
= )moduluss'Young/stress(stress
2
1
´´
= )moduluss'Young/()stress(
2
1 2
´ =
Y2
S
2
24. (c)
25. (c)
For a perfectly rigid body strain produced is zero for
the given force applied, so
Y = stress/strain = ¥
EXERCISE - 2
1. (b) Let W n
ewton be the load suspended. Then
111
1
A
LW
)L/(
)A/W(
Y
ll
==
...(1)
and
222
2
A
LW
)L/(
)A/W(
Y
ll
== ....(2)
Dividing equatio
n (1) by equation (2), we get
÷
ø
ö
ç
è
æ
÷
÷
ø
ö
ç
ç
è
æ
=
÷
÷
ø
ö
ç
ç
è
æ
÷
÷
ø
ö
ç
ç
è
æ
=
1
2
A
A
1
1
2
1
2
1
2
l
l
l
l
\
1
2
2
1
=
l
l
or l
1
: l
2
= 2 : 1
2.
(c) Given that F/A = 4.8 × 10
7
Nm
–2
\F = 4.8 ×10
7
×A or
2
76mv
4.8 10 10 48
r
-
=´´=
or 48
r
rm
22
=
w
or
rm
482
=w
sec/rad416
3.010
48
==÷
ø
ö
ç
è
æ
´
=w
Hints & Solutions

259. 255Mechanical Properties of Solids
3. (b) We know that Y = F L/p r
2
l or r
2
= F L/(Y p l)
\ )Y/(LFR
B
2
B
lp= and )Y/(L FR
S
2
Slp=
or 2
10
102
Y
Y
R
R
10
10
B
S
2
S
2
B
=
´
==
or
2
S
2
B
R2R= o
r
SB
R2R= \
2/RR
BS
=
4. (d)
F = Y A a t = (2.0 × 10
11
) (10
–6
) (1.1 × 10
–5
) (20)
= 44 newton
5. (d)
6. (d) Elastic energy =
xF
2
1
´´
F = 200
N, x = 1 mm = 10
–3
m
\
J1.0101200
2
1
E
3
=´´´=
-
7. (a) U
/ volume =
2
strainY
2
1
´ = 3600 J m
–3
[Str
ain = 0.06 × 10
–2
]
8. (b)
2l
FYA Fr
L
=´´Þµ (Y, l and and L are constant)
If diameter is made four times then force required will
be 16 times, i.e., 16 × 10
3
N
9. (b)
2
8.0
100
1r
´=
q
=f
l
= 0.004 radian
10
. (d)
Q Both wires are same materials so both will have
same Young’s modulus, and let it be Y.
( )ΔL/LA.
F
strain
str
ess
Y == , forceappliedF=
A area of cross-section of wire=
Now,
21
YY=
(
)()
( )( )
2211LA
FL
LA
FL
D
=
D
Þ
Since lo
ad and length are same for both
2
2
21
2
1
LrLrD=DÞ,
4
r
r
L
L
2
1
2
2
1
=
÷
÷
ø
ö
ç
ç
è
æ
=
÷
÷
ø
ö
ç
ç
è
æ
D
D
12
L : L 4:1D D=
11. (d) 2(1)Y= h +s
2.4 2 (1 ) 1.2 1 0.2h= h +sÞ = +sÞs=
12. (b) Bulk modulus
V
PV
)V/V(
P
B
D
-
=
D
-
= ....(1)
and T.V
.3TVV
a=Dg=D or
.T.3
1
V
V
a
=
D
-
...(2)
From e
qs. (1) and (2), )T.3/(PBa= or
B3
P
T
a
=
13. (a)
P
V/V
B
1
K
D
== . Here, P = 100 atm,

K = 4 × 10
–5
and V = 100 cm
3
.
Hence, DV = 0.4 cm
3
14. (a)
I
=d
Y3
W
3
l
, where W = load, l = length of beam and I is
geometrical moment of inertia for rectangular beam,
3
bd
Ι
12
= where b = breadth and d = depth
For square beam b = d
\
4
1
b
Ι
12
=
For a beam of
circular cross-section,
4
2
πr
Ι
4
æö
=ç÷
èø
\
4
3
4
3
1
bY
W4
bY3
12W ll
=
´
=d (for sq. cross section)
and
)r(Y3
W4
)4/r(Y3
W
4
3
4
3
2
p
=
p
=d
ll
(for circular cross-section)
Now
p
=
p
p
=
p
=
d
d 3
)r(
r3
b
r3
22
4
4
4
2
1
(
22
bπr=Q i.e., they have same cross-sectional area)
15. (d)
Weight of load Weight of puller
Compressive stress
area
+
=
5
620A d 10 5 10
1.6 10
area
´´ +´
Þ =´
Where d i
s the density.
35
620A 2.5 10 10 5 10
1.6 10
A
´ ´ ´ +´
=´
ie
, 5× 10
5
= 1.1× 10
6
A or A = 0.45m
2
16. (a)
ll 2
r
C,
2
)rr(
C
4
2
4
1
4
2
1
ph
=
-ph
=
Initial volum
e = Final volume
\ rp=r-p
ll
22
1
2
2
r]rr[
)rr)(rr(rrrr
1212
22
1
2
2
2
-+=Þ-=Þ
)98.702.8)(98.702.8(r
2
-+=Þ
cm8.0rcm64.004.016r
2
=Þ=´=Þ
4
44
4
4
1
4
2
2
1
]8.0[
]98.7[]02.8[
r
rr
C
C -
=
-
=\
17. (a
) Weights without and with liquid proportional to r
b
and r
b
– r
l
.
18. (b)
19. (c)
d
For a beam, the
depression at the centre is given by,
3
fL
4Ybd
æö
d=
ç÷
èø
[f, L, b, d are constants for a particular beam]
i.e.
1
Y
dµ

260. 256 PHYSICS
20. (c)
6
3
250 9.8
F/A50 10
Y
/0.5 10
2
-
-
´
´
==
D ´ll

36
105.0
2
1050
8.9250
--
´
´
´
´
=
Þ
210
m/N106.19´
21. (d) Work don
e by constant force in displacing the object
by a distance l.
= change in potential energy1
2
=´stress × strain ×volume
1
22
=´ ´´´=
llFF
AL
AL
22. (c) Volume V = cross sectional A × length l or V = Al
Strain =
Elongation
Original length
Y
l
=
Young’s modulus Y =
Stress
Strain
Work done, W =
1
stress × strain × volume
2
´
21
(strain)
2
W Y Al=´´´

2
2211
22
y YA
Y Al y Wy
ll
æö æö
=´´ ´= Þµç÷
ç÷
èø èø
23. (a) Energy stored per unit volume =
1
–
22
F l Fl
A L AL
æ öæö
=
ç ÷ç÷
è øèø
24. (c
)
6100
10
0.01/100
K== atm
= 10
11
N/m
2
= 10
12
dyne/cm
2
25. (
b) Stress = 1 kg wt/mm
2
= 9.8 N/mm
2
= 9.8 × 10
6
N/m
2
.
112
Y 1 10 N/m=´ , 100 ?
D
´=
l
l
Stress Stress
Y
Strain /
==
Dll
\
6
11
Stress 9.8 10
Y 1 10
D´
==
´
l
l
116
100 9.8 10 100 10
-D
´=´ ´´
l
l
= 9.8 × 10
–3
= 0.0098 %
26
. (d) At extension l
1
, the stored energy =
2
1
1
2
Kl
At extension l
2
, th e stored energy =
2
2
1
2
Kl
Work done in increasing its extension from l
1
to l
2
22
21
1
(–)
2
Kll=
27. (d)
3
9200 10 10
2 10
/ / 0.1/100
P hg
K
VV VV
D
r ´´
=== =´
DD
28. (a) When same stress is applied at two different
temperatures, the increase in length is more at higher
temperature. Thus T
1
> T
2
.
29. (d)
30. (b)
11
22
W F l mgl= ´ ´=

31
10 10 1 10 0.05
2
J
-
=´´´´ =
31. (c)
We know that Young's modulus
l
L
r
F
Y
2
´
p
=
Since Y, F are same fo
r both the wires, we have,
2
2
2
21
1
2
1
L
r
1L
r
1
ll
= or,
2
2
1
1
2
2
2
1
Lr
Lr
´
´
=
l
l
=
2
2
1
1
2
2
L)2/D(
L)2/D(
´
´
or,
8 1
L2
L
)D2(
D
LD
LD
2
2
2
2
2
2
2
2
1
1
2
2
2
1
=´=
´
´
=
l
l
So, 8:1:
21
=ll
32. (
c)
Y
l
A
Wire (1)
Wire (2)
3A Y
l/3
As shown in the figure, the wires will have the same
Young’s modulus (same material) and the length of
the wire of area of cross-section 3A will be l/3 (same
volume as wire 1).
For wire 1,
/
/
=
Dl
FA
Y
x
...(i)
For wire 2 ,
'/3
/( / 3)
FA
Y
x
=
Dl
...(ii)
From (i) and (ii) ,
'
33
´=´
DD
llFF
AxAx
'9Þ=FF
33. (c) Young’s modulus of rubber,
rubber
Y
F
F YA.
A
D
= ´ Þ=
D
ll
ll
On putting the values from question,
2
268
1010
1051025105
F
-
--
´
´´´´´
=
N6250102525
12
=´´=
-
kinetic energy
= potential energy of rubber
2
mv
2
1
lD=F
2
1
2
3
F 6250 5 10
v
m 5 10
-
-
D ´´
==
´
l
=62500
= 25 × 10 = 25
0 m/s

261. 257Mechanical Properties of Solids
34. (b) The atoms when brought from infinity are attracted
due to interatomic electrostatic force of attraction. At
point B, the potential energy is minimum and force of
attraction is maximum. But if we bring atoms closer
than x = B, force of repulsion between two nuclei starts
and P.E. increases.
x
B
C
P.E.
A
35. (c)
F
x
F
com
F
ext
From the figure, it is clear that
F
com
< F
ext.
36. (d)
4
. Constant
2
r
rC
L
phq
=q==

44
00
( ) ( / 2 ) ( ')
2 2( / 2)
rr
ll
ph q-q p h q -q
Þ=
l/2
l
q
0
q
q¢ = 0
00
0
() 8
2169
q-qq
Þ = Þq=q
37. (a)
6
3
10 100
34
33 10 10
S
Lm
dg
== ==
´´
38. (b)
r 6mm 30
r 0.18
1m
q ´°
q= fÞf= = = °l
l
39. (a)F YA YA
D
= = aDq
l
l
= 2 × 10
11
× (1 × 10
–6
) × 1.1 × 10
–5
× (10) = 22 N
40. (a) 2
mg
Y
( /4)dx
=
p
l
......... (1)
max
dY m dx
2
Y m dx
D D DDæö
=+++
ç÷
èø
l
l
m = 20.0 kg Þ Dm = 0.1 kg
l = 125cm Þ Dl = 1 cm.
d = 0.050 cm. Þ D d = 0.001 cm
x = 0.100 cm. Þ Dx = 0.001 cm.
max
0.1kg 1cm
dY
20.0kg 125cm
100% 6.3%
0.001cm 0.001cmY
2
0.05cm 0.100cm
æö
+
ç÷æö
= ´=
ç÷ ç÷
èø
+´+ç÷
èø
41.
(d) Given,
1
2
,
l
a
l
=
1
2
r
b
r
=,
1 2
Y
c
Y
=
Brass
T
T
2g
Steel
T
2g
Let Young’s modulus of steel be Y
1
, and that of brass
be Y
2
\
11
1
11
Fl
Y
Al
=
D
…(i)
and
22
2
22
Fl
Y
Al
=
D
…(ii)
Dividing equat
ion (i) by equation (ii), we get1 1212
2 2121
Y FAll
Y FAll
D
=
D
…(iii)
Force on st
eel wire from free body diagram
1
(2 ) NewtonTFg==
Force on br
ass wire from free body diagram
21
2 4 NewtonFTTgg== +=
Now putting the value of F
1
, F
2
in equation (iii), we
get
2
12 1 22
22
2 2 111
2 11
...
42
Yr lllg
a
Yglll rb
æö éùæö æöæöpDD æö
== ç÷ êú ç÷ç÷ ç÷ ç÷
èøèøDD èø èøpèø ëû
42. (d)
If side of the cube is L then
3
3
dV dL
VL
VL
=Þ=
\ % change in vo
lume = 3 × (% change in length)
= 3 × 1% = 3%
\ Bulk strain
0.03
V
V
D
=
43. (b)
2
2
4 9.8
19.6 10
2 10
F
K
x
-
´
== =´
´
Wor
k done =
221
19.6 10 (0.05)2.45
2
J´´=
44. (d)
F =
YA
L
× l
So, ext
ension, lµ
L
A
µ
2
L
D
[QF and Y are
constant]
l
1
µ
2
100
1
µ 100 andl
2
µ
2
200
2
µ 50
l
3
µ
2
300
3
µ
100
3
and l
4
µ
50
1
4
µ 200
The r
atio of
2
L
D
is maximum for case (d).
Hence, option (d) is correct.

262. 258 PHYSICS
45. (a)
Percentage elongation in the wire
F
AY
=
l
4
2 11
6.28 10 (1) 1
1000(0.01) 2 10
´´
==
p ´´
46. (
a) Work done
211
Stress Strain (Str ain)
22
Y=´ ´ =´´
Since, ela
sticity of steel is more than copper, hence
more work has to be done in order to stretch the steel.
47. (b) Stress is defined as internal force (restoring force) per
unit area of a body. Also, rubber is less elastic than
steel, because restoring force is less for rubber than
steel.
48. (d) 49. (d) 50. (a)
EXERCISE - 3
Exemplar Questions
1. (b) As liquid is ideal so no. frictional force exists hence,
tangential forces are zero so there is no stress
developed.
2. (d) As we know that,
Breaking stress =
Breaking force
Area of cross-section
..(i)
When length
of the wire changes (or by reducing half)
area of cross-section remains same.
Hence, breaking force will be same because breaking
stress does not depend on length.
3. (d) As we know that, length of a wire when the temperature
increased
0
(1)
tLLT= + aD
where DT is chan ge in the temperature.
L
0 is original length,
a is coefficient of linear expansion and
L
t is length at temperature T.
Now,
00t
LLLLTD = - = aD
Now, Young's modulus (Y)
Stress
Strain
=
00
0
1FL FL
ALALTT
==µ
´D aDD
As,
1
Y
T
µ
D
So, if temperature increass DT increases, hence
Young's modulus of elasticity (Y) decreases.
4. (c) When a spring is stretched by applying a load to its
free end. Clearly the length and shape of the spring
changes. So strain produced when change in length
corresponds to longitudinal strain and change in shape
corresponds to shearing strain.
5. (b) As we know that,
The Young's modulus
Stress/
Strain/
FAFL
Y
LLAL
= = =´
DD
22
4
( / 2)
F L FL
LD DL
= ´=
Dp pD
244FL FL
DD
LY LY
= Þ=
pD pD
If F and
L
LD
are constants.
So,
1
D
Y
µ
Hence, we can fin
d ratio ascopper iron
iron copper
D Y
DY
=
6. (a) Consider
the given diagram
m
90°–q90°–q
qq
x
O
BA C
L L
So, change in length
00
()()L A B AC CBD=+-+
= 2BO – 2AC
= 2 [B
O – AC] (\ AO = BO, AC = CB)

2 2 1/2
2[( )]xLL=+-

1/2
2
2
211
x
L
L
éù
æö
êú=+-ç÷
ç÷
êú
èø
ëû
22
2
1
211
2
xx
LL
LL
éù
D» + -=êú
ëû
[]xL<<Q
22
2
/
Strain =
22 2
LxLx
LL L
D
==
7. (c) Let u
s consider the free body diagram of the
rectangular frame
Tsinq Tsinq
T
TcosqTcosq
qq
m
mg
Net forces actin
g on frame will be zero.
So, Balancing vertical forces
2T sinq – mg = 0 [T is tension in the string]
2T sinq = mg ...(i)
Total horizontal force
cos cos 0TT= q- q=

263. 259Mechanical Properties of Solids
Now from Eq. (i),
2sin
mg
T=
q
max
min2sin
mg
T=
q
As mg is co
nstant then1
sin
Tµ
q
min min
sin00q=Þq=
No option matches with q = 0°
min
max2sin
mg
T=
q
(since
, sinq
max = 1)
maxsin 1 90q =Þq=°
So tension is all three cases are different rejects option
(a)
For minimum tension q must be 90° i.e. sin q = 1
Hence, tension is the least for the case (b).
8. (d) A mass M is attached at the centre or midpoint of rod
of rubber and steel. As the mass is attached to both
the rods, both rod will be elongated as shown in
figures but due to different elastic properties of material
rubber changes shape also.
M M
(Steel)(Rubber)
As the Yo
ung's modulus of rigidity for steel, is larger
than rubber, so strain
L
L
D
for rubber is larger than
steel for same stress.
NEET/AIPMT (2013-2017) Questions
9. (d) F =
YA
L
× l! So, extension, lµ
L
A
µ
2
L
D
[QF and Y are constant]
l
1
µ
2
100
1
µ 100 and l
2
µ
2
200
2
µ 50
l
3
µ
2
300
3
µ
100
3
and l
4
µ
50
1
4
µ 200
The ra
tio of
2
L
D
is maximum for case (d).
Hence, option (d) is correct.
10. (c) From formula,
Increase in length
2
4FL FL
L
AY DY
D==
p
2
S S C CS
C C S SC
L F D YL
L F D YL
æöD
=ç÷
D
èø
=
2
711
5
q
ps
æöæö
´ç÷ç÷
èøèø
=
2
7
(5)
q
sp
11. (b) As Y =
F
FA
AY
ÞD=
D
l
l
l
l
ButV = Al so A =
V
l
Therefore Dl =
2
2F
VY
µ
l
l
Hence graph of Dl versus l
2
will give a straight line.
12. (b) Compressibility of water,
K = 45.4 × 10
–11
Pa
–1
density of water P = 10
3
kg/m
3
depth of ocean, h = 2700 m
We have to find
V
?
V
D
=
As we know, compressibility,
K =
1 ( V / V)
(P gh)
BP
D
= =r
So,(DV/V
) = Krgh
= 45.4 × 10
–11
× 10
3
× 10 × 2700 = 1.2258 × 10
–2
13. (a) Young's modulus Y =
W
.
A
l
lD
12
12
WW
YY
=
[Q A, l, Dl same for both brass and steel]
Brass
A, , lDl
Steel
l, A, Dl
Y
2 Y
1
11
22
WY
2
WY
== [Y
steel
/Y
brass
= 2 give n]
14. (c) Bulk modulus is given by
B =
P
V
V
Dæö
ç÷
èø
or
VP
VB
D
=
RP
3
RB
D
= (here,
R
R
D
= fractional decreases in radius)
RP
R 3B
D
Þ=

264. 260 PHYSICS
FLUIDS
Fluid is s
omething that can flow. All liquids and gases are fluids.
The force exerted normally at a unit area of the surface of a fluid
is called fluid pressure.
i.e., P =
F
a
Its S.I. unit is Nm
–2
or Pascal. Its dimensions are [ML
–1
T
–2
].
PASCAL'S LAW AND ITS APPLICATIONS
Pascal's law : Pressure in a fluid in equilibrium is the same
everywhere, if the effect of gravity is neglected.
Another form of Pascal's law : The excess pressure, applied
anywhere in a mass of confined incompressible fluid is
transmitted by the fluid in all directions without being
diminished in magnitude.
Applications of Pascal's Law
Hydraulic lift : Its working is based on Pascal's law. A piston of
small cross-sectional area (a) exerts a force (f) on the liquid.
Applied force = f
F = pA
f = pa
Area = aArea = A
The pressure is t
ransmitted undiminished to the larger cylinder
of cross-sectional area A.
==
fF
P
aA
Þ=
A
Ff
a
Hydraulic brakes also work on pascal's law.
ATMOSPHERIC, HYDROSTATIC AND GUAGE PRESSURE
Atmospheric pressure : The atmosphere exerts pressure on the
earth's surface. The atmospheric pressure at sea level is given by
P
0
= 1.01 × 10
5
Pa
Hydrostatic pressure : The hydrostatic pressure at a depth h
below the surface of a fluid is given by
P = hrg
where r is the density of the fluid, g acceleration due to gravity
and h is the depth of the liquid column.
Gauge pressure : The pressure at any point in a fluid is equal to
the sum of the atmospheric pressure acting on its surface and the
hydrostatic pressure due to the weight of the fluid above that
point which is at a depth h below the surface of the fluid.
The gauge pressure is given by
P = P
0
+ hrg
or, P – P
0
= hrg
BUOYANCY AND ARCHIMEDES' PRINCIPLE
Buoyancy
If a body is partially or wholly immersed in a fluid, it experiences
an upward force due to the fluid surrounding it. The phenomenon
of force exerted by fluid on the body called buoyancy and the
force is called buoyant force or upthrust.
A body experiences buoyant force whether it floats or sinks, under
its own weight or due to other forces applied on it.
Archimedes' Principle
When any body is immersed (totally or partially) in a liquid it
appears to lose part of its weight and the apparent loss of weight
is equal to the weight of liquid displaced.
Let a body of weight W is immersed in a fluid and W' is upthrust
on it then
(i) if W > W', then body will sink.
(ii) if W = W', then the body floats with whole or some part of
its volume inside the fluid.
Let V be the volume of a body of density d and V' be the
volume of liquid of density r displaced. If the body floats
then Vd = V¢r
\
V
'Vd
=
r
gives the fraction
of the volume inside the liquid
in which the body floats.
Also, body immersed in a fluid experiences an upward buoyantforce equivalent to the weight of the fluid displaced by it.
The proof of this principle is very simple. Imagine a body of
arbitrary shape completely immersed in a liquid of density r as
10
Mechanical
Properties of Fluids

265. 261Mechanical Properties of Fluids
shown in the figure (a). A body is being acted upon by the forces
from all directions. Let us consider a vertical element of height h
and cross-sectional area dA (as shown in the figure (b)).

h
2
dA
h
h
1 F
1
F
2
(a) (b)
Th
e force acting on the upper surface of the element is F
1
(downward) and that on the lower surface is F
2
(upward). Since
F
2
> F
1
, therefore, the net upward force acting on the element is
dF = F
2
– F
1
It can be easily seen from the figure (b), that
11
F ( gh )dA=r and
22
F ( gh )dA=r so dF g(h)dA=r
Also, h
2
– h
1
= h
and d (dA) = dV
\ The net upward force is
F gdV Vg= r =rò
Hence, for
the entire body, the buoyant force is the weight of the
volume of the fluid displaced.
The buoyant force acts through the centre of gravity of the
displaced fluid. Keep in Memory
1.The
pressure is perpendicular to the surface of the fluid.
2.The upthrust on a body immersed in a liquid does notdepend on the mass, density or shape of the body. It onlydepends on the volume of the body.
3.The weight of the plastic bag full of air is same as that ofthe empty bag because the upthrust is equal to the weightof the air enclosed.
4.The cross-section of the water stream from a top decreasesas it goes down in accordance with the equation ofcontinuity.
5.We cannot sip a drink with a straw on the moon, becausethere is no atmosphere on the moon.
6.The line joining the centre of gravity and centre of
buoyancy is called central line.
7. Metacenter - is a point where the vertical line passing
through the centre of buoyancy intersects the central line.
8.The floating body is in stable equilibrium when the
metacenter is above the centre of gravity (centre of gravity
is below the centre of buoyancy).
9.The floating body is in the unstable equilibrium when the
metacenter lies below the centre of gravity (centre of gravity
is above the centre of buoyancy).
10.The floating body is in the neutral equilibrium when centre
of gravity coincides with the metacenter (centre of gravity
coincides with the centre of buoyancy).
11.The wooden rod cannot float vertically in a pond of water
because centre of gravity lies above the metacenter.
12.(i) If a body just floats in a liquid (density of the body is
equal to the density of liquid) then the body sinks if it
is pushed downwards.
(ii) If two bodies have equal upthrust when just immersed
in a liquid, both will have the same volume.
(iii) If a person floats on his back on the surface of water,
the apparent weight of person is zero.
13.The hydrometer can be used to measure density of the
liquid or fluid.
Relative Density (or Specific Gravity)
Liquids may be treated as incompressible. Hence their density
may be assumed to be constant throughout.
Relative density =
Weight of substance in air
Weight of equal volume of water

=
Weight of substance in air
Loss of weightin water

4º
=
Density of substance
Density of water at C
Density in SI system = 1000 × density in the cgs system.
(i) The density of liquid of bulk modulus B at a depth h is
given by 01
h
gh
B
ræö
r=r+
ç÷
èø
where
0
r is the density of
liquid on its surface and r is the average density of liquid.
(ii) The density of liquid changes with pressure as
0
1
h
P
PP
B
Dæö
=+ç÷
èø
where =DP change in pressure and B = bulk modulus of
liquid.
(iii) If two liquids of masses m
1
, m
2
and densities r
1
, r
2
are
mixed together, then the density of the mixture is given by
12
12
12
mm
mm
+
r=
+
rr
And if m
1
= m
2
but dif ferent densities are mixed together,
then the density of the mixture is harmonic mean of the
densities.
i.e.,
21
212
r+r
rr
=r or ú
û
ù
ê
ë
é
r
+
r
=
r
1
21
11
2
1
(iv) If t
wo drops of same volume but different densities are
mixed together, then the density of the mixture is the
arithmetic mean of the densities.
i.e.,
2
21r+r
=r (as
10 20
00
VV
VV
r +r
r=
+
)
SURFACE TE
NSION
It is defined as the force per unit length acting at right angleson either side of an imaginary line drawn on the free surface ofthe liquid.
i.e., =
l
F
S
The surf
ace tension is also defined as the work required to
increase unit area of that liquid film.Its SI unit is N/m or J/m
2
and dimensions are [ML
0
T
–2
].

266. 262 PHYSICS
Keep in Memory
1.The liquid sur
face always acquires minimum surface area
due to surface tension (ST). So, the small droplet of any
liquid is always spherical.
2.The ST is a molecular phenomenon as ST is due to
'cohesion' between the molecules of a liquid.
3.The force of attraction between the molecules of the same
substance is called a cohesive force and that between
molecules of different substance is called adhesive force.
4.The molecular range is the maximum distance (10
–9
m) upto
which the molecules attract each other.
5.In general the ST of liquids decreases with increase in
temperature but the ST of molten Cadmium and Copper
increases with increase in temperature.
6.If the impurity is completely soluble then on mixing it in the
liquid, its surface tension increases. For example on
dissolving ionic salts in small quantities in a liquid, its
surface tension increases. On dissolving salt in water, its
surface tension increases.
7.If the impurity is partially soluble in a liquid, then its surface
tension decreases. For example on mixing detergent or
phenol in water its surface tension decreases.
8.On increasing temperature surface tension decreases. At
critical temperature and boiling point it becomes zero. Surface
tension of water is maximum at 4°C.
S.T.
4°CTemp.
S.T.=0
C
of water
ANGLE
OF CONTACT
The angle between the tangent to the liquid surface and the
tangent to the solid surface at the point of contact (inside the
liquid) is known as angle of contact.
q
q < 90º
Water
Glass Glass
Mercury
q
q > 90°
Some values of angle of contact of solid and liquid :
Pair of s urface Angle of contact
Pure water and glass 0°
Silver an d glas s 90°
Alcohol and glass 138°
Normal water & glass 8°
Mercury & glass 135°
Adhesion > cohesion Adhesion = cohesion Adhesion < cohesion
1. Liquid will wet the solid Critical Liquid will not wet the solid
2. Meniscus is concave Meniscus is plane Meniscus is convex
3. Angle of contact is acute (q < 90°) Angle of contact is 90º Angle of contact is obtuse (q > 90°)
4. Pressure below the menisucs is lesserPressure below the meniscusPressure below the meniscus is more
than above it by (2T/r), i.e. p = p
0
–
2T
r
is same as above it, i.e. p = p
0
then above it by (2T/r), i.e., p = p
0
+
2T
r
5. In capillary tube liquid will ascend.No capillary rise In capillary tube liquid will descend.
Keep in Memory
1.The valu
e of angle of contact lies between 0º and 180º. For
pure water and glass it is 0º, for tap water and glass it is 8º
and for mercury and glass it is 135º.
2.For all those liquids which wet the solid surface and which
rise up in a capillary tube, the angle of contact is an acute
angle (q < 90°), e.g. water and glass.
3.For all those liquids which do not wet a solid surface and
which depress in a capillary tube, the angle of contact is an
obtuse angle (q > 90°), e.g. glass and mercury.
4.For all those liquids which neither rise nor get depressed in
a capillary tube, the angle of contact is right angle (q = 90°),
e.g. silver and water.
5.Angle of contact depends on impurities, water proofing
agent, surface in contact and temperature. Angle of contact
q
C
µ T where T is the temperature.
Capillarity :
The phenomenon of rise or fall of liquids in capillary tube is
known as capillarity.

267. 263Mechanical Properties of Fluids
The rise or fall of a liquid in a capillary tube is given by
2cos2TT
h
rg Rg
q
==
rr

2T
hg
R
Þ r=
q
q
R
r
water
where T = surface tension, q = angle of contact, r = density of
liquid, r = radius of capillary tube, R = radius of meniscus.
(i) If capillary tube is of insufficient length l (i.e. l < h), then the
liquid rises to a full height h with radius R' such that hR = lR'
(ii) When the capillary tube is tilted from vertical by an angle
a, then the vertical height h of liquid column remains the
same. The length of liquid in capillary increases such that
h
h
cos
¢
=a or
a
=¢
cos
h
h .
Accordin
g to Zurin's law capillary rise
1
h
r
µ
where r is the radius of the capillary tube.
Keep in Memory
1.Wor
k done in forming a liquid drop of radius R, surface
tension T is, W = 4pR
2
T.
2.Work done in forming a soap bubble of radius R, surfacetension T is, W = 2×4pR
2
T = 8pR
2
T.
3.When n no. of smaller drops of liquid, each of radius r,surface tension T are combined to form a bigger drop of
radius R then 1/3
Rnr= .
4.The surface are
a of bigger drop = 4pR
2
= 4pn
2/3
r
2
. It is lessthan the area of n smaller drops.
SHAPE OF LIQUID MENISCUS :
The pressure on the concave side is always greater than the
pressure on the convex side.
q < 90º
Water
A
P = Atmospheric
0
pressure
Concave meniscus
q > 90º
Mercury
A
P = Atmospheric
0
pressure
Convex meniscus
P
A
=P
0
–2T/r P
A
=P
0
+2T/r
(r is radius of
meniscus)
Excess pressure of liquid drop and soap bubble :(i) Excess of pressure for spherical soap bubble is
p = 4T/r and excess of pressure for liquid drop and air bubble
in a liquid is p = 2T/r.
(ii) (a) Excess of pressure within a cylindrical liquid drop p =
T/R
(b) Excess of pressure within a cylindrical soap bubble p
= 2T/R
where T = surface tension, R = radius of the cylindricaldrop.
Keep in Memory
1.Wor
k done in breaking a liquid drop of radius R into n
equal small drops = T)1n(R4
3/2
-p
1
; where
T = surface tension.
2.Work done in breaking a soap bubble of radius R into n
equal small drops = T)1n(R8
3/2
-p
1
; where
T = surface tension.
Example 1.
A solid uniform ball having volume V and density
r floats at
the interface of two unmixable liquids as shown in fig. The
densities of the upper and the lower liquids are
r
1
and
r
2
respectively, such that
r
1
<
r < r
2
. What fraction of the
volume of the ball will be in the lower liquid ?
r
r
1
r
2
Solution :
Let V
1
and V
2

be the volumes of the ball in the upper and
lower liquids respectively. So V
1
+ V
2
= V.
As ball is floating in the two liquids ; weight of the ball =
upthrust on ball due to two liquids
i.e., V r g = V
1
r
1
g + V
2
r
2
g ;
or V r = V
1
r
1
+ (V – V
1
) r
2
;
or
VV
21
2
1 ÷
÷
ø
ö
ç
ç
è
æ
r-r
r-r
=
\ Fraction
in the upper liquid =
21
21
V
V
r-r
r-r
=
Fraction in
the lower liquid=
1
V
1
V
-
21
12 12
1
r-r r -r
=-=
r-r r-r
Example 2.
A piece of
cork is embedded inside of block of ice which
floats on water. What will happen to the level of water
when all the ice melts?
Solution :
Let , M = mass of the block of ice, m = mass of piece of cork
and V = Volume of water displaced.
Now (M + m) = V × 1 = V ...(1)
When the ice melts, let it be converted into
V' c.c. of
wa
ter.
Also
M V' 1 V'= ´=
The piece of cork floats on the surface of water when all ice
melts. Let the cork displaces a volume V '' c.c. of water..
Then m V" 1 V"= ´=
If V
1
be the volume of water displaced by melted ice and
cork, then
() 1
Mm V'V"V+=+= ...(2)

268. 264 PHYSICS
From eqns
. (1) and (2), V = V
1
Hence, no change in the level of water.
Example 3.
Two substances of densities
r
1
and
r
2
are mixed in equal
volume and the relative density of mixture is 4. When they
are mixed in equal masses, the relative density of the
mixture is 3. Determine the values of
r
1
and
r
2
.
Solution :
When the substances are mixed in equal volumes, then
4V2VV
21
´=r+r ... (1)
When the two substances are mixed in equal masses, then
3
m2mm
21
=
r
+
r
...(2)
From eq
. (1), 8
21
=r+r ... (3)
From eqn. (2)
3
2
or
3
211
21
21
21
=
rr
r+r
=
r
+
r
or
12or
3 28
21
21
=rr=
rr
. ...(4)
Now
2/1
21
2
21
21
]4)[( rr-r+r=r-r
4]4864[
2/1
=-= ...(5)
Solving eqns. (3) and (5), we get 6
1
=r and
2
ρ2=]
Example 4.
A sealed tank
containing a liquid of density
r moves with
a horizontal acceleration a, as shown in fig. Find the
difference in pressure between the points A and B.
C A
B
h
a
l
Solution :
Since p
oints A and C are in the same horizontal line but
separated by distance l and liquid tank is moving
horizontally with acceleration a, henceP
C
– P
A
= lra or P
C
= P
A
+ lra
Points B and C are vertically separated by h\ P
B
– P
C
= h r g
orP
B
– (P
A
+ l r a) = h r g
or P
B
– P
A
= h r g + l ra
Example 5.
Calculate the excess pressure within a bubble of air ofradius 0.1 mm in water. If the bubble had been formed 10 cmbelow the water surface when the atmospheric pressurewas 1.013 × 10
5
Pa, then what would have been the total
pressure inside the bubble?
Solution :
Excess pressure within air bubble
=
2T
r
=
3
3
2 73 10
0.1 10
-
-
´´
´
= 1460 Pa
The press
ure at a depth d, in liquid P = hdg. Therefore, the
total pressure inside the air bubble is
P
in
= P
atm
+ hdg +
2T
r
orP
in
= 1.013 × 10
5
+ 1 0 × 10
–2
× 10
3
× 9.8 + 1460
= 101300 + 980 + 1460
= 103740 = 1.037 × 10
5
Pa.
Example 6.
A capillary of the shape as shown is dipped in a liquid.
Contact angle between the liquid and the capillary is 0°
and effect of liquid inside the meniscus is to be neglected. T
is the surface tension of the liquid, r is radius of the meniscus,
g is acceleration due to gravity and
r is density of the
liquid then determine the height h in equilibrium.
h
Solution :
As weight
of liquid in capillary is balanced by surface tension,
then
2
1
T2r rhg´p=pr (for a tube of uniform radius r)
1
2T
h
rg
=
r
r
h
1
But weight of liquid in tapered tube is more than uniform tube
of radius r, then in order to balance h < h
1
.
2T
h
rg
<
r

r
Example 7.
A hydraulic
automobile lift is designed to lift car with a
maximum mass of 3000 kg. The area of cross-section of thepiston carrying the load is 425 cm
2
. What maximum pressure
would the smaller piston have to bear ?
Solution :
Here mass of car = 3000 kg.
Area of cross section of larger piston
= 425 cm
2
= 425 × 10
–4
m
2
.
\ The maximum pressure that the smaller piston would
have to bear
=
4
Weight of car 30 00 9.8
Area of cross-section425 10
-
´
=
´
52
6.92 10 Nm
-
=´

269. 265Mechanical Properties of Fluids
FLOW OF LIQUIDS
The motion of fluids are of following four types :
(i) Streamline motion : When fluid in motion, if fluid particles
preceeding or succeeding a fluid particle follow the same
path, then the path is called streamline and the motion of
the fluid is called streamline motion. This type of motion
takes place in non-viscous fluids having very small speed.
Principle of continuity : When incompressible, non-viscous liquid
flows in non-uniform tube then in streamline flow product of
area and velocity at any section remains same.
The mass of liquid flowing in equals the mass flowing out.
i.e., m
1
= m
2
or,v
1
A
1
r
1
Dt = v
2
A
2
r
2
Dt ..... (1)
v
1
A
1
v
2
A
2
P
Q
As we have considered the fluid incompressible thus,
v
1
A
1
= v
2
A
2
or Av = constant....(2)
(Since r
1
= r
2
)
Equations (1) and (2) are said to be as equation of continuity.
(ii) Steady state motion : In a liquid in motion, when liquid
particles, crossing a point, cross it with same velocity, thenthe motion of the liquid is called steady state motion. This
type of motion takes place in non-viscous liquids havingvery small speed.
(iii) Laminar motion : Viscous liquids flow in bounded region
or in a pipe, in layers and when viscous liquid is in motion,different layers have different velocities. The layers incontact with the fixed surface has least velocity and the
velocity of other parallel layers increases uniformly and
continuously with the distance from the fixed surface to
the free surface of the liquid. This is called laminar motion
of the liquid.
(iv) Turbulent motion : When the velocity of a liquid is irregular,
haphazard and large, i.e. Beyond a limiting value called
critical velocity the flow of liquid loses steadiness then
the motion of the liquid is called turbulent motion.
Critical velocity V
c

K
r
h
=
r
Here h is called coefficient of viscosity.
VISCOSITY
The internal friction of the fluid, which tends to oppose relative
motion between different layers of the fluid is called viscosity.
The viscous force between two layers of a fluid of area A having
a velocity gradient
dv
dx
is given by
= -h
dv
FA
dx
where h is called the coefficient of viscosity.
Its S.I. unit is poiseuille or decapoise. The C.G.S. unit is called
poise.
1 decapoise =10 poise.
Effects on Viscosity :
(1) Effect of temperature : On increasing temperature viscosity
of a liquid decreases.
(2) Effect of pressure : On increasing pressure viscosity of a
liquid increases but viscosity of water decreases.
Keep in Memory
1.
The viscosity of gases increases with increase of
temperature, the rate of diffusion increases.
2. The viscosity of liquids decrease with increase of
temperature, because the cohesive force between the liquid
molecules decreases with increase of temperature.
Critical Velocity :It is the maximum velocity of a fluid above which a stream line
flow changes to a turbulent flow, i.e. it is the maximum velocity
of a liquid below which its flow remains streamline.
(i) Reynold's formula for critical velocity is
η
ρ
=
c
N
V
r
or
r
=
h
c
Vr
N
where N =
Reynold's no.,
h = coefficient of viscosity,,
r = density of liquid; r = radius of tube, N = 1000 for narrow
tube.
(ii) (a) If 0 < N < 2000 then the flow is laminar
(b) If 2000 < N < 3000 then flow of liquid is unstable and
may change from laminar to turbulent
(c) If N > 3000, then the flow is turbulent.
(iii) When velocity of fluid is less than its critical velocity then
the flow of liquid is determined by its viscosity, its densityhas no effect on its flow.
(iv) When the velocity of liquid is more than its critical velocity
then its flow is determined by its density, where viscosityhas little effect on its flow. For example lava from volcano ishighly thick, despite that it comes out with high speed.
(v) When
c
VV£ the flow of liquid is streamline and when V
> V
c
then the flow of liquid is turbulent.
1c
c
r
r
V
V 2
2
1
=
for same liqu
id, which is flowing in two tubes of
radii r
1
and r
2
respectively.
(vi) The critical velocity of a liquid with high viscosity and
smaller radius is higher than that of a liquid with low
viscosity and greater radius.
Reynold's Number (N) :
It is pure number which determines the nature of flow of liquid
through a pipe.
2
ηη
===
æö
ç÷
èø
cc
c
VdrV Inertial force/area
N
V Viscous force/area
r
N is a dimensio
nless quantity and carries no unit.

270. 266 PHYSICS
BERNOULLI'S THEOREM
F
or non-viscous, incompressible, streamline flow of fluids the
sum of pressure per unit volume, potential energy per unit volume
and Kinetic energy per unit volume remain constant.
i.e.,
r
p
+ gh +
2
1
v
2
= constant
where r = density
of fluid.
When h = 0 then
2p1
v constant
2
+=
r
Bernoulli's theor
em is strictly applicable for an ideal fluid.
An ideal fluid is one which is (a) incompressible (b) streamline(c) irrotational and (d) non-viscous.
Applications of Bernoulli’s Principle
Dynamic lift :
(i)Wings of aeroplane : The wings of the aeroplane are having
tapering. Due to this specific shape of wings when the
aeroplane runs, air passes at higher speed over it as compared
to its lower surface. This difference of air speeds above and
below the wings, in accordance with Bernoulli's principle,
creates a pressure difference, due to which an upward force
called 'dynamic lift' (= pressure difference area of wing) acts
on the plane. If this force becomes greater than the weight of
the plane, the plane will rise up.

v large, p small
v small, p large
(ii)Ball moving without spin: T he velocity of fluid (air) above
and below the ball at corresponding points is the same
resulting in zero pressure difference. The air therefore, exerts
no upward or downward force on the ball.
(iii)Ball moving with spin: A ball which is spinning drags air
along with it. If the surface is rough more air will be dragged.
The streamlines of air for a ball which is moving and spinning
at the same time. The ball is moving forward and relative to it
the air is moving backwards. Therefore, the velocity of air
above the ball relative to it is larger and below it is smaller.
The stream lines thus get crowded above and rarified below.
This difference in the velocities of air results in the pressure
difference between the lower and upper faces and their is a net
upward force on the ball. This dynamic lift due to spinning is
called Magnus effect.
Some other applications of Bernoulli’s principle :
(i) The action of carburator, sprayer or atomizer based on
Bernoulli’s principle.
(ii) The action of bunsen burner, exhaust pump etc.
(iii) Air foil or lift on aircraft wing works on Bernoulli’s principle.
(iv) Motion of a spinning ball i.e., magnus effect.
(v) Blowing of roofs by wind storms etc. based on Bernoullis
principle.
STOKE’S LAW
When a solid moves through a viscous medium, its motion is
opposed by a viscous force depending on the velocity and shape
and size of the body. The energy of the body continuously
decreases in overcoming the viscous resistance of the medium.
This is why cars, aeroplanes etc. are shaped streamline to minimize
the viscous resistance on them.
The viscous drag on a spherical body of radius r, moving with
velocity v, in a viscous medium of viscosity h is given by
6
= ph
viscous
F rv
This relation is called Stokes' law. .
Importance of Stoke’s law :(i)It is used in the determination of electronic charge with thehelp of milikan’s experiment.
(ii)It accounts the formation of clouds.
(iii)It accounts why the speed of rain drops is less than that ofa body falling freely with a constant velocity from the height
of clouds.
(iv)It helps a man coming down with the help of a parachute.
Terminal Velocity :
When a spherical body is allowed to fall through viscous
medium, its velocity increases till the viscous drag plus upthrust
is equal to the weight of the body. After that body moves with
constant velocity, called terminal velocity.
The terminal velocity is given by
2
2
()
9
= -s
h
R
v dg
whered = density o f body,
s= density of medium,
h
= coefficient of viscosity of medium,
R = radius of the spherical body.
From terminal velocity,
h
a
1
V, i.e. greater the vi
scosity, smaller
is the terminal speed.
Flow of liquid through tube /pipe :
(i) Poiseuille's equation is
4
pr
Q
8
p
=
hl
, where p is the pressure
difference between the two ends of the tubes, r is the
radius, l is the length of the tube and h is the coefficient
of viscosity, Q = rate of flow of liquid.
Equivalent liquid resistance =
2
12
44
12
88hh
+
pp
ll
rr
, when tube
are joined in series.
It means
that liquid flow through capillary tube is similar
to flow of electric current through a conductor i.e., Q (rate
of liquid flow) corresponds to I (rate of flow of charge),
pressure difference similar to potential difference.
(ii) When two tubes are joined in series then the volume of
fluid flowing through the two tubes is the same but the
pressure difference across the two tubes is different. The
total pressure difference, P = P
1
+ P
2
.
(iii) If two tubes are joined in parallel then the pressure
difference across the two tubes is the same but the volume
of fluid flowing through the two tubes is different. The
total volume of the fluid flowing through the tube in one
second is Q = Q
1
+ Q
2
.

271. 267Mechanical Properties of Fluids
VELOCITY OF EFFLUX AND TORRICELLI'S THEOREM
(i)Torricelli's theorem : For liquid filled in a tank upto a
height H having a hole O at a depth h from free level of
liquid through which the liquid is coming out, velocity of
efflux of liquid v = 2gh.
(ii)Time tak
en by the liquid in falling from hole to ground
level
2()-
=
Hh
t
g
H
h
Free
surface
(iii)The horizontal range covered by the liquid
x = horizontal velocity (v) ×time (t)

2()
2
-
=´
Hh
gh
g
2()=-hHh .
For maximum ho
rizontal range, differentiating both side w.r.t.
h and we get h = H/2, the range is maximum.
Keep in Memory
1.T
he cross-section of the water stream from a top decreases
as it goes down in accordance with the equation of
continuity.
2.When a hale blows over a roof, the force on the roof is
upwards.
3.Sudden fall in atmospheric pressure predicts possibility of
a storm.
4. Venturimeter is a device used for the measurement of the
rate of flow of incompressible fluid through a tube. Its
working is based on Bernoulli’s principle.
Example 8.
The vessel of area of cross-section A has liquid to a heightH. There is a hole at the bottom of vessel having area ofcross-section a. Find the time taken to decrease the levelfrom H
1
to H
2
.
Solution :
The average velocity of efflux,
122gH 2gH
v
2
+
=
Let t be the
time taken to empty the tank from level H
1
to
H
2
.
Then,
]HH[Ata
2
Hg2Hg2
21
21
-=´´
+
or t
( )( )
( )( ) ú
ú
û
ù
ê
ê
ë
é
-+
-´-
´
÷
÷
ø
ö
ç
ç
è
æ
´=
2121
2121
HHHH
HHHH
g
2
a
A
( )
21HH
g
2
a
A
-´
÷
÷
ø
ö
ç
ç
è
æ
´=
( )12
A2
HH
ag
æö
=´ ´-
ç÷
èø
Exam
ple 9.
In a test experiment on a model aeroplane in a wind tunnel,
the flow speeds on the upper and lower surfaces of the
wing are 70 m/s and 63 m/s respectively. What is the lift on
the wing if its area is 2.5 m
2
? Take the density of air to be
1.3 kg m
–3
.
Solution :
Let v
1
be the speed and P
1
be the pressure on the upper
surface of the wing, and corresponding values on the lower
surface be v
2
and P
2
respectively.
\v
1
= 70 m/s, v
2
= 63 m/s, A = 2.5 m
2
,
P = 1.3 kg m
–3
.
According to Bernoulli’s theorem
22
1122
11
PvPv
22
+r = +r
22
21 12
1
P P (v v)
2
-=r- =
22
21
1
P P 3 (70 63 )
2
-=´´-
Fo
rce (lift) on the wing =
21
A(P P)-
=
221
2.5 1.3 (70 63 )
2
´´´-
=
1
2.5 1.3 133 7
2
´´´´
3
1.5 10N=´
Example 10.
In Millikan’s oil dr
op experiment, what is the terminal
speed of an uncharged drop of radius 2.0 × 10
–5
m and
density 1.2 × 10
3
kg m
–3
. Take the viscosity of air at the
temperature of the experiment to be 1.8 × 10
–5
Pa s. How
much is the viscous force on the drop at that speed ? Neglect
buoyancy of the drop due to air.
Solution :
Here, r = 2.0 × 10
–5
m, r = 1.2 × 10
3
kg m
–3
,
h = 1.8 × 10
–5
Pa s.
From formula, terminal velocity
2
2r ( )g
V
9
r-s
=
h
Þ
523
5
2 (2 10 )(1.2 10 0) 9.8
V
9 1.8 10
-
-
´´ ´ -´
=
´´

21
5.8 10 ms
--
=´
Now viscous
force on the drop
F = 6phrv
Þ
55222
F 6 (1.8 10 ) (2 10 ) 5.8 10
7
---
=´´´´´´ ´
= 3.93 × 10
–19
N

272. 268 PHYSICS

273. 269Mechanical Properties of Fluids
1.The constant velocity attained by a body while falling
through a viscous medium is termed as
(a) critical velocity (b) terminal velocity
(c) threshold velocity(d) None of these
2.The difference between viscosity and solid friction is/are
(a) viscosity depends on area while solid friction does not
(b) viscosity depends on nature of material but solid
friction does not
(c) both (a) and (b)
(d) neither (a) nor (b)
3.Water is not used in thermometer because
(a) it sticks to glass
(b) its shows anamalous expansion
(c) both (a) and (b)
(d) neither (a) nor (b)
4.Toricelli’s theorem is used to find
(a) the velocity of efflux through an orifice.
(b) the velocity of flow of liquid through a pipe.
(c) terminal velocity
(d) critical velocity
5.Gases do not posses
(a) density (b) surface tension
(c) volume (d) viscosity
6.Paint-gun is based on
(a) Bernoulli’s theorem(b) Archimedes’ principle
(c) Boyle’s law (d) Pascal’s law
7.Water is flowing through a horizontal pipe having a
restriction, then
(a) pressure will be greater at the restriction.
(b) pressure will be greater in the wider portion.
(c) pressure will be same throughout the length of the pipe.
(d) None of these
8.Fevicol is added to paint to be painted on the walls, because
(a) it increases adhesive force between paint and wall.
(b) it decreases adhesive force between paint and wall
molecules.
(c) it decreases cohesive force between paint molecules.
(d) None of these
9.A beaker containing a liquid of density r moves up with an
acceleration a. The pressure due to the liquid at a depth h
below the free surface of the liquid is
(a) h r g (b) h r (g – a)
(c) h r (g + a) (d)
÷
÷
ø
ö
ç
ç
è
æ
-
+
r
ag
ag
gh2
10.The density
of ice is x gram/litre and that of water is y gram/
litre. What is the change in volume when m gram of ice
melts?
(a) mx y (x – y) (b) m/(y – x)
(c)
÷
÷
ø
ö
ç
ç
è
æ
-
x
1
y
1
m
(d) (y – x)/x
11.C
onsider a 1 c.c. sample of air at absolute temperature T
0
at
sea level and another 1 c.c. sample of air at a height where
the pressure is one-third atmosphere. The absolute
temperature T of the sample at the height is
(a) equal to T
0
/3
(b) equal to 3/T
0
(c) equal to T
0
(d) cannot be determined in terms of T
0
from the above data
12.A small ball (menu) falling under graivty in a viscous medium
experiences a drag force proportional to the instantaneous
speed u such that
drag F Ku=. Then the terminal speed of
the ball within the viscous medium is
(a)
mg
K
(b)
K
mg
(c)
K
mg
(d)
2
K
mg
÷
ø
ö
ç
è
æ
13.A cylin
der is filled with non-viscous liquid of density d to a
height h
0
and a hole is made at a height h
1
from the bottom
of the cylinder. The velocity of the liquid issuing out of the
hole is
(a)
0
gh2 (b) )hh(g2
10
-
(c)
1
dgh (d)
0
dgh
14.The termin
al velocity depends upon
(a)
r
1
(b)
2
r
1
(c )
3
r
1
(d)
2
r
15.The velocity of efflux of a liquid through an orifice in the
bottom of the tank does not depend upon
(a) size of orifice
(b) height of liquid
(c) acceleration due to gravity
(d) density of liquid
16.At the boiling point of a liquid, surface tension
(a) is zero
(b) is infinite
(c) is same as that at any other temperature
(d) cannot be determined
17.The surface energy of a liquid drop of radius r is proportional
to
(a)r
3
(b) r
2
(c)r (d)
r
1

274. 270 PHYSICS
18.A liquid is con
tained in a vessel. The liquid-solid adhesive
force is very weak as compared to the cohesive force in the
liquid. The shape of the liquid surface will be
(a) horizontal (b) vertical
(c) concave (d) convex
19.Two liquids drops coalesce to form a large drop. Now,
(a) energy is liberated
(b) energy is neither liberated nor absorbed
(c) some mass gets converted into energy
(d) energy is absorbed
20.A man is sitting in a boat which is floating in pond. If the
man drinks some water from the pond, the level of water in
the pond will
(a) rise a little (b) fall a little
(c) remain stationary(d) none of these
21.A liquid is allowed to flow into a tube of truncated cone
shape. Identify the correct statement from the following.
(a) The speed is high at the wider end and high at the
narrow end.
(b) The speed is low at the wider end and high at the
narrow end.
(c) The speed is same at both ends in a stream line flow.
(d) The liquid flows with uniform velocity in the tube.
22.The rain drops falling from the sky neither injure us nor
make holes on the ground because they move with
(a) constant acceleration
(b) variable acceleration
(c) variable speed
(d) constant terminal velocity
23.The lift of an air plane is based on
(a) Torricelli's theorem
(b) Bernoulli's theorem
(c) Law of gravitation
(d) conservation of linear momentum
24.Surface tension may be defined as
(a) the work done per unit area in increasing the surface
area of a liquid under isothermal condition
(b) the work done per unit area in increasing the surface
area of a liquid under the adiabatic condition
(c) the work done per unit area in increasing the surface
area of liquid under both isothermal and adiabatic condition
(d) free surface energy per unit volume
25.A liquid does not wet the sides of a solid, if the angle of
contact is
(a) Zero (b) Obtuse (more than 90°)
(c) Acute (less than 90°) (d) 90° (right angle)
1.The reading of spring balance when a block is suspended
from it in air is 60 newton. This reading is changed to 40
newton when the block is submerged in water. The specific
gravity of the block must be therefore
(a)3 (b) 2 (c)6 (d) 3/2
2.An ice-berg floating partly immersed in sea water of density
1.03 g/cm
3
. The density of ice is 0.92 g/cm
3
. The fraction of
the total volume of the iceberg above the level of sea water
is
(a) 8.1%(b) 11% (c) 34% (d) 0.8%
3.A boat having a length of 3 metres and breadth 2 metres is
floating on a lake. The boat sinks by one cm when a man
gets on it. The mass of the man is
(a) 60 kg(b) 62 kg (c) 72 kg(d) 128 kg
4.The excess of pressure inside a soap bubble is twice the
excess pressure inside a second soap bubble. The volume
of the first bubble is n times the volume of the second where
n is
(a) 0.125 (b) 0.250(c)1 (d) 2
5.The level of water in a tank is 5m high. A hole of area 1 cm
2
is made in the bottom of the tank. The rate of leakage of
water from the hole (g = 10 m/s
2
) is
(a) 10
–2
m
3
/s (b) 10
–3
m
3
/s
(c) 10
–4
m
3
/s (d) 10
3
m
3
/s
6.A spherical ball of iron of radius 2 mm is falling through a
column of glycerine. If densities of glycerine and iron are
respectively 1.3 × 10
3
kg/m
3
and 8 × 10
3
kg/m
3
. h for glycerine
= 0.83 Nm
–2
sec, then the terminal velocity is
(a) 0.7 m/s (b) 0.07 m/s
(c) 0.007 m/s (d) 0.0007 m/s
7.A cylinder of height 20m is completely filled with water. The
velocity of efflux of water (in ms
–1
) through a small hole on
the side wall or the cylinder near its bottom is
(a) 10 m/s (b) 20 m/s (c) 25.5 m/s (d) 5 m/s
8.8 mercury drops coalesce to form one mercury drop, the
energy changes by a factor of
(a)1 (b) 2 (c)4 (d) 6
9.Water rises to a height of 10 cm in capillary tube and mercury
falls to a depth of 3.1 cm in the same capillary tube. If the
density of mercury is 13.6 and the angle of contact for
mercury is 135°, the approximate ratio of surface tensions of
water and mercury is
(a) 1 : 0.15 (b) 1 : 3(c) 1 : 6(d) 1.5 : 1
10.A vessel with water is placed on a weighing pan and it reads
600 g. Now a ball of mass 40 g and density 0.80 g cm
–3
is
sunk into the water with a pin of negligible volume, as shown
in figure keeping it sunk. The weighting pan will show a
reading
(a) 600 g
(b) 550 g
(c) 650 g
Weighing pan
(d) 632 g

275. 271Mechanical Properties of Fluids
11.Water flows in a stream line manner through a capillary tube
of radius a, the pressure difference being P and the rate flow
Q. If the radius is reduced to
2
a
and th
e pressure is increased
to 2P, the rate of flow becomes
(a) 4Q (b) Q (c)2
Q
(d)
Q
8
12.A rain drop of radius 0.3 mm has a terminal velocity in air =
1 m/s. The viscosity of air is 8 × 10
–5
poise. The viscous
force on it is
(a) 45.2 × 10
–4
dyne (b) 101.73×10
–5
dyne
(c) 16.95 × 10
–4
dyne(d) 16.95 × 10
–5
dyne
13.A big drop of radius R is formed by 1000 small droplets of
water. The radius of small drop is
(a)
10
R
(b)
100
R
(c)
500
R
(d)
1000
R
14.1 m
3
water is b
rought inside the lake upto 200 metres depth
from the surface of the lake. What will be change in the
volume when the bulk modulus of elastically of water is
22000 atmosphere?
(density of water is 1 × 10
3
kg/m
3
atmosphere pressure
= 10
5
N/m
2
and g = 10 m/s
2
)
(a) 8.9 × 10
–3
m
3
(b) 7.8 × 10
–3
m
3
(c) 9.1 × 10
–4
m
3
(d) 8.7 × 10
–4
m
3
15.Horizontal tube of non-uniform cross-section has radii of
0.1 m and 0.05 m respectively at M and N for a streamline
flow of liquid the rate of liquid flow is
M
N
(a) continuously changes with time
(b) greater at M than at N
(c) greater at N than at M
(d) same at M and N
16.There is a hole in the bottom of tank having water. If total
pressure at bottom is 3 atm (1 atm = 10
5
N/m
2
) then the
velocity of water flowing from hole is
(a)400 m/s (b)600 m/s
(c)60 m/s (d) None of these
17.In the figure, the velocity V
3
will be
A = 0.2 m
2
2
V = 4 ms
1
–1
V = 2 ms
2
–1
A = 0.2 m
1
2
A = 0.4 m
3
2V
3
(a) Zero(b) 4 ms
–1
(c) 1 ms
–1
(d) 3 ms
–1
18.1 centipoise is equal to
(a) 1 kg m
–1
s
–1
(b) 1000 kg m
–1
s
–1
(c) 0.1 kg m
–1
s
–1
(d) 0.001 kg m
–1
s
–1
19.If a water drop is kept between two glass plates, then its
shape is
(a) (b)
(c) (d) None of these
20.The fraction of a floating object of volume V
0
and density
d
0
above the surface of liquid of density d will be
(a)
0
0
dd
d
-
(b)
d
dd
0-
(c)
d
d
0
(d)
0
0
dd
dd
+
21.An open v
essel containing water is given a constant
acceleration a in the horizontal direction. Then the free
surface of water gets sloped with the horizontal at an angle
q, given by
(a)
a
g
cos
1-
=q (b)
g
a
tan
1-
=q
(c)
g
a
sin
1-
=q (d)
a
g
tan
1-
=q
22.Two drops o
f the same radius are falling through air with a
steady velocity of 5 cm per sec. If the two drops coalesce,the terminal velocity would be
(a) 10 cm per sec (b) 2.5 cm per sec
(c) 5 × (4)
1/3
cm per sec (d)
35´ cm per sec
23.A tan
k is filled with water upto a height H. Water is allowed
to come out of a hole P in one of the walls at a depth h below
the surface of water (see fig.) Express the horizontal distance
X in terms H and h.
h
P
x
H
(a) )hH(hX-= (b) )hH(
2
h
X -=
(c) )hH(h2X-= (d) )hH(h4X-=
24.A body of dens
ity '
r is dropped from rest at a height h into
a lake of density r where 'r>r neglecting all dissipative
forces, calculate the maximum depth to which the body sinksbefore returning to float on the surface :
(a)
'
h
r-r
(b)
r
r'h
(c)
'
'h
r-r
r
(d)
'
h
r-r
r

276. 272 PHYSICS
25.Two capi
llary of length L and 2L and of radius R and 2R are
connected in series. The net rate of flow of fluid through
them will be (given rate to the flow through single capillary,
X =
L8
PR
4
h
p
)
(a)X
9
8
(b)X
8
9
(c)X
7
5
(d)X
5
7
26.One drop of soap
bubble of diameter D breaks into 27 drops
having surface tension s. The change in surface energy is
(a)2psD
2
(b) 4psD
2
(c)psD
2
(d) 8psD
2
27.In case A, when an 80 kg skydiver falls with arms and legs
fully extended to maximize his surface area, his terminal
velocity is 60 m/s. In Case B, when the same skydiver falls
with arms and legs pulled in and body angled downward to
minimize his surface area, his terminal velocity increases to
80 m/s. In going from Case A to Case B, which of the
following statements most accurately describes what the
skydiver experiences?
(a)F
air resistance
increases and pressure P increases
(b) F
air resistance
increases and pressure P decreases
(c)F
air resistance
decreases and pressure P increases
(d) F
air resistance
remains the same and pressure P increases
28.A ring is cut from a platinum tube 8.5 cm internal and 8.7 cm
external diameter. It is supported horizontally from the pan
of a balance, so that it comes in contact with the water in a
glass vessel. If an extra 3.97. If is required to pull it away
from water, the surface tension of water is
(a) 72 dyne cm
–1
(b) 70.80 dyne cm
–1
(c) 63.35 dyne cm
–1
(d) 60 dyne cm
–1
29.A capillary tube of radius r is immersed in a liquid. The
liquid rises to a height h. The corresponding mass is m.
What mass of water shall rise in the capillary if the radius of
the tube is doubled?
(a)m (b) 2m ( c ) 3m(d) 4m
30.In a satellite moving round any planet, the gravitational
force is effectively balanced. If an ice cube exists there, and
it melts with passage of time, its shape will
(a) remain unchanged
(b) change to spherical
(c) become oval-shaped with long-axis along the orbit
plane
(d) become oval-shaped with long axis perpendicular to
orbit plane
31.An egg when placed in ordinary water sinks but floats when
placed in brine. This is because
(a) density of brine is less than that of ordinary water
(b) density of brine is equal to that of ordinary water
(c) density of brine is greater than that of ordinary water
(d) None of these
32.Two pieces of metals are suspended from the arms of a
balance and are found to be in equilibrium when kept
immersed in water. The mass of one piece is 32 g and its
density 8 g cm
–3
. The density of the other is 5 g per cm
3
.
Then the mass of the other is
(a) 28 g(b) 35 g(c) 21 g(d) 33.6 g
33.Figure shows a weigh-bridge, with a beaker P with water on
one pan and a balancing weight R on the other. A solid ball
Q is hanging with a thread outside water. It has volume 40
cm
3
and weighs 80 g. If this solid is lowered to sink fully in
water, but not touching the beaker anywhere, the balancing
weight R' will be
Q
P
R
(a) same as R (b ) 40 g less than R
(c) 40 g more than R(d) 80 g more than R
34.Figure here shows the vertical cross section of a vesselfilled with a liquid of density r. The normal thrust per unit
area on the walls of the vessel at the point P, as shown, willbe
q
h
P
H
(a)hrg (b) Hrg
(c) (H –
h) rg (d) (H – h) rg cosq
35.Two vessels A and B of cross-sections as shown in figurecontain a liquid up to the same height. As the temperaturerises, the liquid pressure at the bottom (neglecting expansionof the vessels) will
A B
(a) increase in A, decrease in B
(b) increase in B, decrease in A
(c) increase in both A and B
(d) decrease in both A and B

277. 273Mechanical Properties of Fluids
36.A beaker with a liquid of density 1.4 g cm
–3
is in balance
over one pan of a weighing machine. If a solid of mass 10 g
and density 8 g cm
–3
is now hung from the top of that pan
with a thread and sinking fully in the liquid without touching
the bottom, the extra weight to be put on the other pan for
balance will be
(a) 10.0 g
(b) 8.25 g
(c) 11.75 g
(d) –1.75 g
37.The pr
essure energy per unit volume of a liquid is
(a)
r
P
(b) P (c) P
× r(d)
P
r
38.A water tan
k of height 10m, completely filled with water is
placed on a level ground. It has two holes one at 3 m and the
other at 7 m from its base. The water ejecting from
(a) both the holes will fall at the same spot
(b) upper hole will fall farther than that from the lower hole
(c) upper hole will fall closer than that from the lower hole
(d) more information is required
39.A fast train goes past way side station platform at high
speed. A person standing at the edge of the platform is
(a) attracted to train
(b) repelled from train
(c) unaffected by outgoing train
(d) affected only if the train's speed is more than the speed
of sound
40.Three tubes X, Y and Z are connected to a horizontal pipe
in which ideal liquid is flowing. The radii of the tubes X, Y
and Z at the junction are respectively 3 cm, 1 cm and 3 cm. It
can be said
X Y Z
(a) the height of the liquid in the tube A is maximum.
(b) the height of liquid in the tubes Aand B is same.
(c) the height liquid in the tubes A,B and C is same.
(d) the height of the liquid in the tubes A and C is the
same.
41.A sphere of brass released in a long liquid column attains a
terminial speed v
0
. If the terminal speed attained by sphere
of marble of the same radius and released in the same liquid
is nv
0
, then the value of n will be (Given : The specific
gravities of brass, marbles and the liquid are 8.5, 2.5 and 0.8
respectively.)
(a)
17
5
(b)
77
17
(c)
31
11
(d)
5
17
42.Figur
e shows a capillary rise H. If the air is blown through
the horizontal tube in the direction as shown then rise in
capillary tube will be
H
(a) = H (b) > H (c) < H (d) zero
43.Two soap bubbles (surface tension T) coalesce to form abig bubble under isothermal condition. If in this process thechange in volume be V and the surface area be S, then thecorrect relation is (P is atmospheric pressure)
(a) PV + TS = 0 (b) 3PV + 4TS = 0
(c) 3PV + TS = 0 (d) 4PV + 3TS = 0
44.Two liquids of densities d
1
and d
2
are flowing in identical
capillary tubes uder the same pressure difference. If t
1
and
t
2
are time taken for the flow of equal quantities (mass) of
liquids, then the ratio of coefficient of viscosity of liquids
must be
(a)
22
11
td
td
(b)
2 1
t
t
(c)
1 2
1
2
t
t
d
d
(d)
22
11
td td
45.A tank ha
s a small hole at its botom of area of cross-section
a. Liquid is being poured in the tank at the rate Vm
3
/s,
the maximum level of liquid in the container will be (Area of
tank A)
(a)
gaA
V
(b)
gAa2
V
2
(c)
gAa
V
2
(d)
gaA2
V
Direction
s for Qs. (46 to 50) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following-
(a)Statement -1 is false, Statement-2 is true
(b)Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c)Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d)Statement -1 is true, Statement-2 is false
46. Statement 1 : Smaller the droplets of water, spherical they
are.
Statement 2 : Force of surface tension is equal, and opposite
to force of gravity.
47. Statement 1 : If a body is floating in a liquid, the density of
liquid is always greater than the density of solid.
Statement 2 : Surface tension is the property of liquid
surface.
48. Statement 1 : The velocity of flow of a liquid is smaller
when pressure is larger and vice-versa.
Statement 2 : According to Bernoulli’s theorem, for the
stream line flow of an ideal liquid, the total energy per unit
mass remains constant.

278. 274 PHYSICS
Exemplar Questions
1.A tall cylinder is filled with viscous oil. A round pebble is
dropped from the top with zero initial velocity. From the plot
shown in figure, indicate the one that represents the velocity
(v) of the pebble as a function of time (t) .
v
t
(a)
v
t
(b)
v
t
(c)
v
t
(d)
2.Which of the following diagrams does not represent a
streamline flow?
(b)(a)
(d)(c)
3.Along a stream
line,
(a) the velocity of a fluid particle remains constant
(b) the velocity of all fluid particles crossing a given
position is constant
(c) the velocity of all fluid particles at a given instant is
constant
(d) the speed of a fluid particle remains constant
4.An ideal fluid flows through a pipe of circular cross-section
made of two sections with diameters 2.5 cm and 3.75 cm.
The ratio of the velocities in the two pipes is
(a) 9 : 4(b) 3 : 2(c)
3:2(d)2:3
5.The angle of contact at the interface of water-glass is 0°,
ethyl alcohol-glass is 0°, mercury-glass is 140° and
methyliodide-glass is 30°. A glass capillary is put in a trough
containing one of these four liquids. It is observed that the
meniscus is convex. The liquid in the trough is
(a) water (b) ethylalcohol
(c) mercury (d) methyliodide
NEET/AIPMT (2013-2017) Questions
6.The wetability of a surface by a liquid depends primarily on(a) surface tension [2013]
(b) density(c) angle of contact between the surface and the liquid(d) viscosity
7.A fluid is in streamline flow across a horizontal pipe of
variable area of cross section. For this which of the following
statements is correct? [NEET Kar. 2013]
(a) The velocity is minimum at the narrowest part of the
pipe and the pressure is minimum at the widest part of
the pipe
(b) The velocity is maximum at the narrowest part of the
pipe and pressure is maximum at the widest part of the
pipe
(c) Velocity and pressure both are maximum at the
narrowest part of the pipe
(d) Velocity and pressure both are maximum at the widest
part of the pipe
49. Statement 1 : Falli ng raindrops acquire a terminal velocity.
Statement 2 : A constant force in the direction of motion
and a velocity dependent force opposite to the direction of
motion, always result in the acquisition of terminal velocity.
50. Statement 1 : The buoyant force on a submerged rigid object
can be considered to be acting at the centre of mass of the
object.
Statement 2 : For a rigid body a force field distributed
uniformly through its volume can be considered to be acting
at the centre of mass of the body.

279. 275Mechanical Properties of Fluids
8.A certain number of spherical drops of a liquid of radius ‘r’
coalesce to form a single drop of radius ‘R’ and volume ‘V’.
If ‘T’ is the surface tension of the liquid, then : [2014]
(a) energy =
11
4VT
rR
æö
-ç÷
èø
is released
(b
) energy =
11
3VT
rR
æö
+ç÷
èø
is absorbed
(c) energy =
11
3VT
rR
æö
-ç÷
èø
is released
(d
) energy is neither released nor absorbed
9.A wind with speed 40 m/s blows parallel to the roof of a
house. The area of the roof is 250 m
2
. Assuming that the
pressure inside the house is atmospheric pressure, the force
exerted by the wind on the roof and the direction of the
force will be (r
air
= 1.2 kg/m
3
) [2015]
(a) 4.8 × 10
5
N, upwards
(b) 2.4 × 10
5
N, upwards
(c) 2.4 × 10
5
N, downwards
(d) 4.8 × 10
5
N, downwards
10.The cylindrical tube of a spray pump has radius, R, one end
of which has n fine holes, each of radius r. If the speed of the
liquid in the tube is V, the speed of the ejection of the liquid
through the holes is : [2015 RS]
(a)
2
2
VR
nr
(b)
2
32
VR
nr
(c)
2
VR
nr
(d)
2
22
VR
nr
11.Wate r rises to a height 'h' in a capillary tube. If the length of
capaillary tube above the surface of water is made less than'h' then : [2015 RS]
(a) water rises upto the top of capillary tube and stays
there without overflowing
(b) water rises upto a point a little below the top and stays
there
(c) water does not rise at all.
(d) Water rises upto the tip of capillary tube and then starts
overflowing like fountain.
12.Two non-mixing liquids of densities r and nr(n > 1) are put
in a container. The height of each liquid is h. A solid cylinder
of length L and density d is put in this container. The cylinder
floats with its axis vertical and length pL(p < 1) in the denser
liquid. The density d is equal to : [2016]
(a) {1 + (n + 1)p}r (b) {2 + (n + 1)p}r
(c) {2 + (n – 1)p}r (d) {1 + (n – 1)p}r
13.A U tube with both ends open to the atmosphere, is partially
filled with water. Oil, which is immiscible with water, is poured
into one side until it stands at a distance of 10 mm above the
water level on the other side. Meanwhile the water rises by
65 mm from its original level (see diagram). The density of
the oil is [2017]
A
Oil
B C
Water
Initial water level
Final water level
10 mm
PaPa
F
E
D
65 mm
65 mm
(a) 425 kg m
–3
(b) 800 kg m
–3
(c) 928 kg m
–3
(d) 650 kg m
–3

280. 276 PHYSICS
EXERCISE - 1
1. (b) 2. (a) 3. (c) 4. (a) 5. (b)
6. (a) 7. (b) 8. (a)
9. (c) When a beaker containing a liquid of density r moves
up with an acceleration a, it will work as a lift moving
upward with acceleration a. The effective acceleration
due to gravity in lift = (a + g)
\Pressure of liquid of height h = h r (a + g)
10. (c) Volume of m g of ice = m/x and
volume of m g of water = m/y. So change in volume
= ÷
÷
ø
ö
ç
ç
è
æ
-=-
x
1
y
1
m
x
m
y
m
11. (a) TPµ
12. (b)
13. (b
) Velocity of liquid flowing out of hole =
gh2.
Here h = (h
0
– h
1
)
1
4. (d) Terminal velocity,
( )s-r
h
=
9
gr2
v
2
T ,
2
T
rvµ
15. (a)=vvelocity of efflux through an orifice gH2=
0v=
H
v
It is independent of the size of orifice.
16. (a)
17. (b) Surface energy µ surface area = pr
2
1
8. (d)
19. (a) When liquid drops coalesce, there is a decrease of
surface area and therefore decrease of surface energy.
Hence, energy is liberated.
20. (c)
21. (b) The theorem of continuity is valid.
\ A
1
v
1
r = A
2
v
2
r

as the density of the liquid can be
taken as uniform.
A
2
A
1
\ A
1
v
1
= A
2
v
2
Þ Smaller the ar
ea, greater the velocity.
22. (b)23. (b) Apply Bernoulli’s theorem.
24. (a) In Isothermal conditons
W
T
A
=
D
where, T = surface tension, W = work done,
DA = change in area.
25. (b)
EXERCISE - 2
1. (a)
weight of block inair
Specific gravity of block
loss of weight in water
=
= 3
4060
60
=
-
2. (b) Let v be th
e volume of the ice-berg outside the sea
water and V be the total volume of ice-berg. Then as
per question
0.92 V = 1.03 (V – v) or v/V = 1 – 0.92/1.03
= 11/103
\(v/V) × 100 = 11 × 100 / 103 @ 11%
3. (a) Weight of a man = wt. of water displaced
= volume × density =
3
10
100
1
23´´´ = 60 kg
4. (a)
Given,
21 r
T4
2
r
T4
´= or r
2
= 2r
1
3
1
3
2
3
1 )r2(
3
4
nr
3
4
nr
3
4
p´=p´=p or
125.0
8
1
n==
5. (b) Velocit
y of efflux,
v 2gh= ;
volume of liquid f
lowing out per sec
=
vA 2ghA´=´ =
34
10105102
--
=´´´ m
3
/s
6. (b) Termi
nal velocity,
h
r-r
=
9
g)(r2
v
0
2
0
=
83.09
8.910)3.18()102(2
323
´
´´-´´´
-
= 0.07 ms
–1
7.
(b) P.E. = K.E.
20 m = h
v
21
mgh mv
2
=
gh2v= = 20102´´ (Here g = 10 m/s
2
) = 20 m/s
Hints & Solutions

281. 277Mechanical Properties of Fluids
8. (c) Surface energy =suface tension × area of surface
For 1 drop, volume=
3
R
3
4
pif =Rradius of drop.
Total volume of 8 drops
3
R
3 4
.8p=
()
3
R2
3 4
p=
,R2'R
=new radius of big drop
New area = 4p4R
2
= 4 × old area
Energy µ area
E
1
= 4pR
2
.....(1)
E
2
= 4.4pR
2
.....(2)
From equation (1) and (2) we get, E
2
= 4E
1
9. (c)
gr
cos2
h
r
qs
=
h
cos
r
Þsµ
q
mm
m
w
ww
m
w
h
cos
cos
h
r
q
´
q
r
=
s
s
Þ
6.131.3
135cos
0cos
110
´-
°
´
°
´
=

6
1
6.131.3
)707.0(10
»
´-
-´
=
10. (c)
Volume of ball =
340
50 cm
0.8
=
Downth
rust on water = 50 g.
Therefore reading is 650 g.
11. (d)
4
a
(2P)
Q2
Q'
88
æö
pç÷
èø
==
hl

4
Pa
Q
8
éù p
=êú
hêúëû
Q
l
12. (a) F = 6 p h r n
= 6 × 3.14 × (8 ×10
–5
) × 0.03 ×100 = 4.52 ×10
–3
dyne
13. (a)
14. (c) K =
P
V/VD
\ VD =
PV
K
P = hgr = 200 × 10
3
× 10 N/m
2
K = 22000 atm = 22000 × 10
5
N/m
2
V = 1m
3
3
43
5
200 10 10 1
V 9.11 0m
22000 10
-´ ´´
D= =´
´
1
5. (d) According to principle of continuity, for a streamline
flow of fluid through a tube of non-uniform cross-
section the rate of flow of fluid (Q) is same at every
point in the tube.
i.e., Av = constant Þ A
1
v
1
= A
2
v
2
Therefore, the rate of flow of fluid is same at M and N.
16. (a) Pressure at the bottom of tank P =
5
2
3 10
N
hg
m
r=´
Pressur
e due to liquid column
555
1
3 10 1 10 2 10P=´ -´ =´
an
d velocity of water
2v gh=
\
5
1
3
2 2 2 10
400 m/s
10
P
v
´´
===
r
17. (
c) According to equation of continuity
A
1
V
1
= A
2
V
2
+ A
3
V
3
Þ 4 × 0.2 = 2 × 0.2 +0.4 × V
3
Þ V
3
= 1 m/s.
18. (d) 1 centipoise = 10
–2
gcm
–1
s
–1
= 0.001 kg m
–1
s
–1
19. (c) Angle of contact is acute.
20. (b) Let x be the fraction of volume of object floating above
the surface of the liquid.
As weight of liquid displaced = weight of object
\(V
0
– x V
0
)d g = V
0
d
0
g
(1 – x)d = d
0
or
d
dd
d
d
1x
00 -
=-=
21. (d)
a
g
ma
mg
tan ==q or q = tan
–1
g/a
22. (c
) If R is radius of bigger drop formed, then
33
r
3
4
2R
3
4
p´=p or R = 2
1 /3
r
As
2
0
rvµ
\
3/2
2
23/1
2
2
0
01
2
r
)r2(
r
R
v
v
===
or
3/13/2
001
)4(52vv´=´=
23. (c
) Vertical distance covered by water before striking
ground = (H – h). Time taken is,
g/)hH(2t-= :
Horizon
tal velocity of water coming out of hole at P,
gh2u=
\ Horizon
tal range =
ut 2gh 2(H h)/g=´-
= )hH(h2-
24. (c)
The effective acceleration of the body
g
'
1'g
÷
÷
ø
ö
ç
ç
è
æ
r
r
-=
h
Now, the depth to which the body sinks
'g2
gh2
'g2
u
'h
2
=
÷
÷
ø
ö
ç
ç
è
æ
=
ghh'
g''
´r
==
r-r
25. (a) Fluid resistance is given by R =
4
8L
r
h
p
When two capillary tubes of same size are joined in
parallel, then equivalent fluid resistance is
21S
RRR+==
44
8 L 8 2L
R (2R)
h h´
+
pp
=
8
9
R
L8
4
´
÷
÷
ø
ö
ç
ç
è
æ
p
h

282. 278 PHYSICS
Effect
ive weight = (32 – 4) gf = 28 gf
If m be the mass of second body, volume of second
body is
5
m
Now,
m
28 m m 35 g
5
=-Þ=
33. (c)
Upthrust = weight of 40 cm
3
of water
= 40 g = down thrust on water
34. (c) Pressure is proportional to depth from the free surface
and is same in all directions.
35. (a) As temperature rises, the density decreases, height
increases. In A, the top cross-section is smaller.
Therefore h
A
> h
B
.
36. (a) 10g is the force on water = extra wt. on other pan.
37. (b) Bernoulli’s theorem.
38. (a) Velocity of water from hole
A =
gh2v
1
=
Velocity of water fr
om hole B
20v 2g(H h)==-
Time of reaching the ground from hole B
g/)hH(2t
01
-==
Time of reachin
g the ground from hole A
g/h2t
2
==
39. (a) Ap
ply Bernoulli’s theorem.
40. (d) Use the equation of continuity and Bernoulli’s theorem.41. (b) For the same radius, terminal velocity is directly
proportional to density difference.
42. (b) Due to increase in velocity, pressure will be low above
the surface of water.
43. (b) 44. (a)45. (b)46. (c)47. (d)
48. (a) 49. (a) 50. (d)
EXERCISE - 3
Exemplar Questions
1. (c) When the pebble is dropped from the top of cylinder
filled with viscous oil and pebble falls under gravity
with constant acceleration, but as it is dropped it enter
in oil. So dragging or viscous force is
F = 6phrv
where r is radius of the pebble, v is instantaneous speed,
h is coefficient of viscosity.
As the force is variable, hence acceleration is also
variable so v-t graph will not be straight line due to
viscosity of oil. First velocity increases and then
becomes constant known as terminal velocity.
2. (d) In a streamline flow the velocity of fluid particles
remaines constant across any cross-sectional area, then
a point on the area cannot have different velocities at
the same time, hence two streamlines flow layers do
not cross each other.
Rate of flow =
4
S
P PR8
R 8L9
p
=´
h
=
9
8
X
4
PR
asX
8L
éù p
=êú
hêúëû
26. (d) Volume of bigger bubble = volume of 27 smaller
bubbles
33
d
3
4
27D
3
4
p´=pÞ
3
D
d=Þ
Initial sur
face energy
2
i
D4Sp=
s
Final surface energy
2
f
S 27 4d= ´ps
if
SSS-=D and using
3
D
d=
ú
ú
û
ù
ê
ê
ë
é
-´p´s=D
2
2
D
9
D
274S
=
22
D84D2ps=s
´p´
27. (d) For the first part of the question, remember that terminal
velocity means the acceleration experienced becomes
zero.
Since a = 0 m/s
2
, then, 0FFF
wresistance airy
=-=S
mgFF
w resistance air
=
For the second part of the question, while the velocity
is higher, the acceleration is still zero. Therefore, the
F
air resistance
is still equal to the skydiver’s weight.
F
air resistance Case A
= F
air resistance Case B
What has changed is the surface area of the skydiver.
Since pressure is P = F/A, as A decreases, the pressure
experienced increases.
P
A
A
A
= P
B
A
B
= mg
Since A
A
> A
B
, then P
A
< P
B
28. (a)
12
(2r 2r) mgp + p s=
98097.3
2
5.8
2
2
7.8
2 ´=s
ú
û
ù
ê
ë
é
´p+´p
1
72 dyne cm
-
Þs=
29. (b) Mas
s of liquid which rises in the capillary,
r´
r
qs
´p=rp=
gr
cos2
rh)r(m
22
Þ m µ r
30. (b) Becau
se of surface tension.
31. (c) Brine due to its high density exerts an upthrust which
can balance the weight of the egg.
32. (b) Volume of first piece of metal =
332
4cm
8
=
Upthrust = 4 gf

283. 279Mechanical Properties of Fluids
3. (b) In streamline flow, the speed of liquid of each particle
at a point in a particular cross-section is constant,
between two cross-section of a tube of flow because
AV = constant (law of continuity).
4. (a) As given that,
Diameter at 1st section (d
1
) = 2.5.
Diameter at 2nd section (d
2
) = 3.75.
According to equation of continuity,
for cross-sections A
1
and A
2
.
A
1
v
1
= A
2
v
2
2
2
1222
2
2 11 1
()
()
vArr
vAr r
æöp
=== ç÷
p èø

2
2
2
2
2
1
1
2
2
d
d
dd
æö
ç÷
æöèø
== ç÷
èøæö ç÷
èø

21
21
,
22
dd
rr
éù
==
êú
ëû
Q

2
3.759
2.54
æö
==
ç÷
èø
12
: 9:4vv\=
5. (c) W
e observed that meniscus of liquid is convex shape
as shown in figure which is possible if only, the angle
of contact is obtuse. Hence, the combination will be of
case of mercury-glass (140°). Hence verifies the option
(c).
140°
convex
mercury
NEET/AIPMT (2013-2017) Questions
6. (c) Wetability of a surface by a liquid primarily depends
on angle of contact between the surface and liquid.
If angle of contact is acute liquids wet the solid and
vice-versa.
7. (b) According to Bernoulli’s theorem,
21
2
Pv+r = constant and A vv = constant
If A is minimum, v is maximum, P is minimum.
8. (c) As surface area decreases so energy is released.
Energy released = 4pR
2
T[n
1/3
– 1]
where R = n
1/3
r
=
311
4RT
rR
éù
p-
êú
ëû
=
11
3VT
rR
éù
-
êú
ëû
9. (b) According to Bernoulli’s theorem,
12
20
P vP0+r=+
So,DP =
1
2
rv
2
F = DPA =
1
2
rv
2
A P
0
P
=
1
2
× 1.2 × 40 × 40 × 250
= 2.4 × 10
5
N (u
pwards)
10. (a) Inflow rate of volume of the liquid = Outflow rate of
volume of the liquid
pR
2
V = npr
2
(v) Þ v =
22
22
R V VR
n r nr
p
=
p
11. (a) Water rises upto the top of capillary tube and stays
there without overflowing.
12. (d) As we know,
Pressure P = Vdg
r
d
(1 – p)L
nr
pL
Here, L A d g
= (pL) A (nr)g + (1 – p)L A r g
Þd = (1 – p)r + pn r = [1 + (n – 1)p]r
13. (c) Here,
oil oil water water
hg hg´r ´ = ´r ´
r
0
g × 140 × 10
–3
= r
w
g × 130 × 10
–3
r
oil

=
33130
10 928kg / m
140
´» [Q r
w
= 1 kgm
–3
]

284. 280 PHYSICS
TEMPERATURE
The
temperature of a body is that physical quantity which indicates
how much hot or cold the body is? So temperature indicates the
hotness or coldness of a body.
The flow of heat is always from higher temperature to lower
temperature. Temperature is a macroscopic concept.
Two bodies are said to be in thermal equilibrium with each other,
when no heat flows from one body to the other, that is when both
the bodies are at the same temperature.
Temperature measuring device :
Thermometer is a device to measure the temperature.
Thermometers used for measuring very high temperatures are
called pyrometers.
Different types of temperature scales
Name of the
scale
Symbol
for each
degree
Lower
fixed
point
Upper
fixed
point
No. of
divisions on
the scale
Reaumur ºR 0ºR 80ºR 80
Celsius ºC 0ºC 100ºC 100
Fahrenheit ºF 32ºF 212 ºF 180
Rankine Ra 460Ra 672Ra 212
Kelvin K 273K 373K 100
*Lower fixed point (LFP) is the freezing point of water.
*Upper fixed point (UFP) is the boiling point of water.
TºK = (tºC + 273) or tºC = (TK – 273)
Relationship between different temperature scales
32 460 273
80 100 180 212 100
---
====
RCFRaK
The zero
of the Kelvin scale is called absolute zero. The Kelvin
scale is often termed as the absolute scale. In common use the
absolute zero corresponds to –273ºC. However, its exact value is
–273.16º C.
IDEAL GAS EQUATION AND ABSOLUTE TEMPERATURE
The equation,PV = nRT
where, n = number of moles in the sample of gas
R = universal gas constant; (its value is 8.31 J mol
–1
K
–1
), is
known as ideal-gas equation
It is the combination of following three laws :
(i) Boyle's law : When temperature is held constant, the pressure
is inversely proportional to volume.
i.e.,
1
P
V
µ ( at constant temperature)
(ii) Charle's law : When the pressure is held constant, the volume
of the gas is directly porportional to the absolute temperature.
i.e., VTµ(at constant pressure)
(iii) Avogadro's law : When the pressure and temperature are
kept constant, the volume is directly proportional to thenumber of moles of the ideal gas in the container.
i.e.,
Vnµ (at constant pressure and temperature)
Absolute Temperature
The lowest temperature of –273.16 °C at which a gas is supposed
to have zero volume and zero pressure and at which entire
molecular motion stops is called absolute zero temperature. A
new scale of temperature starting with –273.16°C by Lord Kelvin
as zero. This is called Kelvin scale or absolute scale of temperature.
T(K) = t°C + 273.16
Thermometric property : It is the property used to measure the
temperature. Any physical quantity such as length of mercury (in
a glass capillary), pressure of a gas (at constant volume), electrical
resistance (of a metal wire), thermo emf (thermo-couple
thermometer) which changes with a change in temperature can be
used to measure temperature.
The unknown temperature t on a scale by utilising a general
property X of the substance is given by
0
1000
100º
æö-
=´ç÷
-
èø
t
XX
tC
XX
Keep in Memory
1.The mercury
thermometer with cylindrical bulb are more
sensitive than those with spherical bulb.
2.Alcohol thermometer is preferred to the mercury thermometerdue to the larger value of the coefficient of cubical expansion.
3. Properties that make mercury the ideal thermometric
substance are :(i) It does not stick to the glass walls(ii) Available in pure form(iii) Low thermal conductivity and low specific heat(iv) Vapour Pressure is low
11
Thermal Properties
of Matter

285. 281Thermal Properties of Matter
(v) C
oefficient of expansion is uniform
(vi) It shines
4.Gas thermometer is more sensitive than the mercury
thermometer, because its coefficient of cubical expansion is
much larger.
5.At the following temperatures different temperature scales
have the same reading :
(i) Fahreinheit and Reaumur at –25.6ºF = –25.6ºR
(ii) Reaumur and Celsius at 0ºR = OºC
(iii) Celsius and Fahrenheit at –40ºC = –40ºF
(iv) Fahrenheit and Kelvin at 574.25ºF = 574.25K
6.The Celsius scale and the Kelvin cannot have the same
reading at any temperature.
7.Six’s thermometer measures minimum and maximum
temperature during a day. It uses mercury as well as alcohol
as the thermometric substances.
8.Normal human body temperature is 37ºC = 98.6ºF.
9.Clinical thermometers are much shorter than the laboratory
thermometers because they are used to measure a limited
range (96º F to 100º F) of temperature.
10.Rapidly changing temperature is measured by using thermo
electric thermometer.
11.The temperature inside a motor engine is measured using
platinum resistance thermometer.
12.The radiation thermometers can measure temperature from
a distance.
13.Adiabatic demagnetisation can be used to measure
temperature very near to the absolute zero.
14.The Rankine is the smallest temperature range among all the
scales of temperature.
15.When a substance is heated it expands along each
dimension in the same proportion. (if it is uniform)
THERMAL EXPANSION
Thermal Expansion of Substances
Linear expansion : i.e., increase in length with increase in
temperature
T
0
Da=Dll
or )T1(T
000
Da+=ÞDa=-lllll
where l
0
is the initial length, l = final length, DT = change in
temperature, a = coefficient of linear expansion.
Areal expansion : i.e., increase in area with increase in
temperature.
0
D=bDA AT where b = coefficient of area expansion
Volume expansion : i.e. increase in volume with increase in
temperature.
0
D=gDV VT where g = coefficient of volume expansion. The
units of a, b and g is (°C)
–1
or K
–1
.
Graph of a
v
versus T :
500
T (K)
250
(For copper)
a ®
v
(10 K )
–
5
–
1
Keep in Memory
1.Du
ring change in temperature, the mass does not change
but the density decreases due to increase in volume.Therefore
0
1
r
r=
+ gDT
2. Relation between a, b and g
a=g3 and a=b2
3.Water h
as negative coefficient of volume expansion for
temperature range (0°C – 4°C). Therefore water contractswhen the temperature rises from 0°C to 4°C and then expandsas the temperature increases further.
Density of water reaches a maximum value of 1000 kg m
–3
at
4°C. This is the anomalous behaviour of water.
Vo
l
u
m
e
Temperature4°C
V
o
l
u
m
e
Temperature4°C
Thermal Expa
nsion of Liquid (which is heated in a vessel)
T)3(VV
app
Da-g=D
where V
D
app
= apparent ch
ange in volume
V= volume of the liquid initially
g = coefficient of volume expansion of liquid
a = coefficient of linear expansion of vessel
It is clear from the formula that if
a=g3, 0V
app
=D but if a>g3,
app
VD is positiv
e and the liquid rises in the vessel and if
a<g3,
app
VD is negative
and liquid falls in vessel.
Change in density of solids and liquids with temperature.
The density at t°C is given by
r
r=
+g
0
(1)
t
t
where r
0
is the density at 0°C. If r
1
and r
2
are the densities at
temperature t
1
and t
2
respectively, then
r -r
g=
r-
12
221
()tt

286. 282 PHYSICS
Isothermal and adiabatic
elasticities :
(i) Isothermal elasticity,
.P
V
dV
dP
E
iso
=
÷
ø
ö
ç
è
æ
-
º
(since PV = R
T = constant,
V
P
dV
dP
-=)
(ii) Adiabatic e
lasticity,
P
V
dV
dP
E
adia
g=
÷
ø
ö
ç
è
æ
-
= ; where
v
p
C
C
=g
(iii)
isoadia
EEg=
Expansion
of Gases
There are two coefficients of expansion of gases.
(a) Pressure coefficient of expansion of gas (g
P
)
0
0
-
g=
t
P
PP
Pt
where P
0
= Pres sure at 0°C, P
t
= Pressure at t°C
(b) Volume coefficient of expansion of gas (g
v
)
0
0
-
g=
t
v
VV
Vt
where V
0
= Volu me at 0°C, V
t
= Volume at t°C
HEAT
Heat is the form of energy which is transferred from one body to
another body due to temperature difference between two bodies.
Thermal energy of a body is the sum of kinetic energy of random
motion of the molecules/atoms and the potential energy due to
the interatomic forces.
Calorimeter is an instrument used to measure the heat.
The SI unit of heat is joule and its dimensions are [ML
2
T
–2
]. Other
units of heat are :
(i)Calorie: It is the amount of heat required to raise the
temperature of 1g of water from 14.5ºC to 15.5ºC.
1 calorie = 4.2 joule. Calorie is the unit of heat in CGS system.
(ii)Kilocalorie, 1Kcal = 1000 cal
(iii)British Thermal Unit (BTU): It is the fps unit of heat
= 252 cal = amount of heat required to raise the temperature
of 1 lb of water through 1ºF (from 63°F to 64ºF).
SPECIFIC HEAT CAPACITY
It is the amount of heat required to raise the temperature of unit
mass of substance through 1 degree.
Specific heat capacity,
=
´D
Q
c
mt
Its SI unit is J kg
–1
K
–1
C
GS unit cal g
–1
°C
–1
It dimensions are [M
0
L
2
T
–2
q
–1
]
For gases : During the isothermal process
Dt = 0 Þ c = ¥ (infinite)
Thermal Stress
If a rod fixed at two ends is heated or cooled, then stress are
produced in the rod.
Thermal stress = TY
Da´
0
T strain
æöD
=aD=
ç÷
èø
l
l
Effect of Temperature
on Pendulum Clock
A Pendulum clock consists of a metal bar attached with a bob atone end and fixed at the other end. Let pendulum clock read correct
time when its length is l
0
. The time period
t
0

=
l
0
g
2π
Suppose that the te
mperature is raised by DT, then new time period
is
g
t2=p
l

=Þ
l
l0 0
t
t
.....(i)
We have = + aDll
0
(1 t)
Þ = + aD
l
l
0
(1 t)
.....(ii)
From equations
(i) and (ii),

= + aD
0
1
2
t
(1 t)
t
= + aD
1
2
1t (since a is very small)
Þ -= aD
0
t1
1
t2
t

-
Þ= aD
0
0
tt1
t2
t
so, fractional change in time
D
=aD
0
t1
t2
T
Loss of time per day =aD´
1
2
T 86400 second.
(Q no. of seconds in one day = 86400 sec.)
When temperature increases (during summer), the length of thependulum increases due to which the time period increases andthe clock loses time. On the other hand, when temperature
decreases (during winter season), the length decreases and the
time period decrease, the clock in this case, gains time.
Error in scale reading :
As the temperature changes, the length of the metal scale changes
so is the difference in graduation
am
R R (1 T)= + aD
where R
a
= actu al reading, R
m
= measured reading

287. 283Thermal Properties of Matter
and
during the adiabatic process
Q = 0 Þ c = 0 (zero)
Specific heat of gas : The specific heat of a gas varies from zero to
infinity. It may have any positive or negative value.
Following of two specific heats of gas are more significant.
(a)Molar specific heat at constant volume (C
V
) : It is the
amount of heat required to raise the temperature of
1 mole of a gas through 1K or 1°C at constant volume.
1Q
T
æö
=
ç÷
èø
V
d
C
nd
(b)Molar specific heat at constant pressure (C
P
) : It is the
amount of heat required to raise the temperature of 1 mole ofa gas through 1K or 1°C at constant pressure.
1
C
æö
=
ç÷
èø
P
dQ
n dT
Mayer’s relation :C
P
– C
V
= R
[where R is a gas constant. (R = 8.314 Jmol
–1
K
–1
)]
Why C
P
is greater than C
V
?
When heat is supplied to a gas at constant volume entirely usedto raise its temperature. When a gas is heated at constant pressure,some work is done in expansion of gas which is in addition toraise its temperature and hence C
P
> C
V
.
THERMAL CAPACITY OR HEAT CAPACITY
It is the amount of heat required to raise the temperature of
substance through 1ºC or 1K.
Thermal capacity = mass × specific heat.
This is the relation between sp. heat capacity and heat capacity.
Its SI unit is JK
–1
CGS unit cal/°C. Its dimensions are [ML
2
T
–
2
q
–1
]
Water Equivalent
It is the mass of water in gram which would require the same
amount of heat to raise its temperature through 1ºC as the body
when heated through the same temperature. It is measured in
gram. Hence
Water equivalent of a body = mass of the body × specific heat
Principle of Heat (or Calorimetry)
When two bodies at different temperatures are placed in contact
with each other then heat will flow from the body at higher
temperature to the body at lower temperature until both reach to a
common temperature.
i.e.Heat lost by hot body = heat gained by cold body.
It follows the law of conservation of energy.
LATENT HEAT
It is defined as the amount of heat absorbed or given out by a
body during the change of state while its temperature remaining
constant.
It is of two types :
(i)Latent heat of fusion : It is the quantity of heat required to
change unit mass of the solid into liquid at its melting point.
Latent heat of fusion of ice is 80 cal/g.
(ii)Latent heat of vaporisation : It is the quantity of heat required
to convert unit mass of liquid into vapour at its boiling
point.
Latent heat of vaporisation of water is 536 cal/g.
Latent heat increases the intermolecular potential energy of the
molecules while kinetic energy remains constant.
Different Processes of Phase Change or State Change
(i)Sublimation : It is the process of conversion of solids to
gaseous state on heating. On cooling the vapours get
converted back into solids.
(ii)Condensation : It is the process of conversion of gases/
vapours to liquid state on cooling. On heating, these liquids
are converted into vapours/gases.
(iii)Boiling : It is the process of conversion of liquid to gaseous
state on heating. It is a cooling process.
(iv)Melting : It is the process in which solid gets converted
into liquid on heating.
(v)Evaporation : Conversion of liquid into gaseous state at all
temperatures is called evaporation. It is a phenomenon that
occurs at the surface of the liquid.
Melting and boiling occur at definite temperature called melting
point and boiling point respectively. The liquid boils at a
temperature, at which its vapour pressure is equal to the
atmospheric pressure.
How evaporation is different from boiling?
Evaporation is a slow process occurring at all the temperature at
the surface of liquid while boiling is a fast process involving whole
of the liquid heated to a particular temperature called boiling point
of the liquid.
Keep in Memory
1.Th
e heat capacity or thermal capacity depends on nature of
the substance (specific heat capacity) and mass of quantity
of matter of the body.
2.In case of melting and boiling, the temperature does not
change, only intermolecular potential energy of the system
changes.
3.System are in thermal equilibrium when their temperature
are same or average kinetic energy per molecule is same.
4.Internal energy is the sum of all energies in a system which
includes energy of translational, rotational as well as
vibrational motion of the molecules.
5.The molar specific heat capacity depends on the molecular
weight of the substance.
6.The temperature, volume or pressure of a system may remain
constant when it absorbs heat.
7.When we press two block of ice together such that after
releasing the pressure two block join this phenomenon is
called regelation.

288. 284 PHYSICS
Phase Dia
gram
D
B
C
P
E
F
T
O
A
T
CT
tr
P
Liquid
Vapour
Solid
4.58 mm
of merc
ury
273.16ºKwater
(i) The diagram shows vaporisation curve (AB), fusion curve
(CD), sublimation curve (EF).
(ii) For water, AB is called steam line, CD is called ice line and
EF is called hoar-frost line.
(iii) The temperature at which all the three states of matter (solid,
liquid, gas) are mutually in thermal equilibrium is called the
triple point (point P in diagram) of the substance.
(iv) The temperature upto which a gas can be liquified under
pressure alone is called the critical temperature (T
c
).
Example 1.
A solid material is supplied with heat at a constant rate.
The temperature of the material is changing with the heat
input as shown in Fig. Study the graph carefully and answer
the following :
(i) What is represented by horizontal regions AB & CD?
(ii) If CD = 2 AB. What do you infer?
Solution :
A
B
C D
E
Heat Supplied
Temp.
Y
O
X
0ºC(ice)
(0ºC liquid)
(100ºC liquid)
(100ºC Vapour)
Q
1
Q
2
Q
3
Q
4
(i) In the horizon
tal regions AB and CD heat is supplied
but temp. remains constant. Therefore, they represent
change of state. The graph AB represents change from
solid state to liquid state, at constant temp. = melting
point of solid.
AB = latent heat of fusion.
Again CD represents change of state from liquid to
vapour at boiling point of the liquid.
CD = latent heat of vaporisation.
(ii) As CD = 2 AB
\ Latent heat of vaporisation is twice the latent heat
of fusion.
Example 2.
A clock which keeps correct time at 25ºC has a pendulum
made of brass whose coefficient of linear expansion is
0.000019. How many seconds a day will it gain if the
temperature fall to 0ºC?
Solution :
Let L
0
and L
25
be the length of pendulum at 0ºC and 25ºC
respectively. We know that
L
25
= L
0
(1 + aT) = L
0
(1 + 0.000019 × 25) = 1.000475 L
0
If T
25
and T
0
be the time periods at 25ºC and 0ºC respectively,
then
T
25
= 2p
25
L
g
æö
ç÷
èø
and T
0
= 2p
0
L
g
æö
ç÷
èø
\
25
0
T
T
=
25
0
L
L
æö
ç÷
èø
=
0
0
1.000475L
L
æö
ç÷
èø
= (1.000475) = 1.000237
Now
250
0
TT
T
-
= 0.000237
\Gain in tim
e for one vibration = 2 × 0.000237 sec.
Number of vibration in one day =
24 60 60
2
´´
(T = 2 sec.)
Hence the gain in time in one day
= 2 × 0.000237 ×
24 60 60
2
´´
= 20.52sec.
Example
3.
A circular hole in an aluminium plate is 2.54 cm in diameter
at 0ºC. What is the diameter when the temperature of the
plate is raised to 100ºC ? Given
a
A
= 2.3 × 10
– 5
(ºC)
– 1
Solution :
Let D
0
and D
t
be diameters of hole at 0ºC and tºC
respectively.
Circumference of hole at 0ºC
l
0
= 2pr
0
= pD
0
Circumference of hole at t = 100ºC
l
t
= 2pr
t
= pD
t
From relation l
t
= l
0
(1 + a.t), we get
pD
t
= pD
0
(1 + 2.3 × 10
–5
× 100)
D
t
= 2.54 (1 + 0.0023) = 2.5458 cm
Example 4.
If the volume of a block of a metal changes by 0.12% when
it is heated through 20ºC, what is the coefficient of linear
expansion of metal ?
Solution :
Coefficient of cubical expansion of metal is given by
g =
V
Vt
D
Here g =
V 0.12
Vt 100t
D
= , t = 20ºC
\ g =
0.12
100 20´
= 6.0 × 10
–5
per ºC
Co
efficient of linear expansion
a =
5
6.0 10
33
-
g´
= = 2.0 × 10
–5
Per ºC

289. 285Thermal Properties of Matter
Exa
mple 5.
From what height must a block of ice be dropped into a
well of water so that 5% of it may melt? Given : both ice
and water are at 0°C, L = 80 cal g
–1
, J = 4.2 J cal
–1
and
g = 980 cm s
–2
.
Solution :
Let m be the mass of ice. Let h be the height from which
block of ice is dropped. Work done,
W = mgh = m × 980 × h erg
Mass of ice to be melted =
5
m
100
´
Heat required,
5
Q
100
= × m × 80 cal or Q = 4 m cal
[Q L = 80 cal g
–1
] [Q W = JQ]
Now, m × 980 × h = J × 4 m
7
4.2 10 4m
h
m 980
´´
=
´
cm = 1714.
3 m
[Q J = 4.2 × 10
7
erg cal
–1
]
SATURATED AND UNSATURATED AIR
The air is said to be saturated when the maximum possible amount
of water vapours are present in it. The pressure of the water
vapours in the saturated air is called saturation vapour pressure.
If the air contains vapours less than the maximum possible amount
possible in the air, then it (air) is said to be unsaturated.
Vapour Pressure
A vapour differs from a gas in that the former can be liqufied by
pressure alone, wheras the latter cannot be liqufied unless it is
first cooled.
If a liquid is kept in a closed vessel, the space above the liquid
becomes saturated with vapour. The pressure exerted by this
vapour is called saturated vapour pressure. Its value depends
only on temperature. It is independent of any external pressure.
Humidity
The humidity shows the presence of water vapours in the
atmosphere. The amount of water vapours present per unit volume
of the air is called absolute humidity.
Relative humidity (RH) is defined as the ratio of the mass of
water vapour (m) actually present in the given volume of air to
the mass of water vapour (M) required to saturate the same
volume at the same temperature. Normally RH is expressed in
percentage
i.e.,
% 100%=´RH
M
m
RH is also de
fined as the ratio of actual pressure (p) of water
vapour to the saturated vapour pressure (P) of the water at the
same temperature
i.e.,
% ×100%=
p
RH
P
Also,
RH =
Saturatedvapour pressure of water at dew point
Saturated vapour pressure of water at room temperature
Dew Point :
It is the temperature at which the amount of water vapours
actually present in a certain volume of the air which is sufficient
to saturate that volume of air.
At the dew point the actual vapour pressure becomes the
saturated vapour pressure.
It means that vapour pressure at room temperature = saturated
vapour pressure(S.V.P) at dew point
If room teperature = dew point
then Relative humidity (R.H.) = 100%
Keep in Memory
1.
Hygrometry is the branch of thermal physics that deals
with the measurement of amount of water vapours presentin the atmosphere.
2. Hygrometer is an instrument used to determine dew point
and relative humidity.
3.Relative humidity is low when the air is dry.
4.Saturated vapours do not obey (except Dalton’s law) gaslaws, while unsaturated vapours obey them. The S.V.P.varies linearly with temperature (keeping volume constant)and equal to one atmospheric pressure at boilingpoint.(shown in fig.(a)) S.V.P. constant at constanttemperature on increasing or decreasing volume fig. (b).
V = constant
T
Boiling point
1 atm
(S.V.P
.)
(a)
P

T = constant
V
(S.V.P.)
(b)
P
5.At 0ºC, the saturation vapour pessure is 4.6 mm of Hg.
Therefore, water vapours are always present around theice. It increases with the increase in temperature.
6.The relative humidity is 100% when the temperature of the
atmosphere is equal to the dew point.
7. Cryogenics is the study of low temperature.
8.As RH is expressed in percentage, hence the maximumvalue of RH may be 100%.
9.The saturation vapour pressure of water at 100ºC is 760mm of Hg at sea level.
10.As the temperature rises, the absolute humidity mayincrease and the relative humidity may remain constant ormay even decrease.
11.If the relative humidity is high then at the same temperatueone feels hotter.
12.If the air is absolutley dry, the dew point is not observed.
13.The relative humidity decreases with the increase intemperature.
14.If water is sprinkled in the room, both relative humidity aswell as the dew point increase.
15.Dew appears on the leaves of the trees etc. due to thecondensation of saturated vapours.
16.The mist or fog is the small droplets of water formed due tothe condensation of water vapours near the surface ofearth.
17.Mist or fog formed much higher above the surface of earth
is called cloud.
11.1

290. 286 PHYSICS
18.The Rel
ative humidity between 50% and 60% gives
comfortable feeling to human beings.
19.When the relative humidity is 100%, the reading of the dry
and wet bulbs is same.
20.Smaller the difference in the temperature of dry and wet
bulbs, larger is the relative humidity.
21.Water vapours are always seen around the ice, because the
temperature around it is less than the dew point.
HEAT TRANSFER
Three modes of transmission of heat :
(a) Conduction (b) Convection and (c) Radiation
(a)Conduction : In thermal conduction particles of body at
higher temperature transmit heat to the particles at lower
temperature by mutual contact (collision) only and not by
the movement of the particles. All solids are heated by
conduction. Conduction cannot take place in vacuum.
Dx
D
D
Q
t
D
D
Q
t
A
T
1
T
2
T > T
1 2
Figure shows a so
lid of cross-section area A and thickness
Dx. The face are at different temperature T
1
and T
2
(T
1
> T
2
)
The rate of heat flow
÷
ø
ö
ç
è
æ
D
D
t
Q
as found experimentally is
given by
÷
ø
ö
ç
è
æ
D
D
µ
D
D
x
T
A
t
Q
where
21
TTT- =D
Þ
ΔΔ
ΔΔ
æö
=
ç÷
èø
QT
KA
tx
…(1)
where K is pr
oportionality constant called thermal
conductivity.
It is a measure of how quickly heat energy can conduct
through the substance.
Coefficient of thermal conductivity (K) : The coefficient of thermal
conductivity, K, of a material is defined as the amount of heat that
flowing per second through a rod of that material having unit
length and unit area of cross-section in the steady state, when
the difference of temperature between two ends of the rod is 1 ºC
and flow of heat is perpendicular to the end faces of the rod.
Unit of coefficient of thermal conductivity in SI system is
watt/m-K
Dimensions : [M L T
–3
q
–1
]
For a perfect conductor thermal conductivity K is infinite and
for a perfect insulator K is zero.
In general, solids are better conductors than liquids and liquids
are better conductors than gases. (Heat transfer through the
process of conduction is possible in liquids and gases also, if
they are heated from the top.) Metals are much better conductors
than non-metals.
A good conductor of heat is also a good conductor of electricity
The conduction of both heat and electricity is due to the movement
of free electrons.
(b)Convection : It is the process by which heat is transferred
from one place to another in a medium by the movement of
particles of the medium. It occurs due to density difference.
This phenomenon occurs in fluids.
(c)Radiation : It is the process by which heat is transferred
from one place to another without any intervening medium.
The light reaches from Sun to Earth by radiation process.
Keep in Memory
1.The equa
tion
ΔΔ
ΔΔ
æö
=ç÷
èø
QT
KA
tx
is valid for steady state
condition. The condition is said to occur when no part of
heat is used up in raising the temperature of any part of
cross-section of the solid.
2.On comparing equation (1) with the following equation used
for flowing of charge on account of potential difference.
V
1
Dq
V
2
s
l
We find :
l
V
A
t
q D
s=
D
D
The role of resistance (thermal
resistance) is played by
Dx
KA
3. Series combination of conductors
l
1
l
2
K , A
11
T
1 T
T
2
K , A
22
Equivalent
thermal conductivity :
12
12
2
=
+
eq
KK
K
KK

12
12
12
(–)ATT
H
ll
KK
=
+
where H = heat flow per
second
4. Parallel combination of conductors
l
K , A
11
T
1 T
2
K , A
22
Equivalent
thermal conductivity :
K
eq
=
11 22
22
+
+
KA KA
AA
12
12
12
12
12
KK
T+T
ll
H=
KK
+
ll

291. 287Thermal Properties of Matter
5.Dav
y’s safety lamp is based on the conduction. It is used in
mines to know the ignition temperature of gases. Danger of
explosion can be avoided.
6.(i) Principle of chimney used in a kitchen or a factory is
based on the convection.
(ii) Land and sea breezes are due ot the convection.
(iii) Temperature of the upper part of the flame is more than
the temperature on the sides, because the currents of
air carry the heat upwards.
(iv) Radiation can be detected by differential air
thermometer, Bolometer, thermopile, etc.
Example 6.
Two plates of same area are placed in contact. Theirthickness as well as thermal conductivities are in the ratio2 : 3. The outer surface of one plate is maintained at 10ºCand that of other at 0ºC. What is the temp. of the commonsurface?
Solution :
Let q be the temp. of common surface.
As
21
t
Q
t
Q
÷
÷
ø
ö
ç
ç
è
æ
D
D
=
÷
÷
ø
ö
ç
ç
è
æ
D
D
2
22
1
11
x
)0(
AK
x
)10(
AK
-q
=
q-
or
11
22
Kx
(10θ)θ
Kx
-=
K
1
x
1
x
2
K
2
Plate 1 Plate 2
10°C 8°C 0°C
Now,
3
2
K
K
x
x
;AA
2
1
2
1
21 ===;
.Cº5or
3
2
)10(
3
2
=qq=q-
Example 7.
A
slab consists of two parallel layers of two different
materials of same thickness and thermal conductivities
K
1
and K
2
. Find the equivalent thermal conductivity of
the slab.
Solution :
Rate of flow of heat through each layer of slab is same,
because slabs are in series.
x
x
DT
1
DT
2
lll 2
A)TT(KT
AK
T
AK
212
2
1
1
D+D
=
D
=
D
or C
2
)TT(K
TKTK
21
2211
=
D+D
=D=D , say
K
C2
TT,
K
C
T,
K
C
T
21
2
2
1
1 =D+D=D=D\
or
K
C2
K
C
K
C
21
=+ or
21
21
KK
KK2
K
+
=
Example 8.
T
wo walls of thickness d
1
and d
2
and thermal conductivities
K
1
and K
2
are in contact. In the steady state, if the
temperatures at the outer surfaces are T
1
and T
2
then find
the temperature at the common wall.
Solution : K
1
T
1
T
2
T
d
1
d
2
K
2
Let the temperature of common wall be T. Then
2
22
1
11
d
t)TT(AK
d
t)TT(AK-
=
-
or,
2
22
1
11
d
t)TT(K
d
t)TT(K-
=
-
or, )]TT(
K[d)]TT(K[d
221112
-=-
or,
2212121211
TKdTKdTdKdTK-=-
or, ]KddK[TTdKdTK
2121212211
+=-
\
ú
û
ù
ê
ë
é
+
-
=
1221
212211
dKdK
TdKdTK
T
HEAT T
RANSFER BY RADIATION AND NEWTON’S LAW
OF COOLING
Stefan’s Law:
This law is also called Stefan Boltzmann law. This law states that
the power radiated for overall wavelength from a black body is
proportional to the fourth power of thermodynamic temperature T.
44
ET ETµ Þ =s for perfectly black body....(1 )
where E is energy emitted per second from unit sruface area of theblack body, T is temperature and s is Stefan’s constant, and
s = 5.67×10
–8
Wm
–2
K
–4
.
If body is not perfectly black body, then
E = es T
4
....(2)
where e is emissivity. Then energy radiated per second by a body
of area A is
E
1
= es T
4
....(3)
If the body temperature is T and surrounding temperature is T
0
,
then net rate of loss of energy by body through radiation fromequation (3) is
' 44
10
()EeATT=s- ....(4)
Now ra
te of loss of energy in terms of specific heat c is
'
1
dQ dT
mcE
dt dt
== ....(5)
and 44
0
()
dT eA
TT
dt mc
s
=-
....(6)
(T- te
mperature, t - time in second)

292. 288 PHYSICS
So body
cools by radiation and rate of cooling depends on
e (emissivity of body), A (area of cross-section of body),
m (mass of body) and c (specific heat capacity of body).
A black body is defined as a body which completely absorbs all
the heat radiation falling on it without reflecting and
transmitting any of it.
Weins Law :
l
m
T = b,
where b is the Weins constant and b = 2.898 × 10
–3
mK
and T = temperature.
Graph of l
m
versus T
T

2
1
1
2
l
=
l
m
m
T
T
Hence on increasing temperature of a body, its colour changes
gradually from red to orange ®yellow ® green ® blue ®
violet. Thus the temperature of violet star is maximum and
temperature of red star is minimum. Sun is a medium category starwith l
m
= 4753Aº (yellow colour) and temperature 6000K.
Kirchoff’s Law
According to this law, the ratio of emissive power to absorptive
power is same for all surfaces at the same temperature and is
equal to the emissive power of a perfectly black body at the same
temperature.
i.e.
e
EA
a
=
where e = emiss
ive power of a given surface
a = absorptive power of a given surface
E = emissive power of a perfect black body
A = absorptive power of a perfect black body
(i) For a perfect black body, A = 1,
\=
e
E
a
(ii) If emissive and absorptive power are considered for a
particular wavelenth l then
l
l
l
=
e
E
a
(iii)Si nce E
l
is con stant at a given temperature, according to
this law if a surface is good absorber of a particular
wavelength then it is also a good emitter of that wavelength.
Fraunhoffer's Line
(i) Fraunhoffer lines are the dark lines in the spectrum of sun
and are explained on the basis of Kirchoff’s law. When white
light emitted from central core of sun (photosphere) passes
through its atmosphere (chromosphere) radiation of those
wavelength will be absorbed by the gases present there,
which they usually emit (as good emitter is a good absorber)
resulting dark lines in the spectrum of sun.
(ii) On the basis of Kirchoff’s law, Fraunhoffer identified some
of the elements present in the chromosphere. They are
hydrogen, helium, sodium, iron, calcium, etc. Fraunhoffer
had observed about 600 darklines in the spectrum of sun.
Newton's Law of Cooling
The rate of loss of heat of a body is directly proportional to the
temperature difference between the body and the surroundings.
i.e.
0
()
dT
KTT
dt
=--
where T = tem
perature of body, T
0
= temperature of surrounding.
and
mc
eA
K
s
= where, e = e
missivity of body, A = area of surface
of body, s = Stefan’s constant, m = mass of body, c = specific heat
of body.It is a special case of Stefan’s law and above relation is applicableonly when the temperature of body is not much different from thetemperature of surroundings.Solar constant : Solar constant is the solar radiation incident
normally per second on one square metre at the mean distanceof the earth from the sun in free space.It is given by S = 1.34 × 10
9
Jm
–2
s
–1
.
Temperature of sun is given T
4
= S/s(R/r)
2
, where R is mean
distance of the earth from the sun and r is the radius of the sun.
Keep in Memory
1.Good absor
ber is a good emitter.
2.Cooking utensils are provided with wooden or ebonitehandles, since wood or ebonite is a bad conductor of heat.
3.Good conductor of heat are good conductors of electricity,Mica is an exception which being a good conductor of heat
is a bad conductor of electricity.
Example 9.
Compare the rate of loss of heat from a metal sphere oftemperature 827°C, with the rate of loss of heat from thesame sphere at 427 °C, if the temperature of surroundingsis 27°C.
Solution :
Given : T
1
= 827 °C = 1100 K, T
2
= 427 °C = 700 K
and T
0
= 27 °C = 300 K
According to Steafan Boltzmann law of radiation,
dQ
dt
= s AAe (T
4
– T
0
4
)
\
44 44
101
44 44
20
2
dQ
dt (T T) [(1100) (300) ]
dQ (T T ) [(700) (300) ]
dt
æö
ç÷
èø - -
==
æö --
ç÷
èø
= 6.276
or
12
dQ dQ
: 6.276 :1
dt dt
æöæö
=
ç÷ç÷ èøèø
Example 10.
The filament
of an evacuated light bulb has a length 10 cm,
diameter 0.2 mm and emissivity 0.2, calculate the power it
radiates at 2000 K. (
s = 5.67 × 10
–8
W/m
2
K
4
)
Solution :
l = 10 cm = 0.1 m, d = 0.2 mm = 2 × 10
–4
m
r = 0.1 mm = 1 × 10
–4
m,
e = 0.2, T = 2000 K, s = 5.67 × 10
–8
W/m
2
K
4
According to stefan's law of radiation, rate of emission of
heat for an ordinary body,
E = sAeT
4
= s(2 p r l) eT
4
[Q A = 2prl]
= 5.67 × 10
–8
× 2 × 3.14 × 1 × 10
–4
× 0.1 × 0.2 × (2000)
4
= 11.4 W
\ Power radiated by the filament = 11.4 W

293. 289Thermal Properties of Matter

294. 290 PHYSICS
1.The coeffi
cient of thermal conductivity depends upon
(a) temperature difference between the two surfaces.
(b) area of the plate
(c) material of the plate
(d) All of these
2.The wavelength of radiation emitted by a body depends
upon
(a) the nature of its surface
(b) the area of its surface
(c) the temperature of its surface
(d) All of the above
3.A vessel completely filled with a liquid is heated. If a and g
represent coefficient of linear expansion of material of vessel
and coefficient of cubical expansion of liquid respectively,
then the liquid will not overflow if
(a)g = 3 a (b)g > 3 a
(c)g < 3 a (d)g £ 3 a
4.The fastest mode of transfer of heat is
(a) conduction (b) convection
(c) radiation (d) None of these
5.In order that the heat flows from one part of a solid to
another part, what is required?
(a) Uniform density
(b) Temperature gradient
(c) Density gradient
(d) Uniform temperature
6.The sprinkling of water slightly reduces the temperature
of a closed room because
(a) temperature of water is less than that of the room
(b) specific heat of water is high
(c) water has large latent heat of vaporisation
(d) water is a bad conductor of heat
7.Which of the following is not close to a black body?
(a) Black board paint (b) Green leaves
(c) Black holes (d) Red roses
8.According to Newton’s law of cooling, the rate of cooling of
a body is proportional to (Dq)
n
, where Dq is the difference of
the temperature of the body and the surroundings, and n is
equal to
(a) two (b) three
(c) four (d) one
9.A metallic rod l cm long, A square cm in cross-section is
heated through tºC. If Young’s modulus of elasticity of the
metal is E and the mean coefficient of linear expansion is a per
degree celsius, then the compressional force required to
prevent the rod from expanding along its length is
(a) E A a t (b) E A a t/(1 + at)
(c) E A a t/(1 – a t) (d) E l a t
10.A black body rediates energy at the rate of E watt per metre
2
at a high temperature T K. when the temperature is reduced
to (T/2) K, the radiant energy will be
(a) E/16 (b) E/4
(c) E/2 (d) 2E
11.The tempearture of an isolated black body falls from T
1
to
T
2
in time t, then t is (Let c be a constant)
(a)
21
11
tc
TT
æö
=-
ç÷
èø
(b)
22
21
11
tc
TT
æö
=-ç÷
èø
(c)
33
21
11
tc
TT
æö
=-ç÷
èø
(d)
44
21
11
tc
TT
æö
=-ç÷
èø
12.Two straight metallic strips each of thickness t and length l
are rivetted together. Their coefficients of linear expansionsare a
1
and a
2
. If they are heated through temperature DT,
the bimetallic strip will bend to form an arc of radius
(a)
12
t /{
α α ) ΔT}+ (b)
21
t /{(α α ) ΔT}-
(c)
12
t(α α )ΔT- (d)
21
t(α α )ΔT-
13.Two solid spher
es, of radii R
1
and R
2
are made of the same
material and have similar surfaces. The spheres are raised to
the same temperature and then allowed to cool under
identical conditions. Assuming spheres to be perfect
conductors of heat, their initial rates of loss of heat are
(a)
2
2
2
1
/RR (b)
21
/RR
(c)
12
/RR (d)
2
1
2
2
/RR
14.A radiation of energy E falls normally on a perfectly
reflecting surface. The momentum transferred to the surface
is
(a)
Ec (b)2E/c
(c)E/c (d)
2
E/c
15.Which of the following graph correctly represents the
relation between Eln and Tln where E is the amount of
ra
diation emitted per unit time from unit area of body and T
is the absolute temperature
(a)
Eln
Tln
(b)
Eln
Tln
(c)
Eln
Tln
(d)
Eln
Tln

295. 291Thermal Properties of Matter
1.Th
e resistance of a resistance thermometer has values 2.71
and 3.70 ohms at 10ºC and 100ºC respectively. The
temperature at which the resistance is 3.26 ohm is
(a) 40ºC (b) 50ºC
(c) 60ºC (d) 70ºC
2.The temperature of an iron block is 140ºF. Its temperature
on the Celsius scale is
(a) 60º (b) 160º
(c) 140º (d) 132º
3.A pendulum clock is 5 seconds fast at temperature of 15ºC
and 10 seconds slow at a temperature of 30ºC. At what
temperature does it give the correct time? (take time interval =
24 hours)
(a) 18ºC (b) 20ºC
(c) 22ºC (d) 25ºC
4.If a bar is made of copper whose coefficient of linear
expansion is one and a half times that of iron, the ratio of
force developed in the copper bar to the iron bar of identical
lengths and cross-sections, when heated through the same
temperature range (Young’s modulus of copper may be taken
to be equal to that of iron) is
(a) 3/2 (b) 2/3
(c) 9/4 (d) 4/9
5.A metallic bar is heated from 0ºC to 100ºC. The coeficient of
linear expansion is 10
–5
K
–1
. What will be the percentage
increase in length?
(a) 0.01% (b) 0.1%
(c) 1% (d) 10%
16.For the construction of a thermometer, one of the essential
requirements is a thermometric substance which
(a) remains liquid over the entire range of temperatures to
be meaured .
(b) has property that varies linearly with temperature
(c) has a property that varies with temperature
(d) obey Boyle's law
17.The temperature of stars is determined by
(a) Stefan’s law
(b) Wien’s displacement law
(c) Kirchhoff’s law
(d) Ohm’s law
18.The temperature of the Sun is measured with
(a) platinum thermometer
(b) gas thermometer
(c) pyrometer
(d) vapour pressure thermometer.
19.Which of the following circular rods, (given radius r and
length l) each made of the same material and whose ends
are maintained at the same temperature will conduct most
heat?
(a) r = 2r
0
; l = 2l
0
(b) r = 2r
0
; l = l
0
(c) r = r
0
; l = 2l
0
(d) r = r
0
; l = l
0
20.At a certain temperature for given wave length , the ratio of
emissive power of a body to emissive power of black body
in same circumstances is known as
(a) relative emissivity (b) emissivity
(c) absorption coefficient (d) coefficient of reflection
21.If a liquid is heated in weightlessness, the heat is transmitted
through
(a) conduction
(b) convection
(c) radiation
(d) None of these, because the liquid cannot be heated in
weightlessness.
22The rate of heat flow through the cross-section of the rod
shown in figure is (T
2
> T
1
and thermal conductivity of the
material of the rod is K)
r
2
r
1
L
T
1 T
2
(a)
1221
K rr (TT)
L
p-
(b)
2
12 21
K (r r )(T T)
4L
p+-
(c)
2
11 21
K (r r ) (TT )
L
p+-
(d)
2
11 21
K (r r ) (TT )
2L
p+-
23.When a bo
dy is heated, which colour corresponds to high
temperature
(a) red (b) yellow
(c) white (d) orange
24.The two ends of a rod of length L and a uniform cross-
sectional area A are kept at two temperatures T
1
and T
2
(T
1
> T
2
). The rate of heat transfer,
dQ
dt
through the rod in
a steady state is given by
(a)
12
k(T T)dQ
dt LA
-
= (b) 12
dQ
kLA(TT)
dt
=-
(c)
12
kA (T T )dQ
dtL
-
= (d)
12
kL(T T)dQ
dtA
-
=
25.Two rods of t
he same length and areas of cross-section A
1
and A
2
have their ends at the same temperature. K
1
and K
2
are the thremal conductivities of the two rods. The rate of
flow of heat is same in both rods if
(a)
1 1
22
A K
AK
= (b)
1 2
21
A K
AK
=
(c)A
1
A
2
= K
1
K
2
(d) A
1
K
1
2
= A
2
K
2
2

296. 292 PHYSICS
6.Steam is pass
ed into 22 gm of water at 20ºC. The mass of
water that will be present when the water acquires a
temperatue of 90ºC (Latent heat of steam is 540 cal/g) is
(a) 24.83 gm (b) 24 gm
(c) 36.6 gm (d) 30 gm
7.A cylindrical rod of aluminium is of length 20 cms and radius
2 cms. The two ends are maintained at temperatures of 0ºC
and 50ºC the coefficient of thermal conductivity is
0.5 cal
cm×sec × ºC
. Then the thermal resistance of the rod in
cal
sec × ºC
is
(a) 318 (b) 31.
8
(c) 3.18 (d) 0.318
8.A metal ball of surface area 200 square cm, temperature 527ºCis surrounded by a vessel at 27ºC. If the emissivity of themetal is 0.4, then the rate of loss of heat from the ball is
approximately 8
22
joule
5.67 10
m secK
-
s=´
´´
(a) 108 jo
ule (b) 168 joule
(c) 182 joule (d) 192 joule
9.The rate of radiation of a black body at 0ºC is E joule per sec.
Then the rate of radiation of this black body at 273ºc will be
(a) 16 E (b) 8 E
(c) 4 E (d) E
10.Solar radiation emitted by sun resembles that emitted by a
black body at a temperature of 6000 K. Maximum intensity is
emitted at wavelength of about 4800 Å. If the sun were to
cool down from 6000 K to 3000 K, then the peak intensity
would occur at a wavelength
(a) 4800 Å (b) 9600 Å
(c) 2400 Å (d) 19200 Å
11.A bucket full of hot water is kept in a room and it cools from
75ºC to 70ºC in T
1
minutes, from 70ºC to 65ºC in T
2
minutes
and from 65ºC to 60ºC in T
3
minutes. Then
(a)T
1
= T
2
= T
3
(b) T
1
< T
2
< T
3
(c)T
1
> T
2
> T
3
(d) T
1
< T
2
> T
3
12.The maximum energy in the thermal radiation from a hot
source occurs at a wavelength of 11 ×10
–5
cm. According
to Wien’s law, the temperature of this source (on Kelvin
scale) will be n times the temperature of another source (on
Kelvin scale) for which the wavelength at maximum energy
is 5.5 × 10
–5
cm. The value n is
(a)2 (b) 4
(c) 1/2 (d) 1
13.The temperature of a furnace is 2324ºC and the intensity is
maximum in its radiation spectrum nearly at 12000 Å. If the
intensity in the spectrum of a star is maximum nearly at
4800Å, then the surface temperature of the star is
(a) 8400ºC (b) 7200ºC
(c) 6492.5ºK (d) 5900ºC
14.A mountain climber finds that water boils at 80ºC. The
temperature of this boiling water is...... Fahrenheit
(a) 50º (b) 150º
(c) 176º (d) 200º
15.A metal cube of length 10.0 mm at 0°C (273K) is heated to
200°C (473K). Given : its coefficient of linear expansion is
2 × 10
–5
K
–1
. The percent change of its volume is
(a) 0.1 (b) 0.2
(c) 0.4 (d) 1.2
16.No other thermometer is suitable as a platinum resistance
thermometer to measure temperatures in the entire range of
(a) –50ºC to + 350ºC (b) –200ºC to + 600ºC
(c) 0ºC to 100ºC (b) 100ºC to 1500ºC
17.Calculate the surface temperature of the planet, if the energy
radiated by unit area in unit time is 5.67 ×10
4
watt.
(a) 1273°C (b) 1000°C
(c) 727°C (d) 727K
18.In a thermocouple, the temperature of the cold junction and
the neutral temperature are –40°C and 275°C respectively. If
the cold junction temperature is increased by 60°C, the
neutral temperature and temperature of inversion
respectively become
(a) 275°C, 530°C (b) 355°C, 530°C
(c) 375°C, 590°C (d) 355°C, 590°C
19.In a surrounding medium of temperature 10°C, a body takes
7 min for a fall of temperature from 60°C to 40°C. In what time
the temperature of the body will fall from 40°C to 28°C?
(a) 7 min (b) 11 min
(c) 14 min (d) 21 min
20.Two rods P and Q of same length and same diameter having
thermal conductivity ratio 2 : 3 joined end to end. If
temperature at one end of P is 100°C and at one and of Q
0°C, then the temperature of the interface is
(a) 40°C (b) 50°C
(c) 60°C (d) 70°C
21.100 g of ice is mixed with 100 g of water at 100ºC. What will
be the final temperature of the mixture ?
(a) 13.33ºC (b) 20ºC
(c) 23.33ºC (d) 40ºC
22.A crystal has a coefficient of expansion 13×10
–7
in one
direction and 231 × 10
–7
in every direction at right angles to
it. Then the cubical coefficient of expansion is
(a) 462 × 10
–7
(b) 244 × 10
–7
(c) 475 × 10
–7
(d) 257 × 10
–7
23.A glass flask of volume 1 litre is fully filled with mercury at
0ºC. Both the flask and mercury are now heated to 100ºC. If
the coefficient of volume expansion of mercury is 1.82 × 10
–
4
/ºC, volume coefficient of linear expansion of glass is 10 ×
10
–6
/ºC, the amount of mercury which is spilted out is
(a) 15.2 ml (b) 17.2 ml
(c) 19.2 ml (d) 21.2 ml

297. 293Thermal Properties of Matter
24.A re
ctangular block is heated from 0ºC to 100ºC. The
percentage increase in its length is 0.10%. What will be the
percentage increase in its volume?
(a) 0.03% (b) 0.10%
(c) 0.30% (d) None of these
25.The coefficient of apparent expansion of mercury in a glass
vessel is 153 × 10
–6
/ºC and in a steel vessel is 144 × 10
–6
/ºC.
If a for steel is 12 × 10
–6
/ºC, then, that of glass is
(a) 9 × 10
–6
/ºC (b) 6 × 10
–6
/ºC
(c) 36 × 10
–6
/ºC (d) 27 × 10
–6
/ºC
26.A beaker contains 200 gm of water. The heat capacity of the
beaker is equal to that of 20 gm of water. The initial
temperatue of water in the beaker is 20ºC. If 440 gm of hot
water at 92ºC is poured in it, the final temperature, neglecting
radiation loss, will be nearest to
(a) 58ºC (b) 68ºC
(c) 73ºC (d) 78ºC
27.Steam at 100ºC is passed into 1.1 kg of water contained in a
calorimeter of water equivalent 0.02 kg at 15ºC till the
temperature of the calorimeter and its contents rises to
80ºC.The mass of the steam condensed in kg is
(a) 0.130 (b) 0.065
(c) 0.260 (d) 0.135
28.A body of mass 5 kg falls from a height of 20 metres on the
ground and it rebounds to a height of 0.2 m. If the loss in
potential energy is used up by the body, then what will be the
temperature rise?
(specific heat of material = 0.09 cal gm
–1
ºC
–1
)
(a) 0ºC (b) 4ºC
(c) 8ºC (d) None of these
29.80 g of water at 30ºC are poured on a large block of ice at 0ºC.
The mass of ice that melts is
(a) 1600 g (b) 30 g
(c) 150 g (d) 80 g
30.Ice starts forming in a lake with water at 0ºC when the
atomspheric temperature is –10ºC. If the time taken for the
first 1 cm of ice to be formed is 7 hours, then the time taken
for the thickness of ice to change from 1 cm to 2 cm is
(a) 7 hours (b) 14 hours
(c) 21 hours (d) 3.5 hours
31.Two identical rods of copper and iron are coated with wax
uniformly. When one end of each is kept at temperature of
boiling water, the length upto which wax melts are 8.4 cm
amd 4.2 cm, respectively. If thermal conductivity of copper
is 0.92, then thermal conductivity of iron is
(a) 0.23 (b) 0.46
(c) 0.115 (d) 0.69
32.Two vessels of different materials are similar in size in every
respect. The same quantity of ice filled in them gets melted
in 20 min and 35 min, respectivley. The ratio of coefficients
of thermal conduction of the metals is
(a) 4 : 7 (b) 7 : 4
(c) 25 : 16 (d) 16 : 25
33.A partition wall has two layers A and B, in contact, each made
of a different material. They have the same thickness but the
thermal conductivity of layer A is twice that of layer B. If the
steady state temperature difference across the wall is 60 K,
then the corresponding difference across the layer A is
(a) 10 K (b) 20 K
(c) 30 K (d) 40 K
34.The spectral energy distribution of the sun (temperature
6050 K) is maximum at 4753 Å. The temperature of a star for
which this maximum is at 9506 Å is
(a) 6050 K (b) 3025 K
(c) 12100 K (d) 24200 K
35.Two bodies A and B are placed in an evacuated vessel
maintained at a temperature of 27ºC. The temperature of A is
327ºC and that of B is 227ºC. The ratio of heat loss from A
and B is about
(a) 2 : 1 (b) 1 : 2
(c) 4 : 1 (d) 1 : 4
36.The thermal capacity of 40g of aluminium (specific heat =
0.2 cal g
–1
°C
–1
) is
(a) 40 cal °C
–1
(b) 160 cal °C
–1
(c) 200 cal °C
–1
(d) 8 cal °C
–1
37.Which of the following temperatures is the highest?
(a) 100K (b) –13°F
(c) –20°C (d) –23°C.
38.5 kg of water at 10°C is added to 10 kg of water at 40°C.
Neglecting heat capacity of vessel and other losses, the
equilibrium temperature will be
(a) 30°C (b) 25°C
(c) 35°C (d) 33°C
39.A beaker contains 200 g of water. The heat capacity of beaker
is equal to that 20 g of water. The initial temperature of water
in the beaker is 20°C . If 440 g of hot water at 92°C is poured
in, the final temperature, neglecting radiation loss, will be
(a) 58°C (b) 68°C
(c) 73°C (d) 78°C
40.The rectangular surface of area 8 cm × 4 cm of a black body
at temperature 127°C emits energy E per second. If the length
and breadth are reduced to half of the initial value and the
temperature is raised to 327°C, the rate of emission of energy
becomes
(a)
8
3
E (b)
16
81
E
(c)
16
9
E (d)
64
81
E

298. 294 PHYSICS
41.Two ro
ds of same length and transfer a given amount of
heat 12 second, when they are joined as shown in figure (i).
But when they are joined as shwon in figure (ii), then they
will transfer same heat in same conditions in
l
l l
Fig. (i)
Fig. (ii)
(a) 24 s (b) 13 s
(c
) 15 s (d) 48 s
42.A slab consists of two parallel layers of copper and brass ofthe same thickness and having thermal conductivities in
the ratio 1 : 4. If the free face of brass is at 100°C and that of
copper at 0°C, the temperature of interface is
(a) 80°C (b) 20°C
(c) 60°C (d) 40°C
43.540 g of ice at 0°C is mixed with 540 g of water at 80°C. The
final temperature of mixture is
(a) 0°C (b) 40°C
(c) 80°C (d) less than 0°C
44.Three identical rods A, B and C of equal lengths and equal
diameters are joined in series as shown in figure. Their
thermal conductivities are 2k, k and k/2 respectively. The
temperatures at two junction points are
T
1
A B C 0°C
0.5°Ck2k
100°C
T
2
(a) 85.7, 57.1°C (b) 80.85,
50.3°C
(c) 77.3, 48.3°C (d) 75.8, 49.3°C
45.Which one of the following graphs best represents the waysin which the total power P radiated by a black body dependsupon the thermodynamic temperature T of the body?
(a) TO
P
(b)
TO
P
(c)
TO
P
(d)
TO
P
46.A black body at 227°C radiates heat at the rate of 7 cals/
cm
2
s. At a temperature of 727°C, the rate of heat radiated in
the same units will be
(a) 50 (b) 112 (c) 80 (d) 60
47.The top of an insulated cylindrical container is covered by
a disc having emissivity 0.6 and conductivity 0.167
WK
–1
m
–1
and thickness 1 cm. The temperature is maintained
by circulating oil as shown in figure. Find the radiation loss
to the surrounding in Jm
–2
s
–1
if temperature of the upper
surface of the disc is 27°C and temperature of the
surrounding is 27°C.
Oil out
Oil in
(a) 595 Jm
–2
s
–1
(b
) 545 Jm
–2
s
–1
(c) 495 Jm
–2
s
–1
(d) None of these
48.Two marks on a glass rod 10 cm apart are found to increasetheir distance by 0.08 mm when the rod is heated from 0°C to
100°C. A flask made of the same glass as that of rod measures
a volume of 1000 cc at 0°C. The volume it measures at 100°C
in cc is
(a) 1002.4 (b) 1004.2
(c) 1006.4 (d) 1008.2
DIRECTIONS for Qs. 49 to 50 : These are Assertion-Reason
type questions. Each of these question contains two statements:
Statement-1 (Assertion) and Statement-2 (Reason). Answer
these questions from the following four options.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a
correct explanation for Statement -1
(b) Statement-1 is True, Statement -2 is True; Statement-2 is
NOT a correct explanation for Statement - 1
(c) Statement-1 is True, Statement- 2 is False
(d) Statement-1 is False, Statement -2 is True
49. Statement 1 : The equivalent thermal conductivity of two
plates of same thickness in series is less than the smaller
value of thermal conductivity.
Statement 2 : For two plates of equal thickness in series the
equivalent thermal conductivity is given by
21K
1
K
1
K
1
+=
50. Statement 1 : As
the temperature of the black body
increases, the wavelength at which the spectral intensity( E
l
) is maximum decreases.
Statement 2 : The wavelength at which the spectral intensity
will be maximum for a black body is proportional to the fourth
power of its absolute temperature.

299. 295Thermal Properties of Matter
Exemplar Questions
1.A bimetallic strip is made of aluminium and steel (a
Al >
a
steel). On heating, the strip will
(a) remain straight
(b) get twisted
(c) will bend with aluminium on concave side
(d) will bend with steel on concave side
2.A uniform metallic rod rotates about its perpendicular
bisector with constant angular speed. If it is heated uniformly
to raise its temperature slightly
(a) its speed of rotation increases
(b) its speed of rotation decreases
(c) its speed of rotation remains same
(d) its speed increases because its moment of inertia
increases
3.The graph between two temperature scales A and B is shown
in figure between upper fixed point and lower fixed point
there are 150 equal division on scale A and 100 on scale B.
The relationship for conversion between the two scales is
given by
O 100
180
Dt
B
= 100°
Dt
A
= 150°
Temperature (°B)
Temperature (°A)
(a)
180
100 150
AB
tt-
= (b)
30
150 100
AB
tt-
=
(c)
180
150 100
BA
tt-
= (d)
40
100 180
BA
tt-
=
4.An aluminium sphere
is dipped into water. Which of the
following is true?
(a) Buoyancy will be less in water at 0°C than that in water
at 4°C
(b) Buoyancy will be more in water at 0°C than that in water
at 4°C
(c) Buoyancy in water at 0°C will be same as that in water
at 4°C
(d) Buoyancy may be more or less in water at 4°C depending
on the radius of the sphere
5.As the temperature is increased, the period of a pendulum
(a) increases as its effective length increases even though
its centre of mass still remains at the centre of the bob
(b) decreases as its effective length increases even though
its centre of mass still remains at the centre of the bob
(c) increases as its effective length increases due to shifting
to centre of mass below the centre of the bob
(d) decreases as its effective length remains same but the
centre of mass shifts above the centre of the bob
6.Heat is associated with
(a) kinetic energy of random motion of molecules
(b) kinetic energy of orderly motion of molecules
(c) total kinetic energy of random and orderly motion of
molecules
(d) kinetic energy of random motion in some cases and
kinetic energy of orderly motion in other
7.The radius of a metal sphere at room temperature T is R and
the coefficient of linear expansion of the metal is a. The
sphere heated a little by a temperature DT so that its new
temperature is T + DT. The increase in the volume of the
sphere is approximately.
(a)2 RTp aD (b)
2
RTp aD
(c)
3
4 /3RTp aD (d)
3
4RTp aD
8.A sphere, a cube and a thin circular plate, all of same material
and same mass are initially heated to same high temperature.
(a) Plate will cool fastest and cube the slowest
(b) Sphere will cool fastest and cube the slowest
(c) Plate will cool fastest and sphere the slowest
(d) Cube will cool fastest and plate the slowest
Past Years (2013-2017) NEET/AIPMT Questions
9.A piece of iron is heated in a flame. It first becomes dull red
then becomes reddish yellow and finally turns to white hot.
The correct explanation for the above observation is
possible by using [2013]
(a) Wien’s displacement law
(b) Kirchoff’s law
(c) Newton’s law of cooling
(d) Stefan’s law
10.The density of water at 20°C is 998 kg/m
3
and at 40°C 992
kg/m
3
. The coefficient of volume expansion of water is
[NEET Kar. 2013]
(a) 10
–4
/°C (b) 3 × 10
–4
/°C
(c) 2 × 10
–4
/°C (d) 6 × 10
–4
/°C

300. 296 PHYSICS
11.Two me
tal rods 1 and 2 of same lengths have same
temperature difference between their ends. Their thermal
conductivities are K
1
and K
2
and cross sectional areas A
1
and A
2
, respectively. If the rate of heat conduction in rod 1
is four times that in rod 2, then [NEET Kar. 2013]
(a)K
1
A
1
= K
2
A
2
(b)K
1
A
1
= 4K
2
A
2
(c)K
1
A
1
= 2K
2
A
2
(d)4K
1
A
1
= K
2
A
2
12.Certain quantity of water cools from 70°C to 60°C in the first
5 minutes and to 54°C in the next 5 minutes. The temperature
of the surroundings is: [2014]
(a) 45°C (b) 20°C
(c) 42°C (d) 10°C
13.Steam at 100°C is passed into 20 g of water at 10°C. When
water acquires a temperature of 80°C, the mass of water
present will be: [2014]
[Take specific heat of water = 1 cal g
– 1
°C
– 1
and latent heat
of steam = 540 cal g
– 1
]
(a) 24 g (b) 31.5 g
(c) 42.5 g (d) 22.5 g
14.On observing light from three different stars P, Q and R, it
was found that intensity of violet colour is maximum in the
spectrum of P, the intensity of green colour is maximum in
the spectrum of R and the intensity of red colour is maximum
in the spectrum of Q. If T
P
, T
Q
and T
R
are the respective
absolute temperature of P, Q and R, then it can be concluded
from the above observations that [2015]
(a)T
P
> T
R
> T
Q
(b) T
P
< T
R
< T
Q
(c)T
P
< T
Q
< T
R
(d) T
P
> T
Q
> T
R
15.The two ends of a metal rod are maintained at temperatures
100°C and 110°C. The rate of heat flow in the rod is found to
be 4.0 J/s. If the ends are maintained at temperatures 200°C
and 210°C, the rate of heat flow will be [2015]
(a) 16.8 J/s (b) 8.0 J/s
(c) 4.0 J/s (d) 44.0 J/s
16.The value of coefficient of volume expansion of glycerine is
5 × 10
-4
K
-1
. The fractional change in the density of glycerine
for a rise of 40°C in its temperature, is: [2015 RS]
(a) 0.020 (b) 0.025
(c) 0.010 (d) 0.015
17.A piece of ice falls from a height h so that it melts completely.
Only one-quarter of the heat produced is absorbed by the
ice and all energy of ice gets converted into heat during its
fall. The value of h is : [2016]
[Latent heat of ice is 3.4 × 10
5
J/kg and g = 10 N/kg]
(a) 34 km (b) 544 km
(c) 136 km (d) 68 km
18.Coefficient of linear expansion of brass and steel rods are
a
1
and a
2
. Lengths of brass and steel rods are
1
land
2
l
respectively. If
21
()ll-is maintained same at all
temperatures, which one of the following relations holds
good ? [2016]
(a)
2212 21
lla =a (b)
12 21
lla =a
(c)
12 21
lla =a (d)
11 22
a =all
19.A black body is at a temperature of 5760 K. The energy of
radiation emitted by the body at wavelength 250 nm is U
1
,
at wavelength 500 nm is U
2
and that at 1000 nm is U
3
. Wien's
constant, b = 2.88 × 10
6
nmK. Which of the following is
correct ? [2016]
(a)U
1
= 0 (b) U
3
= 0
(c)U
1
> U
2
(d) U
2
> U
1
20.A spherical black body with a radius of 12 cm radiates 450watt power at 500 K. If the radius were halved and thetemperature doubled, the power radiated in watt would be :
(a) 450 (b) 1000 [2017]
(c) 1800 (d) 225
21.Two rods A and B of different materials are welded together
as shown in figure. Their thermal conductivities are K
1
and
K
2
. The thermal conductivity of the composite rod will be :
A K
1
B K
2
T
1
T
2
d
[2017]
(a)
12
3(K K)
2
+
(b) K
1
+ K
2
(c) 2 (K
1
+ K
2
) (d)
12
KK
2
+

301. 297Thermal Properties of Matter
EXERCISE - 1
1. (c) 2. (d) 3. (d) 4.(c)5.(b)
6. (c) 7. (a) 8. (d)
9. (a)
strain
stress
/
A/F
E =
D
=
ll

where Dl=(l'–l) = lat so F = EAat
10. (a) 11. (c)
12. (b) Let the angle subtended by the arc formed be q. Then
21
12
rrr
or
r -
-
=
D
D
=q=q
llll
t
T)(
12Da-a
=q\
l

t
T)(
r
or
12Da-a
=
ll
So,
T)(
t
r
12Da-a
=
13. (a
) Initial rate of loss of heat
2
2
2
1
2
4
1
4
R
R
eAT
eAT
=
´´s
´´s
=
14. (b) M
omentum of photon
E
c
=
Change in momentum
2E
c
=
= momentum transf
erred to the surface
(the photon will reflect with same magnitude of
momentum in opposite direction).
15. (c) We know
4
eTE= Tln4elnEln+=Þ
negativeiseln\
1ebecause<
Eln
Tln
16. (c) The temperature is measured by the value of the
thermodynamic property of a substance i.e., the
property which varies linearly with the temperature.
17. (b) Temperature of stars can be determined by Wein’s
displacement law
=l
m
constant
18. (c)
19. (b) We know that Q =
T T
R
H L
-
Also Thermal resistance, R =
l l
KAKr
=
p
2
Heat flow
will be maximum when thermal resistance is
minimum. From given option
(i) r = 2r
0
, l = 2l
0

\ = =R
K r Kr
2
2 2
0
0
2
0
0
2
l l
p p()
(ii) r = 2r
0
,
l = l
0

\ = =R
K r Kr
l l
0
0
2
0
0
2
2 4p p()
(iii) r = r
0
,
l = 2l
0

\ = =R
Kr Kr
2 2
0
0
2
0
0
2
l l
p p
(iv) r = r
0
, l
= l
0

\ = =R
Kr Kr
l l
0
0
2
0
0
2
p p
It is clear that f
or option (b) resistance is minimum,
hence heat flow will be maximum.
20. (c)
21. (a) In condition of weightlessness, convection is not
possible.
22. (a)
eff 12
r rr=
2 1 1221
KA(T T ) K r r (T T )dQ
dtLL
- p-
==
23
. (c) When a body is heated then relation between colours
and temperature is according to Prevost’s theory of
radiation which states that everybody emitting radiant
energy in all directions at a rate depending only on the
nature of its surface and its temperature e.g., when a
body is placed in an enclosure (furnace) it would acquire
the temperature of furnace and seem white means
radiate white light. So it becomes first dark and then
white.
24. (c)
12
kA (T T )dQ
dtL
-
=
[(T
1
–T
2
) is th
e temperature difference]
25. (b)
EXERCISE - 2
1.
(c) )t1(RR
0t
a+=
)101(R71.2
0
´a+= ... (1)
)1001(R70.3
0
´a+= ... (2)
)t1(R26.3
0
a+= ... (3)
Solve these equations to obtain the value of t.
2. (a)
5
C
9
32140
=
-
or C
9
160700
=
-
or C = 60ºC
3.
(c)
tT
2
1
t ´Da=D
Hints & Solutions

302. 298 PHYSICS
86400)15T(
2
1
5 ´-a=\
and 86400)T30(
2
1
10 ´-a=
4. (a) aµa= ForAtYF
(Q Y t A is same
for both copper and iron)
or
CC
ΙΙ
FαandFαµµ
2
3
1
2/3
F
F
C
==\
I
5. (b)
Δ
αΔT=
l
l

53
10 100 10
--
= ´=
3Δ
100% 10 100
-
´ =´
l
l

1
10 0.1%
-
==
6. (a) Let m be the
mass of steam condensed. Then
m × 540 + m × 10/2 = 22 × 70
\ m = 2.83 gm
Now, total mass = 22 + 2.83 = 24.83 gm
7. (d) Thermal resistance R =l/KA
Where l =20cm. & A(cylindrical rod)
=pr
2
= 40pcm
2
So
==
´p´
20 cal
R 0.318
0.5 40 sec ? C
8. (c) According to Stefan’s Law, the rate of loss of heat is
e)TT(A
t
Q 4
2
4
1
´-s=
here s = 5.67 ×
10
–8
J/m
2
× sec.K
2
,
T
1
= 527+273 = 800K,
T
2
= 27 + 273 = 300K & A = 200×10
–4
m
2
So.
82
44
Q
5.67 10 2 10
t
[(800) (300) ] 0.4
--
= ´ ´´
-´
@ 182 joule
9. (
a) According to Stefan’s Law
energy radiated per sec E = sAT
4
(here e = 1 for black body)
for first case E = sA(273)
4
for second case E
1
=sA(546)
4
so E
1
=16E
10. (b) According to Wein's displacement law,
λT = b.
m
where b =2.884 × 10
–3
mK.
So l =l
mm1212
TT
or 4800 × 6000 = l´
m2
3000 or l=°
m2
9600A
11. (b) The time of cooling increases as the difference between
the temperature of body & surrounding is reduced. So
T
1
< T
2
< T
3
(according to Newton’s Law of cooling).
12. (c)
55
1211.0 10 T 5.5 10 T
--
´ =´
2
1
T
T
2
1
=
13. (c)
14. (c)
5
80
9
32F
=
-
F – 32 = 144
or F = 176ºF.
15. (d) Percentage change in volume = gt × 100
t300100t3
a=´a= 2.1200102300
5
=´´´=
-
16. (b)
17. (c
) According to Stefan's Boltzmann law, the energy
radiated per unit time E = sAT
4
.
\ 5.67 × 10
4
= 5.67 × 10
–8
× 1 × T
4
(\ A = 1 m
2
)
or,
1/4
4
–8
5.67 10
T 1000K
5.67 10
æö´
==ç÷
´èø
or , T = 1000 – 273 = 727°C.
18. (
a) When the temperature of cold junction is increased,
the neutral temperature remains constant for a given
thermocouple and it is independent of the temperature
of the cold junction.
\ q
n
= 275°C, q
i
– q
n
= q
n
– q
c
\ q
i
= 2q
n
– q
c
= 530°C.
19. (a) According to Newton's law of cooling,
12 12
0
K
t2
q -q q +qéù
= -q
êú
ëû
where q
0
is the surrounding temperature.
\
60 40 60 40
K 10
72
-+ æö
=-ç÷
èø
Þ
20
40K
7
= Þ
1
K
14
=
\
40 28 40 28
K 10
t2
-+ éù
=-
êú
ëû
Þ
12
24K
t
=
or
12 12 14
t 7 min
24K 24
´
===
20. (a)
Let q be temperature of interface.
\
QP
K A( 0)KA(100) q--q
=
ll
\
P
Q
K
K 100
q
=
-q
Þ
2
3 100
q
=
-q
or, 200 – 2q = 3q or, 5q = 200 or, q = 40°C
21. (a) Let the final temperature of mixture be T.
Then 100 × 80 + 100 (T – 0) ×½
(as specific heat of ice is 0.5 cal/g C° and specific heat
of water is 1 cal/gC°)
= 100 × 1 × (100 – T)
Solving, we get T = 13.33ºC.
22. (a)
321
a+a+a=g

777
10231102311013
---
´+´+´=

7
10475
-
´=

303. 299Thermal Properties of Matter
23. (
a) T)(VV
gm0
Dg-g=D
100)]1010(31082.1[1
64 --
´´-´=
100)]103.01082.1[1
44 --
´-´= = 15.2 ml
24. (c) Given Δ/ 0.10% 0.001 andΔT 100
ºC===ll
Now
Δ
αΔT=
l
l
or 0.001α100=´
or
5
α10/ºC
-
=
Furthe
r
5
γ 3α310/
ºC
-
= =´
3ΔV
100 (3 10 )(100) 0.30%
V
-
\´=´=
25
. (a) We know that
real apparent vessel
γγγ=+
So, app vessel glass app vessel steel
(γ γ ) ( γ γ )+ =+
(Q
real
g is same in both cases)
or
6
vessel glass
153 10 ( γ )
-
´+
steelvessel
6
)(10
144 g+´=
-
Further
vessel steel
(γ)
66
3 3(121 0) 3610/ºC
--
=a=´ ´
=´
6
vessel glass
153 10 (γ )
-
\´+
66
1036
10144
--
´+´=
Solving we get
6
vessel glass
( ) 27 10 /º C
-
g =´
or
glass 6
910/ºC
3
-
g
a= =´
26.
(b) Let the final temperature be T.
Then 200 × 1 × (T – 20) + 20 × (T – 20)
= 440 (92 – T)
Solving it, we get T = 68ºC.
27. (a) )80100(mmL -+
)1580(02.0)15–80(11.1-´+´´=
30.15.71m20540m
+=+´
80.72m560= \ 130.0m=
28. (d) hmgmghW WW
21
¢-=-= )hh(mg ¢-=
8.19105)2.020(105´´=-´=
joule9901985=
´=
This energy is converted into heat when the ball strikes
the earth. Heat produced is
calorie
2.4
990
Q=
Cº
32
11
09.0500042
10099
mc
Q
T =
´´
´
==D
29. (b) 80
× 1 × 30 = m × 80
Þ m = 30 gm.
30. (c)
3
1. (a) Use
2
2
2
1
2
1
K
K
l
l
=
32. (b
) Here flow of heat, area of vessels & temperature
gradient are same so
1122
dT dT
QKAtKAt
dx dx
==
so
4
7
20
35
t
t
K
K
1
2
2
1
===
33. (b)
34
. (b)
11 22
λT λT=
2
T.950647536050 =´
K3025T
2
=
35. (a)
)TT(
)TT(
E
E
4
0
4
2
4
0
4
1
2
1
-s
-s
=

44
44
)300()500(
)300()600(
-
-
=
36. (d)
Thermal capacity = 40 × 0.2 = 8 cal °C
–1
37. (b) –13°F is (13 + 32)° below ice point on F scale.
38 (a) Mass ratio 1 : 2; hence DT ratio 2 : 1. Therefore
equilibrium temp is 30°C.
39. (b) Heat capacity of cold : hot =1 : 2
So final temp. is
3
1
× (2 × 92 + 1 × 20) = 68°
C.
40. (d)
4
TareaE´´s=; T increases by a factor
2
3
.
Area incr
eases by a factor
4
1
.
41. (d)
2
t , t'
A A/2
µµ
ll
42. (a) q= q-Þ-q=q-4400)0(K)100(K4
80CÞq=°
43. (a)540 × 80 + 540 q = 540 (80 – q)
°=qÞ=qÞq-=q+Þ 0028080
44. (a)
45
. (c) The total power radiated by a black body of area A at
temperature TK is given by P = AsT
4
Where s = Stefan's constant
= 5.7 × 10
–8
W m
–2
K
–4
Which is best represented in graph. (c)
46. (b) According to Stefan’s law
4
E T,=s
T
1
= 500 K
T
2
=
1000 K
4 4
22
11
ET 1000
16
E T 500
æö æö
=== ç÷ç÷
èøèø
\ E
2
= 16 × 7
= 112 cal / cm
2
s
47. (a) The rate of heat loss per unit area due to radiation
= Îs (T
4
–T
0
4
)
= 0.6 × 5.67 × 10
–8
[(400)
4
–(300)
4
] = 595 Jm
–2
s
–1
.

304. 300 PHYSICS
48. (a)
21
121
VV
;
V(T T)
-
g=
-
21
121
(T T)
-
a=
-
ll
l
T
1
= 0°C, T
2
= 100°C
2 1 21
11
VV
;
100V 100
--
g= a=
ll
l
l
1
= 10 cm; l
2
–l
1
= 0.08 mm = 0.008 cm
60.008
8 10 / C;
10 100
-
a= =´°
´
g = 3a = 24 × 10
–6
/ °C
\ 24 × 10
–6

2
V 1000
1000 100
-
=
´
V
2
– 1000 = 24 × 10
–6
× 10
5

= 2.4 \ V
2
= 1002.4 cc
49. (a) For equivalent thermal conductivity, the relation is
21RK
1
K
1
K
1
+=
;If KKK
21
==
2
K
K
K
2
K
1
K
1
K
1
R
R
=Þ=+=
Which is les
s than K.
If
21
KK> suppose xKK
21
+=
21
1 2 12
KK111
K K K KK
+
=+=
22
22
KKx1
K (K x)K
++
Þ=
+
Þ
xK2
xKK
K
2
2
2
2
+
+
=
Now, KK
2
- =
xK2
xKK
K
2
2
2
2
2
+
+
-
)xK2(
xKKxKK2
2
2
2
22
2
2
+
--+
=
2
2
2
K
positive
2Kx
==
+
So, K
2
> K, s
o the value of K is smaller than K
2
and K
1
.
50. (c) From Wein's law l
m
T = constant i.e. peak emission
wavelength l
m
µ
1
T
.
Hence as T increa
ses l
m
decreases.EXERCISE - 3
Exemplar Questions
1. (d) If strips of aluminium and steel are fixed together on
metallic strip and both are heated then (a
Al > a
steel)
aluminium will exapnd more because the metallic strip
with higher coefficient of linear expansion (a
Al) will
expand more Thus it should have larger radius of
curvature. Hence, aluminium will be on convex side.
q
O
Aluminium
Steel
2. (b) On h
eating a uniform metallic rod its length will increase
so moment of inertia of rod increased from I
1 to I
2.
No external torque is acting on the system so angularmomentum should be conserved.
L = Angular momentum = Iw = constant
11 22
IIw=w
Rod
So,
21
12
1
I
I
w
=<
w
w
2 < w
1
(Q Due to expansion of the rod I
2 > I
1)
So, angular veloctiy decreases.
3. (b) In the given graph shows lowest fixed point for scale A
is 30° and lowest point for scale B is 0°. Upper fixed
point for the scale A is 180° and upper fixed point for
scale B is 100°. Hence, formula is
(LEP) (LEP)
(UFP) (LFP) (UFP) (LFP)
AA BB
AA BB
tt--
=
--
where , LFP -
Lower fixed point, UFP - Upper fixed
point.
O 100
180
Dt
B
= 100°
Temperature (°B)
Te mpe rature(°A)
Dt
A
= 150°
30°
90°–q
q
C
B(+A)
O
(+ B)
So,
300
180 30 100 0
AB
tt--
=
--
30
150 100
-
=
AB
tt
4. (a) As we know that, the Buoyant force (F) on a body
volume (V) and density of (r), when immersed in liquid
of density (r
l) is '
l
Vg=r
where V ' = volume of displaced liquid by dipped body
(V).
Let volume of the sphere is V and r is its density, then
we can write buoyant force
s
FVG=r
l
Fµr (for liquid)
44 00
1
CC
CC
F
F
°°
°°
r
=>
r
40
()
CC°°
r >rQ
40CC
FF
°°
>
Hence, buoy
ancy will be less in water at 0°C than that
in water at 4°C .

305. 301Thermal Properties of Matter
5.
(a) As the temperature increased the length (L) of the
pendulum increases due to expansion i.e., linear.
Pendulum
L
So, Time per
iod of pendulum
2
L
T
g
=p
TLµ
Hence on i
ncreasing temperature, time period (T) also
increases.
6. (a) As we know that when the temperature increases
vibration of molecules about their mean position
increases hence the kinetic energy associated with
random motion of molecules increases.
7. (d) Let the radius of the sphere is R. As the temperature
increases radius of the sphere increases as shown.
R
dV
Original volume
34
3
VR=p
Coefficient of linear expansion = a
Coefficient of volume expansion = 3a
As we know that
V
Y
Vt
D
=
D
By putting the value of V, increase in the volume
3V VtÞ D = aD
3
4Rt= p aD
8. (c)Loss of heat temperature on cooling temperature
increase depend on material of object surface area
exposed to surrounding and temperature difference
between body and surrounding.
Let us consider the diagram where all the three objects
are heated to same temperature T. As we know that
density
mass
volume
r=
where r is
same for all the three objects hence, volume
will also be same.
m
T
PlateCubeSphere
T
T
m
m
As thickness of the plate is least so, surface area of the
plate is maximum.
We know that, according to Stefan's law of heat loss
4
H ATµ
where, A is surface area of object and T is temperature.
So, H
sphere : H
cube : H
plate
= A
sphere : A
cube : A
plate
So area of circular plate is maximum.For sphere, as the sphere is having minimum surface
area.
Hence, the sphere cools slowest and circular plate will
cool faster.
Past Years (2013-2017) NEET/AIPMT Questions
9. (a) Wein’s displacement law
According to this law
l
max
µ
1
T
or, l
max
× T = co nstant
So, as the temperature increases l decreases.
10. (b) From question,
Dr = (998 – 992) kg/m
3
= 6 kg/m
3
r =
33998 992
kg/m 995 kg/m
2
+
=
r =
m
V
Þ
V
V
DrD
=-
r
Þ
V
V
DrD
=
r
\ Coefficien
t of volume expansion of water,
11V
Vtt
D Dr
=
D rD
=
46
310 /C
995 20
-
»´°
´
11. (b) Q
1
= 4Q
2
(G
iven)
Þ
11 22
4
KAt KAt
LL
DD
= Þ K
1
A
1
= 4K
2
A
2
.
12. (
a) Let the temperature of surroundings be q
0
By Newton's law of cooling12 12
0
k
t2
q -q q +qéù
= -q
êú
ëû
Þ 0
70 60 70 60
k
52
-+ éù
= -q
êú
ëû
Þ2 = k [65 – q
0
] ...(i)

306. 302 PHYSICS
Similarly
,
0
60 54 60 54
k
52
-+ éù
= -q
êú
ëû
Þ
6
5
= k [57 – q
0
] ...(ii)
B
y dividing (i) by (ii) we have0
0
6510
6 57
-q
=
-q
Þq
0
= 45º
13. (d) Ac
cording to the principle of calorimetry.
Heat lost = Heat gained
mL
v
+ ms
w
Dq = m
w
s
w
Dq
Þm × 540 + m × 1 × (100 – 80)
= 20 × 1 × (80 – 10)
Þm = 2.5 g
Therefore total mass of water at 80°C
= (20 + 2.5) g = 22.5 g
14. (a) From Wein’s displacement law
l
m
× T = constant
P – max. intensity is at violet
Þ l
m
is minimum Þ temp maximum
R – max. intensity is at green
Þ l
m
is moderate Þ temp moderate
Q – max. intensity is at red Þ l
m
is maximum Þ temp.
minimum i.e., T
p
> T
R
> T
Q
15. (c) As the temperature difference DT = 10°C as well as the
thermal resistance is same for both the cases, so thermal
current or rate of heat flow will also be same for both
the cases.
16. (a) From question,
Rise in temperature Dt = 40°C
Fractional change in the density
0
Dr
r
= ?
Coefficient
of volume expansion
g = 5 × 10
–4
K
–1
r = r
0
(1 –gDt)
Þ
0
Dr
r
= gDT = (5 × 10
–4
) (40 ) = 0.02
17. (c) According to question only one-quarter of the heat
produced by falling piece of ice is absorbed in themelting of ice.
i.e.,
mgh
4
= mL
Þh =
5
4L 4 3.4 10
136 km
g 10
´´
== .
18. (d) From
question, (l
2
– l
1
) is maintained same at all
temperatures hence change in length for both rods
should be same
i.e.,Dl
1
= Dl
2
As we know, coefficient of linear expansion,
a =
0
T
D
D
l
l
l
1
a
1
DT = l
2
a
2
DT
l
1
a
1
= l
2
a
2
19. (d) A
ccording to wein's displacement law, maximum amount
of emitted radiation corresponding to l
m
=
b
T
l
m
=
6
2.88 10 nmK
5760K
´
= 500 nm
U
2
U
Emitted
(radiation)
250 nm
500 nm
1000 nm
­
wave length K
From the graph U
1
< U
2
> U
3
20. (c) Given r
1
= 12 cm , r
2
= 6 cm
T
1
= 500 K and T
2
= 2 × 500 = 1000 K
P
1
= 450 watt
Rate of power loss 24
P rTµ
24
1 11
24
222
P rT
PrT
=
24
22
21
24
11
rT
PP
rT
=
Solving we get, P
2
= 1800 w
att
21. (d) Heat current H = H
1
+ H
2
=
1 12 2 12
KA(T T) KA( T T)
dd
--
+
EQ 12 12
12
K 2A(T T) A(T T)
[K K]
dd
- -
=+
Hence equiv
alent thermal conductivities for two rods
of equal area is given by
12
2
EQ
kk
K
+
=

307. THERMAL EQUILIBRIUM AND ZER OTH LAW OF
THERMODYNAMICS
Thermal Equilibrium
Two systems are said to be in thermal equilibrium with each other
if they have the same temperature.
Zeroth Law of Thermodynamics
If objects A and B are separately in thermal equilibrium with a
third object C then objects A and B are in thermal equilibrium
with each other.
FIRST LAW OF THERMODYNAMICS
First law of thermodynamics gives a relationship between
heat, work and internal energy.
(a)Heat : It is the energy which is transferred from a system to
surrounding or vice-versa due to temperature difference
between system and surroundings.
(i) It is a macroscopic quantity.
(ii) It is path dependent i.e., it is not point function.
(iii) If system liberates heat, then by sign convention it is
taken negative, If system absorbs heat, it is positive.
(b)Work : It is the energy that is transmitted from one system
to another by a force moving its points of application. The
expression of work done on a gas or by a gas is
2
1
V
V
W dW PdV==òò
where V
1
is v
olume of gas in initial state and V
2
in final
state.(i) It is also macroscopic and path dependent function.(ii) By sign convention it is +ive if system does work (i.e.,
expands against surrounding) and it is – ive, if work isdone on system (i.e., contracts).
(iii) In cyclic process the work done is equal to area under
the cycle and is negative if cycle is anti-clockwise and+ive if cycle is clockwise (shown in fig.(a) and (b)).
(c)Internal energy : The internal energy of a gas is sum of
internal energy due to moleculer motion (called internal
kinetic energy U
K
) and internal energy due to molecular
configuration (called internal potential energy U
P.E.
)
i.e., U = U
K
+ U
P.E.
……(1)
(i) In ideal gas, as there is no intermolecular attraction,
hence

3
2
K
n
U U RT== ……(2)
(for n mole
of ideal gas)
(ii) Internal energy is path independent i.e., point function.
(iii) In cyclic process, there is no change in internal energy
(shown in fig.)
i.e., dU =U
f
– U
i
= 0
Þ U
f
=U
i
(iv) Internal energy of an ideal gas depends only on
temperature eq.(2).
First law of thermodynamics is a generalisation of the
law of conservation of energy that includes possible
change in internal energy.
First law of thermodynamics “If certain quantity of heat dQ is
added to a system, a part of it is used in increasing the internal
energy by dU and a part is use in performing external work
done dW
i.e.,
dQ dU dW dU dQ dW=+ Þ =-
The
quantity dU (i.e., dQ – dW) is path independent but dQ and
dW individually are not path independent.Applications of First Law of Thermodynamics
(i) In isobaric process P is constant
so
ò
-==
2
1
V
V
12
)VV(PPdVdW
so dQ = dU + dW = n C
P
dT
12
Thermodynamics

308. 304 PHYSICS
For ideal gas
, dQ = 0
dU =
mC
V
dT (for any process)
22
11
VV
VV
K
dW PdV dV
V
g
==òò
(where PV
g
= K = constant)

2 2 11
11
21
()11
11
PV PVK
VV
gg
gg--
æö -
= -=
ç÷
--èø
where PV
g
= constant is applicable only in adiabatic process.
Adiabatic process is called isoentropic process (in these
process entropy is constant).
(iii)Isobaric process : A process taking place at constant
pressure is called an isobaric process. In this process
dQ = n C
p
dT, dU = n C
V
dT and dW = P(V
2
–V
1
)
(iv)Isochoric process : A process taking place at constant
volume is called isochoric process.
In this process, dQ = dU =n C
V
dT and dW = 0
(v)Cyclic process : In this process the inital state and final
state after traversing a cycle (shown in fig.) are same.
In cyclic process, dU = 0 = U
f
– U
i
and dW = area of cycle
= area (abcd)
Slope of adiabatic and isothermal curve :
For isothermal process PV = constant
On differentiating, we get PdV + VdP = 0
And slope of isothermal curves
V
P
dV
dP
isothermal
-=÷
ø
ö
ç
è
æ
... (1)
Isothermal
Adiabatic
V
P
For adiabatic process PV
g
= constant
On differentiation, we get slope of adiabatic curve
æö
= -gç÷
èøadiabatic
P
(P/V)
V
d
d
.... (2)
It is clear from
equation (1) and (2) that the slope of
adiabatic curve is more steeper than isothermal curve as
shown by fig by
g time (g = C
P
/C
V
)
(ii) In cyclic process heat given to the system is equal to work
done (area of cycle).
(iii) In isothermal process temperature T is constant and work
done is
1
2
V
V
e
V
V
LognRTPdVdW
2
1ò
==
Since,
T = constant so for ideal gas dU = 0
Hence,
1
2
e
V
V
LognRTdWdQ== (for ideal gas)
(
iv) In isochoric process W = 0 as V = constant
It means that heat given to system is used in increasing
internal energy of the gas.
(v) In adiabatic process heat given or taken by system from
surrounding is zero i.e., dQ = 0
( )12
nR
dU dW TT
1
éù
=-=--
êú
g-ëû

11 22
(PV PV)
1
éù-
=
êú
g-ëû
It means that if system expands dW is +ive and dU is –ive
(i.e., temperature decrease) and if system contracts dW is–ive and dU is +ive (i.e., temperature increase).
THERMODYNAMIC PROCESSES
(i)Isothermal process : If a thermodynamic system is perfectly
conducting to surroundings and undergoes a physical
change in such a way that temperature remains constant
throughout, then process is said to be isothermal process.
T = constant
V
P
For isothermal proc
ess, the equation of state is
PV = nRT = constant, where n is no. of moles.
For ideal gas, since internal energy depends only ontemperature.

22
11
0
VV
VV
dV
dU dQ dW PdV nRT
V
=Þ=== òò
or
22
10
11
log 2.303 log
e
VV
dQ nRT nRT
VV
==
(ii)Adiabatic process : If system is completely iso
lated from
the surroundings so that no heat flows in or out of it, then
any change that the system undergoes is called an
adiabatic process.
V
P

309. 305Thermodynamics
Graphs of thermodynamic processes :
1.In the figure (i) P–V graph the process ab is isothermal, bc
is isobaric and ca is isochoric.
V
P a
c b
Fig (i)
Th
e fig (ii) is the P–T diagram of fig (i)
T
P a
c b
Fig.(ii)
2.Fi
gure below shows P – V diagrams for two processes.
P I
II
V
The heat absorbed in process I is more than that in II.
Because, area under process I is also more than area underprocess II. The work done in the process I is more than thatin II. Also, the change in internal energy is same in bothcases.
3.The P–V and corresponding V–T diagram for a cyclicprocess abca on a sample of constant mass of ideal gas are
shown below:
V
P
a
c
b

T
V
a
c b
4.For isochoric process, the P–V, V–T and P–T graphs :
P
2
1
V
2
1
V
T
2
1
P
T
5.For isobaric process, the P – V, P – T and V – T graphs :
V
P
P
T
V
T
6.For isothermal process, the P – V, V – T and P – T graphs :
P
V
V
T
P
T
Keep in Memory
1.In th
ermodynamics heat and work are not state variables,
whereas internal energy is a state variable.
2.For ideal-gas
(i)relation between P and V is PV
g
= constant
(ii)relation between V and T is TV
g–1
= constant
(iii)relation between P and T is T
g
P
1–
g
= constant
3.A quasi-static process is an infinitely slow process such
that system remains in thermal and mechanical equilibrium
with the surroundings throughout.
4.Pressure, volume, temperature and mass are state variables.
Heat and work are not state variables.
5.A graphical representation of the state of a system with the
help of two thermodynamical variables is called indicator
diagram.

310. 306 PHYSICS
REVERSIBLE AND IRR
EVERSIBLE PROCESS
Reversible Process :
A process which can proceed in opposite direction in such a
way that the system passes through the same states as in direct
process and finally the system and the surroundings acquire
the intial conditions.
Conditions for a process to be reversible :
(a) The process must be extremely slow.
(b) There should no loss of energy due to conduction, or
radiation. The dissipating forces should not be in the system.
(c) The system must always be in thermal and chemical
equilibrium with the surroundings.
Examples : Fusion of ice, vaporisation of water, etc.
Irreversible Process :
The process which cannot be traced back in the opposite
direction is defined as irreversible process.
Examples : Work done against friction, magnetic hysteresis.
•In nature all process are irreversible, because no natural
process can fulfil the requirement of a reversible process.
HEAT ENGINE
A heat engine is a device which converts heat energy into
mechanical energy.
Hot
Reservoir
T
1
Cold
Reservoir
T
2
Working
Substance
Work (W) = Q – Q
12
Q
1
Q
2
Efficiency of h
eat engine is given by
1
22
11
()
η
()
11
Work done W
Efficiency
Heat taken from source T
QT
QT
=
=- =-
where Q
2
= amo
unt of heat rejected per cycle to the sink
(of temp

T
2
)
Q
1
= amount of heat energy absorbed per cycle from the source
(of temp

T
1
).
The efficiency of heat engine h is never greater than unity,
h =1 only for ideal engine & for practical heat engine h < 1.
REFRIGERATOR AND HEAT PUMP :
Refrigerator or heat pump is a heat engine running in backward
direction i.e. working substance (a gas) takes heat from a cold
body and gives out to a hotter body with the use of external
energy i.e. electrical energy. A heat pump is the same as a
refrigerator.
Hot
Reservoir
T
1
Cold
Reservoir
T
2
Working
Substance
Work (W) = Q – Q
12
Q
1
Q
2
The coefficient of p
erformance of refrigerator or heat pump is
Heat extracted from cold reservoir
Work doneon refrigerator
b= ==
--
22
1 2 12
QT
QQ TT
,
wher e T
2
is tempera
ture of cold body and T
1
is temperature of
hot body.
CARNOT ENGINE
Carnot devised an ideal engine which is based on a reversible
cycle of four operations in succession : isothermal expansion,
adiabatic expansion, isothermal compression and adiabatic
compression.
A
D
B
Q
1
Q
2
C
P
Adiabatic
expansion
Isothermal
expansion
Isothermal
compression
A
d
i
a
b
a
t
i
c
co
m
p
r
e
s
s
i
o
n
T
1
T
2
VV
1
V
2
Efficiency of Carnot engine,
1
W
Q
h= =
24
12
13
2
1
1
VV
µRT In µRT In
VV
V
µRT In
V
æöæö
+
ç÷ ç÷
èø èø
æö
ç÷
èø
The points B and C are conne
cted by an adiabatic path as
are the points D and A. Hence, using this eq
n
. and the
adiabatic gas eq
n
.
T
1
V
2
(g – 1)
= T
2
V
3
(g – 1)
and T
1
V
1
(g – 1)
= T
2
V
4
(g – 1)
.
Combination of the above eq
n
s. gives
32
14
VV
VV
=
, and,
12
1
TT
T
-
h=
=
12
1
QQ
Q
-
or,
22
11
11
QT
QT
h=- =-
.
The percentage effi
ciency of Carnot’s engine,
12
1
100%
TT
T
-
h=´ or,
12
1
100%
QQ
Q
-
h=´
The efficiency
of a Carnot engine is never 100% because it
is 100% only if temperature of sink T
2
= 0 which is impossible.
In a Carnot cycle,
22
11
QT
QT
= or
12
12
QQ
TT
=.

311. 307Thermodynamics
Carnot Theorem : No irreversible engine (I) can have
efficiency greater than Carnot reversible engine (R)
working between same hot and cold reservoirs.
i.e.,
RI
h >h or
22
11
11
TQ
TQ
- >-
SECOND LA
W OF THERMODYNAMICS
It states that it is impossible for a self acting machine unaided
by any external agency, to transfer heat from a body at a lower
temperature to a body at higher temperature.
It is deduced from this law that the efficiency of any heat engine
can never be 100%.
Entropy :
Entropy is a measure of disorder of the molecular motion of a
system. The greater the disorder, the greater is the entropy. The
change in entropy is given by
()
()
Heat absorbedby the system dQ
S
Absolute tempe
rature T
D=

12
dQ
SS
T
-=ò
(here T is not differentiable)
Clausius inequality
dQ
0
T
£òÑ
or,
dQ
dS
T
³ò
or, dQ = TdS ³ dU + PdV
Also, S = K log
e
w
2
e
1
S Klog
w
D=
w
is the microscopic form of entropy, where
K is Boltzmann's constant and w respresents the number of
possible microscopic states.Energy entering a body increases disorder.Energy leaving a body decreases disorder.When a hot body is brought into thermal contact with acold body for a short time, then :
(i) Each body will experience a change in the entropy of
its particle.
(ii) The hot body experiences a decrease in entropy (a
negative change) of magnitude
1
1
Q
S
T
D
D=
(iii)
The cold body experiences an increase in entropy (a
positive change) of magnitude
2
2
Q
S
T
D
D=
(iv) T
he net change in entropy
12
SSSD =D +D
The effect of naturally occurring processes is always to
increase the total entropy (or disorder) of the universe.
Example 1.
A cyclic process is shown in fig. Work done during isobaricexpansion is
1 2 3
BA
D C
2 × 10
2
P
(N/m)
2
10
2
V(m)
3

(a) 1600 J (b) 100 J
(c) 400 J (d) 600 J
Solution : (c)
Isobaric expansion is represented by curve AB;
Work done = area under AB
= 2 × 10
2
× (3 – 1) = 4 × 10
2
= 400 J.
Example 2.
An ideal gas heat engine operates in carnot cycle between
227ºC and 127ºC. It absorbs 6 × 10
4
cal of heat at higher
temp. Amount of heat converted into work is
(a) 1.2 × 10
4
cal (b) 2.4 × 10
4
cal
(c) 6 × 10
4
cal (d) 4.8 × 10
4
cal
Solution : (a)
As
1
2
1
2
T
T
Q
Q
= ;
2
4
Q 127 273 400
227 273 5006 10
+
\==
+´
(As
1
2
1
2
T
T
1
Q
Q
1 -=-=h)

44
2
4
6 10 4.8 10
5
Q cal=´´=´
\ =
- =´ -´ =´
444
12
W Q Q 6 10 4.8 10 1.2 10 cal.
Example 3.
An idea
l carnot engine whose efficiency is 40% receives
heat at 500 K. If its efficiency were 50%, then what would
be intake temp. for same exhaust temp ?
Solution :
From,
h-=-=h 1
T
T
;
T
T
1
1
2
1
2
5
3
100
40
1 =-=;
K300500
5 3
T
5 3
T
12
=´==\
Again
2
1
T
1η'
T'
=- or
1
300 50 1
1
T' 1002
=-=
or
1
T ' 600K=

312. 308 PHYSICS
Example 4.
When a system is taken from state a to state b, in fig. along
the path a
® c ® b, 60 J of heat flow into the system, and
30 J of work are done :
(i) How much heat flows into the system along the path
a
® d ® b if the work is 10 J.
(ii) When the system is returned from b to a along the
curved path, the work done by the system is –20 J.
Does the system absorb or liberate heat, and how
much?
(iii) If, U
a
= 0 and U
d
= 22 J, find the heat absorbed in the
process a
® d and d ® b.
P
V
c
a
b
d
Solution :
For the path a ® c ® b
dU = dQ – dW = 60 – 30 = 30 J or U
b
– U
a
= 30 J
(i) Along the path a ® d ® b
dQ = dU + dW = 30 + 10 = 40 J
(ii) Along the curved path b – a
dQ = (U
a
– U
b
) + W = (–30) + (–20) = –50 J,
heat flows out the system
(iii) Q
ad
= 32 J; Q
db
= 8 J
Example 5.
Two samples of a gas initially at same temperature and pressure are compressed from a volume V to V/2. One sample is compressed isothermally and the other adiabatically. In which sample is the pressure greater?
Solution :
Let initial volume, V
1
= V and final volume, V
2
= V/2
Initial pressure, P
1
= P ; final pressure, P
2
= ?
For isothermal compression
P
2
V
2
= P
1
V
1
or
11
2
2
PV PV
P 2P
V V/2
===
For adiabatic compression
P
2
' V
2
g
= P
1
V
1
g
or P
2
' = P
1
1
2
V
V
g
æö
ç÷
èø
=
V
P
V/2
g
æö
ç÷
èø
orP
2
´ = 2
g
P
Since g > 1\ 2
g
> 2 \ P
2
' > P
2
Pressure during adiabatic compression is greater than the
pressure during isothermal compression.
Example 6.
A Carnot engine working between 300 K and 600 K has a
work output of 800 J per cycle. What is the amount of heat
energy supplied to the engine from source per cycle?
Solution :
W = 800 J, T
1
= 600 K, T
2
= 300 K
\ h = 1 –
1
21
TW
TQ
= =
1
300 800
1
600Q
-= or 0.5 =
1
800
Q
Heat energy supplied by source,
800
Q
0.5
= = 1600 joule per cycle
Example 7.
The temperatures T
1
and T
2
of the two heat reservoirs in
an ideal Carnot engine are 1500°C and 500°C respectively. Which of the following : increasing T
1
by 100°C or
decreasing T
2
by 100°C would result in a greater
improvement in the efficiency of the engine?
Solution :
The efficiency of a Carnot's engine is given by
2
1
T
1
T
h=-
Given T
1
= 1500°C = 1500 + 273 = 1773 K and
T
2
= 500°C = 500 + 273 = 773 K.
When the temperature of the source is increased by 100°C, keeping T
2
unchanged, the new temperature of the source
is T´
1
= 1500 + 100 = 1600°C = 1873 K. The efficiency becomes
2
1
T 773
´ 1 1 0.59
T´ 1873
h=-=-=
On the other hand, if the temperature of the sink is decreased by 100°C, keeping T
1
unchanged, the new temperature of
the sink is T´
2
= 500 – 100 = 400°C = 673 K. The efficiency
now becomes
2
1
T´ 673
´´ 1 1 0.62
T 1773
h=- =- =
Since h´´ is greater than h´, decreasing the temperature of
the sink by 100°C results in a greater efficiency than
increasing the temperature of the source by 100°C.
Example 8.
Calculate the work done when 1 mole of a perfect gas is
compressed adiabatically. The initial pressure and volume
of the gas are 10
5
N/m
2
and 6 litre respectively. The final
volume of the gas is 2 litres. Molar specific heat of the gas
at constant volume is 3R/2. [(3)
5/3

= 6.19]
Solution :
For an adiabatic change PV
g
= constant
P
1
V
1
g
= P
2
V
2
g

313. 309Thermodynamics
As molar specific heat of gas at constant volume
v
3
CR
2
=
C
P
= C
V
+ R =
35
RRR
22
+= ;

P
V
C (5/2)R5
C (3/2)R3
g===
\ From eq
n
. (1)
1
21
2
V
PP
V
g
æö
=
ç÷
èø
=
5/3
526
10 N/m
2
æö
´
ç÷ èø
= (3)
5/3
× 10
5
= 6.
19 × 10
5
N/m
2
Work done =
53531
[6.19 10 210 10 6 10 ]
5
1
3
---
´´´-´´
æö
-ç÷
èø
2
2103
(6.19 3)
2
éù´´
=--êú
êúëû
= – 3 × 10
2
× 3.1 9 = – 957 joule
[–ve sign shows external work done on the gas]
Example 9.
A refrigerator is to maintain eatables kept inside at 90C. If
room temperature is 360C, calculate the coefficient of
performance.
Solution :
Here, T
1
= 36°C = 36 + 273 = 309 K,
T
2
= 10°C = 10 + 273 = 283 K
2
12
T 283 283
COP 10.9
T T 309 283 26
=
= ==
--
Example 10.
One mole of an ideal gas at pressure P
0
and temperature
T
0
is expanded isothermally to twice its volume and then
compressed at constant pressure to (V
0
/2) and the gas is
brought back to original state by a process in which P
µ V
(Pressure is directly proportional to volume). The correct representation of process is
(a)
V
P
V /2
0V
0 2V0
P0 (b)
V
P
V /2
0V
02V0
P
0
(c)
P0
T
P
P0/2
T
0T /4
0
(d)
T
V
T0
V /2
0
V
0
Solution : (c)
Pr
ocess AB is isothermal expansion,
BC is isobaric compression and in process CA
2nRT
P PT
P
µ Þµ
P
0
A
P
P
0/2
T
0T /4
0
B
C
Example 11.
A Carnot’s heat engi
ne works with an ideal monatomic
gas, and an adiabatic expansion ratio 2. Determine its
efficiency.
Solution :
Given,
3
2
V
2
V
r==
and g for a monatomic gas = 5/3.
Using,
1
1
1
g-
æö
h=-
ç÷
rèø
we have, the required efficiency

5
1
31
1 1 0.63 0 .37 or 37%
2
-
æö
h=- =- =
ç÷
èø

314. 310 PHYSICS

315. 311Thermodynamics
1.Which of the following is incorrect regarding first law of
thermodynamics?
(a) It is a restatement of principle of conservation of
energy.
(b) It is applicable to cyclic processes
(c) It introduces the concept of entropy
(d) It introduces the concept of internal energy
2.Choose the incorrect statement related to an isobaric
process.
(a)
constant
V
T
=
(b)W = PDV
(c) H
eat given to a system is used up in raising the
temperature only.
(d)DQ > W
3.The internal energy of an ideal gas does not depend upon
(a) temperature of the gas
(b) pressure of the gas
(c) atomicity of the gas
(d) number of moles of the gas.
4.During isothermal expansion, the slope of P-V graph
(a) decreases (b) increases
(c) remains same (d) may increase or decrease
5.During melting of ice, its entropy
(a) increases (b) decreases
(c) remains same (d) cannot say
6.Which of the following processes is adiabatic ?
(a) Melting of ice
(b) Bursting of tyre
(c) Motion of piston of an engine with constant speed
(d) None of these
7.At a given temperature the internal energy of a substance
(a) in liquid state is equal to that in gaseous state.
(b) in liquid state is less than that in gaseous state.
(c) in liquid state is more than that in gaseous state.
(d) is equal for the three states of matter.
8.Air conditioner is based on the principle of
(a) Carnot cycle
(b) refrigerator
(c) first low of thermodynamics
(d) None of these
9.A mass of ideal gas at pressure P is expanded isothermally
to four times the original volume and then slowly
compressed adiabatically to its original volume. Assuming
g to be 1.5, the new pressure of the gas is
(a) 2 P (b) P
(c) 4 P (d) P/2
10.One mole of an ideal gas at temperature T was cooled
isochorically till the gas pressure fell from P to
n
P
. Then,
by an is
obaric process, the gas was restored to the initial
temperature. The net amount of heat absorbed by the gas
in the process is
(a) nRT (b)
n
RT
(c) RT (1 – n
–1
) (d) RT (n – 1)
11.Ice contained in a beaker starts melting when
(a) the specific heat of the system is zero
(b) internal energy of the system remains constant
(c) temperature remains constant
(d) entropy remains constant
12.A uniform sphere is supplied heat electrically at the centre
at a constant rate. In the steady state, steady temperatures
are established at all radial locations r, heat flows outwards
radial and is ultimately radiated out by the outer surface
isotropically. In this steady state, the temperature gradient
varies with radial distance r according to
(a)r
–1
(b)r
–2
(c)r
–3
(d)r
–3/2
13.For an ideal gas graph is shown for three processes. Process
1, 2 and 3 are respectively.
1
2
3
Work done (magnitude)
Temperature change
DT
(a) Isobaric, adiabatic, isochoric
(b) Adiabatic, isobaric, isochoric
(c) Isochoric, adiabatic, isobaric
(d) Isochoric, isobaric, adiabatic
14.The efficiency of carnot engine when source temperature
is T
1
and sink temperature is T
2
will be
(a)
1
21
T
TT-
(b)
21
2
TT
T
-
(c)
12
2
TT
T
-
(d)
1
2
T
T
15.In the equation PV
g
= constant, the value of g is unity. Then
the process is
(a) isothermal (b) adiabatic
(c) isobaric (d) irreversible

316. 312 PHYSICS
16.For adia
batic processes (Letters have usual meanings)
(a) P
g
V = constant (b) T
g
V = constant
(c) TV
g–1
= constant(d) TV
g
= constant
17.The gas law
T
PV
= constant is true
for
(a) isothermal changes only
(b) adiabatic changes only
(c) both isothermal and adiabatic changes
(d) neither isothermal nor adiabatic change
18.When heat is given to a gas in an isothermal change, the
result will be
(a) external work done
(b) rise in temperature
(c) increase in internal energy
(d) external work done and also rise in temperature
19.Volume of one mole gas changes according to the V = a/T.
If temperature change is DT, then work done will be
(a)RDT (b) – RDT
(c)
R
1g-
DT (d) R (g – 1) DT
20.In changing
the state of thermodynamics from A to B state,
the heat required is Q and the work done by the system isW. The change in its internal energy is
(a)Q + W (b)Q – W
(c)Q (d)
2
QW-
21.If DQ and DW represent the heat supplied to the system
and the work done on the system respectively, then the
first law of thermodynamics can be written as
(a)QUWD =D +D (b)QUWD =D -D
(c)QWUD =D -D (d) –Q WUD = D -D
22.The work done in which of the following processes is equal
to the internal energy of the system?
(a) Adiabatic process(b) Isothermal process
(c) Isochoric process(d) None of these
23.Which of the following processes is reversible?
(a) Transfer of heat by conduction
(b) Transfer of heat by radiation
(c) Isothermal compression
(d) Electrical heating of a nichrome wire
24.In thermodynamic processes which of the following
statements is not true?
(a) In an isochoric process pressure remains constant
(b) In an isothermal process the temperature remains
constant
(c) In an adiabatic process PV
g
= constant
(d) In an adiabatic process the system is insulated from
the surroundings
25.Monatomic, diatomic and polyatomic ideal gases each
undergo slow adiabatic expansions from the same initial
volume and same initial pressure to the same final volume.
The magnitude of the work done by the environment on
the gas is
(a) the greatest for the polyatomic gas
(b) the greatest for the monatomic gas
(c) the greatest for the diatomic gas
(d) the question is irrelevant, there is no meaning of slow
adiabatic expansion
1.A gas at 27ºC and pressure of 30 atm. is allowed to expand to
atmospheric pressure and volume 15 times larger. The final
temperature of the gas is
(a) – 123ºC (b) +123ºC
(c) 273ºC (d) 373ºC
2.A system changes from the state (P
1
, V
1
) to (P
2
, V
2
) as
shown in the figure. What is the work done by the system?
Pressure i
n

N
/
m
2
6×10
5
5×10
5
4×10
5
3×10
5
2×10
5
1×10
5
12345
Volume in metre
3
(P , V )
22
(P, V)
11
(a) 7
.5 × 10
5
joule (b) 7.5 × 10
5
erg
(c) 12 × 10
5
joule (d) 6 × 10
5
joule
3.A refrigerator works between 0ºC and 27ºC. Heat is to beremoved from the refrigerated space at the rate of 50 kcal/minute, the power of the motor of the refrigerator is
(a) 0.346 kW (b) 3.46 kW
(c) 34.6 kW (d) 346 kW
4.A perfect gas goes from a state A to another state B by
absorbing 8 × 10
5
J of heat and doing 6.5 × 10
5
J of external
work. It is now transferred between the same two states in
another process in which it absorbs 10
5
J of heat. In the
second process
(a) work done by gas is 10
5
J
(b) work done on gas is 10
5
J
(c) work done by gas is 0.5 × 10
5
J
(d) work done on the gas is 0.5 × 10
5
J

317. 313Thermodynamics
5.The temperature of 5 moles of a gas which was held at
constant volume was changed from 100º to 120ºC. The
change in the internal energy of the gas was found to be 80
joule, the total heat capacity of the gas at constant volume
will be equal to
(a) 8 joule per K (b) 0.8 joule per K
(c) 4.0 joule per K (d) 0.4 joule per K
6.A polyatomic gas (g = 4/3) is compressed to 1/8th of its
volume adiabatically. If its initial pressure is P
0
, its new
pressure will be
(a) 8P
0
(b) 16P
0
(c) 6P
0
(d) 2P
0
7.The efficiency of a Carnot engine operating with reservoir
temperatures of 100ºC and –23ºC will be
(a)
100
23100+
(b)
100
23100-
(c)
373
23100+
(d)
373
23100-
8.By what
percentage should the pressure of a given mass of
a gas be increased so as to decrease its volume by 10% at
a constant temperature?
(a) 8.1 % (b) 9.1 %
(c) 10.1 % (d) 11.1 %
9.A gas has pressure P and volume V. It is now compressed
adiabatically to 1/32 times the original volume. Given that
(32)
1.4
= 128, the final pressure is (g = 1.4)
(a) P/128 (b) P/32
(c) 32 P (d) 128 P
10.At 27ºC a gas is compressed suddenly such that its pressure
becomes (1/8) of original pressure. Final temperature will
be (g = 5/3)
(a) 450 K (b) 300 K
(c) –142ºC (d) 327ºC
11.A diatomic gas initally at 18ºC is compressed adiabatically
to one eighth of its original volume. The temperature after
compression will be
(a) 18ºC (b) 887ºC
(c) 327ºC (d) None of these
12.Absolute zero is obtained from
(a) P–V graph (b)
V
1
P-graph
(c) P–T graph (d)
V–T graph
13.An ideal gas heat engine operates in Carnot cycle between
227°C and 127°C. It absorbs 6 × 10
4
cals of heat at higher
temperature. Amount of heat converted to work is
(a) 4.8 × 10
4
cal (b) 6 × 10
4
cal
(c) 2.4 × 10
4
cal (d) 1.2 × 10
4
cal
14.Three moles of an ideal gas kept at a constant temperature
at 300 K are compressed from a volume of 4 litre to 1 litre.
The work done in the process is
(a) – 10368 J (b) –110368 J
(c) 12000 J (d) 120368 J
15.The temperature at which speed of sound in air becomes
double of its value at 27° C is
(a) 54°C (b) 327°C
(c) 927°C (d) None of these
16.1 gm of water at a pressure of 1.01 × 10
5
Pa is converted into
steam without any change of temperature. The volume of 1
g of steam is 1671 cc and the latent heat of evaporation is
540 cal. The change in internal energy due to evaporation
of 1 gm of water is
(a)
»167 cal (b)»500 cal
(c) 540
cal (d) 581 cal
17.An ideal refrigerator has a freezer at a temperature of 13ºC.The coefficient of performance of the engine is 5. The
temperature of the air (to which heat is rejected) is
(a) 320ºC (b) 39ºC
(c) 325 K (d) 325ºC
18.One mole of an ideal monoatomic gas is heated at a constant
pressure of one atmosphere from 0ºC to 100ºC. Then the
work done by the gas is
(a) 6.56 joule (b) 8.32 × 10
2
joule
(c) 12.48 × 10
2
joule (d) 20.8 × 10
2
joule
19.The pressure inside a tyre is 4 times that of atmosphere. If the
tyre bursts suddenly at temperature 300 K, what will be the
new temperature?
(a) 300 (4)
7/2
(b) 300 (4)
2/7
(c) 300 (2)
7/2
(d) 300 (4)
–2/7
20.A monatomic ideal gas expands at constant pressure, with
heat Q supplied. The fraction of Q which goes as work
done by the gas is
(a)1 (b)
3
2
(c)
5
3
(d)
5
2
21.A carnot’s engine takes 300 calories of heat at 500 K and
rejects 150 calories of heat to the sink. The temperature of
the sink is
(a) 1000 K (b) 750 K
(c) 250 K (d) 125 K
22.The source and sink temperatures of a Carnot engine are
400 K and 300 K, respectively. What is its efficiency?
(a) 100% (b) 75%
(c) 33.3% (d) 25%
23.The volume of a gas is reduced adiabatically to 1/4 of its
volume at 27ºC. If g = 1.4 the new temperature is
(a) (300) 2
0.4
K (b) (300) 2
1.4
K
(c) 300 (4)
0.4
K (d) 300 (2)
1.4
K
24.In pressure-volume diagram, the isochoric, isothermal,
isobaric and iso-entropic parts respectively, are
A
B
C
D
P
V
(a) BA, AD, DC,CB (b) DC, CB, BA, AD
(c) AB, BC, CD, DA (d) CD, DA, AB, BC

318. 314 PHYSICS
25.Two cy
linders fitted with pistons contain equal amount of
an ideal diatomic gas at 300 K. The piston of A is free to
move, while that of B is held fixed. The same amount of heat
is given to the gas in each cylinder. If the rise in temperature
of the gas in A is 30 K, then the rise in temperature of gas in
B is
(a) 30 K (b) 18 K
(c) 50 K (d) 42 K
26.A Carnot engine works first between 200°C and 0°C and
then between 0°C and –200°C. The ratio of its efficiency in
the two cases is
(a) 1.0 (b) 0.577
(c) 0.34 (d) 0.68
27.A Carnot’s engine works as a refrigerator between 250 K
and 300 K. If it receives 750 calories of heat from the reservoir
at the lower temperature, the amount of heat rejected at the
higher temperature is
(a) 900 calories (b) 625 calories
(c) 750 calories (d) 1000 calories
28.A Carnot engine is working between 127°C and 27°C. The
increase in efficiency will be maximum when the temperature
of
(a) the source is increased by 50°C
(b) the sink is decreased by 50°C
(c) source is increased by 25°C and that of sink is
decreased by 25°C
(d) both source and sink are decreased by 25°C each.
29.During an adiabatic process an object does 100J of work
and its temperature decreases by 5K. During another
process it does 25J of work and its temperature decreases
by 5K. Its heat capacity for 2
nd
process is
(a) 20 J/K (b) 24 J/K
(c) 15 J/K (d) 100 J/K
30.The internal energy change in a system that has absorbed
2 kcals of heat and done 500 J of work is
(a) 6400 J (b) 5400 J
(c) 7900 J (d) 8900 J
31.In an adiabatic process, the pressure is increased by
2
%
3
.
If g =
3
2
, then the volume decreases by nearly
(a)
4
%
9
(b)
2
%
3
(c) 1% (d)
9
%
4
32.A closed gas cylinder is divided into two parts by a piston
held tight. The pressure and volume of gas in two parts
respectively are (P, 5V) and (10P, V). If now the piston is left
free and the system undergoes isothermal process, then
the volumes of the gas in two parts respectively are
(a) 2V, 4V (b) 3V, 3V
( c) 5V, V (d)
10 20
V,V
11 11
33.Figure shows the variation of internal energy (U) with the
pressure (P) of 2.0 mole gas in cyclic process abcda. The
temperature of gas at c and d are 300 K and 500 K. Calculate
the heat absorbed by the gas during the process.
P
U
a d
b c
P
0 2P
0
(a) 400 R ln 2 (b) 200 R ln 2
(c) 100 R ln 2 (d) 300 R ln 2
34.The figure shows the P-V plot of an ideal gas taken through
a cycle ABCDA. The part ABC is a semi-circle and CDA is
half of an ellipse. Then,
D
P
A
B
C
V321
0
1
2
3
(a) the proc
ess during the path A ® B is isothermal
(b) heat flows out of the gas during the path B ® C ® D
(c) work done during the path A ® B ® C is zero
(d) positive work is done by the gas in the cycle ABCDA
35.A thermodynamic system goes from states (i) P
1
, V to 2P
1
,
V (ii) P, V
1
to P, 2V
1
. Then work done in the two cases is
(a) zero, zero (b) zero, PV
1
(c)PV
1
, zero (d)PV
1
, P
1
V
1
36.For an isothermal expansion of a perfect gas, the value of
P
P
D
is equal to
(a)
1/2
–
V
V
D
g (b)–
V
V
D
(c)–
V
V
D
g (d)
2
–
V
V
D
g
37.One mole of an ideal gas
at an initial temperature of T K
does 6R joules of work adiabatically. If the ratio of specific
heats of this gas at constant pressure and at constant
volume is 5/3, the final temperature of gas will be
(a) (T – 4) K (b) (T + 2.4) K
(c) (T – 2.4) K (d) (T + 4) K

319. 315Thermodynamics
38.If UD and WD represent the increase in internal energy
and work done by the system respectively in a
thermodynamical process, which of the following is true?
(a) ,UWD =-D in an adiabatic process
(b) ,UWD =D in an isothermal process
(c) ,UWD =D in an adiabatic process
(d) ,UWD =-D in an isothermal process
39.During an isothermal expansion, a confined ideal gas does
–150 J of work against its surroundings. This implies that
(a) 150 J heat has been removed from the gas
(b) 300 J of heat has been added to the gas
(c) no heat is transferred because the process is isothermal
(d) 150 J of heat has been added to the gas
40.When 1 kg of ice at 0°C melts to water at 0°C, the
resulting change in its entropy, taking latent heat of ice to
be 80 cal/°C, is
(a) 273 cal/K (b) 8 × 104 cal/K
(c) 80 cal/K (d) 293 cal/K
41.A mass of diatomic gas (g = 1.4) at a pressure of 2
atmospheres is compressed adiabatically so that its
temperature rises from 27°C to 927°C. The pressure of the
gas in final state is
(a) 28 atm (b) 68.7 atm
(c) 256 atm (d) 8 atm
42.A thermodynamic system is taken through the cycle ABCD
as shown in figure. Heat rejected by the gas during the
cycle is
D
C
BA
3VV
P
2P
Volume
P
r
e
s
s
u
r
e
(a) 2 PV (b) 4 PV
(c)
1
2
PV (d) P V
43.An ideal gas goes from state A to state B via three different
processes as indicated in the P-V diagram :
P
V
A 1
2
B
3
If Q
1
, Q
2
, Q
3
indicate the heat a absorbed by the gas along
the three processes and D U
1
, DU
2
, DU
3
indicate the change
in internal energy along the three processes respectively, then
(a)Q
1
> Q
2
> Q
3
and DU
1
= DU
2
= DU
3
(b)Q
3
> Q
2
> Q
1
and DU
1
= DU
2
= DU
3
(c)Q
1
= Q
2
= Q
3
and DU
1
> DU
2
> DU
3
(d)Q
3
> Q
2
> Q
1
and DU
1
> DU
2
> DU
3
44.Choose the correct relation between efficiency h of a
Carnot engine and the heat absorbed (q
1
) and released
by the working substance (q
2
).
(a)h =
2
1
1
q
+
q
(b)h =
1
2
1
q
+
q
(c)h =
1
2
1
q
-
q
(d)
2
1
1
q
h=-
q
45.In the given (V – T) diagram, what is the relation between
pressure P
1
and P
2
?
T
V
P
2
P
1
q
1
q
2
(a)P
2
> P
1
(b) P
2
< P
1
(c)P
2
= P
1
(d) Cannot be predicted
46.A system goes from A to B via two processes I and II as shown in figure. If DU
1
and DU
2
are the changes in internal
energies in the processes I and II respectively, then
II
I
BA
p
v
(a) relation between DU
1
and DU
2
can not be determined
(b)DU
1
= DU
2
(c)DU
1
< DU
2
(d)DU
1
> DU
2
47.Which of the following statements about a thermodynamic process is wrong ?
(a) For an adiabatic process DE
int
= – W
(b) For a constant volume process DE
int
= + Q
(c) For a cyclic process DE
int
= 0
(d) For free expansion of a gas DE
int
> 0

320. 316 PHYSICS
48.In a Car
not engine efficiency is 40% at hot reservoir
temperature T. For efficiency 50%, what will be the
temperature of hot reservoir?
(a)T (b)
2
T
3
(c)
4
T
5
(d)
6
T
5
Directions for Qs. (49 to 50) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following.
(a) Statement-1 is false, Statement-2 is true
Exemplar Questions
1.An ideal gas undergoes four different processes from the
same initial state (figure). Four processes are adiabatic,
isothermal, isobaric and isochoric. Out of 1, 2, 3 and 4 which
one is adiabatic?
p
V
1
2
3
4
(a)4 (b) 3
(c)2 (d) 1
2.If a
n average person jogs, he produces 14.5 × 10
3
cal/min.
This is removed by the evaporation of sweat. The amountof sweat evaporated per minute (assuming 1 kg requires580 × 10
3
cal for evaporation) is
(a) 0.025 kg (b) 2.25 kg
(c) 0.05 kg (d) 0.20 kg
3.Consider p-V diagram for an ideal gas shown in figure.
p
V
1
2
p
V
constant
=
Out of the following diagrams, which figure represents the
T-p diagram?
(i)
T
p
1
2
(ii)
T
p
1
2
(iii)
T
p
12
(iv)
T
p
1 2
(a) (iv) (b) (ii)
(
c) (iii) (d) (i)
4.An ideal gas undergoes cyclic process ABCDA as shown
in given p-V diagram. The amount of work done by the gas
is
D
C
BA
p
V
V
0
3V
0
2p
0
p
0
(a)6p
0
V
0
(b) –2p
0
V
0
(c) +2p
0
V
0
(d)
+4p
0
V
0
5.Consider two containers A and B containing identical gases
at the same pressure, volume and temperature. The gas in
container A is compressed to half of its original volume
isothermally while the gas in container B is compressed to
half of its original value adiabatically. The ratio of final
pressure of gas in B to that of gas in A is
(a)
1
2
g- (b)
1
1
2
g-
æö
ç÷
èø
(c)
2
1
1
æö
ç÷
-gèø
(d)
2
1
1
æö
ç÷
g-èø
(b) Statement-1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c) Statement-1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d) Statement-1 is true, Statement-2 is false
49. Statement-1 : At a given temperature the specific heat of a
gas at constant volume is always greater than its specific
heat at constant pressure.
Statement-2 : When a gas is heated at constant volume
some extra heat is needed compared to that at constant
pressure for doing work in expansion.
50. Statement -1 : If an ideal gas expands in vacuum in an
insulated chamber, DQ, DU and DW all are zero.
Statement-2 : Temperature of the gas remains constant.

321. 317Thermodynamics
6.Three copper blocks of masses M
1
, M
2
and M
3
kg
respectively are brought into thermal contact till they reach
equilibrium. Before contact, they were at T
1
, T
2
, T
3
(T
1
> T
2
> T
3
). Assuming there is no heat loss to the surroundings,
the equilibrium temperature T is (s is specific heat of copper)
(a)
123
3
TTT
T
++
=
(b)
11 22 33
123
MT MT MT
T
MMM
++
=
++
(c)
11 22 33
123
3()
MT MT MT
T
MMM
++
=
++
(d)
11 22 33
123
MTs M Ts MTs
T
MMM
++
=
++
NEET/AIPMT (2013-2017) Questions
7.A gas is taken through the cycle A®B®C®A, as
shown i
n figure. What is the net work done by the gas ?
P (10 Pa)
5
V (10 m)
33
A
B
C
7
6
5
4
3
2
1
0 2 64 8
[2013]
(a) 1000 J (b)
zero
(c) – 2000 J (d) 2000 J
8.During an adiabatic process, the pressure of a gas is found
to be proportional to the cube of its temperature. The ratio
of
p
v
C
C
for the gas is [2013]
(a)2 (b)
5
3
(c)
3
2
(d)
4
3
9.A system is taken from state a to state c by two paths adc
and abc as shown in the figure. The internal energy at a is
U
a
= 10 J. Along the path adc the amount of heat absorbed
dQ
1
= 50 J and the work done dW
1
= 20 J whereas along the
path abc the heat absorbed dQ
2
= 36 J. The amount of work
done along the path abc is [NEET Kar. 2013]
a b
cd
p
V
(a) 6 J (b) 10 J
(c
) 12 J (d) 36 J
10.Which of the following relations does not give the equation
of an adiabatic process, where terms have their usual
meaning?
(a)P
g
T
1–g
= constant [NEET Kar. 2013]
(b)P
1–g
T
g
= constant
(c)PV
g
= constant
(d)TV
g–1
= constant
11.Two Carnot engines A and B are operated in series. The
engi ne A receives heat from the source at temperature T
1
and rejects the heat to the sink at temperature T. The second
engine B receives the heat at temperature T and rejects to
its sink at temperature T
2
. For what value of T the
efficiencies of the two engines are equal? [NEET Kar. 2013]
(a)
12
2
TT+
(b)
12
2
TT-
(c)T
1
T
2
(d)
12
TT
12.A thermodynamic system undergoes cyclic process ABCDA
as shown in fig. The work done by the system in the cycle
is : [2014]
(a) P
0
V
0
(b) 2P
0
V
0
(c)
00
PV
2

P
CB
D
A
V
0
2V
0V
P
0
2P
0
3P
0
(d) Zero
13.A mo
noatomic gas at a pressure P, having a volume V
expands isothermally to a volume 2V and then adiabaticallyto a volume 16V. The final pressure of the gas is :(take g =
5
3
) [2014]
(a) 64P (b
) 32P
(c)
P
64
(d) 16P
14.Figu
re below shows two paths that may be taken by a gas
to go from a state A to a state C.
2 × 10 m 4 × 10 m
–
33 –33
2×10 Pa
4
6×10 Pa
4
P
B C
A
V

322. 318 PHYSICS
In process
AB, 400 J of heat is added to the system and in
process BC, 100 J of heat is added to the system. The heat
absorbed by the system in the process AC will be [2015]
(a) 500 J (b) 460 J
(c) 300 J (d) 380 J
15.A carnot engine having an efficiency of
1
10
as heat engine,
is used
as a refrigerator. If the work done on the system is
10 J, the amount of energy absorbed from the reservoir at
lower temperature is :- [ 2015, 2017]
(a) 90 J (b) 99 J
(c) 100 J (d) 1 J
16.The coefficient of performance of a refrigerator is 5. If the
inside temperature of freezer is –20°C, then the temperature
of the surroundings to which it rejects heat is [2015 RS]
(a) 41°C (b) 11°C
(c) 21°C (d) 31°C
17.An ideal gas is compressed to half its initial volume by
means of several processes. Which of the process results
in the maximum work done on the gas? [2015 RS]
(a) Isobaric (b) Isochoric
(c) Isothermal (d) Adiabatic
18.A gas is compressed isothermally to half its initial volume.
The same gas is compressed separately through an adiabatic
process until its volume is again reduced to half. Then :
[2016]
(a) Compressing the gas isothermally will require more
work to be done.
(b) Compressing the gas through adiabatic process will
require more work to be done.
(c) Compressing the gas isothermally or adiabatically will
require the same amount of work.
(d) Which of the case (whether compression through
isothermal or through adiabatic process) requires more
work will depend upon the atomicity of the gas.
19.A refrigerator works between 4°C and 30°C. It is required to
remove 600 calories of heat every second in order to keep
the temperature of the refrigerated space constant. The
power required is: (Take 1 cal = 4.2 joules)[2016]
(a) 2.365 W (b) 23.65 W
(c) 236.5 W (d) 2365 W
20.Thermodynamic processes are indicated in the following
diagram : [2017]
P
i
IV
f
III
f
f
f
I
700k
500k
300k
V
II
Match the following
C
olumn-1 Column-2
P. Process I A. Adiabatic
Q. Process II B. Isobaric
R. Process III C. Isochoric
S. Process IV D. Isothermal
(a) P ® C, Q ® A, R ® D, S ® B
(b) P ® C, Q ® D, R ® B, S ® A
(c) P ® D, Q ® B, R ® A, S ® C
(d) P ® A, Q ® C, R ® D, S ® B

323. 319Thermodynamics
EXERCISE - 1
1. (
c) 2. (c) 3. (b) 4. (a) 5. (a)
6. (b) 7. (b) 8. (b)
9. (a) Let P and V be the initial pressure and volume of ideal
gas. After isothermal expansion, pressure is P/4. So volume
i s 4 V.
Let P
1
be the pressure after adiabatic compression.
Then
gg
= )V4()4/P(VP
1
P2)4()4/P(P
2/3
1
==
10. (c) The temperature remains unchanged therefore
if
UU=.
Also, WQ
D=D .
In the firs
t step which is isochoric, 0W
=D .
In second step
, pressure =
n
P
. Volume V i
s increased
from V on nV.
1
P
W (nV V)
n
n1
PV
n
RT(1n)
-
\=-
-æö
=ç÷
èø
=-
11. (c
) During melting temperature remains constant
12. (b) Flow rate
µ gradient × r
2
.
W
hen flow rate is constant, gradient
µ r
–2
.
13. (d)
Isochoric proceess dV = 0
W = 0 proceess 1
Isobaric : W = P DV = nRDT
Adiabatic | W | =
nRT
1
D
g-
0 < g – 1 < 1
As workdone in case of adiabatic process is more so
process 3 is adiabatic and process 2 is isobaric.
14. (a) Efficiency of carnot engine =
1
21
T
TT-
=h
where
1
T =source temperature

2
T=sink temperature.
15. (a) PV = constant represents isothermal process.
16. (c) 17. (c) 18. (a)
19. (b) PV = RT ;
2
RT RT
P
Va
==
a
V
T
=
\
2
a
dV dT
T
=-
2
2
W P dV
RTa
dT
a T
=
æö
=-
ç÷
èø
ò
ò
W = – R DT
20
. (b)
QUWD =D +D
U Q W QWÞD=D- D=- (using proper sign)
21. (b) From FLOT QUWD =D +D
Q Heat supplied to the system so DQ ® Positive
and work is done on the system so DW® Negative
Hence QUW+D =D -D
22. (a)In adiabatic process
DQ = 0
\DW = – DU
23. (c) For process to be reversible it must be quasi-static.
For quasi static process all changes take place
infinitely slowly. Isothermal process occur very slowly
so it is quasi-static and hence it is reversible.
24. (a) In an isochoric process volume remains constant
whereas pressure remains constant in isobaric
process.
25. (a)
nRdT
W
1
=
g-
g is minimum for a polyatomic gas
Hence, W is greatest for polyatomic gas
EXERCISE - 2
1. (a
) We know that
2
22
1
11
T
VP
T
VP
=
11
122
2
VP
TVP
T=\
Here P
1
= 30
atm., P
2
= 1 atm, V
1
= V (say),
V
2
= 15 V, T
1
= 27ºC = 27 + 273 = 300ºK and T
2
=?
Substituting these values, we get T
2
= 150ºK
\ T
2
= 150ºK = 150 – 273 = –123ºC.
2. (c)
2
)15()106101(
W
55
-´+´
=

5
56104
12 10 joule
2
´´
= =´
3. (a)
W
Q
TT
T
2
21
2
=
-
Hints & Solutions

324. 320 PHYSICS
W
000,50
273300
273
=
-
27 50,000
cal/min
273
W
´
=
4.2 27 50,000
Joule/sec
60 273
W
P
t
´´
==
´
= 346 wat
t = 0.346 kW
4. (d) )105.6108(dWdQdU
55
´-´=-= J105.1
5
´=
55
105.110dUdQdW ´-==-=
5
0.5 10J=-´
– ve sign indicates th
at work done on the gas is
J105.0
5
´ .
5. (c)
vv
dU nC dT or 80 5 C (120 100)= =´-
C
v
= 4.0 joule/K
6. (b)
00
4/3
V4/3 4/3
PV =P P =P8
11
8
æö
Þç÷
èø
= 16P
o
7. (c)
373
23100
373
123
373
250
1
T
T
1
1
2 +
==-=hÞ-=h
8. (d)
T
V)100/90(P
T
VP¢
=
or
90
10
1
90
100
P
P
+==
¢
or %1.11
90
10
P
PP
==
-¢
9. (d) PV
g
=
g
÷
ø
ö
ç
è
æ
32
V
P
1
or
g
===
1.4
1P (32) , P (32) , P 128P
10. (c)
g-gg-g
=
1
22
1
11 PTPT
11. (d)
1
T 18º C (273 18) 291 K= = +=
and 8/VV
12=
We know that
1
VT
-g
= constant
or
,
1
11
1
22 VTVT
-g-g
=
14.1
1
2
1
12
)8(291
V
V
TT
-
-g
´=
÷
÷
ø
ö
ç
ç
è
æ
=\
Cº5.395K5.668==
12. (c) TPµ if V is constant, where P is pressure of certain
amount of gas and T is absolute zero temperature.
13. (d) We know that efficiency of carnot engine = I
T
T
L
H
-
Also, Efficien
cy of Heat engine =
Work output
Heat input
1
1
\-=
æö
Þ=-
ç÷
èø
L
HS
S
H
TW
TQ
T
L
WQ
T

4
4
127 273
6101
227 273
1.2 10 cal
+æö
=´- ç÷
èø +
=´
14. (a) Work d
one in an isothermal process is given by
2
10
1
V
W 2.3026nRT log
V
=
Here, n = 3
, R = 8.31 J mol
–1
degree
–1
T = 300 K, V
1
= 4 litre, V
2
= 1 litre.
Hence, W = 2.3026 × 3 × 8.31 × 300 ×
10
1
log
4
= 17221.15(–2log
10
2)
= – 17221.15 × 2 × 0.3010 = –
10368 J.
15. (c) The speed of sound in air,
vTµ
11
22
vT
vT
=
Þ
1
12
v (27 273)
2vT
+
=
2
1 300
2T
=
T
2
= 1200 K = (1200 – 273)°C = 927°C
1
6. (b) Joule10]11671[1001.1VPdW
65 -
´-´=D=

1.01 167
cal.
4.2
40cal. nearly
´
=
=
ΔQ mL 1 540,
ΔQ ΔW ΔU
= =´
=+
or .cal50040540
U =-=D
17. (b)
2
2
12
T 273 13 260,
T
K
TT
= -=
=
-
1
1
260
5
T 260
or T 260 52
=
-
-=
1
2
T 312K,
T 312 273 39º C
=
=-=

325. 321Thermodynamics
18. (b) p
5
dQ n C dT 1 R 100
2
= =´´
J108.
20J5.2077
2
´==

v
dU n C dT
3
1 R 100
2
1246.5J
=
=´´
=
2
dU dQ dU
8.31 10 J
\ =-
=´
19. (d
) Under adiabatic change
g
g-
÷
÷
ø
ö
ç
ç
è
æ
=
1
2
1
1
2
P
P
T
T
or
g
g-
=
1
2112
)P/P(TT
airfor5/74.1;)1/4(300T
)5/7(
)5/7(1
2
==g=\
-
or
7/2
2
)4(300T
-
=
20. (d) p
QnCT=D and WPΔV nRΔ T.==
For monatomic gas,
2
R5
C
p
=.
W2
Q5
Þ=
21. (c)
1
2
1
2
1
2
1
2
T
T
Q
Q
T
T
1
Q
Q
1 =Þ-=-=h
So K250
300
500150
Q
TQ
T
1
12
2
=
´
=
´
=
22. (d) E
fficiency,
T
2
=1–
T
1
η
T
1
(source tem
p.) = 400 K
T
2
(sink temp.) = 300 K
300 1
25%
400 4
= 1 – ==\h
23. (
c)
)14.1(
2
)14.1(
4
V
.TV)300(
-
-
÷
ø
ö
ç
è
æ
=
4.0
2 )4()300(T=
24. (d) F
rom C to D, V is constant. So process is isochoric.
From D to A, the curve represents constant
temperature. So the process is isothermal.
From A to B, pressure is constant . So, the process is
isobaric.
BC represents constant entropy.
25. (d) In cylinder A, heat is supplied at constant pressure
while in cylinder B heat is supplied at constant volume.
(DQ)
A
= nC
P
(DT)
A
and (DQ)
B
= nC
V
(DT)
B
Given : (DQ)
A
= (DQ)
B
\ (DT)
B
P
A
V
C
( T)
C
1.4 30
42K
=D
=´
=
[Q for diatomic gas P
V
C
1.4
C
=
]
26. (b)
2
1
T
1
T
2730
1
273 200
200
473
h=-
+
=-
+
=
2
1
T
'1
T
(273 200)
1
273 0
200
473
h=-
-
=-
+
=
200 273
' 473 200
0.577
h
=´
h
=
27. (a)
250300
250
W
750
-
=
Heat rejected =
750 + 150 = 900 cal.
28. (b)
1
21
T
TT-
is maximum in case (b).
29. (c) For adiabatic process, dU = – 100 J
which remains same for other processes also.
Let C be the heat capacity of 2nd process then
– (C) 5 = dU + dW
= – 100 + 25 = – 75
\ C = 15 J/K
30. (c) According to first law of thermodynamics
Q = DU + W
DU = Q – W
= 2 × 4.2 × 1000 – 500
= 8400 –500 = 7900 J
31. (a) PV
3/2
= K

326. 322 PHYSICS
3
log P log V logK
2
+=
P 3V
0
P 2V
DD
+=
V 2P
V 3P
DD
=-
or
V 22
V 33
4
9
Dæ öæö
=-ç ÷ç÷
è øèø
=-
32. (a) The
volume on both sides will be so adjusted that the
original pressure × volume is kept constant as the
piston moves slowly (isothermal change)
P5V = P'V' ........... (1)
10PV = P'V'' ........... (2)
From (1) and (2), V'' = 2V'
and from V' + V'' = 6V
V' = 2V, V'' = 4V
33. (a) Change in internal energy for cyclic process (DU) = 0.
For process a ® b, (P-constant)
400
®
=D
=D
=-
ab
W PV
nRT
R
For process b ® c , (T-constant)
bc
W 2R (300) ln 2
®
=-
For process c ® d, (P-constant)
cd
W 400R
®
=+
For process d ® a, (T-co
nstant)
da
W 2R (500) ln 2
®
=+
Net work
abbccdda
(W)WWWW
®®®®
D=+++
W 400Rln2D=
\ dQ = dU + dW, first law
of thermodynamics
\ dQ = 400 R ln 2.
34. (b,d)(a) Process is not isothermal.
(b) Volume decreases and temperature decreases
DU = negative,
So, DQ = negative
(c) Work done in process A ® B ® C is positve
(d) Cycle is clockwise, so work done by the gas is positive
35. (b) (i) Case ® Volume = constant Þ
0PdV=ò
(ii) Case ® P = c
onstant
Þ
22
1
11
11
VV
VV
PdV P dV PV==òò
36. (b) Dif
ferentiate PV = constant w.r.t V
0
–
Þ D+D=
DD
Þ=
PV VP
PV
PV
37. (a)
1
nRTnRT
1
VPVP
W
21
2211
-g
-
=
-g
-
=
1
)TT(nR
21
-g
-
=
n = 1, T
1
= T R6
13/5
)TT(R
2
=
-
-
Þ Þ T
f
= ( T – 4)K
38
. (a) By first law of thermodynamics,
QUWD =D +D
In adiabatic process, QD = 0
\UWD = -D
In isothermal process, UD = 0
\QWD =D
39. (a, d) If a process is expansion then work done is positive
so answer will be (a).
But in question work done by gas is given –150J so
that according to it answer will be (d).
40. (d) Change in entropy is given by
dS =
dQ
T
or
f
mLQ
S
T 273
D
D==
1000 80
S 293cal / K.
273
´
D==
41. (c)T
1
= 273 + 27 = 300K
T
2
=
273 + 927 = 1200K
For adiabatic process,P
1–g
T
g
= constant
Þ P
1
1–g
T
1
g
= P
2
1–g
T
2
g
1
2
1
P
P
-g
æö
Þ
ç÷
èø
=
1 2
T
T
g
æö
ç÷
èø

1
1
2
P
T
-g
æö
Þ
ç÷
èø
=
2 1
T
T
g
æö
ç÷
èø
1 1.4 1.4
1
2
P 1200
P 300
-
æö æö
=
ç÷ç÷
èøèø
0.4
1.41
2
P
(4)
P
-
æö
=
ç÷
èø
0.4
1.42
1
P
4
P
æö
=
ç÷
èø
1.47
0.42
211
P P4 P4
æ ö æö
ç ÷ ç÷
è ø èø
==
= P
1
(2
7
) = 2 × 128 = 256 atm
42. (
a)Q Internal energy is the state function.
\In cyclie process; DU = 0
According to 1st law of thermodynamics
D=D+Q UW
So heat absorbed
DQ
= W = Area under the curve
= – (2V) (P) = – 2PV
So heat rejected = 2PV

327. 323Thermodynamics
43. (a) Initial and final condition is same for all process
DU
1
= DU
2
= DU
3
from first law of thermodynamics
DQ = DU + DW
Work done
DW
1
> DW
2
> DW
3
(Area of P.V. graph)
So DQ
1
> DQ
2
> DQ
3
44. (d)Efficiency
1
W
h=
q
and W = q
1
– q
2
\
12
1
–qq
h=
q
=
2
1
1
q
-
q
45. (b) P
1
> P
2
P
1
P
2
T
V
q
2
q
1
As V = con
stant Þ Pµ T
Hence from V–T graph P
1
> P
2
46. (b) Change in internal energy do not depend upon the
path followed by the process. It only depends on initial
and final states i.e.,
DU
1
= DU
2
47. (a) For adiabatic process Q = 0.
By first law of thermodynamics,
Q = DE + W
ÞDE
int
= – W.
48
. (d) The efficiency of carnot's heat engine
2
1
T
1–
T
h=
where T
2
i
s temperature of sink, and T
1
is temperature
of hot reservoir or source.
When efficiency is 40% i.e. h = 40/100 =
2
1
T
1–
T
or
2
1
T2
1– [T T (given)]
5T
==
\T
2
=
3
T
5
Now, when efficiency is 50%
1
1
3
T
506 5
1– TT
100 5T
h= = \=
49. (
a) 50. (c)
EXERCISE - 3
Exemplar Questions
1. (c) For the straight line in the graph denoted by 4, that
shows pressure is constant, so curve 4 represents an
isobaric process.
p
V
1
2
3
4
For the str
aight line in graph denoted by 1, that shows
volume is constant, so curve 1 represents isochoricprocess. Out of curves 3 and 2, curve 2 is steeper.
Hence, curve 2 is adiabatic and curve 3 is isothermal.
2. (a) As we know that amount of sweat evaporated/minute
Sweat produced/ minute
Number of calories required for evaporation/kg
=
Amount of heat produced per minute in jogging
Latent heat (in cal/kg)
=
580 × 10
3
calories are needed to convert
1 kg H
2
O into stream.
1 cal. will produce sweat = 1 kg/ 580 × 10
3
14.5 × 10
3
cal will produce (sweat)
3
3
14.5 10 145
kg= kg/m
580580 10
´
=
´
= 0.025 kg.
3. (c
) According to given P–V diagram that
pV = constant
So we can say that the gas is going through anisothermal process.
If pressure (P) increase at constant temperature volume
V decreases, the graph (iii) shows that pressure (P) is
smaller at point 2 and larger at point 1 point so the gas
expands and pressure decreases. Hence verifies option
(c).
4. (b) According to the given p-V diagram.
Work done in the process ABCD
= (AB) × BC = (3V
0
– V
0
) × (2p
0
– p
0
)
= 2V
0
× p
0
= 2p
0
V
0
Here the direction of arrow is anti-clockwise, so work
done is negative.
Hence, work done by the gas = –2p
0
V
0
That shows external work done on the system.
5. (a) Let us consider the p-V diagram for container A
(isothermal) and for container B (adiabatic).

328. 324 PHYSICS
2 2
1 1
p p
p
0
p
0
V
0
V
0
2V
0
2V
0
V V
Container
(Isothermal)
A Contai ner
(Adiabatic)
B
In both process compression of the gas.
For isothermal process (gas A) during 1® 2
11 22
pV pV=
1 020
(2,)V VVV==Q
0 0 20
(2) ()p V pV=
20
2pp=
For adiabatic proces
s, (gas B) during (1® 2)
12
12
pV pV
gg
=
1 020(2,)V VVV==Q
0 0 20
(2) ()p V pV
gg
=
0
2 00
0
2
(2)
V
p pp
V
g
g
æö
==
ç÷
èø
So, ratio of fin
al pressure
102
10
(2)()
2
()2
A
pp
pp
g
g-B
æö
= ==
ç÷ èø
where, g is ratio o
f specific heat capacities for the gas.
Hence, verifies the option (a).
6. (b) Consider the equilibrium temperature of the system is
T.
Let us consider, T
1
, T
2
< T < T
3
.
As given that, there is no net loss to the surroundings.
Heat lost by M
3
= Heat gained by M
1
+ Heat gained by M
2
33 1122
()()()M sT T MsT T M sT T-=-+-
3 3 3 1 11M sT M sT M sT M sT-=-
2 22M sT M sT+-
(where, s is spec ified heat of the copper material)
123 331122
[]TM M M MT MT MT++= ++
11 22 33
123
MT MT MT
T
MMM
++
=
++
NEET/AIPMT (2013-2017) Questions
7. (a)W
net
= Area of triangle ABC
=
1
2
AC × BC
=
1
2
× 5 × 10
–3
× 4 × 1 0
5
= 1000 J
8. (c) According to question P µ T
3
But as we know for an adiabatic process the
pressure P µ
1
T
g
g-
.
So,
1
g
g-
= 3 Þ g =
3
2
or, ,
p
v
C
C
=
3
2
9. (a) From first law of thermodynamics
Q
adc
=DU
adc
+ W
adc
50 J =DU
adc
+ 20 J
DU
adc
= 30 J
Again,Q
abc
=DU
abc
+ W
abc
W
abc
=Q
abc
– DU
abc
=Q
abc
– DU
adc
=36 J – 30 J
= 6 J
10. (a) Adiabatic equations of state are
PV
g
= constant
TV
g–1
= constant
P
1–g
T
g
= constant.
11. (d) Efficiency of engine A,
1
1
1,
T
T
h=-
Efficiency of en
gine B,
2
21
T
T
h=-
Here, h
1
= h
2
\
2
1
TT
TT
=
Þ
12
T TT=
12. (d) Wo
rk done by the system in the cycle
= Area under P-V curve and V-axis
=
00 00
1
(2P P )(2V V )
2
- -+
0 0 00
1
(3P 2P )(2V V )
2
éùæö
---ç÷êú
èø
ëû
=
00 00
PV PV
0
22
-=
13. (c) For isothermal process P
1
V
1
= P
2
V
2
ÞPV = P
2
(2V)ÞP
2
=
P
2
For adiabatic process
2323
PV PV
gg
=
Þ 3
P
(2v) P 16v)
2
ggæö
=ç÷
èø
ÞP
3
=
5/3
31P
2 8 64
æö
=
ç÷ èø

329. 325Thermodynamics
14. (b) In cyclic process ABCA
Q
cycle
= W
cycle
Q
AB
+ Q
BC
+ Q
CA
= ar. of DABC
+ 400 + 100 + Q
C®A
=
1
2
(2 × 10
–3
) (4 × 10
4
)
ÞQ
C ® A
= –
460 J
ÞQ
A ® C
= + 460 J
15. (a) Given, efficiency of engine, h =
1
10
work done on system W = 10J
Coefficient of performance of refrigerator
2
Q1
W
-h
b==
h
=
19
1
10 10
11
10 10
-
== 9
Energ
y absorbed from reservoir
Q
2
= bw
Q
2
= 9 × 10 = 90 J
16. (d) Coefficient of performance,
Cop =
2
12
T
TT-
5 =
11
273 20 253
T (273 20) T 253
-
=
---
5T
1
– (5 × 253) = 253
5T
1
= 253 + (5
× 253) = 1518
\ T
1
=
1518
5
= 303.6
or, T
1
= 303.6 – 2
73 = 30.6 @ 31°C
17. (d) Since area under the curve is maximum for adiabatic
process so, work done (W = PdV) on the gas will be
maximum for adiabatic process
P
V
Adiabatic
Isobaric
Isot
hermal
18. (b) W
ext
= negative of area with volume-axis
W(adiabatic) > W(isothermal)
P
O
V
0 2V
0 V
Adiabatic
Isothermal
19. (c)Coefficient of performance of a refrigerator,
b =
22
12
QT
W TT
=
-
(Where Q
2
is h
eat removed)
Given: T
2
= 4°C = 4 + 273 = 277 k
T
1
= 30°C = 30 + 273 = 303 k
\b =
600 4.2 277
W 303 277
´
=
-
ÞW = 236
.5 joule
Power P =
W
t
=
236.5joule
1sec
= 236.5 w
att.
20. (a) Process I volume is constant hence, it is isochoric
In process IV, pressure is constant hence, it is isobaric

330. 326 PHYSICS
BEHAVIOUR OF G
ASES
Ideal gas : In an ideal gas, we assume that molecules are point
masses and there is no mutual attraction between them. The ideal
gas obeys following laws :
(i)Boyle’s law : According to Boyle’s law for a given mass of
ideal gas, the pressure of a ideal gas is inveresly proportional
to the volume at constant temperature
i.e.,
1
PV = constantµÞP
V
(ii)Charle’s law : Fo r a given mass, the volume of a ideal gas
is proportional to temperature at a constant pressure
i.e.,constantµÞ=
V
VT
T
(iii)Gay-Lussac’s law : For a given mass of ideal gas, the
pressure is proportional to temperature at constant volume
i.e., constantµ=
P
PT
T
(iv)Avogadro’s la
w : According to Avogadro’s law, the number
of molecules of all gases are same at same temperature,
pressure and volume
i.e.,
12
=
AA
NN for same P, V and T.
Th
e value of Avagadro number is 6.02 × 10
23
molecules.
(v)Graham’s law :
(a) At constant temperature and pressure, the rms speed
of diffusion of two gases is inversely proportional to
the square root of the relative density
i.e.,v
rms
1
a
d
Þ
()
( )
12
1
2
=
rms
rms
v d
dv
(b) According
to Graham’s law, the rate of diffusion of a
gas is inversely proportional to the square root of its
density, provided pressure and temperature are
constant
1
(rate of diffusion)
ρ
r µ
(vi)Dalton’s law: The
pressure exerted by a gaseous mixture is
equal to sum of partial pressure of each component gasespresent in the mixture,i.e.,P = P
1
+ P
2
+ P
3
+.................. P
n
A relation connecting macroscopic properties P, V and T ofa gas describing the state of the system is called equation
of state.
The equation of state for an ideal gas of n mole is
PV
nR
T
=
where R is u
niversal gas constant whose value is
R = 8.31 J/mol K and R = N
A
k, where k is Boltzmann’s
constant (N
A
is Avagadro number). n is the number of moles
of a gas and
==
A
mN
n
MN
, N
A
is Avagadro's number
where m is the mass of a gas, N is the number of molecules
and M is the molecular weight of a gas.
Equation of Real Gas :The real gas follows Vander Wall’s law. According to this:
2
2
()
æö
+ -=ç÷
èø
na
p V nb nRT
V
;
here a and b are V
ander Wall’s constant
Critical Temperature, Volume and Pressure :
(a) The temperature at or below which a gas can be liquefied
by applying pressure alone is called critical temperature.
It is given by
8
27
=
c
a
T
Rb
(b) The volume of gas at a critical temperature T
C
is called
critical volume V
C
, where V
C
= 3b
(c) The pressure of gas at a critical temperature T
C
is called
critical pressure P
C
, where
2
27
=
C
a
P
b
Gas Equation
(
i) The gases found in nature are real gases.
(ii) The real gas do not obey ideal gas equation but they obey
vander wall's gas equation
( )
2
P V RT
V
a
b
æö
+ -=
ç÷
èø
13
Kinetic Theory

331. 327Kinetic Theory
kTT
N
R
Cm
3
1 2
== (
N
R
k= is Boltzmann co
nstant)
or
kT
2
3
Cm
2
1 2
= ....(vi)
or
2
..
13
22
=
rmsmv kT
or
2
..
3
=
rms
kT
v
m
....(vii)
o
r
..
33
==
rms
nkT RT
v
nmM
(a) It is clear from eq.(vi) that at a given temperature, the average
translational kinetic energy of any gas molecules are equal
i.e., it depends only on temperature.
(b) From eq
n
. (vii) It is clear that
(a)
r.m.s
vTµ
(b)
r.m.s
1
v
M
µ , wher e M is moleculer mass of the gas.
DEGREE OF FREEDOM
The degree of freedom of a particle is the number of independent
modes of exchanging energy or the number of idependent motion,
which the particle can undergo.
For monatomic (such as helium, argon, neon etc.) gas the
molecules can have three independent motion i.e., it has
3 degree of freedom, all translational.
Z
Y
X
For a diatomic gas molecules such as H
2
, O
2
, N
2
,

etc. it has two
independent rotational motion besides of three independenttranslational motion, so it has 5 degree of freedom.
Z Z
Y
Y
X X
Diatomic Triatomic
In polyatomic gas molecules such as CO
2
, it can rotate about any
of three coordinate axes. It has six degree (three translational
+three rotational) of freedom. At high temperature the molecule
can vibrate also and degree of freedom due to vibration also
arises, but we neglect it.
(a) 'a' depends upon the intermolecular force and the
nature of gas.
(b) 'b' depends upon the size of the gas molecules and
represents the volume occupied by the molecules of
the gas.
(iii) The molecules of real gas have potential energy as well as
kinetic energy.
(iv) The real gas can be liquefied and solidified.
(v) The real gases like CO
2
, NH
3
, SO
2
etc. obey Vander Wall's
equ
n
at high pressure and low temperature.
KINETIC THEORY OF AN IDEAL GAS
The basic assumptions of kinetic theory are :
(i) A gas consist of particles called molecules which move
randomly in all directions.
(ii) The volume of molecule is very small in comparison to the
volume occupied by gas i.e., the size of molecule is
infinitesimely small.
(iii) The collision between two molecules or between a molecule
and wall are perfectely elastic and collision time (duration
of collision) is very small.
(iv) The molecules exert no force on each other or on the walls
of containers except during collision.
(v) The total number of molecules are large and they obey
Newtonian mechanics.

21
3
=pV mnC ....(i)
where
2
C is called mean s
quare velocity and
2
C
is called
roo
t mean square velocity
i.e,
222
2 12
...............++
=
n
CCC
C
n
....(ii)
and
2
.s.m.rC=n ....(iii)
whe
re m = mass of one molecule and n = number of
molecules

22
r.m.s
M1
PC ρv
3V3
Þ== ....(iv)
o
r
ú
û
ù
ê
ë
é
r=
2
s.m.r
v
2
1
3
2
P where and
æö
= ´ r=ç÷
èø
M
M mn
V
or P
2
E
3
= ....(v)
wh
ere E is translational kinetic energy per unit volume of
the gas. It is clear that pressure of ideal gas is equal to 2/3
of translational kinetic energy per unit volume.
Kinetic interpretation of temperature : From eq
n
. (iv), we get2
CM
3
1
PV=
For 1 m
ole of a gas at temperature T :
PV = RT so
RTCm
3
1 2
=

332. 328 PHYSICS
MEAN FREE PATH
Th
e distance covered by the molecules between two successive
collisions is called the free path.
The average distance covered by the molecules between two
successive collisions is called the mean free path
i.e.,
2
1
2.nd
l=
p
2
2
B
KT
dP
=
p
where,n= number of
molecules per unit volume
d= diameter of each molecule
K
B
= Boltzmann’s constant
T= temperature
P= pressure
Mean free path depends on the diameter of molecule (d) and thenumber of molecules per unit volume n.
At N.T.P., l for air molecules is 0.01 µm.
LAW OF EQUIPARTITION OF ENERGY
If we dealing with a large number of particles in thermal
equilibrium to which we can apply Newtonian mechanics, the
energy associated with each degree of freedom has the same
average value (i.e., 1
T
2
k), and this average value depends on
temperatur
e.
From the kinetic theory of monatomic ideal gas, we have
213
22
=mC kT ....(i)
or
Energy of molecule
= (number of degree of freedom) × kT
2
1
....(ii)
So it is clear fro
m equation (i) and (ii) that monatomic gas has
three degree of freedom and energy associated per degree offreedom is
kT
2
1
(where k is Boltzmann’
s constant)
Since we know that internal energy of an ideal gas depends only
on temperature and it is purely kinetic energy. Let us consider an
1 mole ideal gas, which has N molecules and f degree of freedom,
then total internal energy
U = (total number of molecules)
× (degree of freedom of one molecule)
× (energy associated with each degree of freedom)
=(N) (f) × (
kT
2
1
)
U = NfkT
2
1
....(iii)
Differen
tiating eq.(iii) w.r.t T, at constant volume, we get,
2
NfK
T
U
V
=÷
ø
ö
ç
è
æ
¶
¶
....(iv)
Now molar, spe
cific heat at constant volume is defined as
VV
V
T
U
T
Q
C ÷
ø
ö
ç
è
æ
¶
¶
=÷
ø
ö
ç
è
æ
¶
¶
= ....(v)
so
V
NKR
C
22
ff
== ....(vi)
where R is
universal gas constant.
Now by Mayer’s formula,
1
2
æö
-=Þ=+ ç÷
èø
PVP
f
CCRCR ... (vii )
So ratio of specific heat
V
P
C
C
=g is
V
P
C
C
f
2
1=+=g ...(viii)
w
here f is degree of freedom of one molecule.
(a)For monatomic gas, f = 3,
3
2
=
vCR ,
5
2
=
pCR
so
2
1 1.67
3
g=+=
(b)For diatomic gas
, f = 5,
5
2
=
v
CR ,
7
2
=
p
CR
4.1
5
2
1=+=g
(c)For Polyatomic
gas, f = 6,
3=
v
CR , 4=
p
CR
33.1
6 2
1=+=g
Keep in Memory
1. Real gases behave li
ke perfect gas at high temperature
and low pressure.
In real gas, we assume that the molecules have finite size
ant intermolecular attraction acts between them.
2.Real gases deviate most from the perfect gas at high pressure
and low temperature.
3.Gaseous state of matter below critical temperature is called
vapours. Below critical temperature gas is vapour and above
critical temperature vapour is gas.
4.Random motion of the consituents of the system involving
exchange of energy due to mutual collisions is called
thermal motion.
5.Total kinetic energy or internal energy or total energy does
not depend on the direction of flow of heat. It is determined
by the temperature alone.
6.The internal energy of a perfect gas consists only of kinetic
energy of the molecules. But in case of the real gas it
consists of both the kinetic energy and potential energy of
inter molecular configuration.
Distribution of Molecular Speeds :The speed of all molecules in a gas is not same but speeds ofindividual molecules vary over a wide range of magnitude . Maxwell derived the molecular distribution law (by which wecan find distribution of molecules in different speeds) for sample

333. 329Kinetic Theory
of a gas containing N molecules, which is
kT2/mv2
2/3
2
ev
kT2
m
N4)v(N
-
÷
ø
ö
ç
è
æ
p
p=
...(1)
where
N(v)dv is the number of molecules in the gas sample having
speeds between v and v + dv.
where T = absolute temperature of the gas
m = mass of molecule
k = Boltzmann’s constant
The total number of molecules N in the gas can be find out by
integrating eqation (1) from 0 to ¥ i.e.,
0
()
¥
=ò
N N v dv ....(2)
T
2
T
1
T
2 > T
1
v
rms
v
Speed (m/sec)
N
o
.

o
f

m
o
le
c
u
le
s

p
e
r
unit

s
p
e
e
d

in
t
e
r
v
a
l
v
v
p
Figure shows Maxwell distribution law for molecules at two
different temperature T
1
and T
2
(T
2
> T
1
).
The number of molecules between v
1
and v
2
equals the area
under the curve between the vertical lines at v
1
and v
2
and the
total number of molecules as given by equation (2) is equal to
area under the distribution curve.
The distribution curve is not symmetrical about most probable
speed, v
P
,(v
P
is the speed, which is possesed in a gas by a large
number of molecules) because the lowest speed must be zero,
whereas there is no limit to the upper speed a molecule can attain.
It is clear from fig.1 that
v
rms
(root mean square) >
v(average speed of molecules)
> (most probable speed) v
P
AVERAGE, ROOT MEAN SQUARE AND MOST PROBABLE
SPEED
Average speed
To find the average speed
v, we multiply the number of particles
in each speed interval by speed v characteristic of that interval.
We sum these products over all speed intervals and divide by
total number of particles
i.e.,
()
¥
=
ò
o
N v vdv
v
N
(where summation is replaced by
integration because N is large)

8
1.59==
p
kT kT
hv
mm
....(1)
Root Mean Square Speed
In this c
ase we multiply the number of particles in each speed
interval by v
2
characterstic of that interval; sum of these products
over all speed interval and divide by N
i.e.,
2
2
()
¥
=
ò
o
N v v dv
v
N
....(2)
Root m
ean square speed is defined as 2
..
3
1.73===
rms
kT kT
vv
mm
Most Probable Speed
It is the speed at which N(v) has its maximum value (or possessed
by large number of molecules), so
Ζ∴ p
d 2kT kT
N(v) 0 v 1.41
du mm
<?<< ....(3)
It is
clear from eq
n
. (1), (2) and (3)..
>>
rmsp
v vv
The root mean square velocity of a particle in thermal
system is given by
33
==
rms
RT kT
C
Mm
where R -universal gas constant
T - temperature of gas
m - mass of the gas
M - molecular weight of gas or mass of one mole
of a gas.
The average speed of the gas molecules is given by

88
==
pp
av
kT RT
C
mM
where M is the mass of one mole and m is the mass of one
particle.Most probable speed is that with which the maximum number
of molecules move. It is given by

2 2RT 2T
V
3M
=
==
p rms
k
c
m
The most probable speed, the average speed as well as
root mean square speed increases with temperature.
Example 1.
The root mean square velocity of the molecules in a sample
of helium is 5/7th that of the molecules in a samzple of
hydrogen. If the temperature of the hydrogen gas is 0ºc,
then find the temperature of helium sample.
Solution :
()
( )
==
HH
HeHe
Hs.m.r
Hes.m.r
m/KT3
)m/KT3(
v
v
He
H
H
He
m
m
T
T
´

273
T
2
1
7
5
4273
1T
HeHe
=Þ
´
´
=
or Cº14.284Kº14.557T
He
==

334. 330 PHYSICS
Example 2.
L
et
rmsP
v,v andv respectively denote the mean speed,
r.m.s. speed, and most probable speed of the molecules in
an ideal monatomic gas at absolute temperature T. The
mass of a molecule is m, then which of the following is
correct ?
(a) No molecule can have a speed greater than
max
2v
(b) No molecule
can have a speed less than
p
v/2
(c)p r.m.s.
v vv<<
(d) None of the
se
Solution : (c)
According to kinetic theory of gases, a molecule of a gascan have speed such that 0<
v<¥, so the alternatives (a)
a
nd (b) can never be correct, Since
. ..
3
rms
kT
v
m
= ;
8
3
av
kT
v
m
= ; and
2
mp
kT
v
m
=
so
. ..rmsavp
v vv>> i.e.
. ..p rms
v vv<< ;
Example 3.
A gas cyl
inder of 50 litres capacity contained helium at
80 atmospheric pressure. Due to slow leakage it was found
after a while that the pressure had dropped to 40
atmosphere. Find the proportion of the gas that has
escaped and also the volume the escaped gas would
occupy at atmospheric pressure.
Solution :
Volume of the remaining gas
= volume of cylinder = 50 l at 40 atmospheric pressure
The volume of this gas at 80 atmospheric pressure, using
Boyle’s law p
1
V
1
= p
2
V
2
=
40 50
80
´
= 25 l
Volume of the gas
(at 80 atmospheric pressure) which has
escaped = 50 – 25 = 25 l
\
Mass of gas escaped251
Original mass of gas50 2
==
Again using Boyle’s law,
1 × V = 80 × 25 or V = 2000 litre
Example 4.
In Vander Wal’s equation the critical pressure P
c
is given
by
(a) 3b (b)
2
a
27b
(c)
2
27a
b
(d)
2
b
a
Solution : (b)
The V
ander Wall’s equation of state is

RT)bV(
V
a
P
2
=-
÷
÷
ø
ö
ç
ç
è
æ
+ ; or
2
V
a
bV
RT
P -
-
=
At the critical poin
t, P = P
c
, V = V
c
, and T = T
c
;
\
2
cc
c
c
V
a
bV
RT
P -
-
= ... (i)
At the crit
ical point on the isothermal,
0
dV
dP
c
c
=
\
23
2
0
()
c
cc
RT a
VbV
-
=-
-
; or
3
c
2
c
c
V
a2
)bV(
RT
=
-
...(ii)
Also at critic
al point,
0
dV
Pd
2
c
c
2
=
\
4
c
3
c
c
V
a6
)bV(
RT2
0 -
-
= ; or
4
c
3
a
c
V
a6
)bV(
RT2
=
-
...(iii)
Dividing eq
n
. (iii) b
y (ii), we get,
cc
V
3
1
)bV(
2
1
=- or b3V
c
= ...(iv)
Putt
ing this value in eq
n
. (ii), we get,
32
b27
a2
b4
RT
= or
Rb27
a8
T
c= ...(v)
Putting the val
ues of V
c
and T
c
in eq
n
. (i), we get,
22
c
b27
a
b9
a
Rb27
a8
b2
R
P =-
÷
÷
ø
ö
ç
ç
è
æ
=
Keep in Memory
1.Brownia
n motion, provides a direct evidence for the
existance of molecules and their motion. The zig-zag motion
of gas molecules is Brownian motion.
2.Average speed
8 88
πππ
= ==
B
kT RT PV
v
mMM
3.Root mean square speed,
3 33
= ==
B
rms
kT RT PV
v
mMM
4.Most probable speed
2 22
= ==
B
mp
kT RT PV
V
mMM
5.V
rms
: V : V
mp
= 1.73 : 1.60 : 1.41
V
rms
> V > V
mp.

335. 331Kinetic Theory

336. 332 PHYSICS
1.The kin
etic theory of gases
(a) explains the behaviour of an ideal gas.
(b) describes the motion of a single atom or molecule.
(c) relates the temperature of the gas with K.E. of atoms
of the gas
(d) All of the above
2.The ratio of principal molar heat capacities of a gas is
maximum for
(a) a diatomic gas
(b) a monatomic gas
(c) a polyatomic gas having linear molecules.
(d) a polyatomic gas having non-linear molecules.
3.The correct statement of the law of equipartition of energy
is
(a) the total energy of a gas is equally divided among all
the molecules.
(b) The gas possess equal energies in all the three
directions x,y and z-axis.
(c) the total energy of a gas is equally divided between
kinetic and potential energies.
(d) the total kinetic energy of a gas molecules is equally
divided among translational and rotational kinetic
energies.
4.The internal energy of an ideal gas is
(a) the sum of total kinetic and potential energies.
(b) the total translational kinetic energy.
(c) the total kinetic energy of randomly moving molecules.
(d) the total kinetic energy of gas molecules.
5.A fixed mass of gas at constant pressure occupies a volume
V. The gas undergoes a rise in temperature so that the root
mean square velocity of its molecules is doubled. The new
volume will be
(a) V/2 (b)
2/V
(c) 2 V (d) 4 V
6.In kinetic
theory of gases, it is assumed that molecules
(a) have same mass but can have different volume
(b) have same volume but mass can be different
(c) have different mass as well as volume
(d) have same mass but negligible volume.
7.Gases exert pressure on the walls of the container because
the gas molecules
(a) possess momentum
(b) collide with each other
(c) have finite volume
(d) obey gas laws.
8.Let v denote the rms speed of the molecules in an ideal
diatomic gas at absolute temperature T.
The mass of a molecule is ‘m’ Neglecting vibrational energy
terms, the false statement is
(a) a molecule can have a speed greater than
2v
(b)v is proportional to T
(c) the average rotational K.E. of a molecule is
21
4
mv
(d) the average K.E. of a molecule is
25
6
mv
9.The adjoining figure shows graph of pressure and volume
of a gas at two tempertures T
1
and T
2
. Which of the following
inferences is correct?
T
2
T
1
V
P
(a)T
1
> T
2
(b) T
1
= T
2
(c)T
1
<
T
2
(d) None of these
10.At constant pressure, the ratio of increase in volume of an
ideal gas per degree rise in kelvin temperature to its original
volume is (T = absolute temperature of the gas) is
(a)T
2
(b)T (c) 1/T (d) 1/T
2
11.The K.E. of one mole of an ideal gas is E = (3/2) RT. Then C
p
will be
(a) 0.5 R (b) 0.1 R
(c) 1.5 R (d) 2.5 R
12.The root mean square speed of the molecules of a diatomic
gas is v. When the temperature is doubled, the molecules
dissociate into two atoms. The new root mean square speed
of the atom is
(a)
2v(b) v (c) 2v (d) 4v
13.Wh
ich of the following formula is wrong?
(a)
–1
V
R
C=
g
(b)
–1
P
R
C
g
=
g
(c)C
p
/ C
V
= g (d)C
p
– C
V
= 2R
14.For a
gas if ratio of specific heats at constant pressure and
volume is g then value of degrees of freedom is
(a)
3 –1
2 –1
g
g
(b)
2
–1g
(c)
9
( – 1)
2
g (d)
25
( – 1)
2
g
15.In the kinetic theor
y of gases, which of these statements
is/are true ?
(i) The pressure of a gas is proportional to the mean speed
of the molecules.
(ii) The root mean square speed of the molecules is
proportional to the pressure.
(iii) The rate of diffusion is proportional to the mean speed
of the molecules.
(iv) The mean translational kinetic energy of a gas is
proportional to its kelvin temperature.
(a) (ii) and (iii) only(b) (i),(ii)and (iv) only
(c) (i) and (iii) only(d) (iii) and (iv) only

337. 333Kinetic Theory
16.A gas mixture consists of molecules of type 1, 2 and 3, with
molar masses m
1
> m
2
> m
3
. v
rms
and Kare the r.m.s. speed
and average kinetic energy of the gases. Which of the
following is true?
(a)(v
rms
)
1
< (v
rms
)
2
< (v
rms
)
3
and
123
() () ()KKK==
(b) (v
rms
)
1
= (v
rms
)
2
= (v
rms
)
3
an
d
123
() () ()KKK=>
(c)(v
rms
)
1
> (v
rms
)
2
> (v
rms
)
3

and
123
() () ()KKK<>
(d) (v
rms
)
1
> (v
rms
)
2
> (v
rms
)
3
an
d
123
() () ()KKK<<
17.A samp
le of an ideal gas occupies a volume of V at a pressure
P and absolute temperature. T. The mass of each molecule
is m. The equation for density is
(a) m k T (b) P/k T
(c) P/(k T V) (d) P m/k T
18.The value of critical temperature in terms of van der Waal’s
constant a and b is given by
(a)
c
8a
T=
27Rb
(b)c
a
T=
27bR
(c)
c
a
T=
2Rb
(d)c
–a
T=
Rb
19.At room temperature, the rms speed of the molecules of a
certain diatomic gas is found to be 1930 m/s. The gas is
(a)H
2
(b) F
2
(c)O
2
(d) Cl
2
20.Two gases A and B having the same temperature T, same
pressure P and same volume V are mixed. If the mixture is at
the same temperature T and occupies a volume V, the
pressure of the mixture is
(a) 2 P (b) P
(c) P/2 (d) 4 P
21.The perfect gas equation for 4 gram of hydrogen gas is
(a) PV = RT (b) PV = 2RT
(c) PV=
2
1
R TT (d) PV =
4RT
22.Maxwell's laws of distribution of velocities shows that(a) the number of molecules with most probable velocity
is very large
(b) the number of molecules with most probable velocity
is very small
(c) the number of molecules with most probable velocity
is zero
(d) the number of molecules with most probable velocity
is exactly equal to 1
23.According to kinetic theory of gases, which one of thefollowing statement(s) is/are correct?(a) Real gas behave as ideal gas at high temperature and
low pressure.
(b) Liquid state of ideal gas is impossible(c) At any temerature and pressure, ideal gas obeys
Boyle's law and charles' law
(d) The molecules of real gas do not exert any force on
one another.
24.For hydrogen gas C
p
– C
v
= a and for oxygen gas
C
p
– C
v
= b. So, the relation between a and b is given by
(a) a = 16 b (b) 16 a = b
(c) a = 4 b (d) a = b
25.The relation between the gas pressure P and average kineticenergy per unit volume E is
(a)
1
2
PE= (b)P = E
(c)
3
2
PE= (d)
2
3
PE=
1.If the critical tempe
rature of a gas is 100ºC, its Boyle
temperature will be approximately(a) 337.5ºC (b) 500ºC
(c) 33.3ºC (d) 1000ºC
2.The r.m.s. velocity of oxygen molecule at 16ºC is474 m/sec. The r.m.s. velocity in m/s of hydrogen moleculeat 127ºC is(a) 1603 (b) 1896
(c) 2230.59 (d) 2730
3.The gases are at absolute temperature 300ºK and 350ºKrespectively. The ratio of average kinetic energy of theirmolecules is(a) 7 : 6 (b) 6 : 7
(c) 36 : 49 (d) 49 : 36
4.The total degree of freedom of a CO
2
gas molecule is
(a)3 (b) 6
(c)5 (d) 4
5.If one mole of a monatomic gas (g = 5/3) is mixed with one
mole of a diatomic gas (g = 7/3), the value of g for the mixture
is
(a) 1.40 (b) 1.50
(c) 1.53 (d) 3.07
6.The molecules of a given mass of gas have a root mean
square velocity of 200m s
–1
at 27°C and 1.0 × 10
5
N m
–2
pressure. When the temperature is 127°C and the pressure
0.5 × 10
5
Nm
–2
, the root mean square velocity in ms
–1
, is
(a)
3
400
(b) 2100
(c)
3
2100
(d)
3
100

338. 334 PHYSICS
7.If masses of
all molecule of a gas are halved and their speed
doubled then the ratio of initial and final pressure will be
(a) 2 : 1 (b) 1 : 2
(c) 4 : 1 (d) 1 : 4
8.On a particular day, the relative humidity is 100% and the
room temperature is 30ºC, then the dew point is
(a) 70ºC (b) 30ºC
(d) 100ºC (d) 0ºC
9.The velocity of the molecules of a gas at temperature 120 K
is v. At what temperature will the velocity be 2 v?
(a) 120 K (b) 240 K(c) 480 K(d) 1120 K
10.The density of a gas is 6 × 10
–2
kg/m
3
and the root mean
square velocity of the gas molecules is 500 m/s. The
pressure exerted by the gas on the walls of the vessel is
(a)
32
5×10 N/m (b)
–42
1.2×10 N/m
(c)
–42
0.83×10 N/m (d)
2
30N/m
11.The temperature of the mixture of one mole of helium and
one mole of hydrogen is increased from 0°C to 100°C at
constant pressure. The amount of heat delivered will be
(a) 600 cal (b) 1200 cal
(c) 1800 cal (d) 3600 cal
12.Helium gas is filled in a closed vessel (having negligible
thermal expansion coefficient) when it is heated from 300 K
to 600 K, then average kinetic energy of helium atom will be
(a)
2 times (b) 2 times
(c
) unchanged (d) half
13.One mole of a gas occupies 22.4 lit at N.T.P. Calculate the
difference between two molar specific heats of the gas.
J = 4200 J/kcal.
(a) 1.979 k cal/kmol K(b) 2.378 k cal/kmol K
(c) 4.569 kcal/kmol K(d) 3.028 k cal/ kmol K
14.Four molecules have speeds 2 km/sec, 3 km/sec, 4 km/sec
and 5 km/sec. The root mean square speed of these
molecules (in km/sec) is
(a)
4/54 (b) 2/54
(c) 3.5 (d)33
15.The density of air at pressure of 10
5
Nm
–2
is 1.2 kg m
–3
.
Under these conditions, the root mean square velocity of
the air molecules in ms
–1
is
(a) 500 (b) 1000
(c) 1500 (d) 3000
16.How many degrees of freedom are associated with 2 grams
of He at NTP ?
(a)3 (b) 3.01 × 10
23
(c) 9.03 × 10
23
(d) 6
17.If 2 mole of an ideal monatomic gas at temperature T
0
is
mixed with 4 moles of another ideal monatomic gas at
temperature 2T
0
, then the temperature of the mixture is
(a)
0
5
T
3
(b) 0
3
T
2
(c) 0
4
T
3
(d) 0
5
T
4
18.The temperature at which the root mean square velocity of
the gas molecules would become twice of its value at 0°C is
(a) 819°C (b) 1092°C
(c) 1100°C (d) 1400°C
19.At what temperature is the r.m.s. velocity of a hydrogen
molecule equal to that of an oxygen molecule at 47ºC
(a) 80 K (b) –73 K
(c) 3 K (d) 20 K
20.One litre of oxygen at a pressure of 1 atm, and 2 litres of
nitrogen at a pressure of 0.5 atm. are introduced in the vessel
of 1 litre capacity, without any change in temperature. The
total pressure would be
(a) 1.5 atm. (b) 0.5 atm.
(c) 2.0 atm. (d) 1.0 atm.
21.The air in a room has 15 gm of water vapours per cubic
metre of its volume. However for saturation one cubic metre
of volume requires 20 gm of water vapour then relative
humidity is
(a) 50% (b) 75%
(c) 20% (d) 25%
22.A vessel contains air at a temperature of 15ºC and 60% R.H.
What will be the R.H. if it is heated to 20ºC? (S.V.P. at 15ºC is
12.67 & at 20ºC is 17.36mm of Hg respectively)
(a) 262% (b) 26.2%
(c) 44.5% (d) 46.2%
23.To what temperature should be the hydrogen at 327°C be
cooked at constant pressure so that the root mean square
velocity of its molecules becomes half of its previous value
(a) –123°C (b) 120°C
(c) –100°C (d) 0°C
24.If pressure of a gas contained in a closed vessel is increased
by 0.4% when heated by 1ºC, the initial temperature must
be
(a) 250 K (b) 250ºC
(c) 2500 K (d) 25ºC
25.N molecules, each of mass m, of gas A and 2 N molecules,
each of mass 2 m, of gas B are contained in the same vessel
which is maintained at a temperature T. The mean square of
the velocity of molecules of B type is denoted by v
2
and
the mean square of the X component of the velocity of A
type is dentoed by w
2
, w
2
/v
2
is
(a)2 (b) 1
(c) 1/3 (d) 2/3
26.At what temperature, pressure remaining constant, will the
r.m.s. velocity of a gas be half of its value at 0ºC?
(a) 0ºC (b) –273ºC
(c) 32ºC (d) –204ºC

339. 335Kinetic Theory
27.Graph of specific heat at constant volume for a monatomic
gas is
(a)
Y
X
O
c
v
T
(b)
Y
X
O
c
v
T
3R
(c)
Y
X
O
c
v
T
3
—
2
R
(d)
Y
X
O
c
v
T
28.The order of magnitude of the number of nitrogen molecules
in an air bubble of diameter 2 mm under ordinary conditions is
(a) 10
5
(b) 10
9
(c) 10
13
(d) 10
17
29.At identical temperatures, the rms speed of hydrogen
molecules is 4 times that for oxygen molecules. In a mixture
of these in mass ratio H
2
: O
2
= 1:8, the rms speed of all molecules
is n times the rms speed for O
2
molecules, where n is
(a)3 (b) 4/3
(c) (8/3)
1/2
(d) (11)
1/2
30.In Vander Waal’s equation the critical pressure P
c
is given by
(a) 3 b (b)
2
b27
a
(c)
2
b
a27
(d)
a
b
2
31.1 mole of a
gas with
g = 7/5 is mixed with 1 mole of a gas with
g = 5/3, then the value of g for the resulting mixture is
(a) 7/5 (b) 2/5
(c) 24/16 (d) 12/7
32.N molecules, each of mass m, of gas A and 2 N molecules,each of mass 2 m, of gas B are contained in the same vesselwhich is maintained at a temperature T. The mean squarevelocity of molecules of B type is denoted by V
2
and the
mean square velocity of A type is denoted by V
1
, then
2
1
V
V
is
(a)2 (b) 1
(c)
1/3 (d) 2/3
33.An ideal gas is found to obey an additional law VP
2
=
constant. The gas is initially at temperature T and volume
V. When it expands to a volume 2 V, the temperature becomes
(a)
2/T (b) 2 T
(c) 2T2 (d) 4 T
34.Two mon
atomic ideal gases 1 and 2 of molecular masses
m
1
and m
2
respectively are enclosed in separate containers
kept at the same temperature. The ratio of the speed of
sound in gas 1 to that in gas 2 is given by
(a)
2
1
m
m
(b)
1
2
m
m
(c)
2
1
m
m
(d)
1
2
m
m
35.If the pote
ntial energy of a gas molecule is
U = M/r
6
– N/r
12
, M and N being positive constants, then
the potential energy at equilibrium must be
(a) zero (b) M
2
/4N
(c)N
2
/4M (d) MN
2
/4
36.Air is pumped into an automobile tube upto a pressure of
200 kPa in the morning when the air temperature is 22°C.
During the day, temperature rises to 42°C and the tube
expands by 2%. The pressure of the air in the tube at this
temperature, will be approximately
(a) 212 kPa (b) 209 kPa (c) 206 kPa (d) 200 kPa
37.Work done by a system under isothermal change from a
volume V
1
to V
2
for a gas which obeys Vander Waal's
equation
2
()
n
V n P nRT
V
æöa
-b+=ç÷
ç÷
èø
is
(a)
22 12
1 12
log
e
V n VV
nRTn
V n VV
æ öæö-b-
+aç ÷ç÷
-b
è øèø
(b)
22 12
10
1 12
log
V n VV
nRTn
V n VV
æ öæö-b-
+aç ÷ç÷
-b
è øèø
(c)
22 12
1 12
log
e
V n VV
nRTn
V n VV
æ öæö-b-
+bç ÷ç÷
-b
è øèø
(d)
21 12
2 12
log
–
e
V n VV
nRTn
V n VV
æöæö-b
+aç÷ç÷
-b
èøèø
38.A gas mixture consists of 2 moles of oxygen and 4 moles of
Argon at temperature T. Neglecting all vibrational moles,
the total internal energy of the system is
(a) 4 RT (b) 15 RT
(c) 9 RT (d) 11RT
39.The molar heat capacities of a mixture of two gases at
constant volume is 13R/6. The ratio of number of moles of
the first gas to the second is 1 : 2. The respective gases may
be
(a)O
2
and N
2
(b) He and Ne
(c) He and N
2
(d) N
2
and He

340. 336 PHYSICS
40.A graph is plotte
d with PV/T on y-axis and mass of the gas
along x-axis for different gases. The graph is
(a) a straight line parallel to x-axis for all the gases
(b) a straight line passing through origin with a slope
having a constant value for all the gases
(c) a straight line passing through origin with a slope
having different values for different gases
(d) a straight line parallel to y-axis for all the gases
41.Four mole of hydrogen, two mole of helium and one mole of
water vapour form an ideal gas mixture. What is the molar
specific heat at constant pressure of mixture?
(a)
16
R
7
(b)
7
R
16
(c)R (d)
23
R
7
42.An ideal gas is taken through a process in which the pressure
and the volume are changed according to the equation
P = kV. Molar heat capacity of the gas for the process is
(a) C = C
v
+ R (b) C = C
v
+
R
2
(c) C = C
v
–
R
2
(d) C = C
v
+ 2R
43.A vessel
has 6g of hydrogen at pressure P and temperature
500K. A small hole is made in it so that hydrogen leaks out.
How much hydrogen leaks out if the final pressure is P/2
and temperature falls to 300 K ?
(a) 2g (b) 3g (c) 4g (d) 1g
44.Figure shows a parabolic graph between T and 1/V for a
mixture of a gas undergoing an adiabatic process. What is
the ratio of V
rms
of molecules and speed of sound in mixture?
(a)
3/2
(b)2
(c)2/3
T
1/V
2T
0
T
0
1/V
04/V
0
(d)3
45.Which one the following graphs represents the behaviour
of an ideal gas ?
(a)
V
PV
(b)
V
PV
(c)
V
PV
(d)
V
PV
46.The molar specific he
at at constant pressure of an ideal gas
is (7/2)R. The ratio of specific heat at constant pressure to
that at constant volume is
(a) 5/7(b) 9/7 (c) 7/5 (d) 8/7
DIRECTIONS for Qs. 47 to 50 : Each of these question
contains two statements: STATEMENT-1 (Assertion) and
STATEMENT-2 (Reason). Answer these questions from the
following four options.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a
correct explanation for Statement -1
(b) Statement-1 is True, Statement -2 is True; Statement-2 is
NOT a correct explanation for Statement - 1
(c) Statement-1 is True, Statement- 2 is False
(d) Statement-1 is False, Statement -2 is True
47. Statement 1 : The ratio
V
P
C
C
for a monatomic gas is less
than for a diatomic gas.
Statement 2 : The molecules of a monatomic gas have more
degrees of freedom than those of a diatomic gas.
48. Statement 1 : Air pressure in a car tyre increases during
driving.
Statement 2 : Absolute zero temperature is not zero energy
temperature.
49. Statement 1 : The total translational kinetic energy of all
the molecules of a given mass of an ideal gas is 1.5 times the
product of its pressure and its volume.
Statement 2 : The molecules of a gas collide with each other
and the velocities of the molecules change due to the
collision.
50. Statement 1 : Mean free path of a gas molecules varies
inversely as density of the gas.
Statement 2 : Mean free path varies inversely as pressure
of the gas.

341. 337Kinetic Theory
Exemplar Questions
1.A cubic vessel (with face horizontal + vertical) contains an
ideal gas at NTP. The vessel is being carried by a rocket
which is moving at a speed of 500 m s
–1 in vertical direction.
The pressure of the gas inside the vessel as observed by
us on the ground
(a) remains the same because 500 ms
–1 is very much
smaller than v
rms of the gas
(b) remains the same because motion of the vessel as a
whole does not affect the relative motion of the gas
molecules and the walls
(c) will increase by a factor equal to
2 22
rms rms
( (500) ) /vv+
where v
rms w
as the original mean square velocity of
the gas
(d) will be different on the top wall and bottom wall of the
vessel
2.1 mole of an ideal gas is contained in a cubical volume V,
ABCDEFGH at 300K (figure). One face of the cube (EFGH)
is made up of a material which totally absorbs any gasmolecule incident on it. At any given time,
C
D
G
H
F
E
A
B
(a) the pressure on EFGH would be zero
(b) the pressure on all the faces will the equal
(c) the pressure of EFGH would be double the pressure
on ABCD
(d) the pressure on EFGH would be half that on ABCD
3.Boyle's law is applicable for an
(a) adiabatic process
(b) isothermal process
(c) isobaric process
(d) isochoric process
4.A cylinder containing an ideal gas is in vertical position
and has a piston of mass M that is able to move up or down
without friction (figure). If the temperature is increased
M
(a) both p and V of the gas will change
(b) only p will increase according to Charles' law
(c)V will change but not p
(d)p will change but not V
5.Volume versus temperature graphs for a given mass of an
ideal gas are shown in figure. At two different values of
constant pressure. What can be inferred about relation
between p
1 and p
2?
40
30
20
10
100200300400500
Vl()
TK()
p
1
p
2
(a)p
1 > p
2 (b)p
1 = p
2
(c)p
1 < p
2 (d)
Data is insufficient
6.1 mole of H
2 gas is contained in a box of volume V = 1.00 m
3
at T =300 K. The gas is heated to a temperature of T = 3000
K and the gas gets converted to a gas of hydrogen atoms.
The final pressure would be (considering all gases to be
ideal)
(a) same as the pressure initially
(b) 2 times the pressure initially
(c) 10 times the pressure initially
(d) 20 times the pressure initially
7.A vessel of volume V contains a mixture of 1 mole of
hydrogen and 1 mole oxygen (both considered as ideal).
Let f
1(v)dv, denote the fraction of molecules with speed
between v and (v + dv ) with f
2(v)dv, similarly for oxygen.
Then,
(a)
12
() () ()f v f v fv+= obeys the Maxwell's distribution
law
(b)f
1(v), f
2(v) will obey the Maxwell's distribution law
separately
(c) neither f
1(v), nor f
2(v) will obey the Maxwell's
distribution law
(d)f
2(v) and f
1(v) will be the same
8.An inflated rubber balloon contains one mole of an ideal
gas, has a pressure p, volume V and temperature T. If the
temperature rises to 1.1 T, and the volume is increased to
1.05 V, the final pressure will be
(a) 1.1 p (b) p
(c) less than p (d) between p and 1.1

342. 338 PHYSICS
NEET/AIPMT (2013-2017) Questions
9.In the given (V – T) diagram, what is the relation between
pressure P
1
and P
2
? [2013]
(a)P
2
> P
1
(b) P
2
< P
1

T
V
P
2
P
1
q
1
q
2(c) Cannot be predicted
(d) P
2
= P
1
10.The amount of heat energy required to raise the temperature
of 1g of Helium at NTP, from T
1
K to T
2
K is [2013]
(a)
3
2
N
a
k
B
(T
2
– T
1
)
(b)
3
4
N
a
k
B
(T
2
– T
1
)
(c)
3
4
N
a
k
B
2
1
T
T
(d)
3
8
N
a
k
B
(T
2
– T
1
)
11.In a vessel, the gas is at a pressure P. If the mass of all the
molecules is halved and their speed is doubled, then the
resultant pressure will be [NEET Kar. 2013]
(a)4P (b) 2P
(c)P (d)P/2
12.The mean free path of molecules of a gas, (radius ‘r’) is
inversely proportional to : [2014]
(a)r
3
(b) r
2
(c)r (d)
r
13.The ratio of the specific heats
p
v
C
C
=g
in terms of degrees
of freedom (n) is given by [2015]
(a)
n
1
3
æö
+
ç÷
èø
(b)
2
1
n
æö
+
ç÷
èø
(c)
n
1
2
æö
+
ç÷
èø
(d)
1
1
n
æö
+
ç÷
èø
14.One mole of an ideal diatomic gas undergoes a transition
from A to B along a path AB as shown in the figure.
P(in kPa)
V(in m)
3
2
5
A
B
4 6
The change in internal energy of the gas during the
transition is: [2015]
(a) – 20 kJ (b) 20 J
(c) –12 kJ (d) 20 kJ
15.Two vessels separately contain two ideal gases A and B at
the same temperature. The pressure of A being twice that of
B. Under such conditions, the density of A is found to be
1.5 times the density of B. The ratio of molecular weight of
A and B is : [2015 RS]
(a)
3
4
(b) 2
(c)
1
2
(d)
2
3
16.4.0 g of a gas occupies 22.4 litres at NTP. The specific heat
capacity of the gas at constant volume is 5.0JK
–1
. If the
speed of sound in this gas at NTP is 952 ms
–1
, then the
heat capacity at constant pressure is (Take gas constant R
= 8.3 JK
–1
mol
–1
) [2015 RS]
(a) 7.5 JK
–1
mol
–1
(b) 7.0 JK
–1
mol
–1
(c) 8.5 JK
–1
mol
–1
(d) 8.0 JK
–1
mol
–1
17.The molecules of a given mass of a gas have r.m.s. velocity
of 200 ms
–1
at 27°C and 1.0 × 10
5
Nm
–2
pressure. When the
temperature and pressure of the gas are respectively, 127°C
and 0.05 × 10
5
Nm
–2
, the r.m.s. velocity of its molecules in
ms
–1
is : [2016]
(a)
1002 (b)
400
3
(c)
100 2
3
(d)
100
3
18.A gas mixture consists of 2 moles of O
2
and 4 moles of Ar at
temperature T. Neglecting all vibrational modes, the total
internal energy of the system is :- [2017]
(a) 15 RT (b) 9 RT
(c) 11 RT (d) 4 RT

343. 339Kinetic Theory
EXERCISE - 1
1. (d)
2. (b) 3. (b) 4. (d)
5. (d)Si nce v
rms
is doubled by increasing the temp. so by
rms
3KT
v
m
= , the temp. increase by four times.
Now for constant pressure
12
12
VV
TT
=
V
1
=V, T
1
=T
ºK, T
2
= 4TºK, V
2
= ?
V
2
= 4V
6. (d) 7. (a) 8. (c)
9. (c) For a given pressure, volume will be more if
temperature is more (Charle's law)
Constant
pressure
T
1
V
P
V
1
V
2
From the graph it is clear that V
2
> V
1
ÞT
1
> T
2
10. (c
) At constant pressure
VT
VT
VT
DD
µÞ=
Hence ra
tio of increase in volume per degree rise in
kelvin temperature to it’s origianl volume
( / )1DD
==
VT
VT
11. (d)v
3
C dE / dT R
2
==
p
35
C R R R 2 .5R
22
=+==
12. (c
)
rms
3RT
v
M
= . Acco rding to problem T will become
2T and M will becomes M/2 so the value of v
rms
will
increase by 42= times i.e. new root mean square
velocity will be 2v.
13. (d) The difference of C
P
and C
V
is equal to R, not 2R.
14. (b)
2 212
1 , –1
2 –1 –1
f
f
ff
g=+ Þg
= Þ = Þ=
gg
15. (d)
16. (a)
1
rmsv
M
µ Þ (v
rms
)
1
< (v
rms
)
2
< (v
r
ms
)
3
also in
mixture temperature of each gas will be same, hence
kinetic energy also remains same.
17. (d) We know that P V = n R T = (m/M) R T
where M = Molecular weight.
Now
TR
M
m
d
m
P ÷
ø
ö
ç
è
æ
=÷
ø
ö
ç
è
æ
´
where d = densit
y of the gas
M
TNk
M
TR
d
P
A
== ... (2)
wher
e R = k N
A
, k is Boltzmann constant.
But
AN
M
= m = mass of eac
h molecule so
Pm
d
kT
´
=
18. (a)
19
. (a) For one mole of gas
M
RT3
v
rms=
3 8.31 300
1930
M
´´
= Þ M = 2×10
–3
k g = 2 g
So that di-atomic gas is hydrogen.
20. (a) Here, initially P
1
= P, V
1
= V + V = 2 V ;
Finally, P
2
= P ; V
2
= V
As P
1
V
1
= P
2
V
2
or
V
VP
P
11
2= = P2
V
V2P
=
´
21. (b)
4
n2
2
==
22. (a) Based on Maxwell's velocity distribution curve.
23. (a, b, c)
24. (d) Both are diatomic gases and C
p
– C
v
= R for all gases.
25. (d)
2
3
PE=
EXERCISE - 2
1. (a)
Boyle temperature is defined as
B cc
a 27 8a 27
T T 3.375 T
Rb 8 27Rb 8
æö
=
= ==ç÷
èø
= 3.375 × 100 = 337.5ºC
2. (c)
32
289R3
v
.oxg
´
=
æö
ç÷
èø
3RT
v=
rms
M
2
400R3
v
H
´
= so v
H
= 2230.59 m/sec
3
. (b) 4. (c)
Hints & Solutions

344. 340 PHYSICS
5. (c)Let g
1

and g
2
denote the ratio of two specific heats of
gas 1 and gas 2 respectively when they are mixed in
the ratio of n
1
and n
2
mole. The resultant value of g is
the weighted mean of g
1
and g
2
i.e.,
11
)5/7(1)3/5(1
nn
nn
21
2211
+
+
=
+
g+g
=g
53.1
30
46
215
2125
==
´
+
=
6. (a)
2
1
c 400 2
c 300 3
== Þ
1
2
2 400
c 200 ms
33
-
=´=
7. (b) Since
V
mNv
3
1
P
2
rms
=
since m is halve
d & speed is doubled so pressure
become twice.
8. (b) Relative humidity (R.H)
S.V.P at dew point
S.V.P at room temp.
=
since here
R.H is 100%
Þroom temp. = dew point = 30ºC
9. (c)vTµ \
v' T'
vT
=
Given 'v =2 v or
,
2 T'
1T
=
\'T = 4 T = 4 × 120K = 480 K
10. (a)
222
)500()106(
3
1
3
1
P ´´´=ur=
-

23
m/N105´=
11. (b)
12
12
12
()
pp
p mix
CC
C
m +m
=
m +m
1
5
( ()
2
p
CHeR= and
2
2
7
())
2
pCHR =
(C
p
)
mix
57
11
22
11
RR´ +´
=
+
= 3R = 3 × 2 = 6 cal/mol.
°C
\ Amount of heat needed to raise the temperature
from 0°C to 100°C
( ) 2 6 100 1200
pp
Q C T calD =mD=´´=
12
. (b) Average K.E. =
213
mc kTT
22
=µ
\
600 K
300 K
(Av.K.E.) 600
2
(Av.K.E.) 300
==
\ At 600K it will be 2 times than that at 300K.
13. (a) V = 22.4 litre = 22.4 × 10
–3
m
3
, J = 4200 J/kcal
by ideal gas equation for one mole of a gas,
53
PV 1.013 10 22.4 10
R
T 273
-
´´´
==
5
pv
R 1.013 10 22.4
CC
J 273 4200
´´
- ==
´
= 1.979 kcal/kmol K
14. (a
)
1/2
2222
rms
(2) (3) (
4) (5)
v
4
éù+++
=êú
êúëû
ú
û
ù
ê
ë
é
=
4
54
15. (a)
5
13P 3 10
c 500 ms
1.2
-´
===
r
16. (c) Mol
es of He =
21
42
=
Molecules =
231
6.02 10
2
´´ = 3.01 × 10
23
As there ar
e 3 degrees of freedom corresponding to 1
molecule of a monatomic gas.
\ Total degrees of freedom = 3 × 3.01 × 10
23
= 9.03 × 10
23
17. (a) Let T be the temperature of the mixture, then
U = U
1
+ U
2
Þ
12
f
(n n ) RT
2
+
= 1020
ff
(n ) (R) (T ) (n) (R) (2T )
22
+
Þ (2 + 4)T = 2
T
0
+ 8T
0
(
Q n
1
= 2, n
2
= 4)
\ T = 0
5
T
3
18. (a)
t
0
c 273 t
c 273
+
= t2732734=-´Þ t 819CÞ=°
19. (d) The r.m.s. velocity of the molecule of a gas of molecular
weight M at Kelvin temperature T is given by,
M
RT3
v
.s.m.r
=
Let M
O
and M
H
are molecu
lar weight of oxygen and
hydrogen and T
O
and T
H
the corresponding Kelvin
temperature at which v
r.m.s.
is same for both,
H
H
O
O
.s.m.r
M
RT3
M
RT3
v ==
Hence,
H
H
O
O
M
T
M
T
=
T
o
= 273º + 47º = 320 K,
M
o
= 32, M
H
= 2
K20320
32
2
T
H =´=

345. 341Kinetic Theory
20. (c) According to Dalton’s Law
P = P
1
+ P
2
.....................
here 1 litre of orygen at a pressure of 1atm, & 2 litre of
nitrogen at a pressure of 5 atm are introduced in a
vessel of 1 litre capacity.
so
11 22
VP PV 1 1 2 .5
P 2atm
volume of vessel 1
+ ´+´
= ==
21. (b) Relative humidity %100
M
m
´=
where
m is the mass of vapour actually present in
certain volume of air & M is the mass required to
saturate the air fully of that volume
so
%75100
20
15
H.R =´=
22. (c)
By definition Þ

100
60
;100
.temproomatP.V.S
.temproomatP.V
H.R ´=

15
(V.P.)
12.67
=
(V.P)
1
5
= 7.6mm of Hg.
Now since unsaturated vapour obeys gas equation &
mass of water remains constant so
nRT
PP αT
V
æö
=Þ
ç÷
èø
15
20
20
(V.P) 273 15
(V.P)
(V.P) 273 20
+
=Þ
+
=7.73 mm of Hg
So
20
20
20
(V.P)
(R.H) 100 44.5%
(S.V.P)
= ´=
23.
(a)
273327
273t
)v(
)v(
1rms
2rms
+
+
= or
1 273
2 600
t+
=
or t = – 123º
C
24. (a)
1
11
T
VP
= constan
t for one mole
&
1
1
T
P
= constan
t for constant volume
so
11
1
11
P 1.004P
T 250K
T (T 1)
= Þ=
+
2
5. (d) Total K.E. of A type molecules
2
m
2
3
w=
Total K.E. of
A type of molecule is[ ]
2
zms.r
2
yms.r
2
xms.rA
)v()v()v(m
2
1
E.K ++=
but w=
xms.r
)
v(
so w==
zms.ryms.r
)v()v(
Total K.E. of B type molecules
221
2mv mv
2
=´=
Now,
223
mωmv
2
´= or
22 2
(ω/v)
3
=
26. (d)
rms
vTµ
At 0ºC
,
rmsv 273µ
At temp.
T,
rms
v
T
2
µ
K69~K2.68
4
273
Tor
273
T
2
1
-===\
Cº20427369T-=-=
27. (c) For a monatomic gas
R
2
3
C
v=
So corre
ct graph is
­
vC
R2/3
®T
28. (d) Molar volume = 22400 cm
3
Bubble volume =
3
34d4
10
383
-
p =p
No. of molec
ules
233
176 10 4 10
10
22400 3
-
´ ´p
=»
´
29. (
d) Molecule number ratio is
3
1
:
3
2
O:H
22
= .
That give
s
212
(c)161
rms
33
=+
æö æö
èø èø
times the
va
lue for O
2
.
30. (b) The Vander waal’s equation of state is
RT)bV(
V
a
P
2
=-
÷
÷
ø
ö
ç
ç
è
æ
+
or
2
V
a
bV
RT
P -
-
=
At the critical po
int, P = P
c
, V = V
c
, and T = T
c
\
2
cc
c
c
V
a
bV
RT
P -
-
= ....(i)

346. 342 PHYSICS
At the critic
al point on the isothermal,
0
dV
dP
c
c
=
\
c
23
cc
RT 2a
0
(Vb)V
-
=+
-
or 3
c
2
c
c
V
a2
)bV(
RT
=
-
....(ii)
Also at c
ritical point,
0
dV
Pd
2
c
c
2
=
\ 4
c
3
c
c
V
a6
)bV(
RT2
0 -
-
=
or
c
34
cc
2RT 6a
(Vb)V
=
-
..(iii)
Dividing (i
i) by (iii), we get cc
V
3
1
)bV(
2
1
=- or V
c
= 3 b ....(iv)
Putti
ng this value in (ii), we get
23
RTc 2a
4b 27b
= or
Rb27
a8
T
c=
...(v)
Putting the
values of V
c
and T
c
in (i), we get
22
c
b27
a
b9
a
Rb27
a8
b2
R
P =-
÷
÷
ø
ö
ç
ç
è
æ
=
31. (c)g for resulting mixture should be in between 7/5 and
5/3.
32 (b) For 1 molecule of a gas, v
rms
=
3KT
m
where m is the mass of one molecule
For N molecule of a gas, v
1
=

3KTN
m
´
For 2N molec
ule of a gas v
2
=
3KIN
N
(2m)
´
´
1
2
v
1
v
\=
33. (a) Since VP
2
= c onstant,
2 '2
VP 2VP=
P
P'
2
\=
As
P
constant or TαP,
T
= thus T bec omes T/Ö2
34. (a) We know that,
÷
ø
ö
ç
è
æg
=
M
RT
C
rms
Here 1
2
rms 2
rms
rms1
C m1
C;
MCm
æö
µ\=
ç÷
èø
35. (b)
3 12 2 13
dU d M N 6M 12N
F
dr dr rR rr
-éùéù
==--=-+
êúêú
ëûëû
In equilibrium position, F = 0
\
2 13
6M 12N
0
rr
-= or,
62N
r
M
=
\ Potential ener
gy at equilibrium position
222
2
M N MMM
U
(2N/ M) 2N 4N 4N(2N / M)
= = =-=
36
. (b)
11 22
12
PV PV
TT
= Here, P
1
= 200kPa
T
1
= 22°C = 295 K T
2
= 4
2°C = 315K
V
2
111
2
V V 1.02V
100
=+=
\
1
2
1
200 315V
P
295 1.02V
´
=
´
209.37kPa=
37. (a) Ac
cording to given Vander Waal's equation
2
2
nRTn
P
Vn V
a
=-
-b
Work done,
222
111
2
2
VVV
VVV
dV dV
W PdV nRT n
Vn V
= = -a
-b
òòò
[ ]
2
2
1
1
21
log()
V
V
e V
V
nRTVnn
V
éù
= - b +a
êú
ëû
22 12
1 12
log
e
V n VV
nRTn
V n VV
æ öéù-b-
= +aç÷ êú
-b
è øëû
38. (d) Internal energy of 2 moles of oxygen
2
55
Uo µ RT 2. RT 5RT
22
æö
= ==
ç÷
èø
Internal
energy of 4 moles of Argon.
Ar
33
U µ RT 4. RT 6RT
22
æö
= ==
ç÷ èø
\Total internal en
ergy
2O Ar
U U U 11RT=+=
39. (c)
1v 2v
12
v
mix
12
nC nC
C
nn
+
=
+
Þ
1v 1v
12
11
nC 2nC13R
6 n 2n
+
=
+

1
2
n1
n2
éù
=
êú
ëû
Q
Þ vv
12
13R
C 2C
2
=+

347. 343Kinetic Theory
Possible values are, vv
12
3R 5R
C ,C
22
==
\ Gase
s are monatomic (like He) and diatomic
(like N
2
)
40. (c)
PVm
nRR
TM
æö
== ç÷
èø
or
PVR
m
TM
æö
=ç÷
èø
i.e.
PV
T
versus m graph is straight line passing
through origin with slope R/M, i.e. the slope depends
on molecular mass of the gas M and is different for
different gases.
41. (d) C
v
for hydrogen =
5
R
2
C
v
for helium =
3R
2
C
v
for water vapour =
6R
2
= 3R
\ (C
v
)
mix
=
53
4R2R13R
1622
R
4217
´ + ´ +´
=
++
\ C
p
= C
v
+
R
p
16
C RR
7
=+ or p
23
CR
7
=
42. (b)P = k
V Þ PV
–1
= k
It is polytropic process (PV
n
= constant)
So n = – 1
So, C = C
V
+
V
RR
C
1n2
=+
-
43. (d)
m
PV RT
M
=
Initially, PV =
6
R 500
M
´
Finally,
P (6 x)
V R 300
2M
-
=´ (if x g ga
s leaks out)
Hence, 65
2
6x3
=´
-
\ x = 1 gram
44.
(b) From graph, T
2
V = const. .....(1)
As we know that TV
g–1
= const
Þ
1
1
VT cons.
g-
= ....(2)
On compari
ng (1) and (2), we get
Þ g = 3/2
Also
rms
3P
v=
r
and
sound
P
v
g
=
r
Þ
rms
sound
v 3
2
v
==
g
45. (b
) For an ideal gas PV = constant i.e. PV doesn’t vary
wi th V.
46. (c) Molar specific heat at constant pressure
7
2
P
CR=
Since, C
P
– C
V
= R Þ C
V
– C
P
– R
75
–
22
RRR==
(7/2)7
.
(5/2)5
P
V
C R
CR
\==
47. (c) For monatomic gas, f = 3,
vp
3R 5R
C ,C
22
== ;
v
p
C3
C5
=
For diatomic gas, f = 5
vp
5R 7R
C ,C
22
== ;
v p
C
5
C7
=
48. (b) When a person is driving a car then the temperature
of air inside the tyre is increased because of motion.
From the Gay Lussac’s law,
P µ T
Hence, when temperature increases the pressure also
increase.
49. (b) Total translational kinetic energy
33
22
nRT PV==
In an ideal gas all mo
lecules moving randomly in all
direction collide and their velocity changes aftercollision.
50. (a) The mean free path of a gas molecule is the averge
distance between two successive collisions. It isrepresented by l.
2
1
2
kT
P
l=
ps
and
2
2
m
d
l=
× ps
Here, s = 0 dia meter of molecule and
k = Boltzmann’s constant.
Þ l µ 1 / d, l µ T and l µ 1 / P.
Hence, mean free path varies inversely as density ofthe gas. It can easily proved that the mean free pathvaries directly as the temperature and inversely as thepressure of the gas.

348. 344 PHYSICS
EXERCISE - 3
Exemplar Questions
1. (b) As the relative velocity of molecule with respect to
the walls of container does not change in rocket, due
to the mass of a molecule is negligible with respect to
the mass of whole system and system of gas moves
as a whole and (g = 0) on molecule energy where.
Hence pressure of the gas inside the vessel, as
observed by us, on the ground remain the same.
2. (d) Pressure on the wall due to force exerted by molecule
on walls due to its rate of transfer of momentum to
wall.
In an ideal gas, when a molecule collides elastically
with a wall, the momentum transferred to each molecule
will be twice the magnitude of its normal momentum is
2 mv. For the wall EFGH, absorbs those molecules.
Which strike to it so rate of change in momentum to it
become only mv so the pressure of EFGH would half
of ABCD.
3. (b) Boyle's law is applicable at constant temperature and
temperature remains constant in isothermal process.
For ideal gas, pV = nRT = constant
So, pV = constant (at constant temperature)
1
p
V
µ
So, this process can be called as isothermal process.
4. (c) Let us consider the given diagram where an ideal gas
is contained in a cylinder, having a piston of mass M.
The pressure on gas does not change.
M
p
p
a
p
p
aMg/A
A
The pressure inside the gas will be
p = p
a + Mg /A
where, p
a = atmospheric pressure
A = area of cross-section of the piston.Mg = weight of piston
Hence, p = constant.
As the piston and cylinder is frictionless so the
equation for ideal gas
pV = nRT, volume (V) increases at constant pressure.
as p, R, n are constant so,
VTµ
so on increasing temperature of system its volume
increased but p will remain constant.
5. (a) As we know that an ideal gas equation,
as the pressure and quantity of gas in system areconstant
nR
pV nRT V T
p
æö
= Þ= ç÷
èø
T
V
p
µ as n, R are constant
1V
Tp
æö
µ
ç÷
èø
Slope of the V – T gr aph,
dV nR
m
dTp
==
[m = slope of V – t graph]
1dV
dTp
µ or
1
m
p
µ [Q nR = constant]
So,
1
p
m
µ
hence,
12
21
=
pm
pm
where, m
1 is slope of the graph corresponding to p
1
and similarly m
2 is slope corresponding to p
2. So slope
of p
1
is smaller than p
2
. Hence, (p
1
> p
2
).
6. (d) Pressure exerted by gas is due to rate of change of
momentum (p) imparted by particles to wall.
When the molecules breaks into atoms, the number ofmoles would become twice.
From ideal gas equation,
pV = nRT
where, p = Pressure of gas, n = Number of moles
R = Gas constant, T = temperature
As gases breaks number of moles becomes twice of
initial, so n
2 = 2n
1
So,
p nTµ
or
2221
1111
(2 )(300
0)
20
(300)
pnTn
pnTn
===
So
, p
2 = 20p
1
Hence, final pressure of the gas would be 20 times tothe initial pressure.
7. (b) For a function f
1(v) the number of molecules (n) which
will have their speeds between v and v + dv.
For each function f
1(v) and f
2(v) number of molecules
remain same 1 mole each but due to mass difference
their speed will be different.
Hence both gases of each function f
1(v) and f
2(v) will
obey the Maxwell's distribution law separately.
8. (d) As we know that an ideal gas equation,
pV = nRT
where, n = Number of moles, p = Pressure,
V = Volume, R = Gas constant,
T = Temperature
pV
n
RT
=
If n, R are const
ant for the system or as number of
moles of the gas remains fixed, hence, we can write
pV
T
= constant
or
11 22
12
pV pV
RT RT
=

349. 345Kinetic Theory
2
2 11
21
()
T
p pV
VT
æö
= ç÷
èø

()()(1.1 )
(1.05) ()
pVT
VT
=
[p
1 = p, V
2 = 1.05 V
and T
2 =
1.1 T]

1.1
1.05
p
æö
=´
ç÷
èø
= p (1.0476)
So, f
inal pressure p
2
will lies between p and 1.1p.
NEET/AIPMT (2013-2017) Questions
9. (b) P
1
> P
2
P
1
P
2
T
V
q
2
q
1
As V = con
stant Þ Pµ T
Hence from V–T graph P
1
> P
2
10. (d) From first law of thermodynamics
DQ = DU + DW =
3
2
.
1
4
R (T
2
– T
1
) + 0
=
3
8
N
a
K
B
(T
2
– T
1
) [Q K =
R
N
]
11. (b)Q
2
rms
1
3
mn
PV
V
=
When mass i
s halved and speed is doubled then
Resultant pressure, P
t
=
2
rms
1
(2)
32
mn
v
V
´´
= 2 P.
12.
(b) Mean free path l
m
=
2
1
2 dnp
where
d = diameter of molecule and d = 2r
\l
m
µ
2
1
r
13. (b)Let ‘n’ be the degree of freedom
p
v
n
1R
C 22
1
nCn
R
2
æö
+ç÷
èø æö
g= = =+ ç÷
èøæö
ç÷
èø
14. (a
) Change in internal energy from A ® B
DU =
f
2
nRDT =
f
2
nR (T
f
– T
i
)
=
5
2
{P
f
V
f
– P
i
V
i
}
(As gas is
diatomic \ f = 5)
=
5
2
{2 × 10
3
× 6 – 5 × 10
3
× 4}
=
5
2
{12 – 20} × 10
3
J = 5 × (–4) × 10
3
J
DU = –20 KJ
15. (a) From PV = nRT
P
A
=
AA
M
RT
r
and
BB
B
M
P
RT
r
=
From ques
tion,
AAAA
BBBB
P
MM
3
2
P M M2
r
= ==
r
So,
A
B
M 3
M4
=
16. (d) Molar mass of the gas = 4g/mol
Speed of sound
V =
RT
m
g
Þ 952 =
3
3.3 273
4 10
-
g´´
´
Þ g = 1
.6 =
68
105
1
=
Also, g =
P
V
C8
C5
=
So, C
P
=
85
5
´
= 8JK
–1
mol
–1
[C
V
= 5.0 JK
–1
g iven]
17. (b) Here v
1
= 200 m/s;
temperature T
1
= 27°C = 27 + 273 = 300 k
temperature T
2
= 127° C = 127 + 273 = 400 k, V = ?
R.M.S. Velocity, V µ T
Þ
v 400
200 300
=
Þv =
200 2
m/s
3
´
Þ v =
400
m/s
3
18. (c
) Internal energy of the system is given by
U =
f
nRT
2
Degree of freedom
F
di
atomic
= 5
f
monoatomic
= 3
and, number of moles
n(O
2
) = 2
n(Ar) = 4
U
total
=
53
(2)RT (4)RT
22
+ = 1 1 R TT

350. 346 PHYSICS
F¬
B O C A
x®
i.e., F µ – x
or, F kx=-
r r
where k is called force constant or spring constant
or,
2
2
0
d x kx
m
mdt
+=
rr
(b)Angular S.H.M.
The restorin
g torque is proporational to the angular
displacement from the mean position.
Ct=-q
rr
where C is called torsional rigidity
or, I
2
2
d
C
dt
q
=-q
r
r
or,
2
2
0.
dC
Idt
q
+ q=
r
r
Terms Related to S.H.M.
(i)Amplitude : The maximum displacement of the oscillating
particle on either side of its mean position is called its
amplitude. It is denoted by A.
(ii)Time period : The time taken by a oscillating particle to
complete one oscillation is called its time period. It is
denoted by T.
(iii)Frequency : It is the number of oscillations completed in
one second. It is denoted by u.
1
T
u=
The S.I. unit of fr equency is s
–1
or Hz.
(iv)Angular frequency
2
2
T
p
w= pu=
The S.I. uni
t of angular frequency is rad/sec.
(v)Phase : The parameter, by which the position of particle
from its mean position is represented, is known as phase.The phase at any instant tells the state of position &direction of motion at that instant. The phase at time t = 0
is known as the initial phase or epoch (e).
PERIODIC AND OSCILLATORY MOTION
Periodic Motion
When a body repeats its motion after regular interval of time, it
is said to be in periodic motion. The path of periodic motion may
be rectilinear, open/ closed curvilinear.
Example :(i) Motion of moon around earth.
(ii) Motion of a piston in a cylinder.
(iii) Motion of a simple pendulum etc.
Oscillatory Motion
If during a periodic motion, the particle moves to and fro on the
same path, the motion is vibratory or oscillatory.
Example : (i) The motion of a ball in bowl
Bowl
Ball
(ii) The needle of a sewing machine
(iii) Vibrations of prongs of tuning fork etc.
(i) All oscillatory motion are periodic but all periodic
motioniiare not oscillatory motion.
(ii) The oscillatory motion which can be expressed in terms of
sine and cosine function, is said to be harmonic motion.
SIMPLE HARMONIC MOTION (S.H.M.)
“If a particle moves up and down (back and forth) about a
mean position (also called equilibrium position) in such a way
that a restoring force/ torque acts on particle, which is
proportional to displacement from mean position, but in
opposite direction from displacement, then motion of the particle
is called simple harmonic motion. If displacement is linear, it is
called linear S.H.M. and if displacement is angular, it is called an
angular S.H.M.
Example :(i) Motion of a body suspended by spring.
(ii) Oscillations of simple pendulum.
Equations of S.H.M.
(a)Linear S.H.M.
The restoring force is proportional to the displacement from
mean position.
14
Oscillations

351. 347Oscillations
(ii)
Potential energy : A particle in S.H.M. possesses potential
energy due to its displacement from the mean position. 2 2211
..
22
PE ky my= =w
(iii)
Total mechanical energy
E = K.E. + P.E.

2 2 2 2211
m(A y) my
22
=w -+w

221
2
E mA=w
21
2
kA=
(iv)The curve
s representing KE, PE and total energy are shown
in figure.
O
K.E.
P.E.
Displacement
– A + A
Total energy
E = K.E. + P.E.
Energy
Keep in Memory
1.
Restoring force F = – Mw
2
x
2. Kinetic energy = (1/2) Mw
2
(A
2
– x
2
)
3. Potential energy = 1/2 Mw
2
x
2
4. Total energy of SHM = 1/2 Mw
2
A
2
Equation a = –
w
2
y shows that if body perform S.H.M. then
acceleration of the body is proportional to displacement,
but in the opposite direction of displacement. It is an
essential requirement for any motion to be S.H.M.
5. The kinetic and potential energy of SHM varies sinusoidally
with a frequency twice that of SHM.
6. Total energy
2 2 2 221
m A 2 mAn
2
= w =p
where n
= frequency of vibration.
7. xa
2
w-= where w is constant
22
xAv-w=
a
x

x
v
8. Geometrically the projection of the body undergoing
uniform circular motion on the diameter of the circle is
SHM.
9. In a non-inertial frame.
eff
g
2T
l
p=
x
22
yeff a)ag(g+-=
(vi)Total phase angl
e : The total angle (wt + q) is known as total
phase angle.
Characteristics of S.H.M.
(i)Displacement : The displacement of a particle in S.H.M. is
given by
sin()=+yAtwf
where A is amplitude, w is angular frequency and (wt + f) is
called the phase of the particle at any instant t.
T/4T/2
3T/4 T
+A
–A
(0,0)
Time (t)D
i
s
p
l
a
c
e
m
e
n
t
(ii)Velocity : The velocity of a particle in S.H.M. is given by
cos()
dy
At
dt
= w w +f or,
22
v Ay=w-
At y = 0 (at me
an position),
max
vA=w
T/4
T/2
3T/4T
(0)
Time (t)
V
e
l
o
c
i
t
y
(iii)Acceleration : The acceleration of a particle in S.H.M. is
given by
2
sin()
dv
a At
dt
= =-w w +f
or,a = – w
2
y
The n
egative sign indicates that the acceleration is directed
towards the mean position
At y = A (at extreme position),
a
max
= – w
2
A
T/4T/23T/4T
Time (t)
A
c
c
e
l
e
r
a
t
i
o
n
Energy in S.H.M. :
(i)Kinetic energy : A particle in S.H.M. possesses kinetic
energy by virtue of its motion.
2 22211
.. ()
22
KE mv m Ay= =w-

352. 348 PHYSICS
SOME SYSTEMS EX
ECUTING S.H.M.
Case 1 Spring mass system :
(i) When two springs having force constants k
1
and k
2
connected in parallel, then
k
2
k
1
M
The force constant of the combination is
k = k
1
+ k
2
and hence T = 2p[M/(k
1
+ k
2
)]
1/2
(ii) When two springs of force constants k
1
and k
2
are
connected in series, then
k
2
k
1
M
The force constant of the combination is
1/k = 1/k
1
+ 1/k
2
. i.e.,
k = k
1
k
2
/(k
1
+ k
2
) Hence 2
m
T
K
=p
(iii) If t
wo mass M
1
and M
2
are connected at the two ends of
the spring, then their period of oscillation is given by
M
2
M
1
T = 2p[m/k)]
1/2
where
12
12
MM
MM
m=
+
is the reduced m ass.
(iv) When the length of spring increases, spring constant
decreases. If the length of spring becomes n times, its spring
constant becomes
n
1
times and theref
ore time period will
be increased by
n times.
(v) If we divid
e the spring into n equal parts, the spring
constant of each part becomes n k. Hence time period when
the same mass is suspended from each part is:
1/2
2
M
T
nk
éù
=p
êú
ëû
Case 2 Simple pendulum : A simple pendulum consists of a
point mass suspended by a weightless inextensible cord from arigid support.
q
x =lq
mg sinqmg
mg cosq
l
l
Let a bob of mas
s m is displaced from its, equilibrium position
and released, then it oscillates in a vertical plane under gravity.
Let q be the angular displacement at any time t, then
corresponding linear displacement along the arc is
x = l q.
It is clear from the diagram that mg sinq, is the restoring force
acting on m tending to return it to mean position. So from
Newton’s second law
q-==sinmg
dt
xmd
F
2
2
...(i)
where n
egative sign indicates that restoring force mg sin q (= F)
is opposite to displacement q. If q is very small, then
sin q » q, so from equation (i)
l
x
mgsinmg
dt
xmd
2 2
-=q-= or ...(ii)
where w
2
= g/l.
This is the equati
on of S.H.M. of the bob with time period
2
2T
g
æöp
= =p ç÷
w èø
l
OM
How to find the time period of a body undergoing S.H.M.?
St
ep 1 : First, find the equilibrium position. Equilibrium position
will be one for which0F
=S and 0St=
Step 2 : Displace the body , from the equilibrium position by x.
Find the restoring force acting on the body F = –kx (for translation)
Find the restoring torque acting on the bodyq-=tk (for
rotational)
Step 3 : Since
2
2
dt
xmd
maF==
\ Use kx
dt
xmd
2
2
-= ... (i) for transla
tional

q-=k
dt
xmd
2 2
… (ii) for rotatio
nal
Step 4 :
k
m
2Tp= (for translationa
l)

k
I
2Tp= (for rotation
al) where, I = moment of inertia
COMMON DEFAULT
Incorrect. The time period of spring mass-system is
dependent on the value of g.
Correct. Time period of spring-mass system shifts only the
equilibrium position. It does not change the time period.
Because of this reason, time period of spring mass system
remains same on plains / mountains / in satellites.
Incorrect. The time taken to cover half the amplitude form
equilibrium position is
8
T
.

353. 349Oscillations
Correct. The a ctual time taken is
12
T
.
Incorrect.In a spring mass system, mass oscillate about
the end of a spring when the spring is in unstretched
condition.
Correct. The mass oscillates about the equilibrium position
which may or may not be at the unstretched length.
Case 3 Liquid in U-tube : A U-tube of uniform cross-sectional
area A has been set up vertically with open ends facing up.
x
x
[Restoring force = –2Adg x]
If m gm of a liquid of density d is poured into it then time periodof oscillation.
Adg2
m
2Tp=
Case 4
Rectangular block in liquid :
d
h'
h
d'
Rectangular
block floating in a liquid,
gd
'd'h
2Tp= ; where
d = de
nsity of liquid, d¢ = density of block, h = height of block
Case 5 : Vibration of gas system in a cylinder with frictionless
piston.
A
m
p
h
Time period,
AP
mh
2Tp=
where
m = mass of gas, A = cross sectional area of piston
P = pressure exerted by gas on the piston, h = height of piston
Case 6: If a tunnel is dig in the earth diametrically or along a
chord then time period,
sec36min84
g
R
2T =p= for a particle
released in the tunnel.
Case 7: The time period of a ball oscillating in the neck of a
chamber
B
mV
A
2
T
p
=
Case 8 : If a dipo
le of dipole moment p is suspended in a uniform
electric field E then time period of oscillation
PE
I
Tp2=
Keep in Memory
1.In S.H.M. the
phase relationship between displacement,
velocity and accleration, is as follows :(a) The velocity is leading the displacement by a phase
2
p
radian
(b) Th
e acceleration is leading the displacement by a
phase p radian
(c) The acceleration is leading the velocity by a phase
2
p
radian.
2.(
a) When
2
A
x=, then
velocity V = 0.86V
max
.
(b) When V = V
max
/2, the displacement x = 0.87A.
(c) When 2
A
x=, the k
inetic energy of S.H.M. is 75% of
the total energy and potential energy 25% of the totalenergy.
(d) When the kinetic energy of S.H.M. is 50% of the total
energy, the displacement is 71% of the amplitude.
3.The time period of a simple pendulum of length l which is
comparable with radius of earth.
1/21/2
RR
T22
R ( R)g
1g
éùéù
=p =p
êúêú
+æö ëû
êú+ç÷
èøêúëû
l
l
l
where R = radius of the earth and g is the acceleration due
to gravity on the surface of the earth.
(a) When l << R, then
g
2T
l
p=
(b) Whe
n l = R, we find
hour1
g2
R
2T
2/1
»
ú
ú
û
ù
ê
ê
ë
é
p=
(c) When l = ¥, then
2/1
g
R
2T
ú
ú
û
ù
ê
ê
ë
é
p= = 84.6 minutes. T
hus maximum of T
is 84.6 minutes.
(d) Under weightlessness or in the freely falling lift
¥=
ú
û
ù
ê
ë
é
p=
2/1
0
2T
l
. This means, the pendulum does
not oscillate at all.

354. 350 PHYSICS
4.Time period o
f a simple pendulum in a train accelerating or
retarding at the rate a is given by
2/1
22
ag
2T
ú
ú
û
ù
ê
ê
ë
é
+
p=
l
5.If a simple pen
dulum whose bob is of density d
o
is made to
oscillate in a liquid of density d, then its time period of
vibration in liquid will increase and is given by
g
d
d
1
2T
o
÷
÷
ø
ö
ç
ç
è
æ
-
p=
l
(where d
0
> d)
6.The time p
eriod of a simple pendulum in a vehicle moving
along a circular path of radius r and with constant velocity
V is given by,
1/2
4
2
2
2T
V
g
r
éù
=p
êú
êú
+
êú
ëû
l
7.If T
1
and T
2
are the t
ime periods of a body oscillating under
the restoring force F
1
and F
2
then the time period of the
body under the influence of the resultant force
12
=+
rrr
FFF
will be
2
2
2
1
21
TT
TT
T
+
=
8.(a) The perce
ntage change in time period of simple
pendulum when its length changes is
%100
2
1
100
T
T
´÷
ø
ö
ç
è
æD
=´
D
l
l
(b) The perce
ntage change in time period of simple
pendulum when g changes but l remains constant is
100
g
g
2
1
100
T
T
´
÷
÷
ø
ö
ç
ç
è
æD
=´
D
%
(c) The perce
ntage change in time peirod of simple
pendulum when both l and g change is %100
g
g
2
1
100
T
T
´
÷
÷
ø
ö
ç
ç
è
æ D
+
D
=´
D
l
l
9.If a wire of length l, area of cross-section A, Young’ss
modulus Y is stretched by suspending a mass m, then the
mass can oscillate with time period:
lm
TA
2Tp=
10.If a simple pendul
um is suspended from the roof of
compartment of a train moving down an inclined plane of
inclination q, then the time period of oscillations 2/1
cosg
2T
ú
û
ù
ê
ë
é
q
p=
l
11.If a ball of radius
r oscillates in a bowl of radius R, then its
time period is given by :
2/1
g
rR
2T
ú
û
ù
ê
ë
é-
p=
12.If a disc of radiu
s r oscillates about a point at its rim, then its
time period is given by:
2/1
g
r
2T
ú
û
ù
ê
ë
é
p=
It behaves as a simple
pendulum of length r.
13.The graph between the length of a simple pendulum and its
time period is a parabola.
14.The graph between the length of a simple pendulum and
the square of its time period is a straight line.
15.The graph between l & T and between l & T
2
intersect at
T = 1 second.
16.The time period of the mass attached to spring does not
change with the change in acceleration due to gravity.
17.If the mass m attached to a spring oscillates in a non viscous
liquid density s, then its time period is given
by
2/1
1
k
m
2T
-
ú
û
ù
ê
ë
é
÷
÷
ø
ö
ç
ç
è
æ
r
s
-p=
where k = force c
onstant and
ris density of the mass
suspended from the spring.
18.The length of second pendulum (T = 2 sec) is 99 cm
Physical Pendulum
T
restoring
= – mgd sin q
If q is small, sin q » q
cm
mg
d
q
\ T
restoring
= – mgdq
And =2
I
T
mgd
p
Let a test-tube of radius r, carrying lead shots of mass m is held
vertically when partly immersed in liquid of density r. On pushing
the tube little into liquid and let it executes S.H.M. of time period
2
2=
m
T
rg
p
pr
Conical Pendulum When the bob of a simple pendulum
moves in a horizontal circle
it is called as conical pendulum.
r
T
S
q
If l is the length of the pendulum and the string makes an angle
q with vertical then time period,
g
cos
2T
q
p=
l

355. 351Oscillations
In
this case electric force qE and gravity force are
opposite.
5.A pendulum clock slows down in summer and goes faster
in winter.
6.Potential energy of a particle executing S.H.M. is equal to
average force × displacement.
i.e., 22
2
P
xm
2
1
x
2
xm0
x
2
F0
U w=
÷
÷
ø
ö
ç
ç
è
æ w+
=
ú
û
ù
ê
ë
é+
= .
7.If the total e
nergy of a particle executing S.H.M. is E, then
its potential energy at displacement x is
E
A
x
U
2
2
P= and kinetic energy E
A
x
1U
2
2
k
÷
÷
ø
ö
ç
ç
è
æ
-=
FREE, DAMPED
, FORCED OSCILLATIONS AND
RESONANCE
Free Oscillation
If a system oscillates on its own and without any external
influence then it is called as free oscillation.
Frequency of free oscillation is called natural frequency. The
equation for free S.H.M. oscillation
2
2
mdx
dt
= F
restoring force
= –kx, whe re k is constant.
The differential equation of harmonic motion in absence of
damping and external force is
2
2
0
2
0
dx
x
dt
+w= , where w
0
is na tural frequency of body..
The time period is
0
2
2==
m
T
k
p
p
w
Damped Oscillation
Oscillation performed under the influence of frictional force is
called as damped oscillation.
(a) In case of damped oscillations the amplitude goes on
decreasing and ultimately the system comes to a rest.
t
x
(b) The damping force (F
damping
µ – v Þ F
damping
= – bv) is
proportional to the speed of particle. Hence the equation of
motion
dV
m kx bv
dt
=--
where b is positive constant and is called damping
coefficient. Then the differential equation of a damped
harmonic oscillation is
Torsional Pendulum
It is an arrangement which consists of a heavy mass suspended
from a long thin wire whose other and is clamped to a rigid support.
Time period
2=T
C
p
I
whereI = mo ment of inertia of body about the suspension wire
as axis of rotation.
C = restoring couple per unit thirst.
Keep in Memory
1.Th
e displacement, velocity and acceleration of S.H.M. vary
simple harmonically with the same time period and
frequency.
2.The kinetic energy and potential energy vary periodically
but not simple harmonically. The time period of kinetic
energy or potential energy is half that of displacement,
velocity and acceleration.
3.The graph between displacement, velocity or acceleration
and t is a sine curve. But the graph between P.E. or K.E. of
S.H.M. and time t is parabola.
D
is
p
la
c
e
m
e
n
t
t
x=A sin tw
V
e
lo
c
it
y
t
tcosA
dt
dx
v ww==
A
c
c
e
le
r
a
t
io
n
x
dt
xd
a
2
2
2
w-==
t
+A
x
O–A
K.E., P.E. E
22
xm
2
1
.E.Pw=
)xA(m
2
1
.E.K
222
-w=
½E
E=½ mAw
22
4.(a) If th
e bob of simple pendulum is -vely charged and a
+vely charged plate is placed below it, then theeffective acceleration on bob increases and
consequently time period decreases.
Time period,
m
qE
g
2T
+
p=
l
In thi
s case electric force q E and gravity force act in
same direction.
(b) If the bob of a simple pendulum is -vely charged and
is made to oscillate above the -vely charged plate,
then the effective acceleration on bob decreases and
the time period increases.
T2
qE
g
m
=p
-
l

356. 352 PHYSICS
2
2
0
2
20
d x dx
Cx
dtdt
+ +w= ...(i)
where 2 C = b/m (C is damping c
onstant) &
2
0
m
k
w=, the
natural freque
ncy of oscillating particle i.e., its frequency
in absence of damping.
In case of over damping the displacement of the particle is
0
exp( / 2 )xAt= -t sin (wt + f) ...(ii)
Þx = A s
in (wt + f) where A
0
= max. amplitude of the oscillator.
22
0
c-w=wand
b
m
=t(relaxation time)
It is cle
ar from the fig & eq
n
. (ii) that the amplitude of damped
harmonic oscillator decreases with time. In this case, the motion
does not repeat itself & is not periodic in usual sense of term.
However it has still a time period,
22
0
c
22
T
-w
p
=
w
p
= , which is
the tim
e interval between its successive passage in same direction
passing the equilibrium point.
A=–Ae
0
–ct
A=Ae
0
–ct
t
D
is
p
la
c
e
m
e
n
t
Forced Oscillation and Resonance :
The oscillation of a system under the action of external periodic
force is called forced oscillation.
External force can maintain the amplitude of damped oscillation.
When the frequency of the external periodic force is equal to the
natural frequency of the system, resonance takes place.
The amplitude of resonant oscillations is very very large. In the
absence of damping, it may tend to infinity.
At resonance, the oscillating system continuously absorbs
energy from the agent applying external periodic force.
In case of forced oscillations, the total force acting on the system
is
0
Re storting
Ext periodic
force Damping
force
force
bdx
F kx F sin pt
dt
=- -+ …… (i)
Then by
Newton’s second law :
Þ
ptsinF
dt
dx
bkx
dt
xd
m
0
2
2
+--=
or tsinfx
dt
dx
C2
dt
xd
0
2
0
2
2
b=w++ …… (ii)
where
m
F
f,
m
k
,
m
b
C2
0
0
2
0
==w=
The equation (ii) is the differenti
al equation of motion of forced
harmonic oscillator.
The amplitude at any time t is
0
2 2 22
0
f
x sin(pt )
( p) 4cp
= -q
éùw-+
ëû
where
22
p
cp2
tan
-w
=q & p is the frequen
cy of external periodic
force.
Example 1.
A spring is stretched by 0.20 metre when a mass of 0.50 kg
is suspended. Calculate the period of the spring when a
mass of 0.25 kg is suspended and put to oscillation.
(g = 10 m/sec
2
).
Solution :
The force constant K of the spring is given by
F 0.5 kg Wt
K
y 0.20m
0.5 10 newton
0.20 m
25 newton / metre
==
´
=
=
÷
ø
ö
ç
è
æ
p=
K
m
2TNow

0.25
2 3.14 0.628 second.
25
æö
=´=
ç÷
èø
Example 2.
What w
ill be the force constant of the spring system shown
in fig?
(a)
1
2
k
k
2
+
(b)
1
12
11
2kk
-
éù
+êú
ëû
k
2
k
1
k
1
(c)
12
11
2kk
éù
+
êú
ëû
(d)
1
12
21
kk
-
éù
+êú
ëû
Solution : (b)
Two s
prings of force constants k
1
and k
2
are in parallel.
Hence
111
k2kkk=+=
¢
The third spring is in series with spring of force constant
k¢.

357. 353Oscillations
1
2121 k
1
k2
1
kor
k
1
k2
1
k
1
-
ú
û
ù
ê
ë
é
+=ú
û
ù
ê
ë
é
+=\
Example 3.
A partic
le starts with S.H.M. from the mean position as
shown in figure below. Its amplitude is A and its time period
is T. At one time, its speed is half that of the maximum
speed. What is this displacement at that time ?
(a)
2A
3
(b)
3A
2
(c)
2A
3
(d)
3A
2
Solution : (b)
We know that
2 2 1/2
vω[A x]=-
1/2
2
2
2
v
xA
ω
éù
\=-êú
êúëû
Given that
max
v Aω
v
22
== .
so, A.
2
3
4
A
A
2/1
2
22
2
=
ú
ú
û
ù
ê
ê
ë
é
w
w
-
Example 4.
A
cylindrical piston of mass M slides smoothly inside a
long cylinder closed at one end, enclosing a certain mass
of gas. The cylinder is kept with its axis horizontal. If the
piston is disturbed from its equilibrium position, it
oscillates simple harmonically. The period of oscillation
will be
P A
M
h
(a)
Mh
T 2π
PA
æö
= ç÷
èø
(b)
MA
T 2π
Ph
æö
= ç÷
èø
(c)
M
T 2π
PAh
æö
= ç÷
èø
(d) T2π(MPh A)=
Solution : (a)
We
know that P V = R T
)V/V(PPor0VPPVD=D=D+D\
or
÷
ø
ö
ç
è
æ
-=
÷
÷
ø
ö
ç
ç
è
æ
-=
h
x
P
hA
xA
P
A
F
÷
ø
ö
ç
è
æ
-=
h
xA
PF

x
hM
AP
M
1
h
xA
Pa
÷
÷
ø
ö
ç
ç
è
æ
-=÷
ø
ö
ç
è
æ
-=\
This gi
ves,
2 2
2
æö
= ==
ç÷
èø
PA Mh
orT
Mh PA
p
wp
w
Example 5.
A simple harmonic oscillator has an amplitude A and time
period T. Determine the time required by it to travel from
x = A to
2
A
x=.
Solution :
Fo
r S.H.M.,
÷
ø
ö
ç
è
æp
= t
T
2
sinAx
When x
= A,
÷
ø
ö
ç
è
æp
= t.
T
2
sinAA
1t.
T
2
sin =÷
ø
ö
ç
è
æp
\ Þ ÷
ø
ö
ç
è
æp
=÷
ø
ö
ç
è
æp
2
sint.
T
2
sin Þ t = (T/4)
When
2
A
x= and ÷
ø
ö
ç
è
æp
= t.
T
2
sinA
2
A
or ÷
ø
ö
ç
è
æp
=
p
t
T
2
sin
6
sin or t = (T/12)
Now, time taken to travel from
x = A to x = A/2 = T/4 – T/12 = T/6
Example 6.
Calculate the increase in velocity of sound for 1?C rise of
temperature, if the velocity of sound at 0?C is 332 m/sec.
Solution :
t1
00
1/2
1 00
0 00
t00
Tv 273t
;
v
T 273
tt
v v1 v1
273 2 273
tt
v1 vv
546 546
t
v v v
= 0.61 m/sec.
546
æö +æö
==
ç÷ç÷
èøèø
æöæö
\ =+ =+ ç÷ç÷
èøèø ´
æ ö æö
= + =+ç
÷ ç÷
è ø èø
æö
\-= ç÷
èø
(As v
0
= 332 m/sec and Dt = 1ºC)
Example 7.
Which one of the following equations does not represent
S.H.M.; where x = displacement and t = time. Parameters
a, b and c are the constants of motion
(a) x = a sin bt
(b) x = a cos bt + c
(c) x = a sin bt + c cos bt
(d) x = a sec bt + c cosec bt
Solution : (d)
Sec bt is not defined for bt = p/2.

358. 354 PHYSICS
x = a sec bt + c cosec
bt =
btcosbtsin
btcoscbtsina+
This e
quation cannot be modified in the form of simple
equation of S.H.M.
i.e. x = a sin (wt + f).
so, it cannot represent S.H.M.
Example 8.
If length of pendulum is increased, time period T increases.
So, frequency decreases or angular velocity decreases. As
maximum velocity
max
vA
ω=, so max. velocity will also
dec
rease. (where A is amplitude of bob).
A pendulum bob has a speed of 3 m/s at its lowest position.
The pendulum is 0.5 m long. What will be the speed of the
bob, when the length makes an angle of 60? to the vertical?
(g = 10 m/s
2
)
Solution :
Here, r w = 3 m/s; l = 0.5 m; K.E. at the lowest point
m
2
9
)3(m
2
1
rm
2
1 222
==w=
Let v be the veloc
ity of the bob at B when ÐOAB = 60° then,
A
O
C B
60º
l
OC = h = (l – l co s q) = (l – l cos 60º) = l /2
If v is the velocity of bob at position B, then using law ofconservation of energy, we have,
211
m 9 m v mg (1 cos θ)
22
´=+- l
or
211
m v m 9 mg (1 cos θ)
22
=´-- l
1/2
1/2
1/2
[9 2 (1 cos )]
[9 2 10 0.5 (1 1/2)]
(9–5) 2/
=--
=-
´´-
==
lvg
ms
q
Example 9.
A small spherical steel ball is placed a little away from the
centre of a large concave mirror whose radius of curvature
R = 2.5 cm. When the ball is released, it begins to oscillate
about the centre. Show that the motion of the ball is simple
harmonic and find the period of motion. Neglect friction
and take g = 10 m/sec
2
.
Solution :
F = – mg sin
q = – mg q (Qq is small)
= – mg (x/R)
\F = – kx, where
k = (mg/R)
Now, a = F/m = – kx/m = –
w
2
x
Hence, moti
on is simple harmonic motion and
T =
2p
w
=
R
2
g
æö
p
ç÷
èø
= 2 × 3.14 ×
2.5
10
æö
ç÷
èø
= 3.142 sec.
Example 10.
A spring of stiffness constant k and natural length
l is cut
into two parts of length 3
l / 4 and l / 4 respectively, and an
arrangement is made as shown in the figure. If the mass is
slightly displaced, find the time period of oscillation.
m
(3/4)l l/4
Solution :
The stiff
ness of a spring is inversely proportional to its
length. Therefore the stiffness of each part is
1
4
kk
3
= and k
2
= 4k
m
kk
1 = (4/3) kk
2 = 4
Time perio
d,
12
m
T2
kk
=p
+
or
3m 3m
T2
16k2k
p
=p=

359. 355Oscillations

360. 356 PHYSICS
1.A child sw
inging on swing in sitting position stands up.
The time period of the swing will
(a) increase
(b) decrease
(c) remain same
(d) increase if the child is tall and decrease if the child is
short.
2.The graph of time period (T) of simple pendulum versus its
length (l) is
(a)
T
O l
(b)
T
lO
(c)
O l
T
(d)
O l
T
3.Which of the following is a simple harmonic motion?
(a) Particle moving through a string fixed at both ends.
(b) Wave moving through a string fixed at both ends.
(c) Earth spinning about its axis.
(d) Ball bouncing between two rigid vertical walls.
4.A rod is hinged vertically at one end and is forced to oscillate
in a vertical plane with hinged end at the top, the motion of
the rod:
(a) is simple harmonic
(b) is oscillatory but not simple harmonic
(c) is pericolic but not oscillatory
(d) may be simple harmonic
5.The graph plotted between the velocity and displacement
from mean position of a particle executing SHM is
(a) circle (b) ellipse
(c) parabola (d) straight line
6.Acceleration of a particle executing SHM, at it’s mean
position is
(a) infinity (b) variable
(c) maximum (d) zero
7.A vertical mass-spring system executes simple harmonic
oscillations with a period of 2s. A quantity of this system
which exhibits simple harmonic variation with a period of
1 s is
(a) velocity
(b) potential energy
(c) phase difference between acceleration and displacement
(d) difference between kinetic energy and potential energy
8.A simple pendulum has a metal bob, which is negatively
charged. If it is allowed to oscillate above a positively charged
metallic plate, then its time period will
(a) increase
(b) decrease
(c) become zero
(d) remain the same
9.In the fig. S1 and S2 are identical springs. The oscillation
frequency of the mass m is f. If one spring is removed, the
frequency will become
S
1 S
2
m
(a)f (b) f × 2
(c) 2f´ (d) 2/f
10.A pendulum is
undergoing S.H.M. The velocity of the bob
in the mean position is v. If now its amplitude is doubled,keeping the length same, its velocity in the mean positionwill be
(a) v/2 (b) v
(c) 2 v (d) 4 v
11.The potential energy of a particle (U
x
) executing S.H.M. is
given by
(a)
2
x
k
U (x a)
2
=- (b)
23
x123
U kx kx kx=++
(c)
bx
x
eAU
-
= (d) U
x
= a co
nstant
12.A particle moves such that its acceleration ‘a’ is given by a
= – bx where x is the displacement from equilibrium position
and b is constant. The period of oscillation is
(a) 2 p/b (b) 2π/b
(c) b/2p (d) b/2p
13.A particle executes S
.H.M. having time period T, then the
time period with which the potential energy changes is
(a)T (b) 2 T
(c) T/2 (d)
¥
14.An instantaneous displacement of a simple harmonic
oscillator is x = A cos (wt + p/4). Its speed will be maximum
at time(a)p/4 w (b)p/2 w
(c)p/w (d) 2 p/w
15.The tension in the string of a simple pendulum is
(a) constant
(b) maximum in the extreme position
(c) zero in the mean position
(d) None of these

361. 357Oscillations
16.A r
ectangular block of mass m and area of cross-section A
floats in a liquid of density
r. If it is given a small vertical
displacement from equilibrium. It undergoes oscillations
with a time period T then
(a)
mTµ
(b)
rµT
(c) A/1
Tµ (d) rµ/1T
17.If the magnitude of displacement is numerically equal to
that of acceleration, then the time period is
(a) 1 second (b)p second
(c)2p second (d) 4p second
18.The graph shown in figure represents
Time
Velocity
(a) motion of a simple pendulum starting from mean
position
(b) motion of a simple pendulum starting from extreme
position
(c) simple pendulum describing a horizontal circle
(d) None of these
19.A particle executing simple harmonic motion along y-axis
has its motion described by the equationsin()yA tBw=+ .
The amplitude o
f the simple harmonic motion is
(a)A (b) B
(c) A + B (d)
AB+
20.Resonance is an example of
(a) tuning fork (b) forced vibration
(c) free vibration (d) damped vibration
21.Which one of the following equations of motion represents
simple harmonic motion?
(a) Acceleration = – k(x + a)
(b) Acceleration = k(x + a)
(c) Acceleration = kx
(d) Acceleration = – k
0
x + k
1
x
2
where k, k
0
, k
1
and a are all postive.
22.Out of the following functions, representing motion of a
particle, which represents SHM?
(A)
y sintcost=w-w
(B) y = sin
3
wt
(C)
3
y 5cos 3t
4
pæö
= -w
ç÷
èø
(D)
22
y1tt= +w +w
(a) Onl
y (A)
(b) Only (D) does not represent SHM
(c) Only (A) and (C)
(d) Only (A) and (B)
23.The total mechanical energy of a spring-mass system in
simple harmonic motion is
22
Am
2
1
E w=. Suppose the
os
cillating particle is replaced by another particle of double
the mass while the amplitude A remains the same. The new
mechanical energy will
(a) become 2E (b) become E/2
(c) become E2 (d) remain E
24.The displacement of aparticle along the x-axis is given by x
= a sin
2
.tw The motion of the particle corresponds to:
(a) simple harmonic motion of frequency /wp
(b) simple harmonic motion of frequency 3 /2wp
(c) non simple harmonic motion
(d) simple harmonic motion of frequency /2wp
25.The period of oscillation of a mass M suspended from a
spring of negligible mass is T. If along with it another mass
M is also suspended, the period of oscillation will now be
(a)T (b)/2T
(c) 2T (c)2T
1.A simple pendu
lum performs S.H.M. about x = 0 with an
amplitude a, and time period T. The speed of the pendulum
at x = a/2 will be
(a)
T
3a
p (b)
T2
3ap
(c)
T
ap
(d)
T
a3
2
p
2.A body of ma
ss 5 gram is executing S.H.M. about a fixed
point O. With an amplitude of 10 cm, its maximum velocity
is 100 cm/s. Its velocity will be 50 cm s
–1
at a distance
(in cm)
(a)5 (b)
25
(c) 35 (d) 210
3.A mass M is suspen
ded from a spring of negligible mass.
The spring is pulled a little and then released so that the
mass executes SHM of time period T. If the mass is increased
by m, the time period becomes
3
T5
. Then th
e ratio of
M
m
is
(a)
9
25
(b)
9
16
(c)
3
5
(d)
5 3

362. 358 PHYSICS
4.A tunnel
has been dug through the centre of the earth and
a ball is released in it. It executes S.H.M. with time period
(a) 42 minutes (b) 1 day
(c) 1 hour (d) 84.6 minutes
5.A particle of mass 1 kg is moving in S.H.M. with an
amplitude 0.02 and a frequency of 60 Hz. The maximum
force acting on the particle is
(a) 144 p
2
(b) 188 p
2
(c) 288 p
2
(d) None of these
6.The length of a second’s pendulum at the surface of earth
is 1 m. The length of second’s pendulum at the surface of
moon where g is 1/6th that at earth’s surface is
(a) 1/6 m (b) 6 m
(c) 1/36 m (d) 36 m
7.The displacement of a S.H.M. doing particle when
K.E. = P.E. (amplitude = 4 cm) is
(a)
cm22 (b) 2 cm
(c )cm
2
1
(d)cm2
8.The equat
ion of SHM of a particle is A + 4p
2
x = 0 where a is
instantaneous linear acceleration at displacement x. The
frequency of motion is
(a) 1 Hz (b) 4p Hz
(c)
Hz
4
1
(d) 4 Hz
9.y = 2 (cm
) sin
ú
û
ù
ê
ë
é
f+
p
2
t
what is the maximum acceleration of
the particle doing the S.H.M.
(a)
2
p
cm/s
2
(b)
2
2
p
cm/s
2
(c)
4
2
p
cm/s
2
(d)
4
p
cm/s
2
10.A simple pendu
lum has time period 't'. Its time period in a
lift which is moving upwards with acceleration 3 ms
–2
is
(a)
8.12
8.9
t (b)
8.9
8.12
t
(c)
8.6
8.9
t (d)
8.9
8.6
t
11.Two particles ar
e oscillating along two close parallel straight
lines side by side, with the same frequency and amplitudes.
They pass each other, moving in opposite directions when
their displacement is half of the amplitude. The mean
positions of the two particles lie on a straight line
perpendicular to the paths of the two particles. The phase
difference is
(a)0 (b) 2p/3
(c)p (d)p/6
12.If a simple pendulum of length l has maximum angular
displacement q, then the maximum K.E. of bob of mass m is
(a)
1
m /g
2
l (b) l2/mg
(c) )cos1(mg q-l (d) /2sin mgql
13.Th
ree masses of 500 g, 300 g and 100 g are suspended at
the end of a spring as shown, and are in equilibrium. Whenthe 500 g mass is removed, the system oscillates with aperiod of 2 second. When the 300 g mass is also removed,it will oscillate with a period of
(a) 2 s
(b) 4 s
(c) 8 s
500 g
300 g
100 g(d) 1 s
14.If the mass s
hown in figure is slightly displaced and then
let go, then the system shall oscillate with a time period of
(a)
k3
m
2p
(b)
k2
m3
2p
(c)
k3
m2
2p
m
kk
k
(d)
m
k3
2p
15.Two oscillator
s are started simultaneously in same phase.
After 50 oscillations of one, they get out of phase by p, that
is half oscillation. The percentage difference of frequencies
of the two oscillators is nearest to
(a) 2% (b) 1%
(c) 0.5% (d) 0.25%
16.Two wires are kept tight between the same pair of supports.
The tensions in the wires are in the ratio 2 : 1, the radii are in
the ratio 3 : 1 and the densities are in the ratio 1 : 2. The ratio
of their fundamental frequencies is
(a) 2 : 3 (b) 2 : 4
(c) 2 : 5 (d) 2 : 6
17.The time period of the oscillating system (see figure) is
(a)
12
m
T2
kk
=p
(b)
12
m
T2
kk
=p
+
M
k
1
k
2
(c)
12
T 2 mkk=p
(d) None of these
18.A p
article of mass 10 gm is describing S.H.M. along a straight
line with period of 2 sec and amplitude of 10 cm. Its kinetic
energy when it is at 5 cm from its equilibrium position is
(a) 37.5
p
2
erg (b) 3.75
p
2
erg
(c) 375
p
2
erg (d) 0.375
p
2
erg

363. 359Oscillations
19.A b
oy is executing simple Harmonic Motion. At a
displacement x its potential energy is E
1
and at a
displacement y its potential energy is E
2
. The potential
energy E at displacement (x + y) is
(a)
12
EEE=- (b)
12
EEE=+
(c)
12
EEE=+ (d)
12
EEE=-
20.A particle unde
rgoes simple harmonic motion having time
period T. The time taken in 3/8th oscillation is
(a)
T
8
3
(b)T
8
5
(c)T
12
5
(d)T
12
7
21.Two pa
rticles are executing S.H.M. of same amplitude and
frequency along the same straight line path. They pass
each other when going in opposite directions, each time
their displacement is half of their amplitude. What is the
phase difference between them ?
(a) 5 p/6 (b) 2 p/3
(c)p/3 (d)p/6
22.Lissajous figure obtained by combining x = a sin wt and
y = a sin (wt + p/4) will be a/an
(a) ellipse (b) straight line
(c) circle (d) parabola
23.A simple spring has length l and force constant K. It is cut
into two springs of lengths l
1
and l
2
such that l
1
= n l
2
(n = an integer). The force constant of spring of length l
1
is
(a) K (1 + n) (b) (K/n) (1 + n)
(c)K (d) K/(n + 1)
24.The spring constant from the adjoining combination of
springs is
m
2K
K K
(a)K (b) 2 K
(
c) 4 K (d) 5 K/2
25.A Second’s pendulum is placed in a space laboratoryorbiting around the earth at a height 3 R from the earth’ssurface where R is earth’s radius. The time period of thependulum will be
(a) zero (b) 23
(c) 4 sec (d) infinite
26.A simple pendulum attached to the roof of a lift has a time
period of 2s in a stationary lift. If the lift is allowed to fall
freely the frequency of oscillations of pendulum will be
(a) zero (b) 2 Hz
(c) 0.5 Hz (d) infinity
27.Two simple pendulums of length 0.5 m and 20 m respectively
are given small linear displacement in one direction at the
same time. They will again be in the phase when the
pendulum of shorter length has completed oscillations
[nT
1
=(n–1)T
2
, where T
1
is time period of shorter length &
T
2
be time period of longer wavelength and n are no. of
oscillations completed]
(a)5 (b) 1
(c)2 (d) 3
28.A clock which keeps correct time at 20ºC, is subjected to
40ºC. If coefficient of linear expansion of the pendulum is
12 × 10
–6
per ºC. How much will it gain or loose in time ?
(a) 10.3 seconds/day (b) 20.6 seconds/day
(c) 5 seconds/day(d) 20 minutes/day
29.A mass is suspended separately by two different springs
in successive order then time periods is t
1
and t
2
respectively. It is connected by both springs as shown in
fig. then time period is t
0
, the correct relation is
k
1 k
2
m
(a)
2
2
2
1
2
0
ttt+= (b)
2
2
2
1
2
0
t
tt
---
+=
(c)
1
2
1
1
1
0
ttt
---
+= (d)
210
ttt+=
30.When an oscillator completes 100 oscillations its amplitude
reduces to
3
1
of its i
nitial value. What will be its amplitude,
when it completes 200 oscillations ?
(a)
8
1
(b)
3 2
(c)
6
1
(d)
9 1
31.A particle of mas
s m is fixed to one end of a light spring of
force constant k and unstretched length l. The system is
rotated about the other end of the spring with an
angular velocity w, in gravity free space. The increase in
length of the spring will be
k
m
w
(a)
k
m
2
lw
(b)
2
2
mk
m
w-
wl
(c)
2
2
mk
m
w+
wl
(d) None of th
ese
32.Frequency of oscillation is proportional to
2kk
m
(a)
m
k3
(b)
m
k
(c )
m
k2
(d)
k3
m

364. 360 PHYSICS
33.A pend
ulum bob is raised to a height h and released from
rest. At what height will it attain half of its maximum speed?
(a)
4
h3
(b)
2
h
(c)
4
h
(d) h707.0
34.A mass m fall on sp
ring of spring constant k and negligible
mass from a height h. Assuming it sticks to the pan and
executes simple harmonic motion, the maximum height upto
which the pan will rise is
(a)
mg
k
(b)
mg 2kh
11
k mg
éù
+-êú
ëû
(c)
mg 2kh
11
k mg
éù
++êú
ëû k
m
h
(d)
mg kh
11
k mg
éù
+-êú
ëû
35.The graph shown in figure represents
(a) S.H.M.
Velocity
Displacement+a–aO
(b) circular m
otion
(c) rectillinear motion
(d) uniform circular motion
36.The time period of a simple pendulum of infinite length is
(R
e
= radius of Earth)
(a)
g
R
2T
e
p= (b)
g
R2
2T
e
p=
(c)
g2
R
2T
e
p= (d)¥=T
37.A block rests on a horizontal table which is executing SHM
in the horizontal plane with an amplitude 'a'. If the coefficientof friction is 'm', then the block just starts to slip when the
frequency of oscillation is
(a)
a
g
2
1m
p
(b)
a
gm
(c)
g
a
2
m
p (d)
g
a
m
38.On Earth, a body suspende
d on a spring of negligible mass
causes extension L and undergoes oscillations along length
of the spring with frequency f. On the Moon, the same
quantities are L/n and f ' respectively. The ratio f '/f is
(a)n (b)
n
1
(c)n
–1/2
(d) 1
39.A unifo
rm pole of length l = 2 L is laid on smooth horizontal
table as shown in figure. The mass of pole is M and it isconnected to a frictionless axis at O.A spring with forceconstant k is connected to the other end. The pole isdisplaced by a small angle q
0
from equilibrium position and
released such that it performs small oscillations. Then
x =0M
O
2L
(a)0
M
3k
w= (b)0
k
3M
w=
(c)0
3k
M
w= (d)0
k
2M
w=
40.A particle of
mass is executing oscillations about the origin
on the x-axis. Its potential energy is V(x) = k | x |
3
, where k is
a positive constant. If the amplitude of oscillation is a, thenits time period T is
(a) proportional to
1
a
(b) proportional to a
(c) independent
3
2
a (d) None of these
41.A c
ircular hoop of radius R is hung over a knife edge. The
period of oscillation is equal to that of a simple pendulum
of length
(a)R (b) 2R
(c) 3R (d)
3R
2
42.In the figure shown, the spring is light and has a force
constant k. The pulley is light and smooth and the strring
is light . The suspended block has a mass m. On giving a
slight displacement vartically to the block in the downward
direction from its equilibrium position the block executes
S.H.M. on being released with time period T. Then
(a)
m
T2
k
=p
(b)
m
T2
2k
=p
m
(c)
2m
T2
k
=p
(d)
m
T4
k
=p

365. 361Oscillations
43.A bod
y of mass m falls from a height h onto a pan (of
negligible mass) of a spring balance as shown. The spring
also possesses negligible mass and has spring constant k.
Just after striking the pan, the body starts socillatory motion
in vertical directioin of amplitude A and energy E. Then
(a)
mg
A
k
=
(b)
mg 2kh
A1
k mg
=+
m
h
(c)
21
E mgh kA
2
=+
(d)
2
2mg
E mgh
2k
æö
=+
ç÷
èø
44.A coin is placed on a horizontal platform which undergoes
vertical simple harmonic motion of angular frequency w.
The amplitude of oscillation is gradually increased. The
coin will leave contact with the platform for the first time
(a) at the mean position of the platform
(b) for an amplitude of
2
g
w
(c) for an a
mplitude of
2
2
g
w
(d) at the highe
st position of the platform
45.A point particle of mass 0.1 kg is executing S.H.M. of
amplitude of 0.1 m. When the particle passes through the
mean position, its kinetic energy is 8 × 10
–3
Joule. Obtain
the equation of motion of this particle if this initial phase of
oscillation is 45º.
(a)
y 0.1sin4t
4
pæö
= ±+ç÷
èø
(b)y 0.2sin4t
4
pæö
= ±+ç÷
èø
(c)y 0.1sin2t
4
pæö
= ±+ç÷
èø
(d)
y 0.2sin2t
4
pæö
= ±+ç÷
èø
46.A body of mass 0.01 kg executes simple harmonic motion
about x = 0 under the influence of a force as shown in
figure. The time period of S.H.M. is
80
–80
0.2
–0.2
F(N)
x(m)
(a) 1.05 s (b) 0.5 2 s
(c) 0.25 s (d) 0.03 s
47.A forced oscillator is acted upon by a force F = F
0
sin wt.
The amplitude of oscillation is given by
2
55
2 369w - w+
.
The resona
nt angular frequency is
(a) 2 unit (b) 9 unit
(c) 18 unit (d) 36 unit
48.A straight rod of negligible mass is mounted on a frictionless
pivot and masses 2.5 kg and 1 kg are suspended at
distances 40 cm and 100 cm respectively from the pivot as
shown. The rod is held at an angle q with the horizontal
and released. Then
/////////////////////////////////////////////////////////////
q
2.5kg
1 kg
40cm
100cm
(a)
the rod executes periodic motion about horizontal
position after the release
(b) the rod remains stationary after the release.
(c) the rod comes to rest in vertical position with 2.5 kg
mass at the lowest point
(d) the rod executes periodic motion about vertical
position after the release
DIRECTIONS for Qs. (49-50) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following-
(a) Statement -1 is false, Statement-2 is true
(b) Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c) Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d) Statement -1 is true, Statement-2 is false
49. Statement 1 : The graph between velocity and displacement
for a harmonic oscillator is an ellipse.
Statement -2 : Velocity does not change uniformly with
displacement in harmonic motion.
50. Statement -1 : If the amplitude of a simple harmonic
oscillator is doubled, its total energy becomes four times.
Statement -2 : The total energy is directly proportional to
the square of the amplitude of vibration of the harmonic
oscillator.

366. 362 PHYSICS
Exemplar Questions
1.The displacement of a particle is represented by the
equation 3cos 2.
4
yt
pæö
= -w
ç÷
èø
The motion of the particle
is
(a) simple harmonic with period 2p/w
(b) simple harmonic with period p/w
(c) periodic but not simple harmonic
(d) non-periodic
2.The displacement of a particle is represented by the
equation y = sin
3 wt. The motion is
(a) non-periodic
(b) periodic but not simple harmonic
(c) simple harmonic with period 2p/w
(d) simple harmonic with period p/w
3.The relation between acceleration and displacement of four
particles are given below
(a)a
x = +2x (b)a
x = +2x
2
(c)a
x = –2x
2 (d)a
x = –2x
Which, one of the particle is exempting simple harmonic
motion?
4.Motion of an oscillating liquid column in a U-tube is
(a) periodic but not simple harmonic
(b) non-periodic
(c) simple harmonic and time period is independent of the
density of the liquid
(d) simple harmonic and time period is directly proportional
to the density of the liquid
5.A particle is acted simultaneously by mutually perpendicular
simple harmonic motion x = a cos wt and y = a sin wt. The
trajectory of motion of the particle will be
(a) an ellipse(b) a parabola
(c) a circle (d) a straight line
6.The displacement of a particle varies with time according to
the relation
sin cos.yatbt=w+w
(a) The motion is oscillatory but not SHM
(b) The motion is SHM with amplitude a + b
(c) The motion is SHM with amplitude a
2 + b
2
(d) The motion is SHM with amplitude 22
ab+
7.Four pendulums A, B, C and D ar
e suspended from the
same
G G
D
C
BA
elastic support as shown in figure. A and C are of the same
length, while B is smaller than A and D is larger than A. If A
is given a transverse displacement,
(a)D will vibrate with maximum amplitude
(b)C will vibrate with maximum amplitude
(c)B will vibrate with maximum amplitude
(d) All the four will oscillate with equal amplitude
8.Figure shows the circular motion of a particle. The radius of
the circle, the period, sense of revolution and the initial
position are indicated on the figure. The simple harmonic
motion of the x-projection of the radius vector of the rotating
particle P is
(a)
2
( ) sin
30
t
xtB
pæö
=
ç÷
èø

y
x
B
p t
T
( = 0)
= 30s
(b)( ) cos
15
t
xtB
pæö
=
ç÷
èø
(c)( ) sin
152
t
xtB
ppæö
=+
ç÷
èø
(d)( ) cos
152
t
xtB
ppæö
=+
ç÷
èø
9.The equation of motion of a particle is x = a cos(at)
2. The
motion is
(a) periodic but not oscillatory
(b) periodic and oscillatory
(c) oscillatory but not periodic
(d) neither periodic nor oscillatory
10.A particle executing SHM maximum speed of 30 cm/s and a
maximum acceleration of 60 cm/s
2. The period of oscillation
is
(a)p sec (b)
2
p
sec
(c)2p sec (d)
t
p
sec
Past Years (2013-2017) NEET/AIPMT Questions
11.A particle of mass m oscillates along x-axis according to
equation x = a sin wt. The nature of the graph between
momentum and displacement of the particle is
[NEET Kar. 2013]
(a) straight line passing through origin(b) circle(c) hyperbola
(d) ellipse

367. 363Oscillations
12.Th
e oscillation of a body on a smooth horizontal surface is
represented by the equation,
X = A cos (wt)
where X = displacement at time t
w = frequency of oscillation
Which one of the following graphs shows correctly the
variation of ‘a’ with ‘t’? [2014]
(a)
a
O
Tt
(b)
a
O
Tt
(c)
a O
Tt
(d)
a O
Tt
13.A particle is executing SHM along a straight line. Its
velocities at distances x
1
and x
2
from the mean position
are V
1
and V
2
, respectively. Its time period is [2015]
(a)
22
21
22
12
x –x
2
V –V
p (b)
22
12
22
12
VV
2
xx
+
p
+
(c)
22
12
22
12
V –V
2
x –x
p (d)
22
12
22
12
x –x
2
V –V
p
14.When two di
splacements represented by y
1
= asin(wt) and
y
2
= b cos(wt) are superimposed the motion is: [2015]
(a) simple harmonic with amplitude
a
b
(b) simple harmonic with amplitude
22
ab+
(c) simpl
e harmonic with amplitude
(a b)
2
+
(d) not a simple harmonic
15.A particle is executing a simple harmonic motion. Its maximum
acceleration is a and maximum velocity is b.
Then its time period of vibration will be : [2015 RS]
(a)
a
b
(b)
2
b
a
(c)
2pb
a
(d)
2
2
b
a
16.A particle executes linear simple harmonic motion with an
amplitude of 3 cm. When the particle is at 2 cm from themean position, the magnitude of its velocity is equal to thatof its acceleration. Then its time period in seconds is [2017]
(a)
5
2p
(b)
4
5
p
(c)
2
3
p
(d)
5
p
17.A spring of force constant k is cut into lengths of ratio 1 : 2
: 3. They are connected in series and the new force constantis k'. Then they are connected in parallel and force constantis k¢¢ . Then k' : k¢¢ is [2017]
(a) 1 : 9 (b) 1 : 11
(c) 1 : 14 (d) 1 : 6

368. 364 PHYSICS
EXERCISE - 1
1. (b) 2. (
a) 3. (b) 4. (d) 5. (b)
6. (d) 7. (b) 8. (b)
9. (d) Here effective spring factor, k
eff
= 2 k ;
Frequency of oscillation,
m
k2
2
1
f
p
= ; when one
spring is remo
ved, then spring factor = k ;
New frequency of oscillation
2
f
m
k
2
1
f =
p
=¢
10. (c)
max
va ω=;
max max
v2a2v2v=w==¢

max
(since v v)=
12. (b)
displacement
T 2π
acceleration
= =
b/2
xb
x
2 p=p
11. (a) P.E. o
f body in S.H.M. at an instant,
22211
U m y ky
22
= w=
If the displacement,
y = (a – x) then2211
U k(a x) k(x a)
22
=-=-
13. (c)
P.E. changes from zero to maximum twice in each
vibration so its time period is T/2
14. (a) Velocity,
)4/t(sinA
dt
dx
v p+ww-==
Velocity will be maximum, w
hen
wt + p/4 = p/2 or wt = p/2 – p/4 = p/4 or t
= p/4w
15. (d) Tension is maximum at the mean position.
16. (a) A rectangular block of mass m and area of cross-
section
A floats in a liquid of density r. If it is given a small
vertical displacement from equilibrium. It undergoes
oscillations with a time period T then : T µ m.
17. (c)
2
22
2
yy,1,1
T
4p
=ww==&& or
22
4Tp=
or T =
2p second.
18. (a) At displacement ± a, the velocity is zero. At zero
displacement, velocity is maximum.
19. (a) The amplitude is a maximum displacement from the
mean position.
20. (b)
21. (a) a = – kX, X = x + a.
In simple harmonic motion acceleration is directly
proportional to the displacement from the mean
position. Also the acceleration is in the opposite
direction of displacement.
22. (c) Only functions given in (A) & (C) represent SHM.
23. (d)
22
kA
2
1
EA
m
k
m
2
1
E =Þ=
Þ E does not depend on m
2
4. (a) x = a sin
2
wt =
2
a
(1 –cos 2wt)
dx
dt
=
2
a
2 sin2tww
Þ
2
2
dx
dt
=
2
4
cos2
2
a
t
w
×w
This represents an S. H. M. of frequency =
w
p
25. (d) T = 2p
m
K
\
1
2
T
T
=
1
2
M
M
\T
2
= T
1
2
1
M
M
= T
1
2M
M
T
2
= T
12 = 2T (where T
1
=T)
EXERCISE - 2
1. (a)
22
yav-w=
At x= 0, v
22
a o a.=w - =w
At x =
2 2
2a a 3a
,v'a
2 24
æö
=w - =wç÷
èø
\
v'3
v2
=
or v'
3
a
2
=w
3a 3a
a
2T
pp
= w=
2
T
pæö
w=ç÷
èø
Q
2. (c) w==a100v
max
; w = 100/
a = 100/10 = 10 rad/s
v
2
= w
2
(a
2
– y
2
) or 50
2
= 10
2
(10
2
– y
2
) or 25 = 100 – y
2
or
3575y== cm.
Hints & Solutions

369. 365Oscillations
3.
(b)
k
M
2Tp=
Mm
T2
k
+
=p¢
k
M
Þ
5 M Mm
22
3kk
+
p =p
Þ
25M Mm
9kk
+
= Þ
9
16
M
m
=
4. (d)
6
R 64 10
T22
g 9.8
´
=p =p =
27
108
7
22
2
3
´
´
´´
=
6049
10008222
´
´´´
min = 84.6 min
5.
(c) Max. force = mass × max. acceleration
= m 4 p
2
n
2
a = 1 × 4 ×p
2
× (60)
2
× 0.02 = 288 p
2
6. (a)
g
2T
l
p= ;
)6/g(
2
g
22
ll ¢
p=p=
Time per
iod will remain constant if on moon,
l' = l/6 = 1/6 m
7. (a) Kinetic energy,
()
2222
xAm
2
1
mv
2
1
E.K -w==
Potential e
nergy,
2 2211
P.E kx mx
22
= =w
putting, E.
PE.K
=
()
22222
xm
2
1
xAm
2
1
w=-w
22
x2A=Þ
.cm22
2
A
x ==
8. (a) Co
mparing A + w
2
x = 0
p=wÞp=w24
22 , Hz1,22
=np=pn
9. (b) ÷
ø
ö
ç
è
æ
f+
p
=
2
t
sin2y
velocity
of particle
÷
ø
ö
ç
è
æ
f+
pp
´=
2
t
cos
2
2
dt
dy
accelera
tion
÷
ø
ö
ç
è
æ
f+
pp
-=
2
t
sin
2dt
yd
22
Th
us
max
a =
2
2
p
10. (a)
11
t , t'
9.8 12.8
µµ ( )g ' 9.8 3 12.8= +=Q
t ' 9.8 9 .8
t't
t 12.8 12.8
\= Þ=
11.
(b) Equation of SHM is given by
x = A sin (wt + d)
(wt + d) is called phase.
When x =
A
2
, then
sin
(wt + d) =
1
2
t
6
p
Þw +d=
or
1
6
p
f=
For second particle,
A
A
x = A/22
1
x=0
2
5
66
pp
f=p-=
21
\f=f -f
=
42
63
pp
=
12. (c
) When the bob moves from maximum angular
displacement q to mean position, then the loss of
gravitational potential energy is mgh
where )cos1(hq-=l
13. (d)
k
100
2'T,
k
400
2T p=p=
T' 100 1 T2
T' 1s
T 4002 22
\= =Þ ===
14. (b) The equivalent situation is a series combination of
two springs of spring constants k and 2k.
If k' is the equivalent spring constant, then
3
k2
k3
)k2)(k(
'k ==
k2
m3
2Tp=Þ
15. (b
) Phase change p in 50 oscillations.
Phase change 2p in 100 oscillations.
So frequency different ~ 1 in 100.

370. 366 PHYSICS
16. (a)
)2(
T
D
1
',
T2
)D3(
1
rp
=n
pr
=n
ll
3
2
2
3
2
'
=´=
n
n
\
17. (b)
18. (c)
Kinetic energy
2221
()
2
K mayw=-
2
22212
10 [10 5 ] 375
22
æö
=´
´ -=
ç÷
èø
erg
p
p
19. (b)
22 12
12
2211
,
22
= Þ= = Þ=
EE
Ekxx E kyy
kk
and
2 2
()
2
1
= + Þ+=
E
E kxy xy
k
12
12
22 2EE E
EEE
kkk
Þ + = Þ+=
20. (
c) Time to complete 1/4th oscillation is
4
T
s. Time to
complete
8 1
th vibration fr
om extreme position is
obtained from
t
T
2
cosatcosa
2
a
y
p
=w== or s
6
T
t=
So time to comple
te 3/8th oscillation
=
12
T5
6
T
4
T
=+
21. (b) )t(sina
y f+w=; when y = a/2,
then
)t(sina
2
a
f+w=
or
6
sin
2
1
)t(sin
p
==f+w or
6
5
sin
p
So phase of two particle
s is p/6 and 5 p/6 radians
Hence phase difference = (5 p/6) – p/6 = 2 p/3
22. (a) sin wt = x/a and
222
a/x1tsin1tcos -=w-=w
y = sin (wt + p/4)
=
sin wt cos p/4 + cos wt sin p/4
=
2
1
a/x1
2
1
a
x 22
´÷
ø
ö
ç
è
æ
-+´
or
22
xaxy2-+=
or
222222
xax2)xa(xy2-+-+=
=
222
xax2a-+
It is an equation of
an ellipse.
23. (b) Let k be the force constant of spring of length l
2
.
Since l
1
= n l
2
, where n is an integer, so the spring is
made of (n + 1) equal parts in length each of length l
2
.
\
k
)1n(
K
1 +
= or k = (n + 1)
K
The spring of length l
1
(= n l
2
) will be equivalent to n
springs connected in series where spring constant
n/K)1n(
n
k
k +==¢ & spring const
ant of length l
2
is K(n+1).
24. (c) Here all the three springs are connected in parallel to
mass m. Hence equivalent spring constant
k = K + K + 2 K = 4 K.
25. (d) The second pendulum placed in a space laboratory
orbiting around the earth is in a weightlessness state.
Hence g = 0 so T = ¥
26. (a) When lift is falling freely, the effective acceleration
due to gravity inside the lift is zero i.e. g' = g – g = 0.
Therefore time period will be infinity and frequency is
zero
27. (b) Let T
1
, T
2
be the time period of shorter length and
longer length pendulums respectively. As per question,
n T
1
= (n – 1) T
2
;
so
g
20
2)1n(
g
5.0
2n p-=p
or 6)1n(40)1n(n-»-=
Hence n = 6/5 » 1
28. (a)
g
2T
l
p= ;
1/2T1
2(/g)/
T2
-D
= p´ ´Dl ll
)
g
2TT(
ll
Q
D+
p=D+
\ qDa=
D
=
D
2
1
2
1
T
T
l
l
=
56
1012)2040(1012
2
1 --
´=-´´´
5
1012TT
-
´´=D = 24 × 60 × 6
0 × 12 × 10
–5
= 10.3 s/day
29. (b)
1
1
k
m
2tp= or
1
2
2
1
k
m4
t
p
= or
2
1
2
1
t
m4
k
p
=
Similarly,
2
2
2
2
t
m4
k
p
= and
2
0
2
21
t
m4
)kk(
p
=+
\
2
2
2
2
1
2
2
0
2
t
m4
t
m4
t
m4 p
+
p
=
p
or
2
2
2
1
2
0 t
1
t
1
t
1
+=
30. (d) It is
a damped oscillation, where amplitude of
oscillation at time t is given by
γt
0
A ae
-
=
where a
0
= initial ampli
tude of oscillation
g = damping constant.

371. 367Oscillations
As per question
,
γ100/ν0
0
a
ae
3
-
= …… (i)
(wh
ere
n is the frequency of oscillation)
and
γ200 /ν
0
A ae
-
= …… (ii)
Fr
om (i) ;
γ100/νo
0
a
ae
3
-´
= …… (iii)
Dividi
ng equation (ii) by (iii), we have
γ200 /ν
γ100 /ν
0
Ae
a (1/ 3)e
-
-´
=
=
γ100/ν1
e
3
-´
=
or 00
a
9
1
3
1
3
1
aA =´´=
31. (b
) Let x = the increase in the length of spring. Then the
particle moves along a circular path of radius (l + x),
and the spring force = kx = centripetal force
\ mw
2
(l + x) = kx or
2
2
mk
m
x
w-
w
=
l
32. (a)
k k2
x
Let mass is displaced towards left by x then force on
mass = – kx – 2kx = – 3kx
[negative sign is taken because force is opposite to
the direction of motion]
F 3kxÞ =- xm
2
w-=
3k
.
m
Þw=
1 3k
f.
2 2m
w
==
pp
Thus it is propo
tional to
m/k3
33. (a
) Maximum velocity can be found out by energy
conservation.
2
mv
2
1
mgh= gh2vor
max
= ;
2
gh
'v
2
v
max
==
'mgh)'v(m
2
1
mgh
2
+= ;
'mgh
2
mgh
2
1
mgh +=
Þ 'mgh
4
mgh3
=
4
h3
'hor=
34. (b)
Loss in PE of mass = gain in PE of the spring\mg (h + x) =
2
kx
2
1
2
kx – 2mgx – 2mgh = 0
Þx =
k2
)mgh2(k4gm4mg2
22
--±
or x =
k
mgh2
k
gm
k
mg
2
22
+±
x = A –
k
mg
=
k
mg
k
mgh2
k
gm
2
22
-+
=
k
mg
ú
ú
û
ù
ê
ê
ë
é
-+1
mg
kh2
1
35. (a
) t = 0, v maximum. The motion begins from mean
position. So it represents S.H.M.
36. (a) The both of the pendulum will more along a straight
line AB.
q q
Re
mg
x
A B
The direction of the Earth's gravitational field is radial
Now, mg
R
mGM
F
2
e
e
==
x
R
mGM
R
x
FcosFF
3
e
e
e
x -=-=q-== –kx
where
e
3
e
GMm
k
R
=
Time pe
riod of a simple harmonic oscillator,
3
ee
R/mGM
m
2
k
m
2T p=p=
or
g
R
2
R
GM
R
2T
e
2
e
e
e
p=p=
37. (a
) For the block is about to slip, ag
2
w=m
or
g
a
m
Þw=
g 1g
2νν
a 2a
mm
Þp= Þ=
p
38
. (d) Oscillations along spring length are independent of
gravitation.

372. 368 PHYSICS
39. (c )
2214
I M(2L) ML
33
==
Force applied by th
e spring is F = – kx
Þ F = – k (2Lq)
(q is the angular displacement from the equilibrium
position). Further

22
| F| 4Lksin 4Lkt=´= q=-q
rr
l
Also,
..
2
I I 4Lkt=a=q=-q
Þ
..
3k
0
M
q+ q=
Þ 0
3k
M
w=
40. (a) V(x) = k
| x |
3
since,
2dV(x)
F 3k|x|
dx
=- =- …… (i)
x = a sin (wt)
Th
is equation always fits to the differential equation
2
2
2
dx
x
dt
= -wor
2
2dx
m mx
dt
=-w
Þ F = – mw
2
x …… (ii)
Equations
(i) and (ii) give
22
3k|x| mx- =-w
Þ
1/23kx 3ka
[sin( t)]
mm
w==w
Þ awµ
Þ
1
T
a
µ
41. (b)
2R
T22
gg
=p =p
l
42. (d) Let the extension in the spring be x
0
at equilibrium. If
F
0
be the tension in the string then F
0
= kx
0
. Further if
T
0
is the tension in the thread then T
0
= mg and
2T
0
= kx
0
.
Let the mass m be displaced through a slight
displacement x downwards. Let the the new tension in
the string and spring be T and F respectively.
Þ
0
x
Fx
2
æö
=ç÷
èø
and F = 2T
Þ
0
x
2T kx
2
æö
=ç÷
èø
Þ
0
kx
(T T)
4
-=
Þ Time period
mm
24
k k
4
=p =p
43. (b)
21
mg(h y) ky
2
+=
Þ
mg mg 2kh
y1
k k mg
=±+
At equilibriu
m
mg = ky
0
Þ
0
mg
y
k
=
Þ Amplitud
e A = y – y
0

mg 2kh
1
k mg
=+
Energy of os
cillation is
2
21 mg
E kA mgh
2 2k
æö
= =+ ç÷
èø
44. (b) For block A to move in S.H.M.
mg
x
N
A
mean
position
mg – N = mw
2
x
whe
re x is the distance from mean position
For block to leave contact N = 0
2
2 g
xxmmg
w
=Þw=Þ
45. (a) The
displacement of a particle in S.H.M. is given by
y = a sin (wt + f)
velocity =
dy
dt
= wa cos (wt + f)
The ve
locity is maximum when the particle passes
through the mean position i.e.,

max
dy
dt
æö
ç÷
èø
= w a

373. 369Oscillations
The
kinetic energy at this instant is given by 1
2
m
2
max
dy
dt
æö
ç÷
èø
=
1
2
mw
2
a
2
= 8 × 10
–3
j oule
or
1
2
× (0.1) w
2
× (0. 1)
2
= 8 × 10
–3
Solving we getw = ± 4
Substituting the values of a, w and f in the equation of
S.H.M., we get
y = 0.1 sin (± 4t + p/4) metre.
46
. (d) Slope of F - x curve = – k =
80
0.2
- Þ k = 400 N /m,
Time period, T = 2p
m
k
= 0.0314 sec.
4
7. (b) At resonance, amplitude of oscillation is maximum
Þ 2w
2
– 36w + 9 is minimum
Þ 4w – 36 = 0 (derivative is zero)
Þ w = 9
48. (b) Torque about hinge
2.5 g × 0.40 cos q – 1g × 1 cos q = 0
49. (c) 50. (b) EXERCISE - 3
Exemplar Questions
1. (b) As given that, 3cos2
4
yt
pæö
= -w
ç÷
èø
Velocity of the particle
3cos2
4
dyd
vt
dt dt
épùæö
= = -w
ç÷êú
èøëû

3(2)s in2
4
t
épùæö
= -w - -w
ç÷êú
èøëû
6sin2
4
t
pæö
=w -w
ç÷
èø
So, acceleration,
6sin2
4
dvd
at
dt dt
é pùæö
= = w -w
ç÷êú
èøëû
(6) (2)cos 2
4
t
pæö
= w´-w -w
ç÷
èø

2
43cos2
4
t
épùæö
=-w -w
ç÷êú
èøëû
2
4ay=-w
In simple harmonic motion acceleration (or force) is
directly proportional to the negative of displacement
of particle
Þ as acceleration, ayµ-
Hence, due to negative sign motion is simple harmonic
motion (SHM.)
A simple harmonic motion is always periodic. So motion
is periodic simple harmonic.
From the given equation,
3cos2
4
yt
pæö
= -w
ç÷
èø
Compare it by standard equation
cos()yat= w +f
So, '2w=w
22
2'
'2
T
T
p pp
=wÞ==
ww
Hence, the motion is SHM with period
p
w
.
2. (b) A moti
on will be harmonic if
aµ displacement and a
simple harmonic motion is always periodic but all
simple harmonic motion are periodic but all periodic
are not harmonic.
As given equation of motion is
3
sin
yt=w
(3sin 4sin3 ) / 4tt=w-w
3
[ sin 3 3sin 4sin ]q=q-qQ
Differentiating both side w.r.t. t
So, (3sin) (4sin3 ) /4
dydd
v tt
dt dt
dt
éù
==w-w
êú
ëû
or
3 (3sin sin3 )
sin
4
q-qéù
q=
êú
ëû
4 3 cos 4 [3 cos3 ]
dy
tt
dt
=ww-´ ww
Again, differentiating both side w.r.t. t
2
2
2
4 3 sin 36 sin 3
dy
tt
dt
´ =-ww+ ww
22
2
3 sin 3 6 sin3
4
dy tt
a
dt
-ww+ww
==
So,
2
2
dy
dt
æö
ç÷
ç÷
èø
is not directly proportional to y.
So motion is not harmonic.
3
()sin,ytt=w
3
()sin ()ytT tT+ = w+

32
sin ()tT
T
péù
=+
êú
ëû

33
sin (2 ) sintt= p+w = w

374. 370 PHYSICS
3. (d) Fo
r motion to be SHM acceleration of the particle must
be opposite of restoring force and proportional to
negative of displacement. So, F = ma = m(–2x)
i.e., F = –2mx, so
Fxµ-
Hence, a
x = –2x
We sh
ould be clear that x has to be linear.
4. (c) Consider a U-tube filled in which a liquid column
oscillates. When liquid column lifted upto height y
from A to B in arm Q. The liquid level in arm P becomes
at C', so the difference between the height of two
columns are = AB + A'C = y + y = 2y.
In this case, restoring force acts on the liquid due to
gravity. Acceleration of the liquid column, can becalculated in terms of restoring force.
P Q
B
A
C
B'
A'
C'
h
y
Equilibrium level
Restoring force,
f =
Weight of liquid column of height 2y
(2)2
f A y g A gy=-´ ´r´=-r
[] vh=rQ
As restoring force at A opposite to gravitational force
as liquid is lifted against ( ) 2tmg k Agw m=r
Then fyµ- so motion is SHM.
Time period
(2)
222
2
mAhh
T
k Agg
´ ´r
=p =p =p
r
2
h
T
g
=p
Time period is indepe
ndent of the density of the liquid
and motion is harmonic.
5. (c) The resultant-displacement can be find by adding x
and y-components.
According to variation of x and y, trajectory will be
predicted, so resultant displacement is y' = (x + y)
As given that,
cosxat=w ...(i)
sinyat=w ...(ii)
So, '(cos sin)ya tat= w+w
' (cos sin)yattÞ= w+w
cos sin
'2
22
tt
ya
wwéù
=+
êú
ëû
' 2(cos cos 45 sin sin 45 )yatt= w °+w°
2 cos( 45 )at= w-°
So, the displacement is not straight line and not
parabola also.
Now, squaring and adding eqs. (i) and (ii),
222
xya+=
22
[ cos sin 1]ttw+ w=Q
This is the equation of a circle, so motion in circular
(independent of time).
Clearly, the locus is a circle of constant radius a.
6. (d) As given that, the displacement is
sin cosyatbt=w+w
Let sinaA=f ...(i)
and cosbA=f ...(ii)
Squaring and
adding (i) and (ii)22 22 22
sin cosabAA+=f+f
22
A ab=+
(amplitude)
sin sin cos cosyAtAt= f×w+ fw
cos()yAt= w -f
sin()
dy
At
dt
=- w w -f
2
22
2
cos()
dy
Aty
dt
= - w w -f = -w
Thus,
2
2
()
dy
y
dt
µ- so, motion is SHM.
Hen
ce, it is an equation of SHM with amplitude
22
.A ab=+
7. (b) Whe
n pendulum vibrate with transverse vibration then
2
l
T
g
=p
l = length of pendu
lum.
G G
Elastic support
A
B
C
D
e
Through the e
lastic rigid support the disturbance is
transferred to all the pendulum A and C are having
length and hence same frequency. They will be in
resonance, because their time period of oscillation.
So, a periodic force of period (T) produces resonance
in A and C and they will vibrate with maximum amplitude
as in resonance.

375. 371Oscillations
8.
(a) As the particle (P) is executing circular motion with
radius B.
Let particle P is at Q at instant any (t), foot of
perpendicular on x axis is at R vector OQ makes
Ðq,
with its zero po
sition not P displacement of particles
for O to R.
Consider angular velocity of the particle executing
circular motion is w and when it is at Q makes and
angle q as shown in the diagram.
B
R
x
y
pt( =0)
Q
qq
r
90–q
O
Clearly, tq=w
Now, we can write
cos(90 – )OR OQ=q ()ORX=Q
sin sinx OQ OQ t=q=w
sinrt=w []OQr=Q
22
sin sin
30
xBtBt
T
pp æö
\== ç÷
èø

2
sin
30
xBt
pæö
= ç÷
èø
Hence, this equation represents SHM.
9. (c) According to the question,
2
cos()xat=a
is a cosine function, so it is an oscillatory motion.
Now, at, t = (t + T). The equation of motion of particle
2
()cos[ ()]xtT a tT+ = a+
2
[ () cos( )]xtat=aQ

22
cos[ 2 ] ()a t T tT xt= a+a+ a¹
where, T is supposed as period of the function w(t).
Hence, it is oscillatory but not periodic.
10. (a) Let us consider a equation of an SHM is represented
by y = a sin wt
cos
dy
v at
dt
==ww
max
( ) 30 cm /secva= w= (given) ...(i)
Accel
eration
2
( ) sin==-ww
dv
a at
dt
2
max
60=w=aa (given) ...(i i)
Eqs. (i) and (ii), we get
( ) 60 (30) 60aww=Þw=
w = 2 rad/s
2
2 rad/s
T
p
=
sec=pT
Past Years (2013-2017) NEET/AIPMT Questions
11. (d) As
22
222
1
vy
aa
+=
w
This is the equation of ellipse.
Hence the graph is an ellipse. P versus x graph is
similar to V versus x graph.
12. (c) Displacement, x = A cos (wt) (given)
Velocity, v =
dx
A sin (t)
dt
=-ww
Acceleration, a =
2dv
A cost
dt
=-ww
Hence graph (c) correctly dipicts the variation of a
with t.
13. (a) As we know, for particle undergoing SHM,
22
V A –X=w
2222
11
V (A –x)=w
2222
22
V (A –x)=w
Substrac
ting we get,22
2212
1222
VV
xx+=+
ww
Þ
22
2212
212
V –V
x –x=
w
Þ
22
12
22
21
V –V
w
x –x
=
Þ
22
21
22
12
x –x
T2
V –V
=p
14. (b
) The two displacements equations are y
1
= a sin(wt)
and y
2
= b cos(wt) = b sin
t
2
pæö
w+
ç÷
èø
y
eq
= y
1
+ y
2
= a s
inwt + b coswt = a sinwt + b sin
t
2
pæö
w+
ç÷
èø
Since the frequencies for both SHMs are same,
resultant motion will be SHM.
Now A
eq
=
22
a b 2ab cos
2
p
++
b
a
2 2
eq
A abÞ = +

376. 372 PHYSICS
15. (c)
As, we know, in SHM
Maximum acceleration of the particle, a = Aw
2
Maximum velocity, b = Aw
Þ w =
a
b
Þ T =
22p pb
=
wa
2
T
péù
w=
êú
ëû
Q
16. (b) Given
, Amplitude A = 3 cm
Which particle is at x = 2 cm
According to question, magnitude of velocity
= acceleration
222
Axxw - =w
22 2
(3) (2) 2
T
pæö
-=
ç÷
èø
4
5
T
p
=Þ T =
4
5
p
17. (b)Let l be the co mplete length of the spring.
Length when cut in ratio, 1 : 2 : 3 are , and
632
lll
Spring constant (k)
1
length()
µ
l
Spring constant for given segments
k
1
= 6k, k
2
= 3k and k
3
= 2k
When they are connected in series
1111
k' 6k 3k 2k
=++
Þ
16
k ' 6k
=
\ Force consta
nt k' = k
And when they are connected in parallel
k" = 6k + 3k + 2k
Þ k" = 11k
Then the ratios
k'1
k" 11
= i.e., k' : k" = 1 : 11

377. (Compression) l
C C C
R R R
l
(Rarefaction)
C
Logitudinal wav
es propagate through medium with the help of
compressions and rarefactions.
Equation of a Harmonic Wave
Harmonic waves are generated by sources that execute simple
harmonic motion.
A harmonic wave travelling along the positive direction of x-axis
is represented by
sin()y A t kx= w-

sin2
tx
A
T
ìüæö
= p-íýç÷
lèøîþ
( )
2
sinA vtx
pìü
=-íý
lîþ
where, y =
displacement of the particle of the medium at a location
x at time t
A = amplitude of the wavel= wavelength
T = time period
v=
,ul wave velocity in the medium
w=
2
,
T
p
angular frequency
k=
2
,
p
l
angular wave number or propagation constant.
If the wave is travelling along the negative direction of
x-axis then
sin()y A t Kx= w+

sin2.
tx
A
T
ìüæö
= p+íýç÷
èølîþ
WAVE MOTION
Wave mo
tion is a type of motion in which the disturbance travels
from one point of the medium to another but the particles of the
medium do not travel from one point to another.
For the propagation of wave, medium must have inertia and
elasticity. These two properties of medium decide the speed of
wave.
There are two types of waves
1.Mechanical waves: These waves require material medium
for their propagation. For example : sound waves, waves in
stretched string etc.
2.Non-mechanical waves or electromagnetic waves : These
waves do not require any material medium for their
propagation. For example : light waves, x-rays etc.
There are two types of mechanical waves
(i)Transverse waves : In the transverse wave, the
particles of medium oscillate in a direction
perpendicular to the direction of wave propagation.
Waves in stretched string, waves on the water surface
are transverse in nature.
Transverse wave can travel only in solids and surface of liquids.
Transverse waves propagate in the form of crests and troughs.
Wavelength
Trough
Wave motion
l
Crest
All electr
omagnetic waves are transverse in nature.
(ii)Longitudinal waves : In longitudinal waves particles
of medium oscillate about their mean position along
the direction of wave propagation.
Sound waves in air are longitudinal. These waves can travel in
solids, liquids and gases.
15
Waves

378. 374 PHYSICS
(ii) Th
e speed of transverse waves on stretched string is given
byT
v=
m
where T is the tension in the string and m is the mass per
unit length of the string.
Speed of Longitudinal Waves :
The speed of longitudinal waves in a medium of elasticity E and
density r is given by
E
v=
r
For solids, E is replaced by Young's modulus (Y)
solid
Y
v=
r
For liquids and gases, E is replaced by bulk modulus of
elasticity (B)
liquid/gas
B
v =
r
The density of a solid is much larger than that of a gas but the
elasticity is larger by a greater factor.
v
solid
> v
liquid
> v
gas
Speed of Sound in a Gas :
Newton's formula.
P
v=
r
where P is the atmospheric pressure and r is the density of air at
ST P.
Laplace's correction
P
v
g
=
r
where g is the ratio of two specific heats C
p
and C
v
Power and Intensity of Wave Motion :
If a wave is travelling in a stretched string, energy is transmitted
along the string.
Power of the wave is given by
221
2
P Av= mw where m is mass pe r unit length.
Intensity is flow of energy per unit area of cross section of the
string per unit time.
Intensity
221
2
I Av= rw
Principle of Superposition of Wave
s :
If two or more waves arrive at a point simultaneously then the
net displacement at that point is the algebraic sum of the
displacement due to individual waves.
y = y
1
+ y
2
+ ............... + y
n
.
where y
1
, y
2
.......... y
n
are the displacement due to individual
waves and y is the resultant displacement.
Differential equation of wave motion :
22
2 22
1dy dy
dx v dt
=
Relation between wave velocity and par
ticle velocity :
The equation of a plane progressive wave is
sin()y A t kx= w- ... (i)
The particle velo
city
p
dy
v A cos( t kx)
dt
= = w w- ... (ii)
Slope o
f displacement curve or strain
cos()
dy
Ak t kx
dx
=- w- ... (iii)
Dividing eq
n
.
(ii) by (iii), we get
dy dt
dy dx k
w
=
-
p
v
v since v, wave velocity
dy dx k
wæö
Þ== ç÷
èø-
.
p
dy
vv
dx
Þ =-
i.e., Particle velocity = – wave velocity
× strain.
Particle velocity changes with the time but the wave velocity is
constant in a medium.
Relation between phase difference, path difference and time
difference :
•Phase difference of 2p radian is equivalent to a path
difference l and a time difference of period T.
•Phase difference =
2p
l
× path difference

2
x
p
f=´
l 2
x
l
Þ = ´f
p
•Phase difference
=
2
T
p
× time difference

2
t
T
p
f=´
2
T
tÞ = ´f
p
•Time difference =
T
l
× path difference

T
tx=´
l
xt
T
l
Þ =´
Speed of Transverse Waves :
(i) The speed of transverse waves in solid is given by
v
h
=
r
where h is the modu lus of rigidity of the solid and r is the
density of material.

379. 375Waves
INTERFERENCE OF WAVES
W
hen two waves of equal frequency and nearly equal amplitude
travelling in same direction having same state of polarisation
in medium superimpose, then intensity is different at different
points. At some points intensity is large, whereas at other points
it is nearly zero.
Consider two waves
y1 = A1sin (wt – kx) and
y2 = A2 sin (wt – kx + f)
By principle of superposition
y = y1 + y2 = A sin (wt – kx + d)
where, A
2
= A1
2
+ A2
2
+ 2A1A2 cos f,
and
2
12
sin
tan
cos
A
AA
f
d=
+f
As intensity I µ A
2
So, resultant intensity I = I1 + I2 + 2
12
cosII f
For constructive interference (maximum intensity) :
Phase difference, f = 2np
and path difference = nl
where n = 0, 1, 2, 3, ...
Þ Amax = A1 + A2 and Imax = I1 + I2 + 2
12
II
=
2
12
()II+
For destructive inter
ference (minimum intensity) :
Phase difference, f = (2n + 1)p,
and path difference = ( )
2n1
2
l
- ; where n = 0, 1, 2, 3, ...
Þ A min = A1 – A2 and Imin= I1 + I2 – 212
II
=
2
12
(–)II
Results :
(1
) The ratio of maximum and minimum intensities in any
interference wave form.
2 2
12max 12
min 1212
III AA
I AAII
æö+ æö+
==ç÷ ç÷
-èø-èø
(2) Ave
rage intensity of interference in wave form :
I
av
=
max min
2
II+
Put the value o
f Imax and Imin
orIav = I1 + I2
If A = A1 = A2 andI1 = I2 = I
then Imax = 4I, Imin = 0 and Iav = 2I
(3) Condition of maximum contrast in interference wave form
A1 = A2 and I1 = I2
then Imax = 4I and Imin = 0
For perfect destructive interference we have a maximum
contrast in interference wave form.
Reflection of Waves :
A mechanical wave is reflected and refracted at a boundary
separating two media according to the usual laws of reflection
and refraction.
When sound wave is reflected from a rigid boundary or denser
medium, the wave suffers a phase reversal of p but the nature
does not change i.e., on reflection the compression is reflected
back as compression and rarefaction as rarefaction.
When sound wave is reflected from an open boundary or rarer
medium, there is no phase change but the nature of wave is
changed i.e., on reflection, the compression is reflected back as
rarefaction and rarefaction as compression.
Keep in Memory
(i
) For a wave, v = f l
(ii) The wave velocity of sound in air
E Y RT
v
M
rg
===
rr
(iii) P
article velocity is given by
dt
dx
v
p=. It chan
ges with time.
The wave velocity is the velocity with which disturbances
travel in the medium and is given by
k
v
w
w
=.
(iv) W
hen a wave reflects from denser medium the phase change
is p and when the wave reflects from rarer medium, the
phase change is zero.
(v) In a tuning fork, the waves produced in the prongs is
transverse whereas in the stem is longitudinal.
(vi) A medium in which the speed of wave is independent of the
frequency of the waves is called non-dispersive. For
example air is a non-dispersive medium for the sound waves.
(vii) Transverse waves can propagate in medium with shear
modulus of elasticity e.g., solid whereas longitudinal waves
need bulk modulus of elasticity hence can propagate in all
media solid, liquid and gas.
Energy Transported by a Hormonic Wave Along a String :Kinetic energy of a small element of length dx is
2
t
y
)dx(
2
1
dk ÷
ø
ö
ç
è
æ
¶
¶
m= where m = mas
s per unit length
2221
cos()
2
dk A kx t dxéù= mw -w
ëû
and potential energy stored
2
22211
() cos()
22
y
dU T dx A kx t dx
x
¶æö
= = mw -w
ç÷
¶èø
Ex
ample 1.
The displacement y (in cm) produced by a simple harmonic
wave is given by y = (10/
p) sin (2000 pt – px /17). What will
be the periodic time and maximum velocity of the particles
in the medium?
Solution :
÷
ø
ö
ç
è
æ p
-p
p
=
17
x
t2000sin
10
y or ÷
ø
ö
ç
è
æ
-p
p
=
34
x
t10002sin
10
y
The stan
dard equation of S.H.M. is,
ú
û
ù
ê
ë
é
l
-p=
x
T
t
2sinay ;

380. 376 PHYSICS
\ By comparison of the standard equation to above given
equation,
we get .sec10
1000
1
T
3-
==
Particle velocity
÷
ø
ö
ç
è
æ p
-pp´
p
==
17
x
t2000cos2000
10
dt
dy
s/m200s/cm000,20
dt
dy
max
==÷
ø
ö
ç
è
æ
Þ
(as v
max
= AAw)
Example 2.
A progressive wave of frequency 500 Hz is travelling with
a velocity of 360 m/s. How far apart are two points 60
o
out
of phase?
Solution :
We known that for a wave v = f l
so l =
v 360
f 500
= = 0.72 m
Now as in a wave path difference is related to phase difference by the relation. Phase difference Df = 60
o
= (p/180) x 60 = (p/3) rad
Path difference Dx =
2
l
p
(Df) =
0.72
23
p
p
= 0.12 m
Example 3.
Determine the change in volume of 6 litres of alcohol if the pressure is decreased from 200 cm of Hg to 75 cm. [velocity of sound in alcohol is 1280 m/s, density of alcohol = 0.81 gm/cc, density of Hg = 13.6 gm/cc and g = 9.81 m/s
2
].
Solution :
For propagation of sound in liquid
v = ( )
B/r,i.e.,B = v
2
r
But by definition B = – V
P
V
D
D
so –V
P
V
D
D
= v
2
r, i.e., DV = 2
V( P)
v
-D
r
Here DP = H
2
rg – H
1
rg = (75 – 200) ´ 13.6 ´ 981
= –1.667 ´ 10
6
dyne/cm
2
so DV =
()()
( )
36
2
5
6 10 1.667 10
0.81 1.280 10
´´
´´
= 0.75 cc
Example 4.
(a) Speed of sound in air is 332 m/s at NTP. What will be the speed of sound in hydrogen at NTP if the density of hydrogen at NTP is (1/16) that of air? (b) Calculate the ratio of the speed of sound in neon to that in water vapour at any temperature. [Molecular weight of neon = 2.02
´ 10
–2
kg/mol and for water vapours = 1.8 ´
10
–2
kg/mol]
Solution :
The velocity of sound in air is given by
v =
E
r
=
Pg
r
=
RT
M
g
(a)In terms of density and pressure
H
air
v
v
=
airH
H air
P
P
r
´
r
=
air
H
r
r
[as P
air
= P
H
]
orv
H
= v
air
´
air
H
r
r
= 332 ´
16
1
= 1328 m/s
(b)In terms of temperature and molecular weight
Ne
W
v
v
=
NeW
NeW
M
M
g
´
g
[as T
N
= T
W
]
Now as neon is monatomic (g = 5/3) while water vapours
polyatomic (g = 4/3) so
Ne
W
v
v
=
( )
( )
2
2
5 / 3 1.8 10
4 / 3 2.02 10
-
-
´´
´´
=
5 1.8
4 2.02
´ = 1.05
BEATS
When two wave trains slightly differing in frequencies travel
along the same straight line in the same direction, then the
resultant amplitude is alternately maximum and minimum at a
point in the medium. This phenomenon of waxing and waning
of sound is called beats.
Let two sound waves of frequencies n
1
and n
2
are propagating
simultaneously and in same direction. Then at x=0
y
1
= A sin 2p n
1
t, and y
2
= A sin 2p n
2
t,
For simplicity we take amplitude of both waves to be same. By principle of superposition, the resultant displacement at any instant is
y = y
1
+

y
2
= 2A cos 2p n
A
t sin 2p n
av
t
where
2
nn
n
21
av
+
= ,
2
nn
n
21
A
-
=
Þy = A
beat
sin 2p n
av
t ..................(i)
It is clear from the above expression (i) that

381. 377Waves
(i)A
be
at
= 2A cos 2p n
A
t, amplitude of resultant wave varies
periodically as frequency ÷
ø
ö
ç
è
æ-
=
2
nn
n
21
A
A is maximum whe
n 1|cos|
max
=q A2|A|
maxbeat
=
A is minimum when 0|cos|
min
=q 0|A|
minbeat
=
(ii) Since lntensity is proportional to amplitude i.e.,
2
beat
AIa
For I
max
cos 2p n
A
t = ± 1 For I
min
i.e., 2p n
A
t = 0,p, 2p 2p n
A
t = p/2, 3p/2
i.e., t = 0, 1/2n
A
, 2/2n
A
t = 1/4n
A
, 3/4n
A
.......
So time interval between two consecutive beat is
÷
÷
ø
ö
ç
ç
è
æ
=-=D
-
A
1nn
n2
1
ttt
Number o
f beats per sec is given by 12
12
2()1
2
2
beat A
nn
n n nn
t
-
= = - =-
D
So beat frequen
cy is equal to the difference of frequency of two
interferring waves.
To hear beats, the number of beats per second should not be
more than 10. (due to hearing capabilities of human beings)
Filing/Loading a Tuning Fork
On filing the prongs of tuning fork, raises its frequency and on
loading it decreases the frequency.
(i) When a tuning fork of frequency n produces Dn beats per
second with a standard tuning fork of frequency n
0
, then
nD±n=n
0
If the beat frequency decreases or reduces to zero or remains
the same on filling the unknown fork, then
nD-n=n
0
(ii) If the beat frequency decreases or reduces to zero or remains
the same on loading the unknown fork with a little wax,
then nD+n=n
0
If the beat frequency increases on loading, then
nD-n=n
0
Example 5.
Tuning fork A has frequency 1% greater than that of
standard fork B, while tuning fork C has frequency 2%
smaller than that of B. When A an C are sounded together,
the number of beats heard per second is 5. What is the
frequency of each fork?
Solution :
Let the frequencies of forks be n
1
, n
2
and n
3
respectively.
Then
n
1
= n
2
(1 + 0.01) = 1.01 n
2
and n
3
= n
2
(1 - 0.02) = 0.98 n
2
Further, n
1
– n
3
= 5
Substituting the values, we get (1.01 n
2
– 0.98 n
2
) = 5
\ n
2
= 166.7 Hz
Now n
1
= 1.01 × 166.7 = 168.3 Hz
and n
3
= 0.98 × 166.7 = 163.3 Hz.
Example 6.
Two tuning forks
A and B sounded together give 6 beats
per sec
ond. With an air resonance tube closed at one end,
the two forks give resonance when the two air columns are24 cm and 25 cm respectively. Calculate the frequencies offorks.
Solution :
Let the frequency of the first fork be f
1
and that of second
be f
2
.
We then have,
1
v
f
4 24
=
´
and 2
v
f
4 25
=
´
We also see that
f
1
> f
2
\ f
1
– f
2
= 6 …(i)
and
1
2
f24
f 25
=
…(ii)
Solving
eq
ns
. (i) and (ii), we get
f
1
= 150 Hz and f
2
= 144 Hz
DOPPLER EFFECT
When a source of sound and an observer or both are in motion
relative to each other there is an apparent change in frequency
of sound as heard by the observer. This phenomenon is called
the Doppler's effect .
Apparent change in frequency :
(a)When source is in motion and observer at rest
(i)when source moving towards observer
10
S
V
VV
æö
n =nç÷
-
èø
(ii)when source moving away from observer
10
S
V
VV
æö
¢n =nç÷
+
èø
Here V = velocity of sound
V
S
= velocity of source
n
0
= source frequency.
(b)When source is at rest and observer in motion
(i)when observer moving towards source
0
20
VV
V
+æö
n=n
ç÷
èø
(ii)when observer m oving away from source and
V
0
= velocity of observer.
0
20
VV
V
-æö
¢n=n
ç÷
èø
(c)When source and observer both are in motion
(i)If source and observer both move away from each
other.
0
30
s
VV
VV
æö-
n=nç÷
-
èø

382. 378 PHYSICS
Solution :
In this
case, we can assume that both the source and
observer are moving towards each other with velocity v. If
c be the velocity of signal, then
n
vc
vc
n ÷
ø
ö
ç
è
æ
-
+
=¢ or n
)vc(
)vc()vc(
n
2÷
÷
ø
ö
ç
ç
è
æ
-
-+
=¢
or
222
222
(cv)c
n n n; (as c v)
c v 2c v c 2cv
-
¢¢= Þ = >>
+--
n
v2c
c
n
÷
÷
ø
ö
ç
ç
è
æ
-
=¢\
STATIONARY OR STANDING WAVES
When tw
o progressive waves having the same amplitude, velocity
and time period but travelling in opposite directions
superimpose, then stationary wave is produced.
Let two waves of same amplitude and frequency travel in opposite
direction at same speed, then
y
1
= A sin (wt –kx) and
y
2
= A sin (wt + kx)
By principle of superposition
y = y
1
+ y
2
= (2A cos kx) sin wt ...(i)

s
y A sin
ωt=
It is clear that amplitud
e of stationary wave A
s
vary with position
(a)A
s
= 0, when cos kx = 0 i.e., kx = p/2, 3p/2............
i.e., x = l/4, 3l/4...................[as k = 2p/l]
These points are called nodes and spacing between two
nodes is l/2.
(b)A
s
is maximum, when cos kx is max
i.e., kx = 0, p , 2p, 3p i.e., x = 0, ll/2, 2l/2....
It is clear that antinode (where A
s
is maximum) are also
equally spaced with spacing l/2.
(c) The distance between node and antinode is l/4 (see figure)Antinode Antinode
segment 1segment 2 segment 3
Node
x
l
l
/2
/4
o
2A
Keep in Memory
1. When a strin
g vibrates in one segment, the sound produced
is called fundamental note. The string is said to vibrate in
fundamental mode.
2. The fundamental note is called first harmonic, and is given
by
l2
v
0
=n, where v= speed of wave.
3.
If the fundamental frequency be
0
n then 2
0
n, 3
0
n, 4
0
n
... are respectively called second third, fourth ... harmonicsrespectively.
4. If an instrument produces notes of frequencies
4321
,,,nnnn.... where
1234
νννν<<< ....., then
2
n is
(ii)If source and obse
rver both move towards each other.
0
30
s
VV
VV
æö+
¢n=nç÷
-
èø
;
When the wind blows i
n the direction of sound, then in all above
formulae V is replaced by (V + W) where W is the velocity of wind.
If the wind blows in the opposite direction to sound then V is replaced
by (V – W).
Keep in Memory
1. The m
otion of the listener causes change in number of
waves received by the listener and this produces an apparentchange in frequency.
2. The motion of the source of sound causes change in
wavelength of the sound waves, which produces apparentchange in frequency.
3. If a star goes away from the earth with velocity v, then the
frequency of the light emitted from it changes from n to n'.
n' = n (1–v/c), where c is the velocity of light and
or
vv
cc
Dn Dl
==
nl
where lD is called Doppler’s shi ft.
If wavelength of the observed waves decreases then theobject from which the waves are coming is moving towards
the listener and vice versa.
Example 7.
Two engines cross each other travelling in oppositedirection at 72 km/hour. One engine sounds a whistle offrequency 1088 cps. What are the frequencies as heard by
an observer on the other engine before and after crossing.
Take the speed of sound as 340 m/s.
Solution :
The apparent frequency before crossing,
÷
÷
ø
ö
ç
ç
è
æ
-
+
=¢
s
0
vv
vv
nn
Here V
o
= 72 km/hour =
20 m/s
v
s
= 72 km/hour = 20 m/s
÷
ø
ö
ç
è
æ
-
+
´=¢\
20340
20340
1088n
Hz1224
320 360
1088 =´=
The apparen
t frequency after crossing
0
s
vv
n''n
vv
æö-
=ç÷
+
èø
÷
ø
ö
ç
è
æ
+
-
´=
20340
20340
1088
Hz11.967
36
32
1088 =´=
Example 8.
A rocket i
s going towards moon with a speed v. The
astronaut in the rocket sends signals of frequency n
towards the moon and receives them back on reflection
from the moon. What will be the frequency of the signal
received by the astronaut? (Take v << c)

383. 379Waves
called f
irst overtone,
3
n is called second overtone,
4
n is
called third overtone ... so on.
5. Harmonics are the integral multiples of the fundamental
frequency. If n
0
be the fundamental frequency, then
0
nn is
the frequency of nth harmonic.
6. Overtones are the notes of frequency higher than the
fundamental frequency actually produced by the instrument.
7. In the strings all harmonics are produced.
Stationary Waves in an Organ Pipe :
In the open organ pipe all the harmonics are produced.
In an open organ pipe, the fundamental frequency or first
harmonic is
0
2
v
n=
l
, where v is velocity of sound and l is the
length of air column [see fig. (a)]
(a)
l
(b)
l
2
l
=l,
1
2l
=l
2
2l
=l ,
2
2l
=l
(c)
l
2
3l
=l ,
3
L3
=l
Similarly the
frequency of second harmonic or first overtone is
[see fig (b)],
01
2
2
v
n=
l
Similarly the frequency of third harmonic and second overtone
is, [(see fig. (c)] 02
3
2
v
n=
l
Similarly
03 04
45
,
22
vv
nn==
ll
..................
In the closed organ pipe only the odd harmonics are produced. In
a closed organ pipe, the fundamental frequency (or first
harmonic) is (see fig. a)
4
c
v
n=
l
(a)
l
(b)
l
(c)
l
4
l
=l
4
3l
=l
4
5l
=l
Similarly the
frequency of third harmonic or first overtone (II
nd
harmonic absent) is (see fig. b)
l4
v3
n
2c=
Similarly
ll 4
v7
n,
4
v5
n
4c3c ==........
End Correction
It is o
bserved that the antinode actually occurs a little above the
open end. A correction is applied for this which is known as end
correction and is denoted by e.
(i) For closed organ pipe : l is replaced by l+ e where
e = 0.3D, D is the diameter of the tube.
(ii) For open organ pipe: l is replaced by l + 2e where
e = 0.3D
In resonance tube, the velocity of sound in air given by
()21
2νv ll=-
where n = frequency of tuning fork, l
l
= 1
st
resonating
length, l
2
= 2
nd
resonating length.
Resonance Tube :
It is used to determine velocity of sound in air with the help of a
tuning fork of known frequency.
n n
A
e
l
1
e
l
2
Let l
1
and l
2
are len gths of first and second resonances then
1e
4
l
+=l and 2
3
e
4
l
+=l
21
2
l
Þ -=ll
21
2()Þl=-ll
Speed of sound in air is
v= ul where u is t he frequency
21
2()v=u-ll

384. 380 PHYSICS
For vibrating strings/ open organ pipe.
Mode of
vibration
First or
Fundamental
FirstFundamental
tone
2l
Second Second First
overtone
Second
overtone
3n
2n
Third Third
Har
monic Tone FrequencyWavelength Shape for string
N N
N
A
A
N
NA
v
2l
n =
2
3
l
2
2
l
For closed organ pipe.
Mode of
vibration
First or
Fundamental
First Fundamental
tone
4l
Second Third First
overtone
Second
overtone
5n
3n
Third fifth
Harmonic To
ne FrequencyWavelength Shape for string
v
4l
n =
4
3
l
4
5
l
Comparision of Progressive (or travelling) and Stationary (or standing) Wave:
SI. Progressive Wave Stationary Wave
1. The wave advances with a constant speed The wave does not advance but remains confined
in a particular region.
2. The amplitude is the same for all the particles in
the path of the wave
The amplitude varies according to position, being
zero at the nodes and maximum at the antinodes.
3. All particles within one wavelength have different
phases.
Phase of all particles between two adjacent nodes is
the same. Particles in adjacent segments of length
p/2 have opposite phase.
4. Energy is transmitted in the direction of propagation
of the wave
Energy is associated with the wave, but there is no
transfer of energy across any section of the medium.

385. 381Waves
Comparativ
e Study of Interference, Beats and Stationary Wave:
Interference
Superposition of two waves
of nearly same amplitude or
same amplitude having a
constant phase difference (or
no phase differe
nce)
travelling in the same
direction and with the same
wavelength show the
phenomenon of interference.
Maxima and minima are
fixed at their locations.
Beats
Superposition of two waves
of same amplitude having
zero phase difference
moving in the same direction
having small difference in
frequency (less than 10 Hz)
show the phenomenon of
beats.
Maxima and minima vary
periodically with time at
every location.
Stationary wave
Superposition of two waves
of same amplitude having a
constant phase difference (or
no phase difference) moving in opposite direction having
same frequency give rise to
stat
ionary waves.
Maxima and minima are
fixed at their locations.
CHARACTERISTICS OF SOUND
1.A musical sound consists of quick, regular and periodic
succession of compressions and rarefactions without a
sudden change in amplitude.
2. A noise, consists of slow, irregular and a periodic succession
of compressions and rarefactions, that may have sudden
changes in amplitude.
3.(i) Pitch, (ii) loudness and (iii) quality are the characteristics
of musical sound.
4.Pitch depends on frequency, loudness depends on intensity
and, quality depends on the number and intensity of
overtones.
5. The ratio of the frequencies of the two notes is called the
interval between them. For example interval between two
notes of frequencies 512 Hz and 1024 Hz is 1 : 2 (or 1/2).
6. Two notes are said to be in unison if their frequencies are
equal, i.e., if the interval between them is 1 : 1.
Some other common intervals, found useful in producing
musical sound are the following:
Octave (1 : 2), majortone (8 : 9), minortone (9 : 10) and
semitone (15 : 16)
7. Major diatonic scale : It consists of eight notes. The
consecutive notes have either of the following three
intervals. They are 8 : 9 ; 9 : 10 and 15 : 16.
Acoustics : The branch of physics that deals with the process of
generation, reception and propagation of sound is called
acoustics.
Acoustics may be studied under the following three subtitles.
(a)Electro acoustics. This branch deals with electrical sound
production with music.
(b)Musical acoustics. This branch deals with the relationship
of sound with music.
(c)Architectural acoustics. This branch deals with the design
and construction of buildings.
REVERBERATION
Multiple reflections which are responsible for a series of waves
falling on listener’s ears, giving the impression of a persistence
or prolongation of the sound are called reverberations.
The time gap between the initial direct note and the reflected note
upto the minimum audibility level is called reverberation time.
Sabine Reverberation Formula for Time
Sabine established that the standard period of reverberation viz.,
the time that the sound takes to fall in intensity by 60 decibels or
to one millionth of its original intensity after it was stopped, is
given by
0.05
ii
V
T
S
=
åa
where V = volu
me of room,
ii
Saå = a
1
S
1
+ a
2
S
2
+ ....
S
1
, S
2
.... are different kinds of surfaces of room and
a
1
, a
2
.... are their respective absorption coefficient.
The above formula was derived by Prof C. Sabine.
Shock waves : The waves produced by a body moving with a
speed greater than the speed of sound are called shock waves.
These waves carry huge amount of energy. It is due to the shock
wave that we have a sudden violent sound called sonic boom
when a supersonic plane passes by.
The rate of speed of the source to that of the speed of sound is
called mach number.
Intensity of sound : The sound intensities that we can hear range
from 10
–12
Wm
–2
to 10
3
Wm
–2
. The intensity level b, measured
in terms of

decibel (dB) is defined as
0
10log
I
I
b=
where
I = measured intensity, I
0
= 10
–12
Wm
–1
At the threshold b = 0
At the max
dB120
10
1
log10
12
==b
-
Lissajo
us Figures: when two simple harmonic waves having
vibrations in mutually perpendicular directions superimpose
on each other, then the resultant motion of the particle is along
a closed path, called the Lissajous figures. These figures can be
of many shapes depending on
(i) ratio of frequencies or time periods of two waves
(ii) ratio of amplitude of two waves
(iii) phase difference between two waves.

386. 382 PHYSICS
Example 9.
The eq
uation of stationary wave in stretched string is
given by y = 5 sin
t40cos
3
x
p
p
where x and y are in cm
and t in second.
Find the separation between two adjacent
nodes.
Solution :
Given that
t40cos
3
sin5y p
p
=
x
We know that
22
y a sin cos vtx
pp
=
ll
6or
3
2
=l
p
=
l
p
\
Now distance between t
wo adjacent nodescm3
2
6
2
==
l
=
Example 10.
For a certain or
gan pipe, three successive resonance
frequencies are observed at 425, 595 and 765 Hz,
respectively. Taking the speed of sound in the air to be 340
metre per second, (i) explain whether the pipe is closed at
one end or open at both the ends and (ii) determine the
fundamental frequency and the length of the pipe.
Solution :
(i) Given ratio of successive frequencies
= 425 : 595 : 765 = 5 : 7 : 9.
So, the harmonics are odd harmonics. Hence the pipe
is closed at one end.
(ii) Fundamental frequency n is given by,
85
5
425
n ==
We known tha
t
n4
v
or
4
v
n == l
l
for closed organ
pipe
at one end.
\ Length of the pipe
metre1
854
340
=
´
=l
Example 11.
The fundamental f
requency of an open organ pipe is 300
Hz. The first overtone of this pipe has same frequency as
first overtone of a closed organ pipe. If speed of sound is
330 m/s, then find the length of closed organ pipe.
Solution :
For open pipe,
v
n
2
=
l
v 330 11
2n 2 300 20
\===
´
l
As frequency of fi
rst overtone of open pipe = frequency of
1st overtone of closed pipe
vv
23
24
\=
¢ll
cm25.41
20
11
4
3
4
3
=´==¢
l
l
Example 12.
String wires of same material of length
l and 2 l vibrate
with frequencies 100 and 150 respectively. Determine the
ratio of their tensions.
Solution :
As
m
T
2
1
n
l
=
m
T
2
1
100
1
l
=\ and
m
T
4
1
150
2
l
=
Divide
2
1
T
T
2
4
150
100
l
l
= or
2
1
T
T
2
3 2
=
9
1
T
T
or
3
1
T
T
2
1
2
1
==\ = 1 : 9

387. 383Waves

388. 384 PHYSICS
1.Frequenc
ies of sound produced from an organ pipe open
at both ends are
(a) only fundamental note (b) only even harmonics
(c) only odd harmonics (d) even and odd harmonics
2.The fundamental frequency of an organ pipe is 512 Hz. If its
length is increased, then frequency will
(a) decrease (b) increase
(c) remains same (d) cannot be predicted
3.The property of a medium necessary for wave propagation is
(a) inertia (b) elasticity
(c) low resistance (d) All of the above
4.Doppler’s effect is not applicable for
(a) audio waves (b) electromagnetic waves
(c) shock waves (d) None of these
5.Two sound waves of equal intensity I produce beats. The
maximum intensity of sound produced in beats will be
(a)I (b) 2I (c)3I (d) 4I
6.Two sinusoidal plane waves of same frequency having
intensities I
0
and 4 I
0
are travelling in the same direction.
The resultant intensity at a point at which waves meet with
a phase difference of zero radian is
(a)I
0
(b) 5 I
0
(c) 9 I
0
(d) 3 I
0
7.If the intensities of two interfering waves be I
1
and I
2
, the
contrast between maximum and minimum intensity is
maximum, when
(a)I
1
> > I
2
(b) I
1
<< I
2
(c)I
1
= I
2
(d) either I
1
or I
2
is zero
8.Sound waves of length l travelling with velocity v in a
medium enter into another medium in which their velocity
is 4 v. The wavelength in 2nd medium is
(a)
4λ (b)l
(c)l/4 (d) 16 l
9.Two per
iodic waves of intensities I
1
and I
2
pass through a
region at the same time in the same direction. The sum of
the maximum and minimum intensities is
(a) 2 (I
1
+ I
2
) (b) I
1
+ I
2
(c)
2
21
)II(+ (d)
2
21
)II(-
10.An open and clo
sed organ pipe have the same length. The
ratio of pth mode of frequency of vibration of two pipes is
(a)1 (b) p
(c) p (2p + 1) (d)
)1p2(
p2
-
11.Sound waves are not tr
ansmitted to long distances
because,
(a) they are absorbed by the atmosphere
(b) they have constant frequency
(c) the height of antenna required, should be very high
(d) velocity of sound waves is very less
12.What is the effect of increase in temperature on the
frequency of sound produced by an organ pipe?
(a) increases (b) decreases
(c) no effect (d) erratic change
13.Shock waves are produced by objects
(a) carrying electric charge and vibrating
(b) vibrating with frequency greater than 20000 Hz
(c) vibrating with very large amplitude
(d) moving with a speed greater than that of sound in the
medium
14.The speed of sound in a medium depends on
(a) the elastic property but not on the inertia property
(b) the inertia property but not on the elastic property
(c) the elastic property as well as the inertia property
(d) neither the elastic property nor the inertia property
15.Consider the three waves z
1
, z
2
and z
3
as
z
1
= A sin (kx – wt)
z
2
= A sin (kx + wt)
z
3
= A sin (ky – wt)
Which of the following represents a standing wave?
(a)z
1
+ z
2
(b) z
2
+ z
3
(c)z
3
+ z
1
(d) z
1
+ z
2
+ z
3
16.Each of the properties of sound listed in column A primarily
depends on one of the quantitites in column B. Choose the
matching pairs from two columns
Column A Column B
Pitch Waveform
Quality Frequency
Loudness Intensity
(a) Pitch-wave form; Quality-frequency; Loudness-
intensity
(b) Pitch-frequency; Quality-wave form; Loudness-
intensity
(c) Pitch-intensity; Quality-wave form; Loudness-
frequency
(d) Pitch-wave form; Quality-intensity; Loudness-
frequency
17.Three transverse waves are represented by
)tkx(cosAy
1 w-=
)tkx(cosAy
2 w+=
)tky(cosAy
3 w-=
The combination of waves which can produce stationary
waves is(a)y
1
and y
2
(b) y
2
and y
3
(c)y
1
and y
3
(d) y
1
, y
2
and y
3
18.Which of the following changes at an antinode in a
stationary wave?
(a) Density only
(b) Pressure only
(c) Both pressure and density
(d) Neither pressure nor density

389. 385Waves
19.Sound wav
es are travelling in a medium whose adiabatic
elasticity is E and isothermal elasticity E'. The velocity of
sound waves is proportional to
(a) E' (b)
E
(c) 'E (d)
'E
E
20.The equation o
f a cylindrical progressive wave is
(a) tsinayw= (b) )krt(sinay-w=
(c)
a
y sin( t kr)
r
= w- (d) )krt(sin
r
a
y -w=
21.A wave y =
a sin (wt – kx) on a string meets with another
wave producing a node at x = 0. Then the equation of the
unknown wave is
(a) y = a sin (wt + kx)(b) y = –a sin (wt + kx)
(c) y = a sin (wt – kx)(d) y = –a sin (wt – kx)
22.A sound wave of frequency f travels horizontally to the
right. It is reflected from a large vertical plane surface moving
to left with a speed v. The speed of sound in medium is c.
Then
(a) the number of wave striking the surface per second is
(c v)
f
c
+
(b) the wavelength of reflected wave is
c(c v)
f (c v)
+
+
(c) the frequency of the reflected wave is
(c v)
f
(c v)
-
-
(d) the number of beats heard by a stationary listener to
the left of the reflecting surface is
vf
cv-
23.The equation )xvt(
2
sinay-
l
p
= is expression for
(a) stationary wave of single frequency along x-axis
(b) a simple harmonic motion
(c) a progressive wave of single frequency along x-axis
(d) the resultant of two SHMs of slightly different
frequencies
24.A whistle S of frequency f revolves in a circle of radius R at
a constant speed v. What is the ratio of largest and smallest
frequency detected by a detector D at rest at a distance 2R
from the centre of circle as shown in figure ?
(take c as speed of sound)
D
2R
R
S
(a)
cv
cv
+æö
ç÷
èø-
(b)
cv
2
cv
+æö
ç÷
èø-
(c)2 (d)
(c v)
c2
+
25.When a string is divided into three segments of length l
1
,
l
2
, and l
3
the fundamental frequencies of these three
segments are v
1
, v
2
and v
3
respectively. The original
fundamental frequency (v) of the string is
(a)
123
vvvv=++
(b)v = v
1
+ v
2
+ v
3
(
c)
123
1111
vvvv
=++
(d)
123
1111
vvvv
=++
1.The equatio n of a plane progressive wave is
ú
û
ù
ê
ë
é
-p=
2
x
t4sin9.0y
. When it is reflected at a rigid
support, its amplitude becomes
3
2
of its pr
evious value.
The equation of the reflected wave is
(a)
ú
û
ù
ê
ë
é
+p=
2
x
t4sin6.0y
(b)
ú
û
ù
ê
ë
é
+p-=
2
x
t4sin6.0y
(c)
ú
û
ù
ê
ë
é
-p-=
2
x
t8sin9.0y
(d) ú
û
ù
ê
ë
é
+p-=
2
x
t4sin6.0y
2.If two waves of s
ame frequency and same amplitude, on
superposition, produce a resultant disturbance of the same
amplitude, the wave differ in phase by
(a)p (b) 2 p/3
(c) Zero (d)p/3
3.Two tones of frequencies n
1
and n
2
are sounded together.
The beats can be heard distinctly when
(a) 10 < (n
1
– n
2
) < 20 (b) 5 < (n
1
– n
2
) > 20
(c) 5 < (n
1
– n
2
) < 20(d) 0 < (n
1
– n
2
) < 10
4.The equation of a spherical progressive wave is
(a) tsinayw= (b) )krt(sinay-w=
(c)
)krt(sin
2
a
y -w=(d) )krt(sin
r
a
y -w=

390. 386 PHYSICS
5.The velo
city of sound in air is 330 m/s. The r.m.s. velocity
of air molecules (g = 1.4) is approximately equal to
(a) 400 m/s (b) 471.4 m/s
(c) 231 m/s (d) 462 m/s
6.The velocity of sound in hydrogen is 1224 m/s. Its velocity
in a mixture of hydrogen and oxygen containing 4 parts by
volume of hydrogen and 1 part oxygen is
(a) 1224 m/s (b) 612 m/s
(c) 2448 m/s (d) 306 m/s
7.A person standing symmetrically between two cliffs claps
his hands and starts hearing a series of echoes at intervals
of 1 sec. If speed of sound in air is 340 m/s, the distance
between the parallel cliffs must be
(a) 340 m (b) 680 m
(c) 1020 m (d) 170 m
8.An echo repeats two syllables. If the velocity of sound is
330 m/s, then the distance of the reflecting surface is
(a) 66.0 m (b) 33.0 m
(c) 99.0 m (d) 16.5 m
9.At room temperaturre, velocity of sound in air at 10
atmospheric pressure and at 1 atmospheric pressure will be
in the ratio
(a) 10 : 1 (b) 1 : 10
(c) 1 : 1 (d) cannot say
10.A series of ocean waves, each 5.0 m from crest to crest,
moving past the observer at a rate of 2 waves per second,
what is the velocity of ocean waves?
(a) 2.5 m/s (b) 5.0 m/s
(c) 8.0 m/s (d) 10.0 m/s
11.If the ratio of maximum to minimum intensity in beats is 49,
then the ratio of amplitudes of two progressive wave trains
is
(a) 7 : 1 (b) 4 : 3
(c) 49 : 1 (d) 16 : 9
12.There are three sources of sound of equal intensities and
frequencies 400, 401 and 402 vibrations per second. The
number of beats/sec is
(a)0 (b) 1
(c)3 (d) 2
13.A fork of unknown frequency gives four beats/sec when
sounded with another of frequency 256. The fork is now
loaded with a piece of wax and again four beats/sec are
heard. Then the frequency of the unknown fork is
(a) 256 Hz (b) 252 Hz
(c) 264 Hz (d) 260 Hz
14.If there are six loops for 1 m length in transverse mode of
Melde’s experiment., the no. of loops in longitudinal mode
under otherwise identical conditions would be
(a)3 (b) 6
(c) 12 (d) 8
15.The apparent wavelength of the light from a star moving
away from the earth is 0.2% more than its actual wavelength.
Then the velocity of the star is
(a) 6 × 10
7
m/sec (b) 6 × 10
6
m/sec
(c) 6 × 10
5
m/sec (d) 6 × 10
4
m/sec
16.In a resonance column, first and second resonance are
obtained at depths 22.7 cm and 70.2 cm. The third resonance
will be obtained at a depth
(a) 117.7 cm (b) 92.9 cm
(c) 115.5 cm (d) 113.5 cm
17.A fork of frequency 256 Hz resonates with a closed organ
pipe of length 25.4 cm. If the length of pipe be increased by
2 mm, the number of beats/sec. will be
(a)4 (b) 1
(c)2 (d) 3
18.A distant star which is moving away with a velocity of 10
6
m/sec is emitting a red line of frequency 4.5 × 10
14
Hz. The
observed frequency of this spectral line is
(a) 4.5 × 10
8
Hz (b) 4.485 × 10
14
Hz
(c) 4.515 × 10
14
Hz (d) 4.5 × 10
14
Hz
19.The speed of sound in air under ordinary conditions is
around 330 m s
–1
. The speed of sound in hydrogen under
similar conditions will be (in m s
–1
) nearest to
(a) 330 (b) 1200
(c) 600 (d) 900
20.A wave disturbance in a medium is described by
)x10cos(
2
t50cos02.0)t,x(y p÷
ø
ö
ç
è
æ p
+p= where x and y are
in metre and t is in second. Which of the following is correct?
(a) A node occurs at x = 0.15 m
(b) An antinode occurs at x = 0.3 m
(c) The speed wave is 5 ms
–1
(d) The wavelength is 0.3 m
21.Two waves represented by tsinay
1
w= and
)tsin(ay
2
f+w= with
2
p
=f are super
posed at any point
at a particular instant. The resultant amplitude is(a)a (b) 4a
(c)
2a (d) zero
22.If the sp
eed of a transverse wave on a stretched string of
length 1 m is 60 m/s, what is the fundamental frequency of
vibration?
(a) 10 Hz (b) 30 Hz
(c) 40 Hz (d) 70 Hz
23.The equation of a progressive wave is
ú
û
ù
ê
ë
é
-p=
30.0
x
01.0
t
2sin02.0y
Here
x and y are in metre and t is in second. The velocity of
propagation of the wave is
(a) 300 m s
–1
(b) 30 m s
–1
(c) 400 m s
–1
(d) 40 m s
–1
24.The equation Y = 0.02 sin (500pt) cos (4.5 x) represents
(a) progressive wave of frequency 250 Hz along x-axis
(b) a stationary wave of wavelength 1.4 m
(c) a transverse progressive wave of amplitude 0.02 m
(d) progressive wave of speed of about 350 m s
–1
25.The intensity level of sound wave is said to be 4 decibel. If
the intensity of wave is doubled, then the intensity level of
sound as expressed in decibel would be
(a)8 (b) 16
(c)7 (d) 14

391. 387Waves
26.The equ
ation )tkx(sinAy
2
w-= represents a wave with
(a) amplitude A, frequency pw2/
(b) amplitude A/2, frequency pw/
(c) amplitude 2A, frequency pw4/
(d) it does not represent a wave motion
27.A toothed wheel is rotating at 240 rpm and a post card is
held against the teeth. If the pitch of the tone is 256 Hz,
then the number of teeth on the rotating wheel is
(a) 256 (b) 128
(c) 64 (d) 32
28.When a tuning fork vibrates with 1.0 m or 1.05 m long wire
of a sonometer, 5 beats per second are produced in each
case. What will be the frequency of the tuning fork?
(a) 195 (b) 200
(c) 205 (d) 210
29.The speed of sound in oxygen (O
2
) at a certain temperature
is 460 ms
–1
. The speed of sound in helium (He) at the same
temperature will be (assume both gases to be ideal)
(a) 460 ms
–1
(b) 500 ms
–1
(c) 650 ms
–1
(d) 330 ms
–1
30.A closed pipe and an open pipe have their first overtones
identical in frequency. Their lengths are in the ratio
(a) 1 : 2 (b) 2 : 3
(c) 3 : 4 (d) 4 : 5
31.The threshold of hearing for the human ear is 10
–12
W m
–2
.
This is taken as the standard level. The intensity of sound
is 1 Wm
–2
. It has intensity (in dB).
(a) 10
12
dB (b) 12 dB
(c) 240 dB (d) 120 dB
32.In expressing sound intensity, we take 10
–12
W m
–2
as the
reference level. For ordinary conversation, the intensity
level is about 10
–6
W m
–2
. Expressed in decibel, this is
(a) 10
6
(b) 6
(c) 60 (d) log
e
(10
6
)
33.If two tuning forks A and B are sounded together, they
produce 4 beats per sec. A is then slightly loaded with wax
and same no. of beats/sec. are produced again. If frequency
of A is 256, the frequency of B would be
(a) 250 (b) 262
(c) 252 (d) 260
34.A uniform wire of length 20 m and weighing 5 kg hangs
vertically. If g = 10 m/s
2
, then the speed of transverse waves
in the middle of the wire is
(a) 10 m/s (b)
210 m/s
(c) 4 m/s (d) zero
35.Th
e fundamental frequency of an open organ pipe is 300
Hz. The first overtone of this pipe has same frequency as
first overtone of a closed organ pipe. If speed of sound is
330 m/s, then the length of closed organ pipe is
(a) 41 cm (b) 37 cm
(c) 31 cm (d) 80 cm
36.A pipe closed at one end produces a fundamental note of
412 Hz. It is cut into two pieces of equal length. the
fundamental frequencies produced by the two pieces are
(a) 206 Hz, 412 Hz(b) 824 Hz, 1648 Hz
(c) 412 Hz, 824 Hz (d) 206 Hz, 824 Hz
37.A train has just completed U-curve in a track which is a semi-
circle. The engine is at the forward end of the semi-circular
part of the track while the last carriage is at the rear end of the
semi circular track. The driver blows a whistle of frequency
200 Hz. Velocity of sound is 340 m/sec Then the apparent
frequency as observed by a passenger in the middle of the
train, when the speed of the train is 30 m/sec is
(a) 181 Hz (b) 200 Hz
(c) 188 Hz (d) 210 Hz
38.The displacement of a particle varies according to the
relation x 4(cos
πt sinπt).=+ The amplitud e of the
particle is
(a) – 4 m (b) 4 m
(c)42m (d) 8 m
39.Two w
aves of wavelengths 99 cm and 100 cm both travelling
with velocity 396 m/s are made to interfere. The number ofbeats produced by them per second is
(a)1 (b) 2
(c)4 (d) 8
40.A whistle of frequency 1000 Hz is sounded on a car travelling
towards a cliff with velocity of 18 m s
–1
normal to the cliff.
If c = 330 m s
–1
, then the apparent frequency of the echo as
heard by the car driver is nearly
(a) 1115 Hz (b) 115 Hz
(c) 67 Hz (d) 47.2 Hz
41.An organ pipe open at one end is vibrating in first overtone
and is in resonance with another pipe open at both ends
and vibrating in third harmonic. The ratio of length of two
pipes is
(a) 1 : 2 (b) 4 : 1
(c) 8 : 3 (d) 3 : 8
42.An organ pipe, open from both end produces 5 beats per
second when vibrated with a source of frequency 200 Hz.
The second harmonic of the same pipes produces 10 beats
per second with a source of frequency 420 Hz. The
fundamental frequency of organ pipe is
(a) 195 Hz (b) 205 Hz
(c) 190 Hz (d) 210 Hz
43.A progressive sound wave of frequency 500 Hz is travelling
through air with a speed of 350 ms
–1
. A compression
maximum appears at a place at a given instant. The minimum
time interval after which the rarefraction maximum occurs at
the same point, is
(a) 200 s (b)
s
250
1
(c) s
500
1
(d) s
1000
1
44.When a sound wave of frequency 300 Hz passes through a
medium, the maximum displacement of a particle of the
medium is 0.1 cm. The maximum velocity of the particle is
equal to
(a) 60p cms
–1
(b) 30p cms
–1
(c) 30 cms
–1

1
scm30
-
(d) 60 cms
–1
45.An engine running at speed
10
v
sounds a whistle of
frequency 600 Hz. A passenger in a train coming from the
opposite side at speed
15
v
experiences this whistle to be of
frequency f. If v is speed of sound in air and there is no
wind, f is near to
(a) 710 Hz (b) 630 Hz
(c) 580 Hz (d) 510 Hz

392. 388 PHYSICS
46.A source o
f sound produces waves of wavelength 60 cm
when it is stationary. If the speed of sound in air is 320 m s
–
1
and source moves with speed 20 m s
–1
, the wavelength of
sound in the forward direction will be nearest to
(a) 56 cm (b) 60 cm
(c) 64 cm (d) 68 cm
47.A person carrying a whistle emitting continuously a note
of 272 Hz is running towards a reflecting surface with a
speed of 18 km h
–1
. The speed of sound in air is 345 m s
–1
.
The number of beats heard by him is
(a)4 (b) 6
(c)8 (d) zero
48.In the sonometer experiment, a tuning fork of frequency
256 Hz is in resonance with 0.4 m length of the wire when
the iron load attached to free end of wire is 2 kg. If the load
is immersed in water, the length of the wire in resonance
would be (specific gravity of iron = 8)
(a) 0.37 m (b) 0.43 m
(c) 0.31 m (d) 0.2 m
49.A longitudinal wave is represented by
÷
ø
ö
ç
è
æ
l
-p=
x
nt2sinxx
0
The maximum particle
velocity will be four times the wave
velocity if
(a)
4
x
0p
=l (b)
0
x2p=l
(c)
2
x
0
p
=l (d)
0
x4p=l
50.Two strings A and B
, made of same material, are stretched
by same tension. The radius of string A is double of radius
of B. A transverse wave travels on A with speed v
A
and on
B with speed v
B
. The ratio v
A
/ v
B
is
(a) 1/2 (b) 2
(c) 1/4 (d) 4
51.A sonometer wire of length l vibrates in fundamental mode
when excited by a tuning fork of frequency 416 Hz. If the
length is doubled keeping other things same, the string will
(a) vibrate with a frequency of 416 Hz
(b) vibrate with a frequency of 208 Hz
(c) vibrate with a frequency of 832 Hz
(d) stop vibrating
52.A source of sound is travelling at
3
100
m s
–1
along a road,
towar
ds a point A. When the source is 3 m away from A, a
person standing at a point O on a road perpendicular to thetrack hears a sound of frequency n'. The distance of O from
A at that time is 4 m. If the original frequency is 640 Hz, thenthe value of n' is (given : velocity of sound = 340 m s
–1
)
(a) 620 Hz
q
S
100
3
ms
–1
3 mA
O
4 m
(b) 680 Hz
(c) 720 H
z
(d) 840 Hz
53.The equation of a stationary wave is :
)t96(cos
15
x
sin4y p÷
ø
ö
ç
è
æp
= . The distan ce between a node
and its next antinode is
(a) 7.5 units (b) 1.5 units
(c) 22.5 units (d) 30 units
54. A wave travelling along the x-axis is described by the
equation y(x, t) = 0.005 cos (a x – bt). If the wavelength and
the time period of the wave are 0.08 m and 2.0s, respectively,
then a and b in appropriate units are
(a) a = 25.00 p , b = p(b)
0.08 2.0
,a= b=
pp
(c)
0.04 1.0
,a= b=
pp
(d)12.50 ,
2.0
p
a= pb=
55.Where
should the two bridges be set in a 110cm long wire
so that it is divided into three parts and the ratio of thefrequencies are 3 : 2 : 1 ?(a) 20cm from one end and 60cm from other end
(b) 30cm from one end and 70cm from other end
(c) 10cm from one end and 50cm from other end
(d) 50cm from one end and 40cm from other end
56.A source X of unknown frequency produces 8 beats per
second with a source of 250 Hz and 12 beats per second
with a source of 270 Hz. The frequency of the source X is
(a) 242 Hz (b) 258 Hz
(c) 282 Hz (d) 262 Hz
57.An isotropic point source S of sound emits constant power.
Two points A and B separated by a distance r are situated
near the source as shown in figure. The difference of the
intensity level of sound at the points A and B is about
r r
S A B
(a) 3 dB (b) 2 dB
(c) 6 dB (d) 12 dB
58.A thick
uniform rope of length L is hanging from a rigid
support. A transverse wave of wavelength l
0
is set up at
the middle of rope as shown in figure. The wavelength ofthe wave as it reaches to the topmost point is /////////////////
L
(a)2l
0
(b)
0
2l (c)
0
2
l
(d)l
0
59.Wave pulse on a strin
g shown in figure is moving to the
right without changing shape. Consider two particles at
positions x
1
= 1.5m and x
2
= 2.5m. Their transverse velocities
at the moment shown in figure are along directions
1 2 3 4 5 6
x(m)
y
v
(a) positive y
-axis and positive y-axis respectively
(b) negative y-axis and positive y-axis respectively
(c) positive y-axis and negative y-axis respectively
(d) negative y-axis and negative y-axis respectively

393. 389Waves
60.The tr
ansverse wave represented by the equation
( )t15x3sin
6
sin4y -÷
ø
ö
ç
è
æp
=
has
(a) amplitu
de = 4
(b) wavelength =
3
4
p
(c) spee
d of propagation = 5
(d) period =
15
p
61.An org
an pipe P
1
closed at one end vibrating in its first
overtone and another pipe P
2
open at both ends vibrating
in third overtone are in resonance with a given tuning fork.
The ratio of the length of P
1
to that of P
2
is
(a) 8/3 (b) 3/8
(c) 1/2 (d) 1/3
62.A car is moving towards a high cliff. The car driver sounds
a horn of frequency f. The reflected sound heard by the
driver has as frequency 2f. If v be the velocity of sound,
then the velocity of the car, in the same velocity units, will
be
(a) v /2 (b) v /Ö2
(c) v /3 (d) v /4
63.The phase difference between two waves, represented by
y
1
= 10
–6
sin{100 t + (x/50) + 0.5} m
y
2
= 10
–6
cos{100 t + (x/50)} m
where x is expressed in metres and t is expressed in
seconds, is approximately
(a) 1.5 radians (b) 1.07 radians
(c) 2.07 radians (d) 0.5 radians
64.Two vibrating tuning forks produce progressive waves
given by Y
1
= 4 sin 500 pt and Y
2
= 2 sin 506 pt. Number of
beats produced per minute is
(a) 360 (b) 180
(c) 60 (d) 3
65.A transverse wave is represented by y = A sin (
wt – kx).
For what
value of the wavelength is the wave velocity equal
to the maximum particle velocity?
(a)
2
Ap
(b)pA (c)2pA (d) A
66.A tun
ing fork of freqqency 512 Hz makes 4 beats per second
with the vibrating string of a piano. The beat frequency
decreases to 2 beats per sec when the tension in the piano
string is slightly increased. The frequency of the piano
string before increasing the tension was
(a) 510 Hz(b) 514 Hz(c) 516 Hz(d) 508 Hz
67.Two waves are represented by the equations
y
1
= a sin (wt + kx + 0.57) m and y
2
= a cos (wt + kx) m, where
x is in meter and t in sec. The phase difference between
them is
(a) 1.0 radian (b) 1.25 radian
(c) 1.57 radian (d) 0.57 radian
68.Sound waves travel at 350 m/s through a warm air and at
3500 m/s through brass. The wavelength of a 700 Hz
acoustic wave as it enters brass from warm air
(a) decreases by a factor 10
(b) increases by a factor 20
(c) increases by a factor 10
(d) decreases by a factor 20
69.Two identical piano wires kept under the same tension T
have a fundamental frequency of 600 Hz. The fractional
increase in the tension of one of the wires which will lead to
occurrence of 6 beats/s when both the wires oscillate
together would be
(a) 0.02 (b) 0.03 (c) 0.04 (d) 0.01
70.Velocity of sound waves in air is 330 m/s. For a particular
sound wave in air, a path difference of 40 cm is equivalent
to phase difference of 1.6p. The frequency of this wave is
(a) 165 Hz (b) 150 Hz
(c) 660 Hz (d) 330 Hz
71.Two sources of sound placed close to each other are
emitting progressive waves given by y
1
= 4 sin 600 pt and
y
2
= 5 sin 608 pt. An observer located near these two sources
of sound will hear
(a) 4 beats per second with intensity ratio 25 : 16 between
waxing and waning.
(b) 8 beats per second with intensity ratio 25 : 16 between
waxing and waning
(c) 8 beats per second with intensity ratio 81 : 1 between
waxing and waning
(d) 4 beats per second with intensity ratio 81 : 1 between
waxing and waning
Directions for Qs. (72 to 75) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following-
(a)Statement -1 is false, Statement-2 is true
(b)Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c)Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d)Statement -1 is true, Statement-2 is false
72. Statement 1 : All points on a wavefront vibrate in same
phase with same frequency.
Statement 2 : Two sources are said to be coherent if they
produce waves of same frequency with a constant phase
difference.
73. Statement 1 : A transverse waves are produced in a very
long string fixed at one end. Only progressive wave is
observed near the free end.
Statement 2 : Energy of reflected wave does not reach the
free end.
74. Statement 1 : Two waves moving in a uniform string having
uniform tension cannot have different velocities.
Statement 2 : Elastic and inertial properties of string are
same for all waves in same string. Moreover speed of wave
in a string depends on its elastic and inertial properties only.
75. Statement 1 : Doppler formula for sound wave is symmetric
with respect to the speed of source and speed of observer.
Statement 2 : Motion of source with respect to stationary
observer is not equivalent to the motion of an observer
with respect to stationary source.

394. 390 PHYSICS
Exemplar Questions
1.Water waves produced by a motorboat sailing in water are
(a) neither longitudinal nor transverse
(b) both longitudinal and transverse
(c) only longitudinal
(d) only transverse
2.Sound waves of wavelength l travelling in a medium with a
speed of v m/ s enter into another medium where its speed
in 2v m/s. Wavelength of sound waves in the second
medium is
(a)l (b)
2
l
(c)2l (d) 4 l
3.Speed of sound wave in
air
(a) is independent of temperature
(b) increases with pressure
(c) increases with increase in humidity
(d) decreases with increase in humidity
4.Change in temperature of the medium changes
(a) frequency of sound waves
(b) amplitude of sound waves
(c) wavelength of sound waves
(d) loudness of sound waves
5.With propagation of longitudinal waves through a medium,
the quantity transmitted is
(a) matter
(b) energy
(c) energy and matter
(d) energy, matter and momentum
6.Which of the following statements are true for wave motion?
(a) Mechanical transverse waves can propagate through
all mediums
(b) Longitudinal waves can propagate through solids only
(c) Mechanical transverse waves can propagate through
solids only
(d) Longitudinal waves can propagate through vacuum
7.A sound wave is passing through air column in the form of
compression and rarefaction. In consecutive compressions
and rarefactions,
(a) density remains constant
(b) Boyle's law is obeyed
(c) bulk modulus of air oscillates
(d) there is no transfer of heat
8.Equation of a plane progressive wave is given by
0.6sin2.
2
x
yt
æö
= p-
ç÷
èø
On reflection from a denser medium
its amplitude becomes
2
3
of the amplitude of the incident
wave. The equation of the reflected wave is
(a) 0.6sin2
2
x
yt
æö
= p+
ç÷
èø
(b) 0.4sin2
2
x
yt
æö
=- p+
ç÷
èø
(c) 0.4sin2
2
x
yt
æö
= p+
ç÷
èø
(d) 0.4sin2
2
x
yt
æö
=- p-
ç÷
èø
9.A string of mass 2.5 kg is under tension of 200 N. The
length of the stretched string is 20.0 m. If the transverse
jerk is struck at one end of the string, the disturbance will
reach the other end in
200 N = T
20 m
(a) 1 s (b) 0.5 s (c) 2 s (d
) data given is insufficient
10.A train whistling at constant frequency is moving towards
a station at a constant speed v. The train goes past a
stationary observer on the station. The frequency n' of the
sound as heard by the observer is plotted as a function oftime t (figure). Identify the expected curve.
n
t
(a)
n
t
(b)
n
t
(c)
n
t
(d)
Past Years (2013-2017) NEET/AIPMT Questions
11.If we study the vibration of a pipe open at both ends, then
which of the following statements is not true ? [2013]
(a) Odd harmonics of the fundamental frequency will be
generated
(b) All harmonics of the fundamental frequency will be
generated
(c) Pressure change will be maximum at both ends
(d) Antinode will be at open end
12.A source of unknown frequency gives 4 beats/s, when
sounded with a source of known frequency 250 Hz. The
second harmonic of the source of unknown frequency gives
five beats per second, when sounded with a source of
frequency 513 Hz. The unknown frequency is [2013]
(a) 246 Hz (b) 240 Hz
(c) 260 Hz (d) 254 Hz
13.A wave travelling in the +ve x-direction having displacement
along y-direction as 1m, wavelength 2p m and frequency
1
p
Hz is represented by [2013]
(a) y = sin (2px – 2pt) (b) y = sin (10px – 20p t)
(c) y = sin (2px + 2pt) (d) y = sin (x – 2t)

395. 391Waves
14.The len
gth of the wire between two ends of a sonometer is
100 cm. What should be the positions of two bridges below
the wire so that the three segments of the wire have their
fundamental frequencies in the ratio of 1 : 3 : 5?
(a)
1500 2000
cm, cm
23 23
[NEE
T Kar. 2013]
(b)1500 500
cm, cm
23 23
(c)
1500 300
cm, cm
23 23
(d)
300 1500
cm, cm
23 23
15.Two sour
ces P and Q produce notes of frequency 660 Hz
each. A listener moves from P to Q with a speed of 1 ms
–1
.
If the speed of sound is 330 m/s, then the number of beats
heard by the listener per second will be [NEET Kar. 2013]
(a) zero (b) 4
(c)8 (d) 2
16.If n
1
, n
2
and n
3
are the fundamental frequencies of three
segments into which a string is divided, then the original
fundamental frequency n of the string is given by : [2014]
(a)
123
1111
nnnn
=++
(b)
123
1111
nnnn
=++
(c)
123
nnnn=++
(d) n = n
1
+
n
2
+ n
3
17.The number of possible natural oscillation of air column ina pipe closed at one end of length 85 cm whose frequencieslie below 1250 Hz are : (velocity of sound = 340 ms
–1
)
(a)4 (b) 5 [2014]
(c)7 (d) 6
18.A speeding motorcyclist sees trafic jam ahead of him. Heslows down to 36 km/hour. He finds that traffic has eased
and a car moving ahead of him at 18 km/hour is honking at
a frequency of 1392 Hz. If the speeds of sound is 343 m/s,
the frequency of the honk as heard by him will be :
(a) 1332 Hz (b) 1372 Hz [2014]
(c) 1412 Hz (d) 1464 Hz
19.The fundamental frequency of a closed organ pipe of length
20 cm is equal to the second overtone of an organ pipe
open at both the ends. The length of organ pipe open at
both the ends is [2015]
(a) 100 cm (b) 120 cm
(c) 140 cm (d) 80 cm
20.A source of sound S emitting waves of frequency 100 Hz
and an observor O are located at some distance from each
other. The source is moving with a speed of 19.4 ms
–1
at an
angle of 60° with the source observer line as shown in the
figure. The observor is at rest. The apparent frequency
observed by the observer is (velocity of sound in air
330 ms
–1
) [2015 RS]
(a) 103 Hz
60°
S
O
(b) 106 Hz
(c
) 97 Hz
(d) 100 Hz
21.A string is stretched between two fixed points separated
by 75.0 cm. It is observed to have resonant frequencies of
420 Hz and 315 Hz. There are no other resonant frequencies
between these two. The lowest resonant frequency for this
string is : [2015 RS]
(a) 205 Hz (b) 10.5 Hz
(c) 105 Hz (d) 155 Hz
22.A siren emitting a sound of frequency 800 Hz moves away
from an observer towards a cliff at a speed of 15ms
–1
. Then,
the frequency of sound that the observer hears in the echo
reflected from the cliff is : [2016]
(Take velocity of sound in air = 330 ms
–1
)
(a) 765 Hz (b) 800 Hz
(c) 838 Hz (b) 885 Hz
23.A uniform rope of length L and mass m
1
hangs vertically
from a rigid support. A block of mass m
2
is attached to the
free end of the rope. A transverse pulse of wavelength l
1
is
produced at the lower end of the rope. The wavelength of
the pulse when it reaches the top of the rope is l
2
the ratio
l
2
/l
1
is [2016]
(a)
1
2
m
m
(b)
12
2
mm
m
+
(c)
2
1
m
m
(d)
12
1
mm
m
+
24.An air
column, closed at one end and open at the other,
resonates with a tuning fork when the smallest length ofthe column is 50 cm. The next larger length of the columnresonating with the same tuning fork is : [2016]
(a) 66.7 cm (b) 100 cm
(c) 150 cm (d) 200 cm
25.The two nearest harmonics of a tube closed at one end andopen at other end are 220 Hz and 260 Hz. What is thefundamental frequency of the system? [2017]
(a) 20 Hz (b) 30 Hz
(c) 40 Hz (d) 10 Hz
26.Two cars moving in opposite directions approach each otherwith speed of 22 m/s and 16.5 m/s respectively. The driverof the first car blows a horn having a frequency 400 Hz. Thefrequency heard by the driver of the second car is [velocityof sound 340 m/s] :- [2017]
(a) 361 Hz (b) 411 Hz
(c) 448 Hz (d) 350 Hz

396. 392 PHYSICS
EXERCISE - 1
1. (d) 2. (a) 3
. (d) 4. (c) 5. (d)
6. (c) Let I
1
= I
0
and I
2
= 4 I
0
Resultant intensity,
cosf
1 2 12
I=I +I +2 I
I
=
00000
I + 4 I + 2 I 4I cos0?= 9I
When f = p, I = I
0
Whe
n f = p/2, I = 5 I
0
7. (c) The contrast will be maximum, when I
1
= I
2
i.e.
a = b. In that event, I
min
= (a – b)
2
= 0, where a and b are
the amplitudes of interfering waves.
8. (a) From v = n l, we find v
µl because freq. n is constant.
Therefore, new wavelength = 4 l.
9. (a) Here,
max 1 2 12
I =I +I +2 I I c os0?
min 1 2 12
I = I +I + 2 II cos180?
\
max min 12
I +I =2(I+ I)
10. (d) For open pipe,
l2
v
pn=
For closed pipe
l4
v
)1p2(n-=¢ \
1p2
p2
n
n
-
=
¢
11. (a) Becau
se they are absorbed by the atmosphere.
12. (a)
Tuµ
13. (d)
14. (c
) Speed of sound is determined by elasticity and inertia.
15. (a) Standing waves are produced when two waves
propagate in opposite direction
As z
1
& z
2
are propagating in +ve x-axis &
–ve x-axis
so, z
1
+ z
2
will represent a standing wave.
16. (b)
17. (a) For producing stationary waves, the two transverse
waves must be travelling in opposite directions.
Therefore, y
1
and y
2
is the only combination.
18. (d)
19. (b)
r
=
r
g
=
EP
v
20. (c) The
surface area of cylindrical surface is 2prl, where r
is radius of base, l is length of cylinder.
We know that with increasing distance from source,
the total energy or power transmitted remains same,
but intensity decreases. For any source of power P,
intensity I at distance r from it will be
I = P/S, where S is surface area.
rl
(i) For spherical wave front Þ S = 4pr
2
(surface)
So
22
P11
II
4πrr
æöæö
= Þµ
ç ÷ç÷
èøèø
Since I a (Ampl
itude A)
2
Þ
1
A
r
µ
r
P
(ii) For cylindrical wave front S = 2prl
So
P11
II
2πrr
æ öæö
= Þµ
ç ÷ç÷
è øèøl
So
1
A
r
µ
21. (b) Equa
tion of a wave
y
1
= a sin (wt – kx) ....(i)
Let equations of another wave may be,
y
2
= a sin (wt + kx) ....(ii)
y
3
= –a sin (wt + kx) ....(iii)
If Eq. (i) propagate with Eq. (ii), we get
y = 2a cos kx sin wt
If Eq. (i), propagate with Eq. (iii), we get
y = –2a sin kx cos wt
At x = 0, y = 0, wave produce node
So, Eq.(iii) is the equation of unknown wave
22. (a) Frequency of reflected wave is,
æö
ç÷
èø
c+v
f'=f
c
This is the number of w
ave striking the surface per
second.
23. (c) The equation of progressive wave propagating in the
positive direction of X-axis is
y = a sin
( )
2
t x or
p
n-
l
y = a sin (wt – kx)
Hints & Solutions

397. 393Waves
24. (a)
vS
c
D
D
S
v
c
Largest frequency (f )
1Lowest frequency (f )
2
Largest frequency will be detected when the source
approaches detector along the line joining and the
smallest frequency will be detected when the source
recedes the detector along the line joining them
1
2
c
f
f cvcv
cf cv
f
cv
æö
ç÷
èø +-
==
-æö
ç÷
èø+
25. (c)
Fundamental frequency is given by
1
2
=
m
T
v
l
Þ
1
µv
l
Þ
1
µP
v
Since, P div ided into l
1
, l
2
and l
3
segments
Herel = l
1
+ l
2
+ l
3
So
123
1111
vvvv
=++
EXERCISE - 2
1. (
b) Amplitude of reflected wave =
6.09.0
3
2
=´
It would trav
el along negative direction of x-axis, and
on reflection at a rigid support, there occurs a phase
change of p.
2. (b) As f++=cosba2baR
222
\ f++=cosa2aaa
2222
,
2
1
cos-=f,
3
2p
=f
3. (d)
As number of beats/sec = diff. in frequencies has to
be less than 10, therefore 0 < (n
1
–n
2
) < 10
4. (d) In the spherical source, the amplitude A of wave is
inversely proportional to the distance r i.e.,
1
A
r
µ
Where r is distance of source from the point of
consideration.
5. (b) r.m.s.
3
cv=
g
=
3
330 471.4 m/s
1.4
´»
6. (b) If r
H

= 1, then
4
)14(
16114
mix =
+
´+´
=r
2
1
4
1
v
v
mix
H
H
mix
==
r
r
=
s/m612
2
1224
2
v
v
H
mix
===
7.
(a) Let x be distance of person from each cliff
\x + x = v × t = 340 × 1 = 340
x = 170 mDistance between two cliffs = 2 x 170 = 340 m
8. (a) Time taken for two syllables
2
5
t= sec.
2
330
5
xx vt+=´=´ \ x = 66 m
9
. (c) At a given temperature, velocity of sound is not
affected by pressure.
10. (d) Here, l = 5.0 m, n = 2
\ v = n l = 2 × 5.0 = 10.0 m/s
11. (b)
2
max
2
min
I (a + b)
= = 49
I (a - b)
\ 7
ba
ba
=
-
+
7a – 7b
= a + b or 6a = 8b or
3 4
6 8
b a
==
12. (b
) No. of beats/sec = 402 – 401 = 401 – 400 = 1
13. (d) Freq. of unknown fork = 252or2604256
=±
As frequency decreases on loading, therefore, orginal
freq. of unknown fork = 260 Hz
14. (a) No. of loops in longitudinal mode = 3
2
6
=
15. (c)
8
103
100
2.0
cv ´´=´
l
lD
= = 6 × 10
5
m
/s
16. (a)
1x 22.7
4
l
+==l equatio
n (1)
2
3
x 70.2
4
l
+==l equatio
n (2)
3
5λ
x
4
+=l equation (3)
From equation (1) and (2)

2
1.682.70
2
3
x
12 -
=
-
=
ll
= cm05.1
2
1.2
=
From equat
ion (2) and (3)
5
x
x
1
3
=
+
+
l
l
l
3
= 5l
1
+ 4x = 5 × 22.7 + 4 × 1.05 =117.7 cm
17. (
c)
4.254
v
4
v
256n
1
1
´
===
l
\ v = 256 × 101.6 cm/s
2
2
v 256 101.6
n 254 Hz
4 4 25.6
´
===
´l
No. of beats/sec = n
1
– n
2
= 256 – 254 = 2

398. 394 PHYSICS
18. (b)
8
614
103
10105.4
c
v
nn
´
´´
=´=D = 1.5 × 10
12
Hz
\ nnn D-=¢ = 4.5 × 10
14
– 1.5 × 10
12
= 4.
485 × 10
14
Hz
19. (b)
12
21
v
v
r
=
r
20. (c) Comparing it with y (x, t) = A cos (wt + p/2) cos kx.
If kx = p/2, a node occurs ; \ 10 px = p/2 Þ x = 0.05 m
If kx = p, an antinode occurs Þ 10px = p
Þ x = 0.1 m
Also speed of wave s/m5
10
50
k/ =
p
p
=w and
m2.010/2k
/2 =pp=p=l
21. (c) Resultant amplitude is
22
aa+
i.e. a2
22. (b)
v 60
30 Hz
2 21
n===
´l
23. (b)
30.0
2
kand
01.0
2 p
=
p
=w
1
sm30
2
30.0
01.0
2
k
v
-
=
p
´
p
=
w
=
24. (b) Equatio
n is of stationary wave. Comparing with the
standard equation
y = 2A sin
2
t
T
pæö
ç÷
èø
cos
2
x
pæö
ç÷
èøl
2
4.5
p
=
l
or
2
1.4m
4.5
p
l==
25. (c) In
tensity doubles, dB level increases by 3 dB.
26. (b) )tkx(sinAy
2
w-= )]tkx(2cos1[
2
A
y w--=Þ
'2,2 '2,'
w
w= w pn= wn=
p
27. (c)
240
256 n 4n n 64
60
=´ = Þ=
28. (c)
29. (N
) The speed of sound in a gas is given by
RT
v
M
g
=
22
2
OO He
He O He
v M
vM
g
\=´
g
1.44
0.3237
32 1.67
=´=
2
O
He
v 460
v 1421 m/s
0.3237 0.3237
\===
None of the
option is correct.
30. (c)
31. (d)
dB12010log1210
I
I
log10
10
0
=´=
÷
÷
ø
ö
ç
ç
è
æ
32. (c)
6
6
12
0
I 10
10log 10log 10log10 60d B
I 10
-
-
= ==
33. (c) Tw
o possible frequencies of B are 2604256
=± or
252. When A is lo
aded its frequency reduces from 256
to 248 say which would produce 4 beats/sec only if
freq. of B is 252
34. (a) Here,
20
5
m=kg/m =
4
1
kg/m
Tension in the midd
le of wire
T = weight of half the wire =
5
2
× g =
2
5
× 10 N
= 25 N
As m/Tv= \ s/m10
4/1
25
v ==
35. (a) For open p
ipe,
l2
v
n=, where n
0
is the
fundamental
frequency of open pipe.
\
20
11
3002
330
n2
v
=
´
==l
As freq. of 1st
overtone of open pipe = freq. of 1st
overtone of closed pipe
\
ll ¢
=
4
v
3
2
v
2
Þ cm25.41
20
11
4
3
4
3
' =´==
l
l
36. (b) 412
4
v
=
l
, when cut into two equal pieces, frequency
of closed pipe of half the length
=
v 2v
2 412 824 Hz
4(/2)4
==´=
ll
frequency of open pipe of half the length
=
v
4 412 1648Hz
2( / 2)
=´=
l
37. (b) As passenger is a part of moving train, there is no relative
motion between source and listener. Therefore, n' = n =
200 Hz
38. (c) x = 4(cos pt + sin pt) = ]tsin)]t
2
[sin(4 p+p-
p
=
ú
ú
ú
ú
û
ù
ê
ê
ê
ê
ë
é
÷
÷
÷
÷
ø
ö
ç
ç
ç
ç
è
æ
p+
p
-p
÷
÷
÷
÷
ø
ö
ç
ç
ç
ç
è
æ
p-
p
-p
´
2
t
2
t
cos
2
t
2
t
sin24

399. 395Waves
= ú
û
ù
ê
ë
é
÷
ø
ö
ç
è
æ
p+
p
-
p
t
4
cos.
4
sin8
[since cos (
–q) =cosq].
=
ú
û
ù
ê
ë
é p
-p
4
tcos.
2
8
= ú
û
ù
ê
ë
é p
-p
4
tcos24
Comparing it wit
h standard equation
x = A cos (
tw – kx) Þ A 42=
39. (c) V
elocity of wave
vn=l
where n = frequency of wave
v
nÞ=
l
2
2
2
2
v 396
n 396 Hz
100 10
-
===
l ´
no. of bea
ts = n
1
–n
2
= 4
40. (a) By the concept of accoustic, the observer and source
are moving towards each other, each with a velocity
of 18 m s
–1
.
330 18
' 1000
330 18
+
\n=´
-
1115 Hz»
41. (a)For 3rd harmonic/2nd over tone of organ pipe open at
ends
4/l 4/ll
2
l
2
2
3ν
n
2
Þ=
l
For 1st overtone of organ pipe open at one end
4/l2/l
1l
Þ
1
1
3ν
n
4
=
l
Given
21
n n=Þ
21
33
24
nn
=
ll
or
2
1
2
1
=
l
l
42. (b) Let t
he fundamental frequency of organ pipe be f
Case I : f = 200
± 5 = 205 Hz or 195 Hz
Case II : fre quency of 2nd harmonic of organ pipe = 2f
(as is clear from the second figure)
2f = 420 ± 10 or f = 210 ±5 or f = 205 or 215
H
ence fundamental frequency of organ pipe = 205 Hz
43. (d)
s
1000
1
2
T
t;s
500
1
T ===
44. (a)
1
a 0.1 2 300 60 cm s
-
w= ´p´ =p
45.
(a)
0
s
vv v v /15
n ' n 600
vv vv/10
éù+ +éù
==êú êú
-- ëûëû
16 10 400 16
600 711
1599
´éù
=´=»
êú
ëû
46
. (a) Doppler effect.
47. (c) Velocity of source = 18 km h
–1
= 5 m s
–1
(i) S moves towards listener (v
S
)
(ii) listener moves towards source (v
L
)
L
S
vv
' 280 Hz
vv
+
n = n=
-
, Beats
= n' – n = 8.
48. (a)
l/T = constant; Tension decreases by a factor
(8 – 1) / 8, length decreases by square root of this i.e.
0.77.
49. (c) Particle velocity
v ú
û
ù
ê
ë
é
÷
ø
ö
ç
è
æ
l
-p=
x
nt2sinx
dt
d
0
÷
ø
ö
ç
è
æ
l
-pp=
x
nt2cosnx2
0
\ Maximum
particle velocity =
0
nx2p
Wave velocity =
n
T
l
=l
Given, l=pn4nx 2
0
Þ
2
x
n4
nx2
00
p
=
p
=l
50. (a)
2
1
D
D
v
v
A
B
B
A
==
51. (a)
Since Tension and mass per unit length remains
unchanged, the frequency will be obtained in different
mode.
52. (b) Effective value of velocity of source,
1
s
sm20
5
3
3
100
cos
3
100
v
-
=´=q=

400. 396 PHYSICS
100
3
m s
–1
q
3
4
5
n
-
=n
svv
v
'
Hz680Hz640
20340
340
' =´
-
=nÞ
53. (a) units30
15
2
15
k =lÞ
p
=
l
p
Þ
p
=
Distance b
etween node and next antinodeunits5.7
4
30
4
==
l
=
54. (a)y(x, t) 0.005 cos( x t)= a -b (Given)
Comparing it w
ith the standard equation of wavey(x, t) a cos(kx t)= -w we get
k=a and w = b
But
2
k
p
=
l
and
2
T
p
w=
2p
Þ =a
l
and
2
T
p
=b
Given that l = 0.08 m and T = 2.0s
2
25
0.08
p
\a= =p and
2
2
p
b= =p
55. (a)
l
1 l
2
l
3
110 cm
n
1
: n
2
: n
3
= 3 : 2 : 1
1
nµ
l
123
111
: : : : 2:3:6
321
==lll
123
110++=lll
Þ 2x + 3x + 6x = 110 Þ x = 1 0
\ The two bridges should be set at 2x i.e, 20 cm from
one end and 6x i.e, 60 cm from the other end.
56. (b)n
X
= 258 Hz or 242 Hz
n
X
= 258 Hz or 282 Hz
Þ frequency is 258 Hz
The total momentum will be zero and hence velocity
will be zero just after collision. The pull of earth will
make it fall down.
57. (c) Intensity at A, I
A
=
2
P
4rp
;
intensity at B, I
B

=
2
P
4 (2r)p
Sound level at A,
A
A
0
I
S 10log ;
I
=
Sound level at B,
B
B
0
I
S 10log
I
=
Difference of sound level a
t A and B isABA
AB
00B
III
S S 10log 10log 10log
III
æö
-= -=
ç÷
èø
= 10 log 4 = 20 log 2 » 6 dB
5
8. (b) Speed of pulse at a distance x
from bottom,
v gx= .
While trav
eling from mid point
/////////////////
x
v
to the top, frequency remains
unchanged.
12
12
vv
=
ll
Þ 20
02
g (L / 2)gL
2= Þl=l
ll
59. (b)
x(m)
y
Dotted shape shows pulse position after a short time
interval. Direction of the velocities are decided
according to direction of displacements of the
particles.
60. (c) Compare the given eqn. with the standard form
ú
û
ù
ê
ë
é p
-
l
p
=
T
t2x2
sinay
0
3
2
=
l
p
, 3/2p=l and 15
T
2
=
p
T = 2 p/15
Speed of propagation, 5
15/2
3/2
T
v =
p
p
=
l
=
61. (b)
0c 2
v
4
4
v
3
ll
´=´
or
8
3
v4
2
4
v3
0
c
=´=
l
l
62. (c)Let f ' be th
e frequency of sound heard by cliff.
c
ν
vv
\=¢
-
f
f ......(1)

401. 397Waves
Now for the r
eflected wave cliff. acts as a source
v
)v(v
2
c+¢
=¢\
f
f
......(2)
c
c
vv
)v(v
2
-
+
=
f
f
Þ 2v – 2 v
c
= v + v
c
or
c
v
3
v
=
63. (b) y
1
= 10
–6
sin
(100 t + x/50 + 0.5)m
= 10
–6
cos (100 t + x/50 – p/2 + 0.5)m
y
2
= 10
–6
cos (100 t + x/50)m
\ f = p/2 – 0.5 = 1.07 rad
64. (b) Equation of progressive wave is given by
Y = A sin2pf t
Given Y
1
= 4sin500 pt and Y
2
= 2sin506pt.
Comparing the given equations with equation of
progressive wave, we get
2f
1
= 500,
Þ f
1
= 250
2f
2
= 506 Þ f
2
= 253
Beat
s = f
2
– f
1
= 253 – 250 = 3 beats/sec
= 3 × 60 = 180 beats/minute.
65. (c)y = A sin (
wt–kx)
Particle ve
locity,
v
p
=
dy
dt
= A wcos (wt – kx)
\v
p max
= A w
wave velocity =
k
w
\Aw =
k
w
i. e., A =
1
k
But k =
2p
l
\l= 2pA
66. (d)
The frequency of the piano string = 512 +
4 = 516 or
508. Wh
en the tension is increased, beat frequency
decreases to 2, it means that frequency of the string is
508 as frequency of string increases with tension.
67. (a) Here, y
1
= a sin (wt + kx + 0.57)
and y
2
= a cos (wt + kx)
= a sin
( t kx)
2
péù
+w+
êú
ëû
Phase difference, Df = f
2
– f
1
= 0.57
2
p
-
=
3.14
0.57
2
-
= 1.57 – 0.57
= 1 radian
68
. (c) We have, v = nl
Þ v µ l (as n remains constant)
Thus, as v increases 10 times, l also increases 10 times.
69. (a) For fundamental mode,
f =
1T
2ml
Taking logarithm on
both sides, we get
log f =
1T
log log
2l
æöæö
+
ç÷ ç÷
èø mèø
=
11T
log log
22
æöæö
+
ç÷ ç÷
èø mèøl
or log f = l
og
1
2
æö
ç÷
èøl
1
[logT log ]
2
+ -m
Differentiating both sides, we get

df 1 dT
f 2T
= (as l and m are constants)
dT df
2
Tf
Þ =´
Here d
f = 6
f = 600 Hz

dT 26
T 600
´
\= = 0.02
70. (c)
From
fD
p
l
=D
2
x ,
m5.0
6.1
)4.0(2x
2 =
p
p
=
fD
D
p=l
Hz660
5.0
330v
==
l
=n
71. (d) 2p f
1
=
600 p
f
1
= 300 ... (1)
2p f
2
= 608 p
f
2
= 304 ...(2)
|f
1
– f
2
| = 4 beats
2 2
max 12
22
m.n 12
() (5 4) 81
1( ) (5 4)
I AA
I AA
+ +
= ==
+-
where A
1
, A
2
are ampl
itudes of given two sound wave.
72. (c) 73. (b)
74. (d) Two waves moving in uniform string with uniform
tension shall have same speed and may be moving in
opposite directions. Hence both waves may have
velocities in opposite direction. Hence statement-1 is
false.
75. (d) Statement-2 is true, Statement-1 is false. In doppler for
sound wave effect due to observer and source motion
are different.

402. 398 PHYSICS
EXERCISE - 3
Exemplar Questions
1. (b) As the water waves are produced by a motor boat on
surfaces of water as well as inside the water are both
longitudinal and transverse, because the waves,
produce transverse as well as lateral vibrations in the
particles of the medium.
2. (c) Let the frequency in the first medium is v and in the
second medium is v'.
As we know that,
When waves passes from one medium to another its
frequency does not change but its velocity and
wavelength changes.
Hence,
'
'
'
vv
vv=Þ=
ll
(frequency are same for both medium)
'
'
v
v
æö
l=l
ç÷
èø
l and l', are wavelengths and v and v' speeds in first
and second medium respectively.
So,
2
'2
v
v
æö
l= l=l
ç÷
èø
3. (c) The speed of sound (longitudinal) wave in air is
p
v
p
=
r
.
The density of
water vapours is small (rises up) than
the air, so on increasing humidity the density of
medium increase the speed of sound in air.
For air l and p are constants,
1
vµ
r
, where r is density of air..
22
11
v
v
r
=
r
where r
1 is density of dry air and r
2 is density of
moist air.
As
2
2 1 21
1
1
v
vv
v
r<r= >Þ>
Hence, speed of sound wave in air increases with
increase in humidity.
4. (c) As we know that,
v
t = v
0 (1 + 0.61t)
Speed of sound wave in a medium vTµ
(where T is t
emperature of the medium)
Clearly, when temperature increase then the speed also
increase as frequency does not change during
propogation of wave by formula.
As, v
= nl
where n is frequency and l is wavelength.
Frequency (n) remains fixed,
vµl or vlµ
So velocity v and w avelength (l) both increases.
5. (b) Only energy is transmitted from one point to another
and during propagation of any longitudinal waves in
a medium transmission of energy through the medium
without matter being transmitted.
6. (c) Mechanical transverse wave can propagates through
a solid medium, the constituent of the medium oscillate
perpendicular to wave motion causing change in
shape. Thus each, element of the medium is subjected
to shearing stress. Solids and strings have shear
modulus, so sustain shearing stress.
Fluids have no shape of their own, they yield to
shearing stress. Hence, transverse waves are possible
in solids and strings but not in fluids.
7. (d)
(a) The density of medium changes due to compression
and rarefraction. At compressed regions density is
maximum and at rarefactions density is minimum.
(b) As density is changing very rapidly so temperature of
medium increases. Hence Boyle's law is not obeyed.
(c) Bulk modulus of air remains same or constant.
(d) The time of compressions and rarefraction is very small
i.e., we can assume adiabatic process, i.e. no transfer
of heat from surroundings.
8. (b) As given that
Amplitude of reflected wave
22
0.6 0.4 units
33
riAA
æö
=´=´=
ç÷
èø
Given in
cident wave
0.6sin2
2
i
x
yt
æö
= p-
ç÷
èø
As we know that the reflected wave equation at denser
medium where phase difference is p
sin2
2
rr
x
yAt
æö
= p + +p
ç÷
èø
The positive sign is due to reversal of direction of
propagation
So, 0.4sin2
2
r
x
yt
æö
=- p+
ç÷
èø
[ sin( ) sin ]p+q=-qQ
9. (b) As given that, Mass M = 2.5 kg
(Mass per unit length)
m =
2.5 kg 125
0.125 kg/m
20 10
M
l
= ==
Speed
200
0.125
T
v
m
==
[speed of transver
se waves in any string]
Distance
l vt=´
200
20
0.125
tÞ=´

403. 399Waves
55
125 25 5
20 20
2 10 2 10
t
´
=´ =´
´´

5
1
20 25
0.4 10
=´´
´

42
1 205
205
4 10 2 10
´
=´=
´´
1
0.5 sec.
2
t==
10. (c)
When observer is at rest and source of sound is moving
towards observer then observed frequency n'.
Let the original frequency of the source is n
0.
Let the speed of sound wave in the medium is v.
As observer is stationary
Observer
v
v
0
Apparen
t frequency
0( ')
s
v
nn
vv
æö
=ç÷
-
èø
(when train is approaching)
00
(')
s
v
nnn
vv
æö
=>ç÷
-
èø
When the train is moving away from the observer.
Apparent frequency
00
'' ('')
s
v
n nnn
vv
æö
=<ç÷
+
èø
Hence, frequencies in both cases are same and n' > n''.
so graph (c) verifies.
Past Years (2017-2013) NEET/AIPMT Questions
11. (c) Pressure change will be minimum at both ends. In fact,
pressure variation is maximum at l/2 because the dis-
placement node is pressure antinode.
12. (d) When sounded with a source of known frequency
fundamental frequency
= 250 ± 4 Hz = 254 Hz or 246 Hz
2
nd
harmonic if unknown frequency (suppose) 254
Hz = 2 × 254 = 508 Hz
As it gives 5 beats
\ 508 + 5 = 513 Hz
Hence, unknown frequency is 254 Hz
13. (d) As Y = A sin (wt – kx + f)
w = 2pf =
2p
p
= 2 [ Qf =
1
p
]
k =
2p
l
=
2
2
p
p
= 1 [ Ql = 2 p]
\ Y =
1 sin (2t – x + f) [Q A = 1 m]
14. (a) From formula,
1T
f
xm
=
Þ
1
l
f
µ
\ l
1
: l
2
: l
3
=
123
111
::
fff
= f
2
f
3
: f
1
f
3
: f
1
f
2
[Gi ven: f
1
: f
2
: f
3
= 1 : 3 : 5]
= 15 : 5 : 3
Therefore the positions of two bridges below the wire are
15 100
cm
1553
´
++
and
15 100 5 100
cm
1553
´ +´
++
i.e.
,
1500 2000
cm, cm
23 23
15.
(b)
f
fC
Dn
=
Þ
(Beats)2
fC
n
=
Þ Beats =
2
4.
fv
C
=
16. (a) Total length of string l = l
1
+ l
2
+ l
3
(As string is divided into three segments)
But frequency
1
length
µ
1T
f
2m
æö
=ç÷
ç÷
èø
Q
l
so
123
1111
nnnn
=++
17. (d
) In case of closed organ pipe frequency,
f
n
= (2n + 1)
4
v
l
for n = 0, f
0
= 100 Hz
n = 1, f
1
= 300 Hz
n = 2, f
2
= 500 Hz
n = 3, f
3
= 700 Hz
n = 4, f
4
= 900 Hz
n = 5, f
5
= 1100 Hz
n = 6, f
6
= 1300 Hz
Hence possible natural oscillation whose frequencies
< 1250 Hz = 6(n = 0, 1, 2, 3, 4, 5)
18. (c) According to Doppler's effect
Apparent frequency
n¢ =
0
s
vv 343 10
n 1392 .
vv 3435
æö+ +æö
=ç÷ ç÷
++ èøèø
1412 Hz=

404. 400 PHYSICS
19. (b
) Fundamental frequency of closed organ pipe
V
c
=
c
V
4l
Fundamental frequency of open organ pipe
V
0
=
0
V
2l
Second overtone frequency of open organ pipe =
0
3V
2l
From question,
c0
V 3V
4l 2l
=
Þl
0
= 6l
c
= 6 × 20 = 12
0 cm
20. (a) Here, original frequency of sound, f
0
= 100 Hz
Speed of source V
s
= 19.4 cos 60° = 9.7
60°
O
19.4
19.4 cos 60° = 9.7
S
From Doppler's f
ormula
f
1
= f
0
0
s
VV
VV
æö-
ç÷
-
èø
f
1
= 100
V0
V ( 9.7)
æö-
ç÷
-+èø
f
1
= 100
V
9.7
V1
V
æö
-ç÷
èø
=
100
9.7
1
330
æö
-ç÷
èø
= 103Hz
Apparent fr
equency f
1
= 103 Hz
21. (c) In a stretched string all multiples of frequencies can
be obtained i.e., if fundamental frequency is n then
higher frequencies will be 2n, 3n, 4n ...75 cm
So, the difference between any two successive
frequencies will be 'n'According to question, n = 420 – 315= 105 HzSo the lowest frequency of the string is105 Hz.
22. (c) According to Doppler's effect in sound
Observersource 15 m/s
Apparent frequency,
n' = 0
s
v
n
vv-
=
()
330 330 800
800
330 15 315
´
=
-
= 838 Hz
The fr
equency of sound observer hears in the echo
reflected from the cliff is 838 Hz.
23. (b) From figure, tension T
1
= m
2
g
T
2
= (m
1
+ m
2
)g
Rigid support
As we know
L m
1
T
1
m
2
T
2
Velocity µ T So,
l µ T
Þ
11
2 2
T
T
l
=
l
Þ
2 12
12
mm
m
l+
=
l
24. (c) Fo
r a closed organ pipe first minimum resonating
length
L
1
=
50 cm
4
l
=
\Next or sec
ond resonating length, L
2
=
3
150 cm
4
l
=
25. (a) Differ
ence in two successive frequencies of closed
pipe
2v
4l
= 260 – 220 = 40 Hz
or
2v
4l
= 40 Hz
v
20Hz
4
Þ=
l
Which is the fundamental frequency of system of
closed organ pipe.
26. (c) As we known from Doppler's Effect
f
apprent
=
0
0
s
vv
f
vv
éù+
êú
-
ëû
=
340 16.5
400
340 22
+éù
êú
-ëû
f
apprent
= 448 Hz
v
s
= 22 m/s
f = 400 Hz
0
v = 16.5 m/s
0
A B

405. ELECTRIC CHARGE
Charge is something associ
ated with matter due to which it
produces and experiences electric and magnetic effects.
There are two types of charges :
(i) Positive charge and (ii) Negative charge
Positive and negative charges : Positive charge means the
deficiency of electons while negative charge means excess of
electrons. In any neutral body the net charge is equal to zero
i.e., the sum of positive charges is equal to the sum of negative
charges.
Mass M'
Electron
< Proton
Positively charged
M' < Mbody
+
+ +
+
+ Mass M.
Electron
= Proton
Uncharged body
M

Electron
> Proton
Mass M''
Negatively charged
bo
dy M'' > M
–
–
–
–
–
––
–
–
–
–
–
––
Charge is a scalar quantity and its SI unit is coulomb (C).
CONDUCTORS AND INSULATORS
The materials which allow electric charge (or electricity) to
flow freely through them are called conductors. Metals are very
good conductors of electricity. Silver, copper and aluminium are
some of the best conductors of electricity. Our skin is also a
conductor of electricity. Graphite is the only non-metal which is a
conductor of electricity.
All metals, alloys and graphite have 'free electrons', which can
move freely throughout the conductor. These free electrons make
metals, alloys and graphite good conductor of electricity.
Aqueous solutions of electrolytes are also conductors.
The materials which do not allow electric charge to flow through
them are called nonconductors or insulators.
For example, most plastics, rubber, non-metals (except graphite),
dry wood, wax, mica, porcelain, dry air etc., are insulators.
Insulators can be charged but do not conduct electric charge.
Insulators do not have 'free electrons' that is why insulators do
not conduct electricity.
Induced charge can be lesser or equal to inducing charge (but
never greater) and its max. value is given by
Q' = – Q (1 – 1/k), where 'Q' is inducing charge and 'K' is the
dielectric const. of the material of the uncharged body.
For metals k = ¥ Þ Q' = – Q.
METHODS OF CHARGING
(i) By friction : By rubbing two suitable bodies, given in box
one is charged by +ve and another by –ve charge in equal
amount.
+ve –ve
Wool Amber
Glass rod S ilk
FurEbonite rod
Dry hair Comb
Note : Electric charges remain confined only to the
rubbed portion of a non-conductor but in case of a conductor,
they spread up throughout the conductor.
(ii) By conduction : Charging a neutral body by touching it
with a charged body is called charging by conduction.
·It is important to note that when the bodies are charged
by conduction, a charged and an uncharged bodies
are brought into contact and then seperated, the two
bodies may or may not have equal charges.
·If the two bodies are identical the charges on the two
will be equal.
·If the two bodies are not identical, the charges will be
different.
·The potential of the two bodies will always be the same.
(iii) By induction : Charging a body without bringing it in
contact with a charged body is called charging by
induction.
A ++
+
+
++
+ ––
–
–
––
++
+
+
+
B
e
–
First rearrangement of charge takes place in metal rod B.
When the rod B is connected to earth, electrons flow fromearth to the rod B thus making it -vely chargedThe magnitude of elementary positive or negative charge(electron) is same and is equal to 1.6 × 10
–19
C.
16
Electric Charges
and Fields

406. 402 PHYSICS
Properties of Elec
tric Charge
(i)Similar charges repel and dissimilar charges attract each
other.
+ + + +
F
F+
+
+
+
R
e
p
u
ls
i
v
e
f
o
r
c
e

+ + + +
F
F
Attractive
force
In rare situation you may find similar charged bodies
attracting each other. Suppose a big positive charged bodyis placed near a small positively charged body then becauseof induction, opposite charge produced on the small bodymakes it to attract the other body.
(ii)A charged body attracts light uncharged bodies, due topolarisation of uncharged body.
-
-
-
-
+
+
+
+
Positively
charged
balloon
Wall
Fig : When a positively charged balloon is placed in
contact with the wall, an opposite charge is induced with
the wall, the balloon stick to the wall due to electrostatic
attraction
(iii)Charge is conserved i.e., the charge can neither be created
nor be destroyed but it may simply be transferred from one
body to the other.
Thus we may say that the total charge in the universe is
constant or we may say that charges can be created or
destroyed in equal and opposite pair. For example
)MeV02.1Energy(³g
Electron Positron
ee
-+
¾¾®+
(Pair-productio
n process)
·Positron is an antiparticle of electron. It has same mass
as that of electron but equal negative charge.
g¾®¾+
+-
ee (Pair-annihilation process)
(iv)Charge is unaffected by motion. This is also called charge
invariance with motionMathematically,(q)
at rest
= (q)
in motion
(v)Quantisation of charge. A charge is an aggregate of small
unit of charges, each unit being known as fundamental orelementary charge which is equal to e = 1.6 × 10
–19
C. This
principle states that charge on any body exists as integral
mutliple of electronic charge.
i.e. q = ne where n is an integer.
According to the concept of quantisation of charges, thecharge q cannot go below e. On macroscopic scale, this isas good as taking limit q
0
® 0.
Quantisation of electric charge is a basic (unexplained) lawof nature. It is important to note that there is no analogouslaw of quantisation of mass.Recent studies on high energy physics have indicated thepresence of graphs with charge 2e/3, e/3. But since thesecannot be isolated and are present in groups with total charge,therefore the concept of elementary charge is still valid.
COULOMB’S LAWThe force of attraction or repulsion between two point charges(q
1
and q
2
) at finite separation (r) is directly proportional to
the product of charges and inversely proportional to the squareof distance between the charges and is directed along the linejoining the two charges.
i.e.,
12
2
qq
F
r
µ or
12
2
1
.
4
qq
F
r
=
pe
q
1 q
2
r
wheree is the permittivity of medium between the charges.
If
0
eis the permittivity of free space, then relative permittivity of
medium or dielectric constant (K), is given by
( )
0
r
orK
e
e=
e
The permittivity of free space

21212
0
mNC1086.8
---
´=e
and
2
0 1086.814.34
1
4
1
-
´´´
=
pe
= 9 × 10
9
Nm
2
C
–2
.
Also 1
4
1
0
=
pe
in CGS system of unit.
Coulomb’s law may also be expressed as

12
2
0
1
.
4
r
qq
F
r
=
pee
Let F
0
be the force between
two charges placed in vacuum then

12
0
2
0
1
.
4
qq
F
r
=
pe
r
+q
1
–q
2
••
Medium with dielectric
constant K
Hence K
F
F
0
=
12
2
0
1
. and
4
eæö
==
ç÷
peeèø
Q
qq
FK
r
Therefore we ca
n conclude that the force between two
charges becomes 1/K times when placed in a medium ofdielectric constant K.The value of K for different media
Medium Dielectric Cons tant (K)
A ir 1.006
Vacuum 1.00
Water 1.00026
Mica 3 to 6
Metals ¥

407. 403Electric Charges and Fields
Dielectric : A dielectric is an insulator. It is of two types -
(i)Polar dielectric and
(ii) Non-polar dielectric.
Significance of Permittivity Constant or Dielectric Constant :
Permittivity constant is a measure of the inverse degree of
permission of the medium for the charges to interact.
Dielectric strength : The maximum value of electric field that can
be applied to the dielectric without its electric breakdown is called
its dielectric strength.
Difference between electrostatic force and gravitational force :
Electrostatic force Gravitational force
1.Much stronger 1.Much weaker as compared
to electrostatic force
2.Can be attractive 2.Only attractive
or repulsive
3.Depends on the nature of3.Does not depend on the
medium between chargesnature of medium
between masses
Both electric and gravitational forces follow inverse
square law.
Vector Form of Coulomb’s Law :
12 12
12 12 1232
00
qq qq11
ˆFrr
4k 4krr
==
pe pe
u uur u ur
SUPERPOSITION PRINCIPLE FOR DISCRETE CHARGE
DISTRIBUTION : FORCE BETWEEN MULTIPLE
CHARGES
q
2
q
4
q
3
q
5
q
6
q
7
q
1
Discrete charge
distribution
r
13
r
17
r
12
O
The electric force
1
F
r
on q
1
due to a number of charges placed in
air or vacuum is given by
n
1 1i 12 132 22
i1
00 1i 12 13
131
i 12
qqqq qq11
ˆ ˆˆ
F r r r .... ..
44 r rr=
éù
=å= ++ êú
pe peêúëû
r
Coulomb's law is valid if
15
10r
-
³ m and if charges
are point charges
FORCE FOR CONTINUOUS CHARGE DISTRIBUTION
A small element having charge dq is considered on the body. The
force on the charge q
1
is calculated as follows :
1
1
2
0
1
ˆ.
4
q dq
dFr
r
=
pÎ
ur
Now the total force1F
r
is calculated by integrating
1dF
r
under
proper limi
ts.
i.e.,
11
11
2
00
qdqq1 dq
ˆˆFd F rr
4 4rr
===
pÎ pÎ
òòò
rr
whererˆ is a variable unit vector which points from each dq,
towards the location of charge q
1
(where dq is a small charge
element)
Types of Charge Distribution
(i) Volume charge distribution : If a charge, Q is uniformly
distributed through a volume V, the charge per unit volume
r (volume charge density) is defined by
Q
V
r=; r has unit coulomb/m
3
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
.
..
dq
V
dQ = dVr
Total charge is
uniformly distributed
in a cube
(ii) Surface charge distribution : If a charge Q is uniformly
distributed on a surface of area A, the surface charge density
s, is defined by the following equation
,
Q
A
s= s has unit coulomb / m
2
dA
dQs=
Total charge Q, which
is uniformly
distributed over disc
(iii) Linear charge distribution : If a charge q is uniformly
distributed along a line of length l, the linear charge density
l, is defined by
Q
l=
l
, l has unit coloumb/m.
dldQl=
dl
Total charge Q uniformly
distributed in a tubeQ
lIf the ch
arge is non uniformly distributed over a volume,
surface, or line we would have to express the charge
densities as
, ,,
dQ dQ dQ
dV dA dl
r= s= l=
w
here dQ is the amount of charge in a small volume, surface
or length element.
lIn general, when there is a distribution of direct andcontinuous charge bodies,we should follow the followingsteps to find force on a charge q due to all the charges :

408. 404 PHYSICS
(1) Fix th
e origin of the coordinate system on charge q.
(2) Draw the forces on q due to the surrounding charges
considering one charge at a time.
(3) Resolve the force in x and y-axis respectively and
find
x
FS and y
FS
(4) The resultant force is
22
()()
xy
FFF= S +S and the
direction is giv
en by
tan
x
y
F
F
S
q=
S
.
Calculation of
electric force in some situations :
(a) Force on one charge due to two other charges :
Resultant force on q due to q
1
and q
2
are obtained by
vector addition of individual forces
q
2
q
1
q
F
1
F
2
q
F
+

12
FFF=+
rrr
22
12
F |F | |F |=+
r rr
The direction of F is given by
1
2
|F|
tan
|F|
q=
r
r
(b) Force due to linear charge distribution :
Let AB is a long (length l) thin rod with uniform
distribution of total charge Q.
dx
A
B
dQ
•
a
x
q
We calculate force of these charges i.e. Q on q which
is situated at a distance a from the edge of rod AB.
Let, dQ is a small charge element in rod AB at a distance
x from q .
The force on q due to this element will be
2
ox
dQ
4
q
dF
pe
=
2
ox
dx
4
qm
pe
=
where m is linear c
harge density i.e., m = Q / l.
so, F =
aa
2
o
aa
qµ
dF . dx
4 x
++
=
pÎ
òò
ll
a
2
0
a
qµ1
dx
4 x
+
=
pÎ
ò
l
newton
Keep in Memory
1. When
the distance between the two charges placed in
vacuum or a medium is increased K-times then the force
between them decreases K
2
-times. i.e., if F
0
and F be the
initial and final forces between them, theno
2
F
F
K
=
2. When the d
istance between the two charges placed in
vacuum or a medium is decreased K-times then the forcebetween them increases K
2
-times. i.e., if F
o
and F be the
initial and final forces then
F = K
2
F
o
3. When a medium of dielectric constant K is placed between
the two charges then the force between them decreases byK-times. i.e., if F
o
and F be the forces in vacuum and the
medium respectively, then
o
F
F
K
=
4. When a medi
um of dielectric constant K between the
charges is replaced by another medium of dielectric constantK' then the force decreases or increases by (K/K') times
according as K' is greater than K or K' is less than K.
Example 1.
If we supply a charge to a soap bubble then it will expand.Why?
Solution :
Since we know like charges repel and try to get away fromeach other which is at outer surface of the conductor. So asoap bubble expand.
Example 2.
Calculate the net charge on a substance consisting of(a) 5 × 10
14
electrons (b) a combination of 7 × 10
13
protons and 4 × 10
13
electrons.
Solution :
(a) The charge of one electron is –1.6×10
–19
C. So net
charge on a substance consisting of 5 × 10
14
electrons
is
5 × 10
14
× (–1.6 × 10
–19
C) .C80C108
5
m-=´-=
-
(b) Similarly the net charge on a substance consisting of a
combination of 7 × 10
13
protons and 4 × 10
13
electron
is
]C106.1(104[)]C106.1(107[
19131913
-´-´+´´´
--
= + 4.8 mC.
(
Q the charge on one proton is + 1.6 × 10
19
C)
Example 3.
Two protons in a molecule is separated by a distance
3 × 10
–10
m. Find the electrostatic force exerted by one
proton on the other.
Solution :
m103
10-
´
1.6 × 10 C
–19
1.6 × 10 C
–19

409. 405Electric Charges and Fields
According to coulomb’s law, the electrostatic force F
between two charges q
1
and q
2
, which are seperated by
distance r is
2
or
qq
4
1
F
21
´
pe
=
Here, q
1
= q
2
= 1
.6 × 10
–19
C , r = 3 × 10
–10
m
so,
192
99
20
(1.6 10 )
F 9 10 2.56 10 C
9 10
-
-
-
´
=´ =´
´
(Repulsiv
e)
Example 4.
When a piece of polythene is rubbed with wool, a charge
of 2 × 10
–7
C is developed on polythene. What is the
amount of mass, which is transferred to polythene?
Solution :
No. of electrons transferred,
q
n
e
=
Mass transferred
ee
q
m nm
e
æö
=´=´ ç÷
èø
÷
÷
ø
ö
ç
ç
è
æ
´
´
´´=
-
-
-
19
7
31
106.1
102
101.9
kg1038.11
19-
´=
Example 5.
T
wo negative charges of unit magnitude each and a
positive charge q are placed along a straight line. At what
position and for what value of q will the system be in
equilibrium? Check whether it is stable, unstable or
neutral equilibrium?
Solution :
Let the charge + q be held at a distance x
1
from unit negative
charge at A, and at a distance x
2
from unit negative charge
at B.
–1 +q –1
A B
x
1 x
2
For equilibrium of q,
2
2o
2
1o x4
)1(q
x4
)1(q
ep
-
=
ep
-
\ x
1
= x
2
i.e.
q must be equidistant from A and B.
For equilibrium of unit negative charge at B.
Force on B due to charge at A + force on B due to q = 0
0
x4
)1(q
)xx(4
)1()1(
2
2o
2
21o
=
ep
-
+
+ep
--
Þ
2
2o
2
2o x4
)q(
)x2(4
1
ep
--
=
ep

12
( x x)=Q
11
q = i.e. th
44
of the magnitude of either unit charge.
Stability : If q is displaced slightly towards A, force of
attraction due to A exceeds the force of attraction due to B.
Therefore, q will get displaced further towards A. Hence
the equilibrium of q is unstable.
However, if q is displaced in a direction
^ to A, net force
would bring q back to its normal position. Therefore, the
equilibrium will be stable.
Example 6.
Four identical point charges each of magnitude q are
placed at the corners of a square of side a. Find the net
electrostatic force on any of the charge.
Solution :
Let the concerned charge be at C then charge at C will
experience the force due to charges at A, B and D. Let these
forces respectively be
A
F
r
,
B
F
r
and
DF
r
thus forces are given as
F
D
F
A
F
B
A B
C
D
q
q
q
q

y
x
2
2
0
A
AC
q
4
1
F
pe
=
r
along AC =
÷
÷
ø
ö
ç
ç
è
æ
-
pe 2
j
ˆ
2
i
ˆ
a24
q
2
0
2
2
2
0
B
BC
q
4
1
F
pe
=
r
along BC = )j
ˆ
(
aπε4
q
2
0
2
-
2
2
0
D
DC
q
4
1
F
pe
=
r
along DC = )i
ˆ
(
aπε4
q
2
0
2
DBAnet
FFFF
rrrr
++=

ú
ú
û
ù
ê
ê
ë
é
÷
÷
ø
ö
ç
ç
è
æ
+-
÷
÷
ø
ö
ç
ç
è
æ
+= 1
22
1
j
ˆ
1
22
1
i
ˆ
aπε4
q
2
0
2
2
0
2
net
a4
q
1
22
1
2F
pe
÷
÷
ø
ö
ç
ç
è
æ
+=
r
2
0
2
a4
q
2
2
1
pe
÷
ø
ö
ç
è
æ
+=
Example 7.
El
ectric force between two point charges q and Q at rest is
F. Now if a charge – q is placed next to q what will be the
(a) force on Q due to q (b) total force on Q ?
Solution :
(a) As electric force between two body interaction, i.e., force
between two particles, is independent of presence or
absence of other particles, the force between Q and q
will remain unchanged, i.e., F.
(b) An electric force is proportional to the magnitude of
charges, total force on Q will be given by
FQqq0
0
FQq qq
¢ ¢¢
= = == [as q' = q + (– q) = 0]
i.e.,
the resultant or total force on Q will be zero.
ELECTRIC FIELD
The space around an electric charge, where it exerts a force on
another charge is an electric field.
Electric force, like the gravitational force acts between the bodies
that are not in contact with each other. To understand these
forces, we involve the concept of force field. When a mass is
present somewhere, the properties of space in vicinity can be
considered to be so altered in such a way that another mass

410. 406 PHYSICS
brought to t
his region will experience a force there. The space
where alteration is caused by a mass is called its Gravitational
field and any other mass is thought of as interacting with the
field and not directly with the mass responsible for it.
Similarly an electric charge produces an electric field around it
so that it interacts with any other charges present there. One
reason it is preferable not to think of two charges as exerting
forces upon each other directly is that if one of them is
changed in magnitude or position, the consequent change in
the forces each experiences does not occur immediately but
takes a definite time to be established. This delay cannot be
understood on the basis of coulomb law but can be explained
by assuming (using field concept) that changes in field travel
with a finite speed. (» 3 × 10
8
m / sec).
Electric field can be represented by field lines or line of force.
The direction of the field at any point is taken as the direction of
the force on a positive charge at the point.
Electric field intensity due to a charge q at any position
(
r
r
) from that charge is defined as
0
2
q0
00
F
E
q
F 1q
ˆE(r)Limitr
q4 r®
=
==
pe
ur
u ur
ur
u ur
q
q
0
O
r
F
whereFis the force experienced by a small positive test charge
q
0
due to charge q. Its SI unit is NC
–1
. It is a vector quantity.
If there are more charges responsible for the field, then
123
........EEEE=+++
uur uur uur uur
where ,........E,E,E
321 are the electric field intensities due to
charges q
1
, q
2
, q
3
......respectively.
ELECTRIC LINES OF FORCE
These are the imaginary lines of force and the tangent at any
point on the lines of force gives the direction of the electric field
at that point.
Properties of Electric Lines of Froce
(i) The lines of force diverge out from a positive charge and
converge at a negative charge. i.e. the lines of force are
always directed from higher to lower potential.
+

–
(ii) The electric lines of force contract length wise indicating
unlike charges attract each other and expand laterallyindicating like charges repel each other.
–q+2q
(iii) The number of lines that originate from or terminate on a
charge is proportional to the magnitude of charge.
i.e.,
11
22
||
||
qN
qN
=
(iv) Two electr
ic lines of force never intersect each other.
(v) They begin from positive charge and end on negative charge
i.e., they do not make closed loop (while magnetic field
lines form closed loop).
–+
(vi)Where the electric lines of force are
(a) close together, the field is strong (see fig.1)
(b) far apart, the field is weak (see fig.2)
Electric field lines
Strong field
Fig. 1
Weak field
Fig. 2
(vii) Electric lines of force generate or terminate at charges
/surfaces at right angles.
-
-
-
-
-
-
-
-
-
-
-
-
d
Fixed point charge
near infinite metal plate
Electric Field for Continuous Charge Distribution :
If the charge distribution is continuous, then the electric field
strength at any point may be calculated by dividing the charge
into infinitesimal elements. If dq is the small element of charge
within the charge distribution, then the electric field
dE
r
at point
P at a distance
r from charge element dq is
^
2
1
4
dq
dEr
r
=
pe
u ur
;

411. 407Electric Charges and Fields
Non conducting sphere (dq is small charge element)
Continuous
charge
r
P
distribution
R
dq = ldl (line ch
arge density)
= s ds (surface charge density)
= rdv (volume charge density)
The net field strength due to entire charge distribution is given
by
^
2
1
4
dq
Er
r
=
pe
ò
u ur
where the
integration extends over the entire charge distribution. Electric field intensity due to a point charge q, at a
distance (r
1
+ r
2
) where r
1
is the thickness of medium of dielectric
constant K
1
and r
2
is the thickness of medium of dielectric constant
K
2
as shown in fig. is given by
r
1 r
2
K
1
K2
q
( )
2
0
1122
1
4
q
E
rK rK
=
pÎ
+
CALCULA
TION OF ELECTRIC FIELD INTENSITY FOR
A DISTRIBUTION OF DIRECT AND CONTINOUS
CHARGE
1. Fix origin of the coordinate system where electric field
intenstiy is to be found.
2. Draw the direction of electric field intensity due to the
surrounding charges considering one charge at a time.
3. Resolve the electric field intensity in x and y-axis
respectively and find SE
x
and SE
y
4. The resultant intensity is
2
y
2
x
)E()E(ES+S=
and
y
x
E
tan
E
S
q=
S
where q is t
he angle between
E
ur
and
x-axis.
5. T
o find the force acting on the charge placed at the origin,
the formula F = qE is used.
Energy density : Energy in unit volume of electric field is called
energy density and is given by
21
2
ouE=e ,
where E = el
ectric field and e
o
= permitivity of vacuum
Electric Field due to Various Charge Distribution :
(i)Electric Field due to an isolated point charge
2
q
Ek
x
=
+q
P
Ex
(ii)A circular ring of radius R with uniformly distributed charge
2 2 3/2
()
Qx
Ek
Rx
=
+
+
+
R
Px
E
Q
When x >
> R,
2
Q
Ek
x
=
[The char
ge on ring behaves
as point charge]
E is max when
2
R
x=± . Also E
max

2
063
Q
R
=
pe
(iii)A circular disc of radiu
s R with uniformly distributed
charge with surface charge density
s
R
E
Q x
p
2
22 22
0
2
11
2
kQxx
E
R xR xR
éùéù
s
= - =-êúêú
eêúêú++ëûëû
(iv)An infinite sheet of uniformly distributed charges with
surface charge density
s
p
E
0
2
E
s
=
e
(v)A finite length of charge with linear charge density l
E
y
E
x
a
b
x P
+
+
+
+
+
+
+
+
+
[sin sin ]
x
k
E
x
l
= a+b and [cos cos ]
y
k
E
x
l
= a-b

412. 408 PHYSICS
Special case :
For
Infinite length of charge,
2
p
=b=a
\
x
k2
E
x
l
= and 0E
y
=
(vi)Due to a spherical shell of uniformly
distributed charges
with surface charge density
s
R
+
+
+
+
+
+
+
+
+
+
+
+
Q

R
E
x
E
in
= 0 (x < R)
E
surface
2
0
Q
k
R
s
==
e

(x = R)
2
=
out
Q
Ek
x
(vii)Due to a solid non conducting sphere of uniformly
distributed
charges with charge density
r
R

E
x
0
centre
E =
3
in
Qx
Ek
R
=
2
surface
Q
Ek
R
=
2out
Q
Ek
x
=
(viii)Due to a solid non-con
ducting
cylinder with linear charge denisty
l
E
axis
= 0,
2in
R
x2
kE
l
= ,
R
R
2
kE
surface
l
= ,
x
2
kE
out
l
=
In above cases,
0
1
k
4
=
pÎ
Keep in Memory
1. If t
he electric lines of force are parallel and equally spaced,
the field is uniform.
2. If E
0
and E be the electric field intensity at a point due to a
point charge or a charge distribution in vacuum and in amedium of dielectric constant K then
E = KE
0
3. If E and E' be the electric field intensity at a point in the two
media having dielectric constant K and K' then
E' K'
EK
=
4. The electr
ic field intensity at a point due to a ring with
uniform charge distribution doesn't depend upon the radius
of the ring if the distance between the point and the centre
of the ring is much greater than the radius of the ring. The
ring simply behaves as a point charge.
5. The electric field intensity inside a hollow sphere is zero
but has a finite value at the surface
2
KQ
R
æö
=
ç÷
èø
and outside it
(
2
KQ
x
= ; x being the distance of the point from the centre of
the sphere).
6. The electric field intensity at a point outside a hollow sphere
(or spherical shell) does not depend upon the radius of thesphere. It just behaves as a point charge.
7. The electric field intensity at the centre of a non-conducting
solid sphere with uniform charge distribution is zero. Atother points inside it, the electric field varies directly with
the distance from the centre (i.e. E µ x; x being the distance
of the point from the centre). On the surface, it is constant
but varies inversely with the square of the distance from
the centre (i.e.
2
1
E
x
µ). Note that the field doesn't depend
on the radius of the sphere for a point outside it. It simply
behaves as a point charge.
8. The electric field intensity at a point on the axis of non-
conducting solid cylinder is zero. It varies directly with the
distance from the axis inside it (i.e. E µ x). On the surface, it
is constant and varies inversely with the distance from the
axis for a point outside it (i.e.
1
E
x
µ).
MOTION OF
A CHARGED PARTICLE IN AN ELECTRIC
FIELD
Let a charged particle of mass m and charge q be placed in an
uniform electric field
E, then electric force on the charge particle
is
F qE=
u ur u ur
\ acceleration,
qE
a
m
=
u ur
ur
(constant)
(a)
The velocity of the charged particle at time t is,
v = u + at = at =
qE
t
m
(Particle initially at rest)
or
qE
vt
m
=
(b) Distance
travelled by particle is
21
2
S at=
21
2
qE
t
m
=
(c) Kinetic ener
gy gained by particle,
2 22
2
1
22
qEt
K mv
m
==

413. 409Electric Charges and Fields
If a charged particle is entering the electric field in perpendicular
direction.
E
Y
O
U
P (x, y)
X
Let j
ˆ
EE= and the particle enters the field with speed u along
x-axis.
Acceleration along Y-axis,
y
qE
a
m
=
The initial
component of velocity along y-axis is zero. Hence the
deflection of the particle along y-axis after time t is
2
y
2
y ta
2
1
0t
m
qE
.
2
1
tuy +=+= ;
2
t
m
qE
.
2
1
y=\ …… (i)
Dist
ance covered by particle in x-axis,
x = ut …… (ii) (Q acceleration a
x
= 0)
Eliminating t from equation (i) & (ii),
2
2
.
2
qEx
y
mu
= i.e. y µ x
2
.
This sho
ws that the path of charged particle in perpendicular
field is parabola.
If the width of the region in which the electric field exists be l
then
(i) the particle will leave the field at
a distance from its original path
in the direction of field, given by
2
2
.
2
qE
y
mu
=
l u
l
E
y
q
v
x
v
y
(ii) The particle will leave the region in the direction of the
tangent drawn to the parabola at the point of escape.
(iii) The velocity of the particle at the point of escape is given
by
22
xyx
v v v , whe re v u and=+=
y yy
qE
vuat0t
m
= + =+

qEt qE
t
mm uu
æö
===
ç÷
èø
ll
Q
2
2qE
vu
mu
æö
\=+
ç÷
èø
l
(iv) The direction of the particle in which it leaves the field is
given by
y
2
x
v qE qE
tan
v mu.u mu
q===
ll
1
2
tan
qE
mu
-æö
Þ q= ç÷
èø
l
Example 8.
P
oint charge q moves from point P to point S along the
path PQRS (as shown in fig.) in a uniform electric field
E pointing co-parallel to the positive direction of X-axis.
The coordinates of the points P, Q, R and S are (a, b, 0),
(2 a, 0, 0), (a, –b, 0) and (0, 0, 0) respectively.
E
Q
R
S
P
Y
The workdone by the field in the above case is given by
the expression
(a) q E A (b) – q E A
(c) q E A 2 (d)
22
qE [(2a)+b]
Solution : (b)
W F.S q Ei.S==
r r rr
Now W
P®Q
= q Ei
r
· (ai
r
+ bj
r
) = q E a
W
Q®R
=
q E
i
r
· (– ai
r
+ bj
r
) = – q E a
(workdone against field is taken as negative)
W
R®S
= q Ei
r
· (– ai
r
+ bj
r
) = – q E a

PQ QR RS
WW WW
qEa qEa qEa qEa
®+®
= ++
= - - =-
Exa
mple 9.
Calculate the electric field strength required to just
support a water drop of mass 10
–7
kg and having charge
1.6 × 10
–19
C.
Solution :
Here, m = 10
–7
kg, q = 1.6 × 10
–19
C
Step 1 : Let E be the electric field strength required to
support the water drop.
Force acting on the water drop due to electric field E is
F = qE = 1.6 × 10
–19
E
Weight of drop acting downward,
W = mg = 10
–7
× 9.8 newton.
Step 2 : Drop will be supported if F and W are equal and
opposite.
i.e., 1.6 × 10
–19
E = 9.8 × 10
–7
or E =
7
19
9.8 10
1.6 10
-
-
´
´
= 6.125 × 10
12
N
C
–1
.
Example 10.
Can a metal sphere of radius 1cm hold a charge of
1 coulomb.

414. 410 PHYSICS
Solution :
El
ectric field at the surface of the sphere.
E =
2
KQ
R
=
9
22
9101
(110)
-
´´
´
= 9 × 10
13

V
m
This field is much greater than the dielectric strength of air
(3 × 10
6
v/m), the air near the sphere will get ionised and
charge will leak out. Thus a sphere of radius 1 cm cannot
hold a charge of 1 coulomb in air.
ELECTRIC DIPOLE
Two equal and opposite charges separated by a finite distance
constitute an electric dipole. If –q and +q are charges at distance
2l apart, then dipole moment,
2pq=´l
2 l
–q +q
Its SI unit is coulomb metre.
Its direction is from –q to +q. It is a vector quantity.
The torque t on a dipole in uniform electric field as shown in
figure is given by, 2 sinqEt=´ql pE=´
u ur u ur
So t is maximum, when dipole is ^ to field & minimum (=0) when
dipole is parallel or antiparallel to field.
If k
ˆ
pj
ˆ
pi
ˆ
pp
zyx++= and k
ˆ
Ej
ˆ
Ei
ˆ
EE
zyx++=
Then
zyx
zyx
EEE
ppp
k
ˆ
j
ˆ
i
ˆ
=t
q
2l
+q
qE
qE
–q E
The work done in r
otating the dipole from equilibrium through
an angle dq is given by
sin
dWdpEd=tq= qq
and from q
1
® q
2
,
òò
q
q
q
q
qq==
2
1
2
1
dsinpEdWW
I
f q
1
= 0 i.e., equilibrium position, then
W =
2
2
0
pE sin d pE(1 cos )
q
qq= -qò
Workdone in rotating an
electric dipole in uniform electric field
from q
1
to q
2
is W = pE (cosq
1
– cosq
2
)
Potential energy of an electric dipole in an electric field is,
E.p–U= i.e. U = –pE cosq
where q is the angle betweenE
r
and p
r
.
We can also wr
ite
zzyyxx
EpEpEpU++=
Electric Field due to an Electric Dipole
(a)Along the axial line (or end-on position)
–q +q
P
E
p
2l
x
pandE are parallel
2 22
0
12
.
4 ()
ax
px
E
x
=
pÎ -l
when x >> l
(b)Along equatorial line (or
broad-side on position)
–q +ql l
E
x
p
2 3/2
0
1
.
4 ()
eq
p
E
x
2
=
pÎ +l
when x >>l
When p
r
andE
r
are anti parallel then,
E
ax
= 2
E
eq
(c)At any point (from the dipole)
–q
+q
E
y E
x
E
b
q
p
x
p
2
3
0
1
3cos1
4
p
E
x
= q+
pÎ
;
1
tan tan
2
b=q
Electric field intensity due to a point charge varies inversely as
cube of the distance and in case of quadrupole it varies inversely
as the fourth power of distance from the quadrupole.
Electric Force between Two Dipoles
The electrostatic force between two dipoles of dipole moments
p
1
and p
2
lying at a seperation r is
12
4
0
61
4
pp
F
r
=
pÎ
when dipoles are placed co-axiall y
12
4
0
31
4
pp
F
r
=
pÎ
when dipoles are placed perpendicul ar to
each other.
Keep in Memory
1. The dip
ole moment of a dipole has a direction from the
negative charge to the positive charge.
2. If the separation between the charges of the dipole is
increased (or decreased) K-times, the dipole momentincreases (or decreases) by K-times.
3. The torque experienced by a dipole placed in a uniform
electric field has value always lying between zero and pE,where p is the dipole moment and E, the uniform electricfield. It varies directly with the separation between thecharges of the dipole.

415. 411Electric Charges and Fields
4. The work done in rotating a dipole in a uniform electric field
varies from zero (minimum) to 2pE (maximum). Also , it varies
directly with the separation between the charges of the
dipole.
5. The potential energy of the dipole in a uniform electric field
always lies between +pE and –pE.
6. The electric field intensity at a point due to an electric dipole
varies inversely with the cube of the distance of the point
from its centre if the distance is much greater than the length
of the dipole.
7. The electric field at a point due to a small dipole in end-on
position is double of its value in broad side-on position,
i.e. E
End-on
= 2E
Broad side-on
8. For a small dipole, the electric field tends from infinity at a
point very close to the axis of the dipole to zero at a point at
infinity.
9. The force between two dipoles increases (or decreases) by
K
4
-times as the distance between them decreases (or
increases) by K-times.
10.Time period of a dipole in uniform electric field is
2T
pE
=p
I
where I = m oment of inertia of the dipole about the axis of
rotation.
Example 11.
Calculate the electric field intensity due to a dipole of
length 10 cm and having a charge of 500
m C at a point on
the axis distant 20 cm from one of the charges in air.
Solution :
Given : q = 500 × 10
–6
C, a = 10 cm or a/2 = 5 × 10
–2
m,
r = (20 + 5) cm = 25 × 10
–2
m,
p = q × a = (500 × 10
–6
) × (10 × 10
–2
) = 5 × 10
–5
C-m
The electric intensity on the axial line of the dipole is given
by
2
2
2
o
axial
2
a
r
pr2
4
1
E
ú
ú
û
ù
ê
ê
ë
é
÷
ø
ö
ç
è
æ
-
´
ep
=
25
9
axial
8 2 22
2 (25 10) (5 10 )
E (9 10)
1
0 [(25) (5) ]
--
-
´ ´ ´´
=´´
-

71
3.25 10 NC
-
=´
Example 12.
Consider two obje
cts of masses M and 3M seperated by a
distance
l. Charge q and –q are placed on them
respectively and they are lying in an electric field E.
Find the angular frequency of oscillation (S.H.M.)
Solution :
cm
3Ma03
xa
4M4
+
== O
3a
4
a/4
(3M)
qE
qE
(M)
q
(Q x
cm
is the position of centre of mass of system)
Since net force is zero, the centre of mass will not move but
the dipole will rotate about the centre of mass due to torque
q+q=t sin)4/a3(qEsin)4/a(qE
t = q Ea sin q …… (1)
Also, t = I
cm
a = –
2
2
cm
dt
d
I
q
…… (2)
For s
mall oscillations sin q = q
q-=
q
Þ sinqEa
dt
d
I
2
2
cm = –qEaq;
Fro
m eq. (1) & (2), 2
2
cm
22
cm
222
2
2
2
d qEa
Idt
a 3a
I3MM
44
3
Ma 9Ma 12Ma 3
Ma
16 164
qEa 4qE 4qE
or
3 3Ma 3Ma
Ma
4
æöq
=-q
ç÷
èø
æö æö
=+ç÷
ç÷
èø èø
+
= ==
\w = = w=
ELECTRIC FLUX
Electric flux is a measure of the number of electric field lines
passing through the surface. If surface is not open & encloses
some net charge, then net number of lines that go through the
surface is proportional to net charge within the surface.
For uniform electric field when the angle between area vector
()A
ur
and electric field()E
ur
has the same value throughout the
area, EAf=×
uur ur
cosEAÞf=q
q
n
^
E
For uniform electric field when the angle between the area vector
and electric field is not constant throughout the area
dA
n
^
E
d E dA E dAf= × Þf= ×ò
uur uur uur uur
cos cosEdA E dAÞf=q=qòò
Keep in Memory
1.
The electric flux is a scalar although it is a product of two
vectors
E
ur
andA
ur
(because it is a scalar product of the two).

416. 412 PHYSICS
2. The e
lectric flux has values lying between –EA and +EA,
where E and A are the electric field and the area of cross-
section of the surface.
GAUSS'S LAW
It states that, the net electric flux through a closed surface in
vacuum is equal to 1/
e
o
times the net charge enclosed within
the surface.
i.e.,
..===òò
uurrr r
ÑÑ
j
e
in
os
Q
E dA E ndA
where Q
in
repr
esents the net charge inside the gaussian surface S.
..
.
.
.
..
.
.
.
.
.
.
..
.
..
.
.
.
.
..
.
..
.
.
.
.
..
..
.
.
.
..
..
..
.
.
.
.
.
.
.
..
.
.
.
.
..
.
.
.
.
.
..
.
.
.
.
.
.
.
..
.
.
.
.
.
.
.
..
.
.
..
.
.
.
.
.
..
.
..
.
.
.
.
.
..
.
.
.
.
..
.
.
.
.
.
.
.
.
.
.
.
.
.
.
..
.
.
.
.
..
.
..
.
.
.
.
..
.
.
.
..
..
.
..
.
.
.
.
.
.
.
..
..
.
..
.
.
.
..
.
.
.
.
..
.
.
.
.
.
.
..
.
.
.
.
.
..
.
.
.
.
.
.
.
.
.
..
.
.
dA
n
E
Gaussian
surface
Closed surface of irregular shape which enclosed total charge Q
in
In principle, Gauss's law can always be used to calculate the
electric field of a system of charges or a continuous distribution
of charge . But in practice it is useful only in a limited number of
situation, where there is a high degree of symmetry such as
spherical, cylindrical etc.
(i) The net electric flux through any closed surface depends
only on the charge inside that surface. In the figures, the
net flux through S is q
1
/e
o
, the net flux through S’ is (q
2
+q
3
)/e
o
and the net flux, through S" is zero.

q
1
(a)
S

q
2
q
3
(b)
S'

(c)
S"
(ii) A point charge Q is located outside a closed surface S. In
this case note that the number of lines entering the surfaceequals to the number of lines leaving the surface. In otherwords the net flux through a closed surface is zero, if thereis no charge inside.
S
Ele
ctric field
lin
es
Q
(iii) The n
et flux across surface A is zero
–q
+q
A
i.e., ò
=
e
==f 0
Q
Sd.E
o
in
because Q
in
= – q + q =
0
Applications of Gauss’s Law :
(i)To determine electric field due to a point charge :
E
dAQ
Gaussian
surface
r
The point char
ge Q is at the centre of spherical surface
shown in figure. Gaussian surface and
E is parallel toAd
(direction normal to Gaussian surface ) at every point on
the Gaussian surface.
so,
c
o
Q
E.dA EdAf= ==
e
òò
ur ur
ÑÑ
2
o
Q
E
4r
Þ=
pe ()
2
A 4r=pQ
(ii)To determ
ine electric field due to a cylindrically symmetric
charge distribution :
We calculate the electric
field at a distance r from a
uniform positive line
charge of infinite length
whose charge per unit
length is l = constant. The
flux through the plane
surfaces of the Gaussian
cylinder is zero, since
r
E is
+
+
+
+
+
+
+
+
dA
dA
E
E
Gaussian
surface
l
parallel to t
he plane of end surface (
r
Eis perpendicular to
r
dA). The total charge inside the Gaussian suface is l l,
where l is linear charge density and l is the length of cylinder.
Now applying Gauss’s law and notingE is parallel toAd
everywhere on cy
lindrical surface, we find that
c
Gaussian
surface
E.dAf==ò
ur ur
Ñ
(flux)
both ends
+ (flux)
cylindrical surface
= 0 + E
. 2A
2E . prl =
in
oo
Q l
=
ee
l
or
r2
E
ope
l
=

417. 413Electric Charges and Fields
Keep in Memory
1.
The closed imaginary surfaces drawn around a charge are
called Gaussian surfaces.
2. If the flux emerging out of a Gaussian surface is zero then it
is not necessary that the intensity of electric field is zero.
3. In the Gauss's law,
in
0
Q
E. dA=
Î
ò
ur u u ur
Ñ
E
ur
is the resultant electric field due to all charges lying
inside or outside the Gaussian surface, but Q
in
is the charge
lying only inside the surface.
4. The net flux of the electric field through a closed surface
due to all the charges lying inside or outside the surface is
equal to the flux due to the charges only enclosed by the
surface.
5. The electric flux through any closed surface does not
depend on the dimensions of the surface but it depends
only on the net charge enclosed by the surface.
Example 13.
A charge q is enclosed in a cube. What is the electric fluxassociated with one of the faces of cube?
Solution :
According to Gauss’s theorem,
Total electric flux f=
1
e
o
× total enclosed charge
1
=´
e
o
q
.
Since cube h
as six faces, hence electric flux linked with
each face = (1/6f) = q/6e
o
.
Example 14.
The following figure shows a surface S which is enclosing
–2q charge. The charge +q is kept outside the surface S.
Calculate the net outward/ inward flux from the surface
S.
S
+q
–2q
Solution :
Ac
cording to Gauss’s law, the net flux is
o
1
e
=f × net char
ge enclosed by closed surface
o
q2
e
-
=
(Because +q is
outside the surface S, so net flux due to +q
is zero)
Example 15.
In which of the following figures, the electric flux ismaximum?
q
A

q
B

q
C
Solution :
Ac
cording to Gauss’s law, the electric flux linked with a
closed surface depends only on net charge enclosed bythat surface. It does not depend on the shape and size ofthat closed surface. Hence electric flux linked in above threefigures are same i.e., f
A
= f
B
= f
C
.
Example 16.
A point charge +q is placed at mid point of a cube of side‘L’. What is the electric flux emerging from the cube ?
Solution :
According to Gauss’s law,
f
net closed surface
=
o
q
e

o
Net charge enclosed by closed surface
=
e

418. 414 PHYSICS
Coulomb’s Law
12
2
0
9 2 2
0
1
4
1
9 10
4
qq
F
r
NmC
-
=
pÎ
=´
pÎ
Charge Due to which matter
produces and experiences
electric and magnetic effects.
ELECTRIC CHARGES
AND FIELDS
Electric field Space surrounding
a charge in which its electrostatic
force can be experienced by any
test charge
2
0
1
ˆ
4
=
=
×
p
Î
F
q
E
r
q
r
E
l
e
c
t
r
i
c

f
i
e
l
d
l
i
n
e
s

I
m
a
g
i
n
a
r
y

l
i
n
e
a
l
o
n
g

w
h
i
c
h
a
p
o
s
i
t
i
v
e

te
s
t

c
h
a
r
g
e
w
i
l
l
m
o
v
e

i
f

l
e
f
t
f
r
e
e
M
e
t
h
o
d
s

o
f
c
h
a
r
g
i
n
g
B
y
i
n
d
u
c
t
i
o
n
B
y
c
o
n
d
u
c
t
i
o
n
By friction
Distribution
of charge
Surface charge
density
Volume charge
Linear charge
density
charge
density =
volume
charge
=
length
charge
=
area
Types of charge
Neutral
No of electrons
= no. of protons
Positive charge
Deficiency of
electrons
Negative charge
Excess of electron
B
a
s
i
c
p
r
o
p
e
r
t
ie
s

o
f
e
l
e
c
t
r
i
c
c
h
a
r
g
e
Quantiz
a
t
io
n
Q = ± n
e
C
o
n
s
e
r
v
a
t
i
o
n
N
e
i
t
h
e
r

c
r
e
a
t
e
d
n
o
r
d
e
s
t
r
o
y
e
d
I
n
v
a
r
i
a
n
t
T
r
a
n
s
f
e
r
a
b
l
e

b
e

t
r
a
n
s
f
e
r
e
d

o
n
e

b
o
d
y
t
o
a
n
o
t
h
e
r
c
a
n
f
r
o
m
A
s
s
o
c
i
a
t
e
d
w
i
t
h
m
a
s
s
M
a
s
s

o
f
e
l
e
c
tr
o
n
M
=

9
.1

×

1
0
k
g
e
–
3
1
T
o
r
q
u
e
a
n
d
p
o
t
e
n
t
i
a
l
e
n
e
r
g
y
o
f

a
d
i
p
o
le
T
o
r
q
u
e

=
P
E

s
i
n

t
q
t

=
P
×

E
P
o
t
e
n
t
i
a
l
e
n
e
r
g
y
U
=

–
P
E
c
o
s

U
=

–
P
.E
.
q
Electric field
due to dipole
At axial position
Due to descrete
distribution of charge
3
0
12
ˆ E=
4
P
r
r pÎ
1
E=
n
i
i
E
=
å
urr
At equatorial
position
Due to continuous
distribution of charge
3
0
1
ˆ.
4
p
Er
r
-
=
pÎ
3
0
1
E
4
dq
r
r
=
pÎ
ò
r
E
l
e
c
t
r
i
c

f
l
u
x

=
E
.A

c
o
s


=
E
. A
f
q
f
G
a
u
s
s
’
s

th
e
o
r
e
m
T
o
t
a
l
f
l
u
x

o
v
e
r

a
c
l
o
s
e
d
s
u
r
f
a
c
e
i
s
A
p
p
l
i
c
a
i
o
n
o
f
G
a
u
s
s
’s

t
h
e
o
r
e
m

d
u
e

to

i
n
f
i
n
i
t
e
l
y

l
o
n
g

s
t
r
a
i
g
h
t
w
i
r
e
I
n
f
i
n
i
te

p
l
a
n
e
s
h
e
e
t
0
1
Î
0
2
l
=
p
e
E
r
0
2
s
=
e
E
t
i
m
e
s
t
h
e

t
o
t
a
l
e
n
c
lo
s
e
d

c
h
a
r
g
e
n
c
l
o
s
e
d
0
.
f
=
=
e
ò
q
E
d
s
Properties of
electric field l
i
n
e
s
Never inters
e
c
t
each other
Never form
closed loops
A
l
w
a
y
s
n
o
r
m
a
l
t
o

c
o
n
d
u
c
t
i
n
g
s
u
r
f
a
c
e
Come out of
positive charge a
n
d
go into negativ
e
charge
C
O
N
C
E
P
T
M
A
P

419. 415Electric Charges and Fields
1.The electric field strength at a distance r from a charge q is
E. What will be electric field strength if the distance of the
observation point is increased to 2 r?
(a) E/2 (b) E/4
(c) E/6 (d) None of these
2.The surface density on the copper sphere is s. The electric
field strength on the surface of the sphere is
(a)s (b)s/2
(c) Q / 2e
o
(d) Q / e
o
3.What is the angle between the electric dipole moment and
the electric field due to it on the axial line?
(a) 0º (b) 90º
(c) 180º (d) None of these
4.In a region of space having a uniform electric field E, a
hemispherical bowl of radius r is placed. The electric flux f
through the bowl is
(a)2pRE (b) 4pR
2
E
(c)2pR
2
E (d)pR
2
E
5.A cylinder of radius R and length l is placed in a uniform
electric field E parallel to the axis of the cylinder. The total
flux over the curved surface of the cylinder is
(a) zero (b)pR
2
E
(c)2pR
2
E (d) E / pR
2
6.An electric dipole when placed in a uniform electric field E
will have minimum potential energy if the dipole moment
makes the following angle with E.
(a)p (b)p/2
(c) zero (d) 3 p/2
7.At the centre of a cubical box + Q charge is placed. The
value of total flux that is coming out a wall is
(a) Q / e
o
(b) Q / 3e
o
(c) Q / 4e
o
(d) Q / 6e
o
8.If a charge is moved against the coulomb force of an electric
field, then
(a) work is done by the electric field
(b) energy is used from some outside source
(c) the strength of the field is decreased
(d) the energy of the system is decreased
9.The charge given to any conductor resides on its outer
surface, because
(a) the free charge tends to be in its minimum potential
energy state
(b) the free charge tends to be in its minimum kinetic energy
state
(c) the free charge tends to be in its maximum potential
energy state.
(d) the free charge tends to be in its maximum kinetic energy
state
10.Identify the wrong statement in the following. Coulomb's
law correctly describes the electric force that
(a) binds the electrons of an atom to its nucleus
(b) binds the protons and neutrons in the nucleus of an
atom
(c) binds atoms together to form molecules
(d) binds atoms and molecules together to form solids
11.An infinite parallel plane sheet of a metal is charged to
charge density s coulomb per square metre in a medium of
dielectric constant K. Intensity of electric field near the
metallic surface will be
(a)
K
E
oe
s
= (b)
o2
E
e
s
=
(c)
K2
E
oe
s
= (d)
o2
K
E
e
s
=
12.In a medium o
f dielectric constant K, the electric field is
E
r
.
If
0
Î is permitt
ivity of the free space, the electric
displacement vector is
(a)
0
EK
Î
r
(b)
0K
E
Î
r
(c)
K
E
0
r
Î
(d) EK
0
r
Î
13.Two con
ducting spheres of radii r
1
and r
2
are charged to
the same surface charge density. The ratio of electric fields
near their surface is
(a)
2
2
2
1
r/r (b)
2
1
2
2
r/r
(c)
21
r/r (d) 1 : 1
14.A charge q is placed at the centre of the open end of a
cylindrical vessel. The flux of the electric field through the
surface of the vessel is
q
(a) zero (b) q/e
o
(c) q/2e
o
(d) 2q/e
o
15.Two thin infinite parallel sheets have uniform surfacedensities of charge + s and – s. Electric field in the space
between the two sheets is
(a)s/e
o
(b)s/2e
o
(c)2s/e
o
(d) zero

420. 416 PHYSICS
16.The E-r curv
e for an infinite linear charge distribution will
be
(a)
E
r
(b) E
r
(c)E
r
(d) E
r
17.If a dipole of dipole moment p
r
is placed in a uniform electric
field E
r
, then torque acting on it is given by
(a) E.p
rrr
=t (b) Ep
rrr
´=t
(c) Ep
rrr
+=t (d) Ep
rrr
-=t
18.A charge Q i
s enclosed by a Gaussian spherical surface of
radius R. If the radius is doubled, then the outward electric
flux will
(a) increase four times(b) be reduced to half
(c) remain the same(d) be doubled
19.What is the value of E in the space outside the sheets?
(a)s/e
o
(b)s/2e
o
(c) E ¹ 0 (d) 2s/e
o
20.Two charges are at a distance d apart. If a copper plate of
thickness
2
d
is kept between them, the effective force will
be
(a) F/2 (b) zero
(c) 2F (d) F2
1.The electric fie
ld intensity just sufficient to balance the
earth’s gravitational attraction on an electron will be: (given
mass and charge of an electron respectively are
31
9.1 10 kg
-
´ and C106.1
19-
´ .)
(a) –5.6 ×
10
–11
N/C (b) –4.8 × 10
–15
N / C
(c) –1.6 × 10
–19
N/C (d) –3.2 × 10
–19
N / C
2.The insulation property of air breaks down when the electricfield is 3 × 10
6
Vm
–1
. The maximum charge that can be
given to a sphere of diameter 5 m is approximately
(a) 2 × 10
–2
C (b) 2 × 10
–3
C
(c) 2 × 10
–4
C (d) 2 × 10
–5
C
3.There is an electric field E in X-direction. If work done in
moving a charge 0.2 C through a distance of 2m along a line
making an angle of 60 degree with X-axis is 4.0 joule, what
is the value of E?
(a)
3 newton per coulomb
(b) 4 netwon per coulomb(c) 5 newton per coulomb(d) None of these
4.From a point charge, there is a fixed point A. At A, there is
an electric field of 500 V/m and potential difference of 3000
V. Distance between point charge and A will be
(a) 6 m (b) 12 m
(c) 16 m (d) 24 m
5.If electric field in a region is radially outward with magnitude
E = Ar, the charge contained in a sphere of radius r centred
at the origin is
(a)
3
o
1
Ar
4pe
(b) 4pe
o
Ar
3
(c)
3
o
1A
4 rpe
(d)
o
3
4A
r
pe
21.When air is replaced by a dielectric medium of force
constant K, the maximum force of attraction between two
charges, separated by a distance
(a) decreases K-times(b) increases K-times
(c) remains unchanged (d) becomes
2
1
K
times
22.A point Q lies on th
e perpendicular bisector of an electrical
dipole of dipole moment p. If the distance of Q from the
dipole is r (much larger than the size of the dipole), then
the electric field at Q is proportional to
(a)p
–1
and r
–2
(b)p and r
–2
(c)p
2
and r
–3
(d)p and r
–3
23.A particle of mass m and charge q is placed at rest in a
uniform electric field E and then released. The kinetic energy
attained by the particle after moving a distance y is
(a) q E y
2
(b) q E
2
y
(c) q E y (d) q
2
E y
24.Intensity of an electric field (E) depends on distance r, due
to a dipole, is related as
(a)
1
E
r
µ (b)
2
1
E
r
µ
(c)
3
1
E
r
µ (d)
4
1
E
r
µ
25.The formation of
a dipole is due to two equal and dissimilar
point charges placed at a
(a) short distance (b) long distance
(c) above each other(d) none of these

421. 417Electric Charges and Fields
6.A hollow insulated conduction sphere is given a positive
charge of 10 mC. What will be the electric field at the centre
of the sphere if its radius is 2 metres?
(a) Zero (b) 5 mCm
–2
(c) 20 mCm
–2
(d) 8 mCm
–2
7.An electric dipole has the magnitude of its charge as q and
its dipole moment is p. It is placed in uniform electric field E.
If its dipole moment is along the direction of the field, the
force on it and its potential energy are respectively.
(a) q.E and max. (b) 2 q.E and min.
(c) q.E and min (d) zero and min.
8.If the dipole of moment 2.57 × 10
–17
cm is placed into an
electric field of magnitude 3.0 × 10
4
N/C such that the fields
lines are aligned at 30° with the line joining P to the dipole,
what torque acts on the dipole?
(a) 7.7 × 10
–13
Nm (b) 3.855 × 10
–13
Nm
(c) 3.855 × 10
–15
Nm (d) 7.7 × 10
–15
Nm
9.An electric dipole is placed at an angle of 30° with an electric
field of intensity 2 × 10
5
NC
–1
, It experiences a torque of
4 Nm. Calculate the charge on the dipole if the dipole length
is 2 cm.
(a) 8 mC (b) 4 mC
(c) 8 mC (d) 2 mC
10.Charge Q is distributed to two different metallic spheres
having radii R and 2R such that both spheres have equal
surface charge density, then charge on large sphere is
(a)
4Q
5
(b)
Q
5
(c)
3Q
5
(d)
5Q
4
11.Two point charges q
1
= 4mC and q
2
= 9 mC are placed 20 cm
apart. The electric field due to them will be zero on the line
joining them at a distance of
(a) 8 cm from q
1
(b) 8 cm form q
2
(c)
1
80
cm from q
13
(d) 2
80
cm from q
13
12.Three char
ge q, Q and 4q are placed in a straight line of
length l at points distant 0,
1
2
and l respectively from one
end. In order to make the net froce on q zero, the charge Qmust be equal to(a) –q (b) – 2q
(c)
q
2
-
(d) q
13.Among two di
scs A and B, first have radius 10 cm and charge
10
–6
µC and second have radius 30 cm and charge 10
–5
µC.
When they are touched, charge on both q
A
and q
B
respectively will, be
(a)
AB
q = 2.75
μC, q =3.15μC
(b)
AB
q = 1.09μC, q = 1.53μC
(c)
AB
q = q = 5.5μC
(d) None of these
14.Two particle of equal mass m and charge q are placed at a
distance of 16 cm. They do not experience any force. The
value of
q
m
is
(a)l (b)
0
G
pe
(c)
0
4
G
pe
(d)
0
4Gpe
15.The spatial distribution of electric field due to charges
(A, B) is shown in figure. Which one of the following
statements is correct ?
A B
(a) A is +ve and B –ve, |A| > |B|
(b) A is –ve and B +ve, |A| = |B|(c) Both are +ve but A > B(d) Both are –ve but A > B
16.A drop of oil of density r and radius r carries a charge q
when placed in an electric field E, it moves upwards with avelocity v. If r
0
is the density of air, h be the viscosity of
the air, then which of the following force is directedupwards?
(a) q E (b)6
πηrv
(c) g)(r
3
4
0
3
r-rp (d) None of thes
e
17.ABC is an equilateral triangle. Charges +q are placed at
each corner as shown as fig. The electric intensity at centre
O will be
A+q
r
O
r
CB
+q +q
r
(a)
r
q
4
1
oÎp
(b)
2
or
q
4
1
Îp
(c)
2
or
q3
4
1
Îp
(d) zero
18.An electri
c dipole is placed along the x-axis at the origin O.
A point P is at a distance of 20 cm from this origin such thatOP makes an angle p/3 with the x-axis. If the electric field at
P makes an angle q with the x-axis, the value of q would be
(a)
3
p
(b)
÷
÷
ø
ö
ç ç
è
æ
+
p -
2
3
tan
3
1
(c)
3
2p
(d)
÷ ÷
ø
ö
ç ç
è
æ
-
2
3
tan
1

422. 418 PHYSICS
19.A simple pe
ndulum has a length
l, mass of bob m. The bob
is given a charge q. The pendulum is suspended between
the vertical plates of the charged parallel plate capacitor. If
E is the field strength between the plates, then time period
T equal to
+
+ -
-
(a)
g
2
l
p (b)
m
qE
g
2
+
p
l
(c)
m
qE
g
2
-
p
l
(d)
2
2
m
qE
g
2
÷
ø
ö
ç
è
æ
+
p
l
20.Two identical th
in rings, each of radius a meter, are coaxially
placed at a distance R meter apart. If Q
1
coulomb and Q
2
coulomb are respectively the charges uniformly spread on
the two rings, the work done in moving a charge q coulomb
from the centre of one ring to that of the other is
(a) Zero (b)
12
0
q(Q Q )( 2 1)
42a
--
pe
(c)
12
0
q 2(Q Q)
4a
+
pe
(d)
12
0
q(Q Q )( 2 1)
42a
++
pe
21.A metallic sphere is place
d in a uniform electric field. The
line of force follow the path (s) shown in the figure as
1
2
3
4
1
2
3
4
(a)1 (b) 2
(c)3 (d) 4
22.A s
oap bubble is given negative charge, its radius will
(a) increase (b) decrease
(c) remain unchanged(d) fluctuate
23.A and B are two identically spherical charged bodies which
repel each other with force F, kept at a finite distance. A
third uncharged sphere of the same size is brought in contact
with sphere B and removed. It is then kept at mid point of A
and B. Find the magnitude of force on C.
(a) 2
F
(b)
8
F
(c)F (d) Zero
24.An e
lectric dipole of moment
P is placed in a uniform
el
ectric field
E such that P points along E. If the dipole
is slightly ro
tated about an axis perpendicular to the plane
containing
E and P and passing through the centre of
the dipole, the dipole executes simple harmonic motion.
Consider I to be the moment of inertia of the dipole about
the axis of rotation. What is the time period of such
oscillation?
(a)
)I/pE( (b) )pE/I(2p
(c) )pE2/I(2p (d) None of these
25.The
re exists a non-uniform electric field along x-axis as
shown in the figure below. The field increases at a uniform
rate along +ve x-axis. A dipole is placed inside the field as
shown. Which one of the following is correct for the dipole?
a
x-axis
–q
+q
(a) Dipole move
s along positive x-axis and undergoes a
clockwise rotation
(b) Dipole moves along negative x-axis and undergoes a
clockwise rotation
(c) Dipole moves along positive x-axis and undergoes a
anticlockwise rotation
(d) Dipole moves along negative x-axis and undergoes a
anticlockwise rotation
26.The electric dipole is situated is an electric field as shownin fig 1. The dipole and electric field are both in the plane ofthe paper. The dipole is rotated about an axis perpendicularto plane of paper passing through A in anticlockwisedirection. If the angle of rotation (q) is measured with respect
to the direction of electric field, then the torque (t)
experienced by the dipole for different values of the angleof rotation q will be represented in fig. 2, by
q
E
–q
+q
Fig. 1
p 2p
3p
(1)
O
t (2)(3)(4)
Fig. 2
(a) curve (1) (b) curve (
2)
(c) curve (3) (d) curve (4)

423. 419Electric Charges and Fields
27.Force between two identical charges placed at a distance
of r in vacuum is F. Now a slab of dielectric of dielectric
contrant 4 is inserted between these two charges. If the
thickness of the slab is r/2, then the force between the
charges will become
(a)F (b)
3
F
5
(c)
4
F
9
(d)
F
2
28.A charge +q is at a distance L/2 above a square of side L.
Then what is the flux linked with the surface?
(a)
0
q
4e
(b)
0
2q
3e
(c)
0
q
6e
(d)
0
6q
e
29.Positive and negative point charges of equal magnitude
are kept at
a
0, 0,
2
æö
ç÷
èø
and
a
0, 0,
2
-æö
ç÷
èø
respectively. The work
done by the electric field when another positive point charge
is moved from (–a, 0, 0) to (0, a, 0) is
(a) positive
(b) negative
(c) zero
(d) depends on the path connecting the initial and final
positions
30.A short electric dipole of dipole moment
p
r
is placed at a
d
istance r from the centre of a solid metallic sphere of radius
a (<< r) as shown in the figure. The electric field intensity atthe centre of sphere C due to induced charge on the sphereis
–q O +q
r
C
a
(a) zero (b)
3
0
1 2p
4rpe
along CO
(c)
3
0
1 2p
4rpe
along OC(d)
3
0
1p
4rpe
along CO
31.The
electrostatic potential (f
r
) of a spherical symmetric
system kept at origin, is shown in the figure, and given as
0
0
()
4
r
q
rR
r
f=³
pe

R
0
r
f
r
0
00
()
4
r
q
rR
R
f=£
pe
Which o
f the following option(s) is/are incorrect
(a) For spherical region
0
rR£ total electrostatic energy
stored is zero
(b) Within
0
2rR= total charge is q.
(c) There will be no charge anywhere except at
0
rR=
(d) None of th
ese
32.A solid sphereical conductor of radius R has a spherical
cavity of radius a (a < R) at its centre. A charge + Q is kept
at the centre. The cahrge at the inner surface, outer and at a
position r (a < r < R) are respectively
(a) + Q, – Q, 0 (b) – Q, + Q, 0
(c) 0, – Q, 0 (d) + Q , 0, 0
33.A glass rod rubbed with silk is used to charge a gold leaf
electroscope and the leaves are observed to diverge. The
electroscope thus charged is exposed to X-rays for a short
period. Then
(a) the divergence of leaves will not be affected
(b) the leaves will diverge further
(c) the leaves will collapse
(d) the leaves will melt
34.A solid conducting sphere of radius a has a net positive
charge 2Q. A conducting spherical shell of inner radius b
and outer radius c is concentric with the solid sphere and
has a net charge – Q. The surface charge density on the
inner and outer surfaces of the spherical shell will be
(a)
22
2
,
44
QQ
bc
-
pp
a
b
c
(b)
22
,
44
QQ
bc
-
pp
(c)
2
0,
4
Q
cp
(d) None of these
35.An uncharged sphere of metal is placed in between charged
plates as shown. The lines of force look like
+++++++
–––––––
A
+++++++
–––––––
B
+++++++
–––––––
C
+++++++
–––––––
D
(a)A (b) B
(c
)C (d) D

424. 420 PHYSICS
36.If a charge q is place
d at the centre of the line joining two
equal charges Q such that the system is in equilibrium then
the value of q is
(a) Q/2 (b) –Q/2
(c) Q/4 (d) –Q/4
37.The electric intensity due to a dipole of length 10 cm and
having a charge of 500 mC, at a point on the axis at a distance
20 cm from one of the charges in air, is
(a) 6.25 × 10
7
N/C (b) 9.28 × 10
7
N/C
(c) 13.1 × 10
11
N/C (d) 20.5 × 10
7
N/C
38.Two positive ions, each carrying a charge q, are separated
by a distance d. If F is the force of repulsion between the
ions, the number of electrons missing from each ion will be
(e being the charge of an electron)
(a)
2
0
2
4Fd
e
pe
(b)
2
0
2
4Fe
d
pe
(c)
2
0
2
4Fd
e
pe
(d)
2
0
2
4Fd
q
pe
39.A square surface of s
ide L meter in the plane of the paper is
placed in a uniform electric field E (volt/m) acting along the
same plane at an angle
qwith the horizontal side of the
square as shown in Figure. The electric flux linked to thesurface, in units of volt. m, is
E
q
(a) EL
2
(b) EL
2
cosq
(c) EL
2
sinq (d) zero
40.The electric
potential V at any point (x, y, z), all in meters in
space is given by V = 4x
2
volt. The electric field at the point
(1, 0, 2) in volt/meter is
(a) 8 along positive X-axis (b) 16 along negative X-axis
(c) 16 along positive X-axis (d) 8 along negative X-axis
41.Three charges, each +q, are placed at the corners of an
isosceles triangle ABC of sides BC and AC, 2a. D and E are
the mid points of BC and CA. The work done in taking a
charge Q from D to E is
(a)
0
3qQ
8apÎ
(b)
0
qQ
4apÎ
A
B
D
C
E
(c) zero (d)
0
3qQ
4apÎ
42.Two metallic sphe
res of radii 1 cm and 3 cm are given
charges of –1×10
–2
C and 5×10
–2
C, respectively. If these
are connected by a conducting wire, the final charge on the
bigger sphere is
(a) 2 × 10
–2
C (b) 3 × 10
–2
C
(c) 4 × 10
–2
C (d) 1 × 10
–2
C
43.An electric dipole of moment ´p´ is placed in an electric
field of intensity ´E´. The dipole acquires a position such
that the axis of the dipole makes an angle q with the direction
of the field. Assuming that the potential energy of the dipole
to be zero when = 90°, the torque and the potential energy
of the dipole will respectively be
(a)p E sin q, – p E cos q(b)p E sin q, –2 p E cos q
(c)p E sin q, 2 p E cos q(d)p E cos q, – p E cos q
44.Figure shows two charges of equal magnitude separated by
a distance 2a. As we move away from the charge situated at
x = 0 to the charge situated at x = 2a, which of the following
graphs shows the correct behaviour of electric field ?
+q +q
xa = 2x = 0
xa =
(a) y
x
2a
a
O
E
®
(b)
y
E
®
O
a 2a
x
(c)
y
E
®
O
a 2a
x
(d)
2aa
xO
y
E
®
DIRECTIONS (for Qs. 45 to 50) : Each question contains
ST
ATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following-
(a)Statement-1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(b)Statement-1 is true, Statement-2 is true; Statement -2 is
NOT a correct explanation for Statement-1
(c)Statement-1 is true, Statement-2 is false
(d)Statement-1 is false, Statement-2 is true
45. Statement-1 : Consider two identical charges placed
distance 2d apart, along x-axis.
The equilibrium of a positive test charge placed at the
point O midway between them is stable for displacementsalong the x-axis.Statement-2: Force on test charge is zero.

425. 421Electric Charges and Fields
46. Statement-1 : A deuteron and an a-particle are placed in an
electric field. If F
1
and F
2
be the forces acting on them and
a
1
and a
2
be their accelerations respectively then, a
1
= a
2
.
Statement-2 : Forces will be same in electric field.
47. Statement-1 : Four point charges q
1
, q
2
, q
3
and q
4
are as
shown in figure. The flux over the shown Gaussian surface
depends only on charges q
1
and q
2
.
Statement-2 : Elec tric field at all points on Gaussian surface
depends only on charges q
1
and q
2
.
Exemplar Questions
1.In figure two positive charges q
2
and q
3
fixed along the y-
axis, exert a net electric force in the + x-direction on a chargeq
1
fixed along the x-axis. If a positive charge Q is added at
(x, 0), the force on q
1
y
q
2
q
1
q
3
x
y
q
2
q
1
q
3
x
Q
O(x, 0)
(i) (ii)
(a) s
hall increase along the positive x-axis
(b) shall decrease along the positive x-axis
(c) shall point along the negative x-axis
(d) shall increase but the direction changes because of
the intersection of Q with q
2
and q
3
2.A point positive charge is brought near an isolated conducting
sphere (figure). The electric field is best given by
+q +q
(a) (b)
+q +q
(c) (d)
3.The electric flux through the surface
s
+ q + q
(i)
s
(ii)
+q
s
(iii)
+q
(iv)
(a) in Fig. (iv) is the largest
(b) in Fig. (iii) is the least(c) in Fig. (ii) is same as Fig. (iii) but is smaller than Fig. (iv)(d) is the same for all the figures
4.Five charges q
1
, q
2
, q
3
,q
4
, and q
5
are fixed at their positions
as shown in Figure, S is a Gaussian surface. The Gauss’ law
is given by
0s
q
E.dS.=
e
ò
Which of t
he following statements
is correct?
q
1
q
2
q
3
q
4q
5
48. Statement-1 : The positive charge particle is placed in front
of a spherical uncharged conductor. The number of lines of
forces terminating on the sphere will be more than those
emerging from it.
Statement-2 : The surface charge density at a point on the
sphere nearest to the point charge will be negative and
maximum in magnitude compared to other points on the
sphere.
49..Statement-1 : A uniformly charged disc has a pin hole at its
centre. The electric field at the centre of the disc is zero.
Statement-2 : Disc can be supposed to be made up of many
rings. Also electric field at the centre of uniformly charged
ring is zero.
50. Statement-1 : When a conductor is placed in an external
electrostatic field, the net electric field inside the conductor
becomes zero after a small instant of time.
Statement-2 : It is not possible to set up an electric field
inside a conductor.

426. 422 PHYSICS
(a) E on the LHS of the above equation will have a
contribution from q
1
, q
5
and q
1
, q
5
and q
3
while q on
the RHS will have a contribution from q
2
and q
4
only
(b) E on the LHS of the above equation will have a
contribution from all charges while q on the RHS will
have a contribution from q
2
and q
4
only
(c) E on the LHS of the above equation will have a
contribution from all charges while q on the RHS will
have a contribution from q
1
, q
3
and q
5
.
(d) Both E on the LHS and q on the RHS will have
contributions from q
2
and q
4
only
5.Figure shows electric field lines in which an electric dipole
P is placed as shown. Which of the following statements is
correct?
–pp
+p
(a) The dipole will not experience any force
(b) The dipole will experience a force towards right
(c) The dipole will experience a force towards left
(d) The dipole will experience a force upwards
6.A point charge + q is placed at a distance d from an isolated
conducting plane. The field at a point P on the other side of
the plane is
(a) directed perpendicular to the plane and away from
the plane
(b) directed perpendicular to the plane but towards the
plane
(c) directed radially away from the point charge
(d) directed radially towards the point charge.
7.A hemisphere is uniformely charged positively. The electric
field at a point on a diameter away from the centre is directed
(a) perpendicular to the diameter
(b) parallel to the diameter
(c) at an angle tilted towards the diameter
(d) at an angle tilted away from the diameter
NEET/AIPMT (2013-2017) Questions
8.Two pith balls carrying equal charges are suspended from
a common point by strings of equal length. The equilibrium
separation between them is r. Now the strings are rigidly
clamped at half the height. The equilibrium separation
between the balls now become [2013]
y/2
y
(a)
3
2
ræö
ç÷
èø
(b)
2
3
ræö
ç÷
èø
(c)
2
3
ræö
ç÷
èø
(d)
2
2
ræö
ç÷
èø
9.A charge ‘q’ is placed at the centre of the line joining two
equal charges ‘Q’. The system of the three charges will be
in equilibrium if ‘q’ is equal to [NEET Kar. 2013]
(a)Q/2 (b) – Q/4
(c)Q/4 (d) – Q/2
10.An electric dipole of dipole moment p is aligned parallel to
a uniform electric field E. The energy required to rotate the
dipole by 90° is [NEET Kar. 2013]
(a)pE
2
(b)p
2
E
(c)pE (d) infinity
11.The electric field in a certain region is acting radially outward
and is given by E = Aa. A charge contained in a sphere of
radius 'a' centred at the origin of the field, will be given by
(a) A e
0
a
2
(b) 4 pe
0
Aa
3
[2015]
(c)e
0
Aa
3
(d) 4 pe
0
Aa
2
12.Two identical charged spheres suspended from a common
point by two massless strings of lengths l, are initially at a
distance d (d << l ) apart because of their mutual repulsion.
The charges begin to leak from both the spheres at a
constant rate. As a result, the spheres approach each other
with a velocity v. Then v varies as a function of the distance
x between the spheres, as : [2016]
(a)
1
2
vxµ (b)vxµ
(c)
1
2
vx
-
µ (d)
1
vx
-
µ
13.Suppose the charge of a proton and an electron differ
slightly. One of them is – e, the other is (e + De). If the net of
electrostatic force and gravitational force between two
hydrogen atoms placed at a distance d (much greater than
atomic size) apart is zero, then De is of the order of [Given
mass of hydrogen m
h
= 1.67 × 10
–27
kg] [2017]
(a) 10
–23
C (b) 10
–37
C
(c) 10
–47
C (d) 10
–20
C

427. 423Electric Charges and Fields
EXERCISE - 1
1. (
b) As new distance = 2 r and electric field due to single
charge,
1
E
2
r
µ,
theref
ore, new intensity = E/4.
2. (d) According to Gauss's theorem,
2
2
0
q/4R
E [q/4 R]
p
\ = p =s
Î
Q
or
o
/Ees=
3
. (a) Electric field
E
r
is along the dipole axis.
\ q = 0º.
4. (c)
22
E(ds) cos E(2 r )cos0º 2 r Ef= q= p = p .
5
. (a) For the curved surface, q = 90ºE ds cos90º0\f== .
6. (c) p
U – • = –pE cos=qpE
(U
p
)
minimum
= –
pE, when q = 0°
7. (d) According to Gauss’ Law
flux
Q
ds.E
o
sur
face closedby enclosed
=
e
=
ò
so total flux = Q/e
o
Since cube has six face, so flux coming out through
one wall or one face is Q/6e
o
.
8. (b) To overcome electrostatic repulsion forces, work will
have to be done by external agency.
9. (a)
10. (b) Nuclear force binds the protons and neutrons in the
nucleus of an atom.
11. (c) Electric field intensity due to thin infinite plane sheet
of charge,
K22
E
0e
s
=
e
s
=
[Where d
ielectric constant,
0
K
e
=
e
]
12. (d) E
lectric displacement vector,
EDe=
As, K
0
e=e EKD
0e=\
13. (d) A
s
21
s=s
2
2
2
2
1
1
r4
Q
r4
Q
p
=
p
\
or
12
22
o1 o2
QQ
4r 4r
=
pe pe
21
EE=
\ or 1E/E
21
=
12
E :E 1:1Þ=
14. (a)
The flux is zero according to Gauss’ Law because it is
a open surface which enclosed a charge q.
15. (a)
÷
÷
ø
ö
ç
ç
è
æ
e
s-
-
e
s
=-=
oo
21
22
EEE /
o
=se
The field intensity in between sheets having equal
and opposite uniform surface densities of charge
becomes constant. ie, an uniform electric field is
produced and it is independent of the distance between
the sheets.
16. (c) The field due to infinite linear charge distribution
ò
pe
=
r
dq
4
1
E
0
.hyperbolaSo
r
1
EµÞ
17. (b)Giv
en : Dipole moment of the dipole = p
r
and uniform
e
lectric field =
®
E.Torque (t) = E ither force ×
perpendicular distance between the two forces =
qaEsinq or q=t sinpE or Ep
rrr
´=t (vector form)
18. (c) By Gauss’s theorem, f =
in
0
Q
Î
Thus, the net flux depends only on the charge
enclosed by the surface. Hence, there will be no effect
on the net flux if the radius of the surface is doubled.
19. (c) Inside a charged conducting surface E = 0, but on or
outside the surface E ¹ 0.
\ Electric intensity is discontinuous across a charged
conducting surface.
20. (b) The dielectric constant for metal is infinity, the force
between the two charges would be reduced to zero.
21. (a) In air, F
air
=
12
2
0
1
4
qq
rpe
In medium, F
m
=
12
2
0
1
4
qq
Krpe
air1
m
m
air
FF
F
FKK
\ =Þ=
(decr
eases K-times)
22. (d)3
04.
p
E
r
=
pe
Apparen
tly,
Epµ and
3
3
1
Er
r
-
µµ .
Hints & Solutions

428. 424 PHYSICS
23. (c
) K.E. acquired = work done
= force × distance = q E × y = q E y
24. (c) Intensity of electric field due to a Dipole
E =
3
0
4
p
rpe
2
3cos1q+
Þ
3
1
E
r
µ
25. (a) Dipole i
s formed when two equal and unlike charges
are placed at a short distance.
EXERCISE - 2
1. (a) – e
E = mg 31
19
9.1 10 10
E
1.6 10
-
-
´´
=-
´
u ur
=
11
5.6 10 N/C
-
-´
2. (b)
2 62
29
kQ E r 3 10 (2.5)
EQ
kr 9 10
´ ´´
=Þ==
´
3
210C
-
=´
3. (d) Work don
e, W = F s cosq = (q E) s cos q
4. (a) Given : Electric field (E) = 500 V/m
and potential difference (V) = 3000 V.
We know that electric field
(E) = 500 =
V
d
or
3000
6m
500
d==
[where d = Distan
ce between point charge and A]
5. (b)
22
o0
qq
E Ar
4r 4r
= Þ=
pe pÎ
3
o
q 4 ArÞ = pe
6. (a) Ch
arge resides on the outer surface of a conducting
hollow sphere of radius R. We consider a spherical
surface of radius r < R.
By Gauss theorem
+
+
+
+
+
+
+
+
+
+
+++
+
+
+
+
+
+
+
+
+
+
+
+
+
++++
+
+
+
R
S
O
r
E
´
e
=ò
0
s
1
sd.E
rr
charge enclosed or
2
0
1
E4r0´p=´
e
0E=Þ
i.e electric field inside a hollow sphere is zero.
7. (d) When the dipole is in the direction of field then net
force is qE + (–qE) = 0
–q a
q
E
and its potential energy is minimum = – p.E.
= – qaE
8. (b)( )
174 N1
2.57 10 Cm 3.0 10
C2
- æ öæö
t=´´ ç ÷ç÷
è øèø

13
3.855 10 Nm.
-
=´
9. (d) Torque, p E pEsint=´=q
rrr
4 = p × 2 × 10
5
× sin 30°
or, p =
5
5
4
4 10 Cm
2 10 sin 30
-
=´
´´°
Dipole
moment, p = q × l5
3p 4 10
q 2 10 C 2mC
0.02
-
-´
== =´=
l
10. (a)
Let q and q' be the charges on spheres of radii R and
2R respectively.Given q + q' = Q …(i)
Surface charge densities are
2
q
4R
s=
p
and 2
q'
4 (2R)
s=
p
Given s = s'
\
22
q q'
4 R 4 (2R)
=
pp
or, q' = 4q
Fr
om eq. (i), q' = Q – q or, 4q = Q – q
or, Q = 5q …(ii)
\ q' = Q – q =
Q 4Q
Q
55
-= .
11. (a) E
P
= 0
Þ
22
49
x (20 x)
=
-
Þ
0x3
x2
2-
=
Þ 40 – 2x = 3x Þ x = 8 cm
12. (a) netq
(F)0=
Þ
2
22
Qq 4q
k k0
2
+=
æö
ç÷
èø
ll
l/2 l/2
q Q 4q
where
0
1
k
4
=
pe
Þ 4Qq + 4q
2
= 0 Þ Q = – q
13
. (c) The charge on disc A is 10
–6

C.m The charge on
disc B is 10 × 10
–6
C.m The total charge on both =
11C.m When touched, this charge will be distributed
equally i.e. 5.5 Cm on each disc.

429. 425Electric Charges and Fields
14. (d) They will not experience any force if | |||
Ge
FF=
rr
Þ
22
0
22 22
0
1
.4
4(16 10
) (16 10 )
m qqGG
m
--
= Þ = pe
pe´´
15. (a)
Since lines of force starts from A and ends at B, so A is
+ve and B is –ve. Lines of forces are more crowded
near A, so A > B.
16. (a) Force due to electric field (F = q E) acts upwards.
17. (d) Unit positive charge at O will be repelled equally by
three charges at the three corners of triangle.
By symmetry, resultant
E
r
at O would be zero.
18. (b) From figure, a+
p
=q
3
, where
3
02
3
1 0
4rE psin1
tan tan
E 2pco s24r
æöæö pÎq
a== =qç÷ç÷
qpÎèøèø
13
tan tan
3 2 32
ppæö
q-==ç÷
èø
or
13
tan
32
-
æöp
q-=
ç÷
èø
or
÷
÷
ø
ö
ç ç
è
æ
=a
-
2
3
tan
1
or
÷ ÷
ø
ö
ç
ç
è
æ
+
p
=q
-
2
3
tan
3
1
O
a
–q +q
E
1
E
2
E
p/3
p/3
P
19. (d) 2
0. (b) 21. (d)
22. (a) For a soap bubble (electrified)
excess pressure
r
T4
PPP
0i
=-==
T = surface
tension, r = radius.
Force due to excess pressure balances surface tension.
When bubble is charged
r
T4
PP
ticelectrostaexcess
+=
42
2
00
2
ticelectrosta
r16
q
2
1
2
P
pe
-=
e
s
-
=
()
2
excess
23
0
4q
P T ....... A
r 128 r
éù
=-êú
peêúëû
Surface ten
sion decreases after electrification of
bubble and therefore pressure decreases.
÷
ø
ö
ç
è
æ
µ
r
1
P
means radi
us increases. In equation
()
2
q,Ais not
af
fected by positive or negative charge hence, whether
it is given a positive or a negative charge it always
expands in radius.
23. (c) Initial force between the two spheres carrying charge
(say q) is
2
2
0r
q
4
1
F
pe
= (r is the distanc
e between them)
Further when an uncharged sphere is kept in touchwith the sphere of charge q, the net charge on both
become
2
q
2
0q
=
+
. Force on th
e 3rd charge, when
placed at center of the 1st two
1 23
2/r 2/r
2/q 2/q
q
2
2
0
2
0
3
2
r
2
q
4
1
2
r
2
q
q
4
1
F
÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
pe
-
÷
ø
ö
ç
è
æ
÷
ø
ö
ç
è
æ
pe
=

22
22
00
1q 1q
[21]F
44rr
= -==
pe pe
24. (b
) The dipole experiences a torque pE sin q tending to
bring itself back in the direction of field.
Therefore, on being released (i.e. rotated) the dipole
oscillates about an axis through its centre of mass and
perpendicular to the field. If I is the moment of inertia
of the dipole about the axis of rotation, then the
equation of motion is q-=qsinpEdt/d.I
22
For small amplitude q»qsin
Thus qw-=q-=q
222
).I/pE(dt/d
where
)I/pE(=w .
This is a S
.H.M., whose period of oscillation is)pE/I(2/2Tp=wp=.
25. (d
) The dipole is placed in a non-uniform field, therefore a
force as well as a couple acts on it. The force on the
negative charge is more (F
µE) and is directed along
negative x-axis. Thus the dipole moves along negativex-axis and rotates in an anticlockwise direction.

430. 426 PHYSICS
a
E
–q
+q
Eq
1
Eq
2
26. (b) q=tsinpE , this is given by the second curve.
27. (c) In vacuum,
2
2
0
1q
F
4r
=
pe
…(i)
Suppose, froce between
the chrages is same when
charges are
r¢distance apart in dielectric.
\
2
2
0
1q
F'
4kr'
=
pe
…(ii)
From (i) and (ii),
kr'
2
= r
2
or,
r kr'=
In the given
situation, force between the charges
would be
22
22
0 0
1 q 4 q 4F
F'
4 99 4rrr
4
22
= ==
pe peæö
+
ç÷
èø
28. (c)
q
L/2
L
L
q
L/2
L
L
The given square of side L may be considered as one
of the faces of a cube with edge L. Then given charge
q will be considered to be placed at the centre of the
cube. Then according to Gauss's theorem, the
magnitude of the electric flux through the faces (six)
of the cube is given by
f = q/e
0
Hence, electric flux through one face of the cube for
the given square will be
0
1q
'
66
f = f=
e
29. (c) The charges make an electric dipole. A and B points
lie on the equitorial plane of the dipole. There forepotential at A = potential at B = 0W = Q (V
A
– V
B
) = q × 0 = 0
(0, a, 0)
–q
(0, 0,–a/2)
–Z
Y
B
A
(–a, 0, 0)+q
(0, 0, a/2)
Z
X
30. (b)
r
C
aO
–q q
Electric field at
C due to electric dipole
=
3
0
1 2p
4rpe
along OC
Elec
tric field at C due to induced charge must be equal
and opposite to electric field due to dipole as net fieldat C is zero.
31. (a, b, c, d)
The potential shown is for charged sphericalconductor.
32. (b) The charge at the inner surface, outer surface and
inside the conductor at P = (– Q, + Q, 0) as shown inthe figure
r
P
–Q
+Q
+
+
+
+
+
+
+
+
+Q
33. (b) Charge on glass rod is positive, so charge on gold
leaves will also be positive. Due to X-rays, moreelectrons from leaves will be emitted, so leavesbecomes more positive and diverge further.
34. (a) Surface charge density (s) =
Charge
Surface area
So inner
2
2
4
Q
b
-
s=
p
ab
c
+2Q
–2Q
– Q + 2Q = Q
and Outer
2
4
s=
p
Q
c
35. (c) Electric lines of force never intersect the conductor.
They are perpendicular and slightly curved near thesurface of conductor.
36. (d) Let q charge is situated at the mid position of the line
AB. The distance between AB is x. A and B be thepositions of charges Q and Q respectively.
A
Q
C B
Q
q
x
x
2
x
2
Let
2
x
BC,
2
x
AC ==

431. 427Electric Charges and Fields
The force on A due to charge q at C,
ACalong
)2/
x(
q.Q
.
4
1
F
2
0
CA
pe
=
®
The force on A due to charge Q at B
2
AB
2
0
1Q
F . along BA
4 x
®®
=
pe
The syste
m is in equilibrium, then two oppositely
directed force must be equal, i.e., total force on A is
equal to zero.
CA AB
FF0
®®
+= Þ
ABCA FF
®®
-=
2
22
00
1 4Q.q –1Q
..
44 xx
=
pe pe

4
Q
q-=Þ
37. (a)Given
: Length of the dipole (2l) = 10cm = 0.1m or
l = 0.05 m
Charge on the dipole (q) = 500 mC = 500 × 10
–6
C and
distance of the point on the axis from the mid-point ofthe dipole (r) = 20 + 5 = 25 cm = 0.25 m.
We know that the electric field intensity due to dipole
on the given point (E)
=
222
0 )(r
r)2.q(2
4
1
l
l
-
´
pe
222
6
9
])05.0()25.0[(
25.0)1.010500(2
109
-
´´´
´´=
-
= C/N1025.6
106.3
10225 7
3
3
´=
´
´
-
(k = 1 for air)
38. (c) Let n be the number of electrons missing.
2
2
0
1
4
q
F
d
=×
pe
Þ
2
0
4q d F ne=pe=
\
2
0
2
4Fd
n
e
pe
=
39. (d) El
ectric flux,
f= EA cos q, where q
= angle between E and normal to the surface.
Here
2
p
q=
Þ 0f=
40. (d)
dv dv dv
ˆˆˆ
E i jk
dx dy dz
éù
=- ++
êú
ëû
r
=
ˆ
–8i´ volt/meter
(
1,0,2)
ˆ
E 8iV/m\ =-
r
41. (c) AC = BC
V
D
= V
E
We have,
W = Q (V
E
– V
D
)
E
q
C
qA
q
B DÞ W = 0
42.
(b) At equilibrium potential of both sphere becomes same
if charge of sp here one x and other sphere
Q – x then
where Q = 4 × 10
–2
C
v
1
= v
2 ()
1cm 3cm
-
=
kx kQx
3x = Q – x Þ 4x = Q
x =
2
24 10
1 10
44
-
-´
= =´
Q
C
Q¢ = Q – x = 3
× 10
– 2
C
43. (a) The torque on the dipole is given as
t= PE sin q
The
potential energy of the dipole in the electric field
is given as
U = – PE cos q
44. (a) For the distances close to the charge at x = 0 the field
is very high and is in positive direction of x-axis. As
we move towards the other charge the net electric
field becomes zero at x = a thereafter the influence of
charge at x = 2a dominates and net field increases in
negative direction of x-axis and grows unboundedly
as we come closer and closer to the charge at x = 2a.
45. (b) If +ve charge is displaced along x-axis, then net force
will always act in a direction opposite to that of
displacement and the test charge will always come
back to its original position.
46. (c) q
d
= e, m
d
= 2m
p
= 2m
q
a
= 2e, m
a
= 4m
p
= 4m
F
1
= F
a
= eE, F
2
= F
a
= 2eE ¹ F
1
Further,
1
1
FeE
a
2m 2m
==
and
2
21
F2eE eE
aa
2m 4m 2m
== ==
47. (c)Electric field at any point depends on presence of all
charges.
48. (d) No. of lines entering the surface = No. of lines leaving
the surface.
49. (a) The electric field due to disc is superposition of electric
field due to its constituent ring as given in statement-
2. Statement-1 is true, statement-2 is true, statement-2
is a correct explanation for statement-1.
50. (c) Statement-1 is correct. The induced field cancels the
external field. Statement-2 is false. When a current is
set up in a conductor, there exists an electric field inside
it.

432. 428 PHYSICS
EXERCISE - 3
Exemplar Questions
1. (a) The force on q
1
depend on the force acting between
q
1
and q
2
and q
1
and q
3
so that the net force acting
on q
1
by q
2
and q
1
by q
3
is along the + x-direction, so
the force acting between q
1
, q
2
and q
1
, q
3
is attractive
force as shown in figure :
+q
2
+q
1
q
1 x
+q
3
The attract
ive force between these charges states that
q
1
is a negative charge (since, q
2
and q
3
are positive).
Then the force acting between q
1
and charge Q
(positive) is also know as attractive force and then the
net force on q
1
by q
2
, q
3
and Q are along the same
direction as shown in the figure.
+q
2
–q
1
+q
3
(x, 0)
+Q
x
The figure shows that the force on q
1
shall increase
along the positive x-axis due to the positive charge Q.
2. (a) If a positive point charge is brought near an isolated
conducting sphere without touching the sphere, thenthe free electrons in the sphere are attracted towardsthe positive Charge and electric field passes througha charged body. This leaves an excess of positivecharge on the (right) surface of sphere due to theinduction process.
Both type of charges are bound in the (isolated
conducting) sphere and cannot escape. They,
therefore, reside on the surface.
Thus, the left surface of sphere has an excess of
negative charge and the right surface of sphere has
an excess of positive charge as shown in figure.
+
+
+
+
+
attracted negative
charge
+q
An electric
field lines start from positive charge and
ends at negative charge.
Also, electric field line emerges from a positive charge,
in case of single charge and ends at infinity shown in
figure (a).
3. (d)By Gauss’s law : The total of the electric flux out of a
closed surface is equal to the charge enclosed devided
by the permittivity i.e., f =
0
Q
.
e
Thus, electric flux through a surface doesn’t depend
on the shape, size or area of a surface but it depends
on the number of charges enclosed by the surface. So
all the given figures have same electric flux as all of
them also has same single positive charge.
4. (b) Gauss's law states that total electric flux of an enclosed
surface is given by,
0s
q
E.dS=
e
òÑ
, includes the sum of
all
charges enclosed by the surface.
The charges may be located anywhere inside thesurface, and out side the surface. Then, the electric
field on the left side of equation is due to all the charges,
both inside and outside S.
So, E on LHS of the above equation will have a
contribution from all charges while q on the RHS will
have a contribution from q
2
and q
4
on l y.
5. (c) The electric field lines, are directed away from
positively charged source and directed toward
negatively charged source. In electric field force are
directly proportional to the electric field strength hence,
higher the electric field strength greater the force and
vice-versa.
The space between the electric field lines is increasing,
from left to right so strength of electric field decreases
with the increase in the space between electric field
lines. Then the force on charges also decreases from
left to right.
Thus, the force on charge – q is greater than force on
charge +q in turn dipole will experience a force towards
left.
6. (a) When a positive point charge +q is placed near an
isolated conducting plane, some negative charge
developes on the surface of the plane towards the
charge and an equal positive charge developes on
opposite side of the plane. This is called induction
process and the electric field on a isolated conducting
plane at point is directly projected in a plane
perpendicular to the field and away from the plane.
7. (a) Consider a point on diameter away from the centre of
hemisphere uniformly positively charged, then the
electric field is perpendicular to the diameter and the
component of electric intensity parallel to the diameter
cancel out.

433. 429Electric Charges and Fields
NEET/AIPMT (2013-2017) Questions
8. (a)
From figure, tan q =
e
F
mg
Þ
r/2
y
=
2
2
kq
r
mg
[QF =
2
2
kq
r
from coulomb’s law]
Þr
3
µ y Þ r'
3
µ
y
2
Þ
r'
r
=
1/3
1
2
Þ r' =
3
r
2
9. (b) The system of three charges will be in equilibrium.
A Bq
Q QOr r
For this, force between charge at A and B + force
between charge at point O and either at A or B is zero.
i.e.,
2
22
0
( / 2)
+=
KQ KQq
rr
By solving we get,
q = .
4
Q
-
10. (c) When electric dipole is aligned parallel
q = 0° and the dipole is rotated by 90° i.e.,
q = 90°.
Energy required to rotate the dipole
W = U
f
– U
i
= (–pE cos 90°) – (–pE cos 0°)
= pE.
11. (b) Net flux emmited from a spherical surface of radius a
according to Gauss’s theorem
in
net
0
q
f=
e
or, (Aa) (4pa
2
) =
in
0
q
e
So,q
in
= 4pe
0
A a
3
12. (c) From figure tan q =
e
F
mg
q;
q
l
q q
x
O
2
2
kqx
2x mg
=
l
orx
3
µ q
2
…(1)
orx
3/2
µ q …(2)
Differentiating eq. (1) w.r.t. time
3x
2
dx
dt
µ 2q
dq
dt
but
dq
dt
is constant
so x
2
(v) µ q Replace q from eq. (2)
x
2
(v) µ x
3/2
or v µ x
–1/2
13. (b) According to question, the net electrostatic force (F
E
)
= gravitational force (F
G
)
F
E
= F
G
or
22
22
0
1 e Gm
4 dd
D
=
pe
Þ De =
G
m
K

9
0
1
9 10
4
k
æö
= =´
ç÷
pe
èø
= 1.67 × 10
–27

11
9
6.67 10
9 10
-
´
´
De » 1.436 × 10
–37
C

434. 430 PHYSICS
ELECTROSTATIC POTEN
TIAL
Electric potential at a point in an electric field is defined as the
amount of work done in bringing a unit positive test charge
from infinity to that point along any arbitrary path. (Infinity is
taken as point of zero potential). It is denoted by V ;
0
==
work done in bringing unit positive
W test charge from infinity to some point
V
q unit positive test c
harge
Its SI unit is JC
–1
or volt. It is a scalar quantity.
Also, electric potential at any point in an electric field is defined
as the negative line integral of the electric field vector
E
r
from a
point infin
itely away from all charges to that point
i.e.
.
r
V E dr
¥
=-ò
uur uur

P +lC
Potential due to a Point Charge
The electric potential due to a point charge q at separation r is
given by
1
.
4
q
V
r
=
pe
(Please note that we have to write q with its sign in this formula)
4F potential difference between two points is the work done in
bringing unit positive charge from one point to another.
r
B
r
A
Q B A
V
AB
= V
B
– V
A .
B
A
E dr=-ò
.
4
oBA
Qll
rr
æö-
=-
ç÷
peèø
J/C
Electric
Potential due to Continuous Charge
Distribution
The potential due to a continuous charge distribution is the sum
of potentials of all the infinitesimal charge elements in which the
distribution may be divided.
dq
P
R
r
i.e.
òò
pe
==
r4
dq
dVV
o
where
r4
dq
dV
ope
=
Potential due to a Sys
tem of Charges
The electric potential due to a system of charges q
1
, q
2
, ...q
n
is
V = V
1
+ V
2
+ ... + V
n

12
12
11
. ...
44
ni
ni
qqqq
rrrr
æö
= +++=
ç÷
pe peèø
å
where r
i
is the p oint from charge q
i
and e is the permitivity of
medium in which the charges are situated.
Potential at any point P due to a point chage q at a distance
(r
1
+ r
2
) where r
1
is the thickness of medium of dielectric constant
x
1
and r
2
is the thickness of the medium of dielectric constant k
2
q P
r
1
r
2
K
1 K
2
1122
q
Vk
rK rK
=
+
where
0
1
4
k=
pÎ
Relation between electric field and potential
The relation between electric field (E) and potential (V) is
dV
E
dr
=-
For 3-D we can
write
x
V
E
x
¶
=-
¶
,
y
V
E
y
¶
¶
-=and
z
V
E
z
¶
¶
-=
So electric field i
s equal to negative potential gradient.
In this relation negative sign indicates that in the direction ofelectric field, potential decreases. Consider two points A and Bsituated in a uniform electric field at a distance d then,
17
Electrostatic Potential
and Capacitance

435. 431Electrostatic Potential and Capacitance
d
A B E
The potential difference between A and B is
EdV
AB
=
Conservative nature of electric field
The electric field is conservative in nature. In figure the work,
W
AB
has the same value whatever path is taken in moving the
test charge.
Q
o
Q
o
V
b
Starting
point
Terminal
point
A
1
2B
so,
AB
AB BA
W
V VV
Q
=-=
has t
he same value for any path between A and B and V
B
and V
A
are unique for the points A and B.We cannot find the absolute value of potential
therefore conventionally, we take infinity as the point of zeropotential. If need arises, we can assume any point to be the pointof zero potential and find the potential of other points on this
basis.
POTENTI
AL ENERGY OF A SYSTEM OF CHARGES
Potential energy can be defined only for those forces, which are
conservative, such as gravitational and electrostatic forces. The
potential energy of a charge between two points is defined as
the amount of work done in bringing the charge from one point
to another.
i.e.
ò
-==-
B
A
ext
AB dr.E
q
W
V
V
A
B
1
2
dl
Q
Calculation of external work done against the field and a point charge
Q in moving a test charge q from A to B. For a conservative field the
work done by any path is same. The sectional force is – qE.
If A is at infinity then at infinity since potential is zero
we assume infinity as reference point,V
A
= 0
ò
¥
-=Þ
B
B d.EqVl
Potential energy of a system of two charges Q
1
and Q
2
is,
Q
1 Q
2
r
12
2
0
1
4
QQ
U
r
=
pÎ
[Please note that in thi
s formula we have to write charges with
sign]
Potential energy of a system of three charges Q
1
, Q
2
and Q
3
Q
1
Q
2Q
3
r
3 r
1
r
2

ú
û
ù
ê
ë
é
++
Îp
=
3
13
2
32
1
21
0 r
QQ
r
QQ
r
QQ
4
1
U
Keep in Memory
1.For
an assembly of n charges [Total number of intersection
2
)1n(n
C
2
n -
= ] the pote
ntial energy is
ú
ú
ú
ú
û
ù
ê
ê
ê
ê
ë
é
=å
¹
n
ji
j,iij
ji
r
qq
k
2
1
U
2.For a system
of two charges.
If U
system
= –ve, then there is net force of attraction between
the charges of the system.
If U
system
= +ve, then there is net force of repulsion between
the charges of the system
U
system
= max for unstable equilibrium
U
system
= min for stable equilibrium
Also
dU
F0
dx
=-=
3.The energy required to take away the charges of a dipole at
infinite distance
l2
q
kU
2
=
4.The work done
when a charge q is moved across a potential
difference of V volt is given by W = qV
5.When one electronic charge (1.6×10
–19
coulomb i.e., charge
of electron) is moved across one volt the work done is
called one electron volt (eV). Thus1eV = (1 volt) × (1.6×10
–19
coulomb) = 1.6×10
–19
joule.

436. 432 PHYSICS
EQUIPOTENTIAL SURFACE
It is t
hat surface where the potential at any point of the surface
has the same value. The electric lines of force and the
equipotential surface are mutually perpendicular to each other.
No work is done in moving a charge from one point to other on an
equipotential surface. Work is done in moving a charge from one
equipotential surface to another.
Spherical equipotential surface
for point charge
r
1
V
1
V
2
r
2
o+ q
V > V
12
V = V
1
V = V
2
Plane equipotential surface for uniform field
·Equipotential surface do not cut each other.
·The density of the equipotential lines gives an idea of thestrength of electric field at that point. Higher the density,larger is the field strength.
Potential Due to Various Charge Distribution
(i)Electric potential due to isolated point charge
+q P
x ()q
Vk
x
+
=
(ii) A circular rin
g of radius R with uniformly distributed charge Q
+
+
+
++
+
+
+
+
+
+
+
+
+++
+
+
+
+
Q
Px
R
22
kQ
V
Rx
=
+
·Potentia
l V does not depend on the way of charge
distriubution on the ring (uniform / non-uniform).
(iii)A circular disc of radius R with uniformly distributed charge
with surface charge density s
Q
P
x
R
+
+
+
+
+
+
+
+
++
+ +
22
0
2
V Rxx
séù
= +-
êúëûe
(iv) A finite len
gth of charge with linear charge density
l
+++++++
x
L
P
log
e
xL
Vk
x
+æö
=l ç÷
èø
(v) Due to a spheri cal shell of uniformly distributed charge
with surface charge density s
V
x = R x
R
Q
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
+
++
in
Q
Vk
R
= , surface
Q
Vk
R
= ,out
Q
Vk
x
=
(vi) Due to a solid
sphere of uniformly distributed charge with
volume charge density r.
V
x = R x

Q
R
3
2
centre
Q
Vk
R
= ,
22
3
(3)
2
in
kQRr
V
R
-
=
surface
Q
Vk
R
=, out
Q
Vk
x
=
Potential due to Electric dipole
(a) A
long axial line :
– q +q
P
x
2l
22axial
p
Vk
xl
=
-
when x > > l
2
axial
kp
V
x
=
(b) Along
equatorial line :
V
eq
= zero
(c) At any point from the dipole :
222
cos
( cos)
p
Vk
xl
q
=
-q

q
P
x
– q + q
p

437. 433Electrostatic Potential and Capacitance
Keep in Memory
1.El
ectric field inside a charged conductor is zero
E = 0
in
+
+
+
+
++
+
+
+
+
+++
Spherical conductor

+
+
+
+++
+
+
+
++
++
+
E = 0
in
Irregular shaped conductor
But in both the cases the potential at all the points of the
surface will remain the same. But charges will have same
distribution on spherical conductor and in case of
irregularly shaped conductor the charge distribution will
be non-uniform. At sharp points, charge density has
greatest value.
2.Electronic lines of force are always perpendicular to the
equipotential surfaces.
3.The work done in moving a charge from a point to the
other on an equipotential surface is zero as the potential
difference between the two points is zero.
4.The electric potential at a point due to a point charge
decreases (or increases) by K-times if the distance between
the charge and the point increases (or decreases) by
K-times.
5.A ring with a charge distribution behaves as a point charge
for the points very far from its centre.
6.The electric potential is constant inside a hollow charged
sphere and it is also equal to its value on the surface but it
varies inversely with the distance outside the sphere.
7.The electric potential at points inside a solid sphere has a
non-zero value and decreases as we go from the centre
outwards. It behaves as a point charge for the points
outside the sphere.
8.The electric potential at a point due to a dipole varies
directly with the dipole moment.
COMMON DEFAULT
Incorrect : Where electric field is zero, electric potential is
also zero.
Correct : It i s not always correct, for example in a charged
conducting shell, electric field inside the shell E = 0 butpotential is not zero.
Incorrect : Where electric potential is zero, electric field is
also zero.
Correct : It is not always correct. In the case of equitorial
plane of an electric dipole the electric potential is zero butthe electric field is non-zero.
Example 1
A uniform electric field pointing in positive y-directionexists in a region. Let A be the origin, B be the point on the
x-axis at x = 2 cm and C be the point on the y-axis at y = 2
cm. Then the potential at the points A, B and C satisfy:
(a) V
A
< V
B
(b) V
A
> V
B
(c) V
A
< V
C
(d) V
A
> V
C
Solution : (d)
As electric field represents the direction of motion of positive
charge, which is from higher potential to lower potential,
therefore, from fig, we find V
A
= V
B
and V
A
> V
C

E
r
Z
A B
X
C
Y
E E
Example 2
A uniform electric field pointing in positive x-direction
exists in a region. Let A be the origin, B be the point on the
x-axis at x = + 1 cm and C be the point on the y-axis at
y = + 1 cm. Then the potential at the points A, B and C
satisfy
(a) V
A
< V
B
(b) V
A
> V
B
(c) V
A
< V
C
(d) V
A
> V
C
Solution : (b)
As
E
u ur
is directed along +ve direction of X-axis, therefore,
V
A
> V
B
C is vertically above A. Therefore, V
A
= V
C
.
V
X
E
C
A B
Example 3
Calculate the maximum voltage upto which a sphere of
radius 2 cm can be charged in air under normal conditions,
assuming that maximum electric intensity in air can be
3 × 10
6
volt/m. Also, find the charge required to be given
to the sphere.
Solution :
We know that
Electric intensity (E) =
Electric Potential (V)
Distance (r)
Þ V = Er …… (1)
Given, E = 3 × 10
6
volt/m, r = 2 cm = 2 × 10
–2
m.
Substituting the above values in eq. (1), we get
V = (3 × 10
6
volt/m) × 2 × 10
–2
m = 6 × 10
4
volt.
Þ V = 60 kV
Als
o, we know that the electric intensity on the surface of
a charged sphere is given by

438. 434 PHYSICS
2
0r
Q
.
4
1
E
Îp
= ; Q = charge on th
e sphere
22
96
)102(
Q
)109(103
-
´
´´=´Þ
7
Q 1.33 10 coulomb.
-
Þ=´
ELECTROSTATICS OF CONDUCTORS
C
onductor is a substance that can be used to carry or conduct
electric charges. Metals like silver. Copper, aluminium etc. are
good conductors of electricity.
Regarding electrostatics of conductors following points are worth
noting.
(i) Inside a conductor, electric field is zero.
(ii) The interior of a conductor can have no excess charge in
static situation.
(iii) Electric field at the surface of a charged conductor is
0
ˆEn
s
=
Î
r
where, s = surface charge density
ˆn = unit vector normal to the surface in the outward
direction.
(iv) Electric field just outside a charged conductor is
perpendicular to the surface of the conductor at every
point.
(v) Electrostatic potential is constant throughout the volume
of the conductor and has the same value as on its surface.
(vi) Surface density of charge is different at different points.
CAPACITORS AND CAPACITANCE
A capacitor or condenser is a device that stores electrical
energy. It generally consists of two conductors carrying equal
but opposite charges.
The ability of a capacitor to hold a charge is measured by a
quantity called the capacitance. Let us consider two uncharged
identical conductors X and Y and create a P.D. (Potential
Difference) V between them by connecting with battery B as
shown in figure.
B
X
Y
+ Q– Q
d
A
Fig- A capacit
or consists of electrically insulated conductors carrying
equal positive and negative charge
After connection with the battery, the two conductors X and Yhave equal but opposite charges. Such a combination of charged
conductors is a device called a capacitor. The P.D. between X
and Y is found to be proportional to the charge Q on capacitor.
The capacitance C, of a capacitor is defined as the ratio of the
magnitude of the charge on either conductor to the magnitude
of P.D. between them.
Q
V Q = constant = (Capacitance).
V
µÞ
Capacitance is alway
s a positive quantity.
The S.I. unit of capacitance is coulomb per volt or farad (F).
Further more, the value of capacitance depends on size, shape,
relative positions of plate, and the medium between the plates.
The value of C does not depend on the charge of the plate or p.d.
between the plates.
ENERGY STORED IN A CAPACITOR
If Q is charge, V is p.d, C is the capacitance of the capacitor then
the energy stored is 2
211
2 22
Q
U CV QV
C
= ==
Sharing of Ch
arges
When the two charged conductors of capacitances C
1
and C
2
at
potentials V
1
and V
2
respectively, are connected by a conducting
wire, the charge flows from higher to lower potential, until the
potentials of the two conductors are equal.V
1
C
1
Q
1 Q2
C
2
V
2
' '
The common potential after sharing of charges,
Net charge
V
Net capacitance
=
21
2211
21
21
CC
VCVC
CC
QQ
+
+
=
+
+
=
The charges af
ter sharing on two conductors will be
VCQandVCQ
2
'
21
'
1
==

i.e.,
111
22 2
QCQ
QC Q
¢
¢
==
There is a l
oss of energy during sharing, converted to heat given
by
finalinitial
UUU-=D
2
21
2
22
2
11
V)CC(
2
1
VC
2
1
VC
2
1
+-
ú
û
ù
ê
ë
é
+=
or, Du
2
1212
12
()
2()
CCVV
CC
-
=
+
PARALLEL PLATE CAPA
CITOR
It consists of two parallel metallic plates of any shape, each of
area A and at a distance d apart.
The capacitance of the capacitor is given by
d
A
C
0e
=
Effect of Dielectric o
n Capacitance
When a dielectric slab is placed
between the plates of a parallel plate
capacitor, the charge induced on its
plates due to polarisation of dielectric
is
÷
ø
ö
ç
è
æ
-=
K
1
1Qq
p
where K = die
lectric constant.
–
–
–
–
–
– q
p
– – –
–
–
–
–
–
+
+
+
+
+
+
+
+
+
+
+
+
+
QQ
+q
p
E
0
E
p
––E = E E
0p
+
When an electri
c field is applied across a dielectric, induced
charges appear on the surface of dielectric which is shown in the
above figure. These induced charges produce their own field
which acts in the opposite direction of the applied field. Hence,
total field is reduced, i.e., EEE
p0
=-, where E
0
is the applied
field, E
p
is the induced field and E is the resultant field.

439. 435Electrostatic Potential and Capacitance
E is given by
K
E
0
, where
K is the dielectric constant.
If medium between the plates is having a dielectric of dielectric
constant K then the capacitance is given
o
KA
C
d
eæö
=ç÷
èø
If the space between the plates is partly filled with dielectric
then the capacitance of the capacitor will be given by,
+
+
+
+
+
+
+
–
–
–
–
–
–
–
t
P1
P2
A
d
K
00
1
1
AA
C
t
dt dt
K K
ee
==
æö
-+ --ç÷
èø
,
where
t is the thickness of the dielectric with dielectric constant K.
Keep in Memory
1.The
unit farad is quite a big unit for practical purposes.
Even the capacitance of a huge body like earth is 711 mF.
2.A capacitor is a device which stores charges and produces
electricity whenever required.
3.If the two plates of a capacitor is connected with a
conducting wire, sparking takes place which shows that
electrical energy is converted into heat and light energy.
4.A capacitor allows A.C. but doesn’t allow D.C. to pass
through it.
5.The capacitance of a capacitor increases with insertion of a
dielectric between its plates and decreases with increase in
the separation between the plates.
6.The capacitance of a capacitor increases K times if a medium
of dielectric constant K is inserted between its plates.
7.The energy of a capacitor for a particular separation between
the plates is the amount of work done in separating the two
plates to that separation if they are made to touch to each
other.
8.The loss of energy when the two charged conductors are
connected by a wire doesn’t depend on the length of the
wire.
Exa
mple 4
A parallel plate capacitor is maintained at a certain potentialdifference. When a 3 mm slab is introduced between the plates,in order to maintain the same potential difference, the distancebetween the plates is increased by 2.4 mm. Find the dielectricconstant of the slab.
Solution :
The capacity of a parallel plate capacitor in air is given by
d
A
C
0e
= ... (1)
By intro
ducing a slab of thickness t, the new capacitance
C´ becomes
e
=
--
0
A
C'
d' t(11/K)
... (2)
The cha
rge (Q = CV) remains the same in both the cases.
Hence
ee
=
--
oo
AA
d d' t(1 1/K)
or
æö
=--ç÷
èø
1
d d't1
K

3
d ' d 2.4 10
-
=+´ m, t = 3 mm = 3 × 10
–3
m.
Substituting these values, we have
÷
ø
ö
ç
è
æ
-´-´+=
--
K
1
1103)104.2(dd
33
or ÷
ø
ö
ç
è
æ
-´=´
--
K
1
1103)104.2(
33
Solvi
ng it, we get K = 5.
SPHERICAL CAPACITOR
+
+
+
+
+
+
+
+
+
++++
+
+
+
+
+
+
+
+
+++
B
R
1
R
2
O
A
It consi
sts of two concentric
spherical conductors of radii R
1
and R
2
. The space between two
conductors is filled by a dielectric
of dielectric constant K.
(a)When outer conductor is earthed,
Capacitance of spherical capacitor,
0 12
21
4 RR
C
RR
pÎ
=
-
(without die
lectric)
12
21
(4)
()
o
K RR
C
RR
pe
=
-
(with dielectri
c)
(b)When inner sphere is earthed,
12
2
21
4
4
o
o
KRR
CR
RR
pe
= + pe
-
Thi
s is because the combination behaves as two capacitors
in parallel, one is a capacitor formed by two concentricspherical shells and the other is an isolated spherical shellof radius R
2
.
CYLINDRICAL CAPACITOR
It consists of two-coaxial cylindrical conductors of radii R
1
and R
2
,
the outer surface of outer conductor being earthed. The space
between the two is filled with a dielectric of dielectric constant K.
R
1
R
2
– +
– +
– +
– +
–
+
– +
– +
– +
– +
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
+ –
AKB
l

440. 436 PHYSICS
The cap
acitance of cylindrical condenser of length l
0
2
1
2
log
e
C
R
R
pÎ
=
æö
ç÷
èø
l
(without dielectric)
2
1
2
log
o
e
K
C
R
R
pÎ
=
æö
ç÷
èø
l
(with dielectric)
COMBINA
TION OF CAPACITORS
Series Combination
(i) In this combination, the positive plate of one capacitor is
connected to the negative plate of the other.
A
Q Q Q
B
V
V
1
V
2
V
3
C
1 C
2
C
3
(ii) The charges of individual capacitor are equal.
(iii) The potential difference is shared by the capacitors in the
inverse ratio of their capacities
i.e. Q = C
1
V
1
= C
2
V
2
= C
3
V
3
Hence V = V
1
+ V
2
+ V
3
(iv) The equivalent capacitance (C) between A and B is
.....
C
1
C
1
C
1
C
1
321
+++=
1
+
C
n
Parallel Combination
q
1
+
+
+
+
+
+
–
–
–
–
–
–
V
A
Q
B
q
2
q
3
c
1
c
2
c
3
(i) In this arrangement, +ve plates
of all the condensers are
connected to one point and
negative plates of all the
condensers are connected to
the other point.
(ii) The Potential difference
across the individual capacitor
is same.
(iii) The total charge shared by the individual capacitor is in
direct ratio of their capacities
i.e.
3
3
2
2
1
1
C
q
C
q
C
q
V ===
Hence, Q = q
1
+ q
2
+ q
3
(i
v) The equivalent capacitance between A and B is
C
eq
= C
1
+ C
2
+ C
3
+ ........+ C
n
Keep in Memory
1.The capacitan
ce of a parallel plate capacitor having a
number of slabs of thickness t
1
, t
2
, t
3
.... and dielectric
constant K
1
, K
2
, K
3
.... respectively between the plates is
÷
÷
ø
ö
ç
ç
è
æ
+++
e
=
.....
K
t
K
t
K
t
A
C
3
3
2
2
1
1
o
–
+K1K
2
t
1t
2
2.When a number of dielectric slabs of same thickness (d)
and different areas of cross-section A
1
, A
2
, A
3
... having
dielectric constants K
1
, K
2
, K
3
, .... respectively are placed
between the plates of a parallel plate capacitor then the
capacitance is given by
A
r
e
a
A
1
A
r
e
a
A
2
A
r
e
a
A
3
d
Electric slabs which fill
the whole space of parallel
plate capacitor
+
–
o 11 22 33
(K A K A K A ... .)
C
d
e +++
=
3.When five capacit
ors are connected in wheatstone bridge
arrangement as shown, such that
4
3
2
1
C
C
C
C
=, the bridge is
balanced and C
5
become
s ineffective. No charge is stored
on C
5
. Therefore C
1
, C
2
and C
3
, C
4
are in series. The two
series combinations are in parallel between A and C. Hence
equivalent capacitance can be calculated.
A
B
C
D
C
1
C2
C
3 C
4
C5
RELATION BETWEEN THREE ELECTRIC VEC
TORS
ur ur
D,PAND
r
E
If an electric field E is applied across a parallel plate capacitor
filled with a dielectric of dielectric constant K (or permittivity e),
then
Polarisation P = induced charge per unit area (opposite to free
charge) =
q'
A
Electric displacement D = eE = e
o
E + P
i.e. Polarisation P = (e – e
o
) E = (Ke
o
– e
o
) E
Electric susceptibility, /
e
PEc=
Relation between diel
ectric constant K and electric susceptibility
c
e
is
0
1
e
K
c
=+
e

441. 437Electrostatic Potential and Capacitance
Effect of filling dielectric With battery connected
When there is no dielectric
Capacitance
d
A
C
0
0
e
=
Potentia
l difference between the plates V
Charge on a plate Q = CVEnergy
2
00
VC
2
1
E=
V
Electric field
d
V
E
0=
When dielectr
ic is inserted
0
0
KC
d
KA
C =
e
=
Q = K C
0
V = KQ
0
V
0
2
0 KEVKC
2
1
U ==
0E
d
V
E ==
Effect of fill
ing a dielectric in a capacitor after
disconnection of battery
Capacitance C = K C
0
Q0 Q = Q0
C0
Charge
V
+ Q – Q
2
000
VC
2
1
U=
+ Q– Q
K
U
K
V
KC
2
1
U
0
2
0
0
==
0
0
0
C
Q
V=
K
V
C
Q
V
00
==P.D
Potenti
al energyCHARGING AND DISCHARGING A CAPACITOR
Charging a Capacitor
When an uncharged capacitor is connected across a source of
constant potential difference such as a cell, it takes a finite time
to get fully charged, although this time interval may be small.
This time-interval depends on the capacity of the capacitor and
the resistance in the circuit.
During the period of charging :
1.The charge on the capacitor increases from ‘zero’ to the
final steady charge.
2.The potential difference developed across the capacitor
opposes the constant potential difference of the source.
3.The charge on the capacitor ‘grows’ only as long as the
potential difference of source is greater than the potential
difference across the capacitor. This transport of the charge
from the source to the capacitor constitutes a transient
current in the circuit.
4.As the charge on the capacitor increases, more energy is
stored in the capacitor.
5.When the capacitor is fully charged, potential difference
across the capacitor is equal to the potential difference of
the source and the transient current tends to zero.
If V
0
= constant potential difference of the source
R = pure resistance in the circuit
C = capacity of the capacitor
Q
0
= final charge on the capacitor, when fully charged
q = charge on the capacitor at time ‘t’ from the starting of
the charging
V = potential difference across the capacitor at time ‘t’
( )·
Source KeyV
0
CR
Then ==
0
0
qQ
C
VV
and i = cur
rent in the circuit at time ‘t’ =
dq
dt
At time ‘t’ by Kirchhoff’s law
- -=
0
dqq
VR0
dtC
i.e. =
-
0
dq 1
dt
CV q CR
Integrat
ing and putting in the initial condition q = 0 at t = 0,
we get
1
–t
CR
q=CV 1–e
0
éù
êú
êú
ëû
Special cases :
(i
) At t = 0, q = 0.
(ii) When t increases, q increases.
(iii) As
00
t , q Q CV®¥®=
1
t
CR
0
q Q 1e
-éù
\=- êú
êú
ëû
(iv) At t = CR [‘CR’ has dimensions of time]
éù
= -=
êú
ëû
00
1
q Q 1 0. 631Q
e
This va
lue of t = CR is called the ‘time constant’ of the (CR)
circuit.
Discharging of a Capacitor
If after charging the capacitor, the source of constant potential
difference is disconnected and the charged capacitor is shorted
through a resistance ‘R’, then by Kirchhoff’s law, at time ‘t’ from
the instant of shorting,
+=
q dq
R0
C dt

442. 438 PHYSICS
Putting,
(i) the
initial condition, q = Q
0
at t = 0 and
(ii) the final condition, q = 0 at
®¥t ,
the solution to the
above equation is
1
–t
CRq=Qe
0
q
t
Q
– 0
Keep in Memory
1.If n small dr
ops each having a charge q, capacity ‘C’ and
potential V coalesc to form a big drop, then
(i) the charge on the big drop = nq
(ii) capacity of big drop = n
1/3
C
(iii) potential of big drop = n
2/3
V
(iv) potential energy of big drop = n
5/3
U
(v) surface density of charge on the big drop = n
1/3
×
surface density of charge on one small drop.
2. Charged soap bubble : Four types of pressure act on a
charged soap bubble.
(i) Pressure due to air outside the bubble P
O
, acting
inwards.
(ii) Pressure due to surface tension of soap solution P
T
,
acting inwards.
(iii) Pressure due to air inside the bubble, P
i
, acting
outwards.
(iv) Electric pressure due to charging, P
e
=
2
o
2
s
e
, acting
outwards.
In
equilibrium, P
i
+ P
e
= P
O
+ P
T
or, P
i
– P
O
= P
T
– P
e
or,

P
excess
= P
T
– P
e
\
o
2
excess
2r
T4
P
e
s
-=
Where T = sur
face tension of soap solution,
s = surface charge density of bubble.
If P
i
= P
O
then P
i
– P
O
= P
T
– P
e
= 0 or P
T
= P
e
2
2
oo
2
r4
q
2
1
2r
T4
÷
÷
ø
ö
ç
ç
è
æ
pe
=
e
s
=
Hence for maintain
ing the equilibrium of charged soap
bubble,
r
T8
oe
=s
rT2r8q
o
ep=
3. Force of attraction between the plates
of a parallel plate
capacitor =
AK2
q
o
2
e
where, A = ar
ea of the plates of capacitor and
K = dielectric constant of the medium filled between the
plates.
In terms of electric field, the force of attraction
2
0
1
F KE
2
=e
4. Uses of capacitor :
=In
LC oscillators=As filter circuits
=Tuner circuit in radio etc.
5.The total energy stored in an array of capacitors
(in series or in parallel) is the sum of the individual energies
stored in each capacitor.
COMBINATION OF CAPACITOR : EQUIVALENT CAPACITANCE
(i)
+
–
K
1
K
2
d º
–
K
1
K
2
2/d
A
KC
0
22
e
=
2/d
A
KC
0
11
e
=

2/d
A
KC
0
22
e
=
2/d
A
KC
0
11
e
=
21eq C
1
C
1
C
1
+=
dd
d and area of each plate = A
22
éù
=+
êú
ëû
(ii)
+
?
K
1
K
2
d º
K
2
1 0
1
A
K
2
C
d
æö
eç÷
èø
=
1 0
1
A
K
2
C
d
æö
eç÷
èø
=
+
?
1 0
1
A
K
2
C
d
æö
eç÷
èø
=
+
?
2 0
2
A
K
2
C
d
æö
eç÷
èø
= ; C = C + C
eq 12
(iii)
–
+
K
1
K
2
K
3
d
º
–
+
d
)2/A(K
C
01
1
e
=
2/d
)2/A(K
C
02
2
e
=
2/d
)2/A(K
C
03
3
e
=
32
32
1
CC
CC
CC
eq
+=

443. 439Electrostatic Potential and Capacitance
(iv)
–
+
d
K
3
K
2
K
1
º
–
+
2/d
)2/A(K
C
02
2
e
=
2/d
A
KC
0
11
e
=
2/d
)2/A(K
C
03
3
e
==
321 CC
1
C
1
C
eq
1
+
+=
(v) º
+
–
B
A
1
2
3
4
º
A B+ –
C
C
C
3
1
34
2
2
C = C + C + C = 3C
eq
(vi)
–
A B º
A
B
1
2
3
4
+ + + +
+ + + +
– – – –
= º
A B
+ –
C
C
1
3 4
2
C = C + C = 2C
eq
=
(vii)
B
A
(+)
(–)
º
B
A1
2
3
4
– – – –
– – – –
– –
– –
+ + + +
+ + + +
+ + + +

A B
+ –
C
C C
1 3 2 4
23
eq
eq
1 1 3C
C or,C
11C 2
CC
=+ =
+
º
SOME METHODS OF FINDING EQUIVALENT
CA
PACITANCE
Method 1 : Successive Reduction
This method is applicable only when the capacitor can be clearly
identified as in series or in parallel.
A A
B B
A AB B
19
26
3
13
3
4
B
A
Þ
2 mF
2 mF
2 mFm
2 mF
2 mF
2 mF
2 mF
2 mF
2 mF2 mF
4 mF2 mF
2 mF
2 mF
3 mF
mFm Fm F
Þ
Þ
Þ
Method 2 : Using Symmetry
YX
A
A
B
D
E
C
C
C
C
C
C
C
CC
The abov
e circuit is symmetrical about XAEBY axis. This is
because the upper part of the circuit is mirror image of lower part.
Therefore V
C
= V
E
= V
D
. The circuit can be redrawn as
X
A B
D
C C
CC
C C
Y
A B
D
2
C
2
C
2
C
2
C
=
C
C

444. 440 PHYSICS
2
C3
A B
=
Method 3 : Wheatston
e bridge
A
B
D
C
1 C
2
C
4
C
5
C
3
C
If
4
2
3
1
C
C
C
C
=
then the wheatsto
ne bridge is balanced. In this case
there will be no charge accumulation in C
5
when battery is
attached across A and B. Therefore the equivalent circuit is the
capacitance C
1
and C
2
are in series. Similarly C
3
and C
4
is in
series. Therefore the equivalent capacitance occurs between A
and B is
111
=+
C C +C C +C
eq 4123
The other forms of wheatston
e bridge are :
B
A
C
1
C
2
C
3
C
4
C
5
or
C
5
C
1
C
3
C
4
C
2
B
A
Method 4 : If none of th e above method works, then we can use
the method of Kirchhoff’s laws - junction law and loop law.
SHARP POINT ACTION (CORONA DISCHARGE)
When the electric field )/(
0
es on a point on the surface of a
conductor exceeds the electric strength of air, then the air becomes
conducting and the surface of conductor loses charge. This action
occurs usually at the sharp points of a conductor as here s is
high, thus creating high electric field. This phenomenon is also
called corona discharge.
VAN DE GRAAF GENERATOR
R.J. Van de Graff in 1931 designed an electrostatic generator
capable of generating very high potential of the order of 5 × 10
6
V,
which was then made use of an accelerating charged particles so
as to carry out nuclear reactions.
Principle : When a charged conductor is placed in contact with
the inside of a hollow conductor, all of the charge of first
conductor is transferred to the hollow conductor. i.e., the charge
on hollow conductor or its potential can be increased by any
limit by repeating that processes.
The basic fact of Van de Graaf generator is described in fig.
(Charge is delivered continuously to a high voltage electrode on
a moving belt of insulating material).
A
High voltage electrode
Belt
Insulator
Grounded grid
+
B
Schematic diagram of a Van de Graaf generator. Charge is transferred
to hollow conductor at the top by means of a rotating belt. The charge is
deposited on the belt at point A and is transferred to hollow conductor
at point B.
The high voltage electrode is a hollow conductor mounted on aninsulating medium. The belt is charged at A by means of corona
discharge between comb-like metallic needles and a grounded
grid. The needles are maintained at a positive potential of typically10
4
eV. The positive charge on the moving belt is transferred to
the high voltage electrode by second comb of needles at B.Since the electric field inside the hollow conductor is negligible,the positive charge on the belt easily transfers to the high- voltage
electrode, regardless of its potential. We can increase the potential
of the high voltage electrode until electrical discharge occur
through the air . The “ breakdown” voltage of air is about 3 × 10
6
V/m.
Example 5
Obtain equivalent capacitance of the following network
as shown in fig. For a 300 volt supply, determine the
charge and voltage across each capacitor.
100 pF
200 pF 200 pF
100 pF
300 V
C
1
C
3
C
4
C
2
+
–

445. 441Electrostatic Potential and Capacitance
Solution :
As it is clear from fig, C
2
and C
3
are in series.
11111
=+=+
C C C 200 200
s 23
\

21
= = C = 100 pF
s
200 100
Þ
Now,
C
s
and C
1
are in parallel.
C
p
= C
s
+ C
1
= 100 + 100 = 200 pF.
Again, C
p
and C
4
are in series. Their combined capacitance
C is
200
3
100
1
200
1
C
1
C
1
C
1
4p
=+=+=
F107.66pF7.66
3
200
C
12-
´===
As C
p
and C
4
ar
e in series.
\ V
p
+ V
4
= 300 volt.
Charge on C
4
, .C10230010
3
200
CVq
812
4
--
´=´´==
Potentia
l difference across C
4
:
.volt200
10100
102
C
q
V
12
8
4
4
4
=
´
´
==
-
-
.v
olt100200300V300V
4p
=-=-=
Potential difference across C
1
= V
1
= V
p
= 100 volt.
Charge on
C
1
, q
1
= C
1
V
1
= 100 × 10
–12
× 100 = 10
–8
C
Potential difference across C
2
and C
3
in series = 100 volt
V
2
= V
3
= 50 volt
Charge on
C
2
= q
2
= C
2
V
2
= 200 × 10
–12
× 50 = 10
–8
C
Charge on
C
3
= q
3
= C
3
V
3
= 200 × 10
–12
× 50 = 10
–8
C
Example 6
Two isolated metallic solid spheres of radii R and 2 R are
charged such that both of these have same charge density
s. The spheres are located far away from each other, and
connected by a thin conducting wire. Find the new charge
density on the bigger sphere.
Solution :
Charge on smaller sphere, q
1
= 4pR
2
s
Charge on bigger sphere, q
2
= 4p(2R)
2
s = 1pR
2
s
\ Total charge, q = q
1
+ q
2
= 20pR
2
s
Combined capacity of two spheres,
C = C
1
+ C
2
= 4pe
o
R + 4pe
o
(2R) = 12pe
o
R
After contact, charge is exchanged and a common potential
V is reached.
C
q
capacitytotal
eargchtotal
V ==

oo
2
3
R5
R12
R20
e
s
=
ep
sp
=
Now, char
ge on bigger sphere, 2
'
22
5 40
4 (2)
33
o
o
RR
qC
VR
s ps
pe
e== ´=
\ Surface density of charge
.
6
5
)R2()4(3
R40
areasurface
q
2
2'
2'
2 s=
p
sp
==s
E
xample 7
Two insulated metal spheres of radii 10 cm and 15 cm
charged to a potential of 150 V and 100 V respectively,
are connected by means of a metallic wire. What is the
charge on the first sphere?
Solution :
Here, r
1
= 10 cm, r
2
= 15cm
V
1
= 150 V, V
2
= 100 V
Common potential 11 22
12
011 22
012
CV CV
V
CC
4 (rV r V)
4 (r r)
120 volt
+
=
+
p
Î+
=
pÎ+
=
11
01
1
9
9
9
q CV
4 rV
10
12C
9 10
12
3 10 esu 4esu
9 10
-
=
= pÎ
=´
´
=´´=
´
E
xample 8
Consider a parallel plate capacitor of capacity 10
mF
with air filled in the gap between the plates. Now one half
of the space between the plates is filled with a dielectric of
dielectric constant K = 4 as shown in fig.
K = 4
The capacity of the capacitor changes to
(a) 25
mF (b) 20 mF
(c) 40
mF (d) 5 mF

446. 442 PHYSICS
Solution : (
a)
The arrangement is equivalent to three capacitors in parallel
F5.2
4
10
d
4/A
C
o
1 m==
e
= ;
F20
2
10
4
d
2/AK
C
o
2 m=´=
e
= ;
F5.2
2
10
d
4/A
C
o
3 m==
e
=
eq 123
C C C C 2.5 20 2.5\ =+ + = ++ = 25mF
Examp
le 9
A parallel plate capacitor is filled with dielectric as shown
in fig. Its capacitance has ratio with that and without of
dielectric as
d
d 2
+
–
K
1
K
2
(a) (K
1
+ K
2
) (b)
æö
ç÷
ç÷
èø
12
12
K +K
K –K
(c)
æö
ç÷
ç÷
èø
12
12
2KK
K +K
(d)
æö
ç÷
ç÷
èø
12
12
K +K
KK
Solution : (c)
Wit
hout dielectric,
o
0
A
C
d
e
=
With dielectric as s
hown,
;CK2
2/d
AK
C;CK2
2/d
AK
C
02
o2
201
o1
1
=
e
==
e
=
As C
1
, C
2
are in
series,
21
12
21s CC
CC
C
1
C
1
C
1 +
=+=\
;
21
021
210
0001
21
21
s
KK
CKK2
)KK(C2
CK2CK2
CC
CC
C
+
=
+
=
+
=\
21
21
0
s
KK
KK2
C
C
+
=
Example 10
A capacitor of capacity 1
mF is connected in closed series
circuit with a resistance of 10
7

W, an open key and a cell
of 2 V with negligible internal resistance.
(i) When the key is switched on at a time t = 0, find :
(a) the time constant for the circuit.
(b) the charge on the capacitor at steady state.
(c) time taken to deposit charge equalling half that
at steady state.
(ii) If after fully charging the capacitor, the cell is shorted
by zero resistance at time t = 0, find the charge on the
capacitor at t = 50 s.
Solution :
(i) (a) Time constant = RC = (10
7
) (10
–6
) = 10 s.
(b) Q
0
= charge on capacitor at steady state
= V
0
C = 2 × 10
–6
= 2 mC
(c)
1
t
CR
00
1
qQQ[1e]
2
-
= =-

t
10
e2Þ= or,
10
t 10 2.306 log 2 6.94 s.=´´=
(ii)
1
t
CR
0
q Qe
-
=

50
6610
5
1
(210)(e ) ( 210)
e
-
-- æö
=´ =´
ç÷
èø
= 1.348 × 10
–8
C ( e 2.718)=Q

447. 443Electrostatic Potential and Capacitance
CONCEPT MAP
ELECTROSTATIC

AND CAPACITANCE
Capacitance
(c) = Charge
(
Q
)

Potential
(
V
)
C
a
p
a
c
i
t
a
n
c
e
o
f

a
p
a
r
a
l
l
e
l

p
l
a
t
e
c
a
p
a
c
i
t
o
r
C
=
K
A
e
0
d
k

=
d
i
e
l
e
c
t
r
i
c

c
o
n
s
t
a
n
t
C
a
p
a
c
i
t
a
n
c
e
w
h
e
n
m
e
t
a
l
l
i
c
s
l
a
b
i
n
s
e
r
t
e
d

b
e
t
w
e
e
n

t
h
e

p
l
a
t
e
s

0
A
C
d

t
e
æ
ö
=
ç
÷
è
ø
E
l
e
c
t
r
o
s
t
a
t
i
c

p
o
t
e
n
t
i
a
l
(
V
)
=
w
o
r
k
d
o
n
e
w
0








c
h
a
r
g
e
(
q

)
¥
0
Electric potential
due to a dipole
On axial line

2
0
1p
4r
V=
pe
On equatorial line
V = 0
E
l
e
c
t
r
o
s
t
a
t
i
c

p
o
t
e
n
t
i
a
l
E
q
u
i
p
o
t
e
n
t
i
a
l
s
u
r
f
a
c
e

d
u
e
t
o
a
p
o
i
n
t
c
h
a
r
g
e
I
m
a
g
i
n
a
r
y

s
u
r
f
a
c
e
j
o
i
n
i
n
g
t
h
e
p
o
i
n
t
s
o
f

s
a
m
e

p
o
t
e
n
t
i
a
l
i
n

a
n

e
l
e
c
t
r
i
c

f
i
e
l
d
q
V=K
r
Electrostatic poten
t
i
a
l

due to a system of
c
h
a
r
g
e
s
V= V + V + V ... +
V


1 2 3
n
1
V
K
n
i i
i
qr
=
=
å
C
o
m
b
i
n
a
t
i
o
n
o
f

c
a
p
a
c
i
t
o
r
s
S
e
r
i
e
s

g
r
o
u
p
i
n
g

o
f
c
a
p
a
c
i
t
o
r
s

E
q
u
i
v
a
l
e
n
t

c
a
p
a
c
i
t
a
n
c
e
s
2
1
n
1
1
1
1
=
+
+
.
...
+
C
C
C
C
P
a
r
a
l
l
e
l
g
r
o
u
p
i
n
g
o
f

c
a
p
a
c
i
t
o
r
s
E
q
u
i
v
a
l
e
n
t

c
a
p
a
c
i
t
a
n
c
e
C
=
C

+

C
+

..
.
C
p
1
Electric potential
due to a charged
conducting spherical
shell
Electric potential due to a
charged non-conducting
sphere
At a point outside the
Energy loss when two
non-conducting sphere
isolated charged conductors
are connected to each other
0
1q
V(r R)
4pr
= >
Î
Energy stored in a capacitor
2
2 1 Q
uCV
22C
= =
At a point on the surface
or inside the sphere
0
q
(r R)
R
£
Î
Capacitance of a spherical capacitor
C = 4rpÎ
0
At a point on the surface
or inside the spherical shell
0
q
(r R)
R
£
Î
At a point outside
the spherical shell
0
1q
V(r R)
4 r
= >
pÎ
2
121 2
1 2
CC(V V) 1
2CC
-
=
+
t

=

t
h
i
c
k
n
e
s
s



o
f
s
l
a
b

2
n
1
V
4
=
p
1
V
4
=
p

448. 444 PHYSICS
1.A parallel plate capac
itor is charged to a certain voltage.
Now, if the dielectric material (with dielectric constant k) is
removed then the
(a) capacitance increases by a factor of k
(b) electric field reduces by a factor k
(c) voltage across the capacitor decreases by a factor k
(d) None of these
2.Two identical conducting balls having positive charges q
1
and q
2
are separated by a distance r.If they are made to
touch each other and then separated to the same distance,
the force between them will be
(a) less than before (b) same as before
(c) more than before (d) zero
3.A sphere of radius R has uniform volume charge density.
The electric potential at a points (r <R) is
(a) due to the charge inside a sphere of radius r only
(b) due to the entire charge of the sphere
(c) due to the charge in the spherical sheel of inner and
outer radii r and R, only
(d) independent of r
4.Eight drops of mercury of equal radius and possessing equal
charge combine to form a big drop. The capacitance of bigger
drop as compared to each small drop is
(a) 16 times (b) 8 times
(c) 4 times (d) 2 times
5.The capacitance of a metallic sphere is F1
m, then it’s radius
is ne
arly
(a) 1.11 m (b) 10 m
(c) 9 km (d) 1.11 cm
6.Three charges 2 q, – q and – q are located at the vertices of
an equilateral triangle. At the centre of the triangle
(a) the field is zero but potential is non-zero
(b) the field is non-zero, but potential is zero
(c) both field and potential are zero
(d) both field and potential are non-zero
7.Two conducting spheres of radii r
1
and r
2
are equally
charged. The ratio of their potentials is
(a)
21
r/r(b)
12
r/r(c)
2
2
2
1
r/r(d)
2
1
2
2
r/r
8.The electric potential due to a small electric dipole at a large
distance r from the centre of the dipole is proportional to
(a)r (b) 1/r (c) 1/r
2
(d) 1/r
3
9.An electron of mass m and charge e is accelerated from rest
through a potential difference V in vacuum. Its final speed
will be
(a)
m
Ve2
(b)
m
Ve
(c) e V/2m(d) e V/m
10.A p
ositive point charge q is carried from a point B to a point
A in the electric field of a point charge + Q at O. If the
permitivity of free space is e
0
, the work done in the process
is given by
(a)
÷
ø
ö
ç
è
æ
+
ep b
1
a
1
4
Qq
o
(b)
÷
ø
ö
ç
è
æ
-
ep b
1
a
1
4
Qq
o
(c) ÷
÷
ø
ö
ç
ç
è
æ
-
ep
22
o b
1
a
1
4
Qq
(d) ÷
÷
ø
ö
ç
ç
è
æ
+
ep
22
o b
1
a
1
4
Qq
11.Two conc
entric, thin metallic spheres of radii R
1
and R
2
(R
1
> R
2
) bear charges Q
1
and Q
2
respectively. Then the
potential at distance r between R
1
and R
2
will be
0
1
k
4
æö
=
ç÷
peèø
(a)
12
QQ
k
r
+æö
ç÷
èø
(b)
12
2
QQ
k
rR
æö
+
ç÷
èø
(c)
21
1
QQ
k
rR
æö
+
ç÷
èø
(d)
12
12
QQ
k
RR
æö
+
ç÷
èø
12.Force between two plates of a capacitor is
(a)
o
Q
Ae
(b)
2
o
Q
2Ae
(c)
2
o
Q
Ae
(d) None of these
13.An
alpha particle is accelerated through a potential
difference of 10
6
volt. Its kinetic energy will be
(a) 1 MeV (b) 2 MeV(c) 4 MeV(d) 8 MeV
14.Two capacitors of capacitances C
1
and C
2
are connected in
parallel across a battery. If Q
1
and Q
2
respectively be the
charges on the capacitors, then
2
1
Q
Q
will be equal to
(a)
1 2
C
C
(b)
2 1
C
C
(c)
2
2
2
1
C
C
(d)
2
1
2
2
C
C
15.A system of two par
allel plates, each of area A, are separated
by distances d
1
and d
2
. The space between them is filled
with dielectrics of permittivities e
1
and e
2
. The permittivity
of free space is e
0
. The equivalent capacitance of the system
is
(a)
A
12
dd
21 12
ee
e +e (b)
A
120
dd
11 22
eee
e +e
(c)
A
0
dd
11 22
e
e +e (d)
A
0
dd
12 21
e
e +e

449. 445Electrostatic Potential and Capacitance
16.A large insulated sphere of radius r charged with Q units of
electricity is placed in contact with a small insulated
uncharged sphere of radius r´ and is then separated. The
charge on smaller sphere will now be
(a) Q (r' + r) (b) Q (r + r')
(c)
Q
r'r+
(d)
Qr'
r'r+
17.The capacitance of the capacitor of plate areas A
1
and A
2
(A
1
< A
2
) at a distance d, as shown in figure is
(a)
012
(A A)
2d
Î+
(b)
02
A
d
Î
(c)
0 12
AA
d
Î
A
1
A
2
d
(d)
01
A
d
Î
18.When air i
s replaced by a dielectric medium of force
constant K, the maximum force of attraction between two
charges, separated by a distance
(a) decreases K-times(b) increases K-times
(c) remains unchanged (d) becomes 2
K
1
times
19.A conduct
or carries a certain charge. When it is connected
to another uncharged conductor of finite capacity, then theenergy of the combined system is(a) more than that of the first conductor
(b) less than that of the first conductor
(c) equal to that of the first conductor
(d) uncertain
20.The magnitude of the electric field E in the annular region
of a charged cylindrical capacitor
(a) is same throughout
(b) is higher near the outer cylinder than near the inner
cylinder
(c) varies as
r
1
, where r
is the distance from the axis
(d) varies as 2
r
1
, where r
is the distance from the axis
21.A parallel plate condenser with oil between the plates
(dielectric constant of oil K = 2) has a capacitance C. If the
oil is removed, then capacitance of the capacitor becomes
(a)
2C(b) 2 C (c)
2
C
(d)
2
C
22.An air cap
acitor C connected to a battery of e.m.f. V acquires
a charge q and energy E. The capacitor is disconnected
from the battery and a dielectric slab is placed between the
plates. Which of the following statements is correct ?
(a) V and q decrease but C and E increase
(b) V remains unchange, but q, E and C increase
(c) q remains unchanged, C increases, V and E decrease
(d) q and C increase but V and E decrease.
23.Two parallel metal plates having charges + Q and – Q face
each other at a certain distance between them. If the plates
are now dipped in kerosene oil tank, the electric field
between the plates will
(a) remain same (b) become zero
(c) increases (d) decrease
24.A parallel plate condenser has a uniform electric field E(V/
m) in the space between the plates. If the distance between
the plates is d(m) and area of each plate is A(m
2
) the energy
(joules) stored in the condenser is
(a)E
2
Ad/
0
Î (b)
2
0
1
E
2
Î
(c)
0
EAdÎ (d)
2
0
1
E Ad
2
Î
25.Which of th
e following figure shows the correct
equipotential surfaces of a system of two positive charges?
(a) ++
(b)
+ +
(c)
+ +
(d)
+ +
1.The positive terminal of 12 V battery is connected to the
ground. Then the negative terminal will be at
(a) – 6 V (b) + 12 V(c) zero (d) – 12 V
2.A hollow metal sphere of radius 5 cm is charged such that
the potential on its surface is 10 V. The potential at a distance
of 2 cm from the centre of the sphere is
(a) zero (b) 10 V (c) 4 V (d) 10/3 V
3.Find the dipole moment of a system where the potential
2.0 × 10
–5
V at a point P, 0.1m from the dipole is 3.0 × 10
4
.
(Use q = 30°).
(a) 2.57 × 10
–17
Cm (b) 1.285 × 10
–15
Cm
(c) 1.285 × 10
–17
Cm (d) 2.57 × 10
–15
Cm

450. 446 PHYSICS
4.A battery of e.
m.f. V volt, resistors R
1
and R
2
, a condenser
C and switches S
1
and S
2
are connected in a circuit shown.
The condenser will get fully charged to V volt when
V C
1R
2R
2S
1S
(a)S
1
and S
2
are both cl osed
(b) S
1
and S
2
are both open
(c)S
1
is open and S
2
is closed
(d) S
1
is closed and S
2
is open
5.The electric potential at the surface of an atomic nucleus
(Z = 50) of radius of 9 × 10
–15
m is
(a) 80 V(b) 8 × 10
6
V (c) 9 V (d) 9 × 10
5
V
6.Three point charges +q , + 2q and – 4q where q = 0.1 mC, are
placed at the vertices of an equilateral triangle of side 10 cm
as shown in figure. The potential energy of the system is
1
0

c
m
– 4q
10 cm
+q10 cm+2q
(a) 3 × 10
–3
J (b) –3 × 10
–3
J
(c
) 9 × 10
–3
J (d) –9 × 10
–3
J
7.The four capacitors, each of 25 m F are connected as shown
in fig. The dc voltmeter reads 200 V. The charge on each
plate of capacitor is
V
+ +– –
+ +– –
(a) C102
3-
´± (b) C105
3-
´±
(c) C102
2-
´
± (d) C105
2-
´±
8.An air capacitor of capacity C = 10 mF is connected to a
constant voltage battery of 12 volt. Now the space betweenthe plates is filled with a liquid of dielectric constant 5. The(additional) charge that flows now from battery to thecapacitor is
(a) 120 m C (b) 600 m C (c) 480 m C (d) 24 m C
9.A capacitor is charged to store an energy U. The charging
battery is disconnected. An identical capacitor is now
connected to the first capacitor in parallel. The energy in
each of the capacitors is
(a) 3 U/2 (b) U (c) U/4 (d) U/2
10.Two capacitors when connected in series have a
capacitance of 3 mF, and when connected in parallel have a
capacitance of 16 mF. Their individual capacities are
(a) 1 mF, 2 mF (b) 6 mF, 2 mF
(c) 12 mF, 4 mF (d) 3 mF, 16 mF
11.The capacity of a parallel plate condenser is 10 mF, when
the distance between its plates is 8 cm. If the distance
between the plates is reduced to 4 cm, then the capacity of
this parallel plate condenser will be
(a) 5 mF (b) 10 mF (c) 20 mF (d) 40 mF
12.The capacitor, whose capacitance
is 6, 6 and 3mF respectively are
connected in series with 20 volt
line. Find the charge on 3mF.
(a) 30 mc
(b) 60 mF
(c) 15 mF
6Fm6Fm 3Fm
20Fm
(d) 90 mF
13.Four metallic
plates each with a surface area of one side A,
are placed at a distance d from each other. The two outer
plates are connected to one point A and the two other inner
plates to another point B as shown in the figure. Then the
capacitance of the system is
A
B
(a)
d
A
0e
(b)
d
A2
0e
(c)
d
A3
0e
(d)
d
A4
0e
14.Two spheric
al conductors A and B of radii a and b (b>a) are
placed concentrically in air. The two are connected by acopper wire as shown in figure. Then the equivalentcapacitance of the system is
(a)
ab
ab
4
0
-
pe
(b) )ba(4
0
+pe
A
B
a
b
(c) b4
0
pe
(d) a4
0
pe
15.A bal
l of mass 1 g carrying a charge 10
–8
C moves from a
point A at potential 600 V to a point B at zero potential. The
change in its K.E. is
(a) – 6 × 10
–6
erg (b) – 6 × 10
–6
J
(c) 6 × 10
–6
J (d) 6 × 10
–6
erg
16.Two capacitors C
1
and C
2
in a circuit are joined as shown in
figure. The potentials of points A and B are V
1
and V
2
respectively; then the potential of point D will be
A B
D
2C
1C1
V
2V
(a)
2
)VV(
21+
(b)
21
2112
CC
VCVC
+
+
(c)
21
2211
CC
VCVC
+
+
(d)
21
2112
CC
VCVC
+
+

451. 447Electrostatic Potential and Capacitance
17.A parallel plate capacitor with air between the plates is
charged to a potential difference of 500V and then insulated.
A plastic plate is inserted between the plates filling the
whole gap. The potential difference between the plates now
becomes 75V. The dielectric constant of plastic is
(a) 10/3 (b) 5 (c) 20/3 (d) 10
18.The plates of a parallel plate capacitor have an area of
90 cm
2
each and are separated by 2.5 mm. The capacitor is
charged by a 400 volt supply. How much electrostatic
energy is stored by the capacitor?
(a)2.55 × 10
–6
J (b) 1.55 × 10
–6
J
(c) 8.15 × 10
–6
J (d) 5.5 × 10
–6
J
19.From a supply of identical capacitors rated 8 mF, 250V, the
minimum number of capacitors required to form a composite
16 mF, 1000V is
(a)2 (b) 4 (c) 16 (d) 32
20.Calculate the area of the plates of a one farad parallel plate
capacitor if separation between plates is 1 mm and plates
are in vacuum
(a) 18 × 10
8
m
2
(b) 0.3 × 10
8
m
2
(c) 1.3 × 10
8
m
2
(d) 1.13 × 10
8
m
2
21.A one microfarad capacitor of a TV is subjected to 4000 V
potential difference. The energy stored in capacitor is
(a) 8 J (b) 16 J
(c) 4 × 10
–3
J (d) 2 × 10
–3
J
22.Two capacitors, C
1
= 2mF and C
2
= 8 mF are connected in
series across a 300 V source. Then
(a) the charge on each capacitor is 4.8×10
–4
C
(b) the potential difference across C
1
is 60 V
(c) the potential difference across C
2
is 240 V
(d) the energy stroed in the system is 5.2 × 10
–2
J
23.Two capacitors C
1
and C
2
= 2C
1
are
Q
C
1
C
2
= 2C
1
R
conne
cted in a circuit with a switch between
them as shown in the figure. Initially the
switch is open and C
1
holds charge Q. The
switch is closed. At steady state, the charge
on each capacitor will be
(a) Q, 2Q (b)
Q 2Q
,
33
(c)
3Q
, 3Q
2
(d)
2Q 4Q
,
33
24.Two capacitors of capacitance C are connected in series. If
one of them is filled with dielectric substance k, what is the
effective capacitance ?
(a)
( )
kC
1k+
(b) C(k + 1)
(
c)
2kC
1k+
(d) None of these
25.The potential at a point x (measured in mm) due to some
charges situated on the x-axis is given by V(x) = 20/(x
2
– 4)
volt
The electric field E at x = 4 mm is given by
(a) (10/9) volt/mm and in the +ve x direction
(b) (5/3) volt/mm and in the –ve x direction
(c) (5/3) volt/mm and in the +ve x direction
(d) (10/9) volt/mm and in the –ve x direction
26.Capacitance (in F) of a spherical conductor with radius 1 m
is
(a) 1.1 × 10
–10
(b) 10
6
(c) 9 × 10
–9
(d) 10
–3
27.Two metal pieces having a potential difference of 800 V are
0.02 m apart horizontally. A particle of mass 1.96 × 10
–15
kg
is suspended in equilibrium between the plates. If e is the
elementary charge, then charge on the particle is
(a)8 (b) 6 (c) 0.1 (d) 3
28.Identical charges – q each are placed at 8 corners of a cube
of each side b. Electrostatic potential energy of a charge
+ q which is placed at the centre of cube will be
(a)
b
q24
o
2
ep
-
(b)
b
q28
o
2
ep
-
(c)
b3
q4
o
2
ep
-
(d)
b
q28
o
2
ep
-
29.A charge +q i
s fixed at each of the points x = x
0
, x = 3x
0
,
x = 5x
0
, .... upto
¥ on X-axis and charge –q is fixed on each
of the points x = 2x
0
, x = 4x
0
, x = 6x
0
, .... upto ¥. Here x
0
is
a p
ositive constant. Take the potential at a point due to a
charge Q at a distance r from it to be r4
Q
0pe
. Then the
potential at t
he origin due to above system of charges will be
(a) zero (b)
2logx8
q
e00pe
(c) infin
ity (d)
00
e
x4
2logq
pe
30.Two equa
lly charged spheres of radii a and b are connected
together. What will be the ratio of electric field intensity on
their surfaces?
(a)
b
a
(b)
2
2
b
a
(c )
a
b
(d)
2
2
a
b
31.In a hollow s
pherical shell, potential (V) changes with respect
to distance (s) from centre as
(a)
V
S
(b)
V
S
(c)
V
S
(d)
S
V
32.A parallel plate capacito
r of capacitance C is connected to
a battery and is charged to a potential difference V. Another
capacitor of capacitance 2C is similary charged to a potential
difference 2V. The charging battery is now disconnected
and the capacitors are connected in parallel to each other in
such a way that the positive terminal of one is connected to
the negative terminal of the other. The final energy of the
configuration is
(a) zero (b)
2
CV
2
3
(c)
2
CV
6
25
(d)
2
CV
2
9

452. 448 PHYSICS
33.A solid c
onducting sphere having a charge Q is surrounding
by an uncharged concentric conducting hollow spherical
shell. Let the potential difference between the surface of
the solid sphere and that of the outer surface of the hollow
shell be V. If the shell is now given a charge of – 3Q, the new
potential difference between the same two surfaces is
(a)V (b) 2 V (c) 4 V (d) – 2 V
34.In the electric field of an point charge
q, a certain charge is carried from point
A to B, C, D and E. Then the work
done is
(a) least along the path AB
(b) least along the path AD
A
B
C D
E
q
+
(c) zero along any one of the path AB, AC, AD andAE
(d) least along AE
35.A circuit is connected as shown in the figure with the switch
S open. When the switch is closed, the total amount of
charge that flows from Y to X is
9 V
S
X
W3 W6
F3m F6m
Y
(a)0 (b) 54 mC (c) 27mC (d) 81 mC
36.If a slab of insulating material 4 × 10
–5
m thick is introduced
between the plates of a parallel plate capacitor, the distancebetween the plates has to be increased by 3.5 × 10
–5
m to
restore the capacity to original value. Then the dielectricconstant of the material of slab is
(a)8 (b) 6 (c) 12 (d) 10
37.Three capacitors each of capacity 4mF are to be connected
in such a way that the effective capacitance is 6 m F. This can
be done by
(a) connecting two in parallel and one in series
(b) connecting all of them in series
(c) connecting them in parallel
(d) connecting two in series and one in parallel
38.If we increase ‘d’ of a parallel plate condenser to ‘2d’ and fill
wax to the whole empty space between its two plate, then
capacitance increase from 1pF to 2pF. What is the dielectric
constant of wax?
(a)2 (b) 4 (c)4 (d) 8
39.Two spherical conductors A and B of radii a and b (b > a)
are placed concentrically in air. B is given charge +Q and A
is earthed. The equivalent capacitance of the system is
(a)
ab
ab
4
0
-
pe
(b) )ba(4
0
+pe
(c) b4
0
pe
+
+
+
+
+
+
+
+
+
+
+
+
+ +
+
+
+
+
+
+
+
+
+
+
–
–
–
–
–
––
–
O
a
A
b
B
(d)
÷
÷
ø
ö
ç
ç
è
æ
-
pe
ab
b
4
2
0
40.The capacitance of
a parallel plate capacitor is C
a
(Fig. a). A
dielectric of dielectric constant K is inserted as shown in
fig (b) and (c). If C
b
and C
c
denote the capacitances in fig
(b) and (c), then
d
d
C
a
C
b
C
c
d/2
K
K
(a) (b)
(c)
(a) both C
b
, C
c
> C
a
(b) C
c
> C
a
while C
b
> C
a
(c) both C
b
, C
c
< C
a
(d) C
a
= C
b
= C
c
41.In the circuit shown, which of the following statements istrue if V
1
(potential across C
1
) is 30 V and V
2
(potential
across C
2
) is 20 V?
V30V
1= V20V
2=
1S
2
S3
S
pF2C
1= pF3C
2=
(a) With S
1
closed , V
1
= 15 V, V
2
= 25 V
(b) With S
3
closed, V
1
= V
2
= 25 V
(c) With S
1
and S
2
closed, V
1
= V
2
= 0
(d) With S
1
and S
3
closed, V
1
= 30 V, V
2
= 20 V
42.A parallel plate capacitor is located horizontally such that
one of the plates is submerged in a liquid while the other is
above the liquid surface. When plates are charged the level
of liquid
(a) rises
(b) falls
(c) remains unchanged
Changed Liquid
(d) may rise or fall depending
on the amount of charge
43.Two small conductors A and B are given charges q
1
and
q
2
respectively. Now they are placed inside a hollow metallic
conductor C carrying a charge Q. If all the three
conductors A, B and C are connected by a conducting wire
as shown, the charges on A, B and C will be respectively
A
B
C
Q
1
q
2
q
(a)
2
qq
21+
,
2
qq
21+
, Q
(b)
3
qqQ
31++
,
3
qqQ
21++
,
3
qqQ
21++
(c)
2
Qqq
21++
,
2
Qqq
21++
, 0
(d) 0, 0,
Q + q
1
+ q
2

453. 449Electrostatic Potential and Capacitance
44.Between the plates of a parallel plate capacitor dielectric
plate is introduced just to fill the space between the plates.
The capacitor is charged and later disconnected from the
battery. The dielectric plate is slowly drawn out of the
capacitor parallel to plates. The plot of the potential
difference V across the plates and the length of the dielectric
plate drawn out is
(a)
x
V (b)
x
V
(c)
x
V (d)
x
V
45.Three capacitors C
1
, C
2
and C
3
are connected to a battery
as shown. With symbols having their usual meanings, the
correct conditions are
V
3V
2
V
1V
3Q
2Q
1
Q
3C
2C
1
C
(a)Q
1
= Q
2
= Q
3
a
nd V
1
= V
2
= V
(b) V
1
= V
2
= V
3
= V
(c)Q
1
= Q
2
+ Q
3
and V = V
1
= V
2
(d) Q
2
= Q
3
and V
2
= V
3
46.Figure (i) shows two capacitors connected in series andconnected by a battery. The graph (ii) shows the variationof potential as one moves from left to right on the branchAB containing the capacitors. Then
V
A B
E
(i) )ii(
1
C
2
C
(a)C
1
= C
2
(b) C
1
<
C
2
(c)C
1
> C
2
(d) C
1
and C
2
cannot be compared
47.Two vertical metallic plates carrying equal and opposite
charges are kept parallel to each other like a parallel plate
capacitor. A small spherical metallic ball is suspended by a
long insulated thread such that it hangs freely in the centre
of the two metallic plates. The ball, which is uncharged, is
taken slowly towards the positively charged plate and is
made to touch that plate. Then the ball will
(a) stick to the positively charged plate
(b) come back to its original position and will remain there
(c) oscillate between the two plates touching each plate
in turn
(d) oscillate between the two plates without touch them
48.Two parallel plate capacitors of capacitances C and 2C are
connected in parallel and charged to a potential difference V.
The battery is then disconnected and the region between the
plates of the capacitor C is completely filled with a material fo
dielectric constant K. The potential difference across the
capacitors now becomes
(a)
3V
K2+
(b) KV (c)
V
K
(d)
3
KV
49.A parallel plate capacitor is connected to a battery. The
quantities charge, voltage, electric field and energy
associated with this capacitor are given by Q
0
, V
0
, E
0
, and
U
0
respectively. A dielectric slab is now introduced to fill
the space between the plates with the battery still in
connection. The corresponding quantities now given by
Q, V, E and U are related to the previous ones as
(a) Q > Q
0
(b) V > V
0
(c) E > E
0
(d) U < U
0
50.The effective capacitance of combination of combination of
equal capacitors between points A and B shown in figure is
CC
C C
C C
C C
C
C
BA
(a)C (b) 2C ( c) 3C (d)
C
2
51.A parallel plate capacitor of plate area A and plate separation
d is charged to potential difference V and then the batteryis disconnected. A slab of dielectric constant K is theninserted between the plates of capacitor so as to fill thespace between the plates. If Q, E and W denote respectively,the magnitude of charge on each plate electric field betweenthe plates (after the slab is inserted), and work done on thesystem, in question, in the process of inserting the slab,then which is wrong ?
(a)
0
AV
Q
d
e
= (b)
0
KAV
Q
d
e
=
(c)
V
E
Kd
= (d)
2
0
AV 1
W1
2dK
e æö
=-
ç÷
èø

454. 450 PHYSICS
52.In the circui
t given below, the charge in mC, on the capacitor
having 5 mF is
6 V
F3m
F2m
F5m
F4m
+
d
c
b
e
f
a
(a) 4.5(b) 9 (c)7 (d) 15
53.If
a charge – 150 nC is given to a concentric spherical shell
and a charge +50 nC is placed at its centre then the charge
on inner and outer surface of the shell is
(a) –50 nC, –100 nC(b) +50 nC, –200 nC
(c) –50 nC, –200 nC(d) 50 nC, 100 nC
54.A battery is used to charge a parallel plate capacitor till the
potential difference between the plates becomes equal to
the electromotive force of the battery. The ratio of the
energy stored in the capacitor and the work done by the
battery will be
(a) 1/2(b) 1 (c)2 (d) 1/4
55.Four point charges q, q, q and – 3q are placed at the vertices
of a regular tetrahedron of side L. The work done by electric
force in taking all the charges to the centre of the tetrahedron
is (where
0
1
k
4
=
pe
)
(a)
2
6kq
L
(b)
2
6kq
L
-
(c)
2
12kq
L
(d) zero
56.Two ide
ntical particles each of mass m and having charges
–q and +q are revolving in a circle of radius r under the
influence of electric attraction. Kinetic energy of each
particle is
0
1
k
4
æö
=
ç÷
peèø
(a) kq
2
/4r (b) kq
2
/2r(c) kq
2
/8r(d) kq
2
/r
57.F
igure shows three circular arcs, each of radius R and total
charge as indicated. The net electric potential at the centre
of curvature is
(a)
0
Q
2Rpe
(b)
0
Q
4Rpe
+Q
–2Q
30°
+3Q
45°
R
(c)
0
2Q
Rpe
(d)
0
Q
Rpe
58.If the potential of a capacitor having capacity 6 mF is increased
from 10 V to 20 V, then increase in its energy will be
(a) 4 × 10
–4
J (b) 4 × 10
–4
J
(c) 9 × 10
–4
J (d) 12 × 10
–6
J
59.In the given circuit with steady current, the potential drop
across the capacitor must be
2V
V
V
R
2R
C
A B
(a)
3
V2
(b)
3
V
(c)
2
V
(d) V
60.A uniformly c
harged thin spherical shell of radius R carries
uniform surface charge density of
sper unit area. It is made
o
f two hemispherical shells, held together by pressing them
with force F (see figure). F is proportional to
F F
(a)
22
0
1
s
e
R
(b)
2
0
1
s
e
R
(c)
2
0
1s
eR
(d)
2
2
0
1s
eR
61.A dielectric slab of thickness d is inserted in a parallel plate
capacitor whose negative plate is at x = 0 and positive plate
is at x = 3d. The slab is equidistant from the plates. the
capacitor is given some charge. As one goes from 0 to 3d
(a) the magnitude of the electric field remains the same
(b) the direction of the electric field remains the same
(c) the electric potential decreases continuously
(d) the electric potential increases at first, then decreases
and again increases
62.In the given circuit if point C is connected to the earth and
a potential of +2000V is given to the point A, the potential
at B is
C
B
A
10Fm 10Fm
5Fm
10Fm
(a) 1500V (b
) 1000 V(c) 500 V (d) 400 V
63.A 4 mF capacitor, a resistance of 2.5 MW is in series with
12V battery. Find the time after which the potential differenceacross the capacitor is 3 times the potential diference across
the resistor. [Given In (b) = 0.693]
(a) 13.86s (b) 6.93 s(c)7s (d) 14 s

455. 451Electrostatic Potential and Capacitance
64.If a capacitor 900 µF is charged to 100 V and its total energy
is transferred to a capacitor of capacitance 100 µF then its
potential is
(a) 200 V (b) 30 V (c) 300 V (d) 400 V
65.What is the effective capacitance between points X and Y?
(a) 24 mF
(b) 18 mF
(c) 12 mF
C= 6F
3 m C= 6F
2 m
C= 6F
1 m
C= 6F
4 m
C= 20F
5 m
X
A C B
D
Y
(d) 6 mF
66.In a par
allel plate capacitor, the distance between the plates
is d and potential difference across plates is V. Energy
stored per unit volume between the plates of capacitor is
(a)
2
2
V2
Q
(b)
2
2
0
d
V
2
1
e
(c) 2
0
2
d
V
2
1
e
(d)
d
V
2
1
2
0e
67.A capacitor C
1
is ch
arged to a potential difference V. The
charging battery is then removed and the capacitor is
connected to an uncharged capacitor C
2
. The potential
difference across the combination is
(a)
)CC(
VC
21
1
+
(b)
÷
÷
ø
ö
ç
ç
è
æ
+
1
2
C
C
1V
(c)
÷
÷
ø
ö
ç
ç
è
æ
+
2
1
C
C
1V (d)
)CC(
VC
21
2
+
68.As per this d
iagram a point charge +q is placed at the origin
O. Work done in taking another point charge
– Q from the point A [coordinates (0, a)] to another point B
[coordinates (a, 0)] along the straight path AB is
(a) zero
(b)
a2
a
1
4
qQ
2
0
÷
÷
ø
ö
ç
ç
è
æ
pe
-
(c)
2
a
.
a
1
4
qQ
2
0
÷
÷
ø
ö
ç
ç
è
æ
pe
x
y
A
B
O
(d)
a2.
a
1
4
qQ
2
0
÷
÷
ø
ö
ç
ç
è
æ
pe
69.Two ch
arges q
1
and q
2
are placed 30 cm apart, as shown in
the figure. A third charge q
3
is moved along the arc of a
circle of radius 40 cm from C to D. The change in the
potential energy of the system is
q
k
3
0
4pÎ
,, where k is
C
A
q
q
q
1
3
2
BD
40 cm
30 cm
(a) 8q
1
(b) 6 q
1
(c) 8q
2
(d) 6q
2
70.A network of four capacitors of capacity equal to C
1
= C,
C
2
= 2C, C
3
= 3C and C
4
= 4C are conducted to a battery as
shown in the figure. The ratio of the charges on C
2
and C
4
is
(a) 4/7(b) 3/22 (c) 7/4 (d) 22/3
71.A series combination of n
1
capacitors, each of value C
1
, is
charged by a source of potential difference 4 V. When
another parallel combination of n
2
capacitors, each of value
C
2
, is charged by a source of potential difference V, it has
the same (total) energy stored in it, as the first combination
has. The value of C
2
, in terms of C
1
, is then
(a)
1
12
2C
nn
(b)
2
1
1
16
n
C
n
(c)
2
1
1
2
n
C
n
(d)
1
12
16C
nn
72.A condens er of capacity C is charged to a potential
difference of V
1
. The plates of the condenser are then
connected to an ideal inductor of inductance L. The currentthrough the inductor when the potential difference acrossthe condenser reduces to V
2
is
(a)
1/2
22
12
()CVV
L
æö-
ç÷
èø
(b)
1/2
2
12
()CVV
L
æö-
ç÷
èø
(c)
22
12
()CVV
L
-
(d)
12
()CVV
L
-
DIRECTIONS (for Q
s. 73 to 75) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer(ONLY ONE option is correct ) from the following-
(a) Statement -1 is false, Statement-2 is true
(b) Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c) Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d) Statement -1 is true, Statement-2 is false
73. Statement 1 : Each of the plates of a parallel-plate capacitor
is given equal positive charge Q. The charges on the facing
surfaces will be same.
Statement 2 : A negative charge (–Q) will be induced on
each of the facing surfaces.
74. Statement 1 : Electric potential and electric potential energy
are different quantities.
Statement 2 : For a system of positive test charge and
point charge electric potential energy = electric potential.
75. Statement I : Two equipotential surfaces cannot cut each
other.
Statement II : Two equipotential surfaces are parallel to
each other.

456. 452 PHYSICS
Exemplar Questions
1.A capacitor of 4 µF is connected as shown in the circuit.
The internal resistance of the battery is 0.5W. The amount
of charge on the capacitor plates will be
4mF
10W
2W
2.5 V
(a) 0 µC (b
) 4 µC (c) 16 µC(d) 8 µC
2.A positively charged particle is released from rest in an uniformelectric field. The electric potential energy of the charge
(a) remains a constant because the electric field is uniform
(b) increases because the charge moves along the electric field
(c) decreases because the charge moves along the electric
field
(d) decreases because the charge moves opposite to the
electric field
3.Figure shows some equipotential lines distributed in space.
A charged object is moved from point A to point B.
(a) The work done in Fig. (i) is the greatest
(b) The work done in Fig. (ii) is least
(c) The work done is the same in Fig. (i), Fig.(ii) and Fig. (iii)
(d) The work done in Fig. (iii) is greater than Fig. (ii) but
equal to that in
A B
10V 20V 30V 40V50V
Fig. (i)
20V40V
A B
10V30V 30V
Fig. (ii)
10V 20V40V
A B
50V
30V
Fig. (iii)
4.The electro
static potential on the surface of a charged
conducting sphere is 100V. Two statements are made in this
regard S
1
at any point inside the sphere, electric intensity
is zero. S
2
at any point inside the sphere, the electrostatic
potential is 100V. Which of the following is a correct
statement?
(a)S
1
is true but S
2
is false
(b) Both S
1
and S
2
are false
(c)S
1
is true, S
2
is also true and S
1
is the cause of S
2
(d) S
1
is true, S
2
is also true but the statements are
independant
5.Equipotentials at a great distance from a collection of
charges whose total sum is not zero are approximately
(a) spheres (b) planes
(c) paraboloids (d) ellipsoids
6.A parallel plate capacitor is made of two dielectric blocks in
series. One of the blocks has thickness d
1
and dielectric
constant K
1
and the other has thickness d
2
and dielectric
constant K
2
as shown in figure. This arrangement can be
thought as a dielectric slab of thickness d (= d
1
+ d
2
) and
effective dielectric constant K. The K is
K
1
K
2
d
1
d
2
(a)
11 22
12
Kd Kd
dd
+
+
(b)
11 22
12
Kd Kd
KK
+
+
(c)
()
( )
1212
11 22
KKdd
Kd Kd
+
+
(d)
12
12
2KK
KK+
NEET/AIPMT (2013-2017) Questions
7.A, B and C are three points in a uniform electric field. The
electric potential is [2013]
A
B
C
E
?
(a) maximum at B
(b) maximum at C
(
c) same at all the three points A, B and C
(d) maximum at A
8.Two thin dielectric slabs of dielectric constants K
1
and K
2
(K
1
< K
2
) are inserted between plates of a parallel plate
capacitor, as shown in the figure. The variation of electric

457. 453Electrostatic Potential and Capacitance
field ‘E’ between the plates with distance ‘d’ as measured
from plate P is correctly shown by : [2014]
+
+
+
+
+
+
+
+
+
P Q
K
1K
2
–
–
–
–
–
–
–
–
–
(a)
E
0
d
(b)
E
0
d
(c)
d
0
E
(d)
E
0
d
9.A conducting sphere of radius R is given a charge Q. The
electric potential and the electric field at the centre of the
sphere respectively are: [2014]
(a) Zero and
2
0
Q
4Rpe
(b)
0
Q
4Rpe
and Zero
(c)
2
0 0
QQ
and
4R 4Rpe pe
(d) Both
are zero
10.In a region, the potential is represented byV(x, y, z) = 6x – 8xy – 8y + 6yz, where V is in volts and x, y, zare in metres. The electric force experienced by a charge of 2coulomb situated at point (1, 1, 1) is : [2014]
(a)
65 N (b) 30 N
(c) 2
4 N (d)
4 35 N
11.A paral
lel plate air capacitor of capacitance C is connected
to a cell of emf V and then disconnected from it. A dielectricslab of dielectric constant K, which can just fill the air gapof the capacitor, is now inserted in it. Which of the followingis incorrect ? [2015]
(a) The energy stored in the capacitor decreases K times.
(b) The chance in energy stored is
211
CV –1
2K
æö
ç÷
èø
(c) The charge on the capacitor is not conserved.
(d) The potential difference between the plates decreases
K times.
12.If potential (in volts) in a region is expressed as V(x, y, z) =
6 xy – y + 2yz, the electric field (in N/C) at point (1, 1, 0) is :
[2015 RS]
(a) ˆˆˆ
(6i 5j2k)- ++ (b) ˆ ˆˆ
(2i 3jk)- ++
(c) ˆ ˆˆ
(6i 9jk)- ++ (d) ˆˆˆ
(3i 5j 3k)- ++
13.A parallel plate air
capacitor has capacity 'C' distance of
separation between plates is 'd' and potential difference 'V'is applied between the plates. Force of attraction betweenthe plates of the parallel plate air capacitor is : [2015 RS]
(a)
2
CV
2d
(b)
2
CV
d
(c)
22
2
CV
2d
(d)
22
2
CV
2d
14.
V
12
2Fm 8Fm
A capacitor of 2mF is charged as shown in the diagram.
When the switch S is turned to position 2, the percentage
of its stored energy dissipated is : [2016]
(a) 0% (b) 20%
(c) 75% (d) 80%
15.A capacitor is charged by a battery. The battery is removed
and another identical uncharged capacitor is connected in
parallel. The total electrostatic energy of resulting system:
(a) decreases by a factor of 2 [2017]
(b) remains the same
(c) increases by a factor of 2
(d) increases by a factor of 4
16.The diagrams below show regions of equipotentials. [2017]
(a)
20V 40V
BA
10 V 30V
(b)
20V 40V
BA
10 V 30V
(c)
10V30V
BA
20V 40V
(d)
40V
BA
30V
10V
20V
A positive charge is moved from A to B in each diagram.
(a) In all the four cases the work done is the same(b) Minimum work is required to move q in figure (a)
(c) Maximum work is required to move q in figure (b)
(d) Maximum work is required to move q in figure (c)

458. 454 PHYSICS
EXERCISE - 1
1. (d) 2. (c) 3
. (a) 4. (d) 5. (c)
6. (b) Potential at the centre of the triangle,
0
r4
qqq2
r4
q
V
00
=
ep
--
=
ep
å
=
Obviously, 0E¹
7. (b) As
r
1
V.e.i
r4
Q
V
0
µ
ep
=
1
2
2
1
r
r
V
V
=\
8. (c) Due to
small dipole,
2
1
V.
r
µ
9. (a) K.E. = Work do
ne = eV 21
m eV
2
n=
2 eV
m
\ n=
10. (b) )VV(
qW
BABA
-=

ú
û
ù
ê
ë
é
ep
-
ep
=
b4
Q
a4
Q
q
00
ú
û
ù
ê
ë
é
-
ep
=
b
1
a
1
4
Qq
0
11. (c)
R
2
R
1
r
Q
2
Q
1
21
r
0 01
QQ
V
4r4R
=+
pe pe
21
r
01
QQ1
V
4 rR
æö
=+ ç÷
pe
èø
12. (b) The magnitude of electric field by any one plate is
ooA2
Q
or
2 ee
s
–Q
E
Q
Now force magnitude is EQ i.e.
o
2
A2
Q
F
e
=
13. (b) Cha
rge on a particle, q = 2 e.
K.E. = work done = q × V = 2e × 10
6
V = 2 MeV.
14. (b) In parallel, potential is same, say V
2
1
2
1
2
1
C
C
VC
VC
Q
Q
==
15. (a)
A A
d
1 d
2
e
1
e
2
111
C CC
s12
=+
11
AAC 12s
dd12
=
ee
+
Þ
11 dd12
CA
s 12
æö
=+ç÷
eeèø
Þ
12
21 12
A
C
s
dd
ee
Þ=
e +e
16. (d) Comm
on potential,
00
Q0
V
4r4r
=+
pe pe
0
Q0
4 (r r ')
+
=
pe+
\ charge o
n smaller sphere of radius r' isrr
rQ
Vr4
0
¢+
¢
=´¢pe
17. (d)
0
A
C
d
Î
=
A ® common area, Here A
= A
1
18. (a) In air F
air
=
2
21
0r
qq
4
1
pe
In medium
F
m
=
2
21
0Kr
qq
4
1
pe
K
F
F
K
1
F
F
air
m
air
m
=Þ=\
(decreases K-ti
mes)
19. (b) Energy will be lost during transfer of charge (heating
effect).
20. (c)
r2
E
0pe
l
µ
hence
r
1
Eµ
21. (d) When oil
is placed between space of plates

d
A2
C
0e
= ... (1)

0
KA
C ,whereK2
d
eéù
==
êú
ëû
Q
When oil is
removed
d
A
'C
0e
= ............. (2)
o
n comparing both equation, weget C' = C/2
Hints & Solutions

459. 455Electrostatic Potential and Capacitance
22. (c) When a battery across the plates of capacitor is
disconnected and dielectric slab is placed in between
the plates, then
(i) capacity C increases
(ii) charge q remains unchanged
(iii) potential V decreases
(iv) energy E decreases
23. (d) Electric field
Q
E
A
==
s
ee
e of kerosine oil is more than that of air..
As e increases, E decreases.
24. (d) U =
21
CV
2
U =
220
0
A11
(Ed) A Ed
2d2
Îæö
=Î
ç÷
èø
25. (
c) Equipotential surfaces are normal to the electric field
lines. The following figure shows the equipotential
surfaces along with electric field lines for a system of
two positive charges.
EXERCISE - 2
1. (d
) When negative terminal is grounded, positive terminal
of battery is at +12 V. When positive terminal is
grounded, the negative terminal will be at –12 V.
2. (b) Potential at any point inside the sphere = potential at
the surface of the sphere = 10V.
3. (a)
( )( )
25
2
17
9
2.0 10 V 0.1m
Vr
qd 2.57 10 C m.
k cos
m3
9.0 10
C 2
V
-
-
´
== =´
q ìü
ïï æöïï
´íý ç÷
æöèø
ïï
ç÷
èøïïîþ
No
te that the units cancel to leave units appropriate
for a dipole moment.
4. (d) When S
1
is closed and S
2
is opened, the capacitor will
get charged to a potential difference of V volts.
5. (b)
o0
q (Ze)
V
4r4r
==
pe pÎ
19
9
15
(50 1.6 10 )
9 10
9 10
-
-
´´
´
´
V108
6
´=
6
. (d)
222
0
1 2q 8q 4q
U
4 aaa
æö
= --ç÷
ç÷pe
èø
Þ
2
0
1 10q
U
4a
æö
=-ç÷
ç÷pe
èø
Þ
9 62
9 10 10 (0.1 10)
U
10
100
-
´´´´
=-
æö
ç÷
èø
Þ U = –
9 × 10
–3
J
7. (b) Charge on each plate of each capacitor
6
Q CV 25 10 200=± =± ´ ´ C105
3-
´ ±=
8. (c) C1201210VCq
11
m=´==
C60012105VKCVCq
122
m=´´=´==
Additional charge that flows
21
q q 600 120 4 80 C.=-=-=m
9. (c)As battery is disconnected, total charge Q is shared
equally by two capacitors.
Energy of each capacitor .U
4
1
C2
Q
4
1
C2
)2/Q(
22
===
10. (c) 3
CC
CC
C
21
21
s =
+
=
48CC16CCC
2121p
=\=+=
21
2
2121 CC4)CC(CC-+=-

86448416
2
==´-=
12
C C 16F+ =m
12
C C 8F- =m
11
2C 24F C 12FÞ =mÞ =m
2
48
C 4F
12
\ = =m
11. (c) C = 10 mF ; d = 8 cm
C' = ? ; d' = 4 cm
C =
0
A
d
Î
Þ
d
1
Ca
If d is halve
d then C will be doubled.
Hence C' = 2C = 2 × 10 mF = 20 mF
12. (a) In series 321C
1
C
1
C
1
C
1
++= and ch
arge on each
capacitor is same.
13. (b) It consists of two capacitors in parallel, therefore, the
total capacitance is =
d
A2
0Î
+
+
+
+
–
–
–
–
+
+
+
+
–
–
–
–
+
+
+
+
–
–
–
–
+
+
+
+
–
–
–
–
+A
–B
(The plates of B, having negative charge do not
constitute a capacitor).
14. (c) All the charge given to inner sphere will pass on to
the outer one. So capacitance that of outer one is
0
4bpÎ.
15. (c)
As work is done by the field, K.E. of the body increases
by

460. 456 PHYSICS
)VV(qW
.E.K
BA
-==
J106)0600(10
68 --
´=-=
16. (c) Consider the potential at D be ‘V’.
Potential drop across C
1
is (V – V
1
) and C
2
is
(V
2
– V)
121 1 22
q C (V V ), q C (V V)\=- =-
As q
1
= q
2
[capacitor
s are in series]
1 1 22
C (V V ) C (V V)\-=-
11 22
12
CV CV
v
CC
+
=
+
17. (c)
0
0
q
V
C
=
q
V
C
= Þ
0
0
CV
VC
=
Þ
0C500 20
C753
==
By definition, C = kC
0
Þ
20
k
3
=
18. (a) Here
, A = 90 cm² = 90 × 10
–4
m
2
;
d = 2.5 mm = 2.5 × 10
–3
m; V = 400 volt
124
0
3
A8.854 10 90 10
C
d 2.5 10
--
-
e ´ ´´
==
´
= 3.187 × 10
–11
F
( )
22 1111
W CV 3.187 10 400
22
-
= =´´´
= 2.55 ×
10
–6
J
19. (d) Let ‘n’ such capacitors are in series and such ‘m’ such
branch are in parallel.
\ 250 × n = 1000 \ n = 4 … (i)
Also
8
m 16
n
´=
16n
m
8
´
= = 8 … (ii)
\ No. of capacitor = 8 4 32´=
20. (d) For a parallel plate capacitor C =
0A
d
e
\A =
0
Cd
e
=
3
12
1 10
8.85 10
-
-
´
´
= 1.13 ×
10
8
m
2
This corresponds to area of square of side 10.6 km
which shows that one farad is very large unit of
capacitance.
21. (a)
2 6211
E CV 1 10 (4000) 8J.
22
-
==´´´=
22. (a)s
28
C 1.6F
28
´
= =m
+
Since, Q = C
s
V
= 1.6 × 10
–6
× 300
Q = 4.8 × 10
–4
C
4
1
6
4.8 10
V 240V
2 10
-
-
´
==
´
4
2
6
4.8 10
V 60V
8 10
-
-
´
==
´
22
12
QQ
U
2C 2C
=+
()
2
4
6
4.8 10
1
U
2 1.6 10
-
-
´
æö
Þ=
ç÷
èø´
ÞU = 3 × 2.4 × 10
–2
J
ÞU = 7.2 × 10
–2
J
23. (b
) In steady state, both the capacitors are at the same
potential,
i.e.,
12 12
21
12
QQ QQ
or or Q 2Q
C C C 2C
===
Also Q
1
+ Q
2
= Q
12
Q 2Q
Q ,Q
33
\==
24. (a)
25. (a)
Here, V(x) =
2
20
volt
x4-
We know that E =
dV
dx
-
2
d 20
dxx4
æö
=-
ç÷
è -ø
or, E =
22
40x
(x 4)
+
-
At x = 4mm,
E =
22
404
(4 4)
´
+
-

160 10
volt / m.
1449
=+ =+ m
Positive sign indicates that E
r
is in +ve x-direction.
26. (a) Capacitance of spherical conductor = 4pe
0
a
where a is radius of conductor.
Therefore,
9
9
10
9
1
1
109
1
C
-
´=´
´
=
F1011.0
9-
´=
10
1.1 10 F
-
=´
27. (d) In equilib
rium,
mg
d
V
)en(EqF===
3
800106.1
02.08.91096.1
eV
dmg
n
19
15
=
´´
´´´
==
-
-
28. (c) Lengt
h of body diagnonal =
b3
\ Distance of centr
e of cube from each corner,
b
2
3
r=
Total P.E. of charge + q at the c
entre

461. 457Electrostatic Potential and Capacitance
)2/b3(4
q8
r4
)q(q8
o
2
o pe
-
=
pe
-
=

b3
q4
o
2
pe
-
=
29. (d
) Potential at origin
= (V
1
+ V
3
+ V
5
+ .....) – (V
2
+ V
4
+ V
6
+ .....)
ú
û
ù
ê
ë
é
¥+-
pe
Þ .....
x3
1
x2
1
x
1
4
q
0000
ú
û
ù
ê
ë
é
¥-+-
pe
Þ .....
4
1
3
1
2
1
1
x4
q
00
2log
x4
q
)11(log
x4
q
e
00
e
00 pe
Þ+
pe
Þ
30. (c)
a b
Let charge on each sphere =q
when th
ey are connected together their potential will
be equal .
Now let charge on
1
qa= and on
1
qq2b-=
ba
VV=Þ
11
oo
q 2qq11
or
4a4b
-
=
pe pe
b
a
qq2
q
1
1
=
-
Þ
2
1
2
ao 1
2
2b1
2
o
q1.
E4 q ba
q1E 2qq a
4b
æöpe
== ç÷
-
èø
pe

a
b
a
b
.
b
a
2
2
==
a:b=
31. (b
) In shell, q charge is uniformly distributed over its
surface, it behaves as a conductor.
+
+
+
+
+
+
+
q
R
+
V= potential at surface
R4
q
0pe
= and inside
R4
q
V
0pe
=
Because of this
it behaves as an equipotential surface.
32. (b) Energy stored,
22
aq net
11
U C V (3 C)V
22
==
2
CV
2
3
=
2V
V
2C
–
+
+
–
C
33. (a)
r
2
r1V
1
V2
Q
Situation 1 :
12
1 2 12
Q Q 11
V V V K K KQ
r r rr
éù
-==-=- êú
ëû
Situation 2 :
12
12 22
KQ 3KQ KQ 3 KQ
V' V' V'
rr rr
éùéù
-==- --
êúêú
ëûëû

12
11
KQV
rr
éù
= -=
êú
ëû
34. (c)
Since, all A, B, C, D and E lie on an equipotential surface
so,
W = 0
35. (c) When steady state is reached, the current I coming
from the battery is given by
9 V
S
X
W3 W6
F3m F6m
Y
+Q
1
–Q
1+Q
2
–Q
2
9 = I (3 + 9) Þ I = 1A
Þ potential difference across 3 W resistance = 3V
and potential difference across 6 W resistance
= 6V
Þ p.d. across 3 mF capacitor = 3V
and p.d. across 6mF capacitor = 6V
\ Charge on 3 mF capacitor, Q
1
= 3 × 3 = 9 mC
Charge on 6m F capacitor, Q
2
= 6 × 6 = 36 mC
Þ Charge (–Q
1
) is shifted from the positive plate of
6mF capacitor. The remaining charge on the positive
plate of 6 mF capacitor is shifted through the switch.
\ Charge passing through the switch
= 36 – 9 = 27 mC
36. (a) As x = t ÷
ø
ö
ç
è
æ
-
K
1
1
, where x is the addition distance of
plate, to restore the capacity of original value.
.
K
1
1104105.3
55
÷
ø
ö
ç
è
æ
-´=´\
--
Solvi
ng, we get, K = 8.
37. (d) For series
21
21
CC
CC
'C
+
´
= F2
44
44
m=
+ ´
=
For para
llel eq2
C C' C 2 4 6F= + =+=m
A B

462. 458 PHYSICS
38. (b)
d
AK
C
oe
=
12 0
1A
1 10
d
- ´Î
´= .......... (1)
12 0KA
2 10
2d
- ´Î
´= .......... (2)
(2)K
2
(1)2
Þ= or K = 4
39. (d)
The charge Q given to outer sphere distributes as
Q
1
outside and Q
2
inside which induces charge – Q
2
on outside of inner sphere, +Q
2
on inside of inner
sphere which is earthed.
The inside of outer and the inner sphere constitute a
spherical condenser having capacitance
0
ab
ba
4pÎ
-
and the outside of the outer constitutes an isolated
sphere of capacitance
0
b4pÎ.
\ the effective capacitance is
00
ab
4 4b
ba
pÎ + pÎ
-
ú
û
ù
ê
ë
é
-
-+
Îp=
ú
û
ù
ê
ë
é
+
-
Îp=
ab
aba
b41
ab
a
b4
00
ab
b
4C
2
0
-
Îp=
40. (a)
d
A
C
0
a
Î
= and
d
)K1(A2
K2
d
2
d
A
C
00
b
+Î
=
+
Î
=
and
d
K
2
A
d
2
A
C
00
c
Î
+
Î
=
0
A
(1 K)
2d
Î
=+
or
a
0
b
C)K1(2
d
A
C >+
Î
=
or a
0
c
C
2
K1
d
A
C >
+Î
=
acb
CCandC>\ .
41. (d)
When S
1
and S
3
is closed V
1
= 30 V and potential
drop across C
2
becomes 20 V.
42. (a) The molecules of liquid will convert into induced
dipole, get oriented along the electric field produced
between the plates and rise due to force of attraction.
43. (d) Charge given to a hollow conductor resides only on
the outer surface.
44. (b) C = C
1
+ KC
2
=
d
Kxb
d
b)x(
00 Î
+
Î-l
0
b
[ x Kx]
d
Î
= -+l
db
x
l
K
0
b
C [ x(K 1)]
d
Î
= +-l
0
q qd
V
C b [ x(K 1)]
==
Î+-l
as x decreases, V increases.
45. (c) C
2
and C
3
are parallel so V
2
= V
3
C
1
and combination of C
2
& C
3
is in series.
So, V = V
2
+ V
1
or V = V
3
+ V
1
and also Q
1
= Q
2
+ Q
3
46. (c) Since, potential difference across C
2
is greater than C
1
.
12
q
C C V and q is same in series
C
éù
Þ>=
êú
ëû
Q
47. (c) The ball
on touching plate A will
get positively charged. It will be
repelled by A and get attracted
towards B. After touching B it
will get negatively charged. It
will now be repelled by B and
get attracted towards A.
–+
+
+
+
+
+ –
–
–
–
–
+
+
++
+
+
A B
Thus it will remain oscillating
and at the extreme position touch the plates.
48. (a) Initial charge on first capacitor is CV = Q
1
.
Initial charge on second capacitor is 2CV = Q
2
.
Final capacitance of first capacitor is KC
If V' is the common potential then
12
12
QQ
V
CC
+
=¢
+¢
Þ
CV 2CV 3V
V
KC2C 2K
+
==¢
++
49. (a) Since battery is still in connection, so,
V = V
0
Þ Q
0
= C
0
V
0
and Q = kC
0
V
0
Þ Q = kQ
0
Since, k > 1
Þ Q > Q
0
Also,
0 00
1
U QV
2
= and
0
1
U QV kU
2
== {\ Q = kQ
0
and V = V
0
}
Hence, U > U
0
50. (b)
A
C C
CC
C C
CC
B
The figure shows two independent balanced
wheatstone Brides connected in parallel each having
a capacitance C. So,
C
net
= C
AB
= 2C

463. 459Electrostatic Potential and Capacitance
51. (b) Ket C
0
be the capacitance initially and C be the
capacitance finally. The
A
0
C
0
d
e
=
Since, Q =
C
0
V
0
Q
AV
d
=
e
Þ
Further,
0
0
EV
E andE
dK
==
V
E
Kd
Þ=
Also, i
f U
i
is the initial energy, then
2
i0
1
U CV
2
=
After t
he introduction of slab if U
f
be the final energy,
then
( )
2
2
fslab0
1 1V
U CV KC
2 2K
æö
==
ç÷
èø
2
0
f
CV1
U
2K
Þ=
21
UUUÞD=-
2
0
11
UCV1
2K
æö
ÞD=- ç÷
èø
Since, work done = Decrease in Potential Energy
WUÞ = -D
2
0
AV11
W1
2dK
e æö
Þ=-
ç÷
èø
52. (b)Potential difference across the branch de is 6 V. Net
capacitance of de branch is 2.1 µF
So, q = CV
Þq = 2.1 × 6 µC
Þq = 12.6 µ C
Po
tential across 3 µF capacitance is
12.6
V 4.2 volt
3
==
Potential ac
ross 2 and 5 combination in parallel is
6 – 4.2 = 1.8 VSo, q' = (1.8) (5) = 9 µC
53. (a) Whenever a charge (+50 nC) is kept inside a hollow
metallic spherical shell, it induces an equal andopposite charge on the inner surface and an equaland same type of charges on the outer surface.\ Inside, induced charge is – 50 nC and outside, +50
nC – 150 nC already present.
54. (a) Required ratio
=
Energy stored in capacitor
Workdone by thebattery

2
2
1
CV
2
Ce
= ,
where
C = Capacitance of capacitor
V = Potential difference,e = emf of battery

2
2
1
Ce
2
Ce
= . (Q V = e)

1
2
=
55. (b) i
( 3q) q (
q) (q)
Uk33
LL
-´éù
= ´+´
êú
ëû
=
2
6kq
L
-
U
f
= 0
Work don
e by electric field = – Change in potential energy
= U
i
– U
f
=
2
6kq
L
-
56. (c)
r
O
+q–q
22
2
mv kq
r(2r)
=
;
2
2kq
mv
4r
=
Kinetic en
ergy of each particle
=
2
21 kq
mv
2 8r
=
57. (a)
123
00
1 Q 1 2Q
VVVV.
4R
4R
-
æö
=++=+
ç÷
èøpÎ pÎ
00
1 3Q 1 2Q
4R4R
æö æö
+=
ç÷ ç÷
èø èøpÎ pÎ
58. (
c) Capacitance of capacitor (C) = 6 mF = 6 × 10
–6
F;
Initial potential (V
1
) = 10 V and final potential
(V
2
) = 20 V. The increase in energy (DU)
)VV(C
2
1 2
1
2
2-=
])10()20[()106(
2
1 226
-´´´=
-
J109300)103(
46 --
´=´´=.
59. (b
) As the capacitors offer infinite resistance to steady
current so, the equivalent circuit is
A B
R
2R
2V
I
V
A
B
VV
C
Using ohm’s law, current in circuit is
2V – V = I (2R + R) Þ
V
I
3R
=
The volta
ge drop across
AB
V
V 2V 2R
3R
=-´ =
4
V
3

464. 460 PHYSICS
ABC
4
V V VV
3
= =+
Þ Voltage drop across C =
1
3
V..
60. (a) Th
e electrostatic pressure at a point on the surface of
a uniformly charged sphere =
2
0
2
s
Î
\ The force on a hemispherical shell =
2
2
0
R
2
s
´p
Î
61. (c) Eve
n after introduction of dielectric slab, direction of
electric field will be perpendicular to the plates and
directed from positive plate to negative plate.
Further, magnitude of electric field in air
0
s
=
e
Magnitude of electric field in dielectric
0
K
s
=
e
Similarly electric lines always flows from higher to lower
potential, therefore, electric potential inceases
continuously as we move from x = 0 to x = 3d.
(+)
x =0x = dx=2dx =d3
(–)
62. (c) Th
e given circuit can be redrawn as follows
2000V
B
CA
5Fm 15Fm
(V
A
– V
B
) =
15
2000 – 1500
5 15
AB
VVV
æö
´Þ=
ç÷
+èø
Þ 2000 – V
B
= 1500V Þ V
B
= 500V
63.
(a)
–
0 10
0
1
44
tt
RC
R
V
VVee= = Þ=
10
4 log 4 10log 4 13.86
10
t
e
t
e tsÞ=Þ=Þ==
(RC = 2.
5 × 10
6
× 4 × 10
–6
= 10)
64. (c)
22
11 22
11
CV CV
22
=
because total energ
y is transferred (given).6 2 6211
900 10 100 100 10 V
22
--
\´´´=´´´
\ V
2
= 90
000
Þ V = 300 V..
6
5. (d)
X Y
A
6mF
6Fm
6mF
6mFC
3
C
1
C
4
C
2
C
5
20mF
=
Equivalent circuit
X Y
6mF
6mF
6mF
6mF
6Fm6Fm
6Fm
C
1
C
2
C
C
4
5
C
3
20mF
As
4
2
3
1
C
C
C
C
=
Hence no ch
arge will flow through 20mF
X Y
C
1
C
2
C
4
C
3

X Y
C'
C''
Þ
C
1
and C
2
are in s
eries, also C
3
and C
4
are in series.
Hence C' = 3 mF, C'' = 3 mF
C' and C'' are in parallel hence net capacitance
= C' + C'' = 3 + 3 = 6 mF
66. (c) Energy stored per unit volume
÷
ø
ö
ç
è
æ
=e=÷
ø
ö
ç
è
æ
e=e=
d
V
E
d
V
2
1
d
V
2
1
E
2
1
2
2
0
2
0
2
0
Q
67. (a) Cha
rge Q = C
1
V
Total capacity of combination (parallel) C = C
1
+ C
2
21
1
CC
VC
C
Q
.D.P
+
==
68. (a) We kn
ow that potential energy of two charge system
is given by
U
qq
r
=
Î
1
4
0
12
p
According to que
stion,
U
A
=
1
4
0
pÎ
+-()()qQ
a

=-
1
4
0
pe
Qq
a
and U
B
=
1
4
0
pÎ
+-()()qQ
a
=-
1
4
0
pe
Qq
a
DU = U
B
–U
A
= 0
When known
that for conservative force,
W = –DU = 0
69. (c) We know that potential energy of discrete system of
charges is given by

465. 461Electrostatic Potential and Capacitance
qq qqqq1 23 3112
U
4 rrr
0 12 23 31
æö
= ++
ç÷
pÎèø
According to question,
U
initial
=
qq qqqq1 23 3112
4 0.3 0.5 0.4
0
æö
++
ç÷
pÎèø
U
final
=
qq qqqq1 23 3112
4 0.3 0.1 0.4
0
æö
++
ç÷
pÎèø
U
final
– U
initial
=
qq qq1 23 23
4 0.1 0.5
0
æö
-
ç÷
pÎèø
=
1
10q q 2q q
23 23
4
0
éù -
ëû
pÎ
=
q
3
(8q )
2
4
0
pÎ
70. (b)
Equivalent capacitance for three capacitors
(C
1
, C
2
& C
3
) in series is given by
1 1 1 1
1 2 3
23 31 12
123
C C C C
CC C C CC
CCC
eq.
= + + =
+ +
Þ C
CCC
CC CC CC
eq.
=
+ +
123
12 23 31
Þ
C
CCC
CC CC CC
C
eq.
()()
( )(
)( )( )
=
+ +
=
23
2 23 3
6
11
Þ Charge on capacitors (C
1
, C
2
& C
3
) in series
= =CV
C
V
eq
6
11
Charge on capacitor C
4
= C
4
V = 4C V
2
4
6C
V
Charge on C 61311
Charge on C 4CV 11 4 22
= = ´=
71.
(d) In series, C
eff
1
1
C
n
=
\Energy sto
red,
21
2
S effS
E CV=
21
1
1
16
2
C
V
n
=
21
1
8
C
V
n
=
In parallel, C
eff
= n
2
C
2
\Energy stored, E
p
=
1
2
n
2
C
2
V
2
\
2
1
1
8VC
n
=
2
22
1
2
nCV Þ
1
2
12
16C
C
nn
=
72. (a)
1
cosq CV wt=
Þ 1
– sin
dq
i Cvt
dt
== ww
Also,
21
LC
=w and
1
cosVVt= w
At
1
tt=,
2
VV= and
11
siniCVt=-ww
\
2
1
1
cos
V
t
V
=w
(–ve sign gives direction)
Hence,
1/2
2
2
1
2
1
1
VC
iV
L V
æö
=- ç÷
èø

1/2
22
12
(CVV
L
æö -
=ç÷
èø
73. (d)The charge on each of two facing surfaces will be zero.
74. (d) Potential and potential energy are different quantities
and cannot be equated.
75. (d) Two equipotential surfaces are not necessarily parallel
to each other.
EXERCISE - 3
Exemplar Questions
1. (d) As capacitor offers infinite resistance in dc-circuit.
So, current flows through 2W resistance from left to
right, given by
I =
V
Rr+
=
2.5V
2 0.5+
=
2.5
1A
2.5
=
So, the poten
tial difference across 2W resistance
V = IR = 1 × 2 = 2 volt.
Since, capacitor is in parallel with 2W resistance, so it
also has 2V potential difference across it.
As current does not flow through capacitor branch so
no potential drop will be accross 10W resistance.
The charge on capacitor
q = CV = (4 µF) × 2V = 8 µC
2. (c) The direction of electric field is always perpendicular
to the direction of electric field and equipotential
surface maintained at high electrostatic potential to
other equipotential surface maintained at low
electrostatic potential.
The positively charged particle experiences the
electrostatic force in the direction of electric field i.e.,
from high electrostatic potential to low electrostatic
potential. Thus, the work done by the electric field on
the positive charge, so electrostatic potential energy
of the positive charge decreases because speed of
charged particle moves in the direction of field due to
force
qE
r
.
3. (c)
The work done (in displacing a charge particle) by a
electric force is given by W
12
= q(V
2
– V
1
). Here initial
and final potentials are same in all three cases are
equal (20V) and same charge is moving from A to B, so
work done is (DVq) same in all three cases.
4. (c) As we know that the relation between electric field
intensity E and electric potential V is
dV
E
dr
=-

466. 462 PHYSICS
Electric field i
ntensity E = 0 then
dV
dr
= 0
This imply that V
= constant
Thus, E = 0 inside the charged conducting sphere
then the constant electrostatic potential 100V at every
where inside the sphere and it verifies the shielding
effect also.
5. (a) Here we have to findout the shape of equipotential
surface, these surface are perpendicular to the field
lines, so there must be electric field which can not be
without charge.
So, the collection of charges, whose total sum is not
zero, with regard to great distance can be considered
as a point charge. The equipotentials due to point
charge are spherical in shape as electric potential due
to point charge q is given by
V =
e
q
K
r
This suggest that electric potentials due to point charge
is same for all equidistant points. The locus of these
equidistant points which are at same potential, form
spherical surface.
The lines of field from point charge are radial. So the
equipotential surface perpendicular to field lines from
a sphere.
6. (c)The capacitance of parallel plate capacitor filled with
dielectric of thickness d
1
and dielectric constant K
1
is
C
1
=
1o
1
KA
d
e
Similarly, capacitance of parallel plate capacitor filled
with dielectric of thickness d
2
and dielectric constant
K
2
is
C
2
=
2o
2
KA
d
e
Since both capacitors are in series combination, then
the equivalent capacitance is
1
C
=
12
11
CC
+
or C =
12
12
CC
CC+
=
10 20
12
10 20
12
KAKA
dd
KAKA
dd
ee
ee
+
C =
1 20
12 21
KKA
Kd Kd
e
+
... (i)
So multiply the n
umerator and denominator of equation
(i) with (d
1
+ d
2
)
C =
( )
()
()
121 20
12 21 12
ddKKA
Kd Kd dd
+e
´
++
=
()
( )( )
1212 0
12 21 12
KKdd A
Kd Kd dd
+ e
´
++
... (ii)
So, the equivalen
t capacitances is
C =
0
12
KA
(d d)
e
+
... (iii)
Comparing, (ii
) and (iii), the dielectric constant of new
capacitor1212
12 21
KK(d d)
K
Kd Kd
+
=
+
NEET/AIPMT (2013-2017) Questions
7. (a) Potential at B, V
B
is maximum
V
B
> V
C
> V
A
As in the direction of electric field potential decreases.
8. (c) Electric field,
1
E
K
µ
As K
1
< K
2
so E
1
> E
2
Hence gr
aph (c) correctly dipicts the variation of
electric field E with distance d.
9. (b) Due to conducting sphere
At centre, electric field E = 0
And electric potential V =
0
Q
4RpÎ
10. (d)
$
VVV
E i jk
xyz
¶¶¶
=---
¶¶¶
r
$$
=
ˆ ˆˆ
[(6 8y)i ( 8x 86z)j (6y)k]--+--++
At (1, 1, 1),
$
E 2i 10j 6k=+-
r
$$
Þ
2 22
(E) 2 10 6 140 2 35=++==
r
\F = qE 2 2 35 4 35=´=
r
11. (c) Capacitance of the capacitor,
Q
C
V
=
After inserting the dielectric, new capacitance
C
1
= K.C
New potential difference
V
1
=
V
K
2
2
i
1Q
u cv
2 2C
== (QQ = cv)
2 2 22
i
f
uQQCV
u
2f 2kc 2KC k
æö
====
ç÷
èø
Du = u
f
– u
i
=
211
cv –1
2k
ìü
íý
îþ
As the capacitor is isolated, so change will remain
conserved p.d. between two plates of the capacitor
L =
QV
KCK
=
12. (a) Potential in a region
V = 6xy – y + 2yz
As we know the relation between electric potential
and electric field isE
ur
=
dV
dx
-
E
ur
=
VVV
ˆˆˆ
i jk
xyz
æö¶¶¶
++
ç÷
¶¶¶èø

467. 463Electrostatic Potential and Capacitance
E
ur
=
ˆ ˆˆ
(6yi (6x1 2z)j (2y)kéù+-++
ëû
(1,1, 0)
ˆˆˆ
E (6i 5 j 2k)
®
=- ++
13.
(a) Force of attraction between the plates,
F = qE
=
00
q
qq
2 2A
s
´=
ÎÎ
=
2 222
0
q c v cv
A 2cd 2d
2d
d
==
Îæö
´
ç÷
èø
Here,
0
A
c , q cv, A area
d
Î
= ==
14. (d)When S
and 1 are connected
The 2mF capacitor gets charged. The potential
difference across its plates will be V.
The potential energy stored in 2 mF capacitor
2 2211
2
22
i
U CV VV= =´´=
When S an
d 2 are connected
The 8mF capacitor also gets charged. During this
charging process current flows in the wire and some
amount of energy is dissipated as heat. The energy
loss is
DU =
212
12
12
C1
()
2
C
VV
CC
-
+
Here, C
1
= 2mF, C
2

8 mF, V
1
= V, V
2
= 0
\
221284
( 0)
2285
U VV
´
D=´ -=
+
Th
e percentage of the energy dissipated
=
100
U
Ui
D
´
2
2
4
5
100 80%
V
V
= ´=
15. (
a) When battery is replaced by another uncharged
capacitor
Cq
C
As uncharged capacitor is connected parallel
So, C' = 2C
and V
c
=
12
12
qq
CC
+
+
V
c
=
q0
CC
+
+
ÞV
c
=
V
2
Initial Energy of system, U
i
=
21
CV
2
… (i)
Final ene
rgy of system, U
f
=
2
1V
(2C)
22
æö
ç÷
èø
=
211
CV
22
æö
ç÷
èø
…(ii)
From equ
ation (i) and (ii)
U
f
=
i
1
U
2
i.e., Total electrostatic energy of resulting system
decreases by a factor of 2
16. (a) As the regions are of equipotentials, so Work done
W = qDV
DV is same in all the cases hence work - done will also
be same in all the cases.

468. 464 PHYSICS
ELECTRIC CURRENT
It is the rate
of flow of charge through any cross section.
i.e.
dq
I=
dt
Conventionally, the direction of flow of positive charge is taken
as the direction of electric current. It is a scalar quantity and its
S.I. unit is ampere (A).
Keep in Memory
(i) Cur
rent carriers in conductor are electrons, (valence e
–
s)
ions in electrolytes, electrons & holes in semiconductor
and positive ions /electrons in gases.
(ii) Charge of electron = 1.6 × 10
–19
c
(iii) 1 ampere = 6.25 × 10
18
electrons/sec
(iv) Though direction is associated with current (opposite to
the motion of electrons), but it is not a vector quantity as it
does not follow rules of vector addition.
(v) For a current to flow through a cross-section, there must be
a net flow of charge through that cross-section.
In a metal like copper there are around 10
28
free electrons
per m
3
moving randomly in all direction with speeds of the
order of 10
6
m/s even in the absence of electric field. But
since the number of electrons passing through a cross-
section from left to right is equal to the number of electrons
passing from right to left in a given time, therefore the net
charge flow is zero and hence the electric current is zero.
(vi) A conductor remains uncharged when current flows in it.
i.e. Net charge in a current carrying conductor is zero.
CURRENT DENSITY
Current density at a point inside a conductor is defined as the
amount of current flowing per unit cross sectional area around
that point of the conductor, provided the area is held in a
direction normal to the direction of current.
i.e. Current density,
I
J=
A
If area is not normal to current, then area normal to current is A' =
A cos q (see the figure)
qq
J
A'
A
cos
I
J
A
=
q
or I = J A cos q or ..I JA JA== ò
rrrrr
Its SI unit is Am
–2
Curre
nt density can also be related to electric field as
E
jE=s=
r
r
rr
where s is conduc tivity of the substance & r is specific
resistance of the substance.
J is a vector quantity and its direction is same as that of E
r
.
Dimensions of
J are [M°L
–2
T°A]
Keep in Memory
(i) Elec
tric current is a macroscopic physical quantity where
as current density is a microscopic physical quantity.
(ii) For a given conductor current does not change with change
in cross-sectional area.
DRIFT VELOCITY ()
u u ur
d
V
When the ends of a conductor are connected to the two terminals
of a battery, an electric field is set up in the conductor from the
positive terminal to the negative terminal. The free electrons in
the conductor experiences a force opposite to the direction of
the electric field and hence get accelerated. However this process
of acceleration is soon interrupted by collision with ions of solid.
The average time for which each electron is accelerated before
suffering a collision is called the mean free time or mean
relaxation time.
Thus, the free electrons within the metal, in addition to its random
motion acquire a small velocity towards the positive end of
conductor. This velocity is called drift velocity. It is given by
v–
d
eE
m
t
=
ur
r
,
where e is t
he charge and m is the mass of electron.
E
ur
is the electric field established in conductor and tis the
average r
elaxation time.
Negative sign is because the directions of
E
r
and
d
v
r
(for electron)
a
re opposite.
V
E
l
=
18
Current Electricity

469. 465Current Electricity
I
V
–
–
+
+
J
V
d
E
e
–
l
where V is the potential difference across ends of the conductor
of length l. The uniform current I, flowing through the conductor
is given by
I = n e A v
d
where n = number of free electrons per unit volume,
A = area of cross-section, v
d
= drift velocity
In vector form, J nevdÞ =-
urr
The negative sign is because the direction of drift velocity of
electron is opposite to J
r
.
Mobility - Dr
ift velocity per unit electric field is called mobility.
It is denoted by µ.
V
d
E
m=
r
Its S.I. unit is m
2
/volt-sec.
Keep in Memory
1.
Drift velocity is very small, it is of the order of 10
–4
m/s
which is negligible as compared to thermal speed of e
–
s at
room temperature (;10
5
m/s)
2. The drift velocity is given by d
J
v,
ne
=
where, J = cu
rrent density
I
A
=
e = electronic charge = 1.6 × 10
–19
C
n = the number of free electrons per unit volume
3. The number of free electrons per unit volume (n) can be
determined by the following relation :
–
No. of free e per atom
n=
Volume
and
0
M
volume= .
Nd
0
Nd
n x,
M
=´
whereN
0
= Avogadro number
d = density of the metal
M = molecular weight
andx = number of free electrons per atom
3. For steady current :
A
I
J=;
A
I
E
s
= ;
neA
I
V
d
=
This means that f
or a given material and steady current in
case of non-uniform cross-section of material
A
1
Jµ;
A
1
Eµ;
A
1
V
d
µ
4. Vari
ation of drift velocity :
l
V
ne
1
ne
E
V
d == ; V
d
µ E
when l
ength is doubled, v
d
becomes half andwhen V is doubled, v
d
becomes twice.
Example 1.
The current in a wire varies with time according to the
relation i = 4 + 2t
2
How many coulomb of charge pass a cross-section of wire
in time interval t = 5s to t = 10s?
Solution :
dt
dq
i= Þ dq = i dt Þ dq = (4
+ 2 t
2
) dt
On integrating
10
10 3
2
5 5
t
q (14 2t
)dt 14t 2
3
éù
= + =+ êú
êúëû
ò
= 603.
33 C
Example 2.
The current density at a point is
4
2
A
ˆ
J210j
m
æö
=´ç÷
èø
r
. Find
the rate of charge flow through a cross-s
ectional area
S
r
such that
(i)
2ˆ
S (2cm)j=
r
,
(ii)
2ˆ
S (4cm)i=
r
and (iii)
2ˆˆ
S (2i 3j cm )=+
r
Solution :
Th
e rate of charge flow = current
i JdS=
rr
(i) Current
4 2 42ˆˆ
i (2 10 A/ m ) j.(2 10 m )j 2A
-
=´ ´=
ˆˆ
[using j.j1]=
(ii) Current
422 ˆˆ
i (2 10A/m)j.(4cm)i 0=´=
ˆˆ
[using j.i 0]=
(iii)Current
4 2 42ˆˆˆ
i (2 10 A /m ) j.(2i 3j) 10 m 6A
-
=´ +´=
OHM’
S LAW AND ELECTRICAL RESISTANCE
When a potential difference is applied across the ends of a
conductor, a current I is set up in the conductor.
According to Ohm’s law “Keeping the given physical conditions
such as temperature, mechanical strain etc. constant, the current
(I) produced in the conductor is directly proportional to the
potential difference (V) applied across the conductor”.
i.e.,
VIµ or KVI= ... (1)
wher
e K is a constant of proportionality called the conductance
of the given conductor.
Alternatively, V
Iµ or V = RI ... (2)
wh
ere the constant R is called the electrical resistance or simply
resistance of the given conductor.

470. 466 PHYSICS
From abo
ve two eqs. it is clear that R = 1/K.
If a substance follows Ohm’s law, then a linear relationship exists
between V & I as shown by figure 1. These substance are called
Ohmic substance. Some substances do not follow Ohm’s law,
these are called non-ohmic substance (shown by figure 2)
Diode valve, triode valve and electrolytes, thermistors are some
examples of non-ohmic conductors.
I
V
Ohmic conductor
or linear conductor
Fig. 1
q

Non-linear conductor
or non-ohmic conductor
Fig. 2
V
I
Slope of V-I Curve of a conductor provides the resistance of the
conductor
slope = tan q =
V
I
The SI unit of re sistance R is volt/ampere = ohm (W)
Electrical Resistance
On application of potential difference across the ends of a
conductor, the free e
–
s of the conductor starts drifting towards
the positive end of the conductor. While drifting they make
collisions with the ions/atoms of the conductor & hence their
motion is obstructed. The net hindrance offered by a conductor
to the flow of free e
–
s or simply current is called electrical
resistance.
It depends upon the size, geometry, temperature and nature of
the conductor.
Resistivity : For a given conductor of uniform cross-section A
and length l, the electrical resistance R is directly proportional to
length l and inversely proportional to cross-sectional area A
i.e.,
µ
l
R
A
or =
l
R
A
r
or
RA
ρ=
l
r is called specific resistance or electrical resistivity.
Also,
2
=
m
ne
r
t
The SI unit of r esistivity is ohm - m.
Conductivity(s) : It is the reciprocal of resistivity i.e.
1
s=
r
.
The SI unit of co
nductivity is Ohm
–1
m
–1
or mho/m.
Ohm’s law may also be expressed as, J = sE
where J = current density and E = electric field strength
Conductivity,
2
ne
m
t
s= where n is fr ee electron density, t is
relaxation time and m is mass of electron.
(i) The value of r is very low for conductor, very high for
insulators & alloys, and in between those of conductors &
insulators for semiconductors.
(ii) Resistance is the property of object while resistivity is
the property of material.
Materials and their resistivity
Material Resistivi ty (r) (at 0°c)
(in W m)
(i) Silver 1.6 × 10
–8
(ii) Copper 1.7 × 10
–8
(iii)Aluminium 2.8 × 10
–8
(iv) Tungsten 5.2 × 10
–8
(v) Platinum 10.6 × 10
–8
(vi) Manganin 42 × 10
–8
(vii)Carbon 35 × 10
–6
(viii)Germanium .46
(ix) Silicon 2300
(x) Glass ~ 10
13
(xi) Mica ~ 2 × 10
15
COMMON DEFAU
LT
Since
RR
A
=rÞµ
l
l
It is incorrect to think that if the length of a resistor is doubled
its resistance will become twice.
If you look by an eye of physicist you will find that when l
change, A will also change. This is discussed in the following
article.
Case of Reshaping a Resistor
On reshaping, volume of a material is constant.
i.e., Initial volume = final volume
or, A
i
l
i
= A
f
l
f
... (i)
where l
i
, A
i
are initial length and area of cross-section of resistor
and l
f
, A
f
are final length and area of cross-section of resistor.
If initial resistance before reshaping is R
i
and final resistance
after reshaping is R
f
then
i
i iif
ff fi
f
RAA
RA
A
r
= =´
r
l
l
ll
... (ii)
From eqs.
(i) and (ii),
2
2
f
i
f
i
R
R
R
l
l
l
µÞ
÷
÷
ø
ö
ç
ç
è
æ
=
This means that res
istance is proportional to the square of the
length during reshaping of a resistor wire.
Also from eqs. (i) and (ii),
2
2
i
f
f
i
A
1
R
A
A
R
R
µÞ
÷
÷
ø
ö
ç
ç
è
æ
=
This means that resistan
ce is inversely proportional to the square
of the area of cross-section during reshaping of resistor.
Since A = p r
2
(for circular cross-section)
4
r
1
Rµ\
where r is radi
us of cross section.
Effect of Temperature on Resistance and Resistivity
Resistance of a conductor is given by R
t
= R
0
(1 + aDt)
Where a = temperature coefficient of resistance and Dt = change
in temperature

471. 467Current Electricity
For metallic conductors : If r
1
and r
2
be resistivity of a
conductor at temperature t
1
and t
2
, then
r
2
= r
1
(1 + a D T)
where a = temperature coefficient of resistivity and
where DT = t
2
– t
1
= change in temperature
The value of a is positive for all metallic conductors.
\ r
2
> r
1
In other words, with rise in temperature, the positive ions of the
metal vibrate with higher amplitude and these obstruct the path
of electrons more frequently. Due to this the mean path decreases
and the relaxation time also decreases. This leads to increase in
resistivity.
Please note that the value of a for most of the metals is
11
K
273
-
For alloys : In case o f alloys, the rate at which the resistance
changes with temperature is less as compared to pure metals.
For example, an alloy manganin has a resistance which is
30-40 times that of copper for the same dimensions.
Also the value of a for manganin is very small » 0.00001°C
–1
.
Due to the above properties manganin is used in preparing wires
for standard resistance (heaters), resistance boxes etc.
Please note that eureka and constantan are other alloys for which
r is high. These are used to detect small temperature, protect
picture tube/ windings of generators, transformers etc.
For semiconductors : The resistivity of semi-conductors
decreases with rise in temperature. For semi conductor the value
of a is negative.
2
m
ne
r=
t
With rise in temperature, the value of n increases. Please note
that t decreases with rise in temperature. But the value of increase
in n is dominating for the value of r in this case.
For electrolytes : The resistivity decreases with rise in
temperature. This is because the viscosity of electrolyte decreaseswith increase in temperature so that ions get more freedom to
move.
For insulators : The resistivity increases nearly exponentially
with decrease in temperature. Conductivity of insulators is almost
zero at 0 K.
Superconductors : There are certain materials for which the
resistance becomes zero below a certain temperature. This
temperature is called the critical temperature. Below critical
temperature the material offers no resistance to the flow of e
–
s.
The material in this case is called a superconductor. The reason
for super conductivity is that the electrons in superconductors
are not mutually independent but are mutually coherent. This
coherent cloud of e
–
s makes no collision with the ions of super-
conductor and hence no resistance is offered to the flow of e
–
s
For example, R = 0 for Hg at 4.2 K and R = 0 for Pb at 7.2 K. These
substances are called superconductors at that critical temperature.
Superconductors are used (a) in making very strong
electromagnets, (b) to produce very high speed computers
(c) in transmission of electric power (d) in the study of high
energy particle physics and material science.
SERIES AND PARALLEL COMBINATION OF RESISTORS
Resistances in Series
V
1
V
2
R
1
R
2
I
V
When a number of resistances are
joined end to end so that same current
flows through each, resistor when some
potential difference is applied across
the combination, the conductor are
said to be connected in series.
The equivalent resistance in series is given by
(R
eq
)
s
= R
1
+ R
2
+ ...+ R
n
Equivalent resistance of same
resistances connected in series
is always greater than the
greatest of individual resistance.
Potential division rule in series
V
R
1
R
2
V
1
V
2
combination :
12
12
12 12
VR VR
V ;V
RR RR
==
++
Resistanc
es in Parallel
Two or more resistors are said to be connected in parallel if the
same potential difference exits across all resistors.
R
1
I
1
I
I
2
R
2
V
The equivalent resistance is given by
eqp 12
1 11
...
(R) RR
=++
1
+
R
n
The equivalent resistance in a parallel combination is always less
than the value of the least individual resistance in the circuits.Current division rule in parallel combination
2
1
12
IR
I
RR
=
+
;
1
2
12
IR
I
RR
=
+
In a given comb
ination of resistors, when you want to detect
whether the resistances are in series or in parallel then see that ifthe same current flows through two resistors then these are inseries and if same potential difference is there across two resistors
then these are in parallel potential diff across each resistor is the
same & is equal to the applied potential difference.
HOW TO FIND EQUIVALENT RESISTANCE ?
Successive Reduction
This method is applicable only when the resistors can be clearly
identified as in series or parallel. Let us take some example to find
resistance between ends A and B

472. 468 PHYSICS
Ex. (i)
A B
3 W
3 W
3 W
3 W
6 W
3 W 6 W
A B
3 W
3 W
3 W
6 W
6 W
6 W=
A B
3
3
3
6
B B
3 W
3 W
3 W 3 W
3 W
6 W
6 W
3 W
6 W
A
=
3 W3 W
6 W
2 W
BA
3 W
3 W
A B
= =
Ex. (ii)3
A
A
B
B10
10
10
10
10
55
7
B
B
B
A
A
A
10
10
10
5
5 5
=
= =
=
Ex. (iii)
=
= =
Ex. (iv) = = =
Please note that all points on the circumference are at same potential as there is no resistance on circumference.
Ex. (vi) Infinite series :
B
A
to ¥
r
r
r r
r
r
We observ
e that there is a repetitive unit extending to infinity
on left hand side. We assume that the equivalent resistance
of all the unit except one (shown dotted) is equal to X ohm.
The equivalent circuit will be as shown below.
X
r
rrr
rr
B
A
The equivalent resistance across A and B is
r
rX
rX
rR
AB
+
+
´
+=
Please note th
at R
AB
can be taken as X because if you add
one unit to the sum of infinite units, then it will be
approximately the same.
rX
rX
r2X
+
´
+=\
Solve the equ
ation as a normal algebraic equation to find
X.
Using Symmetry of the Circuit
Axis symmetry :
Ex.(i)The circuit shown in figure is symmetrical about XAEBY
axis. This is because the upper part of the axis is the mirror
image of lower part (resistors and current direction both)
C
D
r
r
r
rr
r rrr
r
rr
A BEX Y
\I
AC
= I
AD
;I
CB
= I
DB
;
I
AE
= I
EB
( CED
VVV<<Q wheatstone
b
ridge principle)

473. 469Current Electricity
Þ I
CE
= I
ED
= 0
Therefore the circuit can be redrawn. It is now easier to find
resistance between X and Y.
C
r
r r
r
r r
D
A BEX Y
Ex.(ii) The circuit shown is symmetrical about axis XY. Therefore
V
B
= V
H
; V
C
= V
I
= V
G
; V
D
= V
F
X YA
B
C
D
E
F
G
H
r
rr
rr
rr
rr
r
r
r
r
r
r
Ir
Therefore the circuit can be redrawn as
B
H
Y
C D
G
F
I
r
r
r
r
r
r
r
r
r r
r r
rr
r
X
Ex.(iii) The circ
uit is asymmetric about the dotted line
\ I
BG
= I
GC
; I
FG
= I
GE
and I
AG
= I
GB
X Y
A
B C
D
EF
r
rrr
r
rr
rr
rr
r
r
r
r
G
Therefo
re the equivalent circuit is
X Y
A
B C
D
F
r
rr
r
r
r
r
r
r
E
rrr
2. Shifted symmetry :
YA
B
C
D
E
R
1
R
1R
2
I
2
I
1
I
2
I
1
IX
I
3
R
2
2.5 W
5 W5 W
2.5 W
The diagram give
n above is symmetrical but the positions
of the resistances are shifted. Let I be the current in thecircuit from A. The same leaves the circuit at C. Let currentin AB, AD and AE be I
1
, I
2
and I
3
respectively. Since the
same current flows in AE and EC, the detached equivalentcircuit can be drawn as
Y
B
C
E
I
2
I
3
I
1
I
1
IX A D
2.5 W
5 W
5 W
10 W
5 W5 W
10 W
2.5 W
Keep in Memory
Equivalen
t resistance between A and B of the resistors connected
as shown in the figure
BA
R
1
R
2
R
1R
2
Wheatston
e bridge
12
211
AB
R3R
)R3R(R
R
+
+
=
3. Path symmetr
y :
All paths from one point to another which have the samesetting of resistances have the same amount of currents.
Example :
H
G
CB
E
F
D
I
1
I
1
I
1
I
1
I – 2I
1
I – 2I
1
I
I
A
Twelve wir
es each having resistance r are joined to form a cube.
We have to find the equivalent resistance across A and B.By path symmetry, I
AB
= I
BC
= I
AD
= I
DC
= I
\ I
AE
= I– 2I
1
Þ I
GC
= I– 2I
1
,
Since current in AB = current in BCÞ I
BF
= 0
Also I
AD
= I
DC
Þ I
DH
= 0
The equivalent circuit will be as shown. The resistance now clearlyvisible as in series and in parallel.
r
r
r
r
r
r
r
r
r
I
I
r

474. 470 PHYSICS
4. Star -
delta connection :
B
C
A
r
3
r
2
r
1
B
C
AR
a
R
b
R
c
=
321
31
a
rrr
rr
R
++
=
;
321
21
b
rrr
rr
R
++
=
;
321
32
c
rrr
rr
R
++
=
;
B
C
A
r
3
r
2
r
1
B
C
A
R
a
R
b
R
c
=
c
accbba
1
R
RRRRRR
r
++
=
;
a
accbba
2
R
RRRRRR
r
++
=
;
b
accbba
3
R
RRRRRR
r
++
=
;
Using delta to star
conversion
2W 2W
3W
3W
9W
2W 2W
3W
2W 2W
X XY Y
9W
9W
=
If none of the a
bove method works then we may use Kirchhoff’s
method which wi ll be discussed later
COMMON DEFAULTS
1. Resistors are not just in series or in parallel if they look so
geometrically, e.g. the resistors in the diagram are not in
parallel but in series.
A
B
These resistors across A and B are in series, as same current
passes through them.
2.This is a common thinking that current which comes out
from the positive terminal of a battery is used up till it reaches
the negative terminal. But infact the current remains the
same in a branch. In fact a potential drop takes place across
a resistor.
5 W
A
B C
D0V+5V
1 amp 1 amp
I
A
= I
B
= I
C
= I
D
= 1 amp
V
A

= V
B
= +5V
V
C
= V
D
= 0V
This means that a potential drop of 5V takes place acrossthe resistor
3.
OIncorrect : If two resistances are not in series then it
is in parallel and vice-versa.
PCorrect : The above thinking is incorrect. We may
have resistances which are neither in series nor in
parallel.
Colour
Coding for Carbon Resistor and their Standard
Values –
(i) It is a system of colour coding used to indicate the values
of resistors.
(ii) For the fixed, moulded composition resistor, four colour
bands are printed on one end of the outer casing as shown
below.
1
2
3
4
Significant
digits
Multiplier
Tolerence
(
iii) The colour bands are always read left to right from the end
that has the bands closest to it.
(iv) (a) The first and second colour bands, represent the first
and second significant digits respectively, of theresistance value.
(b) The third colour band is for the number of zeros that
follow the second digit.
(c) In case the third band is gold or silver, it represents a
multiplying factor of 0.1 or 0.01.
(d) The fourth band represents the manufacture’s
tolerance. It is a measure of the precision with whichthe resistor was made.
(e) If the fourth band is not present, the tolerance is
assumed to be ± 20%.
(v)Standard value of colour codes for carbon resistors
Colour Digit MultiplierTo lerance
Black 0 10
0
= 1
Brown 1 10
1
= 10
Red 2 10
2
Orange 3 10
3
Yellow 4 10
4
Green 5 10
5
Blue 6 10
6
Violet 7 10
7

475. 471Current Electricity
Grey 8 10
8
White 9 10
9
Gold – 0.1 ± 5%
Silver – 0.
01 ± 10%No colour – – ± 20%
To learn the above table of colour codes of resistors let us
learn this interesting sentence :
BB ROY of Great Britain has a Very Good Wife.
In the above sentence the capital letters have the following
meaning :
B
?Black B?Brown
R?Red O?Orange
Y?Yellow G ?Green
V?Violet G?Grey
W?White
Reme
mber the colour in the above order and the
corresponding digits from 0 to 9 and also the multiplier with
the power to 10 from 0 to 9.
Commercial resistors are of two types
(a)Wire round resistor made by winding of wires of an
alloy manganins, constantan and nichrome.
(b)Carbon resistors have low cost and are compact.
THERMISTOR
A thermistor is a heat sensitive resistor usually made up of
semiconductor. The oxides of various metals such as mickel,
iron, copper etc. temperature coefficient of thermistor is –ve but
is usually large, of the order of 0.04/ºC.
The V–I curve of thermistor is as shown.

V
I
Thermistors are used for resistance thermometer in very low
temperature measurement of the order of 10K and to safeguardelectronic circuits against current jumps because initiallythermistors has high resistance when cold and its resistancedrops appreciably when it heats up.
JOULE’S LAW OF HEATING
It states that the amount of heat produced in a conductor is directly
proportional to the
(i) square of the current flowing through the conductor,
(q, T – constt) i.e. H µ i
2
(ii) resistance of the conductor (i, T – constt.)
i.e. H µ R
(iii) time for which the current is passed (i, R, – constt)
i.e., H µ t
Thus H = i
2
RT joule =
2
4.2
i RT
cal
ELECTRIC POWER
It is defined as the
rate at which work is done in maintaining
the current in electric circuit.
Electric power, P = VI = I
2
R = V
2
/ R watt or joule/second.
Electric energy : The electric energy consumed in a circuit is
defined as the total work done in maintaining the current in an
electric circuit for a given time.
Electric energy = VIt = Pt = I
2
Rt = V
2
t / R
The S.I. unit of electric energy is joule (denoted by J)
where 1 joule = 1 watt × 1 second = 1 volt × 1 ampere × 1 sec.
In household circuits the electrical appliances are connected in
parallel and the electrical energy consumed is measured in kWh
(kilo watt hour).
1 kWh (1 B.O.T. unit) = 1000 Wh = 3.6 × 10
6
J
Keep in Memory
For
Series Combination :
1.If resistances (or electrical appliances) are connected in
series, the current through each resistance is same. Then
power of an electrical appliance
RPµand RVµ )RtiP(
2
=Q
It means in series combination of resistances, the potential
difference and power consumed will be more in larger
resistance.
2.(i) When the appliances of power P
1
, P
2
, P
3
... are
connected in series, the effective power consumed
(P) is
1111 1
= + + ...+
PnPPPP
123
i.
e., the effective power is
less than the power of individual appliance.
(ii) If n appliances, each of equal resistance R, are
connected in series with a voltage source V, the power
dissipated P
s
will be
nR
V
P
2
s
= ... (i)
3.When tw
o lamps of different wattage are connected in series
in a house the lamp of lower wattage glows more brightly.
For Parallel Combination :
1.If resistances (i.e. electrical appliances) are connected in
parallel, the potential difference across each resistance is
same. Then R/1Pµ and R/1µI .
It means in parallel combination of resistances the current
and power consumed will be more in smaller resistance.
2.When the appliances of power P
1
, P
2
, P
3
.... are in parallel,
the effective power consumed (P) is
P = P
1
+ P
2
+ P
3
... +P
n
i.e. the effective power of various electrical appliances is
more than the power of individual appliance.
3.If n appliances, each of resistance R, are connected in parallel
with a voltage source V, the power dissipated P
p
will be
( ) R
nV
n/R
V
P
22
P == ... (ii)
From
eq
ns
. (i) and (ii),
2
S
P
n
P
P
= or P
P
= n
2
P
S

476. 472 PHYSICS
This sho
ws that power consumed by n equal resistances in
parallel is n
2
times that of power consumed in series if
voltage remains same.
4.In parallel grouping of bulbs across a given source of voltage,
the bulb of greater wattage will give more brightness and will
allow more current through it, but will have lesser resistance
and same potential difference across it.
5.For a given voltage V, if resistance is changed from R to
R/n, power consumed changes from P to nP.
P = V
2
/ R; when R´ = R/n,
then P´ = V
2
/ (R/n) = n V
2
/ R = nP.
6.Filament of lower wattage bulb is thinner than that of higher
wattage bulb i.e. filament of 60 watt bulb is thinner than
that of 100 watt bulb.
7.If I is the current through the fuse wire of length l, radius r,
specific resistance r and Q is the rate of loss of heat per
unit area of a fuse wire, then at steady state,
I
2
R = QA or
2
2
I
Q 2r
r
r
= ´p
p
l
l
or
2
2 3 3/22Q
I r Ir
p
= Þa
r
Hence current
capacity of a fuse is independent of its length
and varies with its radius as r
3/2
.
8.If t
1
and t
2
are the time taken by two different coils for
producing same heat with same supply, then(i) If they are connected in series to produce same heat,
time taken, t = t
1
+ t
2
(ii) If they are connected in parallel to produce same heat,
time taken is,
21
21
tt
tt
t
+
=
9.When a bulb glows
the temperature of the filament is of the
order of 3000K.
10.To avoid lengthy calculations use P = I
2
R

in series
combination and
2
=
V
P
R
in parallel combination and
P =
VI when we want to find power of a device and V and I
are known.
11.The resistance of an appliance, can be formed by rated
voltage and power is R =
rated
2
rated
P
V
Now this resi
stance does not change (remember resistance
depends only on the parameters of resistors and not on the
voltage across it or the current flowing through it, if we
neglect the changes occurring due to change in
temperature).
12.The maximum current that can be safely passed through an
appliance (a resistor) is
rated
rated
V
P
I= .
13.Bulbs get
fused sometimes when switched on. This is
because with the rise in temperature the resistance increases
and power decreases (P = V
2
/ R). Therefore the bulb glows
brighter in the beginning and get fused.
14.A lead-acid secondary cell is discharged if the relative
density of electrolyte drop to 1.18 and e.m.f. of 1.8V.
15.Hot wire galvanometer is based on heating effect of current.
Its deflection q is directly proportional to the heating effect
(i.e. I
2
). It works on A.C. as well as D.C.
16.Two wires of same material and same length but having
different diameters connected in parallel produce more heat
than when connected in series. i.e., H
parallel
>H
series.
17.If t
1
and t
2
be the time taken by two heaters to boil a given
mass of a liquid, then the time taken to boil the same amount
of the liquid, when both heaters are connected in parallel is
given by
12
p
12
tt
t
tt
<
∗
.
18.If t
1
and t
2
be the time taken by
two heaters to boil a given
amount of a liquid, then the time taken to boil the same
amount of the liquid, when both heaters are connected in
series is given by
t
s
= t
1
+ t
2
.
19.If I be the current at which a fuse wire of radius R blows,
then
2
3
I
R
constant. i.e., for two fuse wires of radii R
1
and R
2
and maximum bearing current I
1
, and I
2
, we have
22
12
33
12
II
RR
<
.
20.If the two resistors R
1
and R
2
ar
e first connected in series
and then parallel then the ratio of heat produced in the two
cases (series to parallel) is given by
2
s 12
p 12
H (R R)
H RR
∗
<
21.If two resis
tances R
1
and R
2
are connected in parallel and a
current is passed in them such that heat produced in them
is H
1
and H
2
respectively, then12
21
HR
HR
<
.
Example 3.
The t
emperature coefficient of resistance of a wire is
0.00125ºC
–1
.At 300 K its resistance is one ohm. At what
temperature the resistance of the wire will be 2 ohm?
Solution :
From formula R
t
= R
0
(1 + µDt)
300 0
R R(1 27)1= +a´=

t0
R R(1 t)2= +µD´=
\
2
1
t1
271
=
a+
a+
or 2 + 54a = 1 + at
Þ 2 +
54(0.00125) = 1 + (0.00125)t
\ t = 854ºC = 1127 K

477. 473Current Electricity
Example 4.
A cylindrical copper rod is reformed to twice its original
length with no change in volume. The resistance between
its ends before the change was R. Now its resistance will
be
(a) 8 R (b) 6 R
(c) 4 R (d) 2 R
Solution : (c)
2
1
1
r
R
p
r
=
l
...(1)
Now th
e rod is reformed such that
l
2
= 2l
1
and
2
2
21
2
1
rr
llp=p (no change in volume)
or )/()r/r(
12
2
2
2
1 ll=
\
2
2
2
2
r
R
p
r
=
l
...(2)
From e
qns. (1) and (2), we get

2
12 11
2
2 2 221
rR 11
R 22r
<≥<≥<≥
l ll
l ll
\ R
2
= 4 R
Ex
ample 5.
A copper wire is stretched to make 0.1% longer. What isthe percentage change in its resistance?
Solution :
The resistance R of a wire is given by
A
R
lr
=, where
r = speci
fic resistance
Let d and m be the density and mass of the wire, respectivelythen A l d = m or A = m/ld
\
2
2
m
d
m
d
m
d
R l
lll
´÷
ø
ö
ç
è
ær
=
r
=
´r
=
Taking l
og, of both sides, we get e e ee
log log 2 log
d
R
m
ræö
=+ç÷
èø
l
Differentiating
l
ld2
R
dR
=
d
constant
m
ræö
=ç÷
èø
Q
)1.0(2%100
d
2100
R
dR
´=÷
ø
ö
ç
è
æ
´=´
l
l
\ Percenta
ge change in resistance = 2×(0.1) % = 0.2%
So the resistance increases by 0.2%.
Example 6.
Determine the equivalent resistance of the arrangement
of resistances shown in fig between the points A and B.
8W
16W
16W
9W
18W
6W
20W
A B
Solution :
8W, 16W, and 16W r
esistances are connected in parallel.
Their equivalent resistance is given by R’
16
4
16
112
16
1
16
1
8
1
'R
1
=
++
=++=
\ R’ = 4W.
Th
is is in series with 20W. So upper part of AB has a resistance
4 + 20 = 24 W ...(1)
For lower part, 9W and 18W resistances are connected in
parallel. Their equivalent resistance R” is given by
18
3
18
12
18
1
9
1
R
1
=
+
=+=
¢¢
Þ W==¢¢ 63/18R
This is in ser
ies with 6W. So the resistance of lower part of
AB is 6 + 6 = 12W ...(2)
The upper and lower parts of AB are in parallel. Hence the
equivalent resistance between A and B is given by
24
3
24
21
12
1
24
1
R
1
=
+
=+= Þ R = 24/3 = 8W
Example 7.
A wire has a resistance of 10
W. It is stretched by one-tenth
of its original length. Then its resistance will be
(a) 10
W (b) 12.1W
(c) 9W (d) 11W
Solution : (b)
Here volume remains constant. Thus
12 1122
VV AA=Þ´=´ll
)10/11(rr
2
2
2
1
llp=p or )r(
11
10
r
2
1
2
2
p=p
{Q When wire is stretched by 1/10 of its original length,
the new length of wire becomes (11l/10)}
Let the new resistance be R
2
. Then
2
2
2
.R
A
=r
l
2
1
2
2
2
r)11/10(
)10/11(
r
10
11
.
R
p
r
=
p
÷
ø
ö
ç
è
æ
r
=
l
l
=
ú
ú
û
ù
ê
ê
ë
é
p
r
2
1r)11/10(
)10/11(l
=
W==´1.12
10
121
10
)11/10(
)10/11(
Example 8.
What w
ill be the equivalent resistance between the two
points A and D of fig?
10W 10 W
R
1
R
2
R
3
R
5
R
4
10W 10W10W
10W 10W10W
A
B
C
D
R
6
Solution :
Resist
ances R
2
and R
3
are in series.
\ R' = R
2
+ R
3
= 10 + 10 = 20 W

478. 474 PHYSICS
Similar
ly, R
5
and R
6
are in series.
\ R'' = R
5
+ R
6
= 10 + 10 = 20 W
Further, R' and R'' are in parallel
\
W=
+
´
=
¢¢+¢
¢¢¢
=¢¢¢ 10
2020
2020
RR
RR
R
Resistance between A
and D = R
1
+ R'''+ R
4
=
10 + 10 + 10 = 30W
Example 9.
A wire 1 m long has a resistance of 1
W. If it is uniformly
stretched, so that its length increases by 25%, then its
resistance will increase by
(a) 25% (b) 50%
(c) 56.25% (d) 77.33%
Solution : (c)
New length,
;
100
125
100
25
llll=+=¢
Let new area o
f cross-section =
A¢. Then
llll ¢=¢¢¢= /AAordAdA (Volume is con stant)
or A
125 100
100 125
/AA =÷
ø
ö
ç
è
æ
´´=¢ ll
A
Rand
A
R
¢
¢r
=¢
r
=
ll
2
125
125100
R ' 1.5625 R
100 A 100
A
125
æö
r
ç÷
ræöèø
===
ç÷
æö èø
ç÷
èø
l
l
% increase i
n resistance
100
R
RR
´÷
ø
ö
ç
è
æ-¢
= %25.56100
1
15625.1
=´÷
ø
ö
ç
è
æ -
=
Example 10 :
A wir
e has a resistance of 16.0 ohm. It is melted and drawn
to a wire half its initial length. What will be the new
resistance of the wire ?
Solution :
The factor by which the length is changed is
1
n
2
¢
==
l
l
The new resistance R' is given by
2
2 1
RR(n)164
2
æö
= = W =W¢ ç÷
èø
Example 11 :
The r
esistivity of silver at 0°C is 1.6 × 10
–8

W-m. If its
temperature coefficient of resistance is 4.1 × 10
–3
°C
–1
,
find the resistivity of silver at 80°C.
Solution :
Here, =r
0
1.6 × 10
–8
W-m,
=a4.1 × 10
–3
°C
–1
and
Dt = 80°C
Using
0
[1 ( t)]r=r +a D , the resistivity r a t a temperature
80°C will be
8 31
1.6 10 m [1 (4.1 10 C ) (80 C)]
- --
r=´W-+´°°

8
2.1 10 m
-
= ´ W-
Example 12.
A bulb has voltage rating of 220 V and power
rating of
40 W. How can this bulb be made to glow with normal
brightness if a voltage source of e.m.f. 330 V is available?
Solution :
Here V = 220 V; P = 40 W
Resistance of the bulb,
W===1210
40
)220(
P
V
R
22
Current in the circ
uit,
S1210
330
+
=I
Potential differ
ence across the bulb,

330
1210
1210 S
æö
=ç÷
èø+
IR
As per question
,
S1210
1210330
220
+
´
=
On solving,
we get S = 605 W
Hence to glow bulb, a resistance of 605W should be
connected in series.
Example 13.
If two bulbs of wattage 25 and 30 W, each rated at 220
volts are connected in series with a 440 volt supply, which
bulb will fuse?
Solution :
Resistance of 25 W bulb,
W=
25
)220(
R
2
1
;
Current A
220
25
I
1
=
Resistance of 3
0 W bulb,
2
2
(220)
R,
30
=W
I
2
30
Current A
220
=
When bulbs are conn
ected in series, effective resistance is
ú
û
ù
ê
ë
é
+=+=
30
1
25
1
)220(RRR
2
21 W´´=
150
11
220220
When supply volt
age is 440 V, then current is

A
220
27.27
11220220
150440
R
440
I =
´´
´
==¢
As
1
II>¢ but less than
2
I, hen ce the bulb of 25 watt will
fuse.
Example 14.
What will happen when 40 W, 220 V lamp and 100 W,
220 V lamp are connected in series across 440V lamp?
Solution :
Resistance of first lamp,
R
1
= 220 × 220 / 40 = 1210 W.
Resistance of second lamp,
R
2
= 220 × 220 /100 = 484 W.
Total resistance in series = 1210 + 484 = 1694 W.

479. 475Current Electricity
Current in the circuit when supply voltage is 440 V will be,
A26.0
1694
440
I ==
Voltage dr
op across R
1
= 0.26 × 1210 = 314.6 V
Voltage drop across R
2
= 0.26 × 484 = 112.84V
Therefore 40 W bulb will fuse because lamp can tolerate
only 220 V.
Example 15.
A fuse wire with a radius of 1 mm blows at 1.5 A. If the fuse
wire of the same material should blow at 3.0 A, the radius
of the wire must be
(a) 4
1/3
mm (b)
2 mm
(c) 0.5 mm (d) 8.0
mm.
Solution : (a)
The temperature of the wire increases to such a value at
which, the heat produced per second equals heat lost per
second due to radiation i.e.
l
l
r2H
r
2
2
p´=
÷
÷
ø
ö
ç
ç
è
æ
p
r
I
,
where H i
s heat lost per second per unit area due to radiation.
Hence,
32
rµI so
I
I
3
1
3
2
r
r
=
2
1
2
2
or II
2/3
2 121
r r(
/)= 2/3 1/3
1 (3.0/1.5) 4 mm.<≥<
Example 16
A
fuse wire with a circular cross section and having
diameter of 0.4 mm blows with a current of 3 amp. The
value of the current for which another fuse wire made of
the same material but having circular cross-section with
diameter of 0.06 mm will blow is
(a) 3 amp. (b)
3 (3/2)´ amp.
(c) 3×(3/2) amp. (d
)
3/2
3 (3/2)´ amp.
Sol
ution : (d)
For a fuse wire
32
rIµ
\
3
2
3
1
2
2
2
1
r
r
I
I
=
where
02.0
2
04.0
r
1 == cm and 03.0
2
06.0
r
2
==cm.
\
33
2
2
2
3
2
03.0
02.0
I
)3(
÷
ø
ö
ç
è
æ
=÷
ø
ö
ç
è
æ
=
or
3
22
2
2
3
)3(I÷
ø
ö
ç
è
æ
= or
2/3
2
2
3
3I÷
ø
ö
ç
è
æ
´=
ELECTROMOTIVE FO
RCE AND INTERNAL RESISTANCE
OF A CELL
An emf (electromotive force) device has a positive terminal (at
high potential) and a negative terminal (at low potential). This
device is responsible for moving positive charge within itself
from negative terminal to positive terminal.
+
–+
+
H.P
H.P
L.P
L.P
For this to happen, work is done by some agency in the emf
device. The energy required to do this work is chemical energy
(as in a battery), mechanical energy (as in electric generator),
temperature difference (as in a thermopile).
The emf is thus given by the formula
dW
E
dq
=
The S.I
unit of emf is
J
volt
C
<(V)
Keep in Memory
1.
Electromotive force is not a force but a potential difference.
2. E.m.f. can be defined as the work done in moving a charge
once around a closed circuit.
Internal Resistance (r)
The potential difference across a real source of emf is not equal
to its emf. The reason is that the charge which is moving inside
the emf device also suffers resistance. This resistance is called
internal resistance of the emf device.
+ –
I
p.d
R
r
E
V
I
E = IR + Ir = V + Ir
Þ V
= E – Ir
Keep in Memory
1.
For a cell
Circuit Mode Expression
(i)
Discharging V = E – Ir\ V < E
(ii) No currentV = E
(iii)
I
ChargingV = E + Ir\ V > E
2. Emf is the property of a cell but terminal potential difference
depends on the current drawn from the cell.

480. 476 PHYSICS
Short Circuit
ing
When the terminals of an emf device are connected with a
conducting path without any external resistance then
r
I
E
E = IrÞ
r
E
I=
Since intern
al resistance has a very small value, therefore a very
high current flows in the circuit producing a large amount of
heat. This condition is called short circuiting.
During short circuiting, the terminal potential
difference is zero.
CO
MBINATION OF CELLS
Series Combination of Cells
A Br
n
E
1
r
2r
1
E
2
E
n
Equivalent Emf E
AB
= E
1
+ E
2
+ ... + E
n
Equivalent internal resistance, R
AB
= r
1
+ r
2
+ ....... + r
n
Parallel Combination of Cells
A B
E
1
E
2
E
n
r
1
r
2
r
n
Equivalent emf
n21
n
n
2
2
1
1
AB
r
1
.....
r
1
r
1
r
E
........
r
E
r
E
E
++
+++
=
Equivalent inte
rnal resistance
n21AB r
1
.........
r
1
r
1
R
1
+++=
Mixed Gro
uping of Cells :
If the cells are connected as shown below then they are said to
be in mixed grouping.
E E E
EEE
E E E
r
r rr
r
rr
r r r
1
1
2
m
2 n
BA
Equivalent emf E
AB
= nE
Equivale
nt resistance =
nr
m
Where n = no. of cells in a row. and
M = no. of rows
If this equivalent cell is attached to an external resistance R
then
nr
m
nE
R
nr
nE IR
m
æö
=+ç÷
èø

nmE
I
mR nr
Þ=
+
Keep in Memory
(i) The
condition for maximum current through external resistance
R
R
r
n
m
mRnr =Þ= Þ R = nr/m
In other words, when exte
rnal resistance is equal to total
internal resistances of all the cells.
The maximum current
max
nE mE
I or
2R 2r
=
(ii) Maximum po
wer dissipation for the circuit shown in fig.
R
E r
Power
2
22
2
)rR(
RE
R
rR
E
RIP
+
=÷
ø
ö
ç
è
æ
+
==
For maximum po
wer across the resistor,
dP
=0
dR
On solvin
g, we get R = r
This is the condition for maximum power dissipation.
(iii) If identical cells are connected in a loop in order, then emf
between any two points in the loop is zero.
(iv) If n identical cells are connected in series and m are wronglyconnected then E
net
= nE – 2mE

481. 477Current Electricity
FARADAY’S LAW OF ELECTROLYSIS
(i)1st law : The mass of the substance liberated or deposited
at an electrode during electrolysis is directly proportional
to the quantity of charge passed through the electrolyte.
i.e., mass m
µ q = Zq = Z It,
where Z = electrochemical equivalent (E.C.E.) of substance.
(ii)2nd law : When the same amount of charge is passed
through different electrolytes, the masses of the substance
liberated or deposited at the various electrodes are
proportional to their chemical equivalents
i.e.
11
22
=
mE
mE
where m
1
an d m
2
are the masses of the substances liberated
or deposited on electrodes during electrolysis and E
1
and
E
2
are their chemical equivalents.
Faraday's Constant
Faraday constant is equal to the amount of charge required to
liberate the mass of a substance at an electrode during electrolysis,
equal to its chemical equivalent in gram (i.e. one gram equivalent)
One faraday (I F) = 96500 C/gram equivalent.
Keep in Memory
1.If r is t
he density of the material deposited and A is the area
of deposition, then the thickness (d) of the layer deposited
in electroplating process is
A
tIZ
A
m
d
r
=
r
= .
2.The ba
ck e.m.f. for water voltameter is 1.67 V and it is 1.34 V
for CuCl
2
electrolytes voltameter with platinum electrodes.
3.96500 C are required to liberate 1.008 g of hydrogen.
4.2.016 g of hydrogen occupies 22.4 litres at N.T.P.
5.E.C.E. of a substance = E.C.E. of hydrogen × chemical
equivalent of the substance.
Example 17.
Upon a six fold increase in the external resistance of a circuit,
the voltage across the terminals of the battery has increased
from 5 V to 10 V. Find the e.m.f. of battery.
Solution :
V = E – Ir.
E
VEr
(R r)
=-´
+
5r
)rR(
E
E =´
+
- ...(1)
or 5
rR
R
E =÷
ø
ö
ç
è
æ
+
...(2)
10r
)rR6(
E
E =´
+
- ...(3)
or 10
rR6
R6
E =÷
ø
ö
ç
è
æ
+
...(4)
Dividing e
qn. (2) by eqn. (4), we get 2r = 3R
Putting values in eqn. (3), we get
10
rr4
rE
E =
+
´
-
Solving we get E =
12.5 V
Example 18.
Two cells P and Q connected in series have each an emf of
1.5 V and internal resistances 1.0
Wand 0.5W respectively.
Find the current through them and the voltages across
their terminals.
Solution :
For a single closed loop, consisting of cells and resistors
the current i flowing through it is given by
i
i
E
i
rr
S
=
S +S

P
i
Q
E =1.5V
1
E =1.5V
2
r =1.0
1
r =0.5
2
\
1.5 1.5V
i
1.0 0.5
+
=
+W
= 2 . 0 AA
Th
e voltage across the cell P is
p 11
V E ir 1.5V 1.5 (1.0)V zero=-=-=
and
across cell Q is
Q 22
V E ir 1.5 V 1.5 (0.5)V 0.75V=-=-=
Example 19
.
If electrochemical equivalent of hydrogen is Z
H
kg/coul-
equivalent and chemical equivalent of copper is W, then
determine the electrochemical equivalent of copper.
Solution :
HH
Cu
W
W
Z
Z
=
We know th
at, W
H
= 1
\ Z
Cu
= W.Z
H
Example 20.
Find the mass of silver liberated in a silver voltameter
carrying a current of 1.5A, during 15 minutes. The electro
chemical equivalent of silver is 1.12× 10
–6
kg/C.
Solution :
Here, m = ?, i = 1.5 A, Z = 1.12× 10
–6
kg/C
and t = 15 × 30s
Using m = Zit
Mass of silver liberated is
6kg 1.5C
m 1.12 10 (450s )
Cs
-æ öæö
=´ç ÷ç÷
èøèø
= 7.6 × 10
–4
kg = 0.76 g
Example 21.
In a water voltameter, the act of passing a certain amount of
current for a certain time produces of 1.2 g H
2
at STP. Find
the amount of O
2
liberated during that period.
Solution :
Since the same current flows through both the electrodes of
a water voltameter, so the amount of oxygen and hydrogen
liberated (for the same charge) will be directly proportional
to their respective equivalent weights.
By Faradays’ second law of electrolysis,
i.e.,
O
2
H
2
m
8
m1
= or
O H2 2
8
m m 8 1.2g 9.6g
1
æö
= =´=ç÷
èø

482. 478 PHYSICS
SEEBECK/THER
MOELECTRIC EFFECT
When an electric circuit is composed of two dissimilar metals
and the junctions are maintained at different temperature, then
an emf is set up in the circuit. This effect is known as
thermoelectric or seebeck effect.
Thermocouple: It is a device in which heat energy is converted
into electrical energy. Its working is based on seebeck effect. It
has two junctions of two dissimilar metals.
Cu
Fe
T
2
T
1
Some of th
e elements forming thermo-electric series
Sb, Fe, Zn, Cu, Au, Ag, Pb, Al, Hg, Pt, Ni, Bi
(i) Lead (Pb) is thermo-electrically neutral
(ii) At the cold junction, current flows from the element occuring
earlier into the element occuring later in the series.
For example: In Cu–Fe thermo–couple, current flows from
Cu to Fe at hot junction.
Neutral Temperature (T
n
)
It is that temperature of hot junction for which the thermo emf
produced in a thermocouple is maximum.
It depends upon the nature of the material of thermocouple but is
independent of temperature of cold junction.
Temperature of Inversion (T
i
)
It is that temperature of hot junction for which the thermo emf
becomes zero and beyond this temperature, the thermo emf in a
thermocouple reverses its direction.
It depends upon the nature of the material of thermocouple and
temperature of cold junction
Let T
o
, T
n
, T
i
be the temperature of cold junction, neutral
temperature and temperature of inversion then
0
or ( )/2-=- =+
n in n io
T TTT T TT
Thermo emf
T
n T
i
OTo
Temp. difference
With temperature difference T between hot and cold junctions,
the thermo-e.m.f. is given by
E = aT + bT
2
where a and b are Seebeck co-efficients
At T
n
, (dE/dT) = 0
\ T
n
= – a/2band T
i
= – a/b, when T
o
= 0
S = dE/dT is called thermo-electric power.
PELTIER EFFECT
It states that if current is passed through a junction of two
different metals the heat is either evolved or absorbed at that
junction. It is the reverse of seebeck effect.
The quantity of heat evolved or absorbed at a junction due to
Peltier effect is proportional to the quantity of charge crossing
that junction.
Peltier Coefficient (p) :
It is defined as the amount of heat energy evolved or absorbed
per second at a junction of two different metals when a unit
current is passed through it.
The Peltier heat evolved or absorbed at a junction of a
thermocouple = pI t
where I = current passing through the junction for time t.
\ p=
dE
T
dT
where T and (T + dT) are the temperature of cold and hot junctions
of a thermocouple and dE is the thermo emf produced.
p
\==
dE
S
T dT
(Seebeck coefficient)
THOMSON EFFECT
If a
metallic wire has a non uniform temperature and an electric
current is passed through it, heat may be absorbed or produced
in different sections of the wire. This heat is over and above the
joule heat I
2
Rt and is called Thomson heat. The effect is called
Thomson effect.
If a charge DQ is passed through a small section of given wire
having temperature difference DT between the ends,
Thomson heat, DH = s DQ DT
where s is constant for a given metal at a given temperature.
Thomson emf, s DT, is defined as s DT = DH/DQ.
s is positive if heat is absorbed when a current is passed
from low temp. to high temperature. s is numerically equal to P.D.
developed between two points of the conductor differing in temp.
by 1ºC.
Keep in Memory
1. The actual em
f developed in a thermocouple loop is the
algebraic sum of the net Peltier emf and the net Thomsonemf developed in the loop.
B
A
T
()p
ABT
( )Tp
AB0
T
0s
B0
(T–T)
s
A0
(T–T)
AB AB T AB T A o B o
o
E ( ) ( ) (T T) (T T)=p -p +s-- s-
2.If S, p and s are the Seebeck coefficient, Peltier coefficient,
and Thomson coefficient respectively then it is found that

483. 479Current Electricity
(i)
TdT
dE
S
p
==
(ii)
dT
TdS
dT
dE
dT
d
T
dT
Ed
T
2
2
-=÷
ø
ö
ç
è
æ
-=-=s
3.For Pel
tier effect or Thomson effect, the heat evolved or
absorbed is directly proportional to current. But for Joule's
law of heating, the heat produced is directly proportional to
the square of the current flowing through it.
4.Thermo-emf set up in a thermocouple when its junctions
are maintained at temperature T
1
and T
3
(i.e.
3
1
T
T
E) is equal
to the sum of the emfs set up in a thermocouple when its
junctions are maintained first at temperature T
1
and T
2
(i.e.
2
1
T
T
E) and then at T
2
& T
3
(i.e.
3
2
T
T
E) i.e. 3
2
2
1
3
1
T
T
T
T
T
T
EEE+=
It is called law of in termediate temperature.
Example 22.
The temperature of inversion of a thermocouple is 620ºC
and the neutral temperature is 300ºC. What is the
temperature of cold junction ?
Solution :
Let T
i
,T
n
and T
c
be the temperature of inversion, neutral
temperature and temperature of cold junction
respectively, then
T
i
– T
n
= T
n
– T
c
\ 620 – 300 = 300 – Tc
Þ 320 = 300 – TT
c
\ T
c
= 300 – 320 = – 20ºC
Example 23.
The temperature of cold junction of a thermocouple is 0ºCand the temperature of hot junction is TºC. The thermoe.m.f. is given by E = 16 T – 0.04 T
2

m volt
Find (a) the neutral temperature and (b) the temperature of inversion
Solution :
Given that E = 16T – 0.04T
2
\
dE
16 2 0.04T
dT
= -´
(a) At
neutral temperature, dE/dT = 0
\ 16 – 2×0.04 T
n
= 0 or 16 = 0.08 T
n
or
Cº200
08.0
16
T
n
==
(b) At the tempe
rature of inversion, E = 0
\ 16 T
i
– 0.04 T
i
2
= 0 or 16 – 0.04 T
i
= 0
or 16 = 0.04 T
i
\
Cº400
040.0
16
T
i
==
Example 24.
One j
unction of a certain thermocouple is at a fixed
temperature T
r
and the other junction is at a temperature
T. The electromotive force for this is expressed by,
ú
û
ù
ê
ë
é
+--=)TT(
2
1
T)TT(kE
r0r
.
At, temperatur
e T = T
0
/2. Determine the thermoelectric
power.
Solution :
ú
ú
û
ù
ê
ê
ë
é
-+---=
2
T
2
TT
TT
2
TT
2
T
TTkE
2
rr
0r
r
2
0
Hence )TT(k
2
T
2
T
TTk
dT
dE
0
rr
0
-=
ú
û
ù
ê
ë
é
+--=
At tem
perature T = T
0
/2,
Thermoelectric power .2/Tk)2/TT(k
000
=-=
KIRCHOFF’S LAWS AND ELECTRICAL CIRCUIT
Many practical combination of resistors cannot be reduced to
simple series, parallel combinations. For example the resistors in
the figure are neither in series nor in parallel.
b
a
E
R
1
R
2
R
1
R
1
R
2
R
5
I
2
I
1
c
e
I
f d
The use of Oh
m’s law is not sufficient to solve such problems.
Kirchoff’s laws are used in such cases.
We will often use the term junction and loop, so let us first
understand the meaning of these words. A junction in a circuit
is a point where three or more conductors meet. A loop is a
closed conducting path. In the above figure e, f, d, c are junctions.
a, b, are not junctions. The various loops are efde, cdfc, eabcf
and eabcde.
(i) Kirchoff’s junction law : (Based on conservation of
charge) At any junction, the sum of currents entering the
junction must be equal to the sum of currents leaving it.
If this is not so, charges will accumulate at the junction.
This cannot happen as this would mean high/low potential
maintained at a point in a wire without external influence.
When we apply this rule at junction c, we get I = I
1
+ I
2
(ii) Kirchoff’s loop law : (Based on energy conservation)
The algebraic sum of changes in potential around any
closed loop of a circuit must be zero.
Sign convention for using loop law. If we move a loop element
(resistor, emf device, capacitor, inductor etc.) in the direction of
increasing potential, we take the potential difference positive
and vice-versa.
–+
E
Travel
p.d = +E
–+
E
Travel
p.d = –E
Travel
IR
H .P. L. P.
p.d = – IR
Travel
IR
L . P. H.P
p.d = +IR

484. 480 PHYSICS
Problem Sol
ving Tactic for Using Kirchoff’s Law
(i)Draw a circuit diagram large enough to show all resistors,
emf device, capacitors, currents clearly.
(ii)Take into account the resistance of voltmeter/ammeter/
internal resistance of a cell (if given).
(iii) Assume the direction of current in all branches. It may be
noted here that one branch has only one direction of
current.
It is best to use junction law simultaneously while drawing
currents. This helps to reduce the number of unknown
quantities.
R
2
R
1
f
E
1
I
2
dI
1
I
2R
4
c
b
a
E
2
R
2
R
3
I
12
+ I
e
Fig 1
loop
n
dir
loop
direction
R
2
R
1
f
E
1
I
2
dI
1
I
2R
4
c
b
a
E
2
I
1
R
3
I
3
e
Fig 2
In the abov
e circuits we arbitrarily assumed the direction
of current I
1
in branch abcd as anti-clockwiswe and the
direction of current I
2
in branch afed as clockwise.
In figure 1 we have two unknown currents (I
1
, I
2
) whereas
figure 2 we have three unknown currents (I
1
, I
2
and I
3
).
The first figure is a better option for solving problems. In
figure 1 we used junction rule at d simultaneously while
labelling currents.
(iv)In a branch containing a capacitor, the current is zero when
d.c is applied and steady state conditions are achieved.
(v)Now we need as many independent equations as there are
conditions unknowns. If we have to find a particular
unknown, we should ensure that, the unknown appears in
one of the equations made by us.
(vi)For making equations choose the loop and travel the loop
completely. We may travel the loop in clockwise or anti-
clockwise direction. While using second law use sign
conventions properly.
(vii)Solve the equations formed to find the unknown quantities.
If any value of current comes out to be negative then that
particular current is in the opposite direction to that
assumed.
Applications
Let us use second law in the loop abcda of figure 1 taking the
loop in anti-clockwise direction starting from a.
+ E
2
– I
1
R
4
– (I
1
+ I
2
) R
3
= 0
For loop afeda, moving the loop in clockwise direction we get –
E
1
– I
2
R
1
– I
2
R
2
– (I
1
+ I
2
)R
3
= 0
Node method to apply Kirchoff’s law (Open loop method)
Step 1 :We select a reference node and assume its potential
to be (zero/x)V
Step 2 :We calculate the voltage of other selected points w.r.t.
the reference node
Step 3:We find some independent node (whose voltage is
not known). We apply Kirchoff’s law to find the
relevant values.
Example 25.
Find current through branch BD
D
C
BA
5V
3W
2W1W
15V
Solution :
Let V
D
= 0V
\ V
A
= +
5V and V
C
= + 15 V
Let the voltage of B = V
B
Applying Kirchhoff’s junction law at BVolt82.6V0
3
V0
2
V15
1
V
5
B
BBB
=Þ=
-
+
-
+
-
Current through
A27.2
3
082.6
BD =
-
=
Example 26.
Calculate the current
s I
1
, I
2
and I
3
in the circuit shown in
figure.
Solution :
Junction rule at C y
ields
I
1
+ I
2
– I
3
= 0 i.e., I
1
+ I
2
= I
3
.... (1)
while loop for meshes a and b yields respectively :
–14 – 4I
2
+ 6I
1
– 10 = 0
i.e., 3I
1
– 2I
2
= 12 .... (2)
and, 10 – 6I
1
– 2I
3
= 0
i.e., 3I
1
+ I
3
= 5 .... (3)
Substituting I
3
from equation (1) in (3)
4I
1
+ I
2
= 5
Solving equations (2) and (4) for I
1
and I
2
, we find
I
1
= 2A and I
2
= –3A
And hence equation (1) yields, I
3
= –1A
The fact that I
2
and I
3
are negative implies that actual
direction of I
2
and I
3
are opposite to that shown in the
circuit.

485. 481Current Electricity
WHEATSTONE BRIDGE
The condition for balanced wheatstone bridge
AB AB T AB T A o B o
o
E ( ) ( ) (T T) (T T)=p -p + s--s-
G
Q
S
C
P
R
A
D
B
E
PR
=
QS
also
PQ
=
RS
Note that when battery and galvanometer of a Wheatstone bridge
is interchanged, the balance position remains undisturbed, while
sensitivity of the bridge changes.
In the balanced condition, the resistance in the branch BD may
be neglected
Example : Resistance connected to BC may be neglected.
2W
2W 2W
2W
2W
B
A
C D
;
2W
2W
2W
2W
2W
A
B
In a Wheatstone bridge, the deflection in a
galvanometer does not change, if the battery and the
galvanometer are interchanged
Measuring temperature with the help of Wheatstone bridge
At balancing
)T1(S
Q
RR
P
Da+
=
D+
G
Q
S
P
R
DR
When P = Q then DR = S a DT [ R S]?Q
a
D
=D\
S
R
T
COMMON DEFAULT
OInco
rrectIf the current flows in a wire, there has to be a
potential difference. The potential drop takes
place only when current passes through a
resistor.
A B C
D
1
r r
2
r
PCorrectIn the
diagram, the three resistors are in parallel.
The potential at A is equal to the potential at C.Current flows in wire 1 but there is no potnetialdrop across A and C.
OIncorrectIf potential difference between the points is zero,there is zero current between the two points.
PCoorectThere is no p.d. between A and C still current
flows in segment 1.
Example 27.
Calculate the effective resistance between A and B in the
following network.
D
C
4W
A
C
6W
B
3W
2W
7W
Solution :
Th
e circuit can be redrawn as
Q=3W
D
C
7W
BA
R=4W S=6W
P=2W
Here
PR2
QS3
== so bridge is balanced
So the resistance between c and d is non useful.
Equivalent resistance = (P + Q) (R + S)
eq
(P Q)(R S)(2 3)(4 6)
R
PQRS 2346
+ + ++
==
+ + + +++

5 10 10
153
´
= =W
Even if not able to observe balanced wheatstone bridge try
to observe symmetry in network and use plane cuttingmethod.
Example 28.
Find the potential difference between the points A and Bin fig.
5W 5W
5W 5W
5W 5W
2V
+
–
A
B

486. 482 PHYSICS
Solution :
Th
e upper three resistances of the cell are in series. Their
equivalent resistance is 15W. Similarly lower three
resistances are in series. Their equivalent resistance is also
15W. The upper and lower equivalent resistances are
connected in parallel. So, resultant resistance of the circuit
is given by
15
2
15
1
15
1
R
1
=+= or W=
2
15
R
Current from t
he cell,
15
4
)2/15(
2
i ==amp.
In order to ca
lculate potential difference between points A
and B, see fig.
5W 5W 5W
5W5W5W
2V
B
E
F D
C
A
Half of the cur
rent goes to each part i.e., current in each
part is (2/15) amp. Consider the loop AEFDCA, we have
BAAB
VV5
15
2
5
15
2
5
15
2
V -=´+´+´-=
=
3
2
5
15
2
=´volt or volt
3
4
5
15
2
2V
A
=´-=,
volt
3
2
10
15
2
2V
B =´-=so volt
3 2
VVV
BAAB =-=
METER BRI
DGE OR SLIDE WIRE BRIDGE
Principle:Based on balanced Wheatstone bridge principle.
Use : To find unknown resistance
Working :Let P be the unknown resistance.
G
100– ll
QP
At balance point
ll-
=
100
QP
Q is known and l can
be calculated.
POTENTIOMETER :
Principle : The p.d. across a resistance wire is directly
proportional to its length provided I, r and A are constant.
V = IR = I r
A
l
Þ V a l[ I, r and A are c onstant]
Working : PQ is the resistance wire of potentiometer generally
made up of constantan or nichrome. One end P is connected to
the positive terminal of the battery B while negative terminal is
connected to Q through a Rheostat (Rh) and key (K). This is the
main circuit.
A cell whose emf has to be measured is also connected to the
potential wire in such a way that the positive terminal is connected
with P and negative terminal is connected to a galvanometer and
then to a jockey (J) which is free to slide along the wire
A
I
B
I
E
A
J
P
C
Q
E
BKRh
There is a potenti
al drop along PQ.
The potential drop per unit length along PQ is called potential
gradient.When the jockey is pressed on some point, current flows from Eto P (Þ). Also current that comes from B after reaching P divides
into two parts. One part moves towards A and the other towardsE (®). Three cases may arise.
(a)I
B
> I
E
. This happens when V
PC
> E. One side deflection in
galvanometer
(b) I
B
= I
E
. This happens when V
PC
= E, Zero deflection in
galvanometer
(c)I
B
< I
E
. This happens when V
PA
< E. Other side defection
in galvanometer
• At null point since no current flows through E therefore it is
said to be in the condition of open circuit.
•More is the length of potentiometer, higher is the sensitivity ofpotentiometer and smaller is the potential gradient.
• Potentiometer will work only when B > E. Also the positive
terminal of the batteries is connected at P. If any of the above
conditions is not followed, we do not get a null point.
Uses : (i)Comparison of emfs of cells
11
22
El
El
=
(ii)To find internal resistance of a cell
l
rR
l'
æö
=ç÷
èø
emf can be measured by potentiometer and not voltmeter.

487. 483Current Electricity
100
G
Copper strip
A B
x
r
E
Copper
strip
1 2
Sol.Q Wheatstone bridge is in balanced condition
100
100
x
1 2
so
12
100x
100100 x+
=
ll
and Q
1
2
2=
l
l
Þx = 100 W
MEASURING INS
TRUMENTS
Galvanometer
It is an instrument used to detect small currents in a circuit.
The current required for full scale deflection in the galvanometer
is called full scale deflection current and is denoted by I
g
.
Current sensitivity of a galvanometer.
It is defined as the deflection produced in the galvanometer,
when unit current flows through it.
Current sensitivity,
S
NBA
I
IC
q
== and its unit i s r a
d AA
–1
Current sensitivity can be increased either by decreasing C i.e.
restoring torque per unit twist or increasing B.
Voltage sensitivity of a galvanometer
It is defined as the deflection produced in the galvanometer when
a unit voltage is applied across the two terminals of the
galvanometer.
Voltage sensitivity
S
NBA
V
I CR
qq
===
VR
, its unit i s
r a d VV
–1
Ammeter
Ammeter is used to measure current in a circuit. Ammeter is
always connected in series in the circuit as shown.
R
ext
A
I
Example 29.
A 10 m long wire AB of uniform area of cross-section and
20
W resistance is used as a potentiometer wire. This wire
is connected in series with a battery of 5V and a resistor
of 480
W. An unknown emf is balanced at 600 cm of the
wire as shown in the figure.
G
BJA
E
5V
480 W
600 cm
Calculate :
(i)The pote
ntial gradient for the potentiometer wire
(ii)The value of unknown emf E
Solution :
(i)
V = 5V
480 W
I
A B
I
Applying Ohm’s law
V = I (R
AB
+ 480)
5 = I (20 + 480) [R
AB
= 20 W (given)] A01.0
500
5
I ==Þ
Potenti
al gradient of potentiometer wire
1AB
AB
V 0.2
0.02Vm
10
,
< <<
l
(ii)With r
eference to the current given in question, the emf E
and the potential drop across AJ should be equal for thebalancing : E = (potential gradient of potentiometer wire)
× balancing length = 0.02 × 6 = 0.12 V
Example 30.
A uniform potential gradient is established across a
potentiometer wire. Two cells of emf E
1
and E
2
connected
to support and oppose each other are balanced over
l
1
= 6m and
l
2
= 2m. Find E
1
/E
2
.
Solution :
E
1
+ E
2
= xl
1
= 6xand E
1
– E
2
= 2x
12
12
EE 6
EE2
+
=
-
or
1
2
E2
E1
=
Example 31.
In a practical w
heatstone bridge circuit as shown, when
one more resistance of 100
W is connected is parallel with
unknown resistance ‘x’, then ratio
12
/ll become ‘2’.
1
l
is balance length. AB is a uniform wire. Find the value of'
x'.

488. 484 PHYSICS
Conversio
n of a galvanometer into an ammeter
G
I – Ig
I
S
Ig
I
A
For this, we c
onnect a small resistance S (called shunt) in
parallel with the galvanometer.
Mathematically, I
g
× G = (I – I
g
) S
where I is the maximum current which ammeter can measure. G is
the resistance of galvanometer and Ig is the current of full scale
deflection in galvanometer. S is shunt.
The resistance of the ammeter will be
1 11
A
R GS
=+
A
GS
R
GS
´
=
+
Since shunt is a small resistance. Therefore the resistance of
ammeter is very small.
The above arrangement is made so that when we connect
ammeter in series to measure current, it does not change the
original current to a large extent. The change is infact very small.
Also since galvanometer is a sensitive device and cannot take
large currents, this arrangement serves the purpose. Most of
the current entering the ammeter passes through the shunt as
current always prefer low resistance path.
An ideal ammeter is one which has zero resistance.
The range of ammeter can be increased but cannot be decreased
below Ig.
Voltmeter
Voltmeter is used to measure potential difference across a
resistor. Voltmeter is always connected in parallel across a
resistor.
Conversion of a galvanometer into a voltmeter
R
R
ext
ba
V
G
I
Ig
Voltmeter (V)
V
For this, we connect a large resistance R in series with the
galvanometer.
The potential difference which has to be measured is across the
external resistance i.e. across points a and b.
Let it be V. Then
V = I
g
(G + R)
where V is the maximum potential difference that the voltmeter
can measure and R is the large resistance connected in series
with the galvanometer
The resistance of the voltmeter will be R
V
= G + R
Since R is a large resistance. Therefore resistance of
voltmeter is very large.
An ideal voltmeter is one which has infinite resistance.
The range of voltmeter can be increased and decreased.
When ammeter/voltmeter is connected in the circuit,
the current or voltage indicated by these is less than the actual
values in their absence.
Example 32.
A galvanometer of coil resistance 20 ohm, gives a full
scale deflection with a current of 5 mA. What arrangements
should be made in order to measure currents upto 1.0 A ?
Solution :
The upper limiting value of current to be measured is to be
increased by a factor
1.0A
n 200
5A
==
\ Resistance of the shunt required will be
G 20
S 0.1
n1 2001
W
= = »W
--
Hence, a shunt of resistance W1.0 should be connected
in paral
lel across the galvanometer coil.
Example 33.
A voltmeter having 100
W resistance can measure a
potential difference of 25V. What resistance R is required
to be connected in series, to make it read voltage upto
250 V ?
Solution :
The upper limiting value of voltage is to be increased by a
factor
250V
n 10
25V
==
\ Resistance, R = (n – 1) G = (10 – 1) 100 = 900 W
Keep in Memory
1.In order to
increase the range of a voltmeter n times, its
total resistance should also be increased by n times. So
the resistance to be connected in series is
R = (n – 1) G.
2.In order to increase the range of an ammeter n times, the
value of shunt resistance to be connected in parallel is
S = G / (n – 1)
3.For galvanometers with their coil in uniform magnetic field B
C
I
NAB sin
q
=
a
C = torsional rigidity of suspension wire.
For galvanometer with concave pole-pieces – radialmagnetic field is produced, a = 90°
I = (C / NAB)q or I = Kq
All galvanometers used in practice have concave
pole-pieces, for making the radial magnetic field.
Example 34.
A galvanometer has a resistance of 30
W and current of
2 mA is needed to give full scale deflection. What is the resistance needed and how is it to be connected to convert the galvanometer
(a)into an ammeter of 0.3 ampere range
(b)into voltmeter of 0.2 volt range

489. 485Current Electricity
Solution :
(a) We know that i
GS
S
i
g ÷
ø
ö
ç
è
æ
+
=
Substituting the
given values, we get 3.0
30S
S
102
3
´÷
ø
ö
ç
è
æ
+
=´
-
Solving it
, we get S = (30/149) W
So, a resistance of (30/149) W must be connected in
parallel with the galvanometer.
(b) In this case,
÷
ø
ö
ç
è
æ
+
=
GR
V
i
g or G
i
V
R
g
-=
\ W=-=-
´
=
-
703010030
102
2.0
R
3
So, a resistanc
e of 70 W must be connected in series
with the galvanometer. 654
Example 35.
A voltmeter has a resistance of G ohm and range V volt.
What will be the value of resistance used in series to
convert it into voltmeter of range nV volt?
Solution :
We know that,
g
V
RG
Ι
=-
The voltmeter gives full scale deflection for potential
difference V. Its resistance is G
Hence g
Ι(V / G)=
\ G)1n(GGnG
)G/V(
Vn
R -=-=-=
Example 36.
In the circui
t shown in fig., Find the reading of voltmeter.
V
2 V, r = 0
+–
80W
80W20W
Solution :
The
equivalent resistance of the circuit is given by
W=+=
´
´
+= 604020
8080
8080
20R
.eq
Current
through the circuit
V
I
R
==(2/60) = (1/30) amp
Reading of voltmeter
= current × R
T
=
V33.1
30
40
8080
8080
30
1
==÷
ø
ö
ç
è
æ
+
´
´

490. 486 PHYSICS
Electric Current (I) The time
rate of flow of charge (Q)
through any cross-section
CURRENT ELECTRICITY
CONCEP
T
M
A
P
I =
t
Electric cell source of energy that
maintains continuous flow of charge
ina circuit
Ohm's law if the physical
conditions remain same,
current I V V = IR
R-electric resistance
µÞ
Current density (J)
Current per unit cross
sectional area (A)
IE
J==
A
r
r
Kirchhoff's laws
Balanced condition of wheatstone bridge
PR
=
QS
Po
t
e
n
t
i
o
m
e
t
e
r

u
s
e
d

t
o
(i)

C
o
m
p
a
r
e

e
m
f
s

(i
i
)
F
i
n
d

i
n
t
e
r
n
a
l
re
s
i
s
t
a
n
c
e

o
f
c
e
l
l
E
r=

1
S
V
æ
ö
ç
÷
è
ø
1 1
2 2
E
E
l
l
=
Dependence of
resistance
On temperature
R = R (1 + t)
t 0
µ
On length () and l
area of cross-section (A)
R
Rr 1
A
A
l
l
µü
ï
= ý
µ
ï
þ
r = resistivity
Conductivity ( )
Reciprocal of resistance
s
1
r
s=
2nd law Loop
rule Algebraic
sum of changes in
potential around
any closed loop is
zero.
lst law Junction law
Algebraic sum of
all the current
meeting at a junction
is zero i.e. I = 0 S
G
r
o
u
p
i
n
g
s

o
f

c
e
l
l
s
C
e
l
l
s

i
n

s
e
r
i
e
s
C
u
r
r
e
n
t
i
n

t
h
e

,

c
i
r
c
u
i
t
n
I
R
n
r
e
=
+
C
e
l
l
s
i
n

p
a
r
a
l
l
e
l
C
u
r
r
e
n
t

i
n
t
h
e

c
u
r
r
e
n
t

c
i
r
c
u
i
t
I
r
R
m
e
=
+
C
e
l
l
s
i
n

s
e
r
i
e
s
a
n
d

p
a
r
a
l
l
e
l

i
.e
.
m
i
x
e
d

C
u
r
r
e
n
t

i
n
t
h
e
c
i
r
c
u
i
t
,


n
I
n
r
R
m
e
=
+
E
l
e
c
t
r
i
c
a
l

e
n
e
r
g
y
2
H
I
Rt
ü
µ
ï
µ
ýï
µ
þ
J
u
l
e
's

h
e
a
t
i
n
g

l
a
w
H
=

I


R
t
;
E
l
e
c
t
r
i
c
a
l

p
o
w
e
r2
2
V
P
=
R
Resistance (R)Obstruction
to flow offered of electrons
Colour coding of
Resistance R = AB
× C D% A, B – First
two significant figures
of resistance C-multiplier
D-tolerance
±
Grouping of resistance
Series grouping
o
f
resistances Equiv
a
l
e
n
t

resistance, R = R



+

R +...+ R
n
s
1
2
P
a
r
a
l
l
e
l
g
r
o
u
p
i
n
g

o
f
r
e
s
i
s
t
a
n
c
e
s

E
q
u
i
v
a
l
e
n
t

r
e
s
i
s
t
a
n
c
e
,
P
1
2
n
1
1
1
1
=
+
+
.
..
+
R
R
R
R
D
r
i
f
t

v
e
l
o
c
i
t
y

(
V


)
A
v
e
r
a
g
e
u
n
i
f
o
r
m

v
e
l
o
c
i
t
y
a
c
q
u
i
r
e
d

b
y

f
r
e
e

e
l
e
c
t
r
o
n
s
d
d
i
V
=
n
e
A
M
o
b
i
t
i
t
y

(

)

D
r
i
f
t

v
e
l
o
c
i
t
y

p
e
r

u
n
i
t
e
l
e
c
t
r
i
c
f
i
e
l
d
m
d
V
=
E
m
Meter Bridge Based on
Wheatstone bridge
PR
R
=
QS
1
0
0
–
S
l
l
Þ
=
Q
–

491. 487Current Electricity
1.The emf developed by a thermocouple is measured with
the help of a potentiometer and not by a moving coil
millivoltmeter because
(a) the potentiometer is more accurate than the voltmeter
(b) the potentiometer is more sensitive than voltmeter
(c) the potentiometer makes measurement without
drawing any current from the thermocouple
(d) measurement using a potentiometer is simpler than
with a voltmeter
2.Three copper wires of lengths and cross sectional areas are
(l, A), (2 l, A/2) and (l/2, 2A). Resistance is minimum in
(a) wire of cross-sectional area A/2
(b) wire of cross-sectional area A
(c) wire of cross-sectional area 2A
(d) same in all the three cases
3.When a current I is set up in a wire of radius r, the drift
velocity is v
d
. If the same current is set up through a wire of
radius 2 r, the drift velocity will be
(a) 4 v
d
(b) 2 v
d
(c)v
d
/2 (d) v
d
/4
4.In the given circuit, as the sliding contact C is moved from
A to B
A
B
C
V
A
(a) the
readings of both the ammeter and the voltmeter
remain constant
(b) the reading of both the ammeter and the voltmeter
increase
(c) the reading of the ammeter remains constant but
that of the voltmeter increases
(d) the reading of the ammeter remains constant but
that of the voltmeter decreases
5.Potentiometer measures potential more accurately because(a) it measures potential in open circuit
(b) it uses sensitive galvanometer for null deflection
(c) it uses high resistance potentiometer wire
(d) it measures potential in closed circuit
6.Coils in the resistance boxes are made from doubled up
insulated wires
(a) to cancel the effect of self induction
(b) to nullify the heating effect
(c) to nullify the Peltier effect
(d) to reduce effective length of the wire
7.For measuring voltage of any circuit, potentiometer is
preferred to voltmeter because
(a) the potentiometer is cheap and easy to handle.
(b) calibration in the voltmeter is sometimes wrong .
(c) the potential draws no current during measurement.
(d) range of the voltmeter is not as wide as that of the
potentiometer.
8.The infinity resistance plug in a post-office box has
(a) an air gap only
(b) a resistance coil of infinite resistance
(c) largest resistance available in box
(d) resistance of the coil 5000
W
9.In an household electric circuit, which of the following is/
are correct?
(A) All electric appliances drawing power are joined in
parallel
(B) A switch may be either in series or in parallel with the
appliance which it controls
(C) If a switch is in parallel with an appliance, it will draw
power when the switch is in the ‘off’ position (open)
(D) If a switch is in parallel with an appliance, the fuse will
blow (burn out) when the switch is put ‘on’ closed.
(a) A, D (b) A, C, D
(c) B, C, D (d) A, B, D
10.Two identical fuses are rated at 10 A. If they are joined
(A) in parallel, the combination acts as a fuse of rating 20 A
(B) in parallel, the combination acts as a fuse of rating 5 A
(C) in series, the combination acts as a fuse of rating 10 A
(D) in series, the combination acts as a fuse of rating 20 A
Select the correct options.
(a) A, B (b) A, C
(c) B, D (d) B, C, D
11.The resistance of the coil of an ammeter is R. The shunt
required to increase its range n-fold should have a
resistance
(a)
n
R
(b)
1n
R
-
(c)
1n
R
+
(d) nR
12.A cell of in
ternal resistance r is connected across an
external resistance nr. Then the ratio of the terminal voltage
to the emf of the cell is
(a)
n
1
(b)
1n
1
+
(c)
1n
n
+
(d)
n
1n-
13.A battery
of e.m.f. 10 V and internal resistance 0.5 W is
connected across a variable resistance R. The value of R
for which the power delivered in it is maximum is given by
(a) 0.5 W(b) 1.0 W(c) 2.0 W (d) 0.25 W
14.An ammeter has a resistance of G ohm and a range of I amp.
The value of resistance used in parallel to convert it into an
ammeter of range nI amp is
(a) nG (b) (n – 1)G
(c) G/n (d) G/(n – 1)

492. 488 PHYSICS
15.A current
source drives a current in a coil of resistance R
1
for a time t. The same source drives current in another coil
of resistance R
2
for same time. If heat generated is same,
find internal resistance of source.
(a)
21
21
RR
RR
+
(b)
21
RR+
(c) zero (d)
21
RR
16.Two electric bu
lbs whose resistance are in the ratio 1 : 2 are
arranged in parallel to a constant voltage source. The
powers dissipated in them have the ratio
(a) 1 : 2(b) 1 : 1(c) 2 : 1(d) 1 : 4
17.Who among the following scientists made the statement ?
“Chemical change can produce electricity”.
(a) Galvani (b) Faraday
(c) Coulomb (d) Thompson
18.Which of the following is not reversible ?
(a) Joule effect (b) Peltier effect
(c) Seebeck effect (d) Thomson effect
19.In which of the following the power dissipation is
proportional to the square of the current ?
(a) Peltier effect (b) Joule's effect
(c) Thomson effect (d) None of the above
20.When current is passed through a junction of two dissimilar
metals, heat is evolved or absorbed at the junction. This
process is called
(a) Seebeck effect (b) Joule effect
(c) Petlier effect (d) Thomson effect
21.A 4 ohm resistance wire is bent through 180º at its mid point
and the two halves are twisted together. Then the resistance
is
(a) 1 W (b) 2 W (c) 5 W (d) 8 W
22.Is it possible that any battery has some constant value of
e.m.f but the potential difference between the plates is zero?
(a) No
(b) Yes, if another identical battery is joined in series.
(c) Yes, if another identical battery is joined in
opposition in series.
(d) Yes, if another similar battery is joined in parallel.
23.Which of the following in electricity is analogous to
momentum mv in dynamics ?
(a ) IV(b) I L ( c ) Q L (d) IQ
24.You are given a resistance coil and a battery. In which of
the following cases is largest amount of heat generated ?
(a) When the coil is connected to the battery directly
(b) When the coil is divided into two equal parts and
both the parts are connected to the battery in parallel
(c) When the coil is divided into four equal parts and all
the four parts are connected to the battery in parallel
(d) When only half the coil is connected to the battery
25.The electric resistance of a certain wire of iron is R. If its
length and radius are both doubled, then
(a) the resistance and the specific resistance, will both
remain unchanged
(b) the resistance will be doubled and the specific
resistance will be halved
(c) the resistance will be halved and the specific
resistance will remain unchanged
(d) the resistance will be halved and the specific
resistance will be doubled
1.A given resistor has the following colour scheme of the
various strips on it : Brown, black, green and silver. Its
value in ohm is
(a) %10100.1
4
±´ (b) %10100.1
5
±´
(c) %10100.1
6
±´ (d) %10100.1
7
±´
2.If R
1
and R
2
are respectively the filament resistances of a
200 watt bulb and a 100 watt bulb designed to operate on
the same voltage
(a)R
1
is two times R
2
(b) R
2
is two times R
1
(c)R
2
is four times R
1
(d) R
1
is four times R
2
3.A conductor carries a current of 50 m A. If the area of cross-
section of the conductor is 50 mm
2
, then value of the current
density in Am
–2
is
(a) 0.5 (b) 1
(c) 10
–3
(d) 10
–6
4.A cell when balanced with potentiometer gave a balance
length of 50 cm. 4.5 W external resistance is introduced in
the circuit, now it is balanced on 45 cm. The internal
resistance of cell is
(a) 0.25 W (b) 0.5 W
(c) 1.0 W (d) 1.5 W
5.A primary cell has an e.m.f. of 1.5 volt. When short-circuited
it gives a current of 3 ampere. The internal resistance of the
cell is
(a) 4.5 ohm (b) 2 ohm
(c) 0.5 ohm (d) (1/4.5) ohm
6.Two wires A and B of the same material, having radii in the
ratio 1 : 2 and carry currents in the ratio 4 : 1. The ratio of
drift speed of electrons in A and B is
(a) 16 : 1 (b) 1 : 16
(c) 1 : 4 (d) 4 : 1
7.A torch bulb rated as 4.5 W, 1.5 V is connected as shown in
fig. The e.m.f. of the cell, needed to make the bulb glow at
full intensity is
4.5 W,
1.5V
E, r = 2.67 W
0.33 W
2E/9
E/9
E/3
(a) 4.5 V (b) 1. 5 V
(c) 2.67 V (d) 13.5 V

493. 489Current Electricity
8.A flow of 10
7
electrons per second in a conducting wire
constitutes a current of
(a) A106.1
26-
´ (b) A106.1
26
´
(c) A106.1
12-
´ (d) A106.1
12
´
9.A 100-W bulb and a 25-W bulb are designed for the same
voltage. They have filaments of the same length and
material. The ratio of the diameter of the 100-W bulb to that
of the 25-W bulb is
(a) 4 : 1 (b) 2 : 1
(c)
1:2 (d) 1 : 2
10.In a m
etre bridge, the balancing length from the left end
(standard resistance of one ohm is in the right gap) is found tobe 20 cm. The value of the unknown resistance is(a) 0.8 W (b) 0.5 W
(c) 0.4 W (d) 0.25 W
11.An electric fan and a heater are marked as 100 W, 220 V and1000 W, 220 V respectively. The resistance of heater is(a) equal to that of fan (b) lesser than that of fan(c) greater than that of fan (d) zero
12.In a neon gas discharge tube Ne
+
ions moving through a
cross-section of the tube each second to the right is 2.9 ×10
18
, while 1.2 × 10
18
electrons move towards left in the
same time; the electronic charge being 1.6 × 10
–19
C, the
net electric current is(a) 0.27 A to the right (b) 0.66 A to the right(c) 0.66 A to the left(d) zero
13.A capacitor of 10m F has a pot. difference of 40 volts across
it. If it is discharged in 0.2 second, the average currentduring discharge is(a) 2 m A (b) 4 m A
(c) 1 m A (d) 0.5 m A
14.The amount of charge Q passed in time t through a cross-section of a wire is Q = 5 t
2
+ 3 t + 1.
The value of current at time t = 5 s is(a) 9 A (b) 49 A
(c) 53 A (d) None of these
15.A wire X is half the diameter and half the length of a wire Yof similar material. The ratio of resistance of X to that of Y is(a) 8 : 1(b) 4 : 1(c) 2 : 1 (d) 1 : 1
16.2, 4 and 6 S are the conductances of three conductors.When they are joined in series, their equivalent conductancewill be(a) 12 S (b) (1/12) S
(c) (12/11) S (d) (11/12) S
17.2, 4 and 6 S are conductances of three conductors. When theyare joined in parallel, their equivalent conductance will be(a) 12 S (b) (1/12) S
(c) (12/11) S (d) (11/12) S
18.Two identical cells connected in series send 1.0A currentthrough a 5 W resistor. When they are connected in parallel,
they send 0.8 A current through the same resistor. What isthe internal resistance of the cell?(a) 0.5 W(b) 1.0 W(c) 1.5 W (d) 2.5 W
19.If the resistance of a conductor is 5W at 50º C & 7W at
100º C, then mean temperature coefficient of resistance (ofmaterial) is
(a) 0.013/ ºC (b) 0.004/ ºC
(c) 0.006/ ºC (d) 0.008/ ºC
20.If negligibly small current is passed though a wire of length
15 m & resistance of 5W, having uniform cross section of
6 × 10
–7
m
2
, then coefficient of resistivity of material is
(a) 1×10
–7
W–m (b) 2×10
–7
W–m
(c) 3×10
–7
W–m (d) 4×10
–7
W–m
21.A potentiometer consists of a wire of length 4m and
resistance 10W. It is connected to a cell of e.m.f. 3V. The
potential gradient of wire is
(a) 5V/m (b) 2V/m
(c) 5V/m (d) 10V/m
22.The four wires from a larger circuit intersect at junction A
as shown. What is the magnitude and direction of the
current between points A and B ?
5A
4A
6A
A B
]
(a) 2 A f
rom A to B (b) 2A from B to A
(c) 3A from A to B (d) 2A from B to A
23.Potentiometer wire of length 1 m is connected in series with
490W resistance and 2 V battery. If 0.2 mV/cm is the potential
gradient, then resistance of the potentiometer wire is
(a) 4.9 W(b) 7.9 W(c) 5.9 W (d) 6.9 W
24.The deflection in a galvanometer decreases from 25 divisions
to 5 divisions when a resistor of 20W is connected in series.
Find resistance of galvanometer.
(a) 4 W (b) 5 W (c) 6 W (d) 7 W
25.A galvanometer of 50 ohm resistance has 25 divisions. A
current of 4 × 10
–4
ampere gives a deflection of one division.
To convert this galvanometer into a voltmeter having a range
of 25 volts, it should be connected with a resistance of
(a) 2450 W in series.(b) 2500 W in series.
(c) 245 W in series.(d) 2550 W in series.
26.In the equation AB = C, A is the current density, C is the
electric field, Then B is
(a) resistivity (b) conductivity
(c) potential difference(d) resistance
27.A dynamo develops 0.5 A at 6 V. The energy which is
generated in one second is
(a) 0.083 J (b) 3 J
(c) 12 J (d) None of these
28.In an electroplating experiment, m g of silver is deposited
when 4 A of current flows for 2 minutes. The amount in g of
silver deposited by 6 A of current for 40 seconds will be
(a) 4 m (b) 2 m (c) m/2 (d) m/4
29.A steady current of 5 A is maintained for 45 minutes. During
this time it deposits 4.572 g of zinc at the cathode of voltameter.
E.C.E. of zinc is
(a) 3.387 × 10
–4
g/C (b) 3.387 × 10
–4
kg/C
(c) 3.384 × 10
–4
kg/C(d) 3.384 × 10
–3
kg/C

494. 490 PHYSICS
30.What is the equivalent resistance between the points A
and D in given figure?
(a) 10 W (b) 20 W
(c) 30 W (d) 40 W
31.Two wires of same metal have the same length but their
cross-sections are in the ratio 3 : 1. They are joined in series.
The resistance of the thicker wire is 10 W. The total
resistance of the combination is
(a) 5/2 W (b) 40/3 W
(c) 40 W (d) 100 W
32.The resistance of a wire at room temperature 30°C is found
to be 10 W. Now to increase the resistance by 10%, the
temperature of the wire must be [ The temperature coefficient
of resistance of the material of the wire is 0.002 per °C]
(a) 36°C(b) 83°C(c) 63°C (d) 33°C
33.A potentiometer wire, 10 m long, has a resistance of 40W. It
is connected in series with a resistance box and a 2 V storage
cell. If the potential gradient along the wire is 0.1 m V/cm,
the resistance unplugged in the box is
(a) 260 W (b) 760 W
(c) 960 W (d) 1060 W
34.If R
1
and R
2
are the filament resistances of 200 W and a 100
W bulb respectively both designed to run at the same
voltage, then
(a)R
2
is four times of R
1
(b) R
1
is four times of R
2
(c)R
2
is two times of R
1
(d) R
1
is two times of R
2
35.Two heating wires of equal length are first connected in
series and then in parallel to a constant voltage source.
The rate of heat produced in two cases is (parallel to series)
(a) 1 : 4(b) 4 : 1(c) 1 : 2(d) 2 : 1
36.Two identical batteries each of e.m.f. 2 V and internal
resistance 1 W are available to produce heat in an external
resistance by passing a current through it. The maximum
power that can be developed across R using these batteries
is
(a) 3.2 W(b) 2.0 W (c) 1.28 W(d) 8/9 W
37.A heater of 220 V heat a volume of water in 5 minutes time.
A heater of 110 V heats the same volume of water in
(a) 5 minutes (b) 8 minutes
(c) 10 minutes (d) 20 minutes
38.The internal resistance of a primary cell is 4W. It generates
a current of 0.2 A in an external reistance of 21 W. The rate
of chemical energy consumed in providing the current is
(a) 0.42 J s
–1
(b) 0.84 J s
–1
(c) 1 J s
–1
(d) 5 J s
–1
39.Water boils in the electric kettle in 15 minutes after switching
on. If the length of heating wire is decreased to 2/3 of its
initial value, then the same amount of water will boil with
the same supply voltage in
(a) 8 minutes (b) 10 minutes
(c) 12 minutes (d) 15 minutes
40.An electric lamp is marked 60 W, 220 V. The cost of kilo
watt hour of electricity is Rs. 1.25. The cost of using this
lamp on 220 V for 8 hours is
(a) Re 0.25 (b) Re 0.60
(c) Re 1.20 (d) Re 4.00
41.If current flowing in a conductor changes by 1% then power
consumed will change by
(a) 10% (b) 2% (c) 1% (d) 100%
42.If nealy 10
5
coulomb are liberated by 1gm equivalent of
aluminium, then amount of aluminium (equivalent weight 9)
deposited through electrolysis in 20 minutes by a current
of 50 ampere will be :
(a) 0.6 gm. (b) 0.09 gm (c) 5.4 gm (d) 10.8 gm
43.Three equal resistors, connected across a source of e.m.f.
together dissipate 10 watt of power. What will be the power
dissipated in watts if the same resistors are connected in
parallel across the same source of e.m.f.
(a) 10 (b) 10/3(c) 30 (d) 90
44.A wire of radius r and another wire of radius 2r, both of
same material and length are connected in series to each
other. The combination is connected across a battery. The
ratio of the heats produced in the two wires will be
(a) 4.00 (b) 2.00 (c) 0.50 (d) 0.25
45.The electrochemical equivalent of a metal is 3.3 × 10
–7
kg
per coulomb. The mass of the metal liberated at the
cathode when a 3 A current is passed for 2 seconds will
be
(a) 19.8 × 10
–7
kg (b) 9.9 × 10
–7
kg
(c) 6.6 × 10
–7
kg (d) 1.1 × 10
–7
kg
46.An electrical cable of copper has just one wire of radius 9
mm. Its resistance is 5 ohm. This single copper wire of the
cable is replaced by 6 different well insulated copper wires
each of radius 3 mm. The total resistance of the cable will
now be equal to
(a) 7.5 ohm (b) 45 ohm
(c) 90 ohm (d) 270 ohm
47.In an experiment to measure the internal resistance of a cell,
by a potentiometer, it is found that the balance point is at a
length of 2 m, when the cell is shunted by a 5 W resistance
and is at a length of 3 m when the cell is shunted by a 10 W
resistance. The internal resistance of the cell is then
(a) 1.5 W (b) 10 W
(c) 15 W (d) 1 W
48.To get maximum current in a resistance of 3 ohms, one can
use n rows of m cells (connected in series) connected in
parallel. If the total number of cells is 24 and the internal
resistance of a cell is 0.5 ohms then
(a) m = 12, n = 2 (b) m = 8, n = 3
(c) m = 2, n = 12 (d) m = 6, n = 4
49.Each of the resistance in the network shown in fig. is equal
to R. The resistance between the terminals A and B is

495. 491Current Electricity
R
RRR
R
L
K
A
B
M
(a)R (b) 5 R
(
c) 3 R (d) 6 R
50.Four resistors are connected as shown in fig. A 6 V battery
of negligible resistance is connected across terminal AC.
The potential difference across terminals B, D will be
6V
+–
B C D E
A
5 W 15 W 30 W 10 W
(a) 0 V (
b) 1.5 V
(c) 2 V (d) 3 V
51.A non-conducting ring of radius R has charge Q distributedunevenly over it. If it rotates with an angular velocity w, the
equivalent current will be(a) zero (b) Qw
(c)
p
w
2
Q (d)
R2
Q
p
w
52.All the edges of a bloc
k with parallel faces are unequal. Its
longest edge is twice its shortest edge. The ratio of themaximum to minimum resistance between parallel faces is
(a)2 (b) 4
(c)8
(d) indeterminate unless the length of the third edge is
specified
53.In the network shown below, the ring has zero resistance.
The equivalent resistance between the point A and B is
(a) 2R
(b) 4R
(c) 7R
R
A
B
3R
3R
3R
(d) 10R
54.The
belt of an electrostatic generator is 50 cm wide and
travels at 30 cm/sec. The belt carries charge into the sphere
at a rate corresponding to 10
–4
ampere. What is the surface
density of charge on the belt.
(a) s/mC107.6
25--
´ (b) s/mC107.6
24--
´
(c) s/mC107.6
27--
´ (d) s/mC107.6
28--
´
55.A wire has a resistance 12 W. It is bent in the form of a
circle. The effective resistance between two points on any
diameter is
(a) 6 W (b) 3 W (c) 12 W (d) 24 W
56.A 4 m long wire of resistance 8 W is connected in series
with a battery of e.m.f. 2 V and a resistor of 7 W. The internal
resistance of the battery is 1 W. What is the potential
gradient along the wire?
(a) 1.00 V m
–1
(b) 0.75 V m
–1
(c) 0.50 V m
–1
(d) 0.25 V m
–1
57.The length of a given cylindrical wire is increased by 100%.
Due to the consequent decrease in diameter the change in
the resistance of the wire will be
(a) 100%(b) 50% (c) 300% (d) 200%
58.Two resistances R
1
and R
2
are made of different materials.
The temperature coefficient of the material of R
1
is
a and
that of mater
ial of R
2
is –
b. The resistance of the series
combination of R
1
and R
2
will not change with temperature
if
2
1
R
R
equal to
(a)
b
a
(b)
b-a
b+a
(c)
ab
b+a
2
22
(d)
a
b
59.Current I
1
i
n the following circuit is
40W
3
I
40W
30W
40V
80V
I
1
I
2
(a) 0.4A(b) – 0
.4 A (c) 0.8 A (d) – 0.8 A
60.Null point with 1V cell comes out to be 55 cm and with
R = 10
W it is 50 cm. What is the internal resistance of the
cell ?
100 cm
1V
2V
R
(a)0.5W(b)0.4W(c)1W (d)0.2W
61.Three resistances are connected to form a T-shape as shown
in the figure. Then the current in the 4K resistor is:
+2V
– 8V
8K– 4V
4K
2K
(a) 0.93mA (b) 1
.42mA
(c) 2.5mA (d) 1.57mA
62.The equivalent resistance between A and B is
A
B
R
R
R
R R
(a)
5
R8
(b)
8
R5
(c)
8
R3
(d)
8
R7

496. 492 PHYSICS
63.Determine the current in W2 resistor.
W2
W3
W1
6V 2.8 W
(a) 1 A(b) 1.5 A(c) 0.9 A (d) 0.6 A
64.Twelve resistors each of resistance 16W are connected in
the circuit as shown. The net resistance between AB is
(a)1W (b) 2W (c)3W (d) 4W
65.A battery of e.m.f E and internal resistance r is connected
to a variable resistor R as shown. Which one of the
following is true ?
E r
R
(a) Potential difference across the terminals of the battery
is maximum where R = r
(b) Power delivered to resistor is maximum when
R = r
(c) Current in the circuit is maximum when R = r (d) Current in the circuit is maximum when R >> r
66.There is an infinite wire grid with cells in the form of
equilateral triangles. The resistance of each wire between
neighbouring joint connections is R
0
. The net resistance
of the whole grid between the points A and B as shown is
A
B
(a)R
0
(b)
2
R
0
(c)
3
R
0
(d)
4
R
0
67.Twelve indentical resistors each of value 1 W are connected
as shown. Net resistance between C and D (R) is
(a) W=
6
7
R
(b) W=
3
4
R
(c) W=1R
A
B
C
D
E
F
G
H
(d)
3
R
4
=W
68.The numerical value of charge on either plate of capacitor C
shown in figure is
( a ) C E
(b)
rR
CER
1
1
+
(c)
rR
CER
2
2
+
||
1R
2RC
E r
(d)
rR
CER
2
1
+
69.In the circuit shown, the internal resistance of the cell is
negligible. The steady state current in the 2W resistor is
(a) 0.6 A
(b) 0.9 A
(c) 1.2 A
(d) 1.5 A
70.In the network shown, each resistance is equal to R. The
equivalent resistance between adjacent corners A and D is
(a)R
(b)
2
R
3
(c)
3
R
7
(d)
8
R
15
71.Resistances 1 W, 2 W and 3 W are connected to form a
triangle. If a 1.5 V cell of negligible internal resistance is
connected across the 3 W resistor, the current flowing
through this resistor will be
(a) 0.25 A (b) 0.5 A
(c) 1.0 A (d) 1.5 A
72.Seven resistances, each of value 20 W, are connected to a 2
V battery as shown in the figure. The ammeter reading will
be
2V
A
(a) 1/10 A (b) 3/10 A
(c) 4/10 A (d) 7/10 A.

497. 493Current Electricity
73.In the circuit shown below, if the resistance of voltmeter is
4 kW, then the error in the reading of voltmeter will be
(a) 50%(b) 68% (c) 17% (d) 33.3%
74.In the circuit , the galvanometer G shows zero deflection. If
the batteries A and B have negligible internal resistance,
the value of the resistor R will be
G
B AR
2V
12V
W500
(a) W100 (b) W200
(c) W1000 (d) W500
75.A fuse wire with a radius of 1 mm blows at 1.5 A. If the fuse wire of the same material should blow at 3.0 A, the radius of the wire must be
(a)4
1/3
mm (b)
mm2
(c) 0.5 mm (d) 8.0 mm
76.A 4 m F conductor is charged 50 volts and then its plates
are joined through a resistance of 1 k W. The heat produced
in the resistance is
(a) 0.16 J (b) 1.28 J (c) 0.64 J (d) 0.32 J
77.The thermo e.m.f. of a thermocouple is given by
E = 2164 t – 6.2 t
2
. The neutral temperature and a temperature
of inversion are
(a) 349, 174.5 (b) 174.5, 349
(c) 349, 698 (d) 698, 349
78.One junction of a certain thermocouple is at a fixed
temperature T
r
and the other junction is at a temperature T.
The electormotive force for this is expressed by,
ú
û
ù
ê
ë
é
+--=)TT(
2
1
T)TT(kE
r0r
At temperature T = T
0
/2, the thermo electric power is
(a) k T
0
/2 (b) k T
0
(c)
2
Tk
2
0
(d)
1 2
()
0
2
kTT
i
-
79.The e.m.f. developed in a thermo-couple is given by
2
T
2
1
TEb+a=
where T is the temperature of hot junction, cold junction
being at 0ºC. The thermo electric power of the couple is
(a) T
2
b
+a (b) Tb+a
(c)
6
T
2
T
32
b
+
a
(d) ba2/
80.The thermo e.m.f. E in volts of a certain thermocouple is
found to vary with temperature T of hot junction while cold
junction is kept at 0ºC
20
T
T40E
2
-=
The neutral temperature of the couple is (a) 100ºC (b) 200ºC (c) 400ºC (d) 800ºC
81.Two bulbs of 500 W and 200 W are manufactured to operate
on 220 V line. The ratio of heat produced in 500 W and 200
W, in two cases, when firstly they are connected in parallel
and secondary in series will be
(a)
5
2
:
2
5
(b)
2
5
:
2
5
(c)
2
5
:
5 2
(d)
5
2
:
5 2
82.A wire of resistance 20 W is covered with ice and a voltage
of 210 V is applied across the wire, then rate of melting the
ice is
(a) 0.85 g/s (b) 1.92 g/s
(c) 6.56 g/s (d) All of these
83.Two identical batteries, each of e.m.f. 2 volt and internal
resistance 1.0 ohm are available to produce heat in a
resistance R = 0.5 W , by passing a current through it. The
maximum power that can be developed across R using these
batteries is
2V
2V
1W
1W
0.5W
(a) 1.28 W (b) 2.0 W(c) 8/9 W(d) 3.2 W
84.The thermo e.m.f. of a thermocouple is 25mV/ºC at room
temperature. A galvanometer of 40 ohm resistance, capable
of detecting current as low as 10
–5
A, is connected with the
thermocouple. The smallest temperature difference that can
be detected by this system is
(a) 12ºC (b) 0ºC (c) 20ºC (d) 16ºC
85.Two different metals are joined end to end. One end is kept
at constant temperature and other end is heated to a very
high temperature. The graph depicting the thermo e.m.f. is
(a)
T
(b)
T
(c) (d)

498. 494 PHYSICS
86.The cold junction of a thermocouple is maintained at 10ºC.
No thermo e.m.f. is developed when the hot junction is
maintained at 530ºC. The neutral temperature is
(a) 260ºC (b) 265ºC (c) 270ºC (d) 520ºC
87.Three equal resistors connected across a source of e.m.f.
together dissipate 10 watt of power. What will be the power
dissipated in watts if the same resistors are connected in
parallel across the same source of e.m.f.?
(a) 10 (b)
3
10
(c) 30 (d) 90
88.An electric heating element in vacuum is surrounded by a
cavity at constant temperature of 227ºC; it consumes 60W
of power to maintain a temperature of 727ºC. What is the
power consumed by the element to maintain a temperature
of 1227º C?
(a) 101 W(b) 304 W(c) 90 W(d) 320 W
89.Silver and copper voltameters are connected in parallel
with a battery of e.m.f 12 V. In 30 minute 1 g of silver and
1.8 g of copper are liberated. The energy supplied by the
battery is
[ Z
Ag
= 11.2 × 10
–4
gc
–1
; Z
Cu
= 6.6 × 10
–4
gc
–1
]
(a) 720 J (b) 2.41 J
(c) 24.12 J (d) 4.34 × 10
4
J
90.A 5–ampere fuse wire can withstand a maximum power
of 1 watt in the circuit. The resistance of the fuse wire is
(a) 0.04 W(b) 0.2 W(c) 5 W (d) 0.4 W
91.In the Seebeck series Bi occurs first followed by Cu and Fe
among other. The Sb is the last in the series. If z
1
be the
thermo emf at the given temperature difference for Bi – Sb
thermocouple and z, be that for Cu-Fe thermocouple, which
of the following is true?
(a)z
1
= z
2
(b) z
1
< z
2
(c)z
1
> z
2
(d) Data is not sufficient to predict it.
92.50 electric bulbs are connected in series across a 220 V
supply and the illumination produced is I
1
. 5 bulbs are
fused. If the remaining 45 are again connected in series,
the illumination produced is I
2
. Which of the following is
true ?
(a)I
1
= I
2
(b) I
1
< I
2
(c)I
1
> I
2
(d) It will depend on the resistance of each bulb.
93.A leclanche cell supplies a current of one ampere for ten
minutes. The electreochemical equivalent of hydrogen =
0.00001014 gram per coulomb. The mass of hydrogen
liberated is :
(a) 0.00625 g (b) 0.01248 g
(c) 0.01872 g (d) 0.02496 g.
94.Three equal resistors connected across a source of e.m.f.
together dissipate 10 watt of power. What will be the power
dissipated in watt if the same resistors are connected in parallel
across the same source of e.m.f.?
(a) 10/3(b) 10 (c) 30 (d) 90
95.Two 1000 W heaters when connected in parallel across
220 V supply produced heat Q
P
in time t. If they are
connected in series across the same power supply the heat
produced in the same time is Q
S
. What is Q
P
/Q
S
?
(a)4 (b) 2 (c) 0.5 (d) 0.25.
96.In the circuit shown in figure, the 5W resistance
develops 20.00 cal/s due to the current flowing through it.
The heat developed in 2 W resistance (in cal/s) is
(a) 23.8(b) 14.2(c) 11.9(d) 7.1
97.A heating coil is labelled 100 W, 220 V. The coil is cut in half and the two pieces are joined in parallel to the same source. The energy now liberated per second is (a) 25 J(b) 50 J(c) 200 J(d) 400 J
98.Five resistances have been connected as shown in the figure. The effective resistance between A & B is
A B
4W
7W
8W6W
3W
(a) 14/3W (b) 20/3W
(c) 14W (d) 21W
99.If specific resistance of a potentiometer wire is 10
–7
Wm
current flowing through it, is 0.1 amp and cross sectional area of wire is 10
–6
m
2
, then potential gradient will be
(a) 10
–2
volt/m (b) 10
–4
volt/m
(c) 10
–6
volt/m (d) 10
–8
volt/m
100.If 25W, 220 V and 100 W, 220 V bulbs are connected in series across a 440 V line, then (a) only 25W bulb will fuse (b) only 100W bulb will fuse (c) both bulbs will fuse (d) None of these
101.An electric kettle has two heating coils. When one of the coils is connected to an a.c. source, the water in the kettle boils in 10 minutes. When the other coil is used the water boils in 40 minutes. If both the coils are connected in parallel, the time taken by the same quantity of water to boil will be (a) 15 min (b) 8 min(c) 4 min(d) 25 min
102.Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time the combination is connected to a 220 volt a.c. supply line. The power drawn by the combination in each case respectively will be (a) 50 watt, 200 watt(b) 50 watt, 100 watt
(c) 100 watt, 50 watt(d) 200 watt, 150 watt
103.The internal resistance of a 2.1 V cell which gives a current
of 0.2 A through a resistance of 10 W is
(a) 0.5 W (b) 0.8 W
(c) 1.0 W (d) 0.2 W
104.Five equal resistances each of resistance R are connected
as shown in the figure. A battery of V volts is connected
between A and B. The current flowing in AFCEB will be
(a)
R
V2
(b)
R
V3
(c)
R
V
(d)
R2
V
A
B
R
D E
R
R
R
R F
C

499. 495Current Electricity
105.A battery is charged at a potential of 15V for 8 hours when
the current flowing is 10A. The battery on discharge
supplies a current of 5A for 15 hour. The mean terminal
voltage during discharge is 14V. The “Watt-hour”
efficiency of the battery is
(a) 87.5%(b) 82.5% (c) 80% (d) 90%
106.When three identical bulbs of 60 watt, 200 volt rating are
connected in series to a 200 volt supply, the power drawn
by them will be
(a) 20 watt (b) 60 watt (c) 180 watt (d) 10 watt
107.In India electricity is supplied for domestic use at 220 V. It
is supplied at 110 V in USA. If the resistance of a 60 W
bulb for use in India is R, the resistance of a 60 W bulb for
use in USA will be
(a) R/2 (b) R (c) 2R (d) R/4
108.For the network shown in the figure the value of the current
i is
(a)
9
35
V
(b)
18
5
V
(c)
5
9
V
(d)
5
18
V
109.In produc
ing chlorine by electrolysis 100 kW power at 125
V is being consumed. How much chlorine per minute is
liberated? (E.C.E. of chlorine is 0.367×10
–6
kg / C)
(a) 1.76 × 10
–3
kg (b) 9.67 × 10
–3
kg
(c) 17.61 × 10
–3
kg (d) 3.67 × 10
–3
kg
110.In the circuit shown in the figure, if potential at point A is
taken to be zero, the potential at point B is
R
1
2V
B
2A
2AC1A
1V
A
R
2
1A
D
(a) –1V (b) +
2V
(c) –2V (d) + 1V
111.The power dissipated in the circuit shown in the figure is 30
Watts. The value of R is
(a) 20 W
R
10V
5W
(b) 15 W
(c) 10 W
(d) 30 W
11
2.Cell having an emf e and internal resistance r is connected
across a variable external resistance R. As the resistance R
is increased, the plot of potential difference V across R is
given by
(a)
V
R
0
e
(b)
V
R
0
e
(c)
V
R
0
e
(d)
V
R
0
113.A milli voltmete
r of 25 milli volt range is to be converted
into an ammeter of 25 ampere range. The value (in ohm) of
necessary shunt will be
(a) 0.001 (b) 0.01
(c)1 (d) 0.05
114.In the circuit shown the cells A and B have negligible
resistances. For V
A
= 12V, R
1
= 500W and R = 100W the
galvanometer (G) shows no deflection. The value of V
B
is
R
1
V
A R
G
V
B
(a) 4 V (b) 2 V
(c)
12 V (d) 6 V
115.If voltage across a bulb rated 220 Volt-100 Watt drops by
2.5% of its rated value, the percentage of the rated value by
which the power would decrease is
(a) 20% (b) 2.5%
(c) 5% (d) 10%
116.A wire of resistance 4 W is stretched to twice its original
length. The resistance of stretched wire would be
(a) 4 W (b) 8 W
(c) 16 W (d) 2 W
117.The resistance of the four arms P, Q, R and S in a
Wheatstone’s bridge are 10 ohm, 30 ohm, 30 ohm and 90
ohm, respectively. The e.m.f. and internal resistance of the
cell are 7 volt and 5 ohm respectively. If the galvanometer
resistance is 50 ohm, the current drawn from the cell will be
(a) 0.2 A (b) 0.1 A
(c) 2. 0 A (d) 1. 0 A
Directions for Qs. (118 to 125) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following-
(a)Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(b)Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(c)Statement -1 is true, Statement-2 is false
(d)Statement -1 is false, Statement-2 is true
118. Statement 1 : When a battery is short-circuited, the terminal
voltage is zero.
Statement 2 : In the situation of a short-circuit, the current
is zero
119. Statement 1 : A current flows in a conductor only when
there is an electric field within the conductor.
Statement 2 : The drift velocity of electron in presence of
electric field decreases.

500. 496 PHYSICS
120. State
ment 1 : Voltmeter is much better than a potentiometer
for measuring emf of cell.
Statement 2 : A potentiometer draws no current while
measuring emf of a cell.
121. Statement 1 : When current through a bulb decreases by
0.5%, the glow of bulb decreases by 1%.
Statement 2 : Glow (Power) which is directly proportional
to square of current.
122. Statement 1 : Long distance power transmission is done
at high voltage.
Statement 2 : At high voltage supply power losses are
less.
123. Statement 1 : Resistance of 50W bulb is greater than that
of 100 W.
Statement 2 : Resistance of bulb is inversely proportional
to rated power.
124. Statement 1 : 40 W tube light give more light in comparison
to 40 w bulb.
Statement 2 : Light produced is same from same power.
125. Statement 1: In a simple battery circuit, the point of the
lowest potential is negative terminal of the battery.
Statement 2: The current flows towards the point of the
higher potential, as it does in such a circuit from the negative
to the positive terminal.
Exemplar Questions
1.Consider a current carrying wire (current I) in the shape of
a circle.
(a) source of emf
(b) electric field produced by charges accumulated on the
surface of wire
(c) the charges just behind a given segment of wire which
push them just the right way by repulsion
(d) the charges ahead
2.Two batteries of emf e
1
and e
2
(e
2
> e
1
) and internal
resistances r
1
and r
2
respectively are connected in parallel
as shown in figure.
(a) Two equivalent emf e
eq
of the two cells is between e
1
and e
2
, i.e., e
1
< e
eq
< e
2
(b) The equivalent emf e
eq
is smaller than e
1
(c) The e
eq
is given by e
eq
= e
1
+ e
2
always
(d)e
eq
is independent of internal resistances r
1
and r
2
r
1
e
1
A B
r
2
e
2
3.A resistance R is to be measured using a meter bridge,
student chooses the standard resistance S to be 100W. He
finds the null point at l
1
= 2.9 cm. He is told to attempt to
improve the accuracy.Which of the following is a useful way?(a) He should measure I
1
more accurately
(b) He should change 5 to 1000W and repeat the experiment
(c) He should change S to 3W and repeat the experiment
(d) He should given up hope of a more accurate
measurement with a meter bridge
4.Two cells of emfs approximately 5 V and 10 V are to beaccurately compared using a potentiometer of length 400cm.(a) The battery that runs the potentiometer should have
voltage of 8V
(b) The battery of potentiometer can have a voltage of 15
V and R adjusted so that the potential drop across thewire slightly exceeds 10 V
(c) The first portion of 50 cm of wire itself should have a
potential drop of 10 V
(d) Potentiometer is usually used for comparing resistances
and not voltages
5.A metal rod of length 10 cm and a rectangular cross-section
of 1cm ×
1
2
cm is connected to a battery across opposite
faces. The resistance will be
(a) maximum when the battery is connected across 1 cm ×
1
2
cm faces
(b) maximum when
the battery is connected across 10 cm ×
1 cm faces
(c) maximum when the battery is connected across 10 cm ×
1
2
cm faces
(d) same ir
respective of the three faces
6.Which of the following characteristics of electronsdetermines the current in a conductor?(a) Drift velocity alone(b) Thermal velocity alone(c) Both drift velocity and thermal velocity(d) Neither drift nor thermal velocity
NEET/AIPMT (2013-2017) Questions
7.A wire of resistance 4 W is stretched to twice its original
length. The resistance of stretched wire would be [2013]
(a) 4 W (b) 8 W
(c) 16 W (d) 2 W
8.The internal resistance of a 2.1 V cell which gives a current
of 0.2 A through a resistance of 10 W is [2013]
(a) 0.5 W (b) 0.8 W
(c) 1.0 W (d) 0.2 W
9.The resistance of the four arms P, Q, R and S in a
Wheatstone’s bridge are 10 ohm, 30 ohm, 30 ohm and 90
ohm, respectively. The e.m.f. and internal resistance of the
cell are 7 volt and 5 ohm respectively. If the galvanometer
resistance is 50 ohm, the current drawn from the cell will be

501. 497Current Electricity
(a) 0.2 A (b) 0.1 A [2013]
(c) 2. 0 A (d) 1. 0 A
10.A 12 cm wire is given a shape of a right angled triangle ABC
having sides 3 cm, 4 cm and 5 cm as shown in the figure.
The resistance between two ends (AB, BC, CA) of the
respective sides are measured one by one by a multi-meter.
The resistances will be in the ratio of [NEET Kar. 2013]
A
B C
5 cm3 cm
4 cm
(a) 3 : 4 : 5 (b) 9 : 16 : 25
(c) 27 : 32 : 35 (d) 21 : 24 : 25
11.Two rods are joined end to end, as shown. Both have across-sectional area of 0.01 cm
2
. Each is 1 meter long. One
rod is of copper with a resistivity of 1.7 × 10
–6
ohm-centimeter,
the other is of iron with a resistivity of 10
–5
ohm-centimeter.
How much voltage is required to produce a current of 1
ampere in the rods? [NEET Kar. 2013]
V
CuFe
(a) 0.117 V (b ) 0.00145 V
(c) 0.0145 V (d) 1.7 × 10
–6
V
12.Ten identical cells connected in series are needed to heat awire of length one meter and radius ‘r’ by 10ºC in time ‘t’.
How many cells will be required to heat the wire of lengthtwo meter of the same radius by the same temperature intime ‘t’? [NEET Kar. 2013]
(a) 10 (b) 20
(c) 30 (d) 40
13.A potentiometer circuit has been set up for finding theinternal resistance of a given cell. The main battery usedacross the potentiometer wire, has an emf of 2.0 V and anegligible internal resistance. The potentiometer wire itselfis 4m long, When the resistace R, connected across thegiven cell, has values of(i) infinity (ii) 9.5W
The balancing lengths’, on the potentiometer wire are foundto be 3 m and 2.85 m, respectively. The value of internalresistance of the cell is [2014]
(a) 0.25W (b) 0.95W
(c) 0.5W (d) 0.75W
14.In an ammeter 0.2% of main current passes through the
galvanometer. If resistance of galvanometer is G, the
resistance of ammeter will be : [2014]
(a)
1
G
499
(b)
499
G
500
(c)
1
G
500
(d)
500
G
499
15.Two cities are 150 km apart. Electric power is sent from one
city to another city through copper wires. The fall of
potential per km is 8 volt and the average resistance per km
is 0.5 W. The power loss in the wires is : [2014]
(a) 19.2 W (b) 19.2 kW
(c) 19.2 J (d) 12.2 kW
16.The resistances in the two arms of the meter bridge are 5W
and RW, respectively. When the resistance R is shunted
with an equal resistance, the new balance point is at 1.6 l
1
.
The resistance ‘R’ is : [2014]
G
5W RW
A B
l
1 100 – l
1
(a) 10W (b) 15W
(
c) 20W (d) 25W
17.Across a metallic conductor of non-uniform cross section
a constant potential difference is applied. The quantity
which remains constant along the conductor is : [2015]
(a) current (b) drift velocity
(c) electric field(d) current density
18.A potentiometer wire has length 4 m and resistance 8W.
The resistance that must be connected in series with the
wire and an accumulator of e.m.f. 2V, so as to get a potential
gradient 1 mV per cm on the wire is [2015]
(a) 40 W (b) 44 W
(c) 48 W (d) 32 W
19.A, B and C are voltmeters of resistance R, 1.5 R and 3R
respectively as shown in the figure. When some potential
difference is applied between X and Y, the voltmeter readings
are V
A
, V
B
and V
C
respectively. Then [2015]
A
B
C
X Y
(a)V
A
¹ V
B
= V
C
(b) V
A
=
V
B
¹ V
C
(c)V
A
¹ V
B
¹ V
C
(d) V
A
= V
B
= V
C
20.A circuit contains an ammeter, a battery of 30V and aresistance 40.8W all connected in series. If the ammeter has
a coil of resistance 480W and a shunt of 20W, the reading in
the ammeter will be: [2015 RS]
(a) 0.25 A (b) 2A
(c) 1 A (d) 0.5 A
21.Two metal wires of identical dimension are connected inseries. If s
1
and s
2
are the conductivities of the metal wires
respectively, the effective conductivity of the combinationis : [2015 RS]
(a)
12
12
2
s +s
ss
(b)
12
12
s +s
ss
(c)
12
12
ss
s +s
(d)
12
12
2ss
s +s

502. 498 PHYSICS
22.A pote
ntiometer wire of length L and a resistance r are
connected in series with a battery of e.m.f. E
0
and a resistance
r
1
. An unknown e.m.f. E is balanced at a length l of the
potentiometer wire. The e.m.f. E will be given by: [2015 RS]
(a)
0
1
Er
.
(r r)L
l
+
(b)
0
E
L
l
(c)
0
1
LEr
(r r)l+
(d)
0
1
LEr
rl
23.The charge flowing through a resistance R varies with time
t as Q = at – bt
2
, where a and b are positive constants. The
total heat produced in R is: [2016]
(a)
3
aR
6b
(b)
3
aR
3b
(c)
3
aR
2b
(d)
3
aR
b
24.A potentiometer wire is 100 cm long and a constant potential
difference is maintained across it. Two cells are connected
in series first to support one another and then in opposite
direction. The balance points are obtained at 50 cm and 10
cm from the positive end of the wire in the two cases. The
ratio of emf's is : [2016]
(a) 5 : 1 (b) 5 : 4
(c) 3 : 4 (d) 3 : 2
25.A potentiometer is an accurate and versatile device to make
electrical measurements of E.M.F. because the method
involves [2017]
(a) Potential gradients
(b) A condition of no current flow through the
galvanometer
(c) A combination of cells, galvanometer and resistances
(d) Cells
26.The resistance of a wire is 'R' ohm. If it is melted and
stretched to 'n' times its original length, its new resistance
will be :- [2017]
(a)
R
n
(b) n
2
R
(c)
2
R
n
(d) nR

503. 499Current Electricity
EXERCISE - 1
1. (
c)
2. (c)
A
R
l
µ;
So, the resis
tance of the wire will be minimum when
the area of cross-section is maximum and length is
minimum.
3. (d)I = n A e v
d
or
2
d
v 1/rµp
4. (c) As we more from A to B the potential difference across
AC increases and hence the reading of voltmeter alsoincreases. But currnet flowing through ammeterremains almost constant.
5. (a)
6. (a) 7. (c)
8. (a)
9. (b) 10. (b)
11. (b)
(1)1
gg
ggg
IRI R
S SR
nII n
In
=Þ==- --
12. (c) Internal resistance = r, External resistance = nr.
Let terminal voltage = V
then
Er
VEIr VE
(n 1)r
=-Þ =-
+
1n
n
E
V
1n
nE
V
+
=Þ
+
=
13. (a
) According to maximum power theorem, the power in
the circuit is maximum if the value of external resistance
is equal to the internal resistance of battery.
14. (d)
( ) ( )
g
g
I IG
SGG
II nII n1
= ´= ´=
- --
15. (d)Let internal resistance of source = R
Current in coil of resistance
1
R =
1
I =
1RR
V
+
Curren
t in coil of resistance
2
R =
2
I =
2RR
V
+
Further, a
s heat generated is same, so
tRItRI
2
2
21
2
1
=
or
1
2
1
R
RR
V
÷
÷
ø
ö
ç
ç
è
æ
+
= 2
2
2
R
RR
V
÷
÷
ø
ö
ç
ç
è
æ
+
Þ
2
21
)RR(R+ =
2
12
)RR
(R+
Þ
21
2
211
2
RRR2RRRR++
=
22
2 1 2 12
RR +R R +2RRR?
Þ )RR(RR)RR(R
212121
2
-=-
Þ R =
21
RR
16. (c) .
1
2
R
R
P
P
R/1PorR/VP
1
2
2
12
==\µ=
17. (a)
Galvani made the statement “Chemical change can
produce electricity”.
18. (a) Joule effect H = I
2
RT
When current flows heat is produced. But, by heating
conduction current cannot produced.
19. (b)
20. (c)
21. (a) .R
4
1
A2
)2/(
A
R;A/R =
r
=
¢
¢r
=¢r=
ll
l
22. (c)
23
. (b) Iv,Lm
@@
24. (c)
A
R
lr
=
When wir
e is cut into 4 pieces and connected in
parallel.eff.C
R
R P 16P
16
=Þ=
2/R
V
:
16/R
V
:
4/R
V
:
R
V
::P:P:P:P
2222
DCBA
25. (c)
1
1
A
R
lr
= , now l
2
=
2l
1
A
2
= p(r
2
)
2
= p (2r
1
)
2
= 4p r
1
2
= 4A
1
2
R
A2A4
)2(
R
1
1
2 =
r
=
r
=\
ll
\ Resistance is halved, but specific resistance
remains the same.
EXERCISE - 2
1. (c)
Numbers attached for brown, black, green and silver
are 1, 0, 5,
± 10%. Therefore the resistance of given
resistor
%101010
5
±W´= %.10100.1
6
±W´=
2. (b)
2
2
2
1
2
1
P
V
Rand
P
V
R ==
2
100
200
P
P
R
R
2
1
1
2
===\
( Q V = con
stant)
12
R2R=
Hints & Solutions

504. 500 PHYSICS
3. (b)
Current density J = I/A
266
Am11050/1650
---
=´´=
4. (a)
5. (c) r = E / I = 1.5 / 3 = 0.5 ohm.
6. (a) Current flowing through the conductor,
I = n e v A. Hence
11
22
2
dd
2
dd
nev(1)v4 4 1 16
or.
1 v 11nev (2)
p ´
= ==
p
7. (d) Resi
stance of bulb
W==5.0
5.4
)5.1(
R
2
b
Current dra
wn from battery
3
E
33.067.2
E
=
+
=
Share of bul
b =
9
E2
3
E
3
2
=´
5.45.0
9
E2
2
=´÷
ø
ö
ç
è
æ
\ or E = 13.5 V..
8.
(c) 1/106.110t/q
197 -
´´==I A106.1
12-
´=
9. (b)
1
2
2
2
1
2
2
1
P
P
P/V
P/V
R
R
==
W25
W100
d
d
)2/d(/
)2/d(/
2
1
2
2
2
2
2
1
==
pr
pr
=
l
l
1
2
5
10
d
d
1
2
==Þ
10. (d)
( )
P
Q 100
=
-
l
l
or
20
P Q 1 0.25.
100 80
= ´=´=W
-
l
l
11. (b) As R µV
2
/P or Rµ1/P, so resistance of heater is less
than that of fan.
12. (b) Current, )102.1109.2(
1818
´+´=I× 1.6 × 10
–19
towards right.
13. (a) Current
2.0
40)1010(
t
CV
t
q
6
´´
===I A102
3-
´=
14. (c)
dQ
10t3
dt
I= =+
At t = 5s, I = 1 0 × 5 + 3 = 53 A
15. (c) .
D
Ror
)4/D(
R
22
ll
µ
p
r
=
22
yyyxx
22
yyyxy
D /2DR 2
R1D (D / 2)
=´= ´=
ll
ll
16. (c) W=W=W=
6
1
R;
4
1
R;
2
1
R
321
In series;
321
s
RRRR++=
12
11
6
1
4
1
2
1
=++=
÷
ø
ö
ç
è
æ
==s\
11
12
R
1
s
s
17. (a) W=W=W
= 6/1R;4/1R,2/1R
321
In parallel;
12642
R
1
R
1
R
1
R
1
321p
=++=++=
p
p
1
12S
R
\s==
18. (d) C
ase (I) : E + E = (r + r + 5) 1 or 2 E = 2 r + 5 ...(i)
Case (II) :
8.05
rr
rr
E ´÷
ø
ö
ç
è
æ
+
+
´
= or 8.05
2
r
E ÷
ø
ö
ç
è
æ
+=
or E = 0.4 r
+ 4.0 ...(ii)
Multiplying (ii) by 2 and equating with (i), we get
2 r + 5 = 0.8 r + 8 or 1.2 r = 3 or
5.2
2.1
3
r ==
19. (a) [Hin
t Þ R
t
= R
o
(1 + a t)]
5W = R
0
(1 + a × 50) and 7W = R
0
(1 + a × 100)
or
515 02
or 0.0133/ C
7 1 100 150
+a
= a==°
+a
20. (b) (Hin
t Þ
l
A.R
=r= Coefficient o
f resistivity)
21. (a)Hint : Potential gradient =
AB
VVPot.Difference
le
ngth of wire
-
=
l
22. (c) Kirchhoff’s junction rule states that the algebraic sum
of all currents into and out of any branch point is zero :
SI = 0. By convention, the sign of current entering a
junction is positive and current leaving a junction is
negative.
4A + 5 A – 6A + I
AB
= 0, therefore I
AB
= – 3A. The wire
between points A and B carries a current of 3A away
from the junction.
23. (a) Pot. gradient = 0.2mV/cm
=
2
3
10
102.0
-
-
´
=
2
2 10 V/m
-
´
Emf of cell = 2×10
–2
×1m = V102
2-
´ = 0.02 V
A
s per the condition of potentiometer
0.02 (R + 490) = 2 (R) or 1.98 R = 9.8
ÞR =
98.1
8.9
= W9.4

505. 501Current Electricity
24. (b)Case (I) : When resistor is not connected
Using V = IR Þ V = 25 )R(
G
.............. (i)
Case (II) : When resistor is connected
V = 5(20 +
G
R) = 100 + 5
G
R............. (ii)
From (i) and (ii), 20
G
R = 100 Þ
G
R = W5
25. (a)R
g
= 50W, I
g
= 25 × 4 × 10
–4
W = 10
–2
A
Range of V = 25 volts
V = I
g
(HR + R
g
)
W=-=\2450R
I
V
HR
g
g
A B
R
HRR gIg
26. (a)JEJE=s Þ r=
J is current density, E is electric field so B = r = resistivity.
27. (b) Energy produced = V I t = 6 × 0.5 × 1 = 3.0 J.
28. (c) Mass of substance deposited
m Z t Z 4 (2 60) 480 Z and=I=´´´=
m Z 6 40 240 Z or m m /2.=´´==¢¢
29. (a) Electrochemical equivalent,
Ι
4m 4.572
Z 3.387 10 g/C.
t 5 45 60
-
== =´
´´
30. (c) The effective circuit will be as shown in the figure.
Effective resistance of R
2
and R
4
in series,
R' = 10 + 10 = 20 W.
Effective resistance of R
3
and R
5
in series,
R'' = 10 + 10 = 20 W
Net total resistance of R' and R'' in parallel is
1
20 20
R 10.
20 20
´
= =W
+
\Total resistance between A and D
= 10 + 10 + 10 = 30 W.
31. (c) Resistance of a wire = / A.rl
For the same length and same material,
21
21
12
RA 3
or, R 3R
RA1
===
The resistance of thick wire, R
1
= 10 W
The resistance of thin wire = 3R
1
= 3 × 10 = 30 W.
Total resistance = 10 + 30 = 40W.
32. (b) R
t
= R
0
(1 + at)
Initially, R
0
(1 + 30a) = 10 W
Finally, R
0
(1 + at) = 11 W
111t
10 1 30
+a
\=
+a
or, 10 + (10 × 0.002 × t) = 11 + 330 × 0.002
or, 0.02t = 1 + 0.66 = 1.066 or
1.66
t 86 C.
0.02
= =°
33. (b) Potential gradient along wire
potential difference along wire
length of wire
=
3I 40
or, 0.1 10 V / cm
1000
- ´
´=
1
or, Current in wire, I= A
400
21
or, or R 800 40 760
40 R 400
= =-=W
+
34. (c)As R= V
2
/P or Rµ1/P
so R
2
/R
1
= P
1
/P
2
= 200/100 = 2
or R
2
= 2 R
1
.
35. (b) If R is the resistance of each wire, total resistance in
series = R + R = 2 R; and total resistance in parallel
.
2
R
RR
RR
=
+
´
=
Heat produced per second (= V
2
/R) will be four times
in parallel than in series.
36. (b) For maximum current, the two batteries should be
connected in series. The current will be maximum when
external resistance is equal to the total internal
resistance of cells i.e. 2 W. Hence power developed
across the resistance R will be
=
R
r2R
E2
R
2
2
÷
÷
ø
ö
ç
ç
è
æ
+
=I .W22
22
22
=´÷
ø
ö
ç
è
æ
+
´
=
37. (d) Heat produced, .
R
tV
H
2
= When voltage is halved,
the heat produced becomes one fourth. Hence time
taken to heat the water becomes four time.
38. (c) Chemical energy consumed per sec
= heat energy produced per sec.
Ι
-
= += +=
221
(R r) (0.2) (21 4) 1Js .
39. (b) =´´=´
22
VV
H 15 60 t
R (2/3)R
or
2
t 15 60 600 s 10 minutes.
3
=´´==

506. 502 PHYSICS
40. (b
) Energy consumed per day = P × t = 60 × 8 = 480 watt
hour = 480/1000 = 0.48 kWh or unit of electricity.
Hence the cost = 0.48 × 1.25 = Re 0.60.
41. (b) As ΙΙ==
22
1P R, so P (1.01 )R
2
1.02 1.02 .RP= I=
It means % i
ncrease in power
%.21001
P
P
1
=´÷
ø
ö
ç
è
æ
-=
42. (c) (Hin
t : m = Z it, where Z is electro chemical equivalent)
43. (d) (Hint : P = V
2
/R)
44. (a) H = I
2
Rt. Here
1
2
l
R
r
=r
p
and
()
2
2
l
R
2r
=r
p
That is, R
1
= 4R
2
. H ence,
1
2
H
4
H
=
45. (a) m = ZIt.
46
. (a) Let the resistance of single copper wire be R
1
. If r is
the specific resistance of copper wire, then
2
11
1
rA
R
p
´r
=
´r
=
ll
...(1)
When the w
ire is replaced by six wires, let the
resistance of each wire be R
2
. Then
2
22
2
rA
R
p
´r
=
´r
=
ll
...(2)
From eqs. (1) and (2), w
e get
23
23
2
2
1
2
2
2
1
)109(
)103(
R
5
or
r
r
R
R
-
´
´
==
; W=45R
2
These six wires a
re in parallel. Hence the resistance of
the combination would be W=5.7R
2
47. (b) In case of internal resistance measurement by
potentiometer,
)rR(R
)rR(R
)}rR/(RE{
)}rR/(RE{
V
V
12
21
22
11
2
1
2
1
+
+
=
+
+
==
l
l
Here W===5R, m3,m2
121
ll and W=10R
2
)r5(10
)r10(5
3 2
+ +
=\
or 20 + 4 r = 3
0 + 3 r or r = 10 W
48. (a) Let, we connect 24 cells in n rows of m cells, then if I is
the current in external circuit then
Rn/mr
mE
I
+
= ...(1)
For I to be maximum, (mr
+ nR) should be minimum.
It is minimum for
n
mr
R= ...(2)
So maximum curre
nt in external circuit is
R2
mE
I= ...(3)
here R =3, r = 0.5 so equatio
n (2) become
m
=6
n
so n = 2, m = 12
49. (a) The equivalent circuit is shown in fig. Since the
Wheatstone’s bridge is balanced, therefore no current
will flow through the arm KL. Equivalent resistance
between
AKM = R + R = 2 R
Equivalent resistance between ALM = R + R = 2 R
The two resistances are in parallel. Hence equivalent
resistance between A and B is given by
R R
RR
R
A MB
L
K
R
1
R2
2
R2
1
R2
1
R
1
==+=
¢
i.e., RR=¢
50. (a) The given figure is a circuit of balanced Wheatstone’s
bridge. Point B and D would be at the same potentiali.e., potential difference between these points is zero.
51. (c) With each rotation, charge Q crosses any fixed point
P near the ring. Number of rotations per second = w/2p.
\ charge crossing P per second = current =
p
w
2
Q
52. (b) Let th
e edges be 2l, a, and l, in decreasing order.
a2a2
R,
a
2
a
2
R
minmax
r
=r=
r
=r=
l
l
l
l
;
.4
R
R
min
max
=
53. (a) As
the ring has no resistance, the three resistances of
3R each are in parallel.
1111
R 3R 3R 3R
Þ=++
¢

1
RR
R
¢=Þ=
\ between point A and B equivalent resistance
= R+ R = 2R

507. 503Current Electricity
54. (b) ).50.030.0/(10A/J
4
´=I=
-

2424
Am107.6s/mC107.6
----
´=´=
55. (b) Resistance of the wire of a semicircle = 12/2 = 6W
For equivalent resistance between two points on any
diameter, 6W and 6W are in parallel.
or
If a wire of resistance R is bent in the form of a circle,
the effective resistance between the ends of a diameter
= R/4.
56. (d) Current in the potentiometer,
.A
8
1
178
2
=
++
=I
Voltage dro
p across potentiometer wire
V18
8
1
=´=
\ Potentia
l gradient of potentiometer wire
m/V25.0
4
1
==
57. (c)
2 2
11
22
R R1
R R2
æö æö
= Þ= ç÷ç÷
èø¢èø
l
l
R3RorR4R=D=¢Þ
\ % chan
ge in resistance of wire
%,300100
R
R3
=´=
58. (d) R
1

+ R
2
= Constant, R
1
will increase, R
2
will decrease.
1
R TR T0aD-bD=
12
RTRTÞ aD = bD
1 2
R
R b
\=
a
59. (b)
A B
F C
E D
80V
1
I
2I 3
I
W40
W40
40V
W30
Applying 1
st
law of Ki rchoff’s, I
3
= I
1
+ I
2
… (i)
Applying 2nd law of Kirchhoff
in Mesh ABCFA
13
30I 40 40I 0-+-= … (ii)
in Mesh
FCDEF
32
40I 40 8040I 0--+= … (iii)
putti
ng
3
I from (i) in (ii) and (iii) we get
0I40I4040I30
211
=--+-
40I40I70,or
21
-=--
or,
12
7I4I4+= … (iv)
and 0I
408040I40I40
221
=+--+
120I80I40,or
21
=+ 12I8I4,or
21
=+
()v..............6I4I2,or
21
=+
()() 2I5givesviv
1
-=- 4.0I
1
-= AA
60. (c) Length of wire 100cm.=
=V2voltage of battery, =V1voltage of a cell
When current flows along wire electric potential falls
continuously along wire.
V2V
A
E1V
B
100cm
Potent
ial gradient = fall in potential length,
l
E
K=
because
in null point position, VE
=
\
()1..............
R
R
l
l
E
E
2
1
2
1
2
1
==
For inte
rnal resistance,
21
Er,EE==
iriRE +=
( )R/ViirVE ==- Q R1
V
E
r ÷
ø
ö
ç
è
æ
-=
.R1
l
l
r
2
1
÷
÷
ø
ö
ç
ç
è
æ
-=Þ ÷
ø
ö
ç
è
æ
=÷
ø
ö
ç
è
æ
-=
50
5
101
50
55
10 W=1
61. (d) Le
t potential of X be x.
Now current from AX and BX will pass through XC.+2V
–8V
8K–4V
4K
2KX
A B
C
so
4
8x
8
x4
2
x2 +
=
--
+
-
4
2
x
4
x4
x2 +=
--
+- 2
2
x
4
x
1x +=---

508. 504 PHYSICS
3
2
x
4
x5
=--
7
12
xor3
4
x7
-=Þ=-
So current i
n 4K will be
57.1
7
11
28
44
4
8
7
12
===
+-
A
62. (b) The equ
ivalent circuit can be redrawn as

BA
· ·
RR
R
R
R
º·
R
R
R
2R
º
R
R
3/R2
º
R
3/R5
º
8/R5
63. (c) At ste
ady state the capacitor will be fully charged and
thus there will be no current in the
W1 resistance. So
the ef
fective circuit becomes
W2
W3
W8.2
1
I
2
I
I
V6
BA
Net current fro
m the 6V battery,
1
8.2
32
32
6
I
+÷
ø
ö
ç
è
æ
+
´
= A5.1
2
3
8.22.1
6
==
+
=
Between A and B, volt
age is same in both resistances,
21
I3I2= where 5.1III
21
==+
Þ A9.0I)I5.1(3I2
111
=Þ-=
64. (d)
R
R
R
R
R
R
R
R
R
R
R
A
B
R/3
R/3 R/3
R/3A B
R
net
between AB = W==
+
´
4
R4
R
3
R
3
R3
3
R
3
R3
2
65. (b,c)
rR2)rR(
E
rR
E
I
2
+-
=
+
=
I is maximum when R = r
RIP
2
=
, when I is max, P is also max.
RIP
2
maxmax
= .
66. (c) By principle of symmetry and superposition,
0
0 eq. eq.
RI
2RIRR
63
´´= Þ=
(Current
6
I
in AB is due to divis
ion in current entering
at A and current
6
I
is due to cu
rrent returning from
infinity of grid).
67. (d) Current in arm AE and FB is zero.
So,
A
B
C D
E
F
G H
68. (b) Current in branch of capacitor is zero.
R
net
= R + r
rR
E
I
+
= potential acro
ss capacitor is equal to
potential across resistance R
1
.
1
1
1 R
rR
E
V
+
=
So charge = CV
1
=
rR
ECR
1
+
69. (b) Give
n Capacitance of the capacitor, C = 0.2 µF and
e.m.f. of cell, E = 6V.
Reactance of a capacitor for a cell, which is a DC source,
is infinity. Therefore, no current flows in 4W resistance.

509. 505Current Electricity
Resistances 2 W and 3 W (both in upper arm) are
connected in parallel. Therefore, their equivalent
resistance
()
23
R 1.2
23
´
¢= =W
+
Now, R¢ and 2.8 W are in series combination.
Therefore, equivalent resistance of the circuit,
R = R¢ + 2.8 = 1.2 + 2.8 = 4 W
Current drawn in the circuit,
E6
I 1.5A
R4
= ==
Therefore, potential difference across 2 W resistance,
V = IR = 1.5 × 1.2 = 1.8 V
Thus, current in 2 W resistance ()1
V 1.8
I 0.9A
22
===
70. (d) The equivalent circuit is as shown in figure.
The resistance of arm AOD (= R + R) is in parallel to
the resistance R of arm AD.
Their effective resistance
1
2RR2
RR
2RR3
´
==
´
The resistance of arms AB, BC and CD is
2
28
RRRRR
33
=+ +=
The resistance R
1
and R
2
are in parallel. The effective
resistance between A and D is
12
3
12
28
RR
RR 833
R R.
28R R 15
RR
33
´
´
===
+
+
71. (b) Equivalent resistance between A and B = series
combination of 1 W and 2 W in parallel with 3 W
resistor..
1112
R333
=+= or R = 1.5 W.
\ Current in the circuit is I = V/R = 1.5/1.5 = 1A.
Since the resistance in arm ACB = resistance in arm AB
= 3 W, the current divides equally in the two arms. Hence
the current through the 3 W resistor = I/2 = 0.5 A.
72. (c) 73. (d)
74. (a)
A
R
12V
2V
i
W500
12 – 2 = i)500(W Þ
50
1
500
10
i ==
Again, i =
50
1
R500
12
=
+
Þ 500 + R = 600 Þ R = 100W
75. (a) The temperature of the wire increases to such a value
at which, the heat produced per second equals heat
lost per second due to radiation i.e.
l
l
r2H
r
2
2
p´=
÷
÷
ø
ö
ç
ç
è
æ
p
r
I , where H is heat lost per
second per unit area due to radiation. Hence,
32
rµI
so ( )
3/2
1212
3
2
3
1
2
2
2
1
/rror
r
r
II==
I
I
.mm4)5.1/0.3(1
3/13/2
=´=
76. (d) The energy stored in the capacitor
2
CV
2
1
= ; This
energy will be converted into heat in the resistor.
77. (b) At neutral temperature, dE/dt = 0;
so 2164 – 6.2 × 2 × t
n
= 0
or t
n
= 174.5ºC.
At temperature of inversion, E = 0
Þ 2164 t
i
– 6.2
2
i
t= 0 Hence t
i
= 349ºC.
78. (a)
éù
= --- ++êú
ëû
22
r rr
0 r0
TT TT TT
E k TT TT
22 22
Hence )TT(k
2
T
2
T
TTk
dT
dE
0
rr
0
-=
ú
û
ù
ê
ë
é
+--=
At temp.T = T
0
/2,
thermoelectric power = k (T
0
– T
0
/2) = k T
0
/2.

510. 506 PHYSICS
79. (b)
Thermo-electric power
T
dT
dE
S b+a==
80. (c) Given E = 40T
–
2
T
.
20
At neutral temperature,
odET
0or 40 0or T 400 C .
dT 10
= -==
81. (a) In p
arallel,
2
5
200
500
P
P
tP
tP
H
H
2
1
2
1
2
1
====
In series,
2 2
11 11
22
2222
RtH R V /P
H R t R V /P
I
= ==
I
.
5 2
500 200
P
P
1
2
===
82. (c)
JRL
tV
morRJ/tVLmH
2
2
===
.s/g56.6
80202.4
1)210(
2
=
´´
´
=
83. (b) Total in
ternal resistance of two cells
W=
+
´
= 5.0
11
11
.
Since internal r
esistance of coil is equal to external
resistance (= 0.5 W), hence power developed is
maximum by cells in circuit.
Current through
A2
5.05.0
2
R =
+
=
Power .W25.
0)2(
2
=´=
84. (d) V1040IRV
5
ggg
-
´==;
.Cº16
1025
104
V
V
6
4
g
=
´
´
==q
-
-
85. (d) At the heigh
est point O
­E
B A
temperature
=E maximum and A shows temperature of inversion
at which emf changes in sign.
86. (c) Neutral temperature,
.Cº270
2
10530
2
0i
n =
+
=
q+q
=q
87. (d) In serie
s, Equivalent resistance = 3R
R3
V
Power
2
= Þ Þ=
R3
V
10
2
V
2
= 30R
In parallel,
R
1
R
1
R
1
R
1
++=
¢
=
R
3
\Equivalent resis
tance
3
R
R=¢
\
3/R
R30
R
V
Power
2
=
¢
= = 90 W
88. (d)==s-
44
12
H
P (T T)
t
T
1
= 727 + 273 = 1000 K, T
2
= 227 + 273 = 500 K,
P = 60 watt
In
the second case, T
1
'
= 1227 + 273 = 1500 K,
2
T' = 500 K, P' = ?
¢ --
==
--
4 4 44
4 4 44
P (1500) (500) (50 0) [3 1]
P(1000) (500) (500) [2 1]
Þ
15
80
ÞP'

= 320
15
80
60 =´watt.
89. (d) m
Cu
= Z
Cu
I
Cu
t
m
Ag
= Z
Ag
I
Ag
t
tZ
m
tZ
m
III
Ag
Ag
Cu
Cu
AgCu
+=+=
ú
ú
û
ù
ê
ê
ë
é
+===
Ag
Ag
Cu
Cu
Z
m
Z
m
VVItEnergyPt
44
1.81
12
6.6 10 11.2 10
--
éù
=+
êú
´´ëû
=12 × 10
4
[0.362] = 4. 34 × 10
4
J.
90. (a)R
P
I
= = =
2
1
25
0 04.W
91. (c)
92. (b
) Resistance of 45 bulbs in series is less than that of 50
bulbs. Since illumination is proportional to the heat
produced
2
Vt
R
æö
ç÷
ç÷
èø
, therefore it will be more with 45
bulbs.
93. (a) m = ZIt = 0.0000104 × 1 × 600 g
94. (d) R
s
= 3R and
P
R
R
3
=
2
S
V
P 10W
3R
== and
P
P
= V
2
/(R / 3) =
3V
2
/R = 9P
S
= 90 W.
95. (a) P
P
= P
1
+ P
2
and P
S
= P
1
P
2
/ (P
1
+ P
2
)
Heat produced = Pt.
Hence,
pP
SS
PQ 2000W
4.
Q P 500W
===

511. 507Current Electricity
96. (b) Let I
1
be the current throug 5 W resistance, I
2
through
(6 + 9) W resistance. Then as per question,
2
11
I 5 20 or,I 2A.´==
Poten
tial difference across C and D = 2 × 5 = 10V
Current
2
102
I A.
693
==
+
Heat produc
ed per second in 2 W
2
28
I R 2 14. 2cal / s.
3
æö
= ´=
ç÷
èø
97. (
d) Power of heating coil = 100 W and voltage
(V) = 220 volts. When the heating coil is cut into two
equal parts and these parts are joined in parallel, then
the resistance of the coil is reduced to one-fourth of
the previous value. Therefore energy liberated per
second becomes 4 times i.e., 4 × 100 = 400 J.
98. (a)Hint : The wheatstone bridge is balanced, when
P/Q = R/S, In the this case 34
68
=, so bridge is balanced
& 7W resistance is not effective)
99. (a) Potential gradient
7
AB
6
VV i 0.1 10
A 10
-
-
- ´r´
= ==
l
2
10 V/m
-
=
100. (
a) As for an electric appliance
2
s
R (V /P)= , so for same
specified voltage V
s
,
4
25
100
R
R
100
25
==
i.e, RRwith
R4R
10025
==
Now, in series potential divides in proportion to
resistance.
So,V
)RR(
R
V
21
1
1
+
= i.e., V352440
5
4
V
25 =´=
and, V
)RR(
R
V
21
2
2
+
= i.e., V88440
5 1
V
100 =´=
From this it is c
lear that voltage across 100 W bulb
(= 88 V) is lesser than specified (220 V) while across 25
W bulb (=352 V) is greater than specified (220 V), so
25 W bulb will fuse.
101. (b)
min8
50
400
4010
4010
Time ==
+
´
=
102. (a)
Resistance
1
Powerµ
In series c
ombination, Resistance doubles
Hence, Power will be halved.
In parallel combination, resistance halved
Hence, power will be double.
103. (a)
104. (d) A balanced Wheatstone’s bridge exists between
A & B. R
eq
= R
current through circuit is V/R
current through AFCEB = V/2R
105. (a) Efficiency is given by
output Power
input Power
h= 75.8
15810
14155
=
´´
´´
= or 87.5 %
106.
(a)
60
3
P
1
or
P
1
P
1
P
1
P
1
eq321eq
=++= Þ P
eq
= 20 watt.
10
7. (d)
60
)110(4
60
)220(
P
V
R
R
V
P
2222
===Þ=;
4
R
60
)110(
R
2
==¢
108. (d)
It is balanced Wheatstone bridge. Hence bridge 4W
can be eliminated.
\
R
eq
=
´
+
=
69
69
18
5
\
I
V
R
V
eq
= =
5
18
109. (c)
I =
P
V
=
3
100 10
125
A
´
5
10
60
A=
E.C.E. = 0.3
67 × 10
–6
kg C
–1
Charge per minute
= (I × 60) C =
5
10 60
125
C
´
6
6 10
125
C
´
=
\ Mass liberated =
6
66 10
0.367 10
125
-´
´´
=
3
6 1000 0.367 10
125
-
´´´
= 17.
616 × 10
–3
kg
110. (d) Current from D to C = 1A
\ V
D
– V
C
= 2 × 1 = 2V
V
A
= 0 \ V
C
= 1V, \ V
D
– V
C
= 2
Þ V
D
– 1 = 2 \ V
D
= 3V
\ V
D
– V
B
= 2 \ 3 – V
B
= 2 \ V
B
= 1V
111. (c) The power dissipated in the circuit.
2
eq
V
P
R
=
...(i
)
v = 10 volt
eq
1 11
5RR
=+ =
5
5
+R
R
R
eq
=
5
5
æö
ç÷
èø+
R
R
P = 30 W

512. 508 PHYSICS
Substituting the values in equation (i)

2
(10)
30
5
5
=
æö
ç÷
èø+
R
R
15
10
5
=
+
R
R
Þ 15R = 50 + 10R
5R = 50 Þ R = 10 W
112. (c) The current through the resistance R

eæö
=ç÷
èø+
I
Rr
The potential difference across R
eæö
==ç÷
èø+
VIRR
Rr
rI
e
R

1
e
=
æö
+ç÷
èø
V
r
R

V
R
0
e
when R = 0, V = 0,
R = ¥, v = e
Thus V increases as R increases upto certain limit, but
it does not increase further.
113. (a) Galvanometer is converted into ammeter, by connected
a shunt, in parallel with it.
I
G
S
3
25 10
25
G
VGS
G + SI
-
´
==
0.001
GS
G + S
=W
Here S << G so S = 0.001 W
114. (b) R
1
V
A
G
V
B
R
Since deflection in galvanometer is zero so current
will flow as shown in the above diagram.
current
1
=
A
V
I
RR+
=
12 12
500 100 600
=
+
So V
B
= IR =
12
100 2
600
V´=
115. (c) Resistance of bulb is constant
2
V
P
R
= Þ
2p VR
pVR
D DD
=+
p
p
D
= 2 × 2.5 + 0 = 5%
116. (c) Resistance R =
A
rl
Q l¢ = 2 l
\ A¢ =
A
2
\ R¢
2
A
2
r
l
= 4R = 4 × 4 W = 16 W
Therefore the resistance of new wire becomes 16 W
117. (a) Given : V = 7 V
r = 5W
P Q
5W
7V
R
eq
=
40 120
40 120
´
+
W
I =
V
R
=
7
40 120
5
40 120
´
+
+
=
7
5 30+
=
1
5
= 0.2 A.A.
118. (c) In the case of a short-circuited battery, the current
E (emf of the battery)
I0
r (internal resistance)
=¹
Terminal voltage V = IR = I (i) = I (0) = 0
where R = external resistance = 0
119. (c) Before the presence of electric field, the free electrons
move randomly in the conductor, so their drift velocity
is zero and therefore there is no current in the
conductor. In the presence of electric field, each
electron in the conductor experience a force in a
direction opposite to the electric field. Now the free
electrons are accelerated from negative and to the
positive end of the conductor and hence a current
starts to flow from the conductor.

513. 509Current Electricity
120. (d) Potentiometer is preferred for measurement of emf of
cell. Potentiometer draws no current from cell while
voltmeter draws some current, therefore emf measured
by voltmeter is slightly less than actual value of emf
of cell. Further the potentiometer is used with a
galvanometer which is set to null reading when the
experiment is performed. The method of null reading
avoids many errors.
121. (a) Glow = Power (P) = I
2
R
%15.02
I
dI
2
P
dP
=´=÷
ø
ö
ç
è
æ
=\
122. (
a) Power loss = i²R =
2
P
R
V
æö
ç÷
èø
[P = Transmitt
ed power]
123. (a)
2
V
P
R
=;
1
R
P
µ (same rated voltage)
124. (c) In tube light majority portion of radiation comes under
visible region while bulb radiation consists of visible,
ultraviolet, infrared radiation giving less visible part.
125. (c) Positive terminal of a battery is point of highest
potential and current flows from highest to lowest
potential i.e. from +ve to –ve potential.
EXERCISE - 3
Exemplar Questions
1. (b) As we know, electric current per unit area
I/A, is called current density j i.e.,
I
j
A
=
The SI units of the current density are A/m
2
.
The current density is also directed along E and isalso a vector and the relationship is
j =sE
Current density changes due to electric field producedby charges accumulated on the surface of wire.
2. (a) As we know the equivalent emf (e
eq
) in the parallel
combination
e
eq
=
21 12
12
rr
rr
e +e
+
So acco
rding to formula the equivalent emf e
eq
of the
two cells in parallel combination is between e
1
and e
2
.
Thus (e
1
< e
eq
< e
2
).
3. (c) Adjusting the blance point near the middle of the
bridge, i.e.. when l
1
is close to 50 cm. requires a suitable
choice of S, R is unknown resistance :
Since,
R
S
=
( )
1
1
R
R 100-
l
l
R
S
=
1
1100-
l
l
or R =
1
1
S
100
éù
êú
-
ëû
l
l
R =
2.9
S
97.1
éù
êú
ëû
So, here, R : S = 2.9 : 97.1 implies that the S is nearly 33
times to that of R. In orded to make this ratio 1 : 1 it is
necessary to reduce the value of S nearly
1
33
times
i.e., ne
arly 3 W,
4. (b) The potential drop across wires of potentiometer
should be more than emfs of primary cells. Here,
values of emfs of two cells are given as 5V and 10V, so
the potential drop along the potentiometer wire must
be more than 10V. So battery should be of 15V and
about 4V potential is droped by using variable
resistance.
5. (a) As we known that the resistance of wire is R =
A
r
l
For maximum value of R, l must be higher and A should
be lower and it is possible only when the battery is
connected across area of cross section =
1
1cm
2
æö
´ç÷
èø
cm.
6. (a
) We know that the relationship between current and
drift speed is
I = ne Av
d
Where, I is the current and V
d
is the drift velocity.
So, I µ V
d
Hence, only drift velocity determines the current in a
conductor.
NEET/AIPMT (2013-2017) Questions
7. (c) Resistance R =
A
rl
Q l¢ = 2 l
\ A¢ =
A
2
\ R =
2
A
2
r
l
= 4R = 4 × 4 W = 16 W
Therefore the resistance of new wire becomes 16 W
8. (a) Given : emf e = 2.1 V
I = 0.2 A, R = 10W
Internal resistance r = ?
From formula.
e – Ir = V = IR
2.1 – 0.2r = 0.2 × 10
2.1 – 0.2 r = 2 or 0.2 r = 0.1
0.1
r
0.2
Þ= = 0.5 W
ALTERNA
TE : i =
rR
e
+
Þ 0.2 =
2.1
r 10+
Þ 2.1 = 0.2 r + 2Þr =
1
2
= 0.5 W

514. 510 PHYSICS
9. (a) Given : V = 7 V
r = 5W
P Q
5W
7V
R
eq
=
40 120
40 120
´
+
W
I =
V
R
=
7
40 120
5
40 120
´
+
+
=
7
5 30+
=
1
5
= 0.2 A.A.
10. (c) Resistance is directly proportional to length
111
3 45
AB
R
=+
+
=
(45)3
(3)(4 5)
++
+
R
AB
=
3 (4 5) 27
3 (4 5) 12
´+
=
++
Similarly,
R
BC
=
4 (3 5) 32
4 (3 5) 12
´+
=
++
R
AC
=
5 (3 4) 35
5 (3 4) 12
´+
=
++
\ R
AB
: R
BC
: R
AC
= 27 : 32 : 35
11. (a) Copper rod and iron rod are joined in series.
\ R = R
Cu
+ R
Fe
= (r
1
+ r
2
)
A
l
R
A
æö
=rç÷
èø
l
Q
From ohm’s law V = RI
= (1.7 × 10
–6
× 10
–2
+ 10
–5
× 10
–2
) ¸ 0.01 × 10
–4
volt
= 0.117 volt (Q I = 1A)
12. (b) Resistance is directly proportionl to length of the wire.
As length is doubled so mass is doubled and
resistance is doubled.
We have
22
(10) ()
, Now (2 )
2
E nEt
t mST mST
RR
=D =D
Þ
22 22
10
2
2
nEt Et
RR
=
Þ n = 20
13. (c) Internal resistance of the cell,
r =
EV
R
V
-æö
ç÷
èø
=
12
2
R
æö-
ç÷
èø
ll
l
=
3 2.85
(9.5) 0.5
2.85
-æö
´ W=W
ç÷
èø
14. (c) As 0.2% of main current passes through the
galvanometer hence
998
I
1000
current through the
shunt.
G
S
I 2I
1000
998I
1000
2I 998I
GS
1000 1000
æöæö
=
ç÷ç÷
èøèø
Þ S =
G
499
Total resistance of Ammeter
R =
G
G
SGG 499
GS G 500
G
499
æö
ç÷
èø
==
+ æö
+
ç÷
èø
15. (b) Total resistance R = (0.5 W/km) × (150 km)
= 75 W
Total voltage drop = (8 V/km) × (150 km)
= 1200 V
Power loss =
22
( V) 1200)
W
R 75
D(
=
= 19200 W = 19.2 kW
16. (b) This is a balanced wheatstone bridge condition,
11
11
1.655
and
R 100 R / 2 100 1.6
==
--
ll
ll
ÞR = 15 W
17. (a) Here, metallic conductor can be considered as the
combination of various conductors connected in
series. And in series combination current remains
same.
i
V
18. (d) Total potential difference across potentiometer wire
= 10
–3
× 400 volt = 0.4 volt
potential gradient =
1mv
cm
= 10
–3
v/cm = 10
–1
v
m
Let resistance of RW connected in series.
2V
i
+0.4V
RW 8W

515. 511Current Electricity
So,
–1
2 10 41
R 8 8 20
´
==
+
ÞR + 8 = 4
0 or,R = 32 W
19. (d) Effective resistance of B and
2
BC
BC
RR 1.5R 3R 4.5R
R
R R 1.5R 3R 4.5R
× ´
= = ==
++
i.e., equal to r
esistance of voltmeter A.
1.5R
B
C
3R
A
R
In parallel potential difference is same so,
V
B
= V
C
and in series current is same
So, V
A
= V
B
= V
C
20. (d) From circuit diagram
20W
40.8W
480W
A
E = 30V
Resistance of ammeter =
480 20
480 20
´
+
= 19.2W.
T
otal resistance R = 40.8 + 19.2 = 60W
Reading in the ammeter i =
V
R
=
30
40.8 19.2+
= 0.5A
21.
(d) In figure, two metal wires of identical dimension are
connected in series
A
l l
s
1
s
2
R
eq
=
eq
1 2 eq eq
AAA
lll
+=
sss
eq
2
A
l
s
=
A
l12
12
æös +s
ç÷
ss
èø
\ s
eq
=
12
12
2ss
s +s
22. (a)EMF, E = Kl where K =
V
L
potential gradient
K =
V
L
=
iR
L
=
0
1
Er
rrL
læö
ç÷
+èø
So, E = Kl =
0
1
Er
(r r )L
l
+
23. (a)Give
n: Charge Q = at – bt
2
\Current i =
Q
t
¶
=
¶
a – 2bt
{for i = 0Þt = a
2b
}
From jo
ule's law of heating, heat produced
dH = i
2
Rdt
H =
( )
a/2b
2
0
a 2bt Rdt-
ò
H =
( )
a
3 32b
0
a 2bt R aR
3 2b 6b
-
=
-´
24. (d)
When two cells are connected in series i.e., E
1
+ E
2
the balance point is at 50 cm. And when two cells are
connected in opposite direction i.e., E
1
– E
2
the balance
point is at 10 cm. According to principle of potential
12
12
EE 50
E E 10
+
=
-
Þ
1 2
2E
50 10
2E 50 10
+
=
-
Þ
1 2
E
3
E2
=
25. (b)Reading of potentiometer is accurate because during
taking reading it does not draw any current from the
circuit.
26. (b) We know that, R =
A
rl
or R =
2
2
R
Volume
r
Þµ
l
l
According to question l
2
= nl
1
2
1
R
R
=
22
1
2
1
nl
l
or,
22
1
R
n
R
=
ÞR
2
= n
2
R
1

516. 512 PHYSICS
OERSTED EXPERIMENT
In 180
2 Gian Domenico Romagnosi, an Italian lawyer and judge,
found that a steady electric current flowing in wire affected a
magnetic needle placed near it. He published his observation in a
local newspaper (called Gazetta di Trentino). But nobody noticed
this phenomenon.
In 1820 Hans Christian Oersted (a Danish Physicist) rediscovered
this phenomenon. He noted that a magnetic compass needle,
brought close to a straight wire carrying a steady electric current,
aligned itself perpendicular to the wire i.e., the direction of magnetic
field
B
r
is tangential to a circle which has the wire as centre, and
which has its plane perpendicular to the wire (Fig 1-a). Oersted
also noticed that on reversing the direction of current; the direction
of magnetic field is reversed.
Anticlockwise
Fig. 1(b)
i
B
Clockwise
Fig. 1(a)
i
B
In fi
rst case when current is in upward direction, magnetic field is
clockwise (Fig 1-a) and when the current is downward, magneticfield is anticlockwise (Fig. 1-b).
MAGNETIC FIELD
It is a region of space around a magnet or current carrying
conductor or a moving charge in which its magnetic effect can
be felt.
The conductor carrying current is electrically neutral but a
magnetic field is associated with it.
The SI unit of magnetic field induction is tesla (T) or weber/m
2
and cgs unit is gauss. 1 gauss = 10
–4
T
Comparison between electric field and magnetic field
Electric field Magnetic field
1. Sou
rce is an electric1. Source is a current element
charge (q).
()Id
u ur
l.
2. Isolated charge exists2.
Isolated poles do not exist.
3. Electric field at a point3. Magnetic field at a point due to
due to a point charge is a current element is perpendicular
in the plane containing to the plane containing the point
the point and the charge.and the current element.
4. It obeys inverse square4. It also obeys inverse square law
law (a long range force).(a long range force).
5. It obeys principle of 5. It also obeys principle of
superposition as the fieldsuperposition.
is linear related to charge.
6. Angle dependence is not6. Angle dependence is present.
present.
7. Line of electric lines of force7. Lines of magnetic lines of
do not form closed loops.force form closed loops.
8. Electric field changes8. Magnetic field does not
kinetic energy of a chargedchange kinetic energy of a
particle. charged particle.
9. A charged particle whether9. A charged particle at rest
at rest or in motion in ando not experience force
electric field experiencesdue to magnetic field.a force due to electric field.
MAGNETIC FIEL
D DUE TO CURRENT CARRYING
CONDUCTOR, BIOT-SAVART’S LAW
The magnetic induction
dB
uuur
at any point outside
the current path due to a small current element of
length
uur
ld (in the direction of the current) is given
by Biot-Savart's law
0
3
()
4
m ´
=
p
u ur ur
uuur
lIdr
dB
r
r
A
B
i P
dl
q
or,
2
0
r
sind
4
|dB|
q
p
m
=
lI
Also,
0
3
()
4
m ´
=
p
ur ur
uur
qvr
dB
r
where v is the drift velocity of charge
where m
0
= 4p × 10
–7
TmA
–1
.
19
Moving Charges
and Magnetism

517. 513Moving Charges and Magnetism
Direction of
u u ur
dB
The direction of dB
uur
is perpendicular to both
uur
ld and r
r
, governed
by the right hand thumb rule of the cross-product of
uur
ld and r
r
.
The magnetic fie
lds going into the page and coming out of the
page are represented as follows :
Magnetic field Mag netic field
going into the page coming out of the page.
Magnetic Field due to Various Current Carrying
Conductors
Magnetic field due to finite sized conductor :
I
O
P
a q
1
q
2

k
j
i
^
^
^
(at P)
I
0 ˆ
B (sin sin )( k)
12
2a
m
= q+q-
p
r

0
12
2I
B (sin sin )
4a
m
Þ= q+q
p
Elucidat
ion
( )
3
I
0
dB dr
4r
m
=´
p
ur rr
l r = a sec q, l = a tanq
22 ˆ
d a sec d d a sec d jÞ = q qÞ = qq
r
ll
ˆˆ
r a tanj ai=- q+
ur
2
0
33
Iasecd
ˆ ˆˆ
dB j ( a t an j ai)
4 a sec
m qq
Þ = ´
- q+
pq
ur
)k
ˆ
(
a4
dcosI
)k
ˆ
(
seca4
dsecI
0
3
2
0
-
p
qqm
=
-
qp
qqm
=
òò
q
q-
q
q-
qq
p
m
=qq
p
m
=\
2
1
2
1
)(
00
dcos
a4
I
dcos
a4
I
B
]sin[sin
a4
I
B
12
0
q+q
p
m
=Þ (Pointing into the plane of paper)
Magnetic field near the end of a finite sized conductor :
0
()
I
sin
4
m
=q
p
ur
atP
B
a
I
a
P
q
Magnetic field due to an infinitely long conductor :
0
(at P)
I
ˆ
B ( k)
2a
m
=-
p
ur

02
4
m
Þ=
p
I
B
a
I
a
P
Elucidation
Magnetic field in the case of infinitely long wire
ò
p
p-
qq
p
m
=
2/
2/
0
dcos
a4
I
dB
ˆ
I()
0
2
m-
Þ=
p
r k
B
a
a
I
Magnetic field near the end of a long conductor :
0
()
I
4
m
=
patP
B
a
I
a
P
Elucidation
r = a cosec q, l = a cot q,
dl = a cosec
2
q dq
P
I
l
r
a
90°
O
q
A
0
3
2
0
32
00
3
dlr
dB
4r
ˆ
Ia cosec d
j
ˆ
(aj – a cot j)
4 a cosec
IIcos ec
ˆˆ
d ( k) dB sin d ( k)
4a 4acos ec
m´
=
p
m qq
=
´q
pq
mm q
= q- Þ = qq-
pp q
ur
r
uur


00
/2
II
ˆˆ
B [ cos ] ( k) 0 ( k)
4a 4a
p
p
mm æö
= -q -= --
ç÷
èøpp
u ur
)k
ˆ
(
a4
I
B
0
-
p
m
=Þ
Magnetic field d
ue to a current carrying coil :
(i)Magnetic field at a point on the axis of symmetry of a
circular coil, at a distance “x” from its centre :
B = m
0
NI a
2
/2(a
2
+ x
2
)
3/2
or,

2
2 2 3/2
2I
4()
m p
=
p+
o Na
B
ax
I
x
a
N = tota
l number of turns
a = coil radius
The direction of
B
r
is given by Right hand screw rule.
Right hand screw rule : If direction of rotation of right handed
screw-head is the directon of current in a circular conductor then
the direction of its advance is the direction of magnetic field. This
is applicable even if the current, magnetic field are interchanged, as
in case of current flowing through a straight conductor.

518. 514 PHYSICS
Elucidat
ion
x
r
y
z
q
q
dl
x
a
Let for a particular angle, position of small length element dl is
given by its coordinates as
q=q-=sinay,cosaz .
Now,
ˆˆ
a acosk as inj=-q+q
r
,
ˆˆˆˆ
r a xi xi asin j a cos k=-+=- q+ q
rr
Also we have q=Þq = rddrll .
Now ad^l at any instant
)j
ˆ
cosk
ˆ
(sindd q+q=\ll , ( )
0
3
I
dB dr
4r
m
=´
p
uur urr
l
0
3
220
2 2 3/2
0
2 2 3/2
Ird
ˆˆ ˆˆˆ
dB (sin k cos j) (xi a sin j a cos k)
4r
Ird
ˆˆˆˆ
(x sin j a sin i xcos k a cos i)
4(x a)
Ird
ˆ ˆˆ
[x(sin
j cos k) ai]
4(x a)
mq
Þ = q+ q´ - q+ q
p
mq
= q+q+q+q
p+
mq
= q+ q+
p+
u ur
2
0
02 2 3/2
0
2 2 3/2
B dB
Ia
ˆˆˆ
[x | cos j sin k | a(2 0)i]
4(x a)
ˆ
I2 a(i)
B
4(
x a)
p
Þ=
m
= - q + q + p-
p+
mp
=
p+
ò
uur uur
u ur
If number of t
urns of coil are N, then
00
2 23/2 2 23/2
0
2 2 3/2
ˆˆ
I2 Na (i) INa (i)
|B|
4(xa) 2(xa)
2 NIa
B
4(x a)
mpm
==
p++
m p
Þ=
p+
r
(ii)At the centre of a circular coil, B = m
0
NI / 2a
02
4
mp
=
p
NI
a
(iii)Magnetic field at the centre of a circular arc carrying
current
0
()
ˆ
()
2 360
m q
=´-
ur
atP
I
Bk
a
0
4
mq
Þ=
p
I
B
a P
a
q
where q is in radian.
In this case the direction of magnetic field B
r
is into the
page.
Ma
gnetic field inside a current carrying solenoid
(i)Finite size solenoid
0
12
()
nI
(cos cos )
2
m
= q-q
atP
B
q
2
q
1
P
(ii)Near the end of a finite
solenoid
0
12
nI
cos ; ( & / 2)
2
m
= q q =q q =pB
(iii)In t
he middle of a very long solenoid, B = m
0
n I
(iv)Near the end of a very long solenoid0
2
m
=
nI
B
q
P
n is the number of turns per unit length of solenoid.
(v)Magnetic field in the endless solenoid (toroid) is same
throughout and is m
0
nI.
(vi)Magnetic field outside a solenoid or toroid is zero.
AMPERE’S CIRCUITAL LAW
The line integral of magnetic field across a closed loop is equal to
40 times the net correct inside the loop
i.e.,
0
.=mò
uur u ur
Ñ
BdlI
where I is the net current inside the loop.
(1) The direction of the magnetic field at a point on one side of
a conductor of any shape is equal in magnitude but opposite
in direction of the field at an equidistant point on the other
side of the conductor.
(2) If the magnetic field at a point due to a conductor of any
shape is B
o
if it is placed in vacuum then the magnetic field
at the same point in a medium of relative permeability m
r
is
given by mBB
ro
= .
(3) If the distan
ce between the point and an infinitely long
conductor is decreased (or increased) by K-times then themagnetic field at the point increases (or decreases) by K-times.
(4) The magnetic field at the centre of a circular coil of radius
smaller than other similar coil with greater radius is more
than that of the latter.
(5) For two circular coils of radii R
1
and R
2
having same current
and same number of turns, we have
,
12
=
12
BR
RB
where B
1
and
B
2
are the magnetic fields at their
centres.
(6) The magnetic field at a point outside a thick straight wire
carrying current is inversely proportional to the distancebut magnetic field at a point inside the wire is directly
proportional to the distance.

519. 515Moving Charges and Magnetism
Keep in Memory
1.If in a coil
the current is clockwise, it acts as a South-pole. If
the current is anticlockwise, it acts as North-pole.
2.No magnet ic field occurs at point P, Q and R due to a thin
current element dlI.
P QR
dlI
3.Magnetic f
ield intensity in a thick current carrying conductor
at any point x is
Rxfor
R
Ix2
4
B
2
0
inside <
p
m
=
R
x
P
Rxfor
R
I2
4
B
0
surface =
p
m
=
Rxfor
x
I2
4
B
0
outside >
p
m
=
4.Graph o
f magnetic field
B
ur
versus x
B
x
5.Magnetic field is zero at all points inside a current carrying
hollow conductor.
Magnitude and direction of magnetic field due to different configuration of current carrying conductor.
Configuration of
curr
ent carrying conductor
Point of observation
Magnetic field
Magnitude Direction
1.
2.
3.
4.
Wire 1 Wire 2
I I
r
P P''
P'
xx‘
Two long linear and parallel
current carrying conductors
Square l
oop At the centre.
At the centre.
At the centre.
Perpendicular to the
plane of paper
inwards.
Normal to the plane of
paper, inwards.
Normal to the plane of
paper, outwards.
Normal to the plane of
paper, inwards.
B
= 0
ú
û
ù
ê
ë
é
+
+
p
m
=
xr
1
x
1
I2
4
B
0
ú
û
ù
ê
ë
é
-
-p
m
=
'x
1
'xr
1
I2
4
B
0
45°45°
a/2
I
I
I
I
a
O
a
O
n
1
b
n
2
Two concentric circular
coils
having turns n and n
12
Straight wire and loop
If
b
n
a
n
21
>
I IO
B = 0
S.No.
At P, the mid point between

t
he two wires.
The distance of P from each
wire is r/2.
At P', distant x from wire 2 as
shown.
At P'', distant x' from wire 1
as s
hown.
( )
0I
B44a/2
sin45sin45
mé ù
ê ú
=p
ê ú
°+ °ê úë û
0 1 2
nn
B 2I
4 ab
m é ù
= p -
ê ú
pë û

520. 516 PHYSICS
FORCE ON A CONDUCTOR
Th
e force on a conductor is given by
F = BIl sin a
where l is the length of the conductor in meter; B is the flux
density of field in tesla (Wb/m
2
); I is the current in ampere and
a is the angle which the conductor makes with the direction of the
field.
Special case :
If a = 90°, then F = BIl
The direction of the force is given by Fleming's left hand rule.
TORQUE ON A COIL
The torque acting on a rectangular coil placed with its plane parallel
to a uniform magnetic field of flux density B is given by
t = BINA
where N is the number of turns in the coil, A is the area and I is the
current.
If the plane of the coil makes an angle a with the direction of the
field, then
t = BINA cos a.
Example 1.
The field normal to the plane of a wire of n turns and radius
r which carries a current i is measured on the axis of the
coil at a small distance h from the centre of the coil. By
what fraction this is smaller than the field at the centre?
5.
6.
7.
At the centre.
At the centre of loop.
At the common centre.
Straight wire & semi-circular loop
Two concentric circular arcs
I I
O
O
a
a
I
4
B
00p
p
mm
=
Normal to the plane of paper,
inwards.
Normal to the plane of
paper, i
nwards.
Normal to the plane of paper,
outwards.
a
I I
I
00
mm
ú
û
ù
ê
ë
é
-
p
p
=
a
I2
a
I2
4
B
q
a
b
O
00
mm
ú
û
ù
ê
ë
é
-
p
= q
b
1
a
1
I
4
B
Circular loop
8.
9.
At the centre of the semi-
ci
rcle.
At the common centre.
Semi-circular area and straight
conductors
Two concentric coils mutually
normal to eac
h other.
Normal to the plane of paper,
outwards.
According to law of vectors
addition.
I
I
I
r
O
I
1
I
2 B
B
2
B
1
b
a
O
n
1
r
I
4r
I
4
B
00
p
m
+
p
p
m
=
2
2
2
1
BBB +=
where
a
In2
4
B
10
1
p
p
m
=
b
In2
4
B
20
2
p
p
m
=

521. 517Moving Charges and Magnetism
Solution :
The magnetic field on the axis of a current i carrying coil of
turns n, radius r and at a distance h from the centre of the
coil
2/322
2
0
)hr(
nir2
4
B
+
p
´
p
m
= .....(1)
T
he field at the centre is given by
r
ni2
4
B
0
centre
´p
´
p
m
= ..
..(2) (
Q at centre h = 0)
3
2 2 3/2
centre
Br
B (r h)
=
+
3
3/2
2
3
2
r
h
r1
r
=
éù
+êú
ëû
=
÷
÷
ø
ö
ç
ç
è
æ
+
2
2
r
h
2
3
1
1
or
centre
2
2
B
r
h
2
3
1B =
÷
÷
ø
ö
ç
ç
è
æ
+
\
2
2
centre
r
h
2
3
B/)
BB(=-
Example 2.
In fig., there are two semi-circles of radii r
1
and r
2
in which
a current i is flowing. Find the magnetic induction at
centre O.
O
r
1
r
2
i
Solution :
B = B
1
+
B
2
=
2
0
1
0
r
i
4r
i
4
p
´
p
m
+
p
´
p
m
= ú
û
ù
ê
ë
é+m
=ú
û
ù
ê
ë
é
+
m
21
210
21
0
rr
rr
4
i
r
1
r
1
4
i
Example 3.
T
wo circular coils X and Y having equal number of turns and
carry equal currents in the same sense and subtend same
solid angle at point O. If the smaller coil X is midway between
O and Y, then if we represent the magnetic induction due to
bigger coil Y at O as B
Y
and due to smaller coil X at O as B
X
then find
Y
X
B
B
.
Y X
O
d/2
d
2r
r
Solution :
Magnetic
induction at O due to coil Y is given by
2/322
2
0
Y
])d()r2[(
)r2(2
4
B
+
Ip
´
p
m
=
...(1)
Similar
ly, the magnetic induction at O due to coil X is given
by
2/322
2
0
X
])2/d()r[(
)r(2
4
B
+
Ip
´
p
m
= ...(2)
From eq
s. (1) and (2),
2
1
B
B
X
Y
=
Example 4.
A ce
ll is connected between two points of a uniformly thick
circular conductor. i
1
and i
2
are the currents flowing in
two parts of the circular conductor of radius a. What will
be the magnetic field at the centre of the loop?
Solution :
Let l
1
, l
2
be the lengths of the two parts PRQ and PSQ of the
conductor and r be the resistance per unit length of the
conductor. The resistance of the portion PRQ will be R
1
= l
1
r
QS O
R
P
i
2
i
1
The res
istance of the portion PSQ will be R
2
= l
2
r
Pot. diff. across P and Q = i
1
R
1
= i
2
R
2
or i
1
l
1
r = i
2
l
2
r or i
1
l
1
= i
2
l
2
…… (i)
Magnetic field induction at the centre O due to currents
through circular conductors PRQ and PSQ will be
B
1
– B
2
=
0011 22
22
i sin90º
i sin90º
0
44 rr
mm
-=
pp
ll
Example 5.
A
current passing through a circular coil of two turns
produces magnetic field B at its centre. The coil is then
rewound so as to have four turns and the same current is
passed through it.The magnetic field at its centre now is
(a) 2 B(b)
B
2
(c)
B
4
(d) 4 B
Sol
ution : (d)
r
n2
4
B
0 Ip
p
m
= i.e.
r
n
Bµ ;
Given,
L = 2pr
1
× 2 = 2pr
2
×4 ; r
1
/r
2
= 4/2 = 2
42
2
4
r
r
n
n
B
B
2
1
1
2
1
2
=´=´=
or B
2
= 4B
1
= 4B.
Exa
mple 6.
Compute the flux density in air at a point of 9 cm from the
long straight wire carrying a current of 6A.

522. 518 PHYSICS
Solution :
Giv
en : a = 9 cm = 9 × 10
–2
m, I = 6A
7
o
2
I4106
B
2a2 9 10
-
-
m p´´
==
p p´´
= 1.33 × 10
–5
T
Example 7.
Calc
ulate the flux density at a distance of 1 cm from a very
long straight wire carrying a current of 5A. At what distance
from the wire will the field flux density neutralize that due
to the earth's horizontal component flux density 2 × 10
–5
T ? (
m
0
= 4
p × 10
–7
Hm
–1
)
Solution :
7
4o
2
I4105
B 10T
2a2 1 10
-
-
-
m p´´
===
p p´´

For th
e second part,7
2o
5
I4105
a 5 10 5 cm
2B2 2 10
-
-
-
m p´´
= = =´=
p p´´
Ex
ample 8.
A wire 28m long is bent into N turns of circular coil ofdiameter 14 cm forming a solenoid of length 60 cm.Calculate the flux density inside it when a current of 5 amppassed through it. (
m
0
= 12.57 × 10
–7
m
–1
)
Solution :
Given : d = 14cm = 0.14m l = 60cm = 0.6 m
By the question, N × pd = 28 m.
N × p × 0.14 = 28

28
N
0.14
\=
´p
= 63.66 turns.

7
oo
N 63.66
B nI I 12.57 10 5
0.6
,
<λ<λ<???
l
= 6.67 ×10
–
4
T
Example 9.
A vertical conductor X carries a downward current of 5A.(a) What is the flux density due to the current alone at a
point P 10 cm due east of X?
(b) If the earth's horizontal magnetic flux density has a
value 4 × 10
–5
T, calculate the resultant flux density
a t P.Is the resultant flux density at a point 10cm due northof X greater or less than at P? Explain your answer.
B
B
e
I
W
N
S
E
Solution :
(a) I = 5A
, a = 10cm = 0.1 m
7
5o
I4105
B 110T
2 a 2 0.1
-
-m p´´
= = =´
p p´
At
P, the earth's horizontal magnetic flux density,
B
e
= 4 × 10
–5
T (from South to North)
The direction of B is from north to south.
\ Resultant intensity at
P = 4 × 10
–5
– 1 × 10
–5
T
= 3 × 10
–5
T (From south to north)
For a point 10 cm , north of X the flux density due to the
current in X = 1 ×10
–5
T (due east)
(b) The flux density due to the horizontal component of
the earth's field = 4× 10
–3
T (due north)
Be BR
B
\Resultant Intensity
22
R e
B B B.=+ =
52 52
(4 10 ) (1 10 )
--
´ +´
5
17 10
-
=´
= 4.1 × 10
–5
T
which is greater than th
e flux density at P.
Example 10.
A horizontal wire, of lenth 5 cm and carrying a current of
2A placed in the middle of a long solenoid and right anglesto its axis. The solenoid has 1000 turns per metre and carriesa steady current I. Calculate I if the force on the wire isvertically downwards and equal to 10
–4
N.
Solution :
l = 5cm = 5 × 10
–2
m, I
wire
= 2A, n = 1000 m
–1
,

F = 10
–4
N
If I be the current through the solenoid, then
B = m
0
nI
Force = BI
wire
× l

or 10
–4
= 0 wire
NIIlm´´
or
47 25
10 4 10 1000 I 2 5 10 4 10 I
-- --
=p´ ´ ´´´´
=p´
4
5
10 10
I 0.8A
44 10
-
-
\= ==
pp´
Example 11.
Two long
parallel conductors carry currents of 12A and 8A
respectively in the same direction. If the wires are 10cm
apart, find where the third parallel wire also carrying a
current must be placed so that the force experienced by it
shall be zero.
Solution :
For the force on the third conductor to be zero, the direction of
the flux density due to the current flowing in the two wires
must be opposite in the position of the wire.
\ Third wire must be placed between the wire. Let the third
wire placed at a distance x m from the wire carrying 12A,
then, B
1
= B
2

523. 519Moving Charges and Magnetism
12A 8A
X
0.lm

o1 o2
II
.
2 x 2 (0.1 x)
mm
=
p p-
or
128 32
or
x 0.1x x 0.1x
==
--
or 0.3 = 5x =
0.3
5
= 0.06 m
FORCE ACTING ON A CHARGED PARTICLE MOVING
IN A UNIFORM MAGNETIC FIELD
The force acting on a particle having a charge q and moving with
velocity v
r
in a uniform magnetic field B
r
is given by
( ) sin=´Þ=q
ur ur ur
FqvB Fqv B ,
where q is the angle between v
r
and B
r
Case (i)If q = 0, F = 0. Also if q = 180°, F = 0
If a charged particle enters a uniform magnetic field in the direction
of magnetic field or in the opposite direction of magnetic field, the
force acting on the charged particle is zero.
Case (ii)If q = 90°, F = qvB
In this case the force acting on the
particle is maximum and this force acts
as centripetal force which makes the
charged particle move in a circular path.
2
mv
F qvB
r
\==
v
F
v
v
q
×
×
×
×
×
×
×
×
× × × × ×
×
×
× × × × ×
×
×
× × × × ×
×
B
mv
r=
qB
Þ
where
r is the radius of the circular path.
It is important to note that this force cannot change the speed of
the charged particle and hence its kinetic energy. But it changes
the velocity of charged particle (due to change in direction) and
hence also causes a change in momentum.
Also the work done by the force is zero as the force is acting
perpendicular to the direction of motion.
Case (iii) If q = a is any other angle then the path taken is helical.
The velocity of the charged particle can be split into two parts for
better understnading.
a
v sin a
vcosa
B
v
(a)v cos a : The f orce on charged particle due to this component is
zero. This component is responsible in moving the charged
particle uniformly in the direction of B.
(b)v sin a : The
force acting on charged particle due to this
component is
q(vsin )Bsin90 qvBsina °=a . This acts
as th
e centripetal force and moves the particle in a circular
path.
The combined effect of these two is a helical path.
A charged particle entering a uniform magnetic field at an angle
executes helical path.
Radius of the helix,
mvsin
R
qB
a
=
Angul
ar frequency of rotation, m/qB)T/2(=p=w
Pitch of the helix =
2 mvcos
(vcos )T .
qB
pa
a=
Direction o
f force
F
ur
:We can use the rule of cross product. The
direction of F is perpendicular to the plane containing v and B
and can be fo
und by right hand thumb rule. It is important to note
that if q is positive, we will get the correct direction of
F by right
hand thumb rule. But if q is negative, we have to reverse the
direction of force.
Flemings left hand rule : It states that if the fore finger, the central
finger and the thumb of the left hand are stretched at right angles
to each other then if the central finger represents the direction of
current and fore finger represents field, the thumb will represent
the direction of motion or force experienced by the current
carrying conductor.
FORCE BETWEEN TWO PARALLEL CURRENTS
When a current flows in a conductor, the free charges (electrons
in case of a metal wire) move. Each free charge movement generates
a force which adds up to give the force on the conductor.
Force between infinitely long conductors placed parallel to each
other at distance d.
d
I
2
I
1
F
B
1
Force per unit length =
012
2
4
m
=
pl
IIF
d
where l is the length of wire.
If currents are pointing in same direction, the force is of attractive
nature and if currents are oppositely directed the force is of
repulsive nature.
Lorentz Force Equation
For a charged particle q moving in a region of simultaneously
applied electric field
E
u ur
and magnetic field B
u ur
, the force
experienced by it is given by
[ ( )]= +´
u ur u urur u ur
F qE vB
Torque on a cur rent loop in uniform magnetic field B
u ur
is given
by ()t=´
ur u u uru ur
MB where M
uur
is the magnetic moment of coil.
ˆI=
u u ur
M N An where n
r
is the unit vector normal to the plane of
the loop.

524. 520 PHYSICS
Keep in Memory
1.No for
ce acts on a charged particle if it enters a magnetic
field in a direction parallel or antiparallel to the field.
2.A finite force acts on a charged particle if it enters a uniform
magnetic field in a direction with finite angle with the field.
3.If two charged particles of masses m
1
and m
2
and charges
q
1
and q
2
are projected in a uniform magnetic field with
same constant velocity in a direction perpendicular to the
field then the ratio of their radii (R
1
: R
2
) is given by
1 12
2 21
R mq
R mq
=´
4.The force on a con
ductor carrying current in a magnetic
field is directly proportional to the current, the length ofconductor and the magnetic field.
5.If the distance between the two parallel conductors isdecreased (or increased) by k-times then the force betweenthem increases (or decreases) k-times.
6.The momentum of the charged particle moving along thedirection of magnetic field does not change, since the forceacting on it due to magnetic field is zero.
7.Lorentz force between two charges q
1
and q
2
moving with
velocity v
1
, v
2
separated by distance r is given by:
0 11 22
m 2
(qv)(q v)
F.
4 r
m
=
p
8.If the charges move, the electric as well as magnetic fields
are produced. In case the charges move with speedscomparable to the speed of light, magnetic and electric forcebetween them would become comparable.
9.A current carrying coil is in stable equilibrium if the magnetic
dipole moment
M
r
, is parallel to B
r
and is in unstable
e
quilibrium when
M
r
is antiparallel to B
r
.
10.Magnetic moment is inde
pendent of the shape of the loop.
It depends on the area of the loop.
11.A straight conductor and a conductor of any shape in the
same plane and between the same two end points carrying
equal current in the same direction, when placed in the same
magnetic field experience the same force.
12.There is net repulsion between two similar charges moving
parallel to each other inspite of attractive magnetic force
between them. This is because of electric force of repulsion
which is much more stronger than the magnetic force.
13.The speed of the charged particle can only be changed by
an electric force.
MOVING COIL GALVANOMETER
The moving coil galvanometer was first devised by Kelvin and
later on modified by D’Arsonaval. This is used for detection and
measurement of small electric current.
The principle of a moving coil galvanometer is based on the fact
that when a current carrying coil is placed in a magnetic field, it
experiences a torque.
Construction: A moving coil ballistic galvanometer is shown in
figure.
M
T
1
N S
P
S
Q
R
Spring
T
2
Torsiona
l
head
Phosphor
bronze wire
Mirror
L
N S
Soft iron core
Showing radial
field
It essentially co
nsists of a rectangular coil PQRS or a cylindrical
coil of large number of turns of fine insulated wire wound over anon-conducting frame of ivory or bamboo. This coil is suspendedby means of phosphor bronze wire between the pole pieces of a
powerful horse shoe magnet NS. The poles of the magnet are
curved to make the field radial. The lower end of the coil, is attached
to a spring of phosphor-bronze wire. The spring and the free ends
of phosphor bronze wire are joined to two terminals T
2
and T
1
respectively on the top of the case of the instrument. L is a soft
iron core. A small mirror M is attached on the suspension wire.
Using lamp and scale arrangement, the deflection of the coil can
be recorded. The whole arrangement is enclosed in a non-metallic
case.
Theory : Let the coil be suspended freely in the magnetic field.
Suppose, n = number of turns in the coil
A = area of the coil
B = magnetic field induction of radial magnetic field in which
the coil is suspended.
Here, the magnetic field is radial, i.e., the plane of the coil always
remains parallel to the direction of magnetic field, and hence the
torque acting on the coil
t = niAB … (1)
Due to this torque, the coil rotates. As a result, the suspension
wire gets twisted. Now a restoring torque is developed in the
suspension wire. The coil will rotate till the deflecting torque acting
on the coil due to flow of current through it is balanced by the
restoring torque developed in the suspension wire due to twisting.
Let C be the restoring couple for unit twist in the suspension wire
and q be the angle through which the coil has turned. The couple
for this twist q is Cq.
In equilibrium, deflecting couple = restoring couple
\ ni AB = Cq or i = Cq/ (nAB)
or i = Kq (where C/nAB = K) … (2)
K is a constant for galvanometer and is known as
galvanometer constant.
Hence i
µq
Therefore, the deflection produced in the galvanometer is directly
proportional to the current flowing through it.
Current sensitivity of the galvanometer : The current sensitivity
of a galvanometer is defined as the deflection produced in the
galvanometer when a unit current is passed through it.

525. 521Moving Charges and Magnetism
We know that, niAB = Cq
\Current sensitivity, i
s
=
nAB
iC
q
=
where C = r
estoring couple per unit twist
The SI unit of current sensitivity is radian per ampere or deflection
per ampere.
Voltage sensitivity of the galvanometer : The voltage sensitivity
of the galvanometer is defined as the deflection produced in the
galvanometer when a unit voltage is applied across the terminals
of the galvanometer.
\Voltage sensitivity, s
V
V
q
=
If R be the resistance of the galvanometer and a current is passed
through it, then
V = iR
\Voltage sensitivity, V
s
=
nAB
iR CR
q
=
The SI. unit of v
oltage sensitivity is radian per ampere or deflection
per ampere.
Conditions for sensitivity : A galvanometer is said to be more
sensitive if it shows a large deflection even for a small value of
current.
We know that,
nAB
i
C
q=
For a given value of i, q w
ill be large if (i) n is large, (ii) A is large,
(iii) B is large, and (iv) C is small.
Regarding above factors, n and A cannot be increased beyond a
certain limit. By increasing n, the resistance of the galvanometer
will increase and by increasing A, the size of the galvanometer will
increase. So, the sensitivity will decrease. Therefore, B is increased.
The value of B can be increased by using strong horse shoe
magnet. Further, the value of C can be decreased. The value of C
for quartz and phosphor-bronze is very small. So, the suspension
wire of quartz or phosphor-bronze is used. The value of C is further
decreased if the wire is hammered into a flat strip.
HALL EFFECT
When a current passes through a slab of material in the presence
of a transverse magnetic field, a small potential difference is
established in a direction perpendicular to both, the current
flow and the magnetic field. This effect is called Hall effect
The voltage thus developed is called Hall voltage.
X
V
Z
B
F
P
m
1
P2
Y
Hall effect enables us to :
(i) Determine the sign of charge carriers inside the conductor.(ii) Calculate the number of charge carriers per unit volume.
Explanation : Let us consider a conductor carrying current in +X
direction. The magnetic field is applied along +Y direction.
Consider two points P
1
and P
2
on the conductor and connect a
voltmeter between these points. If no magnetic field is applied
across the conductor, then the points P
1
and P
2
will be at same
potential and there will be no deflection in the galvanometer.
However, if a magnetic field is applied as shown in the figure,
then the Lorentz force acts on electrons as shown in the figure.
The Lorentz force on electrons F
m
= –e
d
(v B)
®®
´ acts in the
down
wards direction.
Now there may be two cases:
Case I : If the charge particles are negatively charged, then
these negative charges will accumulate at the point P
2
and
therefore P
2
will be at lower potential than P
1
.
Case II : If the charged particles are positively charged, then the
point P
2
will be at higher potential than P
1
.
Magnitude of hall voltage :
Let w be width and A be the cross-sectional area of the conductor.
If e is magnitude of charge or the current carrier (electron or
hole).
The force on the current carrier due to magnetic field B,
F
m
= Bev
d
Here, v
d
is drift velocity of the current carries.
Due to the force F
m
, the opposite charges build up at the points
P
1
and P
2
of the conductor.
If V
H
is Hall voltage developed across the two faces, then the
strength of electric field due to Hall voltage is given by
E
H
=
H
V
w
.
Here w =
P
1
P
2
.
This electric field exerts an electric force on the current carries in
a direction opposite to that of magnetic force. The magnitude of
this force is
H
eH
V
FEee
w
==
In equilibrium c
ondition, F
e
= F
m
,
or
HV
w
e = B e v
d
, or V
H
= B v
d
w
N
ow, drift velocity of current carrier is given by, v
d
=
j
ne
where n is the number of current carries per unit volume of the
strip.
Hall resistance,
1
H
H
V Bwj
R
I neI
æö
== ç÷
èø
\Hall voltage
H
Bwj
V
ne
=
19.1

526. 522 PHYSICS
Keep in Memory
1.Hall eff
ect can determine nature of current (charge) carriers in
the material. i.e. whether the charge is +ve or –ve.
2.Hall voltage
Aen
IBb
V
H=
where n is the den
sity of charge carriers
b
I
I
X B
Y
Z
O
V
H
b = thickness of plate, B = magnetic field, I = current flowing
through plate, A = area of cross-section of plates
Example 12.
Two particles X and Y having equal charges, after being
accelerated throught the same potential difference, enter
a region of uniform magnetic field and describe circular
paths of radii R
1
and R
2
respectively. Find the ratio of
mass of X to that of Y.
Solution :

qVmv
2
1
2
=
2/1
)m/qV2(v=\
(where V is the po
tential difference)
Centripetal force
2
mv
qvB
R
<
R
m
qB
v ÷
ø
ö
ç
è
æ
=\
Hence R
m
qB
m
qV2
2/1
÷
ø
ö
ç
è
æ
=÷
ø
ö
ç
è
æ
or
B
1
q
mV2
R
2/1
´
÷
÷
ø
ö
ç
ç
è
æ
=
Here V, q and B a
re constant. Hence
Rmµ
\
2
2
1
2
1
R
R
m
m
÷
÷
ø
ö
ç
ç
è
æ
=
Example 13.
A rectangular loop of sides 25 cm and 10 cm
carrying a
current of 15 A is placed with its longer side parallel to a
long straight conductor 2.0 cm apart carrying a current of
25 A. What is the net force on the loop?
Solution :
Consider a rectangular loop PQRS placed near a long straight
conductor AB as shown in Fig. Due to the interaction of
currents, the arm PQ of the loop will get attracted while arm
RS will get repelled. Forces on the arms QR and SP will be
equal and opposite and hence cancel out.
Here,PQ = 25 cm = 25 × 10
–2
m,
A
P S
RQ
B
I
=
2
5
A
1 I
=
1
5
A
2
I
=
1
5
A
1
r
2
r
1
PS = 10 cm =
10 × 10
–2
m
Distance of PQ from AB,
r
1
= 2.0 cm = 2.0 × 10
–2
m
Distance of RS from AB,
r
2
= 2.0 + 10 = 12.0 cm = 12.0 × 10
–2
m
Current through long wire AB, I
1
= 25 A
Current through rectangular loop, I
2
= 15 A
\ Force on the arm PQ,
PQlength
r
2
4
F
1
210
1 ´
p
m
=
II
2
21
1
100.2
10251525210
F
-
--
´
´´´´´
=
= 9.375 × 1
0
–4
(towards AB)
Force on the arm RS,
0 12
2
2
2
F length RS
4r
m
=´
p
II

72
2
10 2 25 15 25 10
12 10
--
-
´´´´´
=
´

-4
= 1.563×10 (away from AB)
\ Effective force on the loop,
N10812.710563.110375.9FFF
444
21 ´=´-´=-=
--
(towards AB)
Example 14.
An electron beam moving with a velocity of 10
6
ms
–1
through a uniform magnetic field of 0.1T, which is
perpendicular to the direction of the beam. Calculate the
force on an electron if the electron charge is 1.6 ×10
–19
C.
Solution :
v=10
–6
ms
–1
, B = 0.IT, q = 1.6 × 10
– 19
C
F = Bqv = 0.1 × 10
6
× 10
19
= 1.6 × 10
–14
N
Example 15.
A narrow vertical rectangular coil is suspended from the
middle of its upper side with its plane parallel to a uniform
horizontal magnetic field of 0.02 T. The coil has 10 turns
and the lengths of its vertical and horizontal sides are 0.1
m and 0.05 m respectively. Calculate the torque on the coil
when a current of 5A is passed into it.
What would be the new value of the torque if the plane of
the vertical coil was initially at 60
o
to the magnetic field
and a current of 5A was passed into the coil.

527. 523Moving Charges and Magnetism
Solution :
B = 0.02 T, N = 10 turns
A = 1 × b = 0.1 × 0.05 = 0.005 m
2
I = 5A
Torque = BINA = 0.02 × 5 × 10 × 0.005
= 0.005 Nm = 5 × 10
–3
Nm
New value of the torque when the plane of the vertical coil
was at 60° to the magnetic field.
= BINA cos q = 5 × 10
–3
cos 60°
= 5× 10
–3
×
1
2
= 2.5 × 10
–3
Nm.
Example 16
.
A rectangular coil of 50 turns hungs vertically in a uniform
magnetic field of magnitude 10
–2
T, so that the plane of the
coil is parallel to the field, the mean height of the coil is
5cm and its mean width is 2cm. Calculate the strength of
the current the must pass through the coil in order to
deflect it 30° if the torsional constant of the suspension is
10
–9
Nm per degree.
Solution :
N = 50 turns, B = 10
–2
T,
π= 30°
C = 10
–9
Nm per
degree,
A = 5 × 2 cm
2
= 10 × 10
–4
m
2
= 10
–3
m
2
Torque = BINA cos q = Cq
9
5
2 3o
C 10 30
I = 6.9 × 1 0
BNA cos10 50 10 cos30
-
-
--
q´
\==
q ´´

A = 69mA
Example 17
.
A copper wire has 1.0 × 10
29
free electrons per cubic metre,
a cross sectional area of 2.0 mm
2
and carries a current of
5.0A. Calculate the force acting on each electron if thewire is now placed in a magnetic field of flux density 0.15
T which is perpendicular to the wire (e=1.6 × 10
19
C)
Solution :
n = 10 × 10
29

m
–3
, A = 2.0 mm
2
= 2 × 10
–6
m
2
I = 5.0A, B = 0.15 T, F = ?.
I = nevA or v =
I
nev
F = Bev
Be I BI
neA nA
´
==
296
0.15 5
1.0 10 2 10
-
´
=
´ ´´
= 3.7
5×10
–24
N
Example 18.
If the coil of a moving coil galvanometer having 10 turns
and of resistance 4
W is removed and is replaced by a second
coil having 100 turns and of resistance 160
W. Calculate
(a) the factor by which the current sensitivity changes and
(b) the factor by which the voltage sensitivity changes.
Solution :
Given : N
1
= 10 turns, R = 4W
N
2
= 100 turns, R
2
= 160W
(a) Current sensitivity with the Ist coil,
1
1
qæö
=ç÷
èø
N AB
IC
Current sensitivity with the 2nd coil
2
2
N AB
IC
qæö
=ç÷ èø
22
1
1
I N100
10
N 10
I
qæö
ç÷
èø
\ ===
qæö
ç÷
èø
(b) Vo
ltage sensitivity with the Ist coil
1
11
N AB
V CR
qæö
=ç÷ èø
Voltage
sensitivity with the 2nd coil
2
22
N AB
V CR
qæö
=ç÷ èø
2 21
12
1
v NR 100 4 1
N R 10 160 4
v
qæö
ç÷
èø
\ = =´=
qæö
ç÷
èø

528. 524 PHYSICS
MOVING CH
A
R
G
E
S
AND MAGN
E
T
I
S
M
Motion of a ch
a
r
g
e
d
particle in a un
i
f
o
r
m
magnetic field
f
o
l
l
o
w
s
a circular path,

r
a
d
i
u
s
M
V
s
i
n
B
q
r
q
=
F
o
r
c
e

o
n
a
c
o
n
d
u
c
t
o
r

c
a
r
r
y
i
n
g

c
u
r
r
e
n
t
i
n

a
u
n
i
f
o
r
m
m
a
g
n
e
t
i
c

f
i
e
l
d
,
F
=

I
B

s
i
n

F
=

I
(
B

×

)
l
l
q
Magnetic field Space in
the surrounding of a mag
n
e
t

or any current carrying c
o
n
d
u
c
t
o
r
in which its magnetic inf
l
u
e
n
c
e

can be experienced
Direction of magnetic field-Depends
upon the direction of current. Right
hand thumb rule-Thumb points in the
direction of current, curling of fingers
represents direction of magnetic field .
A
m
p
e
r
e
's
c
i
r
c
u
i
t
a
l
l
a
w

0
=
m
ò
B
d
l
I
M
a
g
n
e
t
i
c

f
i
e
l
d
d
u
e
t
o
a

s
o
l
e
n
o
i
d
.
I
n
s
i
d
e
a
l
o
n
g
s
o
l
e
n
o
i
d
B

=
µ
n
I
A
t

a
p
o
i
n
t

o
n
o
n
e

e
n
d
0
0
n
I
B
=
2
m
M
a
g
n
e
t
i
c

f
i
e
l
d
d
u
e
t
o
a

t
o
r
o
i
d
i
n
s
i
d
e
t
h
e
t
u
r
n
s
B
=
n
I
m
0
F
o
r
c
e

a
c
t
i
n
g

o
n
a
c
h
a
r
g
e
d
p
a
r
t
i
c
l
e
m
o
v
i
n
g
i
n
a

u
n
i
f
o
r
m
m
a
g
n
e
t
i
c

f
i
e
l
d
F
=

q
V
B
s
i
n

=

q
(
V

×

B
)
q
L
o
r
e
n
t
z

f
o
r
c
e
F
=

q
(
E
×
V

×

B
)
Biot-sava
r
t
's

l
a
w
magnetic

f
i
e
l
d
d
u
e

to current
c
a
r
r
y
i
n
g
element,
0
2
sin
dB=
4
mq
p
Idl
r
Magnetic field due
to a current carrying
circular loop
On the axis of
circular loop
2
0
22 3/2
NIa
2(r a)
m
B=
+
At the centre
of circular loop

N
I
2R
m
B=
For
c
e

b
e
t
w
e
e
n
t
w
o
par
a
l
l
e
l

c
u
r
r
e
n
t

c
a
r
r
y
i
n
g
con
d
u
c
t
o
r
s
0
1
2
2
I
I
.
4
r
m
=
p
F
Torque experienced by
a current carrying loop
in a uniform magnetic
field T = MB sin
= M × B
qˆn
Magnetic field due to
a straight current carrying
conductor of infinite lenght
0
I
2R
B
m
=
p
Galvanometer to ammeter
conversion : Low resistance
or shunt connected in parallel
g
g
I
S= G
I–I
æ ö
ç ÷
ç ÷
è ø
Galvanometer to voltmeter
conversion : High resistance
in series
CONCEPT MAP
g
V
R G
I
=-

529. 525Moving Charges and Magnetism
1.A current carrying coil is subjected to a uniform magnetic
field. The coil will orient so that its plane becomes
(a) inclined at 45° to the magnetic field
(b) inclined at any arbitrary angle to the magnetic field
(c) parallel to the magnetic field
(d) perpendicular to the magnetic field
2.An electron enters a region where magnetic field (B) and
electric field (E) are mutually perpendicular, then
(a) it will always move in the direction of B
(b) it will always move in the direction of E
(c) it always possesses circular motion
(d) it can go undeflected also
3.A current carrying conductor placed in a magnetic field
experiences maximum force when angle between current and
magnetic field is
(a) 3 p/4 (b)p/2 (c)p/4 (d) zero
4.Two concentric circular coils of ten turns each are situated
in the same plane. Their radii are 20 and 40 cm and they carry
respectively 0.2 and 0.4 ampere current in opposite direction.
The magnetic field in weber/m
2
at the centre is
(a)m
0
/80 (b) 7m
0
/80(c) (5/4) m
0
(d) zero
5.A wire of length L metre carrying a current I ampere is bent
in the form of a circle. Its magnitude of magnetic moment
will be
(a) IL/4p(b) I
2
L
2
/4p(c) IL
2
/4p(d) IL
2
/8p
6.Two straight long conductors AOB and COD are
perpendicular to each other and carry currents I
1
and I
2
.
The magnitude of the magnetic induction at a point P at a
distance a from the point O in a direction perpendicular to
the plane ABCD is
(a)
)II(
a2
21
0
+
p
m
(b) )II(
a2
21
0
-
p
m
(c)
2
1
2
2
2
1
0
)II(
a2
+
p
m
(d)
21
210
II
II
a2+p
m
7.A prot
on, deutron and an a-particle enter a magnetic field
perpendicular to field with same velocity. What is the ratio of
the radii of circular paths?
(a) 1 : 2 : 2 (b) 2 : 1 : 1
(c) 1 : 1 : 2 (d) 1 : 2 : 1
8.If an electron and a proton having same momenta enter
perpendicular to a magnetic field, then
(a) curved path of electron and proton will be same
(ignoring the sense of revolution)
(b) they will move undeflected
(c) curved path of electron is more curved than that of the
proton
(d) path of proton is more curved
9.In cyclotron the gyro radius is
(a) proportional to momentum
(b) proportional to energy
(c) inversely proportional to momentum
(d) inversely proportional to energy
10.The current sensitivity of a moving coil galvanometer
depends on
(a) the number of turns in the coil
(b) moment of inertia of the coil
(c) current sent through galvanometer
(d) eddy current in Al frame
11.Current i is flowing in a coil of area A & number of turns N,
then magnetic moment of the coil is M is equal to
(a) NiA (b) Ni/A (c)
Ni/A (d) N
2
Ai
12.1 Wbm
–2
is equ
al to
(a) 10
4
G (b) 10
2
G(c) 10
–2
G(d) 10
–4
G
13. The radius of motion of a charged particle oscillating in a
magnetic field is
(a)
qv
mB
(b)
qB
mv
(c )
vB
mq
(d)
mB
qv
14.Magneti c effect of current was discovered by
(a) Faraday (b) Oersted
(c) Kirchhoff (d) Joule
15.In cyclotron the charged particle may be accelerated upto
energies
(a) Several eV (b) MeV
(c) BeV (d) Kev
16.In cyclotron the resonance condition is
(a) the frequency of revolution of charged particle is
equal to the frequency of A.C. voltage sources
(b) the frequency of revolution of charged particle is
equal to the frequency of applied magnetic field
(c) the frequency of revolution of charged particle is
equal to the frequency of rotation of earth
(d) the frequency of revolution of charged particle,
frequency of A.C. source and frequency of magnetic
field are equal
17.Two parallel wires carrying currents in the same direction
attract each other because of
(a) mutual inductance between them
(b) potential difference between them
(c) electric forces between them
(d) magnetic forces between them

530. 526 PHYSICS
18.If we dou
ble the radius of a coil keeping the current through
it unchanged, then the magnetic field at any point at a large
distance from the centre becomes approximately
(a) double (b) three times
(c) four times (d) one-fourth
19.To convert a galvanometer into an ammeter, one needs to
connect a
(a) low resistance in parallel
(b) high resistance in parallel
(c) low resistance in series
(d) high resistance in series.
20.If a current is passed through a spring then the spring will
(a) expand (b) compress
(c) remains same (d) None of these.
21.A charged particle moves through a magnetic field in a
direction perpendicular to it. Then the
(a) velocity remains unchanged
(b) speed of the particle remains unchanged
(c) direction of the particle remains unchanged
(d) acceleration remains unchanged
22.A long solenoid carrying a current produces a magnetic
field B along its axis. If the current is double and the number
of turns per cm is halved, the new value of the magnetic
field is
(a) 4 B (b) B/2
(c)B (d) 2 B
23.The total charge induced in a conducting loop when it is
moved in a magnetic field depends on
(a) the rate of change of magnetic flux
(b) initial magnetic flux only
(c) the total change in magnetic flux
(d) final magnetic flux only
24.Energy in a current carrying coil is stored in the form of
(a) electric field (b) magnetic field
(c) dielectric strength(d) heat
25.Tesla is the unit of
(a) magnetic flux (b) magnetic field
(c) magnetic induction(d)magnetic moment
1.A portion of a conductive wire is bent in the form of a
semicircle of radius r as shown below in fig. At the centre of
semicircle, the magnetic induction will be
i
O
r
i
(a) zero (b) inf
inite
(c)
0
μπi
. gauss
4πr
(d)
0
μπi
. tesla
4πr
2.A helium nucleus
makes a full rotation in a circle of radius
0.8 meter in 2 sec. The value of the magnetic field inductionB in tesla at the centre of circle will be
(a)
0
19
102m´
-
(b)
0
19
/10m
-
(c)
0
19
10m
-
(d)
0
20
/102m´
-
3.A solenoid of length 1.5 m and 4 cm diameter possesses 10
turns per cm. A current of 5A is flowing through it, the
magnetic induction at axis inside the solenoid is
7 11
0
( 4 10 weber amp m )
- --
m = p´
(a) gauss104
5-
´p (b) ga
uss102
5-
´p
(c) tesla104
5-
´p (d) tesla102
5-
´p
4.An electron (mass = 9 × 10
–31
kg, charge = 1.6 × 10
–19
C)
moving with a velocity of 10
6
m/s enters a magnetic field. If it
describes a circle of radius 0.1m, then strength of magnetic
field must be
(a) 4.5 × 10
–5
T (b) 1.4 × 10
–5
T
(c) 5.5 × 10
–5
T (d) 2.6 × 10
–5
T
5.An electron moving with kinetic energy 6×10
–16
joules
enters a field of magnetic induction 6 × 10
–3
weber/m
2
at
right angle to its motion. The radius of its path is
(a) 3.42 cm (b) 4.23 cm
(c) 5.17 cm (d) 7.7 cm
6.An electron moves in a circular arc of radius 10 m at a contant
speed of 2 × 10
7
ms
–1
with its plane of motion normal to a
magnetic flux density of 10
–5
T. What will be the value of
specific charge of the electron?
(a) 2 × 10
4
C kg
–1
(b) 2 × 10
5
C kg
–1
(c) 5 × 10
6
C kg
–1
(d) 2 × 10
11
C kg
–1
7.A current of 3 A is flowing in a linear conductor having a
length of 40 cm. The conductor is placed in a magnetic field
of strength 500 gauss and makes an angle of 30º with the
direction of the field. It experiences a force of magnitude
(a) 3 × 10
–4
N (b) 3 × 10
–2
N
(c) 3 × 10
2
N (d) 3 × 10
4
N
8.A cathode ray beam is bent in a circle of radius 2 cm by a
magnetic induction 4.5 × 10
–3
weber/m
2
. The velocity of
electron is
(a) 3.43 × 10
7
m/s (b) 5.37 × 10
7
m/s
(c) 1.23 × 10
7
m/s (d) 1.58 × 10
7
m/s

531. 527Moving Charges and Magnetism
9.Two long parallel wires P and Q are held perpendicular to
the plane of paper with distance of 5 m between them. If P
and Q carry current of 2.5 amp. and 5 amp. respectively in
the same direction, then the magnetic field at a point half-
way between the wires is
(a)
0
/m17 (b)
0
/mp32
(c)
0
/mp2 (d)
0
/mp32
10.A charge d particle with velocity 2 × 10
3
m/s passes
undeflected through electric and magnetic field. Magnetic
field is 1.5 tesla. The electric field intensity would be
(a) 2 × 10
3
N/C (b) 1.5 × 10
3
N/C
(c) 3 × 10
3
N/C (d) 4/3 × 10
–3
N/C
11.If in a circular coil A of radius R, current I is flowing and in
another coil B of radius 2R a current 2I is flowing, then the
ratio of the magnetic fields B
A
and B
B
, produced by them
will be
(a)1 (b) 2 (c) 1/2 (d) 4
12.A circular loop of area 0.02 m
2
carrying a current of 10A, is
held with its plane perpendicular to a magnetic field
induction 0.2 T. The torque acting on the loop is
(a) 0.01 Nm (b) 0.001 Nm
(c) zero (d) 0.8 Nm
13.Through two parallel wires A and B, 10A and 2A of currents
are passed respectively in opposite directions. If the wire A
is infinitely long and the length of the wire B is 2m, then
force on the conductor B, which is situated at 10 cm distance
from A, will be
(a) 8 × 10
–7
N (b) 8 × 10
–5
N
(c) 4 × 10
–7
N (d) 4 × 10
–5
N
14.The magnetic flux density B at a distance r from a long
straight wire carrying a steady current varies with r as
(a)
B
r
(b)B
r
(c)B
r
(d)B
r
15.A current of I ampere flows in a wire forming a circular arc of
radius r metres subtending an angle q at the centre as shown.
The magnetic field at the centre O in tesla is
I
O
q
(a)
0
I
4r
mq
p
(b)
0
I
2r
mq
p
(c)
0
I
2r
mq
(d)
0
I
4r
mq
16.A uniform electric field and a uniform magnetic field exist in
a region in the same direction. An electron is projected with
velocity pointed in the same direction. The electron will
(a) turn to its right
(b) turn to its left
(c) keep moving in the same direction but its speed will
increase
(d) keep moving in the same direction but its speed will
decrease
17.A current of I ampere flows along an infinitely long straight
thin walled hollow metallic cylinder of radius r. The magnetic
field at any point inside the cylinder at a distance x from the
axis of the cylinder is
(a)
¥ (b)
0
I
2r
m
p
(c)
0
I
2x
m
p
(d) zero
18.Two
particles X and Y having equal charge, after being
accelerated through the same potential difference enter a
region of uniform magnetic field and describe circular paths
of radii R
1
and R
2
respectively. The ratio of the mass of X to
that of Y is
(a)
1
2
R
R
(b)
2
2
1
R
R
æö
ç÷
èø
(c)
2
1 2
R
Ræö
ç÷
èø
(d)
2
1
R
R
19.A square coil of side a carries a current I. The magnetic field
at the centre of the coil is
a O
(a)
0
I
a
m
p
(b)
0
21
a
m
p
(c)
0
1
2a
m
p
(d)
0
22I
a
m
p
20.Protons and a-particles of equal momenta enter a uniform
magnetic field normally. The radii of their orbits will have
the ratio.
(a)1 (b) 2 (c) 0.5 (d) 4
21.Under the influence of a uniform magnetic field a charged
particle is moving in a circle of radius R with constant speed
v. The time period of the motion
(a) depends on both R and v
(b) is independent of both R and v
(c) depends on R and not v
(d) depends on v and not on R

532. 528 PHYSICS
22.What is cy
clotron frequency of an electron with an energy
of 100 e V in the earth's magnetic field of 1 × 10
–4
weber / m
2
if its velocity is perpendicular to magnetic field?
(a) 0.7 MHz (b) 2.8 MHz
(c) 1.4 MHz (d) 2.1 MHz
23.A circular loop of area 0.02 m
2
carrying a current of 10A, is
held with its plane perpendicular to a magnetic field
induction 0.2 T. The torque acting on the loop is
(a) 0.01 Nm (b) 0.001 Nm
(c) zero (d) 0.8 Nm
24.Two thin, long, parallel wires, separated by a distance ‘d’
carry a current of ‘i’ A in the same direction. They will
(a) repel each other with a force of )d2/(i
2
0
pm
(b) attract each other with a force of )d2/(i
2
0
pm
(c) repel each other with a force of )d2/(i
22
0
pm
(d) attract each other with a force of )d2/(i
22
0
pm
25.A horizontal overhead powerline is at height of 4m from
the ground and carries a current of 100A from east to west.
The magnetic field directly below it on the ground is
(m
0
= 4p × 10
–7
Tm A
–1
)
(a) 2.5×10
–7
T southward (b) 5 × 10
–6
T northward
(c) 5 × 10
–6
T southward (d) 2.5 × 10
–7
T northward
26.If an electron describes half a revolution in a circle of radius
r in a magnetic field B, the energy acquired by it is
(a) zero (b)
21
mv
2
(c)
21
mv
4
(d)r Bevp´
27.The orbital speed of electron orbiting around a nucleus in a
circular orbit of radius 50 pm is 2.2 × 10
6
ms
–1
. Then the
magnetic dipole moment of an electron is
(a) 1.6 × 10
–19
Am
2
(b) 5.3 × 10
–21
Am
2
(c) 8.8 × 10
–24
Am
2
(d) 8.8 × 10
–26
Am
2
28.A deutron of kinetic energy 50 keV is describing a circular
orbit of radius 0.5m, in a plane perpendicular to magnetic
field B
r
. The kinetic energy of a proton that discribes a
circular orbit of radius 0.5m in the same plane with the samemagnetic field
B
r
is
(a) 200 keV
(b) 50 keV(c) 100 keV (d) 25 keV
29.A proton and an a-particle enter a uniform magnetic field
perpendicularly with the same speed. If proton takes 25 m
second to make 5 revolutions, then the time period for the
a-particle would be
(a) 50 m sec (b) 25 m sec
(c) 10 m sec (d) 5 m sec
30.A cell is connected between two points of a uniformly thick
circular conductor and i
1
and i
2
are the currents flowing in
two parts of the circular conductor of radius a. The magnetic
field at the centre of the loop will be
(a) zero (b)
)II(
4
21
0
-
p
m
(c) )II(
a2
21
0
+
m
(d) )II(
a
21
0
+
m
31.In fig, what is the magnetic field induction at point O
(a)
0
i
4r
m
p
(b)
00
ii
4r 2r
mm
+
p
(c)
00
ii
4r 4r
mm
+
p
O
r
i
(d)
00
ii
4r 4r
mm
-
p
32.The field B at the centre of a circular coil of radius r is p times
that due to a long straight wire at a distance r from it, for
equal currents. Fig. shows three cases:
P P
P
(a) (b) (c)
in all cases the circular part has radius r and straight ones are
infinitely long. For same current the field B at the centre P incases 1, 2, 3 has the ratio
(a)
÷
ø
ö
ç
è
æ
-
pp
÷
ø
ö
ç
è
æp
-
2
1
4
3
:
2
:
2
(b) ÷
ø
ö
ç
è
æ
+
p
÷
ø
ö
ç
è
æ
+
p
÷
ø
ö
ç
è
æ
+
p
-
2
1
4
3
:1
2
:1
2
(c)
3
::
224
ppp
-
(d)
1 31
1::
2 44 42
pppæ öæ öæ ö
--++ç ÷ç ÷ç ÷
è øè øè ø
33.An infinite straight conductor carrying current 2 I is split
into a loop of radius r as shown in fig. The magnetic field at
the centre of the coil is
(a)
r
)1(2
4
0 +p
p
m
(b)
r
)1(2
4
0 -p
p
m
O
2I2I
I
I
(c)
r
)1(
4
0+p
p
m
(d) zero
34.A long wire is bent into
shape ABCDE as shown in fig., with
BCD being a semicircle with centre O and radius r metre. A
current of I amp. flows through it in the direction
A ® B ® C ® D ® E. Then the magnetic induction at the
point O of the figure in vacuum is
(a)
0
2 r 4r
éùII
m+
êú
pëû
(b)0
2 r 4r
éùII
m-
êú
pëû
A
E
I
B
O C
D
r
I
I
×B
(c)
0
/4rmI
(d)
0
/rmIp

533. 529Moving Charges and Magnetism
35.Three wires are situated at the same distance. A current of
1A, 2A, 3A flows through these wires in the same direction.
What is ratio of F
1
/F
2
,
where F
1
is force on 1 and F
2
on 2?
(a) 7/8
(b) 1
(c) 9/8
1A2A3A(d) None of these
36.A conducting circular loop of radius r carries a constant
current i. It is placed in a uniform magnetic field
0B
r
such
that
0B
r
is perpendicular to the plane of the loop. The
magnetic force acting on the loop is
(a)ir B
0
(b) 2p ir B
0
(c) zero (d)p ir B
0
37.An electron traveling with a speed u along the positive
x-axis enters into a region of magnetic field where B = –B
0
k
ˆ
( x > 0)
. It comes out of the region with speed v then
x ®
y
e
–
u
B×
(a) v = u at y > 0 (b) v = u at y < 0
(c) v > u at y > 0 (d) v > u at y < 0
38.A wire ABCDEF is bent in the form as shown in figure. Thewire carries a current I and is placed in a uniform magneticfield of induction B parallel to positive Y-axis. If each side isof length L, the force experienced by the wire will be
AB
C
D
Z
X
Y
E F
L
L
L
L
L
B
I
O
(a) IBL along the positive Z-direction
(b) IBL along the negative Z-direction(c) 2IBL along the positive Z-direction(d) 2IBL along the negative Z-direction
39.Four wires, each of length 2.0 m, are bent into four loops P,Q, R and S and then suspended in a uniform magnetic field.If the same current is passed in each, then the torque will be
maximum on the loop
P
Q
R
S
(a)P (b) Q
(c
)R (d) S
40.A charged particle (charge q) is moving in a circle of radius
R with uniform speed v. The associated magnetic moment µ
is given by
(a) qvR
2
(b) qvR
2
/2 (c) qvR (d) qvR/2
41.A straight wire of diameter 0.5 mm, carrying a current of 1A
is replaced by another wire of 1mm diameter carrying the
same current. The strength of magnetic field far away is
(a) unchanged
(b) quarter of its earlier value
(c) half of the earlier value
(d) twice the earlier value
42.Two equal electric currents are flowing perpendicular to each
other as shown in figure. AB and CD are perpendicular to
each other and symmetrically placed with respect to the
currents. Where do we expect the resultant magnetic field
to be zero?
(a) on AB
(b) on CD
I
O
I
BC
A D
(c) on both AB & CD
(d) on both OD & BO
43.The magnetic field (dB) due to small element (dl) at a distance
)r(
r
from element carrying current i, is
(a)
0
dlr
dBi
4r
®æö
m´
ç÷
=
èøp
r
(b)
20
2
dlr
dBi
4 r
®æö
m´
ç÷
=
èøp
r
(c)
20
2
dlr
dBi
4 r
®æö
m´
ç÷
=
èøp
r
(d)
0
3
dlr
dBi
4 r
®æö
m´
ç÷
=
èøp
r
44.A 10eV electron is circulating in a plane at right angles to a
uniform field at a magnetic induction 10
–4
Wb/m
2
(= 1.0
gauss). The orbital radius of the electron is(a) 12 cm (b) 16 cm(c) 11 cm (d) 18 cm
45.A coil of one turn is made of a wire of certain length andthen from the same length a coil of two turns is made. If thesame current is passed in both the cases, then the ratio ofthe magnetic inductions at their centres will be(a) 2 : 1(b) 1 : 4(c) 4 : 1(d) 1 : 2
46.A very long straight wire carries a current I. At the instant
when a charge + Q at point P has velocity
v
®
, as shown,
the force on the charge is
I
P
Q
v
®
o
y
x
(a) along oy (b) opposite to oy
(c) along ox (d) opposite to ox

534. 530 PHYSICS
47.A square curren
t carrying loop is suspended in a uniform
magnetic field acting in the plane of the loop. If the force on
one arm of the loop is
F
r
, the net force on the remaining
three arms of the loop is
(a) 3 F
r
(b) – F
r
(c) – 3 F
r
(d)F
r
48.A current loop consists of two identical semicircular parts
each of radius R, one lying in the x-y plane and the other in
x-z plane. If the current in the loop is i., the resultant magnetic
field due to the two semicircular parts at their common centre is
(a)
0
2
i
R
m
(b)
0
22
i
R
m
(c)
0
2
i
R
m
(d)
0
4
i
R
m
49.A current carrying loop in the form of a right angle isosceles
triangle ABC is placed in a uniform magnetic field acting
along AB. If the magnetic force on the arm BC is F, what is
the force on the arm AC?
(a)
2F-
r A
B C
(b)
F-
r
(c)
F
r
(d)2F
r
50.A uniform electric field and uniform magnetic field are acting
along the same direction in a certain region. If an electron is
projected in the region such that its velocity is pointed along
the direction of fields, then the electron
(a) will turn towards right of direction of motion
(b) speed will decrease
(c) speed will increase
(d) will turn towards left direction of motion
51.A square loop, carrying a steady current I, is placed in a
horizontal plane near a long straight conductor carrying a
steady current I
1
at a distance d from the conductor as
shown in figure. The loop will experience
d
I
1
I
I
(a) a net repulsive force away from the conductor
(b) a net torque acting upward perpendicular to the
horizontal plane
(c) a net torque acting downward normal to the horizontal
plane
(d) a net attractive force towards the conductor
52.Charge q is uniformly spread on a thin ring of radius R. The
ring rotates about its axis with a uniform frequency f Hz.
The magnitude of magnetic induction at the centre of the
ring is
(a)
0
qf
2R
m
(b)
0
q
2fR
m
(c)
0
q
2 fR
m
p
(d)
0
qf
2R
m
p
53.Two similar coils of radius R are lying concentrically with
their planes at right angles to each other. The currents
flowing in them are I and 2 I, respectively. The resultant
magnetic field induction at the centre will be
(a)
0
5
2
I
R
m
(b)
0
3
2
I
R
m
(c)
0
2
I
R
m
(d)
0
I
R
m
54.An alternating electric field, of frequency v, is applied across
the dees (radius = R) of a cyclotron that is being used to
accelerate protons (mass = m). The operating magnetic field
(B) used in the cyclotron and the kinetic energy (K) of the
proton beam, produced by it, are given by
(a)
m
B=
e
n
and K = 2mp
2
n
2
R
2
(b)
2m
B
e
pn
= and K = m
2
pnR
2
(c)
2m
B
e
pn
= and K = 2mp
2
n
2
R
2
(d)
m
B
e
n
= and K = m
2
pnR
2
55.A proton
carrying 1 MeV kinetic energy is moving in a
circular path of radius R in uniform magnetic field. What
should be the energy of an a-particle to describe a circle of
same radius in the same field?
(a) 2 MeV (b) 1 MeV
(c) 0.5 MeV (d) 4 MeV
56.A magnetic needle suspended parallel to a magnetic field
requires
3J of work to turn it through 60°. The torque
needed to maintain the needle in this position will be
(a)2 3J(b) 3J (c)3J (d)
3
J
2
57.A current loop in a magnetic field
(a) can be in equilibrium in one orientation
(b) can be in equilibrium in two orientations, both the
equilibrium states are unstable
(c) can be in equilibrium in two orientations, one stable
while the other is unstable
(d) experiences a torque whether the field is uniform or
non-uniform in all orientations

535. 531Moving Charges and Magnetism
58.A charged particle moves through a magnetic field in a
direction perpendicular to it. Then the
(a) velocity remains unchanged
(b) speed of the particle remains unchanged
(c) direction of the particle remains unchanged
(d) acceleration remains unchanged
59.Electron move at right angle to a magnetic field of 1.5 × 10
–
2
tesla with speed of 6 × 10
7
m/s. If the specific charge of the
electron is 1.7 × 10
11
C/kg. The radius of circular path will be
(a) 3.31 cm (b) 4.31cm
(c) 1.31 cm (d) 2.35 cm
60.A conducting circular loop of radius r carries a constant
current i. It is placed in a uniform magnetic field B such that
B is perpendicular to the plane of the loop. The magnetic
force acting on the loop is :
(a)ir B (b) 2priB
(c) zero (d)priB
61.An infinitely long straight wire contains a uniformly
continuous current of 10A. The radius of the wire is 4× 10
–
2
m. The magnetic field at 2 × 10
–2
m from the centre of the
wire will be:
(a)0 (b) 2.5 × 10
– 5
T
(c) 5.0 × 10
– 5
T (d) none of these.
62.A proton moving vertically downward enters a magnetic
field pointing towards north. In which direction proton will
deflect?
(a) East (b) West
(c) North (d) South
63.When a charged particle moving with velocity
r
v is subjected
to a magnetic field of induction B
ur
, the force on it is non-
zero. This implies that
(a) angle between
r
v and B
ur
is necessarily 90°
(b) angle between
r
v and B
ur
can have any value other
than 90°
(c) angle between
r
v and B
ur
can have any value other
than zero and 180°
(d) angle between
r
v and B
ur
is either zero or 180°
64.A coil carrying electric current is placed in uniform magnetic
field, then
(a) torque is formed
(b) e.m.f is induced
(c) both (a) and (b) are correct
(d) none of the above
65.A charge moving with velocity v in X-direction is subjected
to a field of magnetic induction in negative X-direction. As
a result, the charge will
(a) remain unaffected
(b) start moving in a circular path Y–Z plane
(c) retard along X-axis
(d) move along a helical path around X-axis
66.The magnetic field at a distance r from a long wire carrying
current i is 0.4 tesla. The magnetic field at a distance 2r is
(a) 0.2 tesla (b) 0.8 tesla
(c) 0.1 tesla (d) 1.6 tesla
67.A straight wire of length 0.5 metre and carrying a current of
1.2 ampere is placed in uniform magnetic field of induction 2
tesla. The magnetic field is perpendicular to the length of
the wire. The force on the wire is
(a) 2.4 N (b) 1.2 N
(c) 3.0 N (d) 2.0 N
68.A moving coil galvanometer has a resistance of 900 W. In
order to send only 10% of the main current through this
galvanometer, the resistance of the required shunt is
(a) 0.9 W(b) 100 W (c) 405 W (d) 90 W
69.A uniform magnetic field acts at right angles to the direction
of motion of electron. As a result, the electron moves in a
circular path of radius 2cm. If the speed of electron is
doubled, then the radius of the circular path will be
(a) 2.0 cm (b) 0.5 cm
(c) 4.0 cm (d) 1.0 cm
70.An electron moves in a circular orbit with a uniform speed
v. It produces a magnetic field B at the centre of the
circle. The radius of the circle is proportional to
(a)
B
v
(b)
B
v
(c)
v
B
(d)
v
B
71.The magnetic induction at a point P which is at a distance of
4 cm from a long current carrying wire is 10
–3
T. The field of
induction at a distance 12 cm from the current will be
(a) 3.33 × 10
–4
T(b) 1.11 × 10
–4
T
(c) 3 × 10
–3
T (d) 9 × 10
–3
T
Directions for Qs. (72 to 75) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following.
(a)Statement -1 is false, Statement-2 is true
(b)Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c)Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d)Statement -1 is true, Statement-2 is false
72. Statement 1 : If the current in a solenoid is reversed in
direction while keeping the same magnitude, the magnetic
field energy stored in the solenoid decreases.
Statement 2 : Magnetic field energy density is proportional
to square of current.

536. 532 PHYSICS
Exemplar Questions
1.Two charged particles traverse identical helical paths in a
completely opposite sense in a uniform magnetic field
0
ˆB=Bk .
(a) They have equ
al z-components of momenta
(b) They must have equal charges
(c) They necessarily represent a particle, anti-particle pair
(d) The charge to mass ratio satisfy
21
ee
0
mm
æöæö
+=ç÷ç÷
èøèø
2.Biot-Savart law indic
ates that the moving electrons (velocity
v) produce a magnetic field B such that
(a) B is perpendicular of
(b) B is parallel to v
(c) it obeys inverse cube law
(d) it is along the line joining the electron and point of
observationt.
3.A current carrying circular loop of radius R is placed in the x-
y plane with centre at the origin. Half of the loop with x > 0 is
now bent so that it now lies in the y-z plane.
(a) The magnitude of magnetic moment now diminishes
(b) The magnetic moment does not change
(c) The magnitude of B at (0, 0, z), z > R increases
(d) The magnitude of B at (0, 0, z), z >> R is unchanged
4.An electron is projected with uniform velocity along the
axis of a current carrying long solenoid. Which of the
following is true?
(a) The electron will be accelerated along the axis
(b) The electron path will be circular about the axis
(c) The electron will experience a force at 45° to the axis and
hence execute a helical path
(d) The electron will continue to move with uniform velocity
along the axis of the solenoid
5.In a cyclotron, a charged particle
(a) undergoes acceleration all the time
(b) speeds up between the dees because of the magnetic field
(c) speeds up in a dees
(d) slows down within a dee and speeds up between dees
NEET/AIPMT (2013-2017) Questions
6.A current loop in a magnetic field [2013]
(a) can be in equilibrium in one orientation
(b) can be in equilibrium in two orientations, both the equi-
librium states are unstable
(c) can be in equilibrium in two orientations, one stable
while the other is unstable
(d) experiences a torque whether the field is uniform or
non-uniform in all orientations
7.When a proton is released from rest in a room, it starts with
an initial acceleration a
0
towards west. When it is projected
towards north with a speed v
0
it moves with an initial
acceleration 3a
0
towards west. The electric and magnetic
fields in the room are respectively [2013]
(a)
0
ma
e
west,
0
0
2ma
ev
down
(b)
0
ma
e
east,
0
0
3ma
ev
up
(c)
0
ma
e
east,
0
0
3ma
ev
down
(d)
0
ma
e
west,
0
0
2ma
ev
up
73. Statement 1 : If a c
harged particle is released from rest in a
region of uniform electric and magnetic fields parallel to
each other, it will move in a straight line.
Statement 2 : The electric field exerts no force on the particle
but the magnetic field does.
74. Statement 1 : A cyclotron cannot accelerate neutrons.
Statement 2 : Neutrons are neutral.
75. Statement 1 : The magnetic field at the centre of the circular
coil in the following figure due to the currents I
1
and I
2
is
zero.
I
1
I
2
II
q
Statement 2 : I
1

= I
2
implies that the fields due to the current
I
1
and I
2
will be balanced.

537. 533Moving Charges and Magnetism
8.A long straight wire carries a certain current and produces a
magnetic field of 2 × 10
–4

2
weber
m
at a perpe
ndicular distance
of 5 cm from the wire. An electron situated at 5 cm from the
wire moves with a velocity 10
7
m/s towards the wire along
perpendicular to it. The force experienced by the electron
will be [NEET Kar. 2013]
(charge on electron =1.6 × 10
–19
C)
(a) Zero (b) 3.2 N
(c) 3.2 × 10
–16
N(d) 1.6 × 10
–16
N
9.A circular coil ABCD carrying a current i is placed in a uniform
magnetic field. If the magnetic force on the segment AB is
F
r
, the force on the remaining segment BCDA is
[NEET Kar. 2013]
A
D B
C
i
(a)
F
r
(b)
F-
r
(c)3F
r
(d)3F-
r
10.Two iden tical long conducting wires AOB and COD are
placed at right angle to each other, with one above othersuch that ‘O’ is their common point for the two. The wires
carry I
1
and I
2
currents respectively. Point ‘P’ is lying at
distance ‘d’ from ‘O’ along a direction perpendicular to the
plane containing the wires. The magnetic field at the point
‘P’ will be : [2014]
(a)
01
2
I
2dI
æöm
ç÷
pèø
(b)
0
12
(I I)
2d
m
+
p
(c)
220
12
(I I)
2d
m
-
p
(d)
2 2 1/20
12
(I I)
2d
m
´
p
11.An electron
moving in a circular orbit of radius r makes n
rotations per second. The magnetic field produced at the
centre has magnitude: [2015]
(a) Zero (b)
2
0
ne
r
m
(c)
0
ne
2r
m
(d)
0
ne
2r
m
p
12.A wire carrying current I has the shape as shown in adjoining
figure. Linear parts of the wire are very long and parallel toX-axis while semicircular portion of radius R is lying in Y-Z
plane. Magnetic field at point O is : [2015]
I
I
X
R
Z
Y
O
(a)
$
()
0I
B – i 2k
4R
m
= m´
p
ur
$
(b)
$
( )
0I
B – i 2k
4R
m
= p+
p
ur
$
(c)
$
()
0I
B i – 2k
4R
m
=p
p
ur
$
(d)
$
()
0I
B i 2k
4R
m
= p+
p
ur
$
13.A proto
n and an alpha particle both enter a region of uniform
magnetic field B, moving at right angles to field B. If the
radius of circular orbits for both the particles is equal and
the kinetic energy acquired by proton is 1 MeV the energy
acquired by the alpha particle will be: [2015 RS]
(a) 0.5 MeV (b) 1.5 MeV
(c) 1 MeV (d) 4 MeV
14.A rectangular coil of length 0.12 m and width 0.1 m having 50
turns of wire is suspended vertically in a uniform magnetic field
of strength 0.2 weber/m
2
. The coil carries a current of 2A. If
the plane of the coil is inclined at an angle of 30° with the
direction of the field, the torque required to keep the coil in
stable equilibrium will be : [2015 RS]
(a) 0.20 Nm (b) 0.24 Nm
(c) 0.12 Nm (d) 0.15 Nm
15.A square loop ABCD carrying a current i, is placed near and
coplanar with a long straight conductor XY carrying a
current I, the net force on the loop will be : [2016]
Y
I
X A D
LL/2
L1
B C
(a)
0
2 Ii
3
m
p
(b)
0Ii
2
m
p
(c)
0
2 IiL
3
m
p
(d)
0
IiL
2
m
p

538. 534 PHYSICS
16.A long straight wire o
f radius a carries a steady current I.
The current is uniformly distributed over its cross-section.
The ratio of the magnetic fields B and B', at radial distances
a
2
and 2a respectively, from the axis of the wire is :[2016]
(a)
1
4
(b)
1
2
(c)1 (d) 4
17.A long sol
enoid has 1000 turns. When a current of 4A flows
through it, the magnetic flux linked with each turn of thesolenoid is 4 × 10
–3
Wb. The self inductance of the solenoid
is : [2016]
(a) 4H (b) 3H
(c) 2H (d) 1H
18.A 250-turn rectangular coil of length 2.1 cm and width 1.25cm carries a current of 85 mA and subjected to magnetic field
of strength 0.85 T. Work done for rotating the coil by 180ºagainst the torque is [2017]
(a) 4.55 mJ (b) 2.3 mJ
(c) 1.15 mJ (d) 9.1 mJ
19.An arrangement of three parallel straight wires placed
perpendicular to plane of paper carrying same current 'I
along the same direction is shown in fig. Magnitude of force
per unit length on the middle wire 'B' is given by [2017]
dB
A
d
90°
C
(a)
2
0
2i
d
m
p
(b)
2
0
2i
d
m
p
(c)
2
0
i
2d
m
p
(d)
2
0
i
2d
m
p

539. 535Moving Charges and Magnetism
EXERCISE - 1
1. (d) 2. (d)
3. (b) F = iB l sin q. This is maximum when sin q = 1
or q = p/2.
4. (d)
2
20
1
10
r
ni2
.
4r
ni2
.
4
B
p
p
m
-
p
p
m
=
ú
û
ù
ê
ë
é
-
m
=
2
2
1
10
r
ni
r
ni
2
5. (c) I
f r is the radius of the circle,
then L = 2pr or,
L
r
2
=
p
Area =
2 2 22
r L/4 L/4p=pp=p
6. (c)The point P is lying symmetrically w.r.t. the two long
straight current carrying conductors. The magnetic
fields at P due to these current carrying conductors are
mutually perpendicular.
7. (a)
mvm
r or,r
Bqq
=µ for th e same value of v and B.
8. (a) r = mv/Bq is same for both.
9. (a) Bqv =
r
mv
2

Bq
mv
r=Þ
10. (a
) Current sensitivity =
K
nBA
where
K is constant of torsional rigidity.
11. (a)
12. (a)
13. (b) 14. (b)
15. (b) In cyclotron, energy with which aceleration takes place
is in term of MeV.
16. (a)
17. (d)
18. (c) B
axis
=
2
3
0
R
x2
NI
÷
÷
ø
ö
ç
ç
è
æm
2
RBµ
So, when radius is doubled, magnetic field becomes
four times.
19. (a) To convert a galvanometer into an ammeter, one needs
to connect a low resistance in parallel so that maximum
current passes through the shunt wire and ammeter remains
protected.
20. (b) It will compress due to the force of attraction between
two adjacent coils carrying current in the same
direction.
21. (b) Magnetic force acts perpendicular to the velocity.
Hence speed remains constant.
22. (c)
00
B Ni=m ;
0
1 0 00
N
B () (2 i) NiB
2
æö
=m =m=
ç÷
èø
Þ B
1
= B
23. (c)
1
;
d ed
ei
dt R R dt
ff
= ==
Total c
harge induced =
1
.
d
i dt dt
R dt
f
=òò

2
1
21
11
()d
RR
f
f
= f= f -fò
24. (b
) Energy is stored in magnetic field.
25. (b) Tesla is the unit of magnetic field.
EXERCISE - 2
1.
(d) The straight part will not contribute magnetic field at the
centre of the semicircle because every element of the
straight part will be 0º or 180º with the line joining the
centre and the element
Due to circular portion, the field is
r4
i
r2
i
2
1
00
m
=
m
Hence
total field at
0
i
O tesla
4r
m
=
2. (c) where
r
i2
4
B
0p
p
m
=
A106.1
2
106.12
t
e2
i
19
19
-
-
´=
´´
==
19
190 0
0
i 1.6 10
B 10T
2r 2 0.8
-
-m m´´
\
= = =m´
´
3. (d) 510104nB
7
0 ´´´p=Im=
- 5
2 10 T.
-
= p´
4. (c)
2 316
19
mv mv (9 10 ) 10
Bqv orB
r rq 0.1 (1.6 10 )
-
-
´´
= ==
´´
T105.5
5-
´
=
5. (a)
2
kk
1
E mv or mv 2 E m and
2
==
k
2Emmv
r
Bq Bq
==
6. (d) Bq
v = mv
2
/r or q/m = v /rB.
7. (b)F = Il B sin q = 3 ×0.40 × (500 × 10
–4
) × sin 30º
= 3 × 10
–2
N.
8. (d)
Bqr
v
m
=
31
2193
101.9
102106.1105.4
-
---
´
´´´´´
=
s/m1058.1
7
´=
9. (c) )ii(
r
4
4)2/r(
i2
4)2/r(
i2
4
B
12
01020
-
p
m
=
m
m
-
p
m
=
.
2
)5.25(
5
4
4
00
p
m
=-
p
m
=
10. (c)
33
E vB 2 10 1.5 3 10 V / m.= =´ ´ =´
1
1. (a) In coil A,
.
R
2
4
B
0Ip
p
m
=
I
B
R
\µ ;
Hints & Solutions

540. 536 PHYSICS
Hence,
1
2
2R
RB
B
2
2
1
1
2
1
==×=
I
I
12. (c) Torq
ue on loop t = nIAB cos q; Here q = 90º
\ t = 0.
13. (b)
0 12
2II
F
4r
m
=´
p
l N1082
1.0
210210
5
7
-
-
´=´
´´´
=
14. (c)
15. (a)
00
µII
B
2r 2 4r
mqq
= ´=
pp
16. (d) No magnetic force acts on the electron and force due
to electric field will act opposite to its initial direction
of motion. Hence its velocity decreases in magnitude.
17. (d) Since no current is enclosed inside the hollow
conductor. Hence B
inside
= 0.
18. (c)
1 2mV
r
Bq
=
1x
2y
Rm
Rm
=
2
x1
y2
mR
mR
æö
Þ= ç÷
èø
19. (d) B
total
= 4B
side
0
total
I
B 4 sin sin
a 44
2
2
m ppéù
=+
êú
æöëû
pç÷
èø
0
total
22I
B
a
m
=
p
20. (b)
mvp
r
qB qB
==

p
p
rq2e2
r q e1
a
a
Þ= ==
21. (b) In a uniform magnetic field, a charged particle is moving
in a circle of radius R with constant speed v.
\
2
mv mv
Bqv or, R
R Bq
== .....(1)
Time perio
d,
2R 2mv 2m
T
v Bqv Bq
ppp
=== .....(2)
Time peri
od T does not depend on both R and v
because when v is changed, R is also changed
proportionately and for period, it is R/v that is taken.
22. (b) 23. (c)
24. (b)
F
i i
d2
iiF
210
p
m
=
l
=
d2
i
2
0
p
m
(attractive as current
is in the same direction)
25. (c) The magnetic field is
02I
B
4r
m
=
p
72 100
10
4
- ´
=´ = 5 × 10
–6
T
E
N
100A
4m
Ground
S
B
W
According to right hand palm rule, the magnetic field is
directed towards south.
26. (a) Since magnetic force is always perpendicular to the
velocity of electron, so it can only change the direction
of velocity of electron, but it (the magnetic force) cannot
accelerate or deaccelerate the electron.
27. (c) Magnetic dipole moment
22e e erv
miA r r.
T(2r/v)2
= = ´p = ´p =
p
19 12 6
1.6 10 50 10 2.2 10
2
--
´ ´´ ´´
=
242
8.8 10 Am .
-
=´
28. (c) So
Bq
Em2
r
dd
deutron= ;
Bq
Em2
r
pp
proton
=
For same r
adius, B and q
m
p
E
p
= m
d
E
d
d
pd
p
m2
E E 50 100keV
m1
Þ = =´=
29. (c)
Time taken by proton to make one revolution
.sec5
5
25
m==
As
2
1
1
2
1
2
q
q
m
m
T
T
so;
qB
m2
T ´=
p
=
or
q2
q
m
m45
qm
qm
TT
1
1
21
12
12 ´
´
== = 10 m sec.
30. (a)
Let l
1
, l
2
be the lengths of the two parts PRQ and PSQ of
the conductor and r be the resistance per unit length of
the conductor. The resistance of the portion PRQ will be
R
1
= l
1
r
O
I
2
S Q
P
R
I
1

541. 537Moving Charges and Magnetism
The resistance of the portion PSQ will be R
2
= l
2
r
Pot. diff. across P and Q = I
1
R
1
= I
2
R
2
or I
1
l
1
r = I
2
l
2
r or I
1
l
1
= I
2
l
2
...(1)
Magnetic field induction at the centre O due to currents
through circular conductors PRQ and PSQ will be
= B
1
– B
2

0011 22
22
sin 90º sin 90º
0.
44 rr
mmII
=-=
pp
ll
31. (c) B at O will be due to the following portions
(i) Vertical straight portion. This is zero.
(ii) Circular portion. This is given by
00
circular
1ii
B
2 2r 4r
mm
==
(iii) Straight h
orizontal portion. This is given by
0
straight
i
B
4r
m
=
p
00
Total
ii
B
4r 4r
mm
\ =+
p
32. (a)Fo r case (a) magnetic field due to straight portions is
cancelled & the magnetic field due to semi circular arcof radius r at P is
p´÷
ø
ö
ç
è
æ
p
m
=
p´
p
m
=
r4
i
r
i
4
B
oo
a
It is in upward dire
ction & we take upward direction
negative, So
p÷
ø
ö
ç
è
æ
p
m
-= .
r4
i
B
o
a
r
For case (b) Due to straight portion the magnetic field
is zero so the magnetic field due to semi circular arc is
p´÷
ø
ö
ç
è
æ
p
m
=
r4
i
B
o
b
r
(in down wards direction so +ive sign)
For case (c) Magnetic field due to straight portion is
r
i
4
o
p
m
-= (upward dir
ection)
Magnetic field due to circular arc which substand an
angle 3p/2 at centre is
2
3
r4
i
o p
´÷
ø
ö
ç
è
æ
p
m
= (down war
d direction)
so
o
c
i3
B1
4r2
m pæöæö
=- ç÷ç÷
èøèøp
r
so
pæö
=-pp-ç÷
èø
rrr
a bc
3
B :B :B : : 1
2
31
::
2 2 42
-pppæöæö
=-ç÷ç÷
èøèø
33. (d) Here, the wire does not produce any magnetic field at O
because the conductor lies on the line of O. Also, the
loop does not produce magnetic field at O.
34. (a)
Ι
0 II
B
4rrr
m péù
= ++
êú
pëû
02
4rr
mI pIéù
+
êú
pëû
0
2 r 4r
éùII
=m+
êú
pëû
(The direction of B
r
is into the page.)
35. (a) Due to flow of current in same direction at adjacent
side, an attractive magnetic force will be produced.
36. (c) The magnetic field is perpendicular to the plane of the
paper. Let us consider two diametrically opposite
elements. By Fleming's Left hand rule on element AB
the direction of force will be Leftwards and the
magnitude will be
dF = Idl B sin 90° = IdlB
x
x
x
x
x
x x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
A
B
C
D
dF
dl
dldF
I
On eleme
nt CD, the direction of force will be towards
right on the plane of the papper and the magnitude will
be dF = IdlB.
37. (b) From Lorentz equation
)k
ˆ
(Bi
ˆ
euF
0
-´-= j
ˆ
euB
0
-=
hence it will complete a semicircular arc and comes out
of the region at a position y, such that y < 0
38. (a)
39. (d) For a given perimeter the area of circle is maximum. So
magnetic moment of (S) is greatest.
40. (d) Magnetic moment µ = IA
2R
SinceT
v
p
=
Also,
q qv
I
T 2R
==
p
\ ()
2qv qvR
R.
2R2
æö
m= p=
ç÷
pèø
41. (a
) [Hint
r2
i
B
0
p
m
=Þ, where r is
distance of point from the
wire, where we want to calculate the magnetic field. It isclear from expression that B is independent of thicknessof wire.]
42. (a) The direction of the magnetic field due to a current is
given by right hand curled fingers rule. Therefore atAB axis, the components of magnetic field will canceleach other and the resultant magnetic field will be zeroon AB.
43. (d)
44. (b) [Hint
qvB
r
mv
2
=Þ].
45. (b)
Let l be length of wire

542. 538 PHYSICS
Ist ca
se : l = 2pr Þ
p
=
2
r
l
l
I
r2
I
B
00m
=
p
m
=
2nd Case : )r2(2¢p=l Þ
p
=¢
4
r
l
2
I2
4
2
In
B
00
ll
m
=
p
p
m
=¢ (where n =
2)
on putting the value of B Þ B¢ =
B4
I
4
0
=÷
ø
ö
ç
è
æm
l
46. (a) The
direction of B is along ()-
Ù
k
\ The magnetic force
j
ˆ
QvB)k
ˆ
(B)i
ˆ
v(Q)Bv(QF =-´=´=
Þ a l o n g O Y..
4
7. (b) The force on the two arms parallel to the field is zero.
<
<
<
<
– F
F
B
\ Force on remaining arms = – F
48. (b) Magnetic fields due to the two
parts at their common centre are
respectively,
0
4
y
i
B
R
m
= and
0
4
z
i
B
R
m
=
i
i
z
y
i
Resultant field =
22
yz
BB+
2
2
00
44
ii
RR
mmæöæö
=+ç÷ç÷
èøèø
=
00
2.
422
ii
R R
mm
=
49. (b) Le
t a current i be flowing in the loop ABC in the
direction shown in the figure. If the length of each ofthe sides AB and BC be x then|F|
r
= i x B
Direction of
magnetic field
B
A
C
where B is the magnitude of the magnetic force.
The direction of F
r
will be in the direction perpendicular
to the plane of the paper and going into it.
By Pythagorus theorem,
AC =
22
x x 2x+=
\ Magnitude
of force on AC = i
2x B sin 45°
1
i 2xB
2
=´ = ixB = |F|
r
The direction of the force on AC is perpendicular to
the plane of the paper and going out of it. Hence, force
on AC = F-
r
50. (b)v
r
and B
r
are in same direction so that magnetic force
on electron becomes zero, only electric force acts. But
force on electron due to electric field is opposite to the
direction of velocity.
51. (d)
I
1
F
4
F
2
F
3
F
1
I
F
1
> F
2
as
1
F,
d
µ and F
3
and F
4
are e qual and opposite.
Hence, the net attraction force will be towards theconductor.
52. (a) Magnetic field at the centre of the ring is
0
qf
2R
m
53. (a)
B
1
B
2
The magnetic field, due the coil, carrying current I
Ampere
0
1
2
I
B
R
m
=
The magnetic field due
to the coil, carrying current 2I
Ampere0
2
(2)
2
I
B
R
m
=
The resultant B

543. 539Moving Charges and Magnetism
22
net 1 2 12
2 cos,B B B BB=++q q = 90°
22 0
net 12
(2)
14
2
I
B BB
R
m
=+=+
0
5
2
I
R
m
=
54. (c) Time period of cyclotron is
12m
T
eB
p
==
u
;
2m
B
e
p
=u ;
mp
R
eB eB
u
==
ÞP = eBR =
2m
e
e
pu
´ R = 2pmuR
K.E. =
22
(2)
22
p mR
mm
pu
= = 2p
2
mu
2
R
2
55. (b) According to the principal of circular motion in a
magnetic field
F
c
= F
m
Þ
2
mv
qVB
R
=
Þ
2.
===
mv P mk
R
qB qB qB
2(4)'
2
a
=
mK
R
qB

'
a
=
RK
RK
but R = R
a
(given)
Thus K = K¢ = 1 MeV
56. (b) According to work energy theorem
W =U
final
– U
initial
= MB (cos 0 – cos 60°)
3J
2
MB
W== ...(i)
3
sin 60
2
MB
M B MB
æö
t= ´ = °=
ç÷
èø
rr
...(ii)
From equation (i) and (ii)

233
2
´
t= = 3J
57. (c) A current loop in a magnetic field is in equilibrium in
two orientations one is stable and another unstable.
Qt
r
= MB´
u ur ur
= M B sin q
If q = 0° Þ t = 0 (stable)
If q = p Þ t = 0 (unstable)
Do not experience a torque in some orientations
Hence option (c) is correct.
58. (b) Magnetic force acts perpendicular to the velocity. Hence
speed remains constant.
59. (d) B = 1.5 × 10
–2
T, q = 90°, sin q = 1
v = 6 × 10
7
m/s,
e
m
= 1.7 × 10
11
C/kg
r =
mv
Be
=
7
2 11
6 10
1.5 10 1.7 10
-
´
´ ´´
= 2.35 × 10
–2
m = 2.35 cm
60. (c) The force on each point on loop is radially outward and so
net force = 0
61. (b)
0
2
ir
B.
2R
m
=
p
=
2
7
2
2 10
2 10 10
(4 10 2)
-
- ´
´ ´´
´-

5
2.510T
-
=´
62. (a) Proton will represent the direction of current, so the direction
of current is vertically downward. By Fleming's left hand
rule, force acting on the proton in east direction.
63. (c) As F qVBsin=q
r ur ur
F is zero for sin 0° or sin 180° and is non-zero for angle
between VandB
ur ur
any value other than zero and 180°.
64. (a) A current carrying coil has magnetic dipole moment.
Hence, a torque m
pB´
rr
acts on it in magnetic field.
65. (a) The force acting on a charged particle in magnetic field
is given by
F
r
= q (vB´
rr
) or F = qvB sin q,
When angle between v and B is 180°,
F = 0
66. (a)
0 1
or
2
i
BB
rr
m
=µ
p
When r is doubled, the magnetic field becomes half,
i.e., now the magnetic field will be 0.2 T.
67. (b)F = Bil = 2 ×1.2 × 0.5 = 1.2 N
68. (b)g s g gs
I 0.1I, I 0.9 I ; S I R / I===
0.1 900 / 0.9 100 .=´ =W
69. (c) or
mv
r rv
qB
=µ
As v is doubled, the radius also becomes double.
Hence, radius = 2 × 2 = 4 cm
70. (d)r
mv
qB
= Þ r µ
v
B
71. (a)
0 1
2
I
BB
rr
m
= Þµ
p
As the distance is increased to three times, the magnetic induction reduces to one third. Hence,
341
10 tesla 3.33 10 tesla
3
B
--
=´ =´
72. (a) Reversing the direction of the current reverses the
direction of the magnetic field. However, it has no effect on the magnetic-field energy density, which is proportional to the square of the magnitude of the magnetic field.

544. 540 PHYSICS
73. (c) Due to electric field, the force is F qE=
rr
in the direction
of E
r
. Since E
r
is parallel to B
r
, the particle velocity v
r
(acquired due to force F
r
) is parallel to B
r
. Hence B
r
will not exert any force since vB0´=
rr
and the motion
of the particle is not affected by B
r
.
74. (b) Neutrons are neutral.
75. (d)
1
12
2
I2
II(2)
I
p-q
= Þ q= p-q
q
........... (1)
01
1
I
B.
2 2R
mq
=
p
and
02
2
I2
B.
2 2R
mp-q
=
p
Using (1), we get B
1
= B
2
.
EXERCISE - 3
Exemplar Questions
1. (d) As we know that the uniqueness of helical path is
determined by its pitch
P(Pitch) =
2 mvcos
Bq
pq
Where q is angle of velocity of charge particle with x-
axis
For the given pitch d correspond to charge particle, we
have
q
m
=
2 vcos
constant
BP
pq
=
If motion is not helical, (q = 0)
As charged particles traverse identical helical paths in a completely opposite direction in a same magnetic field B, LHS for two particles should be same and of
opposite sign.
\
12
ee
mm
æöæö
+ç÷ç÷
èøèø
=0
2. (a) By Biot-Savart law
dB =
2
Id sin
r
ql
=
Id
r
´æö
ç÷
èø
l
In Biot-Savat’s law, magnetic field B||idl × r and idl due
to flow of electron is in opposite direction of v and by direction of cross product of two vectors
B ^ V
So, the magnetic field is ^ to the direction of flow of
charge.
3. (a) As the direction of magnetic moment of circular loop
of radius R placed in the x-y plane is along z-direction and given by M = I (pr
2
), when half of the loop with x >
0 is now bent so that it now lies in the y-z plane, the magnitudes of magnetic field moment of each semicircular loop of radius R lie in the x-y plane and the y-z plane is M' = I(pr
2
)/4 and the direction of magnetic
field moments are along z-directlon and x-direction respectively.
Then resultant is :
M
net
=
22
MM¢¢+
=2M¢ = ()
2
2Ir4p
So, M
net
< M or M diminishes.
Hence, the magnitude of magnetic moment is now
diminishes.
4. (d) Magnetic Lorentz force :
F = qVB sin q
Magnetic Lorentz force electron is projected with
uniform velocity along the axis of a current carrying
long solenoid F = –qvB sin 180° = 0(q = 0°) as magnetic
field and velocity are parallel and electric field is zero (E
= 0) due to this magnetic field (B) perpendicular to the
direction of motion (V). So it will not affect the velocity
of moving charge particle. So the electron will continue
to move with uniform velocity along the axis of the
solenoid
5. (a) There is crossed electric and magnetic field between
dees so the charged particle accelerates by electric field
between dees towards other dees.
So, the charged particle undergoes acceleration as
(i) speeds up between the dees because of the oscillating
electric field.
(ii) speed remain the same inside the dees because of the
magnetic field but direction undergoes change
continuously.
Hence, the charge particle accelerates inside and
between Dees always.
NEET/AIPMT (2013-2017) Questions
6. (c) A current loop in a magnetic field is in equilibrium in
two orientations one is stable and another unstable.
Qt
r
= MB´
u ur ur
= M B sin q
If q = 0° Þ t = 0 (stable)
If q = p Þ t = 0 (unstable)
Do not experience a torque in some orientations
Hence option (c) is correct.

545. 541Moving Charges and Magnetism
7. (a)
When moves with an acceleration a
0
towards west,
electric field
E =
F
q
=
0
ma
e
(West)
When moves with an acceleration 3a
0
towards east,
magnetic field
B =
0
0
2ma
ev
(downward)
8. (c) Given:
Magnetic field B = 2 × 10
–4
weber/m
2
Velocity of electron, v = 10
7
m/s
Lorentz force F = qvB sin q
= 1.6 × 10
–19
× 10
7
× 2 × 10
–4
(Q q = 90°)
= 3.2 × 10
–16
N
9. (b) Here, 0+=
rrr
AB BCDA
FF
Þ =- =-
r rr
BCDA AB
F FF
()
AB
FF=
r
Q
10. (d) Net magnetic field, B =
22
12
BB+
=
22
01 02
II
2d 2d
mmæöæö
+
ç÷ç÷
ppèøèø

01 02
12
II
B andB
2d 2d
mmæö
==
ç÷
ppèø
Q
=
220
12
II
2d
m
+
p
11. (c) Radius of circular orbit = r
No. of rotations per second = n
i.e., T =
1
n
O
r
Magnetic field at its centre, B
c
=?
As we know, current
ee
i
T (1/ n)
==
= en = equivalent current
Magnetic field at the centre of circular orbit,
00
c
i ne
B
2r 2r
mm
==
12. (b) Magnetic field due to segment ‘1’
$0
1
I
B [sin90 sin 0 ] (–k)
4R
m
= °+°
p
u ur
=
$
()
0
3
–I
kB
4R
m
=
p
ur
Magnetic field due to segment 2
()()
00
2
I –I
B–ii
4R 4R
mm
= =p
p
$$
I
X
Z
Y
2
1
O
3
I
\
B
ur
at centre
c123BBBB=++
ur ur ur ur
=
$
()
0
–I
i 2k
4R
m
p+
p
$
13. (c) As we know, F = qvB =
2
mv
R
\ R =
2m(kE)mv
qB qB
=
Since R is same so, KE µ
2
q
m
Therefore KE of a particle
=
2
q
m
=
()
2
2
4
= 1 MeV
14. (a) Here, number of turns of coil, N = 50
Current through the coil I = 2A
Area A = l × b = 0.12 × 0.1m
2
Magnetic field B
r
= 0.2 w/m
2
30°
60°
B
M
Torque required to keep the coil in stable equilibrium.
MBt=´
rr
= MB sin 60° = Ni AB sin 60°
= 50×2×0.12×0.1×0.2×
3
2
=
2
12 3 10
-
´ = 0.20784 Nm

546. 542 PHYSICS
15. (a)
The direction of current in conductor
F
BC
F
CD
C
D
F
AD
A
F
AB
B
i
Y
X
I L
L
XY and AB is same
\F
AB
=
ilB (attractive)
F
AB
= i(L).
() ()
00
I iI
L
2
2
mm
¬=¬
pæö
pç÷
èø
F
BC
opposite to F
AD
F
BC
(­) and F
AD
(¯)
Þcan
cels each other
F
CD
= ilB (repulsive)
F
CD
= i(L)
() ()
00
I iI
3L 3
2
2
mm
®=®
pæö
pç÷ èø
Therefore the n
et force on the loop
F
net
= F
AB
+ F
BC
+ F
CD
+ F
AD
ÞF
net
=
ooo
iI iI 2 iI
33
mmm
-=
ppp
16. (c) For points inside the wire i.e., (r < R)
Magnetic field
B =
0
2
Ir
2R
m
p
For points outside th
e wire (r >
R)
Magnetic field
, B' =
0
I
2R
m
p
\
B
B'
=
()
()
0
2
0
I a/2
2a
1:1
I
2 2a
m
p
=
m
p
17. (d) H
ere, number of turns n = 100; current through the
solenoid i = 4A; flux linked with each turn = 4 × 10
–3
Wb
\Total flux linked, and total
= 1000[4 × 10
–3
] = 4 Wb
f
total
= 4ÞL i = 4
ÞL = 1 H
18. (d) Work done, W = MB(cosq
1
–cosq
2
)
When it is rotated by angle 180° then
W = MB (cos0° – cos 180°) = MB (1 + 1)
W = 2MB
W = 2 (NIA) B
= 2 × 250 × 85 × 10
–6
[1.25 × 2.1 × 10
–4
] × 85 × 10
–2
= 9.1 mJ
19. (c) Force per unit length between two parallel current car-
rying conductors,
F =
012
ii
2d
m
p
Since same current flowing through both the wires
i
i
= i
2
= i
so F
1

2
0
2
i
F
2d
m
==
p
F [due to wire A]
1
F [due to wire C]
2
\ Magnitude of force per unit length on the middle
wire 'B'
F
net
=
2
22 0
12
i
FF
2d
m
+=
p

547. NATURAL MAGNET
A natural magnet is an ore of iron (Fe
3
O
4
) which
(i) attracts small pieces of iron, cobalt and nickel towards it.
(ii) when suspendeded freely, comes to rest along north-south
direction.
The magnets which are obtained artificially are called artificial
magnets, e.g. a bar magnet, a magnetic needle, horse shoe magnet
etc.
BAR MAGNET
A bar magnet consists of two equal and opposite magnetic poles
separated by a small distance. Poles are not exactly at the ends.
The shortest distance between two poles is called effective length
(L
e
) and is less than its geometric length (L
g
).
For bar magnet L
e
= 2l and L
e
=(5/6) L
g
.
For semicircular magnet L
g
= pR and L
e
= 2R.
L =2
e l
L
g
S N
Bar magnet
Pr
operties of Magnets
(i)Attractive property : A magnet attracts small pieces of iron,
cobalt, nickel, etc. and other magnetic subsances.
(ii)Directive property : A freely suspended magnet aligns itself
nearly in the geographical north-south direction.
(iii)Law of magnetic poles : Like magnetic poles repel, and unlike
magnetic poles attract each other.
According to Gauss’s theorem in magnetism, surface
integral of magnetic field intensity over a surface (closed or
open) is always zero i.e.
.(or .)0 .Bds Bds =òò
u ur ur uur ur
Ñ
This theorem establishes that the poles always exist in equal
and unlike pairs.
(iv)Magnetic poles exist in pairs : Isolated magnetic poles do
not exist. If we break a magnet into two pieces, we get two
smaller dipole magnets.
(v) Repulsion is a sure test of magnetism.
MAGNETIC FIELD
The space around a magnet within which its influence can be
experienced is called its magnetic field.
Uniform magnetic field : A uniform magnetic field is one where the
strength of the magnetic field is the same at all points of the field. In
a uniform field, all the magnetic lines of force are parallel to one
another. But in non-uniform magnetic field the strength of
magnetic field is not same at all points of the field and also the
magnetic lines of force are not parallel.
Atom as a Magnetic Dipole
Every atom of a magnetic material behaves as a magnetic dipole,
because electrons in the atom revolve round the nucleus. The
magnetic moment M associated with an atomic dipole as
4
= =m
p
B
neh
Mn
m
where n = 1, 2, 3 ..... de
notes the no. of orbits and
B
eh
4m
m=
p
.
Least v
alue of dipole moment of atom = 9.27 × 10
–24
Am
2
.
B
m is called Bohr magneton.
Most of the magnetic moment is produced due to electron spin,
the contribution of the orbital revolution is very small.
MAGNETIC LINES OF FORCE
Magnetic line of force is an imaginary curve the tangent to which
at a point gives the direction of magnetic field at that point or
the magnetic field line is the imaginary path along which an
isolated north pole will tend to move if it is free to do so.
Properties of Magnetic Lines of Force
(i) Magnetic lines of force are hypothetical lines use to depict
magnetic field in a region and understand certain
phenomenon in magnetism.
Magnetic field lines
in a bar magnet
Direction of magnetic
lines inside the body
of magnet (from south
pole to north pole)
Direction of magnetic
lines outside the body
of magne
t (from north
pole to south pole)
SN
20
Magnetism and
Matter

548. 544 PHYSICS
Magnetic length : The s
hortest distance between the two poles
of a magnet is called its magnetic length. It is less than the
geometrical length of the magnet. This magnetic length is also
called an effective length.
84.0
lengthlGeometric
a
lengthMagnetic
=

Geometrical length
Magnetic length
S N
MAGNETIC MOMENT
The magnetic moment of a magnet in magnitude is equal to the
product
of pole strength with effective length (i.e. magnetic
length). Its direction is along the axis of magnet from south pole
to north pole.
M m 2 (n) | M | 2m=´´Þ=
rr r
ll
If the same bar magnet is bent in a semicircle then
l2r=pÞ
p
=
l2
r
r
mm
Net magnetic moment
2
22¢=´ =´´
p
l
M m rm
42
=´=
pp
lM
m
Where m is pole s
trength, 2l is effective length and
n
r
is unit
vector havin
g a direction from S-pole to N-pole.
The SI unit of
M
r
is A m
2
, which is equivalent to J/T..
Circular current loop as a magnet : A small plane loop of current
behaves as a magnet with a definite dipole moment given by
ˆI=
u u ur
M An
where A is the area of the loop, I the current in the loop and nˆ is
a unit vecto
r perpendicular to the plane of the loop, and its direction
is decided by the sense of flow of current I using the Fleming’s
right hand rule.
Relation between magnetic moment and angular momentum
2
=
u u ur u ur
q
ML
m
Where q is the total charge on a body of mass m rotaing about a
fixed axis.
(ii) Tangent to field line at a point gives us the direction of
magnetic field intensity B
u ur
at that point. No two magnetic
lines of force can inter
sect each other because magnetic
field will have two directions at the point of intersection.
(iii) Magnetic lines of force are continuous curve from north to
south, outside the body of the magnet and from south to
north inside the body of the magnet.
(iv) The number of lines originating or terminating on a pole is
proportional to its pole strength.
Magnetic flux = number of magnetic lines of force
= µ
0
× m
Where µ
0
is number of lines associated with unit pole.
(v) The number of lines of force per unit area at a point gives
magnitude of field at that point. The crowded lines show a
strong field while distant lines represent a weak field.
(vi) The magnetic lines of force have a tendency to contract
longitudinally like a stretched elastic string producing
attraction between opposite pole.
N S
Longitudnal contraction (attraction)
(vii) The magnetic lines of force have a tendency to repel each
other laterally resulting in repulsion between similar poles.
N S NN X
Longitudnal contraction (attraction) Lateral expansion (repulsion)
(v
iii) The region of space with no magnetic field has no lines of
force. At neutral point where resultant field is zero therecannot be any line of force.
SOME TERMS RELATED TO MAGNETISM
Magnetic poles : These are the regions of apparently concentrated
magnetic strength where the magnetic attraction is maximum.
It means that pole of a magnet is located not at a point but over a
region. Magnetic poles exist in pairs. An isolated magnetic pole
(north or south) does not exist. If a magnet is cut into two pieces,
then instead of obtaining separate N-pole and S-pole, each of the
two parts are found to behave as complete magnets.
Magnetic axis : The line passing through the poles of a magnet is
called its magnetic axis.
Magnetic equator : The line passing through the centre of the
magnet and at right angles ot the magnetic axis is called the
magnetic equator of the magnet.

549. 545Magnetism and Matter
Magnetic moment, pole strength and effective length when a magnet is cut
Magnet placed after Pole strength Effective length Magnetic moment
cutting after cutting
N
Length
SBreadth
m 2 l M = 2ml

N S
X X´
m/2 2 l
2
M
2.
2
m
M
1
==l
NS
Y
Y´
m l
2
M
mM
2 ==l

N S
Y
Y´
X X´
m/2 l
4
M
2
m
M
3 ==
l
2l
m m
m 2l l2mM´=
COULOMB’S LAW OF MAGNETIC FORCE
It states that :
(i)The force of attraction or repulsion between two magnetic
poles is directly proportional to the product of their pole
strengths.
(ii)The force of attraction or repulsion between two magnetic
poles is inversely proportional to the square of the distance
between them. This law is also known as inverse square law.
i.e., F µ m
1
m
2
and F µ
1
r
or,
012
2
.
4
m
=
p
mm
F
r
where m
1
an d m
2
are the pole strengths of the two magnetic
poles, r is the distance between them and m
0
is the
permeability of free space.
Unit magnetic pole : A unit magnetic pole may be defined as the
pole which when placed in vacuum at a distance of one metre from
an identical pole, repels it with a force of 10
–7
newton.
TORQUE ON A MAGNET IN A MAGNETIC FIELD
N
S
q
mmB
mB
B
(South pol
e)
m(North pole)
Bar magnet at
an angle . with
the magnetic field B
q
A magnet of
dipole moment M suspended freely in a magnetic
field B experiences a torque
t given by
t=´
ur u u uru ur
MB ; sint=qMB
where q i
s the angle between
uur
M and
ur
B
It is clear from the expression that MB||
max
=t
r
i.e., when dipole is perpendicular to field the torque is maximum
and when they are parallel, the torque is minimum
(for q = 0 or 180º ® ||0t=
r
).
The net fo
rce acting on a bar magnet placed
® in a uniform magnetic field is zero
® in a non-uniform magnetic field is non-zero
Let the length of a bar magnet be 2l and pole strength be m,
the magnetic field is B, and the angle between B and bar
magnet is q. Force on north pole is mB along the field and
that on south pole is mB opposite to the field.
q
mB
mB
B
2l
lsinq
N
S
O

550. 546 PHYSICS
The torq
ue of these two forces about O is
t = 2mBlsinq = MB sinqMBt=´
u u r uur
where M is the magnetic moment of the magnet.
( M 2m)=Ql
This torque tries to align the magnet with the field.
Work Done by External Agent in Rotating the Magnet
If an external agent rotates the magnet slowly, the agent has to
exert a torque MBsinq opposite to that exerted by the field.
Work done by the agent in changing the angle from q to q + dq is
dW = (MBsinq) dq
ò
q
q
qq=
0
d)sinMB(W
ext
)cos(cosMBW
0ext
q-q=
W
ext
is stored as potential energy of the field-magnet system.
Thus
00
U( ) U( ) MB(cos cos )q-q= q-q
If we take
00
U( ) 0 for 90 , thenq = q=
o
Potential energy
( ) ( ) (90 ) c os – .q = q - ° =- q=
uur uur
U U U MB MB
(i) When q = 0, U = – M B (minimum PE)
(ii) When 90ºq= , U = 0
(iii) When 180ºq= , U = MB (maximum PE)
N
S
U = 0
= 90°q
q = 90°
M
B

NS
U = – MB (min)
= 0°q
M
B
Stable equilibrium
N S
U = + MB (max)
= 180°q
M
B
Unstable equilibrium
Work done in Rotating a Uniform Magnetic Dipole
in a Magnetic Field
Work done in deflecting the dipole through an angle q is,
W = MB (1 – cos q)
If q = 0, cos q = 1 then W = MB (1 – 1) = 0
If q = 90°, cos q = 0 then W = MB
If q = 180°, cos q = –1, then W = 2MB
GAUSS'S LAW IN MAGNETISM
The surface integral of magnetic field
r
B over a closed surface S
is alway
s zero.
Mathematically
.0
®®
=òÑ
S
B da
(1)Isolated magnetic pole
s do not exist is a direct consequence
of gauss's law in magnetism.
(2)The total magnetic flux linked with a closed surface is always
zero.
(3)If a number of magnetic field lines are leaving a closed
surface, an equal number of field lines must also be entering
the surface.
Magnetic Flux
The magnetic flux through a given area may be defined as the
total number of magnetic lines of force passing through this
area. It is equal to the product of the normal components of the
magnetic field B and the area over which it is uniform. In general,
Magnetic flux,
. cosf==qòò
u ur u u ur
AA
B dA BdA , where q is an gle
between normal to the area dA with magnetic field B.
Magnetic flux linked with a closed surface is zero i.e., 0dA.B
s
=
ò
The S.I. Unit of mag
netic flux is weber (Wb) : If a magnetic field of
1 tesla passes normally through a surface of area 1 square metre,
then the magnetic flux linked with this surface is said to be 1
weber.
Oscillations of a Bar Magnet in a Magnetic Field
A freely suspended magnet of magnetic moment M and of moment
of inertia I oscillates simple harmonically in a magnetic field B with
frequency
B
N
S
–m
+m
mB
mB
q
Freely suspended bar
magnet, at an angle q
with the magnetic field B
1I
, ,2
2I
n= \ =p
p
MB
Time period T
MB
MAGNETIC FIELD DUE TO A BAR MAGNET (i)Magnetic field intensi
ty B
1
due to a bar magnet at any point
on the axial line of the magnet is
0
1
2 22
2
4()
m
=
p -l
Md
B
d
S N
d
2l
B
1
P

551. 547Magnetism and Matter
where d = distance of the point from the centre of the magnet.
The direction of B
1
is along SN.
0
N 2
m
B
4(d)
m
=
p-l
where, m is pole strength.
B
N
is magnetic field due to north pole, it is directed away from the
magnet.
2
0
S
)d(
m
4
B
l+p
m
=
it is directed tow
ards the magnet.
NS
BBB\=-
ú
ú
û
ù
ê
ê
ë
é
+
-
-p
m
=Þ
22
0
)d(
1
)d(
1
4
m
B
ll
222
0
222
0
)d(
Md2
4)d(4
)d4(m
ll
l
-p
m
=
-p
m
=
[ M m(2L)]=Q
If d > > l,
0
3
2M
B
4d
m
=
p
(ii) Magnetic fie
ld intensity (B
2
) due to A bar magnet at any
point on the equatorial line of the bar magnet is
0
2
2 2 3/2
4()
m
=
p +l
M
B
d
The direc
tion of B
2
is along a line parallel to NS.
S
N
2l
O
B
P
d
3
0
N
NP
NPm
4
B
p
m
= ,
3
0
S
PS
PSm
4
B
p
m
=
Now,
2 2 3/2
NPPS( d)==+ l
Resultant field at P is,
NS
BBB=+
uur uuuruuur
)NS(
)d(
m
.
4
)PSNP(
)d(
m
.
4
B
2/322
0
2/322
0
ll +p
m
=+
+p
m
=Þ
2/322
0
2/322
0
)d(
M
4)d(
ml2
4
|B|
ll +p
m
=
+p
m
=
If d > > l,
0
3
M
|B|
4d
m
=
p
u ur
The magnetic field at any point having polar coordinates (r, q)
relative to centre of magnet or loop
B =
20
3
(13cos)
4
m
+q
p
M
r
and direction is given by
1
tan tan
2
a=q
MAGNETIC POTENTIAL
The magnetic potential at a point is defined as the work done in
carrying a unit N-pole from infinity to that point against the
field. It may also be defined as the quantity whose space rate of
variation in any direction gives the intensity of the magnetic field
i.e.,
=-
B
dV
B
dx
(i)Magnetic potential due to a point dipole, at a distance r
from the pole of strength m is given by
0
. (joule/Wb)
4
m
=
p
B
m
V
r
B due t
o a pole of Pole strength m at a distance r is given by
0
2
4
m
=
p
m
B
r
Now V
0
at A i s the work done is bringing a unit pole from
infinity to A.
r
0
A 2
m
V V a dr
4r
¥
¥
m
\-=
p
ò
0
4
m
Þ=
p
A
m
V
r
(ii)Potential due to a magnetic dipole at a point in end-side on
position is
0
22
.
4
mæö
=
ç÷
p -èøl
B
m
V
r
, wher e M = 2ml.
S N
O
P
r
If l
2
< < r
2
t hen
0
2
4
m
=
p
B
m
V
r
(iii)Potential due to a magnetic pole at a point in the broad side-
on position. Net potential at P = 0. The potential at any point
lying on the magnetic equator of a magnet is zero in CGS
and MKS system.
SN
2l
P
r
(iv)The magnetic potential at a point lying on a line passing
through the centre and making angle q with the axis
0
22
cos
4()
m q
=
p -l
B
M
V
r
and for small dipole, r >> l
0
2
cos
4
m q
=
p
B
M
V
r

552. 548 PHYSICS
Example 1.
T
wo identical thin bar magnets each of length
l and pole
strength m are placed at right angles to each other, with
north pole of one touching south pole of the other, then
find the magnetic moment of the system.
S
1
N
2
S
2
N
1
2
Solution :
Initial m
agnetic moment of each magnet = m × l.
As is clear from fig., S
1
and N
2
neutralize each other.
Effective distance between
N
1
and
22
2
S2= +=lll \ 2¢=lMm .
Example 2.
A steel wire
of length
l has a magnetic moment M. It is then
bent into a semi-circular arc. Find the new magnetic
moment.
Solution :
Let d be the diameter of semi-circle.
\ l = (p d/2) or d = (2 l/p)
New magnetic moment = m × d = m × (2 l/p)
= 2m l/p = (2 M/p)
Example 3.
Work done in turning a magnet of magnetic moment M by an
angle 90º from the magnetic meridian is n times the
corresponding work done to turn through an angle of 60º,
where n is
(a) 1/2 (b) 2
(c) 1/4 (d) 1
Solution : (b)
W
1
= – MB (cos 90º – cos 0º) = MB
W
2
= – MB (cos 60º – cos 0º)

1
111
MB 1 MB W
222
æö
=- -==ç÷
èø
As W
1
= n W
2
; \ n
= 2
Example 4.
In a hydrogen atom, an electron revolves with a frequency
of 6.8 × 10
9
megahertz in an orbit of diameter 1.06 Å. What
will be the equivalent magnetic moment?
Solution :
n = 6.8 × 10
9
MHz = 6.8 × 10
15
Hz,

Å53.0
2
06.1
r == = 0.53 × 10
–10
m
22
rer
/1
e
AIM pn=p÷
ø
ö
ç
è
æ
n
==

2101519
)1053.0(
7
22
)1080.6()106.1(
--
´´´´=

224
mA107.9
-
´=
Example 5.
The time period of oscillation of
a magnet in a vibration
magnetometer is 1.5 sec. What will be the time period of
oscillation of another magnet similar in size, shape and
mass but having 1/4 magnetic moment than that of the 1st
magnet oscillating at the same place?
Solution :
2
M
4
1
M
M
M
T
T
1
1
2
1
1
2
===; s3T2T
12
==\.
Example 6.
The time period of oscillation of a
magnet is 2 sec. When it
is remagnetised so that its pole strength is 4 times, what
will be its period?
Solution :
l
l
2m
2m
M
M
T
T
2
1
2
1
1
2
´
´
==
;
2
1
m4
m
T
T
1
1
1
2
==
; T
2
= 1 sec.
Exa
mple 7.
A thin rectangular magnet suspended freely has a period
of oscillation of 4s. If it is broken into two equal halves,
what will be the period of oscillation of each half ?
Solution :
For each half, mass is half and length is half;
As
12
m
I
2
l
=..M \ M.I. becomes 1/
8th.
Also M becomes 1/2
As
T 2 M.I / MB=p .
\ T becomes
2/1
8/1
times =
2
1
times
New time per
iod
s2s4
2
1
=´=
EARTH’S MAGNETISM
Magnetic fie
ld of earth extends nearly upto five times the radius
of earth i.e., 3.2 × 10
4
km.
The magnetic field of earth is fairly uniform and can be represented
by equidistant parallel lines.
Magnetic axis
Imaginary
bar
magnet in
the core of Earth
to depict earth's
magnetic field
Geographic axis
Magnetic equator
Geographic equator
Earth's magnetic
li
nes of force
NG
S
m
N
mS
G
Geographic meridi
an : The geographic meridian at a place is the
vertical plane passing through the geographic north & south at
that place.

553. 549Magnetism and Matter
Magnetic meridian : The magnetic meridian at a place is the vertical
plane passing through the magnetic axis of a freely suspended
small magnet. The earth’s magnetic field acts in the plane of
magnetic meridian.
Vertical plane
passing through
the magnetic axis
Magnetic meridian
S
N
S
N
Elements of Earth’s Magnetic Field
The earth’s magnetic field at a place can be completely described
by three independent parameters called elements of earth’s
magnetic field. These are :
a
Geographic
meridian
Magnetic
mer
idian
d
B
B
v
Bcos d

Magnetic
meridian
B
v
BH
BI
d
90°–d
Magnetic
meridian
d
S
N
Axis of a freely
suspended magnet
Horizontal
line
(
i) Magnetic declination (q) : The angle between the
geographic meridian and the magnetic meridian at a place
is called the magnetic declination at that place.
(ii) Angle of dip (d) : The angle made by the earth’s total
magnetic field
u ur
B with the horizontal is called angle of
dip at any place.
tand=
H
V
B
B
True and apparent dip
When the plane of the dip circle is in the magnetic meridian, thenthe needle stops in the actual direction of the Earth's magneticfield. The angle made by the needle with the horizontal in thiscondition is called true dip.
In case, the plane of the dip circle is not in the magnetic meridianthen the angle made by the needle with the horizontal is calledapparent dip. In this case the vertical component of earth's magnetic
field remains the same but the effective horizontal componentB'
H
= B
H
cos q
qPlane in
which the dip
circle is
present
Mag
netic
meridian
B'
H
B
H
tan
cos
¢
d¢==
¢ q
VV
HH
BB
BB
where d¢ = appare
nt dip
d = true dip
If the dip circle is rotated by 90°, the new apparent dip d¢¢ and d¢
and d are related as ¢¢d+¢d=d
222
cotcotcot
(iii) Horizontal component of earth’s magnetic field (B
H
) : It is
the component of the earth’s total magnetic field
u ur
B in the
horizontal diection and is give
n by, B
H
= B cos d
Keep in Memory
1.Dip c
ircle is an instrument used to measure angle of dip at a
place.
2.At poles total magnetic intensity is 0.66 oersted and at
equator it is 0.33 oersted
The total magnetic intensity at a particular latitude is
l+=
2
0
sin31II
where l is the angle of latitude
d = 90°
d = 90°
d = 0°
Magnetic equator
Magnetic axis
The angle of dip d = 0° at magnetic equator
and d = 90° at magnetic poles.
3.A spectacular effect due to earth's magnetism is observednear the magnetic poles of earth. This effect is called aurora-
borealis in the north and aurora - austorlis in south. It is
shown by patterns of coloured lights.
Magnetic Maps
Magnetic surveys all over the earth have been carried out and
magnetic maps have been prepared which show the values of
magnetic element throughout the world. Lines can be drawn to
join places having the same value of a particular magnetic element.
(i)Isogonic lines : These join places of equal declination. A
line joining places of zero declination is called agonic line.
(ii)Isoclinic lines : These join places of equal dip.
A line joining the places of zero-dip is called aclinic line.
(iii)Isodynamic lines : These join places of equal horizontal
component.
Shielding from magnetic fields : For shielding a certain region of
space from magnetic field, we surround the region by soft iron
rings. Magnetic field lines will be drawn into the rings and the
space enclosed will be free of magnetic field.
Neutral Points
Neutral points are the points where the net field intensity due to
the field of the bar magnet and field of earth is zero. When magnet

554. 550 PHYSICS
is placed with its
north pole towards geographic north, neutral
points lie on equatorial line of the magnet. At each neutral point,
0
2
2 2 3/2
4()
m
==
p+l
M
BH
d
where H = horizontal component of earth’s magnetic field.
When the bar magnet is placed with its north pole towards
geographic south, the neutral points lie on the axial line of the
magnet. At each neutral point,
0
1
2 22
4()
m
==
p -l
M
BH
d
Relation between the units of quantities associated with magnetic
field :
1A = 1JT
–1
m
–2
= 1J Wb
–1
1T = 1JA
–1
m
–2
= 1Wb m
–2
1Wb = 1JA
–1
= 1Tm
2
[B] = NA
–1
m
–1
= T = Wb m
–2
[M] = A m
2
= JT
–1
= J m
2
Wb
–1
[m
0
] = NA
–2
= T
2
m
2
N
–1
= Wb
2
J
–1
m
–1
Example 8.
A circular coil of radius 0.157 m has 50 turns. It is placedsuch that its axis is in magnetic meridian. A dip needle issupported at the centre of the coil with its axis of rotationhorizontal and in the plane of the coil. The angle of dip is30º, when a current flows through the coil. The angle ofdip becomes 60º on reversing the current. Find the currentin the coil assuming that magnetic field due to the coil issmaller than the horizontal component of earth’s field. TakeH = 3 × 10
–5
T.
Solution :
If H is horizontal component and V is vertical component atthe place, then true value of dip (d) at the place is given by,
H
V
tan=d
If B is magnetic f
ield intensity at the centre of coil due to current
I in circular coil and B is along H, then
BH
V
º30tan
+
= .
On reversin
g the direction of current, the direction of B is
reversed.
\
BH
V
º60tan
-
=
Dividing, we get 3
3/1
3
BH
BH
º30tan
º60tan
==
-
+
=
\ B = H/2
Þ T105.1
2
103
B
5
5
-
-
´=
´
=
For a circular co
il,
o
nI
B
2r
m
=
\ 1.5 × 10
–5
=
7
(4 10 ) 50 I
2 0.157
-
p´ ´´
´
\ A75.0
50104
157.02105.1
I
7
5
=
´´p
´´´
=
-
-
.
Example 9.
If
q
1
and
q
2
are angles of dip in two vertical planes at
right angle to each other and
q is true dip then prove
cot
2
q = cot
2q
1
+ cot
2
q
2
.
Solution :
If the vertical plane in which dip is q
1
subtends an angle a
with meridian than other vertical plane in which dip is q
2
and
is perpendicular to first will make an angle of 90° – a with
magnetic meridian. If q
1
and q
2
are apparent dips then
V
1
H
B
tan
B cos
q=
a
;
VV
2
HH
BB
tan
B cos(90 ) B sin
q==
-aa
22
12
22
12
11
cot cot
(tan ) (tan )
q+q=+
qq
222222
HHH 2
22 VV
Bcos Bsin B B c
os
cot
BsinBB
a+a qæö
= == =qç÷
èøq
s o cot
2
q
1
+ cot
2
q
2
= cot
2
q
TERM
S RELATED TO MAGNETISM
Magnetic intensity
)H( : When a magnetic material is placed in
a magnetic field, it becomes magnetised. The capability of the
magnetic field to magnetise a material is expressed by means of
a magnetic vector
uur
H, called the `magnetic intensity’ of the field.
The relation between magnetic induction B and magnetising field
H =
m
B
H , m being perm eability of medium.
Intensity of magnetisation (I) : When a material is placed in a
magnetising field, it acquires magnetic moment M. The intensity
of magnetisation is defined as the magnetic moment per unit
volume i.e.,
m
I=
M
V
V being volume of mateiral. If the material is in the form of a bar of
cross-sectional area A, length 2l and pole strength m, then
M = m × 2l; V = A × 2l\
m
.2
I
.2
\===
l
l
Mmm
VAA
Magnetic susceptibility (c) : The m agnetic susceptibility is defined
as the intensity of magnetisation per unit magentising field.
i.e.,
m
I
c=
H
Magnetic permeability (m) : The magnetic permeability of a material
is the measure of degree to which the material can be permited by
a magnetic field and is defined as the ratio of magnetic induction
(B) in the material to the magnetising field i.e. m=
B
H
Relation between Magnetic Susceptibility and
Permeability
We have magnetic induction in mateiral, B = mH
Also B = B
0
+ B
m

555. 551Magnetism and Matter
The characteristics of paramagnetic substances are
(a) They are attracted by a strong magnet
(b) Their susceptibility is positive but very small ( c > 0)
(c) The
ir relative permeability is slightly greater than unity.
(m > 1)
(d) Their susceptibility and permeability do not change
with the variation of magnetising field.
(e) Their susceptibility is inversely proportional to
temperature,
÷
ø
ö
ç
è
æ
ac
T
1
.e.i.
(f) The
y are found in those material which have atoms
containing odd number of electrons
3. Ferromagnetic Substances : These are the substances which
are strongly magnetised by relatively weak magnetising
field in the same sense as the magnetising field. The
examples are Ni, Co, iron and their alloys.
The characteristics of ferromagnetic substances are
(a) They are attracted even by a weak magnet.
(b) The susceptibility is very large and positive.
(
c >> 0)
(c) The
relative permeability is very high (of the order of
hundreds and thousands). (m >> 1)
(d) The intensity of magnetisation is proportional to the
magnetising field H for smaller values, varies rapidly
for moderate values and attains a constant value for
larger values of H.
(e) The susceptibility of a ferromagnetic substance is
inversely proportional to temperature i.e.,
1/Tcµ
C
;
T
Þc= C curie constant= .
This is called C
urie law. At a temperature called curie
temperature, ferromagnetic substance becomes
paramagnetic. The curie temperatures for Ni, Fe and
Co are 360ºC, 740ºC and 1100ºC respectively.
(f) They are found in those material which have domains
and can be converted into strong magnetsKeep in Memory
1.Diama
gnetism is universal. It is present in all materials. But
it is weak and hard to detect if substance is para or
feromagnetic.
2.I – H curve for different materials
Ferromagnetic
Paramagnetic
I
H
Diamagnetic
3.Curve for magnetic susceptibility and temperature for a
paramagnetic and ferromagnetic material.
1
T
cµ
X
T
where B
0
= ma gnetic induction in vacuum produced by
magnetising field
B
m
= magnetic induction due to magnetisation of material.
But B
0
= m
0
H and B
m
= m
0Im
Þ ]IH[B
m0
+m=
00
I
BH[1]B[ 1]
H
\ =m + = +c
m
;
0
B/B[1]\ = +c
00
0
H
B/B/
H
m
\ = =mm
m
= µ
r
, the rela
tive magnetic permeability
c+=m\1
r
This is required rela
tion.
CLASSIFICATION OF MATERIALS
According to the behaviour of substances in magnetic field, they
are classified into three categories:
1. Diamagnetic Substances : These are the substances which
when placed in a strong magnetic field acquire a feeble
magnetism opposite to the direction of magnetising field.
The examples are copper, gold, antimony, bismuth, alcohol,
water, quartz, hydrogen, etc.
N S

N S
Diamagnetic
B<B
0
B
o

N S
Behaviou
r of diamagnetic substance in an external magnetic field
Bo
r
The characteristics of diamagnetic substances are
(a) They are feebly repelled by a strong magnet
(b) Their susceptibility is negative (i.e. c < 0)
(c) T
heir relative permeability is less than 1. (i.e. m
r
< 1)
(d) Their susceptibility is independent of magnetising field
and temperature (except for Bismuth at low
temperature)
2. Paramagnetic Substances : These are the materials which
when placed in a strong magnetic field acquire a feeble
magnetism in the same sense as the applied magnetic field.
The examples are platinum, aluminium, chromium,
manganese, CuSO
4
, O
2
, air, etc.
N S
N S
Paramagnetic
B > B
0
B
o
N S
Behaviour of paramagn
etic substance in an external field oB
r

556. 552 PHYSICS
HYSTERESIS
Wh
en a bar of ferromagnetic material is magnetised by a varying
magnetic field and the intensity of magnetisation
Im
induced is
measured f
or different values of magnetising field H, the graph of
I versus H is as shown in fig

A
D
F
C
I
m
HGO
B
The graph shows :
(
i) When magnetising field is increased from O the intensity of
magnetisation
Im
increases and becomes m
aximum (i.e.
point A). This maximum value is called the saturation value.
(ii) When H is reduced,
Im
reduces b
ut is not zero when H = 0.
The remainder value OB of magnetisation when H = 0 is
called the residual magnetism or retentivity. OB is retentivity.
(iii) When magnetic field H is reversed, the magnetisaiton
decreases and for a particular value of H, it becomes zero
i.e., for H = OC, I = 0. This value of H is called the coercivity.
(iv) When field H is further increased in reverse direction, the
intensity of magnetisation attains saturation value in reverse
direction (i.e., point D).
(v) When H is decreased to zero and changed direction in steps,
we get the part DFGA.
Properties of Soft Iron and Steel
For soft iron, the susceptibility, permeability and retentivity are
greater while coercivity and hysteresis loss per cycle are smaller
than those of steel.
H
B
Soft magnetic
material
Hysteresis
c
urve

H
B
Hard magnetic
material
Hyster
esis
curve
PERMANENT MAGNETS AND ELECTROMAGNETS
P
ermanent magnets are made of steel and cobalt while
electromagnets are made of soft iron.
An electromagnet is made by inserting a soft iron core into the
interior of a solenoid. Soft iron does not retain a significant
permanent magnetization when the solenoid’s field is turned off–
soft iron does not make a good permanent magnet. When current
flows in the solenoid, magnetic dipoles in the iron tend to line up
with the field due to the solenoid. The net effect is that the field
inside the iron is intensified by a factor known as the relative
permeability. The relative permeability is analogous in magnetism
to the dielectric constant in electricity. However, the dielectric
constant is the factor by which the electric field is weakened,
while the relative permeability is the factor by which the magnetic
field is strengthened. The reactive permeability of a ferromagnet
can be in the hundreds or even thousands–the intensification of
the magnetic field is significant. Not only that, but in an
electromagnet the strength and even direction of the magnetic
field can be changed by changing the current in the solenoid.
Keep in Memory
1.By alloyin
g soft-iron with 4% silicon ‘transformer steel’ is
produced. It has a higher relative permeability and is an ideal
material for cores of transformers. Alloys of iron and nickel
called ‘permalloys’, also have very large permeabilities.
2.Energy spent per unit volume of specimen is complete cycle
of magnetisation is numerically equal to area of I – H loop

Perme-
ability
Suscep-
tibility
Intensity of
magnetisation
Reten-
tivity
Coerc-
ivity
Hysteresis
loss
Soft ironhigh high high high low low
Steel low low low low high high
Þ Steel is most suitable for making parmanent magnet
Þ Soft iron is most suitable for making core of an
electromagnet.
Example 10.
A magnetising field of 1600 Am
–1
produces a magnetic
flux of 2.4 × 10
–5
weber in a bar of iron of cross section 0.2
cm
2
. Calculate permeability and susceptibility of the bar.
Solution :
Here, H = 1600 Am
–1
, f = 2.4 ×10
–5
Wb.
A = 0.2 cm
2
= 0.2 × 10
–4
m
2
, m = ? c
m
= ?
5
2
4
2.4 10
B 1.2 weber / m
A0.2 10
-
-
f´
===
´
;
mTA105.7
1600
2.1
H
B 14--
´===m ;
[Q As )1(
m0
c+m=
m ]
\
1
104
105.7
1
7
4
0
m
-
´p
´
=-
m
m
=c
-
-
;
1.59611.5971
4
105.7
3
m
=-=-
p
´
=c
Example 11.
A solenoid of 500 turns/m is carrying a current of 3A. It
s
core is made of iron which has a relative permeability of
5000. Determine the magnitude of magnetic intensity,
magnetisation and magnetic field inside the core.
Solution :
Magnetic intensity
H = ni = 500 × 3 = 1500 A/m
µ
r
= 1 + c
m
so c
m
= µ
r
– 1 = 4999 » 5000
Intensity of magnetisation
I = cH = 5000 × 1500 = 7.5 × 10
6
A/m
Magnetic field B = µ
r
µ
0
H = 5000 × 4p × 10
–7
× 1500
= 9.4 tesla.
TANGENT GALVANOMETER
It is an instrument used for measuring small current. It is based on
tangent law. It is a moving magnet and fixed coil type galvanometer.
Tangent Law : If a small magnetic needle is under the influence
of two crossed magnetic fields (B) and (H) and suffers a deflection
q from field H, then by tangent law, B = H tan q.

557. 553Magnetism and Matter
Formula for current : If a current passing through the coil of n
turns and mean radius r of a tangent galvanometer placed in
magnetic meridian causes a deflection q in the magnetic needle
kept at the centre of the coil, then
0
2rH
I tan
n
æö
=q
ç÷
mèø
; I = K t
anq where
0
2rH
K
n
=
m
and is called the
re
duction factor.
DEFLECTION MAGNETOMETER
It's working is based on the principle of tangent law.
(a)Tan A Position : In this position the magneto- meter is set
perpendicular to magnetic meridian so that, magnetic field
due to magnet is in axial position and perpendicular to earth’s
field and hence
0
2 22
2
tan
4()
m
=q
p -l
Md
H
d
where d = d
istance of needle from centre of magnet and
2l = length of magnet.
(b)TanB position : The arms of magnetometer are set in
magnetic meridian, so that the field is at equatorial position
and hence, H
0
2 2 3/2
tan
4()
m
q=
p -l
M
d
Magnetic field o
f earth extends nearly upto five times the
radius of earth i.e., 3.2 × 10
4
km.
The magnetic field of earth is fairly uniform and can be
represented by equidistant parallel lines.
VIBRATION MAGNETOMETER
It is an instrument for comparing the magnetic moments of two
magnets and for comparing their magnetic fields.
The time period of a bar magnet vibrating in the vibration
magnetometer kept in magnetic meridian is given by
2=p
I
T
MH

2
2
4p
\=
I
M
TH
where
22
m( b)
I
12
æö+
=ç÷
èø
l
is the mo
ment of inertia of the vibrating
magnet, m = mass of magnet, l = length of magnet, b = breadth of
magnet.
Example 12.
A vibration magnetometer consists of two identical bar
magnets, placed one over the other, such that they are
mutually perpendicular and bisect each other. The time
period of oscillation in a horizontal magnetic field is 4
second. If one of the magnets is taken away, find the period
of oscillation of the other in the same field.
Sol.For a vibration magnetometer, we know that
2/=pIT MH
Let M be the magnetic moment and M.I, moment of inertia
of each magnet,
\ 22
M M M M2= +=¢
and n
et M.
Ι= M.Ι+ M.Ι= 2M .Ι¢
\
MH
M.I.2
2
H2M
M.I.2
2T ´=p=¢ ....(1)
When one o
f the magnets is taken away,
M² = M, MI² = MI,
\
M.Ι
T2
MH
=p¢¢ ...(2)
Divide
eqn. (2) by (1),

4/1
)2(
1
T
T
=
¢
¢¢
or s34.3
2
4
)2(
T
T
4/14/1
==
¢
=¢¢
Examples 13.
The m
agnetic needle of an oscillation magnetometer makes
10 oscillations per minute under the action of earth’s
magnetic field alone. When a bar magnet is placed at some
distance along the axis of the needle, it makes 14
oscillations per minute. If the bar magnet is turned so that
its poles interchange their positions, then what will be the
new frequency of oscillation of the needle?
Sol.
1 MH
10
2I
=
p
...(i)
1 M (H F)
14
2I
+
=
p
...(ii)
1 M (H F)
n
2I
-
=
p
...(iii)
Divide e
qs. (ii) by (i),
5
7
H
F
1
H
FH
10
14
=+=
+
=
25
24
H
F
=\
Divid
e eqs. (iii) by (i),
5
1
H
F
1
H
FH
10
n
=-=
-
=
or, 2
5
10
n == vibs /minute.
Example 14.
The period of oscillation of a magnet in a vibration
magnetometer is 2 sec. What will be the period of oscillation
of a magnet whose magnetic moment is four times that of
the first magnet?
Sol.
H
Ι
T2
MB
æö
=p
ç÷
èø
HH
Ι1Ι
T'22
4MB 2 ( MB)
éùæö
=p =p êúç÷
èø
ëû
= 12
2
1
=´ second.

558. 554 PHYSICS
Example 15.
A magnet is suspende
d in such a way that it oscillates in
the horizontal plane. It makes 20 oscillations per minute
at a place where dip angle is 30º and 15 oscillations per
minute at a place where dip angle is 60º. What will be the
ratio of the total earth’s magnetic field at the two places?
Sol.Let the total magnetic fields due to earth at the two places
be B
1
and B
2
. If horizontal components be (B
H
)
1
and (B
H
)
2
respectively, then
(B
H
)
1
= B
1
cos 30º and (B
H
)
2
= B
2
cos 60º
Here T
1
= 3 sec. and T
2
= 4 sec.
1
1
Ι
T 32
M B cos 30
æö
= =p
ç÷
èø
and
2
2
Ι
T 42
M B cos60º
æö
= =p
ç÷
èø
2/1
1
2
30cosB
60cosB
4
3
÷
÷
ø
ö
ç
ç
è
æ
=\ or
30cos
60cos
9
16
B
B
2
1
´=
or
1
2
16 1 2 16
92 3 93
=´´=
B
B
or
12
: 16:93=BB

559. 555Magnetism and Matter
CONCEPT MAP
MAGNETISM AND
MATTER
Magne
t
i
c

f
ie
l
d
l
in
e
s
Imagin
a
r
y

li
n
e
s
i
n

a
magne
t
i
c

f
i
e
l
d
w
h
i
c
h
continu
o
u
s
ly

r
e
p
r
e
s
e
n
t

the dire
c
t
i
o
n
o
f
m
a
g
n
e
t
i
c
field
P
r
o
p
e
r
t
i
e
s
o
f

m
a
g
n
e
ti
c
f
i
e
l
d

l
i
n
e
s
M
a
g
n
e
t
ic

f
ie
l
d
l
i
n
e
s
d
o

n
o
t
i
n
t
e
r
s
e
c
t
e
a
c
h

o
t
h
e
r

C
o
m
e
o
u
t
o
f

s
u
r
f
a
c
e
a
t
a
n
y
a
n
g
l
e
Magnetism Property of
attracting a piece of iron,
cobalt, nickel or steel
Properties of magnet
Directive
A freely suspended magnet always
points in north-south direction
P
o
l
e

o
f
a
m
a
g
n
e
t
a
l
w
a
y
s

e
x
i
s
t
i
n
p
a
i
r
F
o
r
c
e

b
e
t
w
e
e
n
t
w
o
m
a
g
n
e
t
i
c
p
o
l
e
s

0
1
2
2
m
m
F
4
r
m
=
p
M
a
g
n
e
t
ic
d
i
p
o
l
e
m
o
m
e
n
t

M
=

N
I
A
Earth
's

m
a
g
n
e
t
i
c
eleme
n
ts
A
n
g
l
e

o
f
d
i
p
o
r
i
n
c
l
i
n
a
t
i
o
n

(
)
A
n
g
l
e

m
a
d
e
b
y

d
i
r
e
c
t
i
o
n
o
f

e
a
r
t
h
's
m
a
g
n
e
t
i
c

f
i
e
l
d
w
i
t
h

t
h
e
h
o
r
i
z
o
n
ta
l

e
q
u
a
t
o
r

=
0
;
p
o
l
e
=
9
0
º
d
d
d
H
o
r
i
z
o
n
t
a
l

c
o
m
p
o
n
e
n
t
B
=
B

s
i
n

B
=
B

c
o
s

V H
qq
A
n
g
l
e

o
f
d
e
c
l
i
n
a
t
i
o
n

A
n
g
l
e

b
e
t
w
e
e
n
m
a
g
n
e
t
i
c
m
e
r
i
d
i
a
n
a
n
d
g
e
o
g
r
a
p
h
i
c

m
e
r
i
d
i
a
n

I
n
a
u
n
i
f
o
r
m

m
a
g
n
e
t
i
c
f
i
e
l
d
t
i
m
e
p
e
r
i
o
d
o
f
o
s
c
i
l
la
t
io
n

o
f
a
f
r
e
e
l
y
s
u
s
p
e
n
d
e
d

m
a
g
n
e
t
I
T
2
M
B
=
p
T
a
n
g
e
n
t

t
o

t
h
e

f
i
e
l
d
l
i
n
e

a
t

a

g
i
v
e
n

p
o
i
n
t
r
e
p
r
e
s
e
n
ts
t
h
e
d
i
r
e
c
t
i
o
n
o
f

t
h
e

n
e
t

m
a
g
n
e
t
i
c
f
i
e
l
d
Form
c
o
n
t
i
n
u
o
u
s

c
l
o
s
e
d
loops
s
t
a
r
t

f
r
o
m
N
-
p
o
l
e

end S-
p
o
l
e
o
u
t
s
i
d
e

th
e

magne
t
a
n
d
i
t
s

o
p
p
o
s
i
t
e

inside
t
h
e
m
a
g
n
e
t
Repulsive Like poles
always repel one
another. It is sure
test of magnet
Magnet field due to a bar magnet
Magnetic Materials
Paramagnetics Magnetist
in the direction of magnetic
field e.g., Al, Mn µr, I, > 1 c
m
Ferromagnetics Strongly
magnetised in the direction
of magnetic field e.g., Fe, Co,
Ni µr, I, >> 1 c
m
Diamagnetics Magnetised
in a direction opposite to the
direction of magnetic field
e.g., Bi, Cu, Hg µr, I and
are negative
c
m
At a poin
t

o
n
e
q
u
a
t
o
r
i
a
l

line
0
2 2
M
B
4(r)3/2
m
=
p+l
At a point on axial
line
0
3
2M
B
4d
m
=
p
Magnetic
intensity
0B
H=
m
Magnetic permeability
µ = µ (1 + )
0
c
m
Magnetic
susceptibility
I
=
H
c
m
M
I=
V

560. 556 PHYSICS
1.The mai
n difference between electric lines of force and
magnetic lines of force is
(a) electric lines of force are closed curves whereas magnetic
lines of force are open curves
(b) electric lines of force are open curves whereas magnetic
lines of force are closed curves
(c) magnetic lines of force cut each other whereas electric
lines of force do not cut
(d) electric lines of force cut each other whereas magnetic
lines of force do not cut
2.Current i is flowing in a coil of area A and number of turns N,
then magnetic moment of the coil, M is
(a) NiA(b)
A
Ni
(c)
A
Ni
(d) N
2
Ai
3.Nickel shows ferr
omagnetic property at room temperature. If
the temperature is increased beyond Curie temperature, then
it will show
(a) anti ferromagnetism(b) no magnetic property
(c) diamagnetism (d) paramagnetism
4.The line on the earth surface joining the point where the field
is horizontal, is called
(a) magnetic equator(b) magnetic line
(c) magnetic axis (d) magnetic inertia
5.When a ferromagnetic material is heated to temperature above
its Curie temperature, the material
(a) is permanently magnetized
(b) remains ferromagnetic
(c) behaves like a diamagnetic material
(d) behaves like a paramagnetic material
6.The force between two short bar magnets with magnetic
moments M
1
and M
2
whose centres are r metres apart is 8 N
when their axes are in same line. if the separation is increased
to 2 r, the force between them in reduced to
(a) 4 N (b) 2 N (c) 1 N (d) 0.5 N
7.The magnet of pole strength m and magnetic moment M is
cut into two pieces along its axis. Its pole strength and
magnetic moment now becomes
(a)
2
M
,
2
m
(b)
2
M
,m(c) M,
2
m
(d) m, M
8.A bar magnet o
f magnetic moment M and length L is cut
into two equal parts each of length L/2. The magnetic moment
of each part will be
(a)M (b) M/4 (c)
M2(d) M/2
9.A supercond
uctor exhibits perfect :
(a) ferrimagnetism (b) ferromagnetism
(c) paramagnetism (d) diamagnetism
10.In end on and broadside on position of a deflection
magnetometer, if q
1
and q
2
are the deflections produced by
short magnets at equal distances, then tan q
1
/ tanq
2
is
(a) 2 : 1 (b) 1 : 2
(c) 1 : 1 (d) None of these
11.A watch glass containing some powdered substance is
placed between the pole pieces of a magnet. Deep concavity
is observed at the centre. The substance in the watch glass
is
(a) iron (b) chromium
(c) carbon (d) wood
12.Needles N
1
, N
2
and N
3
are made of a ferromagnetic, a
paramagnetic and a diamagnetic substance respectively. A
magnet when brought close to them will
(a) attract N
1
and N
2
strongly but repel N
3
(b) attract N
1
strongly, N
2
weakly and repel N
3
weakly
(c) attract N
1
strongly, but repel N
2
and N
3
weakly
(d) attract all three of them
13.A bar magnet is oscillating in the Earth’s magnetic field
with a period T. What happens to its period and motion if
its mass is quadrupled?
(a) Motion remains S.H. and period remains nearly
constant
(b) Motion remains S.H. with time period =
2
T
(c) Motion r
emains S.H. with time period = 2T
(d) Motion remains S.H. with time period = 4T
14.Two magnets of magnetic moments M and 2M are placed
in a vibration magnetometer, with the identical poles in the
same direction. The time period of vibration is T
1
. If the
magnets are placed with opposite poles together and vibrate
with time period T
2
, then
(a)T
2
is infinite (b) T
2
= T
1
(c)T
2
> T
1
(d) T
2
< T
1
15.If horizontal and vertical components of earths magnetic
field are equal, then angle of dip is
(a) 60° (b) 45° (c) 30° (d) 90°
16.The magnetic materials having negative magnetic
susceptibility are
(a) non-magnetic (b) para magnetic
(c) dia magnetic (d) ferro magnetic
17.For protecting a sensitive equipment from the external
electric arc, it should be
(a) wrapped with insulation around it when passing
current through it
(b) placed inside an iron can
(c) surrounded with fine copper sheet
(d) placed inside an aluminium can
18.If a diamagnetic substance is brought near north or south
pole of a bar magnet, it is
(a) attracted by the poles
(b) repelled by the poles
(c) repelled by north pole and attracted by the south pole
(d) attracted by the north pole and repelled by the south
pole

561. 557Magnetism and Matter
19.A bar magnet, of magnetic moment M, is placed in a
magnetic field of induction B. The torque exerted on it is
(a) B.M(b) –B.M(c) BM´(d) MB´
20.Current i is f
lowing in a coil of area A and number of turns
N, then magnetic moment of the coil M is
(a) NiA (b)
A
Ni
(c)
A
Ni
(d) N
2
Ai
21.A diamagnetic
material in a magnetic field moves
(a) perpendicular to the field
(b) from stronger to the weaker parts of the field
(c) from weaker to the stronger parts of the field
(d) None of these
22.According to Curie’s law, the magnetic susceptibility of a
substance at an absolute temperature T is proportional to
(a)T
2
(b)
T
1
(c)T (d)
2
T
1
23.The magnetic mo
ment of a diamagnetic atom is
(a) equal to zero
(b) much greater than one
(c) unity
(d) between zero and one
24.There are four light-weight-rod samples A,B,C,D separately
suspended by threads. A bar magnet is slowly brought near
each sample and the following observations are noted
(i) A is feebly repelled
(ii) B is feebly attracted
(iii) C is strongly attracted
(iv) D remains unaffected
Which one of the following is true ?
(a) B is of a paramagnetic material
(b) C is of a diamagnetic material
(c) D is of a ferromagnetic material
(d) A is of a non-magnetic material
25.If the magnetic dipole moment of an atom of diamagnetic
material, paramagnetic material and ferromagnetic material
are denoted by m
d
, m
p
and m
f
respectively, then
(a)m
d
= 0 and m
p
¹ 0 (b)m
d
¹ 0 and m
p
= 0
(c)m
p
= 0 and m
f
¹ 0 (d)m
d
¹ 0 and m
f
¹ 0
1.A bar magnet 8 cms long is placed in the magnetic merdian
with the N-pole pointing towards geographical north. Two
netural points separated by a distance of 6 cms are obtained
on the equatorial axis of the magnet. If horizontal component
of earth’s field = 3.2 × 10
–5
T, then pole strength of magnet
is
(a) 5 ab-amp × c (b) 10 ab-amp × cm
(c) 2.5 ab-amp × cm (d) 20 ab-amp × cm
2.Two tangent galvanometers having coils of the same radius
are connected in series. A current flowing in them produces
deflections of 60º and 45º respectively. The ratio of the number
of turns in the coils is
(a) 4/3 (b)
1
13+
(c)
13
13
-
+
(d)
1
3
3.Two ide
ntical magnetic dipoles of magnetic moments
1.0 A-m
2
each, placed at a separation of 2 m with their axis
perpendicular to each other. The resultant magnetic field at
point midway between the dipole is
(a) 5 × 10
–7
T (b)
5 × 10
–7
T
(c) 10
–7
T (d
) 2 × 10
–7
T
4.Two isolated point poles of strength 30 A-m and 60 A-m are
placed at a distance of 0.3m. The force of repulsion is
(a) 2 × 10
–3
N (b) 2 × 10
–4
N
(c) 2 × 10
5
N (d) 2 × 10
–5
N
5.The magnetic moment of a magnet is 0.1 amp × m
2
. It is
suspended in a magnetic field of intensity 3 × 10
–4
weber/m
2
.
The couple acting upon it when deflected by 30º from the
magnetic field is
(a) 1 × 10
–5
N m (b) 1.5 × 10
–5
N m
(c) 2 × 10
–5
N m (d) 2.5 × 10
–5
N m
6.When 2 ampere current is passed through a tangent
galvanometer, it gives a deflection of 30º. For 60º deflection,
the current must be
(a) 1 amp. (b)
32 amp.
(c) 4 amp. (d) 6 amp.
7.A cu
rve between magnetic moment and temperature of
magnet is
(a)
M
T
O
(b) M
T
O
(c) M
T
O
(d) M
T
O

562. 558 PHYSICS
8.The variatio
n of magnetic susceptibility (x) with temperature
for a diamagnetic substance is best represented by
(a)
TO
(b)
TO
(c)
T
O
(d)
T
O
9.At a temperatur of 30°C, the susceptibility of a ferromagnetic
material is found to be c. Its susceptibility at 333°C is
(a)c (b) 0.5c (c)2c (d) 11.1c
10.Of the following fig., the lines of magnetic induction due to
a magnet SN, are given by
N
S
(1)
N
S
(2)
N
S
(3)
N
S
(4)
(a)1 (b) 2 (c)3 (d) 4
11.Th
e B – H curve (i) and (ii) shown in fig. associated with
B
(i)
(ii)
H
(a) (i) diamagn
etic and (ii) paramagnetic substance
(b) (i) paramagnetic and (ii) ferromagnetic substance
(c) (i) soft iron and (ii) steel respectively
(d) (i) steel and (ii) soft iron respectively
12.A thin bar magnet of length 2 l and breadth 2 b pole strength
m and magnetic moment M is divided into four equal parts
with length and breadth of each part being half of original
magnet. Then the pole strength of each part is
(a)m (b) m/2 (c) 2 m (d) m/4
13.In the above question, magnetic moment of each part is
(a) M/4 (b) M (c) M/2 (d) 2 M
14.Two points A and B are situated at a distance x and 2x
respectively from the nearer pole of a magnet 2 cm long. The
ratio of magnetic field at A and B is
(a) 4 : 1 exactly (b) 4 : 1 approximately
(c) 8 : 1 approximately(d) 1 : 1 approximately
15.If a magnet is suspended at angle 30º to the magnetic meridian,
the dip needle makes an angle of 45º with the horizontal.
The real dip is
(a)
)2/3(tan
1-
(b) )3(tan
1-
(c) )2/3(tan
1-
(d) )3/2(tan
1-
16.Two bar magn
ets of the same mass, same length and breadth
but having magnetic moments M and 2M are joined together
pole for pole and suspended by a string. The time period of
assembly in a magnetic field of strength H is 3 seconds. If
now the polarity of one of the magnets is reversed and
combination is again made to oscillate in the same field, the
time of oscillation is
(a)
3sec (b)33 sec
(c) 3 sec (d) 6 sec
17.A com
pass needle placed at a distance r from a short magnet
in Tan A position shows a deflection of 60º. If the distance is
increased to r (3)
1/3
, then deflection of compass needle is
(a) 30º (b)
3
1
360´
(c)
3
2
360´ (d)
3 3
360´
18.Two short magnets
have equal pole strengths but one is
twice as long as the other. The shorter magnet is placed 20
cms in tan A position from the compass needle. The longer
magnet must be placed on the other side of the magnetometer
for no deflection at a distance equal to
(a) 20 cms (b) 20 (2)
1/3
cms
(c) 20 (2)
2/3
cms (d) 20 (2)
3/3
cms
19.A dip needle lies initially in the magnetic meridian when it
shows an angle of dip q at a place. The dip circle is rotated
through an angle x in the horizontal plane and then it shows
an angle of dip q'.
Then
q
q
tan
'tan
is
(a)
xcos
1
(b)
xsin
1
(c)
xtan
1
(d) cos x

563. 559Magnetism and Matter
20.A dip circle is so set that its needle moves freely in the
magnetic meridian. In this position, the angle of dip is 40º.
Now the dip circle is rotated so that the plane in which the
needle moves makes an angle of 30º with the magnetic
meridian. In this position, the needle will dip by an angle
(a) 40º (b) 30º
(c) more than 40º (d) less than 40º
21.Work done in turning a magnet of magnetic moment M by
an angle of 90º from the mgnetic meridian is n times the
corresponding work done to turn through an angle of 60º,
where n is
(a) 1/2 (b) 2 (c) 1/4 (d) 1
22.Two magnets are held together in a vibration magnetometer
and are allowed to oscillate in the earth’s magnetic field with
like poles together. 12 oscillations per minute are made but
for unlike poles together only 4 oscillations per minute are
executed. The ratio of their magnetic moments is
(a) 3 : 1(b) 1 : 3(c) 3 : 5(d) 5 : 4
23.At a certain place, horizontal component is
3 times the
vertic
al component. The angle of dip at this place is
(a)0 (b)p/3
(c)p/6 (d) None of these
24.A freely suspended magnet oscillates with period T in earth’shorizontal magnetic field. When a bar magnet is brought
near it, such that the magnetic field created by bar magnet is
in same direction as earth’s horizontal magnetic field, the
period decreases to
2
T
. The ratio of the
field of the magnet
F to the earth’s magnetic field (H) is
(a) 1 : 3(b) 1 : 1(c) 3 : 1(d) 9 : 1
25.If relative permeability of iron is 2000. Its absolute
permeability in S.I. units is
(a)8p × 10
–4
(b) 8p × 10
–3
(c) 800/p (d) 8p × 10
9
/p
26.A steel wire of length
l has a magnetic moment M. It is then
bent into a semicircular arc. The new magnetic moment is
(a)
p
M
(b)
p
M2
(c)
p
M3
(d)
p
M4
27.The magnetic mome
nt of atomic neon is equal to
(a) zero(b)
B
µ
2
1
(c)
B
µ (d)
B
µ
2
3
28.Torques
1
t an
d
2
t are required for a magnetic needle to
remain perpendicular to the magnetic fields at two different
places. The magnetic fields at those places are B
1
and B
2
respectively; then ratio
2
1
B
B
is
(a)
1
2
t
t
(b)
2
1
t
t
(c)
21 21
t-t
t+t
(d)
21 21
t+t
t-t
29.The net mag
netic moment of two identical magnets each of
magnetic moment M
0
, inclined at 60° with each other is
(a)M
0
(b)
2M
0
(c)3M
0
N
NS
60°
(d) 2M
0
30.A magne
tic needle vibrates in a vertical plane parallel to the
magnetic meridian about a horizontal axis passing through
its centre. Its frequency is n. If the plane of oscillation is
turned about a vertical axis by 90°C, the frequency of its
oscillation in vertical plane will be
(a)n (b) zero
(c) less than n (d) more than n
31.A thin rectangular magnet suspended freely has a period of
oscillation of 4 s. If it is broken into two halves (each having
half the original length) and one of the pieces is suspended
similarly. The period of its oscillation will be
(a) 4 s (b) 2 s
(c) 0.5 s (d) 0.25 s
32.A steel wire of length
l has a magnetic moment M. It is bent
in L-shape (Figure). The new magnetic moment is
(a)M
(b)
2
M
2
l
2
l(c)
2
M
(d) 2M
33.The time per
iod of oscillation of a magnet in a vibration
magnetometer is 1.5 sec. The time period of oscillation of
another magnet similar in size, shape and mass but having
1/4 magnetic moment than that of the 1st magnet oscillating
at the same place will be
(a) 0.75 sec (b) 1.5 sec
(c) 3.0 sec (d) 6.0 sec
34.Time periods of vibation of two bar magnets in sum and
difference positions are 4 sec and 6 sec respectively. The
ratio of their magnetic moments M
1
/ M
2
is
(a) 6 : 4 (b) 30 : 16
(c) 2 . 6 : 1 (d) 1 . 5 : 1
35.Horizontal component of earth's field at a height of 1 m from
the surface of earth is H. Its value at a height of 10 m from
surface of earth is
(a) H/10 (b) H/9
(c) H/100 (d) H
36.If a toroid uses bismuth for its core, the field in the core
compared to that in empty core will be slightly
(a) greater (b) smaller
(c) equal (d) None of these
37.The relative permeability of a medium is 0.075. What is its
magnetic susceptibility?
(a) 0.925(b) – 0.925 (c) 1.075(d) –1.075

564. 560 PHYSICS
38.Relative pe
rmittivity and permeability of a material
r
eand
r
m, respectively. Which of the following values of these
quantities are allowed for a diamagnetic material?
(a)
r
e = 0.5,
r
m= 1.5 (b)
r
e = 1.5,
r
m= 0.5
(c)
r
e = 0.5,
r
m= 0.5 (d)
r
e = 1.5,
r
m= 1.5
39.The momen
t of a magnet (15 cm × 2 cm × 1 cm) is 1.2 A-m
2
.
What is its intensity of magnetisation?
(a) 4 × 10
4
A m
–1
(b) 2 × 10
4
A m
–1
(c) 10
4
A m
–1
(d) None of these
40.The work done in turning a magnet of magnetic moment M
by an angle of 90° from the meridian, is n times the
corresponding work done to turn it through an angle of
60°. The value of n is given by
(a)2 (b) 1 (c) 0.5 (d) 0.25
41.At a certain place, the angle of dip is 30º and the horizontal
component of earth’s magnetic field is 0.50 oerested. The
earth’s total magnetic field (in oerested) is
(a)
3 (b) 1 (c)
3
1
(d)
2
1
42.A coil in the sha
pe of an equilateral triangle of side l is
suspended between the pole pieces of a permanent magnet
such that
B
®
is in plane of the coil. If due to a current i in
the triangle a torque t acts on it, the side l of the triangle is
(a)
2
1
i.B3
2
÷
ø
ö
ç
è
æt
(b)
2
1
i.B3
2
÷
÷
ø
ö
ç
ç
è
æt
(c) ÷
ø
ö
ç
è
æt
i.B3
2
(d)
i.B3
1t
43.Iron is ferromagnet
ic
(a) above 770°C (b) below 770°C
(c) at all temperature(d) above 1100°C
44.A vibration magnetometer placed in magnetic meridian has
a small bar magnet. The magnet executes oscillations with a
time period of 2 sec in earth's horizontal magnetic field of 24
microtesla. When a horizontal field of 18 microtesla is
produced opposite to the earth's field by placing a current
carrying wire, the new time period of magnet will be
(a) 1 s (b) 2 s (c) 3 s (d) 4 s
45.Two identical bar magnets are fixed with their centres at a
distance d apart. A stationary charge Q is placed at P in
between the gap of the two magnets at a distance D from
the centre O as shown in the figure.
S N N S
P
D
O
d
The force on the charge Q is
(a) directed perpendicular to the plane of paper
(b) zero
(c) directed along OP
(d) directed along PO
46.A short bar magnet of magnetic moment 0.4J T
–1
is placed in
a uniform magnetic field of 0.16 T. The magnet is in stable
equilibrium when the potential energy is
(a) – 0.064 J (b) zero
(c) – 0.082 J (d) 0.064 J
DIRECTIONS (Qs. 47 to 50) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following-
(a)Statement -1 is false, Statement-2 is true
(b)Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c)Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d)Statement -1 is true, Statement-2 is false
47. Statement-1 : The ferromagnetic substance do not obey
Curie’s law.
Statement-2 : At Curie point a ferromagnetic substance start
behaving as a paramagnetic substance.
48. Statement-1 : Magnetism is relativistic.
Statement-2 : When we move along with the charge so that
there is no motion relative to us, we find no magnetic field
associated with the charge.
49. Statement-1 : A paramagnetic sample display greater
magnetisation (for the same magnetic field) when cooled.
Statement-2 : The magnetisation does not depend on
temperature.
50. Statement-1 : Electromagnetic are made of soft iron.
Statement-2 : Coercivity of soft iron is small.

565. 561Magnetism and Matter
Exemplar Questions
1.A toroid of n turns, mean radius R and cross-sectional radius
a carries current I. It is placed on a horizontal table taken as
xy-plane. Its magnetic moment m
(a) is non-zero and points in the z-direction by symmetry
(b) points along the axis of the toroid (m = mf )
(c) is zero, otherwise there would be a field falling as
3
1
r
at
large distances outside the toroid
(d) is pointing radially outwards
2.The magnetic field of the earth can be modelled by that of a
point dipole placed at the centre of the earth. The dipole axis
makes an angle of 11.3° with the axis of the earth. At Mumbai,
declination is nearly zero. Then,
(a) the declination varies between 11.3° W to 11.3° E
(b) the least declination is 0°
(c) the plane defined by dipole axis and the earth axis passes
through Greenwich
(d) declination averaged over the earth must be always
negative
3.In a permanent magnet at room temperature.
(a) magnetic moment of each molecule is zero
(b) the individual molecules have non-zero magnetic
moment which are all perfectly aligned
(c) domains are partially aligned
(d) domains are all perfectly aligned
4.Consider the two idealised systems (i) a parallel plate
capacitor with large plates and small separation and (ii) a
long solenoid of length L >> R, radius of cross-section. In
(i) E is ideally treated as a constant between plates and zero
outside. In (ii) magnetic field is constant inside the solenoid
and zero outside. These idealised assumptions, however,
contradict fundamental laws as below
(a) case (i) contradicts Gauss’ law for electrostatic fields
(b) case (ii) contradicts Gauss’ law for magnetic fields
(c) case (i) agrees with
E.dl 0.=òÑ
(d) case (ii) contradicts
en
H.dlI=òÑ
5.A paramagnetic sample shows a net magnetisation of 8 Am
–1
when placed in an external magnetic field of 0.6 T at a
temperature of 4 K. When the same sample is placed in an
external magnetic field of 0.2 T at a temperature of 16 K, the
magnetisation will be
(a)
132
Am
3
-
(b)
12
Am
3
-
(c)
1
6Am
-
(d)
1
2.4Am
-
NEET/AIPMT (2013-2017) Questions
6.A bar magnet of length ‘l’ and magnetic dipole moment ‘M’
is bent in the form of an arc as shown in figure. The new
magnetic dipole moment will be [2013]
(a)
3
p
M (b)
2
p
M
(c)
2
M
(d) M
7.A bar magnet of magnetic moment M is placed at right angles
to a magnetic induction B. If a force F is experienced by
each pole of the magnet, the length of the magnet will be
[NEET Kar. 2013]
(a)F/MB (b)MB/F
(c)BF/M (d)MF/B
8.Following figures show the arrangement of bar magnets in
different configurations. Each magnet has magnet ic dipole
moment m
r
. Which configuration has highest net magnetic
dipole moment ? [2014]
A.
N
SS N
B.
N
N
S
S
C.
S N
N
30ºD.
N
NS
60º
(a)A (b) B
(c)C (d) D
9.The magnetic susceptibility is negative for : [2016]
(a) diamagnetic material only (b) paramagnetic material only (c) ferromagnetic material only (d) paramagnetic and ferromagnetic materials
10.If q
1
and q
2
be the apparent angles of dip observed in two
vertical planes at right angles to each other, then the true angle of dip q is given by :- [2017]
(a) tan
2
q = tan
2
q
1
+ tan
2
q
2
(b) cot
2
q = cot
2
q
1
– cot
2
q
2
(c) tan
2
q = tan
2
q
1
– tan
2
q
2
(d) cot
2
q = cot
2
q
1
+ cot
2
q
2

566. 562 PHYSICS
EXERCISE - 1
1. (b) 2. (a) 3. (d) 4. (a)
5. (d)
6. (d) As
4
r
1
Fµ and r becomes twice, t
herefore, F becomes
16
1
2
1
4
=times
\ .N5.08
16
1
=´
7. (a) When cut
along the axis, area of cross-section
becomes half. Therefore, pole strength is halved and
M = m (2 l), is also halved.
8. (d) As magnetic moment = pole strength x length and
length is halved without affecting pole strength,
therefore, magnetic moment becomes half.
9. (d) A superconductor exhibits perfect diamagnetism.
10. (a)
1
2
tan 2
tan1
q
=
q
11. (a) Iron
is ferromagnetic.
12. (b) Ferromagnetic substance has magnetic domains
whereas paramagnetic substances have magnetic
dipoles which get attracted to a magnetic field.
Diamagnetic substances do not have magnetic dipole
but in the presence of external magnetic field due to
their orbital motion these substance are repelled.
13. (c)
ITµ; MTMIaÞa
2
1
2
1
M
M
T
T
=
Þ T
2
= 2T
1
= 2T
14. (c)
MH3
K
2
H)M2M(
KK
2T
21
1
p=
+
+
p=
MH
K
2
H)MM2(
K
2T
2
p=
-
p=
Obviously T
2
> T
1
15
. (b)
()
V
VH
H
1
B
tan 1,BB
B
tan 1 45
-
q===
q= =°
16. (c)c
m
is negativ e for diamagnetic materials.
17. (b) The iron can produces a magnetic screening for the
equipment as lines of magnetic force can not enter iron
enclosure.
18. (b) Diamagnetic substances do not have any unpaired
electron. and they magnetised in direction opposite to
that of magnetic field. Hence when they are brought to
North or South pole of Bar magnet, they are repelled
by poles.
19. (c) We know that when a bar magnet is placed in the
magnetic field at an angle q, then torque acting on the
bar magnet
(t) =
BMsinMB ´=q
Note : This torq
ue t has a tendency to make the axis of
the magnet parallel to the direction of the magnetic
field.
20. (a) Magnetic moment linked with one turn = iA
Magnetic moment linked with N turn
= iNA amp-m
2.
Here A = Area of current loop.
21. (b) A diamagnetic material in a magnetic field moves, from
stronger to the weaker parts of the field.
22. (b) According to Curie’s law,
m
1
T
cµ
23. (a) The magnetic moment of a diamagnetic atom is equal
to zero.
24. (a) A ® diamagnetic B ® paramagnetic
C ® Ferromagnetic D ® Non-magnetic
25. (a) The magnetic dipole moment of diamagnetic material is
zero as each of its pair of electrons have opposite spins,
i.e., m
d
= 0.
Paramagnetic substances have dipole moment > 0, i.e.
m
p
¹0, because of excess of electrons in its molecules
spinning in the same direction.
Ferro-magnetic substances are very strong magnets
and they also have permanent magnetic moment, i.e.
m
f
¹0.
EXERCISE - 2
1. (a) H
ere,
.cm3
2
6
d,cm4,cm82====ll
At neutral point,
2/322
0
)d(
M
4
BH
l+p
m
==

7
23
MM
10
1250(510)
-
-
==
´
52
M 1250H 1250 3.2 10 A m
-
\ = = ´´
Hints & Solutions

567. 563Magnetism and Matter
5
2
M 1250 3.2
10m A m.
2 8 10
-
-
´´
==
´l

1
0.5 Am 0.5 abamp 100 cm
10
==´´
= 5 a
b-amp cm.
2. (d) In series, same current flows through two tangent
galvanometers.
3. (b) As the axes are perpendicular, mid point lies on axial
line of one magnet and on equatorial line of other
magnet.
7
70
1 33
2M 10 21
B 2 10
4d1
-
-m ´´
\ = = =´
p
and
70
2 3
M
B 10
4d
-m
==
p
\ Resulta
nt field =
T105BB
72
2
2
1
-
´=+
4. (a) .N102
)3.0(
6030
10
r
mm
4
F
3
2
7
2
210 --
´=
´
´=
p
m
=
5. (b)t = MB si
n q = 0.1 × 3 × 10
–4
sin 30º
or t = 1.5 × 10
–5
N–m.
6. (d) As
1
2
1
2
tan
tan
i
i
q
q
=
7. (c) 8. (b)
9
. (b) According to Curie's law,
0
m
C
T
m
c=
where
C is Curie constant, T = temperature
m
1
T
\ca
1
2
m 2
m1
T273 333 606
2
T 273 30 303
c +
== ==
c+
()2 11 1
m mm m
/ 2 0.5
0.5 .cc
\c =c c =c== Q
10. (a
) As lines of magnetic induction B are continuous curves,
they run continuously through the bar and outside, as
shown in Fig. (1).
11. (c) The loop (i) is for soft iron and the loop (ii) is for steel
in Fig.
12. (b) As breadth of each part is half the original breadth,
therefore, pole strength becomes half (i.e. m/2).
13. (a) As length of each part also becomes half, therefore
magnetic moment M = pole strength × length111
th
224
Þ ´= i.e. M/4.
1
4. (c) Taking distances from the centre of the magnet,
.elyapproximat,1:8
1
x
1x2
x
x
B
B
33
1
2
2
1
=÷
ø
ö
ç
è
æ
+
+
=
÷
÷
ø
ö
ç
ç
è
æ
=
15. (d) Angle of dip, d = 45°
\
tan
tan
cos
d
d=¢
q 3
2
2/3
1
º30cos
45tan
===
1
Real dip tan (2 / 3)
-
\ d=¢
16.
(b)
3
MM2
MM2
T
T
2
1
2
2
=
-
+
= .s333TT
12
==\
17. (a)
3 3
21
3 1/33
1 2
ta
nd
r1
tan3d [r(3)]
q
===
q
3
1
3
3
3
60tan
tan
3
1
tan
12 ===q=q 2
30º\q=
18. (b
) Here, d
1
= 20 cm, M
2
= 2 M
1
, d
2
= ? 2
d
d
M
M
3
1
3
2
1 2
== Þ
1/3 1/3
21
d 2 d 20(2) cm==
19. (a)
xcosH
V
tan,
H
V
tan =q¢=q ;
xcos
1
tan
tan
=
q
q¢
20. (d)
12
40º, 30º, ?d = d = d=
22
12
cot cot cotd= d+d
22
cot 40º cot 30º=+
2
cot 1.19 3 2.1d= +=
25º i.e.40º.\ d= d<
21
. (b) W
1
= – MB (cos 90º – cos 0º) = MB
W
2
= – MB (cos 60º – cos 0º)

1W
2 1
MB
2 1
1
2 1
MB ==÷
ø
ö
ç
è
æ
--=
As W
1
= n W
2
\ n
= 2
22. (d) Here,
,s5
12
60
T
1
== s15
4
60
T
2
==
22
22
2
1
2
2
2
1
2
2
2
1
515
515
TT
TT
M
M
-
+
=
-
+
=
250 5
200 4
==

568. 564 PHYSICS
23. (c)
3
1
V3
V
H
V
tan ===d
30º / 6 radian\ d= =p
24. (c)
HM
I
2T
1
p= ,
M)FH(
I
2T
2
+
p=
2 2
1
22
2
THFT4
H F 4H
H1T T /4
+
= = = Þ +=
Þ 3H
= F
25. (a)
74
0r
(4 10 ) 2000 8 10
--
m =m m = p´ ´ = p´ S.I. units
26. (b) Let pole strength = m
So, M = ml
When wire is in form of arc, then the distance between
poles =
p
l2
So,
p
=
p
=
M22m
'M
l
27. (a) Magnetic moment is cancelled and m
net
= 0.
28. (c) q=tsinMB (q = 90°)
2
1
2
1
B
B
MB
t
t
=Þ=t
(since magnetic mome
nt is same)
29. (c) M
net
=
222
000
M M 2M cos60++°
2
00
3M 3M==
30. (c)
I
MB
2
1
n
p
=
When it is turned b
y an angle 90° the effective field is
vertical = V and B > V
So, new frequency < n.
31. (b)
MB
I
2Tp=
8
I
'I
22
m
I
2
=Þ÷
ø
ö
ç
è
æ
=
l
2
M
'M=
So, .sec2
2
T
'T
MB4
I
2'T ==Þp=
32. (b) Ma
gnetic moment, M = ml m
M
=
l
, where m is the polestrength.
Therefore distance between poles
()()
22
22=+ll =
2
l
So,
2
M
2
m
'M==
l
33. (c)
211
12
1
TMM
2
1TM
M
4
===
\ T
2
= 2T
1
= 3 s
34. (c) ()
22 22
1 21
2 2 22
2 21
M TT 6 4 52
2.6 :1
M 20TT 64
+ +
= = ==
--
35. (d)
The value of H is fairly uniform.
36. (b) Field in the core with Bismuth will be smaller because
bismuth is diamagnetic.
37. (b) From
rm
1;m = +c
Magnetic suscaptibility, mr
1c =m-
m
0.075 1 0.925.c = - =-
38. (b
) For a diamagnetic material, the value of µ
r
is less than
one. For any material, the value of
r
Îis always greater
than 1.
39. (a) Intensity of magnetisation
41
m 6
M 1.2
Ι 4 10 Am
V(15 2 1)10
-
-
= = =´
´´
40. (a) M
agnetic moment = M; Initial angle through which
magnet is turned (q
1
) = 90º and final angle which
magnet is turned (q
2
)= 60º. Work done in turning the
magnet through
90º(W
1
) = MB (cos 0º – cos 90º)= MB (1–0) = MB.
Similarly, W
2
= MB (cos 0º – cos 60º)
2
MB
2
1
1MB=÷
ø
ö
ç
è
æ
-= .
\ W
1
= 2W
2
or n = 2.
41. (
c)
H 0.50 0.50 2
B 1/3
cos cos30º 3
´
==
==
q
42. (b)t = MB sinq, t = iAB sin90º
A
B C
l
D

569. 565Magnetism and Matter
\ A
iB
t
== 1/2 (BC) (AD)
B
ut
2
22113
(BC)(AD)( )
2 2 24
æö
= -= ç÷
èø
l
lll
Þ
23
()
4 Bi
t
=l
\
1
2
2
3 B.i
æöt
=
ç÷
èø
l
43. (b)
44.
(d) Time period of a vibration magnetometer,
1
T
B
µ Þ
1
2
T
T
=
2
1
B
B
Þ
1
21
2
B
TT
B
=

6
6
24 10
24
6 10
-
-
´
==
´
s
45. (b)Force on a charged particle is given by F = qvB. Here
v = 0 and also resultant B is zero.
\ Force = 0
46. (a) For stable equilibrium
U = –MB = – (0.4) (0.16) = – 0.064 J
47. (c) The susceptibility of ferromagnetic substance
decreases with the rise of temperature in a complicated
manner. After Curies point in the susceptibility of
ferromagnetic substance varies inversely with its
absolute tempearture. Ferromagnetic substance obey’s
Curie’s law only above its Curie point.
48. (b) A magnetic field is produced by the motion of electric
charge. Since motion is relative, the magnetic field is
also relative.
49. (d) A paramagnetic sample display greater magnetisation
when cooled, this is because at lower temperature, the
tendency to disrupt the alignment of dipoles (due to
magnetising field) decreases on account of reduced
random thermal motion.
50. (b) Electromagnets are magnets, which can be turnd on
and off by switching the current on and off.
As the material in electromagnets is subjected to cyclic
changes (magnification and demangetisation), the
hysteresis loss of the material must be small. The
material should attain high value of I and B with low
value of magnetising field intensity H. As soft iron has
small coercivity, so it is a best choice for this purpose.
EXERCISE - 3
Exemplar Questions
1. (c) Toroid is a hollow circular ring on which a large number
of turns of a wire are closely wound. Thus, in this case
magnetic field is only confined inside the body of toroid.
So no magnetic field outside the toroid and magnetic
field only inside the toroid.
In case of toroid, the magnetic field is in the form of
concentric magnetic lines of force and there is no
magnetic field outside the body of toroid. This is
because the loop encloses no current. Thus, the
magnetic moment of toroid is zero.
In other case, if we take r as a large distance outside
the toroid, then
3
1
m
r
µ. Which is not possible.
2. (a) Magnetic declination is an angle between angle of
magnetic meridian and the geographic meridian.
As the earth’s magnetism, the magnetic field lines of
the earth resemble that of a hypothetical magnetic
dipole located at the centre of the earth.
The axis of the dipole does not coincide with the axis of
rotation of the earth but is presently tilted by 11.3°
(approx) with respect to geographical of axis earth. This
results into two situations as given in the figure.
N
EW
S
S
11.3º
11.3º
N
N
EW
S
N
S1
1
.3
º
So, the declination varies between 11.3° W to 11.3° E .
3. (d) We know that a permanent magnet is a substance which
at room temperature retain ferromagnetic property for along period of time. The individual atoms in aferromagnetic material possess a dipole moment as in aparamagnetic material. However, they interact with oneanother in such a way that they spontaneously alignthemselves in a common direction over a macroscopicvolume i.e., domain.
Hence, in a permanent magnet at room temperature,
domains are all perfectly aligned.
4. (b) The electric field lines, do not form a continuous path
while the magnetic field lines form the closed paths.
Gauss’s law states that,
0s
q
E.ds=
e
òÑ
for ele
ctrostatic
field. So, it does not contradict for electrostatic fields as
the electric field lines do not form closed continuous
path.
According to Gauss’ law in magnetic field,
s
E.ds0=
òÑ

570. 566 PHYSICS
It contr
adicts for magnetic field, because there is a
magnetic field inside the solenoid and no field outside
the solenoid carrying current but the magnetic field lines
form the closed path.
5. (b) According to the Curie law, the intensity of
magnetisation (I) is directly proportional to the magnetic
field induction and inversely proportional to the
temperature (t) in kelvin.
So, I magnetisation
µ
B (magnetic field induction)
t(temperature in kelvin)
Þ
2
1
I
I
=
21
12
Bt
Bt
´
... (i)
As given that :I
1
=
8 Am
–1
, I
2
= ?
B
1
= 0.6 T, t
1
= 4K
B
2
= 0.2 T, t
2
= 16K
by putting the value of B
1
, B
2
, t
1
, t
2
I
1
in equation (i)
So,
0.24
0.6 16
´ =
2
I
8
We get, I
2
=
1
8
12
´
I
2
=
2
A/m
3
NEET/AIPMT (2013-2017) Questions
6. (a) Magnetic dipole moment
M = m × l M' = m × r
From figure
60º
30º
30º
r sin 30ºr sin 30º
r
l
l =
r
3
p
or r =
3
p
l
so, M' = m × r =
m3´
p
l
=
3
p
M
7. (b)FL = MB (= To
rque) Þ L =
MB
F
8. (c) Net magnetic dipole moment = 2 Mcos
2
q
As value of cos
2
q
is maximum in case (c) hence net
magnetic dipole moment is maximum for option (c).
9. (a) Magnetic susceptibility c for dia-magnetic materials
only is negative and low |c| = –1; for paramagnetic
substances low but positive |c| = 1 and for
ferromagnetic substances positive and high |c| = 10
2
.
10. (d)If q
1
and q
2
are apparent angles of dip
Let a be the angle which one of the plane make with
the magnetic meridian.
1
tan
cos
v
H
q=
a
i.e.,
1
cos
tan
v
H
a=
q
…(i)
2
tan,
sin
v
H
q=
a
,
i.e.,
2
sin
tan
v
H
a=
q
…(ii)
Squaring and adding
(i) and (ii), we get
2
22
22
12
11
cos sin
tan tan
V
H
æöæö
a+a =+ç÷ç÷
èø qq
èø
i.e.,
2
22
12
2
1 cot cot
V
H
éù= q+q
ëû
or
2
22
12
2
cot cot
H
V
=q+q
i.e.,
222
12
cot cot cotq=q+q

571. THE EXPERIMENTS OF FARADAY AND HENRY
The discovery and understanding of electromagnetic induction
are based on a long series of experiments carried out by Faraday
and Henry. These experiments are illustrated by the following
figures.
When the bar magnet is pushed towards the coil, the pointer in
the galvanometer G deflects.
Current is induced in coil C
1
due to motion of the current carrying
coil C
2

S. No. Experiment Observation
1.Place a magnet near a conducting No current flows through the galvanometer.
loop with a galvanometer in the
circuit.
2.Move the ma gnet towards theThe galvanometer register a current.
loop.
3.Rev
erse the direction of motionThe galvanometer deflection reverses.
of the magnet.
4.Reverse the polarity of theThe galvanometer deflection reverses.
magnet and move the magnet
towards the loop.
5.Keep magnet fixed and moveThe galvanometer register a
the coil towards the magnet.current.
21
Electromagnetic
Induction

572. 568 PHYSICS
6.Increases
the speed of the magnet. The deflection in the galvanometer increases.
7.Increase the strength of the magnet. The deflection in the galvanometer increases.
8.Increase the diameter of the coil. The deflection in the galvanometer increases.
9.Fix the speed of the magnet butThe deflection in the galvanometer increases.
repeat the experiment with the
magnet closer to the coil.
10.Move the magnet at an angle toDeflection decreases, it is maximum when the magnet moves perpendicular to the
the plane of the coil. plane of the coil and is zero when the magnet moves parallel to the plane of the coil.
11.Increase the number of turns ofMagnitude of current increases.
the coil.
MAGNETIC FLUX
The number of magnetic lines of force crossi
ng a surface is called
magnetic flux linked with the surface.
It is represented by
f.
q
B
n
^
Magnetic flux . cosf==q
ur ur
B A BA
where B is s
trength of magnetic field, A is area of the surface and
q is the angle which normal to the area (unit area vector) makes
with the direction of magnetic field.
The S.I. unit of magnetic flux is weber which is the amount of
magnetic flux over an area of 1 m
2
held normal to a uniform
magnetic field of one tesla.
The c.g.s. unit of f is maxwell.
1 weber = 10
8
maxwell.
FARADAY’S LAW OF ELECTROMAGNETIC
INDUCTION
Whenever the number of magnetic lines of force (flux) linked
with any closed circuit change, an induced current flows through
the circuit which lasts only so long as the change lasts. An
increase in the number of lines of force produces an inverse
current, while a decrease of such lines produces a direct current.
The induced emf is equal to the negative rate of change of
magnetic flux.
i.e.
dt
d
e
f-
=
The -ve sign sh
ows that the induced emf opposes the change in
magnetic flux (Lenz’s law).
LENZ’S LAW
The direction of induced e.m.f. is given by Lenz’s law. According
to this law, the direction of induced e.m.f. in a circuit is always
such that it opposes the every cause which produces it.
Thus,
-f
=
d
e
dt
Lenz’s law is in accordance with the principle of conservation
of energy. Infact, work done in moving the magnet w.r.t. the coil
changes into electric energy producing induced current.
There is also another law for finding the direction of induced
current. This is Fleming’s right hand rule. According to this
rule, if we stretch the right-hand thumb and two nearby fingers
perpendicular to one another such that the first finger points in
the direction of magnetic field and the thumb in the direction of
motion of the conductor, then the middle finger will point in the
direction of the induced current.
Thumb
(motion)
First finger
(field)
Centra
l finger
(current)
A
(a)

S
B
N
Direction of induced
current inwards
Directi
on of motion
of the conductor
Application of Fleming’s
right-hand rule
(b)
Total flow of charge due to change of flux (Df):´
=Df=
(No. of turns change in magnetic flux)
Q N /R
Resistance
METHODS OF INDUCING E.M. F.
As is known, e.m.f. is induced in a circuit only when amount of
magnetic flux linked with the circuit changes. As f = BA cos q,
therefore three methods of producing induced e.m.f. :
(i) By changing B, (ii) By changing A and, (iii) By changing
q
(orientation of the coil). When a conductor of length l moves
with a velocity v in a magnetic field of strength B so that magnetic
flux linked with the circuit changes, the e.m.f. induced (e) is given
by
e = B l v.
Induced e.m.f. and its direction
Case (i) In conducting rod: The induced e.m.f. is generated
because of rotation of a conducting rod in a
perpendicular magnetic field
2
B
e
2
w
-=
l
also, e = – BAf
where f = frequency of rotation and
A = pr
2
, where r is the radius of circle in which this rod
moves, hence r = l. w = angular velocity, l = length of
conducting rod.

573. 569Electromagnetic Induction
Case (ii) In disc: Induced e.m.f generated in a disc rotating with
a constant angular velocity in a perpendicular magnetic
field
w
= =- p =-
2
2 Br
e –BAf Brf
2
where A = area of disc = pr
2
, r = radius of disc,
w = angular velocity of disc.
Case (iii) In two coils: When two coils are arranged as shown in
the figure
+-
Q
P
K
(a) if key K is closed then current in P will flow in
clockwise direction and consequently inducedcurrent in Q will flow in anticlockwise direction.
(see fig. a)
(b) when key K is opened then current in P falls from
maximum to zero and consequently induced
current in Q will flow in clockwise direction. (see
fig. b)
O
P
Q
(a)
O
P
Q
(b)
Case (
iv) In three coils arranged coaxially : Three coils P, Q
and R are arranged coaxially as shown in figure. Equalcurrents are flowing in coils P and R . Coils Q and R arefixed. Coil P is moved towards Q. The induced current
in Q will be in anti-clockwise direction so that it may
oppose the approach of P according to Lenz’s law. As
the face of P towards Q is a south pole hence plane of
Q towards P will also be a south pole.
Observer
P
Induced current
AxisRQ
coils
As there is no relative motion between Q and R, hence
no current is induced in Q due to R.
Case (v) Current increases in straight conductor : When
current in the straight conductor is increased then
(a) the direction of induced current in the loop will
be clockwise so that it may oppose the increase of
magnetic flux in the loop in downward direction.
O
P
A B
(b) the direction of induced current in the loop will
be anti-clockwise so that it may oppose the
increase of magnetic flux in the loop in upward
direction.
O
P
A B
Case (vi) Magnet dropped freely in long vertical copper tube:
The resistance of copper tube is quite negligible andhence maximum induced current are generated in itdue to the motion of the magnet. Due to these inducedcurrent the motion of magnet is opposed to maximum.Consequently the acceleration of the magnet will be
zero (a = g – g = 0).
S
N
g
g
a = g – g = 0
Case (vii) Magnet dropped freely into a long solenoid of copper
wire: The resistance of copper solenoid is much higher
than that of copper tube. Hence the induced currentin it, due to motion of magnet, will be much less than
that in the tube. Consequently the opposition to the
motion of magnet will be less and the magnet will fall
with an acceleration (a) less than g. (i.e. a < g).
S
a < g
N
Case (viii) Motional EMF: Induced emf in a conducting rod moving
perpendicular through a uniform magnetic field asshown
×
×
×
×
×
×
×
×
×
×
×
×
v
l
+ +
– –
B
The induced emf produced across the rod
= =ò´
lrr
ll
0
( ).e Bv v Bd
This is also called motional emf and it develops when
a metal rod cuts magnetic lines of force.

574. 570 PHYSICS
Special cas
e : If the rod moves in the magnetic field making an
angle q with it, then induced emf
==qll sin
n
e B v Bv .
COMMON DEFAULT
û
Incorrect. When there is no change in magnetic flux no
induced current is produced.
ü
Correct. Consider the case (viii) discussed above. There is
no change in the magnetic flux throgh the rod, still induced
emf is produced.
Case (ix)A straight conductor (slider) moving with velocity v
on a U shaped wire placed in a uniform magnetic field.
The induced current produced is
=
lBv
I
R
Case (x)When a rect angular loop perpendicular to the magnetic
field is pulled out, then forces
1F and
3F being equal
and o
pposite cancel out.
F
2
F
1
F
3
v
B
>
l
I
æö
=== ç÷
èø
ll
ll
22
2
Bv Bv
FBIB
RR
Power required to move the loop out
P= F
2
× v =
l
222
Bv
R
Case (xi) The magnet is stationary and the loop is moving
towards the magnet.
S N
Stationary magnet
v
I
Moving loop
The induced emf or current I is shown which is in
accordance to Lenz's law. In this case the magnetic
force causes the charge to move. We know that if a
charged particle is in motion in a field it experiences a
magnetic force. This is because when charged particle
moves it creates its own magnetic field which interacts
with the existing magnetic field.
Case (xii) The magnet is moving towards the loop which is
stationary.
S N
Moving magnet
v
I
Stationary loop
The induced emf or current I is shown which is in
accordance to Lenz’s law. Here the varying magnetic field
at the location of loop (due to the movement of magnet)
creates an electric field.
We should remember certain points regarding the induced
electric field produced due to changing magnetic field.
·Induced electric field lines form closed loops (different
from the electric field lines used to depict electric field
produced due to charges)
·Induced electric field is non-conservative in nature
(again a difference from the electric field produced by
electric charges)
Mathematically,
ò
¹
f
-==0
dt
d
dl.Ee
1. An emf is induced in a circuit where the magnetic flux is
changing even if the circuit is open. But obviously no
current will flow. If we close the circuit, the current will start
flowing.
2. In a loop moving in a uniform magnetic field, when the loop
remains in the field, the net emf induced is zero.

v B
l Bvl Bvl
Example 1.
A copper rod
of length
l is rotated about one end
perpendicular to the uniform magnietic field B with
constant angular velocity
w. What will be the induced
e.m.f. between two ends ?
Solution :
Consider a small element of the rod of length dx at a distance
x from the centre O.
O
l
w
dx
x
Let v be the linear velocity of the element at right angles to the magnetic field B. The e.m.f. developed across the element is d e = B v dx = B (w x) dx (
Q v = w x)
The e.m.f. a
cross the entire rod of length l is given by
l
l
0
2
0 2
x
BdxxBdee
ú
ú
û
ù
ê
ê
ë
é
w=w==ò ò
=
22
B
2
1
)2/(Bll w=w

575. 571Electromagnetic Induction
Example 2.
A conductor of length 10 cm is moved parallel to itself with
a speed of 10 m/s at right angles to a uniform magnetic
induction 10
–4
Wb/m². What is the induced e.m.f. in it?
Solution :
Given :l = 10 cm = 0.1 m, v = 10 m/s
B = 10
–4
Wb/m
2
e.m.f. induced in conductor
e = B l V = 10
–4
× 0.1 × 10 = 10
–4
V
Example 3.
A metal rod of length 1 m is rotated about one of its ends in
a plane right angles to a field of inductance 2.5 × 10
–3
Wb/m². If it makes 1800 revolutions/min. Calculate induced
e.m.f. between its ends.
Solution :
Given :l = 1m, B = 5 × 10
–3
Wb/m
2
f =
1800
60
= 30 rotations/sec
In one rotation, the moving rod of the metal traces a circle of
radius r = l
\Area swept in one rotation = pr
2
d
dt
f
=()
d dA
BA B.
dt dt
=
2
Br
T
p
= = B f p r
2
=(5 × 10
–3
) × 3.14
× 30 × 1 = 0.471 V
\e.m.f. induced in a metal rod = 0.471 V
Example 4.
A coil having 100 turns and area 0.001 metre
2
is free to
rotate about an axis. The coil is placed perpendicular to a
magnetic field of 1.0 weber/metre
2
. If the coil is rotate
rapidly through an angle of 180°, how much charge will
flow through the coil? The resistance of the coil is 10 ohm.
Solution :
The flux linked with the coil when the plane of the coil is
perpendicular to the magnetic field is
f = nAB cos q = nAB.
Change in flux on rotating the coil by 180° is
df = nAB – (–nAB) = 2nAB
\induced charge = d
R
f
=
2nAB
dt
=
2 100 0.001 1
10
´´´
= 0.01 coulomb
Example 5.
Predict the direction of induced current in the situations
described by the following fig. (1) to (5).
a b N S
Fig -1
a bN Sc d
Fig.2
Fig. 3
a
c
b
a b
Fig. 4
Decreasing at
a stead rate
Fig.5
I
Soluti
on :
Applying Lenz’s law
Fig. (1) along a ® b
Fig. (2) along b ® a
Fig. (3) along c ® a
Fig. (4) along a ® b
Fig. (5) no induced current since field lines lie in the plane
of the loop.
EDDY CURRENTS
The induced circulating currents produced in a metal itself due
to change in magnetic flux linked with the metal are called
eddy currents. These currents were discovered by Foucault, so
they are also known as Foucault Currents.
The direction of eddy currents is given by Lenz’s law.
Eddy currents produced in a metallic block moving in a
non-uniform magnetic field is shown in fig.
Applications of Eddy Current
Like friction, eddy currents are helpful in some fields and have tobe increased, while in some other fields they are undesirable and
have to be minimised.
(1) Dead beat galvanometer. (2) Energy meter.
(3) Speedometer. (4) Electric brakes.
(5) Single phase AC motor. (6) Induction furnace.
(7) Diathermy

576. 572 PHYSICS
In a moving coil galvanometer, damping is necessary
to avoid oscillation of display needle. This is brought into practice
with the help of eddy currents. The winding of the coil of
galvanometer is done on a metallic frame. When the coil rotates
the magnetic flux linked with the metallic frame changes due to
which eddy currents are developed which oppose the rotation of
the coil. This is called dead beat galvanometer.
SELF INDUCTANCE AND MUTUAL INDUCTANCE
Self Inductance
The property of a coil by virtue of which the coil opposes any
change in the strength of the current flowing through it, by
inducing an e.m.f. in itself is called self inductance.
Coil
Direction of
induced e.m.f.
(e)
K
When a current I flows through a coil, the magnetic flux f linked
with the coil is f = LI, where L is coefficient of self inductance of
the coil.
On differentiating, we get
f
= =-
dd
Le
dt dt
I
If dI / dt = 1; L = – e.
Hen
ce coefficient of self inductance of a coil is equal to e.m.f.
induced in the coil when rate of change of current through the
same coil is unity. Coefficient of self induction of a coil is also
defined as the magnetic flux linked with a coil when 1 ampere
current flows through the same coil.
The value of L depends on geometry of the coil and is given by
2
0
m
=
l
NA
L .
where l is le
ngth of the coil (solenoid), N is total number of turns
of solenoid and A is area of cross section of the solenoid.The S.I. unit of L is henry. Coefficient of self induction of a coil is
said to be one henry when a current change at the rate of 1
ampere/sec. in the coil induces an e.m.f. of one volt in the coil.
Keep in Memory
1.Energy sto
red in a coil (inductor) =
2
Li
2
1
where L is the
self-inductance and i current flowing through
the inductor.The energy stored in the magnetic field of the coil.

2
22
0
0
11
()
22
æö
= =m
ç÷
mèø
l
B
E Li nA
n
22
00
B
volume
22
æö æö
= =´ç÷ ç÷
mmèø èø
l
B
A
2.Th
e self inductance is a measure of the coil to oppose the
flow of current through it. The role of self-inductance in an
electrical circuit is the same as that of the inertia in
mechanics. Therefore it is called electrical inertia .
3.The magnetic energy density (energy stored per unit
volume) in a solenoid
0
2
2
B
m
=
Mutual Inductance
Mutual i
nduction is the property of two coils by virtue of which
each opposes any change in the strength of current flowingthrough the other by developing an induced e.m.f.
First coil
Second
coil
I
1
N
1
N
2
Mutual
induction
Coefficient of mutual inductance (M) of two coils is said to be
one henry, when a current change at the rate of 1 ampere/sec. inone coil induces an e.m.f. of one volt in the other coil. The value
of M depends on geometry of two coils, distance between two
coils, relative placement of two coils etc.
The coefficient of mutual inductance of two long co-axial
solenoids, each of length l, area of across section A, wound on
an air core is
012
m
=
l
NNA
M ] … (1)
where N
1
and N
2
ar
e total number of turns of the two solenoids.
The mutual inductance M is defined by the equation
N
2
f
2
= MI
1
where I
1
is the current in coil 1, due to which flux f
2
is linked
with each turn of secondary coil.Now we can calculate, e.m.f. e
2
induced in secondary by a
changing current in first coil. From Faraday‘s law
)N(
dt
d
e
22
2
f-=
1
I
=-
d
M
dt
If Me1
dt
Id
2
1
-=Þ= ...(2)
The two defin
itions for M defined by equations (1) and (2) are
equivalent. We can express these two equations in words as :
(i) M is numerically equal to the flux-linkage in one circuit,
when unit current flows through the other. (we use this
definition to calculate M)
(ii) M is numerically equal to the e.m.f. induced in one circuit,
when the current changes in the other at the rate of one
ampere in each second. (it is used to describe the mutual
behavior of two circuits).
For a pair of coils, M
12
= M
21
= m
0
N
1
N
2
A/l, when wound on
one another.

577. 573Electromagnetic Induction
Keep in Memory
1.
Coefficient of self inductance of two coils in series :
L
1
L
2
L= L+
s 1 L
2
The effective self inductance is L
s
= L
1
+ L
2
If M is the coefficient of mutual inductance between the
two coils when they have flux linkage in the same sense,then
L = L
1
+ L
2
+ 2M
L
1 L
2
L
1
L
2
And for flux linkage in opposite direction
L = L
1
+ L
2
– 2M
2. Coefficient of self inductance of two coils in parallel :
L
1
L
2
L
p
21p
L
1
L
1
L
1
+=
(i) The c
oefficient of coupling between two coils having
self inductance L
1
& L
2
and coefficient of mutual
inductance M is
21
LL
M
K
±
=
(ii) G
enerally the value of K is less than 1.
(iii) If K is 1, then the coupling of two coils is tight while if
K < 1, then coupling is loose.
·Inductance is pure geometrical factor,and is
independent of current or applied e.m.f.
·If the angle between the axis of two closely placed
coil is q then
qµcosM.
AC GENERATOR/D
YNAMO/ALTERNATOR
An electrical machine used to convert mechanical energy into
electrical energy is known as AC generator/alternator or
dynamo.
Principle : It works on the principle of electromagnetic induction,
i.e., when a coil is rotated in uniform magnetic field, an induced
emf is produced in it.
Working :
N S
B C
A D
OutputR
L
R
1
R
2
B
1
B
2
When the armature coil ABCD rotates in the magnetic field
provided by the strong field magnet, it cuts the magnetic lines of
force. Thus the magnetic flux linked with the coil changes and
hence induced emf is set up in the coil. The direction of the
induced emf or the current in the coil is determined by the
Fleming’s right hand rule.
The current flows out through the brush B
1
in one direction of
half of the revolution and through the brush B
2
in the next half
revolution in the reverse direction. This process is repeated.
Therefore, emf produced is of alternating nature.
0
Nd
e NBAs int esint
dt
f
=-=ww =w , where e
0
= NBAw
0
0
ee
I sint I sint
RR
== w=w , R ® resistance of the circuit
DC MOTOR
A D.C. motor converts direct current energy from a battery into
mechanical energy of rotation.
Principle : It is based on the fact that when a coil carrying current
is held in a magnetic field, it experiences a torque, which rotates
the coil.
Working :
A D
CB
B
1
B
2
R
1R
2
V
N
S

D A
BC
B
1
B
2
R
2
R
1
V
N S
The battery sends current through the armature coil in the
direction shown in fig. Applying Fleming’s left hand rule, CD
experiences a force directed inwards and perpendicular to the
plane of the coil. Similarly, AB experiences a force directed
outwards and perpendicular to the plane of the coil. These two
forces being equal, unlike and parallel form a couple. The couple
rotates the armature coil in the anticlockwise direction. After the
coil has rotated through 180°, the direction of the current in AB
and CD is reversed, fig. Now CD experiences an outward force
and AB experiences an inward force. The armature coil thus
continues rotating in the same i.e., anticlockwise direction.
Efficiency of the d.c. motor : Since the current I is being supplied
to the armature coil by the external source of e.m.f. V, therefore,
Input electric power = VI
According to Joule’s law of heating,
Power lost in the form of heat in the coil = I
2
R
If we assume that there is no other loss of power, then
Power converted into external work
i.e., Output mechanical power = VI – I
2
R = (V – IR) I = EI
\Efficiency of the d.c. motor
Output mechanical power
Input electric power
h=
or
...
...
h= ==
EI E Back e m f
VI V Applied em f

578. 574 PHYSICS
Uses of D
.C Motor
1.The D.C. motors are used in D.C. fans (exhaust, ceiling or
table) for cooling and ventilation.
2.They are used for pumping water.
3.Big D.C. motors are used for running tram-cars and even
trains.
Example 6.
Two coils are wound on the same iron rod so that the flux
generated by one also passes through the other. The
primary has 100 loops and secondary has 200 loops. When
a current of 2 A flows through the primary the flux in it is
25 x 10
–4
Wb. Determine value of M between the coils.
Solution :
dt
d
Ne
s
ss
f
= and
dt
di
Me
p
s= ;
\
dt
di
M
dt
d
N
ps
s =
f
or
)02(
)0105.2(200
di
d
NM
4
p
s
s
-
-´
=
f
=
-
= 2.5×10
–2
=
25 mH
Example 7.
A long solenoid of length L, cross section A having N
1
turns has wound about its centre is small coil of N
2
turns
as shown in fig. Then find the mutual inductance of two
circuits.
L
N
2
N
1
1
2
Solution :
Magnetic flux
at the centre of solenoid
1101
i)L/N(Bm=
Magnetic flux through each turn of the coil of area A,
A
L
iN
AB
110
11 ´
m
==f
Magnetic flux linke
d with the coil of turns N
2
,
L
AiNN
N
1210
212
m
=´f=f
According to
the definition of mutual inductance
12
Mi=f
\
A
L
iNN
Mi
1210
1
m
= or
L
ANN
M
210
m
=
Example 8.
A small coil of radius r is placed at the centre of a l
arge
coil of radius R, where R >>r. The two coils are coplanar.
The mutual induction between the coils is proportional to
(a) r/R (b) r
2
/R
(c) r
2
/R
2
(d) r/R
2
Solution : (b)
Let I be the current flows in the large coil.
\ Mag. field at the centre of coil
0
B
2R
mI
=
Mag. flux linked with
smaller coil
22 0
rBr
2R
æömI
f=p =p ç÷
èø
But f = MI \
2
0r
M
2R
pmf
==
I
or R/rM
2
µ
Example
9.
The mutual inductance of a pair of coils is 0.75 H. If current
in the primary coil changes from 0.5 A to zero in 0.01 s find
average induced e.m.f. in secondary coil.
Solution :
Given : M = 0.75 H and
dI 0.5 0
50 A/s
dt 0.01
-
==
\ Average ind
uced e.m.f. in secondary coil,
dI
e M 0.75 50 37.5 V
dt
= = ´=
Example 10.
Find
the self inductance of a coil in which an e.m.f. of 10 V is
induced when the current in the circuit changes uniformly
from 1 A to 0.5 A in 0.2 sec.
Solution :
Given : e = 10 V and
dI 1 0.5 0.5
2.5A/s
dt 0.2 0.2
-
= ==
Self
inductance of coil
e 10
L 4H
dI / dt 2.5
= ==
dI
e L (ConsideringMagnitude only)
dt
éù
=
êú
ëû
Q

579. 575Electromagnetic Induction
CONCEPT MAP
ELECTROMAGNETIC
INDUCTION (EMI)
Generation of current or emf
by changing magnetic field
Faraday's laws of electromagnetic
Induction
lst law When magnetic
flux linked with the
circuit changes an emf is
induced in the circuit
2nd law Induced emf
rate of change of
magnetic flux
µ
–d
e
dt
f
=
Moti
o
n
a
l

e
m
f

d
e
–
–
B
l
v
d
tf
=
=
Across the end of r
o
d
2
1
e
B
l
2
=
w
B B.A=
B
A
c
o
s
f=
q
Magnetic
f
l
u
x

L
e
n
z
's

l
a
w

D
i
r
e
c
t
i
o
n
o
f

i
n
d
u
c
e
d
e
m
f
o
r

c
u
r
r
e
n
t
i
s
a
l
w
a
y
s
i
n
s
u
c
h

a

w
a
y
t
h
a
t

i
t

o
p
p
o
s
e
s

c
a
u
s
e
d
u
e
t
o

w
h
i
c
h

i
t

i
s
p
r
o
d
u
c
e
d
.
I
t
i
s
i
n
a
c
c
o
r
d
a
n
c
e
w
i
t
h
c
o
n
s
e
r
v
a
t
i
o
n

o
f
e
n
e
r
g
y
Ac Generator or Dyn
a
m
o

Produces electrical en
e
r
g
y

from mechanical ene
r
g
y
.
I
t
works on EMI princi
p
l
e
In
d
u
c
t
a
n
c
e
A
m
e
a
s
u
r
e
of

t
h
e

r
a
t
i
o
o
f

t
h
e

f
l
u
x

to

t
h
e
c
u
r
r
e
n
t
S
e
l
f

i
n
d
u
c
t
a
n
c
e
I
n
e
r
t
i
a
o
f

e
l
e
c
t
r
i
c
i
t
y
.
C
o
e
f
f
i
c
i
e
n
t
o
f

s
e
l
f
i
n
d
u
c
t
a
n
c
e

i
f
=
B
L
M
u
t
u
a
l
I
n
d
u
c
t
a
n
c
e
I
n
d
u
c
e
d

e
m
f
i
n
a

c
i
r
c
u
i
t
d
u
e

t
o

c
h
a
n
g
e

i
n

m
a
g
n
e
t
i
c
f
l
u
x
i
n
i
t
s

n
e
i
g
h
b
o
u
r
i
n
g
c
i
r
c
u
i
t
.
C
o
e
f
f
i
c
i
e
n
t
s
o
f

m
u
t
u
a
l
i
n
d
u
c
t
a
n
c
e


C
o
e
f
f
i
c
i
e
n
t
o
f

m
u
t
u
a
l

i
n
d
u
c
t
a
n
c
e

b
e
t
w
e
e
n
t
w
o

l
o
n
g
s
o
l
e
n
o
i
d
s

0
1
2
N
N
A
M
l
m
=
S
e
l
f

i
n
d
u
c
t
a
n
c
e
o
f
a

l
o
n
g

s
o
l
e
n
o
i
d

2
0
N
A
L
l
m
=
M
f
=
I
Eddy current Induced,
when magnetic flux linked
with the conductor changes
Electromagnetic
damping
Induction furnace
Magnetic braking
Electric power meter
Direction of induced current Fleming’s Right Hand Rule:
Thumb, forefinger, central
finger of right hand stretched
perpendicular to each other
then if thumb direction
of motion; forefinger
direction of magnetic field
then central finger induced
current
®
®
®
Induced current in a coil rotated
in uniform magnetic field
NBA sint
I
R
w w
=
®
®
®
®

580. 576 PHYSICS
1.Eddy curre
nts are produced when
(a) a metal is kept in varying magnetic field
(b) a metal is kept in steady magnetic field
(c) a circular coil is placed in a magnetic field
(d) through a circular coil, current is passed
2.An inductor may store energy in
(a) its electric field
(b) its coils
(c) its magnetic field
(d) both in electric and magnetic fields
3.If N is the number of turns in a coil, the value of self
inductance varies as
(a)N
0
(b) N (c)N
2
(d) N
–2
4.A coil having an area A
0
is placed in a magnetic field which
changes from B
0
to 4 B
0
in time interval t. The e.m.f. induced
in the coil will be
(a) t/BA3
00
(b) t/BA4
00
(c) tA/B3
00
(d) tB/A4
00
5.An electron moves along the line PQ which lies in the same
plane as a circular loop of conducting wire as shown in figure.
What will be the direction of the induced current in the loop ?
(a) Anticlockwise
(b) Clockwise
(c) Alternating
loop
P Q
(d) No current will be induced
6.Induced emf in the coil depends upon(a) conductivity of coil(b) amount of flux(c) rate of change of linked flux(d) resistance of coil
7.Two identical coaxial circular loops carry current i each
circulating in the clockwise direction. If the loops are
approaching each other, then
(a) current in each loop increases
(b) current in each loop remains the same
(c) current in each loop decreases
(d) current in one-loop increases and in the other it decreases
8.The mutual inductance of a pair of coils, each of N turns, is
M henry. If a current of I ampere in one of the coils is
brought to zero in t second, the emf induced per turn in the
other coil, in volt, will be
(a)
t
MI
(b)
t
NMI
(c)
It
MN
(d)
Nt
MI
9.A rectangular c
oil of single turn, having area A, rotates in
a uniform magnetic field B with an angular velocity
w about
an axis perpen
dicular to the field. If initially the plane of
the coil is perpendicular to the field, then the average
induced emf when it has rotated through 90° is
(a)
p
wBA
(b)
p
w
2
BA
(c)
p
w
4
BA
(d)
p
wBA2
10.According to Far
aday’s law of electromagnetic induction
(a) electric field is produced by time varying magnetic flux.
(b) magnetic field is produced by time varying electric flux.
(c) magnetic field is associated with a moving charge.
(d) None of these
11.Two solenoids of same cross-sectional area have their
lengths and number of turns in ratio of 1 : 2. The ratio of
self-inductance of two solenoids is
(a) 1 : 1(b) 1 : 2(c) 2 : 1 (d) 1 : 4
12.The back e.m.f. in a d.c. motor is maximum, when
(a) the motor has picked up max speed
(b) the motor has just started moving
(c) the speed of motor is still on the increase
(d) the motor has just been switched off
13.The mutual inductance between two coils depends on
(a) medium between the coils
(b) separation between the two coils
(c) orientation of the two coils
(d) All of the above
14.If coefficient of self induction of a coil is 1 H, an e.m.f. of 1V
is induced, if
(a) current flowing is 1A
(b) current variation rate is 1 As
–1
(c) current of 1A flows for one sec.
(d) None of these
15.Which of the following units denotes the dimension
2
2
Q
ML
,
where Q denotes th
e electric charge?
(a) Wb/m
2
(b) henry (H)
(c) H/m
2
(d) weber (Wb)
16.In an AC generator, a coil with N turns, all of the same area
A and total resistance R, rotates with frequency w in a
magnetic field B. The maximum value of emf generated in
the coil is
(a) N.A.B.R.w (b) N.A.B.
(c) N.A.B.R. (d) N.A.B.w
17.A metal rod moves at a constant velocity in a direction
perpendicular to its length. A constant uniform magnetic
field exists in space in a direction perpendicular to the rod
as well its velocity. Select correct statements (s) from the
following.
(a) The entire rod is at the same potential
(b) There is an electric field in the rod
(c) The electric potential is highest at the centre
(d) The electric potential is lowest at its centre and
increases towards its ends

581. 577Electromagnetic Induction
18.A small square loop of wire of side l is placed inside a
large square loop of side L (L >>l). The loop are coplanar
and their centres coincide. The mutual inductance of the
system is proportional is
(a)
L
l
(b)
L
2
l
(c)
l
L
(d)
l
2
L
19.As a result of
change in the magnetic flux linked to the
closed loop shown in the figure, an e.m.f. V volt is induced
in the loop.
The work done (in joule) in taking a charge Q coulomb once
along the loop is
( a ) QV (b) 2QV (c) QV/2 (d) zero
20.A wire loop is rotated in a uniform magnetic field about an
axis perpendicular to the field. The direction of the current
induced in the loop reverses once each
(a) quarter revolution(b) half revolution
(c) full revolution(d) two revolutions
21.If the number of turns per unit length of a coil of solenoid is
doubled, the self-inductance of the solenoid will
(a) remain unchanged(b) be halved
(c) be doubled (d) become four times
22.The total charge induced in a conducting loop when it is
moved in a magnetic field depend on
(a) the rate of change of magnetic flux
(b) initial magnetic flux only
(c) the total change in magnetic flux
(d) final magnetic flux only
23.Lenz’s law is consequence of the law of conservation of
(a) energy (b) momentum
(c) charge (d) mass
24.If rotational velocity of a dynamo armature is doubled, then
induced e.m.f. will become
(a) half (b) two times
(c) four times (d) unchanged
25.Choke coil works on the principle of
(a) transient current(b) self induction
(c) mutual induction(d) wattless current
1.A current i = 2 sin (pt/3) amp is flowing in an inductor of 2
henry. The amount of work done in increasing the current
from 1.0 amp to 2.0 amp is
(a) 1 J(b) 2 J (c) 3 J (d) 4 J
2.Fig shown below represents an area A = 0.5 m
2
situated in a
uniform magnetic field B = 2.0 weber/m
2
and making an
angle of 60º with respect to magnetic field.
B
60
The value of the magnetic flux through the area would be
equal to
(a) 2.0 weber (b)3 weber
(c) 2/3 weber (d) 0.5 webe r
3.In a coil of area 10 cm
2
and 10 turns with magnetic field
directed perpendicular to the plane and is changing at the
rate of 10
8
Gauss/second. The resistance of the coil is 20W.
The current in the coil will be
(a) 0.5 A (b) 5 A
(c) 50 A (d) 5 × 10
8
A
4.A generator has an e.m.f. of 440 Volt and internal resistance
of 4000 hm. Its terminals are connected to a load of 4000
ohm. The voltage across the load is
(a) 220 volt (b) 440 volt
(c) 200 volt (d) 400 volt
5.When the current in a coil changes from 2 amp. to 4 amp. in
0.05 sec., an e.m.f. of 8 volt is induced in the coil. The
coefficient of self inductance of the coil is
(a) 0.1 henry (b) 0.2 henry
(c) 0.4 henry (d) 0.8 henry
6.A copper disc of radius 0.1 m rotated about its centre with
10 revolutions per second in a uniform magnetic field of 0.1
tesla with its plane perpendicular to the field. The e.m.f.
induced across the radius of disc is
(a)
volt
10
p
(b)
2
volt
10
p
(c) volt10
2-
´p (d) vo
lt102
2-
´p
7.A coil has 200 turns and area of 70 cm
2
. The magnetic field
perpendicular to the plane of the coil is 0.3 Wb/m
2
and take 0.1
sec to rotate through 180º.The value of the induced e.m.f. will
be
(a) 8.4 V (b) 84 V
(c) 42 V (d) 4.2 V

582. 578 PHYSICS
8.If a cu
rrent increases from zero to one ampere in 0.1 second
in a coil of 5 mH, then the magnitude of the induced e.m.f.
will be
(a) 0.005 volt (b) 0.5 volt
(c) 0.05 volt (d) 5 volt
9.A 100 millihenry coil carries a current of 1 ampere. Energy
stored in its magnetic field is
(a) 0.5 J(b) 1 J(c) 0.05 J (d) 0.1 J
10.The armature of a dc motor has 20W resistance. It draws a
current of 1.5 A when run by a 220 V dc supply. The value
of the back emf induced in it is
(a) 150 V (b) 170 V(c) 180 V (d) 190 V
11.In the figure the flux through the loop perpendicular to the
plane of the coil and directed into the paper is varying
according to the relation f = 6t
2
+ 7t + 1 where f is in
milliweber and t is in second. The magnitude of the emf
induced in the loop at t = 2 s and the direction of induce
current through R are
(a) 39 mV; right to left
(b) 39 mV; left to right
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R
(c) 31 mV; right to left
(d) 31 mV; left to right
12.A coil having 500 square loops each of side 10 cm is placed
normal to a magnetic field which increases at the rate of
1 Wb/m
2
. The induced e.m.f. is
(a) 0.1 V (b) 5.0 V (c) 0.5 V (d) 1.0 V
13.A circular coil and a bar magnet placed nearby are made to
move in the same direction. If the coil covers a distance of
1 m in 0.5. sec and the magnet a distance of 2 m in 1 sec, the
induced e.m.f. produced in the coil is
(a) zero(b) 0.5 V(c) 1 V (d) 2 V.
14.Magnetic flux f in weber in a closed circuit of resistance
10W varies with time f (sec) as f = 6t
2
– 5t + 1. The magnitude
of induced current at t = 0.25s is
(a) 0.2 A(b) 0.6 A (c) 1.2 A (d) 0.8 A
15.The current in a coil of L = 40 mH is to be increased uniformly
from 1A to 11A in 4 milli sec. The induced e.m.f. will be
(a) 100 V (b) 0.4 V (c) 440 V (d) 40 V
16.The self inductance of the motor of an electric fan is 10 H.
In order to impart maximum power at 50 Hz, it should be
connected to a capacitance of
(a)8F
m (b) 4Fm (c)2Fm (d) 1Fm
17.The flux linked with a coil at any instant 't' is given by
f = 10t
2
– 50t + 250. The induced emf at t = 3s is
(a) –190 V (b) –10 V(c) 10 V (d) 190 V
18.A conducting square loop of side L and resistance R moves
in its plane with a uniform velocity v perpendicular to one
of its side. A magnetic induction B constant in time and
space, pointing perpendicular and into the plane of the
loop exists everywhere.
x x x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x x
x
x
x
x
x
x
x
x
x
x
x
Bv
The current induced in the loop is
(a)
Bv
clockwise
R
l
(b)
Bv
anticlockwise
R
l
(c)
2B
anticlockwise
R
vl
(d) zero
19.The two rails
of a railway track, insulated from each other
and the ground, are connected to millivoltmeter. What is
the reading of the millivoltmeter when a train passes at a
speed of 180 km/hr along the track, given that the vertical
component of earth’s magnetic field is 0.2 × 10
–4
wb/m
2
and
rails are separated by 1 metre
(a) 10
–2
volt (b) 10 mV
(c) 1 volt (d) 1 mV
20.A long solenoid having 200 turns per cm carries a current of
1.5 amp. At the centre of it is placed a coil of 100 turns of
cross-sectional area 3.14 × 10
–4
m
2
having its axis parallel to
the field produced by the solenoid. When the direction of
current in the solenoid is reversed within 0.05 sec, the induced
e.m.f. in the coil is
(a) 0.48 V (b) 0.048 V
(c) 0.0048 V (d) 48 V
21.Two coils have a mutual inductance 0.005H. The current
changes in first coil according to equation I = I
0
sin wt
where I
0
= 10A and w = 100p radian/sec. The max. value of
e.m.f. in second coil is
(a)2p (b) 5p
(c)p (d) 4p
22.A metal conductor of length 1 m rotates vertically about one of
its ends at angular velocity 5 radians per second. If the
horizontal component of earth’s magnetic field is
0.2 × 10
–4
T, then the e.m.f. developed between the two ends of
the conductor is
(a) 5 mV (b) 50 mV
(c) 5 mV (d) 50 mV
23.Two identical induction coils each of inductance L are
jointed in series are placed very close to each other such
that the winding direction of one is exactly opposite to that
of the other, what is the net inductance?
(a)L
2
(b) 2 L
(c) L /2 (d) zero
24.A thin circular ring of area A is held perpendicular to a
uniform magnetic field of induction B. A small cut is made in
the ring and a galvanometer is connected across the ends
such that the total resistance of the circuit is R. When the
ring is suddenly squeezed to zero area, the charge flowing
through the galvanometer is
(a)
A
BR
(b)
R
AB
(c) ABR (d)
2
2
R
AB

583. 579Electromagnetic Induction
25.Consider the situation shown. The wire AB is sliding on
fixed rails with a constant velocity. If the wire AB is replaced
by semi-circular wire, the magnitude of induced e.m.f. will
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Ä
Ä
A
B
R
v
(a) incr
ease
(b) decrease
(c) remain the same
(d) increase or decrease depending on whether the semi-
circle buldges towards the resistance or away from it.
26.A coil is wound on a frame of rectangular cross-section. If
all the linear dimensions of the frame are increased by a
factor 2 and the number of turns per unit length of the coil
remains the same, self-inductance of the coil increases by
a factor of
(a) 4 (b) 8(c) 12 (d) 16
27.A horizontal telegraph wire 0.5 km long running
east and west in a part of a circuit whose resistance is 2.5
W. The wire falls to g = 10.0 m/s
2
and B = 2 × 10
–5
weber/
m
2
,
then the current induced in the circuit is
(a) 0.7 amp (b) 0.04 amp
(c) 0.02 amp (d) 0.01 amp
28.A conductor of length 0.4 m is moving with a speed of
7 m/s perpendicular to a magnetic field of intensity
0.9 Wb/m
2
. The induced e.m.f. across the conductor is
(a) 1.26 V(b) 2.52 V (c) 5.04 V (d) 25.2 V
29.The inductance between A and D is
(a) 3.66 H
(b) 9 H
(c) 0.66 H
DA 3 H 3 H 3 H
(d) 1 H
30.A square
metal loop of side 10 cm and resistance 1 W is
moved with a constant velocity partly inside a uniform
magnetic field of 2 Wbm
–2
, directed into the paper, as shown
in the figure. The loop is connected to a network of five
resistors each of value 3W. If a steady current of 1 mA
flows in the loop, then the speed of the loop is
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v
(a) 0.5 cms
–1
(b) 1 cm s
–1
(c) 2 cms
–1
(d) 4 cms
–1
31.Two identical circular loops of metal wire are lying on a
table without touching each other. Loop A carries a current
which increases with time. In response the loop B
(a) remains stationary
(b) is attracted by loop A
(c) is repelled by loop A
(d) rotates about is CM with CM fixed
32.A square loop of side a is rotating about its diagonal with
angular velocity w in a perpendicular magnetic field
B
r
. It
has 10 turns. The e
mf induced is
× × × ×B
a
(a) B a
2
sin wt (b) B a
2
cos wt
(c) 5 2 B a
2
(d) 10 B a
2
sin wt
33.In fig., final value of current in 10W resistor, when plug of
key K is inserted is
(a)
3
A
10
(b)
3
A
20
10 W
30 W
1H
3V K
(c)
3
A
11
(d) zero
34.In a cir
cuit given in figure 1 and 2 are ammeters. Just after
key K is pressed to complete the circuit, the reading is
(a) zero in both 1 and 2
(b) maximum in both 1 and 2
1
C R
1
R
2L
+–
2
K
(c) zero in 1 and maximum in 2
(d) maximum in 1 and zero in 2
35.A solenoid has 2000 turns wound over a length of
0.3 m. Its cross-sectional area is 1.2 × 10
–3
m
2
. Around its
central section a coil of 300 turns is wound. If an initial
current of 2 A flowing in the solenoid is reversed in 0.25 s,
the emf induced in the coil will be
(a) 2.4 × 10
–4
V (b) 2.4 × 10
–2
V
(c) 4.8 × 10
–4
V (d) 4.8 × 10
–2
V
36.Two coaxial solenoids are made by winding thin insulated
wire over a pipe of cross-sectional area A = 10 cm
2
and
length = 20 cm. If one of the solenoid has 300 turns and the
other 400 turns, their mutual inductance is
(m
0
= 4p × 10
–7
T m A
–1
)
(a) 2.4p × 10
–5
H (b) 4.8p × 10
–4
H
(c) 4.8p × 10
–5
H (d) 2.4p × 10
–4
H

584. 580 PHYSICS
37.A varyi
ng current in a coil change from 10A to zero in 0.5
sec. If the average e.m.f induced in the coil is 220V, the self-
inductance of the coil is
(a) 5 H (b) 6 H(c) 11 H(d) 12 H
38.In an inductor of self-inductance L = 2 mH, current changes
with time according to relation i = t
2
e
–t
. At what time emf is
zero?
(a) 4s (b) 3s(c) 2s (d) 1s
39.The magnetic flux through a circuit of resistance R changes
by an amount Df in a time Dt. Then the total quantity of
electric charge Q that passes any point in the circuit during
the time Dt is represented by
(a)
t
.RQ
D
fD
= (b)
t
.
R
1
Q
D
fD
=
(c)
R
Q
fD
= (d)
t
Q
D
fD
=
40.A conducting circ
ular loop is placed in a uniform magnetic
field, B = 0.025 T with its plane perpendicular to the loop.
The radius of the loop is made to shrink at a constant rate
of 1 mm s
–1
. The induced e.m.f. when the radius is 2 cm, is
(a)2
Vpm (b)Vpm
(c)
2
V
p
m (d)2Vm
41.The current i in a coil varies with time as shown in the
figure. The variation of induced emf with time would be
T/4T/23T/4T
t0
i
(a)
emf
0
T/4
T/23T/4T
t(b)
emf
0
T/4T/23T/4T
t
(c)
emf
0
T/4T/23T/4T
t(d)
emf
0
T/23T/4T
T/4
t
42.In a coil of resistance 10 W, t he
induced current developed by
changing magnetic flux through
it, is shown in figure as a
function of time. The
magnitude of change in flux
i(amp)
ts()
0.10
4
through the c
oil in weber is
(a)8 (b) 2 (c)6 (d) 4
43.A coil of resistance 400W is placed in a magnetic field. If
the magnetic flux f (wb) linked with the coil varies with time
t (sec) as f = 50t
2
+ 4. The current in the coil at t = 2 sec is
(a) 0.5 A(b) 0.1 A (c) 2 A(d) 1 A
44.A coil of self-inductance L is connected in series with abulb B and an AC source. Brightness of the bulb decreases
when
(a) number of turns in the coil is reduced
(b) a capacitance of reactance X
C
= X
L
is included in
the same circuit
(c) an iron rod is inserted in the coil
(d) frequency of the AC source is decreased
45.A magnetic field of 2 × 10
–2
T acts at right angles to a coil of
area 100 cm
2
, with 50 turns. The average e.m.f. induced in
the coil is 0.1 V, when it is removed from the field in t sec.
The value of t is
(a) 10 s (b) 0.1 s
(c) 0.01 s (d) 1 s
46.A rectangular coil of 20 turns and area of cross-section 25
sq. cm has a resistance of 100W. If a magnetic field which is
perpendicular to the plane of coil changes at a rate of 1000
tesla per second, the current in the coil is
(a) 1 A (b) 50 A
(c) 0.5 A (d) 5 A
DIRECTIONS for Qs. (47 to 50) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following.
(a)Statement -1 is false, Statement-2 is true
(b)Statement -2 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c)Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d)Statement -2 is true, Statement-2 is false
47. Statement 1 : An induced emf appears in any coil in which
the current is changing.
Statement 2 : Self induction phenomenon obeys Faraday's
law of induction.
48. Statement 1 : Lenz's law violates the principle of
conservation of energy.
Statement 2 : Induced emf always opposes the change in
magnetic flux responsible for its production.
49. Statement 1 : When number of turns in a coil is doubled,
coefficient of self-inductance of the coil becomes 4 times.
Statement 2 : This is because L µ N
2
.
50. Statement 1 : An induced current has a direction such that
the magnetic field due to the current opposes the change in
the magnetic flux that induces the current.
Statement 2 : Above statement is in accordance with
conservation of energy.

585. 581Electromagnetic Induction
Exemplar Questions
1.A square of side L metres lies in the xy-plane in a region,
where the magnetic field is given by B = B
0

ˆˆˆ
(2i + 3j + 4k) T ,,
where B
0
is co
nstant. The magnitude of flux passing through
the square is
(a) 2B
0
L
2
Wb (b) 3B
0
L
2
Wb
(c) 4B
0
L
2
Wb (d)
2
0
29B L Wb
2.A loop, made of straight edges has six corners at A (0, 0, 0),
B (L, 0, 0), C(L, L, 0), D (0, L, 0), E(0, L, L) and F (0, 0, L). A
magnetic field B = B
0()ˆˆ
ik+ T is present in the region. The
flux passing through the loop ABCDEFA (in that order) is
(a)B
0
L
2
Wb (b) 2B
0
L
2
Wb
(c)Ö2B
0
L
2
Wb (d) 4B
0
L
2
Wb
3.A cylindrical bar magnet is rotated about its axis. A wire is
connected from the axis and is made to touch the cylindrical
surface through a contact. Then,
(a) a direct current flows in the ammeter A
(b) no current flows through the ammeter A
(c) an alternating sinusoidal current flows through the
ammeter A with a time period T =
2p
w
(d) a time varying non-sinusoidal current flows through
the ammeter A.
4.There are two coils A and B as shown in figure a current
starts flowing in B as shown, when A is moved towards B
and stops when A stops moving. The current in A is counter
clockwise. B is kept stationary when A moves. We can infer
that
(a) there is a constant current in the clockwise direction
in A
(b) there is a varying current in A
(c) there is no current in A
(d) there is a constant current in the counter clockwise
direction in A
v
A B
5.Same as problem 4 except the coil A is made to rotate about
a vertical axis (figure). No current flows in B if A is at rest.
The current in coil A, when the current in B (at t = 0) is
counter-clockwise and the coil A is as shown at this instant,
t = 0, is
(a) constant current clockwise
(b) varying current clockwise
(c) varying current counter clockwise
(d) constant current counter clockwise
A
B
w
6.The self in
ductance L of a solenoid of length l and area of
cross-section A, with a fixed number of turns N increases
as
(a)l and A increase
(b)l decreases and A increases
(c)l increases and A decreases
(d) both l and A decrease
NEET/AIPMT (2013-2017) Questions
7.A wire loop is rotated in a magnetic field. The frequency of
change of direction of the induced e.m.f. is [2013]
(a) twice per revolution
(b) four times per revolution
(c) six times per revolution
(d) once per revolution
8.A current of 2.5 A flows through a coil of inductance 5 H.
The magnetic flux linked with the coil is [NEET Kar. 2013]
(a) 2 Wb (b) 0.5 Wb
(c) 12.5 Wb (d) Zero
9.A thin semicircular conducting ring (PQR) of radius ‘r’ is
falling with its plane vertical in a horizontal magnetic field
B, as shown in figure. The potential difference developed
across the ring when its speed is v, is : [2014]
Q
B
RP
r
(a) Zero
(b) Bv
pr
2
/2 and P is at higher potnetial
(c)prBv and R is at higher potnetial
(d) 2rBv and R is at higher potential

586. 582 PHYSICS
10.A conduc
ting square frame of side ‘a’ and a long staight
wire carrying current I are located in the same plane as
shown in the figure. The frame moves to the right with a
constant velocity ‘V’. The emf induced in the frame will be
proportional to [2015]
X
l
V
a
(a) 2
1
(2x – a)
(b) 2
1
(2x a)+
(c)
1
(2x – a)(2x + a)
(d)
2
1
x
11.An electron moves on a straight line path XY as shown.
The abcd is a coil adjacent to the path of electron. Whatwill be the direction of current if any, induced in the coil?
[2015 RS]
a
c
b d
X Yelectron
(a) adcb
(b)
The current will reverse its direction as the electron
goes past the coil
(c) No current induced
(d) abcd
12.A long solenoid of diameter 0.1 m has 2 × 10
4
turns per
meter. At the centre of the solenoid, a coil of 100 turns and
radius 0.01 m is placed with its axis coinciding with the
solenoid axis. The current in the solenoid reduces at a
constant rate to 0A from 4 A in 0.05 s. If the resistance of the
coil is 10p
2
W. the total charge flowing through the coil during
this time is :- [2017]
(a) 16 mC (b) 32 mC
(c) 16 p mC (d) 32 p mC

587. 583Electromagnetic Induction
EXERCISE - 1
1. (
a) 2. (c) 3. (c)
4. (a) Induced e.m.f.
f
e
d dBA dB
= = =A
0
dt dt dt
t/BA3
t
BB4
A
00
00
0
=÷
ø
ö
ç
è
æ -
=
5. (a) 6.
(c) 7. (c)
8. (a)
dt
dI
NME)NMI(
dt
d
E =Þ=
NMI
E=
t
Þ
emf induc
ed per unit turn =
t
MI
N
E
=
9. (d) Initially
flux, BA0cosBA
==f
After rotating through an angle 90°.
Flux through the coil is zero.
So, BA=fD
Angular speed = w, so, time period = T=
w
p2
4
T
is time taken to rotate 90°.
So,
Δ BA 2BAω
==
ΔtT/4π
f
10. (a) Farady's law states that time varying magnetic flux can
induce an e.m.f.
11. (b) From
2 2
0NA N
L
ll
m
=a
we get,
()
2
1
2
1/2L 1
L 1/22
==
12. (a) The back e.m.f. in a motor is induced e.m.f., which is
maximum, when speed of rotation of the coil is maximum.
13. (d) Mutual inductance between two coils depends on all the three
factors given here.
14. (b) From e = LdI / dt, dI / dt
1e1
1As
L1
-
= ==
15. (b) Mutua
l inductance =
I
BA
I
=
f
22
1
211
QML
]QT[
]LQMT[
]Henry[
-
-
--
==
16.
(d)
d d(NB.A)
e
dt dt
f
=- =-
ur ur
tsinNBA)tcosBA(
dt
d
N ww=w-=
w=ÞNBAe
max
17. (b) Due to shifting of electrons, one end of the rod becomes
positive and the other end negative. This developes a
electric field in the rod.
18. (b)
19. (a)
= Þ =V
W
Q
WQV
20. (b
) It is because after every 1/2 revolution the current
becomes zero and mode of change in flux changes
thereafter (If before the current becomes zero, the mode
of flux change was from left to right then after the current
becomes zero the mode of flux change becomes right to
left).
21. (d) Self inductance of a solenoid =
2
nAm
l
So, self induction µ n
2
So, inductance becomes 4 times when n is doubled.
22. (c)
11 d
q idt edt dt
R R dt
-fæö
=== ç÷
èøòòò

1
d
R
=fò
(taking only magnitude of e)
Hence, total charge induced in the conducting loopdepends upon the total change in magnetic flux.
23. (a)
24. (b)e µ w
25. (b)
EXERCISE - 2
1. (c)
2
. (d) º60cos5.00.2cosBA
´´=q=f
2.0 0.5
0.5 weber.
2
´
==
3. (b)
dt
dB
An
dt
d
=
f
=e
44
10 (10 10 ) (10 )
-
\e=´´
84
(10 Gauss/sec=10 T/s)
= 100 V.
Ιe= ( / R)= (100 / 20) = 5amp.
4. (d) Total resistance of the circuit = 4000 + 400 = 4400 W
Current flowing .amp1.0
4400
440
R
V
i ===
Voltage across load = R i = 4000 × 0.1 = 400 volt.
5. (b) ú
û
ù
ê
ë
é-
==e
05.0
)24(
M8or
dt
di
M
henry2.0
2
05.08
M =
´
=\
Hints & Solutions

588. 584 PHYSICS
6. (c) e.
m.f. induced
2211
BR BR (2 n)
22
=w=p
21
(0.1) (0.1)2 10
2
= ´ ´ ´ p´ = (0.1)
2
pvolts
7. (a) Ch
ange in flux = 2 B A N
\ Induced e.m.f.
1.0
10702003.02
4-
´´´´
=
8. (c) V05.0)1
.0/1()105(
3
=´=e
-
.
9. (c) Energy stored U is given by .J05.0)1()10100(
2
1
iL
2
1
U
232
=´´==
-
10. (d)
11. (d) 1t
7t6
2
++=f
d
12t7
dt
f
Þ=+
At time, t = 2 sec.
d
24 7 31 volt
dt
f
= +=
Direction of
current is from left to right according to
Flemmings right hand rule.
12. (b)
( )
d d dB
e NBA NA
dt dt dt
f
=== = 500×10
–2
× 1 = 5
.0 V
13. (a) Vel. of coil
1
2m/s
0.5
==
velocity of
magnet
2
2m / s.
1
==
As they are made to mov
e in the same direction, their
relative velocity is zero. Therefore, induced e.m.f. = 0.
14. (a)
()
2dd
e 6t 5t 1 12t 5
dt dt
-f-
= = -+=-+
e = – 12 (0.25) + 5 = 2 volt
e2
i 0.2A.
R 10
===
15. (a)
( )
3
3
40 10 11 1LdI
e 100V
dt 4 10
-
-
´-
= ==
´
16. (d) For maximum po
wer,
CL
XX=, which yields
1050504
1
L)n2(
1
C
22
´´´p
=
p
=
\ C = F1F101.0
5
m =´
-
17. (b)
2
10t 50t 250f= -+
)50t20(
dt
d
e --=
f
-=
V10e
3t
-=
=
18. (d) Sinc
e the magnetic field is uniform the flux f through the
square loop at any time t is constant, because
f = B × A = B × L
2
= constant
zero
dt
d
=
f
-=e\
19. (d) ε=Blv mV1V10)18/5180()1()102.0(
34
==´´=
--
20. (b) inB
0
m= 5.
1)10200()104(
27
´´´p=
--
= 3.8 × 10
–2
Wb / m
2
Magnetic flux through each turn of the coil
f = BA = (3.8 × 10
–2
) (3.14 × 10
–4
) = 1. 2 × 10
–5
weber
When the current in the solenoid is reversed, the
change in magnetic flux
weber104.2)102.1(2
55 --
´=´´=
Induced e.m.f. .V048.0
05.0
104.2
100
dt
d
N
5
=
´
´=
f
=
-
21. (b)
M
ε= dI=0.005×I cosωt×ω
0
dt
and e
max
= 0.005 × I
0

×
w = 5p
22. (b) l = 1m, w = 5 rad/s, T102.0B
4-
´=
4
Bω 0.2 10 5 1
50V
22
-
´ ´´
e= = =m
l
23. (d) When two in
ductance coil are joined in series, such that the
winding of one is exactly opposite to each other the emf
produced in the two coils are out of phase such that they
cancel out.
24. (b) The individual emf produced in the coil dt
d
e
f-
=
\ The curren
t induced will be
dt
d
R
1
i
R
|e|
i
f
=Þ=
But
dt
dq
i=
dq 1 d 1 BA
dq dq
dt R dt R R
f
Þ = Þ = fÞ=òò
25.
(c) E.m.f. will remain same because change in area per unit
time will be same in both cases.
26. (b) Self inductance = L)b(nALn
2
0
2
0
´´m=m
l
n = Total number of turns/length
L = Length of inductor
l = Length of rectangular cross section
b = breadth of rectangular cross-section
So, when all linear dimensions are increased by a factor
of 2. The new self inductance becomes L' = 8L.
27. (c)
1d
i
R R dt
ef
==
Here df = B × A
53
(2 10 ) (0.5 10 5)
-+
=´ ´´´
dt = time taken by the wire to fall at ground

589. 585Electromagnetic Induction
1/2 1/2
(2h / g)(10/10) 1sec.===
53
1 (2 10 ) (0.5 10 5)
i
2.51
-éù´ ´ ´´
\= êú
êúëû
= 0.02 amp.
28
. (b) Length of conductor (l) = 0.4 m; Speed (v) = 7 m/s and
magnetic field (B) = 0.9 Wb/ m
2
. Induced e.m.f.
(e) = Blv cos q = 0.9 × 0.4 × 7 × cos 0º = 2.52 V.
29. (d) The given circuit clearly shows that the inductors are in
parallel we have,
3
1
3
1
3
1
L
1
++=or L= 1H
30. (c
)
1
Bv210 v0. 2v
-
e= =´ ´=l
330.2v
I 10 10
R4
--e
==Þ=
[Sin
ce effective resistance R of bridge is
66
R3
66
´
= =W
+
so total resis
tance = 1 + 3 = 4W]
Þ v = 2 cm s
–1
31. (c) An opposite current induced in B in accordance to Lenz's
law. So the two loops repel each other.
32. (d)f = n BA cos q = 10 B a
2
cos wt
( ) ()
22dd
e 10Ba co s t 10Ba sin t .
dt dt
f
=- =- w = w w
33. (d)As resistance of 1 H coil is zero, the entire current flows
through the coil. Current through 10W resistance is zero.
34. (c) Capacitor is a dc blocking element and hence no current
flow in (1).
An inductor offers a zero resistance path to flow of dc
and hence maximum current flows through (2).
35. (b)
N 2000 20000
n
0.33
===
l
( )
d dB
NBA NA
dt dt
x==
Since
B = µ
0
nI
( )
dt
NAn
dt
Þ x=m 0.024VÞ x=
36. (d)
74
012
NNA 4 10 300 400 100 10
M
0.2
--
m p´ ´´´´
==
l
4
2.4 10H
-
= p´
37. (c
) Initial current (I
1
) = 10 A; Final current (I
2
)= 0; Time (t)
= 0.5 sec
and induced e.m.f. (e) = 220 V.
L20
5.0
)100(
L
t
)II(
L
dt
dI
L
12
=
-
-=
-
-=-
or
220
L= =11H
20
(where L = Sel
f inductance of coil)
38. (c) L = 2mH, i = t
2
e
–t
E =
]te2et[L
dt di
L
tt2--
+--=-
when E = 0
–e
–t
t
2
+ 2te
–t
=
0
2t e
–t
= e
–t
t
2
t = 2 sec.
39. (c)
R)ti(iR
t
D=fDÞ=e=
D
fD
= QR
R
Q
fD
=Þ
40. (b)
Magnetic flux linked with the loop is
2
Brf=p
||
d
e
dt
f
= = 2
dr
Br
dt
p×
When r = 2 cm,
dr
dt
= 1 mm s
–1
e = 0.025× p ×2 ×
2 ×10
–2
×10
–3
= 0.100 × p × 10
–5
= p × 10
–6
V = pmV
41. (a)
di
eL
dt
=-
During 0 to
T di
,
4 dt
= const.
\ e = – ve
D
uring
T T di
to,0
4 2 dt
=
\ e = 0
During
T 3T di
to,
2 4 dt
= const.
\ e = +v
e
Thus graph given in option (a) represents the variation
of induced emf with time.
42. (b) The charge through the coil = area of current-time
(i – t) graph
1
0.14
2
=´´q = 0.2 C
q
R
Df
= Q Change in flux (Df) = q × R
0.2
10
Df
==q
Df = 2 weber
4
3. (a) According, to Faraday’s law of induction
Induced e.m.f.
(100 )
d
t
dt
f
e=- =-
Induce
d current i at t = 2 sec.

590. 586 PHYSICS
=
100 2
0.5Amp
400
e´
=+ =+
R
44. (c) By
inserting iron rod in the coil,
L ­ z ­ I ¯ so brightness ¯
45. (b)
21() (0)NBA NBA
e
t tt
- f -f --
===
–2 –2
50 2 10 10
0.1
0.1
NBA
ts
e
´´´
= ==
46. (c)
nAdB
e dt
i
RR
==
A5.0
100
1000)1025(20
4
=
´´´
=
-
47. (b)
48. (a) Le
nz's law (that the direction of induced emf is always
such as to oppose the change that cause it) is direct
consequence of the law of conservation of energy.
49. (b) 50. (b)
EXERCISE - 3
Exemplar Questions
1. (c) As we know that, the magnetic flux linked with uniform
surface of area A in uniform magnetic field is
f = B.A
The direction of A is perpendicular to the plane of
square and square line in x-y plane in a region.
A = L
2
k
As given that, B =
( )
0
ˆˆˆ
B 2i 3j 4k++
So, f = B.A = ( )
2
0
ˆˆˆˆ
B 2i 3j 4k .L k++
=
2
0
4B L Wb
2. (b) The loop can be considered in two planes, Plane of
ABCDA lies x-y plane whose area vector A
1
= |A|
ˆ
k,
A
1
= L
2ˆ
k
whereas plane of ADEFA lies in y-z plane whose area
vector A
2
= |A|
ˆ
i, AA
2
= L
2ˆ
i.
Then the magnetic flux
linked with uniform surface of
area A in uniform magnetic field is
Z
X
A
F
L
L
L
B
C
D
E
Y
(0,0,0)
(0,0,L)
(L,0,0)
(L,L,0)
(0,L,0)
(0,L,L)
f = B.A
A = A
1
+ A
2
= ( )
22ˆˆ
Lk Li+
and B = ()
0
ˆˆ
Bik+
Now, f = B.A =()( )
22
0
ˆˆ ˆˆ
B i k Lk Li+×+
= 2 B
0
L
2
Wb
3. (b) Induc
ed current flow only when circuit is complete
and there is a variation about circuit this problem is
associated with the phenomenon of electromagnetic
induction.
If there is a symmetry in magnetic field of cylindrical
bar magnet is rotated about its axis, no change in flux
linked with the circuit takes place, consequently no
emf induces and hence, no current flows in the ammeter
(A).
S
N
w
Axis
A
Bar
magnet
w
4. (d) When the coil A stops moving the current in B b ecome
ze ro, it possible only if the current in A is constant. If
the current in A would be variable, there must be an
induced emf (current) in B even if the A stops moving.
So there is a constant current in same direction or
counter clockwise direction in A as in B by lenz's law.
5. (a) By Lenz's law, at (t = 0) the current in B is counter-
clockwise and the coil A is considered above to it. The
counterclockwise flow of the current in B is equivalent
to north pole of magnet and magnetic field lines are
emanating upward to coil A.
When coil A start rotating at t = 0, the current in A is
constant along clockwise direction by Lenz’s rule. As
flux changes across coil A by rotating it near the N-
pole formed by flowing current in B, in anticlockwise.
6. (b) The self-inductance of a long solenoid of cross-
sectional area A and length l, having n turns per unit
length, filled the inside of the solenoid with a material
of relative permeability is given by
L = µ
r
µ
0
n
2
Al
\n = N/l
L = µ
r
µ
0
2
N .A
.
éù
êú
ëû
l
l.l
L = µ
r
µ
0
[N
2
A/l]
1
L A,L
æö
µµç÷
èø l
As µ
r
and N are constant here so, to increase L for a
coil, area A must be increased and l must be decreased.

591. 587Electromagnetic Induction
NEET/AIPMT (2013-2017) Questions
7. (a) This is the case of periodic EMI
E
t
From graph, it is clear that direction is changing
once in
1
2
cycle.
8. (c) Given: current I = 2.5 A
Inductance, L = 5H
Magnatic flux, f = ?
We know, f = LI Þ 5 × 2.5 Wb = 12.5 Wb
9. (d) Rate of decreasing of area of semicircular ring
=
dA
(2r)V
dt
=
From Faraday’s law of electromagnetic induction
e =
d dA
B B(2rV)
dt dt
q
- =- =-
As induced current in ring produces magnetic field in
upward direction hence R is at higher potential.
10. (c) Emf induced in side 1 of frame e
1
= B
1
Vl
o
1
I
B
2 (x – a/ 2)
m
=
p
Emf induced in side 2 of frame e
2
= B
2
Vl
o
2
I
B
2 (x a/ 2)
m
=
p+
a
x
v
I
1 2
a
x–
2
a
x
2
+
Emf induced in square frame
e = B
1
Vl – B
2
Vl
=
00
II
v–v
2 (x – a / 2) 2 (x a/ 2)
mm
p p+
ll
or, e µ
1
(2x – a)(2x a)+
11. (b) Current will be induced,
when e
–
comes closer the induced current will be
anticlockwise
when e
–
comes farther induced current will be
clockwise
e
–
e
–e
–
e
–
12. (b) Given, no. of turns N = 100
radius, r = 0.01 m
resistance, R = 10p
2
W, n = 2 × 10
4
As we know,
e =
d
N
dt
f
-
Nd
R R dt
ef
=-
Nd
I
R dt
f
D =-
qN
t Rt
D Df
=-
DD
N
qt
Rt
é Dfùæö
D=-D
ç÷êú
Dèøëû
'–' ve sign shows that induced emf opposes the
change of flux.
2
2 0
0
nNrii1
qnNrt
tRR
m pDé Dù æö
D = m p D=ç÷êú
èøDëû
7 24
2
4 10 100 4 (0.01) 2 10
q
10
-
p´ ´ ´´p´ ´´
D=
p
Dq = 32m C

592. 588 PHYSICS
ALTERNATING AN
D DIRECT CURRENT
An alternating current (A.C.) is one which periodically changes
in magnitude and direction. It increases from zero to a maximum
value, then decreases to zero and reverses in direction, increases
to a maximum in this direction and then decreases to zero.
The source of alternating emf may be a dynamo or an electronic
oscillator.
The alternating emf E at any instant may be expressed as
E = E
0
sinwt
where w is the angular frequency of alternating emf and E
0
is the
peak value of emf.
II
0
T = 0T
—
2
T t3T
–—
2
(A.C.)
Direct c
urrent (D.C.) is that current which may or may not
change in magnitude but it does not change its direction.
I
t
(D.C.)
Advantages of A.C. over D.C.
(i) The generation of A.C. is cheaper than that of D.C.(ii) Alternating voltage can be easily stepped up or stepped
down by using a transformer.
(iii) A.C. can be easily converted into D.C. by rectifier. D.C. is
converted to A.C. by an inverter.
(iv) A.C. can be transmitted to a long distance without
appreciable loss.
AVERAGE AND RMS VALUE OF ALTERNATING
CURRENT
The average value of AC over one full cycle is zero since there are
equal positive and negative half cycles.
The average current for half cycle is 2I
0
/p where I
0
is the peak
value of current.
The root mean square (rms) value of AC is
0
rms
2
=
I
I
where I
0
is the peak o
r maximum value of alternating current.
The rms value of alternating current can also be defined as the
direct current which produces the same heating effect in a given
resistor in a given time as is produced by the given A.C. flowing
through same resistor for the same time. Due to this reason the
rms value of current is also known as effective or virtual value of
current.
\
0
effective virtual rms
I
I II
2
= ==
Similarly the r
ms value of alternating voltgae is called the effective
or virtual value of alternating voltage (or emf).
\
0
effective virtual rms
E
E EE
2
= ==
Keep in Memory
(1)Time perio
d : The time taken by A.C. to go through one
cycle of changes is called its period. It is given as
w
p
=
2
T
(2)Phase : It is that property o
f wave motion which tells us
the position of the particle at any instant as well as its
direction of motion. It is measured either by the angle which
the particle makes with the mean position or by fraction of
time period.
(3)Phase angle : Angle associated with the wave motion (sine
or cosine) is called phase angle.
(4)Lead : Out of the current and emf the one having greater
phase angle will lead the other e.g., in equation
i = i
0
sin
÷
ø
ö
ç
è
æ p
+w
2
t and e = e
0
sin wt,
t
he current leads the emf by an angle
2
p
.
(5)Lag : Out of current
and emf the one having smaller phase
angle will lag the other. In the above equations, the emflags the current by
2
p
.
22
Alternating Current

593. 589Alternating Current
RESISTANCE OFFERED BY VARIOUS ELEMENTS
(INDUCTOR, RESISTOR AND CAPACITOR) TO A.C.
Alternating current in a circuit may be controlled by resistance,
inductance and capacitance, while the direct current is controlled
only by resistance.
(i)Impedance (Z) : In alternating current circuit, the ratio of
emf applied and consequent current produced is called
the impedance and is denoted by Z,
i.e.,
0 rms
0 rms
===
EEE
Z
III
Phys
ically impedance of ac circuit is the hindrance offered
by resistance along with either inductance or capacitance orboth in the circuit to the flow of ac through it. Its unit is ohm.
(ii)Reactance (X) : The hindrance offered by inductance or
capacitance or both to the flow of ac in an ac circuit iscalled reactance and is denoted by X. Thus when there is
no ohmic resitance in the cirucit, the reactance is equal toimpedance. The reactance due to inductance alone is calledinductive reactance and is denoted by X
L
, while the
reactance due to capacitance alone is called the capacitivereactance and is denoted by X
C
. Its unit is also ohm.
(iii)Admittance (Y) : The inverse of impedance is called the
admittance and is denoted by Y, i.e.,
1
=Y
Z
Its SI unit is ohm
–1
.
IMPEDANCES AND PHASES OF AC CIRCUITCONTAINING DIFFERENT ELEMENTS
As already pointed out that in an ac circuit the current and applied
emfs are not necessarily in same phase. The applied emf (E) and
current produced (I) may be expresed as
E = E
0
sin wt and I = I
0
sin (wt + f) with I
0
= E
0
/ Z
where E
0
and I
0
are peak values of alternating emf and current.
Circuit Containing only Resistor (R)
Consider a pure ohmic resistor (zero inductance) of resistance R
connected to an alternating source of emf E

= E
0
sinwt.
E = E sint
0
w
R
Then current I in the circuit is
0
0
EsintE
I Isint
RR
w
== =w , where I
0
= E
0
/R
Comparing this with standard equation, we get that
impedance of circuit, Z = R and phase difference between current
& emf = 0.
Hence we conclude that in a purely resistive ac circuit the current
and voltage are in same phase and impedance of circuit is equal
to the ohmic resistance.
Phasor diagram :
EI
X
Graph of emf or current versus wt :
I
wt
E
or
I
E
Circuit Contian
ing only Inductor (L)
Consider a pure inductor (zero ohmic resistance) of inductance L
connected to an alternating source of emf E = E
0
sin wt.
E = E sin t
0
w
L
Then current I in the circuit is
0
sin
2
pæö
= w-ç÷
èø
IIt wher e 0
0=
w
E
I
L
Comparing this with standard equation, we get
Z = w L and phase difference f = p/2.
Hence we conclude that in a purely inductive circuit the current
lags behind the applied voltage by an angle
p/2 and the
impedance to the circuit is
wL and this is called as inductive
reactance.
Graph of emf or current versus wt
IE
or
I
E
wt
Phasor diagram Gra ph between X
L
and f

X
EO
Y
I
90º

X
L
f
Circuit Containing only Capacitor
Consider a capacitor of capacitance C connected to an alternating
source of emf, E = E
0
sin wt.
E = E sin t
0w
C
V
C
Then the current through capacitor is given by,
0
sin
2
pæö
= w+ç÷
èø
IIt

594. 590 PHYSICS
Comparing this
with standard equation, we find that capacitive
reactance X
C
= 1/wC and phase difference f = + p/2
Phasor diagram Graph between X
C
and f
V
C
I
C
p/2

X
C
f
Hence we conclude that in a purely capacitive circuit the current
leads the applied emf by an angle
p/2 and the impedance of the
circuit is 1/
wC and this is known as capacitive reactance
1
==
w
C
ZX
C
.
Graph of emf or cu
rrent versus wt
I
wt
E
or
I
E
3p
2pp/2
Circuit Containing Resistance and Inductance in
Series (LR Series Circuit)
Consider a circuit containing resistance R and inductance L in
series having an alternating emf E = E
0
sin wt.
E = E sin t
0
w
R L
V
R V
L
Let I be the current flowing in the circuit and V
R
(= IR) the potential
difference across resistance and V
L
(= wL.I) the potential
difference across inductance.
The current I and the potential difference V
R
are always in phase
but the potential difference V
L
across inductance leads the current
I by an anlgle p/2.
Phasor diagram

V
L
I
E
V
R
L
2
R
2
VVE+=
f
From phasor diagrom, resultant voltage is given by,

22 22
E (V V ) (R) ( L.)
RL
= + = +w II
Graph of emf or current versus wt
E
or
I
emf
Current
wt
\
22
()éù= +w
ëû
E
RL
I
\Impedance of R – L circuit,
22
()==+
L
E
Z RX
I
where X
L
= wL
It is ob
vious that the current lags behind the emf by angle f
given by,
f
æö
=
èø
L
R
V
V
–1
= tan ()
wæö
ç÷
èø
L
X L
R R
–1 –1
tan = tan
Circuit Con
taining Resistance and Capacitance in
Series (C–R Series Circuit)
Consider a circuit containing resistance R and capacitance C in
series having an alternating emf E = E
0
sin wt.
E = E sin t
0
w
R C
V
R
V
C
Let I be the current flowing in the circuit, V
R
the potential
difference across resistance and V
C
the potential difference across
capacitance.
Phasor diagram
f
V
R
V
C
E
I
From phasor diagram the resultant emf is given by
])X()R()VV(E
2
C
22
C
2
R
II+=+=
\Impedance, )XR(/EZ
2
C
2
+==I , where
÷
ø
ö
ç
è
æ
w
=
C
1
X
C
The potential
difference V
R
and current I are in same phase and
the potential difference V
C
lags behind the current I (and hence
V
R
) by angle p/2
The current leads the applied emf by an angle f given by
tanj===
I
I
CCC
R
VXX
VRR
or
111/1
tan tan tan
--wæö æ ö æö
f= Þf= =
ç ÷ ç÷ç÷
è ø èøèø w
C
X C
R R CR
Graph of emf or current versus wt
E
or
I
emf
Current
wt

595. 591Alternating Current
Circuit Containing Inductance and Capacitance in
Series (Series LC Circuit)
Consider a circuit containing inductance L and capacitance C in
series having an alternating emf
E = E
0
sin wt.
E = E sin t
0
w
L C
V
C
V
L
Let I be the current flowing in circuit, V
L
the potential difference
across inductance L and V
C
the p.d. across capacitance C.
Phasor diagram :
Y
V > V
LC
V – V
LC
O
X

Y
'
V > V
CL
V – V
CL
O X
E

V
C
E=V
C
–V
L
V
L
I
p/2
p/2
The p.d. V
C
lags beh
ind the current by angle p/2 and the p.d. V
L
leads the current by angle p/2.
\Resultant applied emf, E = V
C
– V
L
= X
C
I – X
L
I
\Reactance of circuit,
1
/
æö
= = - = -w ç÷
èøw
I
CL
XEXXL
C
The curr
ent leads applied emf by 2/p=f.
In case of X
C
= X
L
, Z = 0, then
11
or
()
=w w=
w
L
C LC
\Frequency
1
/2
2()
f
LC
=w p ==
p
At c
ertain frequency the impedance of the circuit is minimum and
the current is maximum.
This frequency is called the resonant frequency.
Circuit Containing Resistance, Inductance and
Capacitance in Series (Series LCR Circuit)
Consider a circuit containing a resistance R, inductance L and
capacitance C in series having an alternating emf
E = E
0
sin wt.
E = E sin t
0
w
R L
V
R
V
L
V
C
C
Let I be the current flowing in circuit. V
R
, V
L
and V
C
are respective
potential differences across resistance R, inductance L and
capacitance C.
Phasor diagram :
V
L
V – V
CL
(if V > V
)
CL
V
C
E
V
R
f
I
The p.d. V
R
is in phase with current I . The p.d. V
C
lags behind
the current by angle p/2. The p.d. V
L
leads the current by angle p/
2.
\Resultant applied emf, ])VV(V[E
2
LC
2
R
-+=
i.e.,{ }
22
CLE (R) (X X)= +-III
\Impedan
ce,
{ }
22
()== +-
I
CL
E
Z RXX
T
he phase leads of current over applied emf is given by
CL C L CL
R
1 CL
VV IXIX XX
tan
V RIR
XX
i.e
., tan
R
-
---
j===
-æö
j=
ç÷
èø
It is concluded that :
(a) If X
C
> X
L
, the value of f is positive, i.e., current leads the
applied emf.
(b) If X
C
< X
L
, the value of f is negative, i.e., current lags
behind the applied emf.
(c) If X
C
= X
L
, the value of f is zero, i.e., current and emf are in
same phase. This is called the case of resonance and
resonant frequency for condition X
C
= X
L
, is given by :
Z
R
f
o
f
Current
E/R
O f
o
Frequency
E
2R
L
C
1
w=
w
i.e.,
LC
1
=w
\
1
/2
2()
=w p=
p
o
f
LC
.
Thus th
e resonant frequency depends on the product of L
and C and is independent of R.
At resonance, impedance is minimum, Z
min
= R and current
is maximum
R
E
Z
E
min
max ==I

596. 592 PHYSICS
w
Z = RminR
Z
Resonance frequency
Circuit impedence in series RLC circuit
wwp = =2f
00
w
A
B
I max
I
Resonance frequency
Small R higher
Q i.e., sharper resonance
High R, small Q i.e.,
no sharp resonance
Rapid fall of current
in A in comparison
to B c
urve
Current amplitude in series RLC circuit
wwp = =2f
00
It is interesting to note that before resonance the current
leads the applied emf, at resonance it is in phase, and after
resonance it lags behind the emf. LCR series circuit is also
called as acceptor circuit and parallel LCR circuit is called
rejector circuit.
COMMON DEFAULT
O
Incorrect. Adding impedances / reactances /resistors
algebrically.
P
Correct. For these physical quantities, vector additon must
be done
O
Incorrect. Kirchoff's laws are applicable in D.C. circuit only
P
Correct. Kirchoff's laws are applicable in A.C. circuit also
(which may include inductor and capacitor).
PARALLEL RESONANT CIRCUIT
A parallel resonant circuit consists of an inductance L and a
capacitance C in parallel as shown in fig.
E=E sin t
0w
R L
C
The condition of resonance is again that the current and applied
emf must be in same phase. The condition gives angular resonantfrequency.
2
r 2
1R
LCL
w=-
\ Resonant frequ
ency
2
r
r 2
1 1R
f
2 2 LC L
w
==-
pp
The impedan
ce at resonance,
2 22
R LL
Z
R RC
+w
==
In parallel resonan
t circuit the impedance is maximum and the
current is minimum.
If 0R
®, then
)LC(2
1
f
r
p
= and ¥®Z .
Q - FACTOR
The sharpn
ess of tuning at resonance is measured by
Q-factor or quality factor of the circuit and is given by

1
=
L
Q
RC
Higher the value of Q-factor, sharper is the resonance i.e. more
rapid is the fall of current from maximum value (I
0
) with slight
change in frequency from the resonance value.
It is clear from the figure that at low value of q, the resonance is
poor. However the bandwidth increases
IHigh Q
Low Q
w
0 w
The figure given below explains the concept of bandwidth and
cut-off frequency.
w
0
w
Upper cut -off frequency
0.707 I
max
I
max
Lower cut off
fre
quency
0.707 I
max
Band width
LC0
CL0
LC0
XX when
XX when
XX
when
=w=w
>w>w
<w<w
Bandwidth : It is the band of allowed frequencies and is defined
as the difference between upper and lower cut-off frequencies,
the frequency at which power becomes half of maximum value
and current becomes I
max
/
2.
POWER IN AN A.C. CIRCUIT
The power is defined as the rate at whic
h work is being done in
the circuit. In ac circuit, the current and emf are not necessarily
in the same phase, therefore we write
E = E
0
sin wt & I = I
0
sin (wt + f).
The instaneous power, P = EI
= E
0
sin wt I
0
sin (wt + f),
The average power P
av
= E
rms
I
rms
cos f
\
00
cos
22
=f
av
EI
P

597. 593Alternating Current
In this expression cos f is known as power factor. The value of
cos f depends on the nature of the circuit. For L, C and
L-C circuit, the power factor is zero ( Q f = 90º); for R-circuit
cos f = 1 (Q f = 0) and for all other circuit cos f = R/Z, where
Z = impedance.
If R = 0, cos f = 0 and P
av
= 0 i.e., in a circuit with no resistance, the
power loss is zero. Such a circuit is called the wattless circuit
and the current flowing is called the wattless current.
Power is of two types
reactiverms rms
active rms rms
(i) Reactive power P V I sin
This is also called wattless power.
It is not read by energy meter
(ii) Active power P V I cos
It is read by energy meter
=f
=f
Half Power Points
The values of w at which the average power is half of its maximum
value (at resonant frequency) are called half power points.
w
P
av
w=
0r
w
Small R
higher Q
Large R,
low Q
w
2
w
1
Dw
Plot of aver
age power versus frequency for a series RLC circuit.
The upper curve is for a small R & lower broad curve is for large
value of R.
It is clear from the figure that for smaller R, value of Q
0
is high (Q
0
is Quality factor of circuit) & hence sharper resonance i.e. greater
rate of fall of average power maximum average power P
av
changes
with slight change in frequency from resonant frequency.
The Quality factor, Q
0
is defined as,
0
0
Q
w
=
Dw
Where Dw = w
2
– w
1
and w
2
& w
1
are half power points.
Now, since
L
R
»wD; so
R
L
Q
0
0
w
»
Whereas w
1
= w
0
o
0
Q
w
-; w
2
= w
0

0
0
Q
w
+
In concise
term, we can write as,
÷
÷
ø
ö
ç
ç
è
æ
±w=w
0
r
Q
1
1
Keep in Memory
1.
Unless mentioned otherwise, all a.c. currents and voltages
are r.m.s. values.
2. For resonance to occur, the presence of both L and C
elements in the circuit is a must.
3. In series resonant circuit, current is maximum at resonance.
In a parallel resonant circuit, current is minimum (or zero)
at resonance but p.d across the combination is maximum.
4. To depict oscillatory motion mathematically we may use
sines, cosines or their linear combination. This is because
changing the zero position transforms one into another.
5. While adding voltage across different elements in an a.c.
circuit we should take care of their phases.
6. The average current over a complete cycle in an a.c circuit
is zero but the average power is not zero.
7. An inductor offers negligibly low resistance path to d.c.
and a resistive path for a.c.
8. A capacitor acts as a block for d.c and a low resistance path
to a.c.
9.
L
C
L
C
L
X L 2 fL 11
X
C 2
fCXf
1
X
f
Current through pure Current through pure
inductor lags behind capacitor leads
emf by 90 emf by 90
For d.c f 0 X 0 For d.c f
=w=p
==
wpÞµ
Þµ
°°
=\=
Inductive reactance Capacitive reactance
C
LC
0X
For a.c as
f increases For a.c as f increases
X increases X decreases
= Þ =¥
10.
LC
LC
2
rr 2
2
22
LC 2
11
XX
XX
1 1 1R
2 LC2LCL
111
ZR(XX)C
ZLR
==
n
= n=-
pp
æö
= + - = + w-ç÷
èø w
Series Reson
ant circuit Parallel resosant
circuit
11. The principle of electric meter is heating effect of current.
These meters give the reading of I
rms
. It is important to note
that these meters can measure D.C. as well as A.C.
12. D.C. flows through the cross-section of the conductor
whereas A.C. flows mainly along the surface of the
conductor. This is also known as Skin Effect. The skin
effect is directly proportional to the frequency.

598. 594 PHYSICS
Example 1.
Calc
ulate the r.m.s. value of e.m.f. given by
E = 8 sin
w t + 6 sin 2 w t volts.
Solution :
The mean square value is given by
2
EE=
\
22
)t2sin6tsin8(
E w+w=
=
22
64sin t 96sin t.sin 2 t 36sin 2 tw+ w w+ w
We know that ,
2
1
tsin
2
=w ,
2
1
t2sin
2
=w and
0t2sin.tsin=ww
\
2
1
36096
2
1
64E
2
´+´+´= = 32 + 18 = 50
or
2
r.m.s.E (E ) 50 7.07 vol t= ==
Example 2.
When 100 vol
t D.C. is applied across a solenoid a current
of 1.0 amp flows in it. When 100 volt A.C. is applied across
the same coil, the current drops to 0.5 amp. If the frequency
of the A.C. source is 50 Hz, then determine the impedance
and inductance of the solenoid.
Solution :
In case of D.C., w = 0 and hence Z = R
\
E 100
ZR
1
====
I
100Ω
For A.C.,
1/2
22
Z R (2 n L)éù= +p
ëû
=
2/122
])L100( )100[( p+
(wherew = 2pn & n is frequency of AC source)
\
2/122
])L100()100[(200 p+=
þ
ý
ü
î
í
ì
W==200
5.0
100
ZQ
Solving we ge
t L = 0.55 henry
Example 3.
A coil has an inductance of 0.7 henry and is joined in
series with a resistance of 220
W. When an alternating
e.m.f. of 220 V at 50 cycles per second, is applied to it, then
what will be the wattless component of current in the
circuit?
Solution :
Here, X
L
= wL = 2 p n L = 2 p × 50 × 0.7 × = 220 W
R = 220 W
.ohm2220220220XRZ
222
L
2
=+=+=
\ wattless compon
ent of current is
O
E 220 1
0.707A
Z220 2 2
== ==I
Example
4.
A 60 volt-10 watt bulb is operated at 100 volt-60 Hz a.c.
The inductance required is
(a) 2.56 H (b) 0.32 H
(c) 0.64 H (d) 1.28 H
Solution : (d)
A
6
1
60
10
V
P
===I ; W=
´
== 360
10
6060
P
V
R
2
;
W===600
6/1
100V
Z
Ι
)360600()360600(360600RZX
22222
L
-+=-=-=
W=´=´=4802240240960X
L
480XnL2L
L
==p=w
L
X 480
L 1.28H
2 n 120
===
pp
Example
5.
An a.c. circuit consists of only an inductor of inductance
2H. If the current is represented by a sine wave of ampli-
tude 0.25 amp. and frequency 60 Hz, calculate the effective
potential difference across the inductor.
Solution :
The effective potential difference across the inductor is
given by
V
eff
= I
eff
. X
L
=
0
I
2
.2 p f L; VV
eff
= V
rms
Give
n thatI
0
= 0.25 amp, f = 60 Hz, L = 2H
\V
eff
=
0.25
2
× 2 × 3.14 × 60 × 2 = 133.2 Volt
Exa
mple 6.
If a domestic appliance draws 2.5 A from a 220-V, 60- Hz
A.C. power supply, then find
(a) the average current
(b) the average of the square of the current
(c) the current amplitude
(d) the supply voltage amplitude.
Solution :
(a) The average of sinusoidal AC values over any whole
number of cycles is zero.
(b) RMS value of current = I
rms
= 2.5 A
\ 2 22
av rms
(I ) (I ) 6.25 A==
(c)
m
rms
I
I
2
=
\ Current amplitu
de
rms
2I 2(2.5A)== = 3.5 AA
(d)
m
rms
V
V 220V
2
==
\ Supply voltage ampli
tude
2( ) 2(220 )
m rms
VVV== = 3 1 1 V..
Exam
ple 7.
A 100
mF capacitor in series with a 40 W resistance is
connected to a 110 V, 60 Hz supply. (a) What is the maximum current in the circuit?
(b) What is the time lag between current maximum and
voltage maximum?

599. 595Alternating Current
Solution :
(a) Here, C = 100 mF = 100 × 10
–6
F, R = 40 W,
V
rms
= 110 V, f = 60 Hz
Peak voltage,
0 rms
V 2 . V 10 0 2 155.54 V= ==
Circuit
impedance, Z =
2
22
1
R
C
+
w
=
2
62
1
40
(2 60 100 10 )
-
+
´p´´´
= 1600 703.60+ = 2303.60 = 48 W
Henc
e, maximum current in coil,
0
0
V155.54
I 3.24 A
Z 48
===
(b) Ph
ase lead angle (for current),11
6
11
tan tan
CR 2 3.14 60 10010 40
--
-
q==
w ´´´´´
= tan
–1
0.66315 = 33° 33’ (taken 33.5°)
Time lead,
33.5
t 0.001551
2 360 60
qq
====
wpn´
sec
= 1.551 × 10
–3
sec
Voltage will lag current by = 1.551 ms.
Example 8.
30.0 ?F capacitor is connected to a 220 V, 50 Hz source.
Find the capacitive reactance and the current (rms and
peak) in the circuit. If the frequency is doubled, what
happens to the capacitive reactance and the current?
Solution :
The capacitive reactance is C
1
X 106
2 fC
= =W
p
The rm
s current is
rms
rms
C
V
I 2.08A
X
==
The pea
k current is
m rms
I 2I 2.96A==
This current os
cillates between 2.96A and – 2.96A and is
ahead of the voltage by 90º.If the frequency is doubled, the capacitive reactance is
halved and consequently, the current is doubled.
VARYING CURRENT
When the key in a D.C. circuit (containing a D.C. source of emf,
inductance coil, resistance and capacitor) is closed or opened,
the current in the circuit varies. This is known as varying current
as it varies w.r.t. time and takes a final value after a short while.
Growth of Current
If K is closed at t = 0 so at t = 0, current in the circuit I = 0
After closing the key K

at time t let current in the circuit = I
and for small time in the circuit, current varies with time,
K
E
I
bR L
a c
so if rate of change of current with time =
dI
dt
then due to phenomenon of self induction, induced emf across
inductance
dI
L
dt
=-
Potentia
l difference across the resistance = IR
During growth of current in L-R circuit, if we applying Kirchhoff’s
loop rule then
d
ELR
dt
Iæö
+ - =Iç÷
èø
On solv
ing it we get the value of current at any time t during
growth of current in LR-circuit.
R
t
L
0
I I 1e
-æö
=-ç÷
ç÷
èø
Graph showing how current varies with time
I = E/R
0
I
0.63 I
0
t
L
t
Time Constant
L
R
has dimensions of time. It is called inductive time constant of
LR-circuit.
At t =
L
R
;
RL
.
LR
0
(1e)
-
I=I- = I
0
(1
– e
–1
) =
0
e1
I
e
-æö
ç÷
èø
= 0
2.71 1
I
2.71
-æö
ç÷
èø
= 0.632 I
0
The indu
ctive time constant of an LR-circuit is the time in which
the current grows from zero to 0.632 (or 63.2%) of its maximum
value. When t ® ¥.
( ) ( )
R
.
L
0 00
I I 1e I e I10
-¥
-¥
æö
=- =I-=-ç÷
èø
Potential
difference across resistance :
V
R
= E
R
t
L
1e
-æö
-ç÷
èø
; V
L
= L
dI
dt
R
t
L
00
I I Ie
-
=-
;
E
V
L
tdI
dt
= 0 – I
0

R
t
L
R
e
L
-æö
-
ç÷
èø
V
L
= E

R
t
L
e
-

600. 596 PHYSICS
Initial
ly, an inductor acts to oppose changes in the current through
it. A long time later, it acts like ordinary connecting wire.
DECAY OF CURRENT
Let the current has reached its steady state value I
0
through
inductor. Now switch K in the circuit shown in fig. has been
closed.
Let this time is t = 0.
Let at t = 0 current in the circuit (which is maximum) = I
0
After time t current in the circuit = I
Applying Kirchhoff’s loop rule to this circuit
dI
0 L IR
dt
æö
+-=ç÷
èø
(since there
is no source of e.m.f.)
ordI
L IR
dt
=-or
dIR
dt
IL
=-
The eq
n
. gives t
he value of current at any time t during decay of
current in LR-circuit.
I = E/R
0
I
0.37 I
0
t
L t
Again, dimensions o
f
L
R
are same as that of time
The in
ductive time constant of the LR-circuit can also be defined
by using equation
Setting
L
t
R
= in equation., we get
.
1
0 00
1
-
-
= ==
RL
LR
I Ie Ie I
e
or I @ 0.37 I
0
.
As t ® ¥, I ® 0
V
R
= IR
E
V
R
0 t
or
R
t
L
R
Ve
-
=
V
L
=
dI
L
dt
I = I
0

R
t
L
e
-
R
t
L
0
dIR
Ie
dtL
-
æö
=-
ç÷
èø
or
R
t
L
L
V Ee
-
=-
t
V
L
0
LC OSCILLATIONS
If a ch
arged capacitor C is short-circuited though an inductor L,
the charge and current in the circuit start oscillating simple
harmonically. If the resistance of the circuit is zero, no energy is
dissipated as heat. Assume an idealized situation in which energy
is not radiated away from the circuit. With these idealizations-
zero resistance and no radiation, the oscillations in the circuit
persist indefinitely and the energy is transferred from the
capacitor’s electric field to the inductor’s magnetic field back and
forth. The total energy associated with the circuit is constant.
This is analogous to the transfer of energy in an oscillating
mechanical system from potential energy to kinetic energy and
back, with constant total energy.
Let us now derive an equation for the oscillations in an L-C circuit.

Refer figure (a) : The capacitor is charged to a potential difference
V such that charge on capacitor q
0
= CV
Here q
0
is the maximum charge on the capacitor. At time t = 0, it is
connected to an inductor through a switch S. At time t = 0, switch
S is closed.
Refer figure (b) : When the switch is closed, the capacitor starts
discharging. Let at time t charge on the capacitor is q (< q
0
) and
since, it is further decreasing there is a current i in the circuit in
the direction shown in figure.
The potential difference across capacitor = potential difference
across inductor,
orV
b
– V
a
= V
c
– V
d
\
q di
L
C dt
æö
=ç÷
èø
...(i)
Now, as the char
ge is decreasing, dq
i
dt
-æö
=ç÷
èø
or
2
2
di dq
dtdt
=-

601. 597Alternating Current
Substituting this value of
di
dt
in equation (i), we get
2
2
q dq
L
C dt
æö
=-
ç÷
èø
or
2
2
dq1
q
LCdt
æö
=-
ç÷
èø
...(ii)
This is the standard e
quation of simple harmonic motion
2
2
2
æö
= -w
ç÷
èø
dx
x
dt
Here
1
w=
LC
or
1
2
=
p
f
LC
...(iii)
The ge
neral solution of equation (ii),
is
0
qqcos(t)= w ±f
In case f = 0 as q = q
0
at t = 0.
Thus, we can say that charge in the circuit oscillates with angular
frequency given by equation (iii). Thus,
In L-C oscillations, q, i and
di
dt
all oscillate harmonically with
same angular frequency w. But the phase difference between q
and i or between i and
di
dt
is p/2. Their amplitudes are q
0
, q
0
w and
w
2
q
0
respectively. So
q = q
0
coswt, then
0
dq
i q sint
dt
= =-ww ;
2
0
di
q cost
dt
=-ww
Similarly potential energy across capacitor (U
C
) and across
inductor (U
L
) also oscillate with double the frequency 2w.
TRANSFORMER
A transformer is a device for converting high voltage into low
voltage and vice versa, without change in power.
There are two types of transformers.
(a)Step up transformer : It converts low voltage into high
voltage.
(b)Step down transformer : It converts high voltage into low
voltage.
The principle of a transformer is based on mutual
induction and a transformer always works on AC. The input
is appleid across primary terminals and output is obtained
across secondary terminals.
The ratio of number of turns in secondary and primary is
called the turn ratio
i.e.,
S
P
n
n
= turn ratio K.
If E
P
an
d E
S
are alternating voltages, I
P
and I
S
the alternating
currents across primary and secondary terminals
respectively then,
= ==
SS P
PPS
En I
K
EnI
.
Efficiency of transfo
rmer, h= ==
out SS
in PP
P EIOutput power
Input power P E I
Comparative study of step-up transformer and step-down
transformer.
spsP
sp sp
Sp sp
sp sp
1.
EE1. E E
2. N N 2. N N
3. I I 3. I I
4. Z Z 4. Z Z
5. k 1 5. k 1
<>
><
<>
<<
><
Step - up transformer Step - down transformer
Power losses in a transformer :
(a)Copper loss. This is due to resistance of the winding of
primary and secondary coil (I
2
R)
(b)Iron loss or Eddy current loss.
(c)Loss due to leakage of magnetic flux.
(d)Hysteresis : Due to repeated magnetisation and
demagnetisation of iron core.
(e) Humming loss : Due to vibration.
Inspite of all these losses, we have transformers with efficiency
of 70% – 90%.
Example 9.
An ideal choke takes a current of 10 ampere when connected
to an A.C. supply of 125 volt and 50 Hz. A pure resistor
under the same conditions takes a current of 12.5 ampere.
If the two are connected to an A.C. supply of
100 2 volt
and 40 hert
z, then find the current in a series combination
of the above resistor and inductor.
Solution :
For series combination,
])X(R[Z
2
L
2
+=
,10
5.12
125
R W== w L = 2 p f L = V/I
\ 2 p×50×L = 125/10 = 12.5 or 2 p L = 0.25
For 40 Hz frequency, X
L
= 2 p L×f = 0.25 × 40 = 10 W
Now 210])10()10[(Z
22
=+=;
Current
0
=
I
Z
= A10
210
2100
=
Example10.
A low
loss transformer has 230 V applied to primary and
gives 4.6 V in secondary. The secondary is connected to a
load which draws 5 A current. Find the current in primary.
Solution :
Assuming no loss of power E
p
I
p
= E
s
I
s
A1.0
230
5
6.4
E
E
p
ss
p =´==\
I
I

602. 598 PHYSICS
CONCEPT MAP
ALTERNATING CURRENT
Direction of current
Changes alternatively and
its magnitude changes
continuously
AC Circuit
Capacitive (C) Circ
u
i
t
Alternating current(I)
and alternating voltage (V)
I = I sin t; V = V sin t 00 ww
RMS value of alternating
current and voltage
0
rms
I
I=
2
0
rms
V
V=
2
Mean or average value of
alternating current and voltage
0
mean
2I
I =
π
0
mean
2V
V =
π
Peak current (I) and voltage V : The
maximum value of current and voltage
I =
00
0
Power in an Ac. Circuit
Pav =V I cos rmsrmsq
Transformer Device Change
s
a low voltage of high curren
t
into a high voltage of low
current and vice-versa
S
t
e
p
-
d
o
w
n
t
r
a
n
s
f
o
r
m
e
r
K
<
1

p
p
s
s
s
p
N
E
I
K
=
=
=
N
E
I
S
t
e
p
-
u
p

t
r
a
n
s
f
o
r
m
e
r

K
>

1
p
s
s p
s
p
I
N
E
K
=
=
=
N
E
I
LCR

s
e
r
i
e
s

C
i
r
c
u
i
t
Curre
n
t
:

I
=
I

s
i
n
(
t
±

)
:
V
o
l
t
a
g
e
:
0
w
f
2
2
C
L
V
=
R
+
(
V
–
V
)
I
m
p
e
d
a
n
c
e

2
2
L
C
Z
=
R
+
(
X
–
X
)
;
P
h
a
s
e
d
i
f
f
e
r
e
n
c
e

L
C
X
–
X
R
A
t
r
e
s
o
n
a
n
c
e
X
=

X
Z
m
i
n
=
R
L
C
Þ
B
a
n
d
-
w
i
d
t
h
:

R
=
L
D
w
;
Q
u
a
l
i
t
y
f
a
c
t
o
r

1
L
=
R
C
Q
R
e
s
i
s
t
i
v
e

(
R
)

C
i
r
c
u
i
t
C
u
r
r
e
n
t

:

I
=
I

s
i
n
t
P
h
a
s
e
d
i
f
f
e
r
e
n
c
e
b
e
t
w
e
e
n
V
a
n
d

I
:
=
0
º
P
o
w
e
r
f
a
c
t
o
r
:

c
o
s

=

1
P
o
w
e
r

:


0
w
f
f
0
0
V
I
P
=
2
P
h
a
s
o
r

:

C
u
r
r
e
n
t

a
n
d

v
o
l
t
a
g
e
b
o
th

i
n
s
a
m
e

p
h
a
s
e
LC Circuit
Current: I = Isin
0
t
2
p æ ö
w±
ç ÷
è ø
Voltage: V= V – V
Impedance : Z = X – X
L C
L C
Phase difference: = 90º
Power factor :
Leading quantity:
Either voltage or current
f
cos f = 0
LR Circuit
Current: I = Isin (t + )
Voltage : V =
0
w f
2 2
R LV V+
Impedance:
2 2
L
Z= R +X
Phase difference:
Leading quantity : voltage
–1L
= tan
R
w
f
Power factor :
2 2
L
R
cos
R +X
f=
RC Circuit
Current: I = I sin (t + )
0
w f
2 2
R C V= V +V Voltage:
Impedance:
2 2
C
Z= R +X
Phase difference: tan
–
1
Power factor: cos f
2
2 2
C
R
=
R +X
Leading quantity: Current
Inductive (L) Circuit
Current : I = I sin
0
–
2
t
p
æ
ö
w
ç
÷
è
ø
Phase difference betwe
e
n
V and I : = 90º or /2
Power factor : cos = 0
Power : P = 0
Phasor : Voltage leads
t
h
e

current by
f
p
p
f
/2
1
CRw
rms
2I
0rms
V2V =
Current : I = I sin (t + /2)
Phase difference between V and
I : = 90º or – /2 Power factor :
cos = 0 Power : P = 0
Phasor : Current leads the voltage
by /2
0
w p
f p
f
p

603. 599Alternating Current
1.The resistance of a coil for dc is in ohms. In ac, the resistance
will
(a) be zero (b) decrease
(c) increase (d) remain same
2.In an a.c. circuit, the r.m.s. value of current, I
rms
is related to
the peak current, I
0
by the relation
(a)
0rms
I2I= (b)
0rms
IIp
=
(c)
0rms
I
1
I
p
= (d) 0rms I
2
1
I=
3.In a RLC c
ircuit capacitance is changed from C to 2 C. For
the resonant frequency to remain unchanged, the
inductance should be changed from L to
(a) 4 L(b) 2 L (c) L/2 (d) L/4
4.An LCR series circuit, connected to a source E, is at
resonance. Then the voltage across
(a) R is zero
(b) R equals applied voltage
(c) C is zero
(d) L equals applied voltage
5.In a LCR circuit at resonance which of these will effect the
current in circuit
(a) R only (b) L and R only
(c) R and C only (d) all L, C and R
6.Fleming's left and right hand rules are used in
(a) DC motor and AC generator
(b) DC generator and AC motor
(c) DC motor and DC generator
(d) Both rules are same, any one can be used
7.The time taken by the current to rise to 0.63 of its maximum
value in a d.c. circuit containing inductance (L) and
resistance (R) depends on
(a) L only (b) R only(c)
R
L
(d) LR
8.A bul
b and a capacitor are connected in series to a source of
alternating current. If its frequency is increased, while
keeping the voltage of the source constant, then bulb will
(a) give more intense light
(b) give less intense light
(c) give light of same intensity before
(d) stop radiating light
9.In LCR circuit if resistance increases quality factor
(a) increases finitely(b) decreases finitely
(c) remains constant(d) None of these
10.In an A.C. circuit with phase voltage V and current I, the
power dissipated is
(a) VI (b) V
2
I (c) VI
2
(d) V
2
I
2
11.Which of the following will have the dimensions of time
(a ) LC(b) R/L (c) L/R (d) C/L
12.In an oscillating LC circuit the max. charge on the capacitor
is Q. The charge on capacitor when the energy is stored
equally between electric and magnetic field is
(a) Q/2(b)
3/Q(c) 2/Q(d) Q
13.The power
factor of an AC circuit having resistance (R) and
inductance (L) connected in series and an angular velocity
w is
(a) R/wL (b) R/(R
2
+ w
2
L
2
)
1/2
(c)wL/R (d) R/(R
2
– w
2
L
2
)
1/2
14.A.C. power is transmitted from a power house at a high
voltage as
(a) the rate of transmission is faster at high voltages
(b) it is more economical due to less power loss
(c) power cannot be transmitted at low voltages
(d) a precaution against theft of transmission lines
15.In a pure capacitive A.C. circuit current and voltage differ in
phase by
(a) 0° (b) 45° (c) 90° (d) 180°
16.Which of the following statement is incorrect ?
(a) In LCR series ac circuit, as the frequency of the source
increases, the impedence of the circuit first decreases
and then increases.
(b) If the net reactance of an LCR series ac circuit is same
as its resistance, then the current lags behind the voltage
by 45°.
(c) At resonance, the impedence of an ac circuit becomes
purely resistive.
(d) Below resonance, voltage leads the current while above
it, current leads the voltage.
17.Resonance frequency of LCR series a.c. circuit is f
0
. Now
the capacitance is made 4 times, then the new resonance
frequency will become
(a)f
0
/4 (b) 2f
0
(c)f
0
(d) f
0
/2.
18.A capacitor has capacitance C and reactance X, if
capacitance and frequency become double, then reactance
will be
(a) 4X (b) X/2 (c) X/4 (d) 2X
19.In a series resonant circuit, having L,C and R as its elements,
the resonant current is i. The power dissipated in circuit at
resonance is
(a)
2
iR
( L 1/ C)w-w
(b) zero
(c)i
2
wL (d)
i
2
R.
Whereas w is angular resonant frequency
20.An inductance L having a resistance R is connected to an
alternating source of angular frequency w. The Quality factor
Q of inductance is
(a) R/ wL(b) (wL/R)
2
(c) (R /wL)
½
(d)wL/R

604. 600 PHYSICS
1.In an A.C. circuit, the c
urrent flowing in inductance is
I = 5 sin (100 t – p/2) amperes and the potential difference is
V = 200 sin (100 t) volts. The power consumption is equal to
(a) 1000 watt (b) 40 watt
(c) 20 watt (d) Zero
2.If resistance of 100W, and inductance of 0.5 henry and
capacitance of 10 × 10
6
farad are connected in series through
50 Hz A.C. supply, then impedance is
(a) 1.8765 W (b) 18.76 W
(c) 187.6 W (d) 101.3 W
3.Using an A.C. voltmeter the potential difference in the
electrical line in a house is read to be 234 volt. If the line
frequency is known to be 50 cycles/second, the equation
for the line voltage is
(a) V = 165 sin (100 p t)(b) V = 331 sin (100 p t)
(c) V = 220 sin (100 p t)(d) V = 440 sin (100 p t)
4.An inductance of negligible resistance whose reactance is 22
W at 200 Hz is connected to 200 volts, 50 Hz power line. The
value of inductance is
(a) 0.0175 henry (b) 0.175 henry
(c) 1.75 henry (d) 17.5 henry
5.An inductive circuit contains resistance of 10 ohms and an
inductance of 2 henry. If an A.C. voltage of 120 Volts and
frequency 60 Hz is applied to this circuit, the current would
be nearly
(a) 0.32 Amp (b) 0.16 Amp
(c) 0.48 Amp (d) 0.80 Amp
6.In an a.c. circuit V and I are given by
V = 100 sin (100 t) volts
I = 100 sin (100 t + p/3) mA
the power dissipated in the circuit is
(a) 10
4
watt (b) 10 watt (c) 2.5 watt (d) 5.0 watt
7.The primary winding of a transformer has 100 turns and its
secondary winding has 200 turns. The primary is connected
to an A.C. supply of 120 V and the current flowing in it is 10
A. The voltage and the current in the secondary are
(a) 240 V, 5 A (b) 240 V, 10 A
(c) 60 V, 20 A (d) 120 V, 20 A
8.A step down transformer is connected to 2400 volts line and
80 amperes of current is found to flow in output load. The
ratio of the turns in primary and secondary coil is 20 : 1. If
transformer efficiency is 100%, then the current flowing in
the primary coil will be
(a) 1600 amp (b) 20 amp
(c) 4 amp (d) 1.5 amp
9.In the circuit shown in fig, the resonant frequency is
(a) 75 kc/s
(b) 750 kc/s
(c) 7.5 kc/s
5Fm
0.1H 5W
(d) 75 mc/s
10.An alte
rnating voltage E (in volts) = 200
2 sin 100 t is
con
nected to one micro farad capacitor through an a.c.
ammeter. The reading of the ammeter shall be
(a) 100 mA (b) 20 mA(c) 40 mA(d) 80 mA
11.The r.m.s value of an a.c. of 50 Hz is 10 amp. The time taken
by the alternating current in reaching from zero to maximum
value and the peak value of current will be
(a) 2 × 10
–2
sec and 14.14 amp
(b) 1 × 10
–2
sec and 7.07 amp
(c) 5 × 10
–3
sec and 7.07 amp
(d) 5 × 10
–3
sec and 14.14 amp
21.The core of any transformer is laminated so as to
(a) reduce the energy loss due to eddy currents
(b) make it light weight
(c) make it robust and sturdy
(d) increase secondary voltage
22.The time constant of C–R circuit is
(a) 1/CR (b) C/R
(c) CR (d) R/C
23.In the circuit of Fig, the bulb will become suddenly bright if
+–
L B
K
B
(a) contact is made or broken
(b) contact is made
(c) contact is broken
(d) won't become bright at all
24.Energy in a current carrying coil is stored in the form of
(a) electric field(b) magnetic field
(c) dielectric strength (d) heat
25.In a circuit L, C and R are connected in series with an
alternating voltage source of frequency f. The current leads
the voltage by 45°. The value of C is
(a)
( )
1
f2fLRp p-
(b)
1
2f(2fL R)p p-
(c)
1
f(2 fL R)p p+
(d)
1
2f(2fL R)p p+

605. 601Alternating Current
12.The frequency of A.C. mains in India is
(a) 30 c/s (b) 50 c/s
(c) 60 c/s (d) 120 c/s
13.A 12 W resistor and a 0.21 henry inductor are connected in
series to an a.c. source operating at 20 volt, 50 cycle. The
phase angle between the current and source voltage is
(a) 30º(b) 40º (c) 80º (d) 90º
14.A step down transformer reduces 220 V to 110 V. The primary
draws 5 ampere of current and secondary supplies 9 ampere.
The efficiency of transformer is
(a) 20%(b) 44% (c) 90% (d) 100%
15.An alternating current is given by
i = i
1
coswt + i
2
sinwt
The rms current is given by
(a)
2
ii
21+
(b)
2
ii
21+
(c)
2
ii
2
2
2
1+
(d)
2
ii
2 2
2
1+
16.The impedance in
a circuit containing a resistance of 1 W
and an inductance of 0.1 H in series, for AC of 50 Hz, is
(a)
W10100 (b) W1010
(c)W100 (d) W10
17.The primary
winding of transformer has 500 turns whereas
its secondary has 5000 turns. The primary is connected to
an A.C. supply of 20 V, 50 Hz. The secondary will have an
output of
(a) 2 V, 5 Hz (b) 200 V, 500 Hz
(c) 2V, 50 Hz (d) 200 V, 50 Hz
18.Determine the rms value of the emf given by
E (in volt) = 8 sin (
wt) + 6sin (2wt)
(a) V25(b) V27(c) 10 V (d) V210
19.A transf
ormer is used to light a 140 W, 24 V bulb from a
240 V a.c. mains. The current in the main cable is 0.7 A. The
efficiency of the transformer is
(a) 63.8 % (b) 83.3 % (c) 16.7 % (d) 36.2 %
20.In the given circuit, the current drawn from the source is
~
W=20R
W=10X
L
W= 20X
C
)
t
1
0
0
s
i
n
(
x
1
0
0
V
p
=
(a) 20 A(b) 10 A (c) 5 A (d)A25
21.An AC volt
age source has an output of V =
200sin 2ftp.
This source i
s connected to a 100 W resistor. RMS current
in the resistance is
(a) 1.41 A (b) 2.41 A (c) 3.41 A (d) 0.71 A
22.An alternating voltage V = V
0
sin wt is applied across a
circuit. As a result, a current I = I
0
sin (wt – p/2) flows in it.
The power consumed per cycle is
(a) zero (b) 0.5 V
0
I
0
(c) 0.707 V
0
I
0
(d) 1.414 V
0
I
0
23.In an A.C. circuit, a resistance of R ohm is connected in
series with an inductance L. If phase angle between voltage
and current be 45°, the value of inductive reactance will be
(a) R/4
(b) R/2
(c)R
(d) cannot be found with given data
24.The ratio of mean value over half cycle to r.m.s. value of
A.C. is
(a) 2 : p (b)
p:22(c) p:2(d) 1:2
25.For the
circuit shown in the fig., the current through the
inductor is 0.9 A while the current through the condenser is
0.4 A. Then
(a) current drawn from
generator I = 1.13 A
(b)w = 1/(1.5 L C)
(c)I = 0.5 A
~
C
L
V = V sin t
0w
(d) I = 0.6 A
26.I
n series combination of R, L and C with an A.C. source at
resonance, if R = 20 ohm, then impedence Z of the
combination is
(a) 20 ohm (b) zero (c) 10 ohm (d) 400 ohm
27.In an LR circuit f = 50 Hz, L=2H, E=5 volts, R=1
W then
ene
rgy stored in inductor is
(a) 50 J (b) 25 J
(c) 100 J (d) None of these
28.A capacitor in an ideal LC circuit is fully charged by a DC
source, then it is disconnected from DC source, the current
in the circuit
(a) becomes zero instantaneously
(b) grows , monotonically
(c) decays monotonically
(d) oscillate infinitely
29.In a circuit inductance L and capacitance C are connected
as shown in figure. A
1
and A
2
are ammeters.
R
1
R
2
K
L
A
1
A
2
C
Battery
When key K is pressed to complete the circuit, then just
after closing key (K), the readings of A
1
and A
2
will be
(a) zero in both A
1
and A
2
(b) maximum in both A
1
a nd A
2
(c) zero in A
1
and maximum in A
2
(d) maximum in A
1
and zero in A
2

606. 602 PHYSICS
30.The tun
ing circuit of a radio receiver has a resistance of
50
W, an inductor of 10 mH and a variable capacitor. AA
1 MHz radio wave produces a potential difference of
0.1 mV. The values of the capacitor to produce resonance is
(Take p
2
= 10)
(a) 2.5 pF (b) 5.0 pF(c) 25 pF(d) 50 pF
31.Which one of the following curves represents the variation
of impedance (Z) with frequency f in series LCR circuit?
(a)
Z
f
(b)
Z
f
(c)
Z
f
(d)
Z
f
32.Two coils A and
B are connected in series across a 240 V, 50
Hz supply. The resistance of A is 5
W and the inductance of
B is 0.02 H. The power consumed is 3 kW and the power
factor is 0.75. The impedance of the circuit is
(a) 0.144 W(b) 1.44 W(c) 14.4 W(d) 144 W
33.In LCR series circuit fed by a DC source, how does the
amplitude of charge oscillations vary with time during
discharge ?
(a)
O
t
q
o
q (b)
O
t
q
(c)
O
t
q
o
q
(d)
O
t
q
oq
34.A steady potential difference of 10 V produces heat at a rate
x in a resistor. The peak value of the alternating voltage
which will produce heat at a rate
2
x
in the same resisto
r is
(a) 5 V(b)
V25(c) 10 V (d)10 2V
35.In the circuit shown, when the switch is closed, the capacitor
charges with a time constant
(a ) RC
(b) 2RC
(c)
RC
2
1
B
+
C R
(d) RC ln 2
36.In the question 86
, if the switch is opened after the capacitor
has been charged, it will discharges with a time constant
(a ) RC(b) 2RC (c)
RC
2
1
(d) RC ln 2
37.An alternating v
oltage of 220 V, 50 Hz frequency is applied
across a capacitor of capacitance 2 µF. The impedence of
the circuit is
(a)
5000
p
(b)
1000
p
(c) 500 p (d)
5000
p
38.An inductive coil has a resistance of 100 W. When an a.c.
signal of requency 1000 Hz is fed to the coil, the applied
voltage leads the current by 45°. What is the inductance of
the coil ?
(a) 10 mH (b) 12 mH (c) 16 mH(d) 20mH.
39.The primary of a transformer has 400 turns while the
secondary has 2000 turns. If the power output from the
secondary at 1000 V is 12 kW, what is the primary voltage?
(a) 200 V (b) 300 V (c) 400 V (d) 500 V
40.An inductor of self inductance 100 mH and a resistor of
resistance 50W, are connected to a 2 V battery. The time
required for the current to half its steady value is
(a) 2 milli second (b)2 ln (0.5) milli second
(c) 2 ln (3) milli second(d) 2 ln (2) milli second
41.The instantaneous voltage through a device of impedance
20 W is e = 80 sin 100 pt. The effective value of the current is
(a) 3 A(b) 2.828 A (c) 1.732 A(d) 4 A
42.A transformer has an efficiency of 80%. It works at 4 kW
and 100 V. If secondary voltage is 240 V, the current in primary
coil is
(a) 0.4 A(b) 4 A (c) 10 A (d) 40 A
43.The primary winding of transformers has 500 turns whereas
its secondary has 5000 turns. The primary is connected to
an A.C. supply of 20 V, 50 Hz. The secondary will have an
output of
(a)2V, 5Hz (b) 200 V, 500 Hz
(c) 2V, 50 Hz (d) 200 V, 50Hz
44.A step up transformer operates on a 230 V line and supplies
a current of 2 ampere. The ratio of primary and secondary
winding is 1:25 . The current in primary is
(a) 25 A(b) 50 A (c) 15 A (d) 12.5 A
45.A step-up transformer has transformation ratio of 3 : 2. What
is the voltage in secondary, if voltage in primary is 30 V?
(a) 45 V(b) 15 V (c) 90 V (d) 300 V
46.In an experiment, 200 V A.C. is applied at the ends of an
LCR circuit. The circuit consists of an inductive reactance
(X
L
) = 50 W, capacitive reactance (X
C
) = 50 W and ohmic
resistance (R) = 10 W. The impedance of the circuit is
(a) 10W (b) 20W (c) 30W (d) 40W
47.In a region of uniform magnetic induction
B = 10
–2
tesla, a circular coil of radius 30 cm and resistance
p
2
ohm is rotated about an axis which is perpendicular to the
direction of B and which forms a diameter of the coil. If the
coil rotates at 200 rpm the amplitude of the alternating current
induced in the coil is
(a)4p
2
mA (b) 30 mA (c) 6 mA (d) 200 mA

607. 603Alternating Current
48.In the given circuit the reading of voltmeter V
1
and V
2
are
300 volt each. The reading of the voltmeter V
3
and ammeter
A are respectively
A
1
V
2
V
~
L C 100R= W
3
V
220 , 50 V Hz
(a) 150 V
and 2.2 A (b) 220 V and 2.2 A
(c) 220 V and 2.0 A (d) 100 V and 2.0 A
Directions for Qs. (49 to 50) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following-
(a)Statement -1 is false, Statement-2 is true
Exemplar Questions
1.If the rms current in a 50 Hz AC circuit is 5 A, the value of
the current 1/300 s after its value becomes zero is
(a)5Ö2A (b) 5Ö3/2 A
(c) 5/6 A (d) 5/Ö2A
2.An alternating current generator has an internal reactance
R
g
and an internal reactance X
g
. It is used to supply power
to a passive load consisting of a resistance R
g
and a
reactance X
L
. For maximum power to be delivered from the
generator to the load, the value of X
L
is equal to
(a) zero (b) X
g
(c) –X
g
(d) R
g
3.When a voltage measuring device is connected to AC mains,
the meter shows the steady input voltage of 220 V. This
means
(a) input voltage cannot be AC voltage, but a DC voltage
(b) maximum input voltage is 220 V
(c) the meter reads not v but < v
2
> and is calibrated to read
2
v<>
(d) The pointer of the meter is stuck by some mechanical
defect
4.To reduce the resonant frequency in an L-C-R series circuit
with a generator(a) the generator frequency should be reduced(b) another capacitor should be added in parallel to the
first
(c) the iron core of the inductor should be removed(d) dielectric in the capacitor should be removed
5.Which of the following combinations should be selectedfor better tuning of an L-C-R circuit used for communication?
(a) R = 20 W, L =1.5 H, C = 35µF
(b) R=25W, L = 2.5 H, C = 45 µF
(c) R =15W, L = 3.5 H, C = 30 µF
(d) R = 25W, L = 1.5 H, C = 45 µF
6.An inductor of reactance 1W and a resistor of 2W are
connected in series to the terminals of a 6V (rms) AC source.
The power dissipated in the circuit is
(a) 8 W (b) 12 W
(c) 14.4 W (d) 18 W
7.The output of a step-down transformer is measured to be
24 V when connected to a 12 W light bulb. The value of the
peak current is
(a) 1 /
2A (b) 2A
(c) 2 A (d) 22A
NEET/AIPMT (2013-2017) Questions
8.A coil of self-inductance L is connected in series with a
bulb B and an AC source. Brightness of the bulb decreases
when [2013]
(a) number of turns in the coil is reduced
(b) a capacitance of reactance X
C
= X
L
is included in the
same circuit
(c) an iron rod is inserted in the coil
(d) frequency of the AC source is decreased
9.The primary of a transformer when connected to a dc battery
of 10 volt draws a current of 1 mA. The number of turns of
the primary and secondary windings are 50 and 100
respectively. The voltage in the secondary and the current
drawn by the circuit in the secondary are respectively
(a) 20 V and 0.5 mA [NEET Kar. 2013]
(b) 20 V and 2.0 mA
(c) 10 V and 0.5 mA
(d) Zero and therefore no current
(b)Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c)Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d)Statement -1 is true, Statement-2 is false
49. Statement - 1 : A capacitor blocks direct current in the steady
state.
Statement - 2 : The capacitive reactance of the capacitor is
inversely proportional to frequency f of the source of emf.
50. Statement - 1 : In the purely resistive element of a series
LCR, AC circuit the maximum value of rms current increases
with increase in the angular frequency of the applied emf.
Statement - 2 :
2
2max
max
1
I ,z
RL
zC
e
æö
= = +w-
ç÷
èøw
,
where I
max
i
s the peak current in a cycle.

608. 604 PHYSICS
10.A transfo
rmer having efficiency of 90% is working on 200V
and 3kW power supply. If the current in the secondary coil
is 6A, the voltage across the secondary coil and the current
in the primary coil respectively are : [2014]
(a) 300 V, 15A (b) 450 V, 15A
(c) 450V, 13.5A (d) 600V, 15A
11.A resistance 'R' draws power 'P' when connected to an AC
source. If an inductance is now placed in series with the
resistance, such that the impedance of the circuit becomes
'Z', the power drawn will be [2015]
(a)
R
P
Z
(b)
R
P
Z
æö
ç÷
èø
(c)P (d)
2
R
P
Z
æö
ç÷
èø
12.A series R-C circuit is connected to an alternating voltage
source. Consider two situations: [2015 RS]
(A) When capacitor is air filled.
(B) When capacitor is mica filled.
Current through resistor is i and voltage across capacitor is
V then :
(a) V
a
> V
b
(b) i
a
> i
b
(c) V
a
= V
b
(d) V
a
< V
b
13.An inductor 20 mH, a capacitor 50 mF and a resistor 40W are
connected in series across a source of emf V = 10 sin 340 t.
The power loss in A.C. circuit is : [2016]
(a) 0.51 W (b) 0.67 W
(c) 0.76 W (d) 0.89 W
14.A small signal voltage V(t) = V
0
sin wt is applied across an
ideal capacitor C : [2016]
(a) Current I (t), lags voltage V(t) by 90°.
(b) Over a full cycle the capacitor C does not consume
any energy from the voltage source.
(c) Current I (t) is in phase with voltage V(t).
(d) Current I (t) leads voltage V(t) by 180°.
15.Figure shows a circuit that contains three identical resistors
with resistance R = 9.0 W each, two identical inductors with
inductance L = 2.0 mH each, and an ideal battery with emf e
= 18 V. The current 'i' through the battery just after the
switch closed is [2017]
CL
RR
L
R
e
+
–
(a) 0.2 A (b) 2 A
(c) 0 (d)
2 mA

609. 605Alternating Current
EXERCISE - 1
1. (c
) The coil has inductance L besides the resistance R.
Hence for ac its effective resistance
22
L
RX+ will
be larg
er than its resistance R for dc.
2. (d)
3. (c) We know that
,
)LC(2
1
f
p
=
when C
is doubled, L should be halved so that resonant
frequency remains unchanged.
4. (b)
5. (a) At resonance, L
w =
C
1
w
Hence
the impedance of the circuit would be just equal
to R (minimum). In other words, the LCR-series circuit
will behave as a purely resistive circuit. Due to this
the current is maximum. This condition is known as
resonance
\ Z = R, Current =
R
V
6. (c) 7
. (c) 8. (a) 9. (b) 10. (a)
11. (c)
12. (c) In the case of maximum charge on capacitor, the whole
energy is, stored in capacitor in the form of electric
field which is
C
Q
2
1
U
2
=
When energ
y in distributed equally between electric
and magnetic field , then energy stored in electric field
i.e. in capacitor is
2
1
C
Q
2
1
2
U
U
2
1
÷
÷
ø
ö
ç
ç
è
æ
==
At that time if c
harge on capacitor is Q
1
, then
2/QQ
C
Q
4
1
2
U
C
Q
2
1
U
1
22
1
1 =Þ===
13. (b) 1
4. (b)15. (c)
16. (d) Option (d) is false because the reason why the voltage
leads the current is because
1
L
C
>w
w
and if the
v
oltage lags, the inductive reactance is greater than
the capacitive reactance.
17. (d) In LCR series circuit, resonance frequency f
0
is given
by
2
0
111
L 2f
C LC LC
w= Þw= \w= =p
w
00
11
f orf
2LCC
\ =a
p
When th
e capacitance of the circuit is made 4 times,
its resonant frequency become
'
0
f
'
'00
0
0
ffC
orf
f24C
\ ==
18. (c) The reactance of capacitor X =
C
1
w
where
w is
frequency and C is the capacitance of capacitor.
19. (d) At resonance wL= 1/wC
and i = E/R , So power dissipated in circuit is P = i
2
R.
20. (d)Potential drop across capacitor or inductor
Q
Potential drop a
cross R.
L
R
=
w
=
21. (a)
22
. (c) The time constant for resonance circuit,
= CR
Growth of charge in a circuit containing capacitance
and resistance is given by the formula,
/
0
(1)
-
=-
t CR
qqe
CR is kno
wn as time constant in this formula.
23. (c) When a circuit is broken, the induced e.m.f. is largest.
So the answer is (c).
24. (b) Energy stored in a coil =
21
2
Li
where, L is the self-inductance and i the current flowing
through the inductor. Thus, energy is stored in themagnetic field of the coil.
25. (d) From figure,
tan 45º =
1
L
C
R
-w
w
Þ
1
L
C
-w
w
= R
O
45º
R
1
w
w
C
L-–
Þ
1
RL
C
= +w
w
ÞC =
1
w w( L)R+
=
1
2 2p pfR fL( )+
Hints & Solutions

610. 606 PHYSICS
EXERCISE - 2
1. (d) P
ower, f´´=cosVP
s.m.rs.m.r
Ι
In the given problem, the phase difference between
voltage and current is p/2. Hence
.0)2/cos(VP
s.m.rs.m.r=p´´=Ι
2. (c)
2
2
C
1
LRZ
÷
÷
ø
ö
ç
ç
è
æ
w
-w+=
Here R = 100
W, L = 0.5 henry, C = 10 × 10
6
farad
w = 2 p p = 100 .p
3. (b) V= V
0
sin w t
Voltage in r.m.s. v
alue
0V 2 234V 331 volt=´=
and t100t502tn
2t
p=´´p=p=w
Thus, the equation of line voltage is given by
V = 331 sin (100 p t)
4. (a) Ln2LX
L
p
=w=
L
X 227
L H 0.0175H
2 n 2 22 200
´
\===
p ´´
5
. (b)
6. (c) f´I´=cosVP
s.m.rs.m.r

fI=cosV
2
1
00
31
100 (100 10 )cos / 3 2.5 W
2
-
=´ ´ ´ p=
7. (a)
÷
÷
ø
ö
ç
ç
è
æ
´==
p
s
ps
p
s
p
s
n
n
EEor
n
n
E
E
V240
100
200
120E
s
=÷
ø
ö
ç
è
æ
´=\
pp s
sp
s ps
nn
or
nn
æö
==
ç÷
èø
Ι
ΙΙ
Ι
\ I
s
=
100
10
200
æö
ç÷
èø
= 5 amp
8. (c)
s
p
p
s
n
n
=
Ι
Ι
; .amp4or
1
2080
p
p
== Ι
Ι
9. (a)
10. (b)
÷
÷
ø
ö
ç
ç
è
æ
w´=w==C
2
E
CE
X
E
0
C
Ι
200
I 120 240V
100
æö
\=´= ç÷
èø
.amp1020
3-
´=
11. (d)
12. (b)
13. (c) The phase angle is given by
5.5
12
21.0502
R
L
tan =
´´p
=
w
=f
1
tan 5.5 80º
-
f==
14. (c)
E 110 9ss
0.9 100% 90%
E 220 5
pp
I ´
h= \h= =
´=
I´
15. (c) 16. (b)
17. (d) The transformer converts A.C. high voltage into A.C.
low voltage, but it does not cause any change in
frequency.
sss
sp
ppp
ENN 5000
E E 20 200V
E N N 500
= Þ= = ´=
Th
us output has voltage 200 V and frequency 50 Hz.
18. (a) t2sin6tsin8Ew+w=
V1068E
22
peak =+=Þ
rms
10
E 5 2V
2
==
19. (b) Po
wer of source = EI = 240 × 0.7 = 166
%3.83
166
140
Efficiency =hÞ=Þ
20. (d
) 21. (a)
22. (a) The phase angle between voltage V and current I is
p/2. Therefore, power factor cos f = cos (p/2) = 0. Hence
the power consumed is zero.
23. (c)
L
XL
tan
RR
w
f==
Given f = 45°
. Hence X
L
= R.
24. (b) We know that
2/
0smrΙΙ= and p=/2
0m
ΙΙ
p
=\
22
smr
m
Ι
Ι
25. (c) The c
urrent drawn by inductor and capacitor will be in
opposite phase. Hence net current drawn from
generator
= I
L
– I
C
= 0.9 – 0.4 = 0.5 amp.
26. (a)
27. (d) 5E,H2L
== volts, W=1R
E
LR
444844476
Z

611. 607Alternating Current
Energy in inductor
2
LI
2
1
=
Z
E
I=
()
22
5
I
RL
=
+w
22
5
14 504
=
+p´ ´

( )
2
5
1 200
=
+p

p
=
200
5
Energy
2
1 55
2
2 200 200
´
= ´´
´p
= 6.33 × 10
–5
joules
2
8. (a) In ideal condition of LC circuit 0R
=and LC oscillation
continue indefinitely. Energy being shunted back and
forth between electric field of capacitor and magnetic
field of inductor. As capacitor is fully charged current
in L is zero and
C
q
2
1
2
0
energy
is stored in electric
field. Then capacitor begins to discharge through L
causing a current to flow and build up a magnetic
field, around L. Therefore, energy stored.
Now in
2
0LI
2
1
L= when C is full
y discharged, V
across the plate reduces to zero.
\ Electric field energy is transferred to magnetic
field and vice-versa.
29. (d) Initially there is no D.C. current in inductive circuit
and maximum D.C.current is in capacitive current.
Hence, the current is zero in A
2
and maximum in A
1
.
30. (a) L = 10 mHz = 10
–2
Hz
f = 1MHz = 10
6
Hz
LC2
1
f
p
=
;
LC4
1
f
2
2
p
=
pF5.2
4
10
1010104
1
Lf4
1
C
12
12222
==
´´´
=
p
=Þ
-
-
31. (c)
Impedance at resonant frequency is minimum in series
LCR circuit.
So,
2
2
fC2
1
fL2RZ÷
ø
ö
ç
è
æ
p
-p+=
When freq
uency is increased or decreased, Z
increases.
32. (c)
Z
cosE
P
2
vf
=
Z
)75.0()240(
3000P
2
== d W=Þ 4.14Z
33. (c) 34.
(c)
35. (a) The resistance in the middle plays no part in the charging
process of C, as it does not alter either the potential
difference across the RC combination or the current
through it.
36. (b) C discharges through both resistance in series.
37. (d) Impedence of a capacitor is X
C
= 1/wC
C
6
1 1 5000
X.
2 fC2 50 2 10
-
===
pp p´ ´´
38. (c
)
39. (a) N
P
= 400, N
S
= 2000 and V
S
= 1000 V.
PP
SS
VN
VN
=
of,
SP
P
S
VN 1000 400
V 200V.
N 2000
´ ´
===
40.
(d) The time constant of the circuit is 3
3L 100 10
2 10
R 50
-
-´
t= = =´ s=
2 milli second.
Current at time t is given by I = I
0
e
–t/t
where I
0
is the steady current. Therefore, time for I to
fall to I
0
/2 is
–t/1
e
2
t
= or, e
t/t
= 2, t = tln(2) = 2ln(2) milli second.
41. (b) Given equation, e = 80 sin 100pt …(i)
Standard equation of instantaneous voltage is given
by e = e
m
sinwt …(ii)
Compare (i) and (ii), we get e
m
= 80 V
where e
m
is the voltage amplitude.
Current amplitude
m
m
e
I
Z
= where Z = impendence
= 80/20 = 4 A.
r .m.s
4 42
I 2 2 2.8 28 A.
22
== ==
42.
(d) As
ipp
PE=Ι
.A40
100
4000
E
P
p
i
p
===\Ι
43. (d)
44
. (b)
25
1
E
E
n
n
s
p
s
p
==
ps
E25E=\
But
SS
p
EI
E I E I I I 50A
sspppp
E
´
= Þ= Þ=
4
5. (a) Transformation ratio
S
P
N
k
N
=
Since
SSS
SP
PPP
VNN
VV
VNN
= \=´
S
3
V 30 45V
2
=´=

612. 608 PHYSICS
46. (a)
Given : Supply voltage (V
ac
) = 200 V
Inductive reactance (X
L
) = 50 W
Capacitive reactance (X
C
) = 50 W
Ohmic resistance (R) = 10 W.
We know that impedance of the LCR circuit
(Z) =
W=+-=+-10})10()5050{(}R)XX{(
2222
CL
47. (c)
0
0
EnBA
I
RR
w
==
Given, n = 1, B = 1
0
–2
T,
A = p(0.3)
2
m
2
, R = p
2
f = (200/60) and w = 2p(200/60)
Substituting these values and solving, we get
I
0
= 6 × 10
–3
A = 6mA
48. (b) As V
L
= V
C
= 300 V, resonance will take place
\ V
R
= 220 V
Current
, I =
220
2.2
100
A=
\ reading of V
3
= 2 20 V
and reading of A = 2.2 A
49. (b) 50. (c)
EXERCISE - 3
Exemplar Questions
1. (b) As given that, v = 50 Hz, I
rms
= 5A
t =
1
s
300
As we know that I
rms
=
0I
2
I
0
= Peak value =
rms
2.I = 25´
I
0
= 5 2A
at, t =
1
sec
300
, I = I
0
sinwt = 5 2sin2 vtp
=
1
5 2sin2 50
300
p´´
I = 5 2 sin
3
p
=
3
52
2
´=5 3 2 Amp
3
sin
32
æöp
\=ç÷
èø
I =
3
5 Amp
2
æö
ç÷
ç÷
èø
2. (c) To de
liver maximum power from the generator to the
load, total internal reactance must be equal to conjugate
of total external reactance.
So,X
int
= X
ext
X
g
= (X
L
) = –X
L
Hence, X
L
= –X
g
(Reactance in external
circuit)
3. (c) As we know that,
The voltmeter in AC reads rms values of voltage
I
rms
=
0
2I and VV
rms
=
0
2v
The voltmeter in AC circuit connected to AC mains
reads mean value (<v
2
>) and is calibrated in such a
way that it gives rms value of <v
2
>, which is multiplied
by form factor Ö2 to give rms value V
rms
.
4. (b) As we know that,
The resonant frequency in an L-C-R series circuit is
n
0
=
1
2 LCp
So, to reduce n
0
either
increase L or increase C.
To increase capacitance, another capacitor must be
connect in parallel with the first capacitor.
5. (c) As we know that, Quality factor (Q) of an
L-C-R circuit must be higher so Q is
Q =
1L
RC
where R is resistance, L is inductance and C is
capacitance of the circuit.
So, for higher Q, L must be large, and C and R should
be low.
Hence, option (c) is verify.
6. (c) As given that,
X
L
= 1W, R = 2W, E
rms
= 6V, P
av
= ?
The average power dissipated in the L, R, series circuit
with AC source
ThenP
av
= E
rms
I
rms
cos f... (i)
I
rms
=
0 rmsIE
Z2
=
Z =
22
L
RX+ =415+=
I
rms
=
6
A
5
cos f =
R2
Z5
=
By putting the value o
f I
rms
, E
rms
, cos f in equation (i),
then,
P
av
=
62
6
55
´´ =
72
55
=
72
14.4 watt
5
=

613. 609Alternating Current
7. (a) As given that,
Secondary voltage (V
S
) is :
V
S
= 24 Volt
Power associated with secondary is :
P
S
= 12 Watt
As we know that P
S
= V
S
I
S
I
S
=
S
S
P12
V 24
= =
1
A 0.5 Amp
2
=
Peak value of the current in the secondary
I
0
=
S
I2 = 0.52
=
5
2
10
Þ0
1
I Amp
2
éù
=
êú
ëû
NEET/AIPMT (2013-2017) Questions
8. (c) By inserting iron rod in the coil,
L ­ z ­ I ¯ so brightness ¯
9. (d) A transformer is essentially an AC device. DC source
so no mutual induction between coils
Þ E
2
= 0 and I
2
= 0
10. (b) Efficiency h =
sss
3
pp
V I V (6)
0.9
VI 3 10
Þ=
´
ÞV
s
= 450 V
As V
p
I
p
= 3000 so
I
p
=
p
3000 3000
A 15A
V 200
==
11. (d)
V
R
V
s
V
R
V
L
Pure resistor L-R series circuit
q
R
Z
X
L
Phasor diagram
Z = impedance
R
cos
Z
q=
For pure resistor circuit, power
2
2V
P V PR
R
= Þ=
For L-R series circuit, power
222
1
2
VVRPRR
P cos . .RP
ZZ ZZZ
æö
=q===
ç÷
èø
12. (a) For series R – C circuit, capacitive reactance,
Z
c
=
2
21
R
C
æö
+
ç÷
wèø
X = 1/c
c
w
R
~
C
AC Source
Current i =
c
V
Z
=
2
2
V
1
R
C
æö
+
ç÷
wèø
V
c
= iX
c
=
2
2
V1
C
1
R
C
´
w
æö
+
ç÷
wèø
V
c
=
2
V
(RC)1w+
If we fill a di-electric material like mica instead of air then capacitance C­ Þ V
c
¯
So, V
a
> V
b
13. (a) Given: L = 20 mH; C = 50 mF; R = 40 W
V = 10 sin 340 t
\V
runs
=
10
2
X
C
=
6
11
C340 50 10
-
=
w ´´
= 58.8 W
X
L
= wL = 340 × 20 × 10
–3
= 6.8 W
Impedance, Z = ( )
22
CL
R XX+-
= ( )
22
40 58.8 6.8 4304+-=W
Power loss in A.C. circuit,
P =
2
2 rms
rms
V
iRR
Z
æö
=ç÷
èø
=
2
10 / 2 50 40
40 0.51 W
43044304
æö ´
´=ç÷
èø
;
14. (b) As we know, power P = V
rms
· I
rms
cosf
as cosf = 0 ( Q f = 90°)
\Power consumed = 0 (in one complete cycle)
15. (b)
At t = 0, no current flows through R
1
and R
3
\
Current through battery just after the switch closed is
2
i
R
e
=

18
9
= = 2A

614. 610 PHYSICS
DISPLACEMENT CURRENT
It is tha
t current which comes into play in a region, where electric
field (and hence the electric flux) is changing with time.
The displacement current is given by,
f
=e
E
Do
d
I
dt
where
o
e= absolute permittivity (or permitivity of free space) and
f
E
d
dt
= rate of change of electric flux.
In case of a steady electric flux linked with a region, the
displacement current is maximum.
The current in the electric circuit which arises due to flow of
electrons in the connecting wires of the circuit, in a defined
closed path is called conduction current.
MAXWELL’S MODIFICATION OF AMPERE’S CIRCUITAL
LAW
In 1864, Maxwell showed that Ampere’s circuital law is logically
inconsistent for non-steady currents. He modified Ampere’s law
as
00
.
fæö
=m +e
ç÷
èøò
ur u u ur
l
Ñ
E
d
BdI
dt
(Sum of conduction current and
displacement current)
The term
d 0E
(d /dt)=efI is displacement curr ent. It is that
current which comes into existence, in addition to the conduction
current, whenever the electric field and hence the electric flux
changes with time.
Maxwell’s Equation
Maxwell found that all the basic principles of electromagnetism can
be formulated in terms of four fundamental equations, called
Maxwell’s equation. These are :
(i) Gauss's law for electrostatics : According to Gauss's law
the total electric flux through any closed surface is equal to
the net charge inside that surface divided by e
o
i.e.,
.
o
Q
E dA=
eò
uurr
Ñ
This law relates the
electric field to charge distribution,
whereas electric field lines orignate from positive (+ive)charge and terminate on negative (–ive) charge. The
differential form of this law is
0
.
r
Ñ=
e
rr
E
where r is volume ch
arge density and e
o
is the permittivity
of free space.
(ii) Gauss's law for magnetism : It states that the net magnetic
flux through a closed surface is zero. It means that number
of magnetic field lines that enter a closed volume must be
equal to number of field lines leaving that volume
i.e.,
.0=
ò
rr
Ñ
B dA
It means that magnetic mono pole
cannot exists in nature.
The differential form of this law is
.0BÑ=
rr
(iii) Faraday’s law : Th is law states that the line integral of the
electric field around any closed path (which equals the e.m.f.)
equals the rate of change of magnetic flux through any
surface area bounded by that path.
i.e.,.
f
=ò
rr
lÑ
m
d
Ed
dt
The consequence of this law is that, if we keep any
conducting loop in a time varying magnetic field, then an
induced current, flows in that conducting loop. The
differential form of this law is Ñ´ =-
rrdB
E
dt
This law decribes a relationship between an electric field
and a changing magnetic flux.
(iv) Modified Ampere-Maxwell law : It states that “ the line
integral of magnetic field around any closed path is
determined by the sum of the net conduction current through
that path and the rate of change of electric flux through any
surface bounded by that path
i.e., 00
.
f
=m +emò
rr
Ñ
e
o
d
BdSI
dt
It describes the relationship between magnetic and electric
fields and electric current. The differential form of this law is
0 00
Ñ´ =m +m
rr dE
BJE
dt
ELECTROMAGNETIC WAVES
Maxwel
l-on the basis of four basic equations of electromagnetism
theoretically predicted the existence of electromagnetic waves.
An electromagentic wave is the one constituted by oscillating
electric and magnetic fields which oscillate in two mutually
perpendicular planes. The wave itself propagates in a direction
perpendicular to both of the directions of oscillations of electric
and magnetic fields.
23
Electromagnetic
Waves

615. 611Electromagnetic Waves
Properties of Electromagnetic Waves
(i) The direction of oscillations of E and B fields are
perpendicular to each other as well as to the direction of
propagation. Electromagnetic waves are transverse in nature.
(ii) The electric and magnetic field oscillate in same phase.
(iii) The electromagnetic waves travel through vacuum with the
same speed of light
18
o0
ms103
1
c
-
´=
em
=
(iv)The energy d
ensity of electric field is
21
2
e
o
E and that of
magneti
c field is
2
0
1
2m
B
, so the energy density of the
electromagnetic wave is
2
2
0
1
2
éù
=e+êú
mêúëû
o
B
uE
where E a
nd B are the instantaneous values of the electric
and magnetic field vectors.
(v) The ratio
0
1
==
me
o
E
c
B
.
Product
ion of Electromagnetic Waves
An accelerated charge emits electromagnetic waves. An oscillating
charge, as in an LC-circuit has non-zero acceleration, it continues
to emit electromagnetic waves. The frequency of electromagnetic
waves is the same as that of the oscillating charge.
Hertz’s experiment : In 1888, Hertz succeeded in experimentally
confirming the existence of electromagnetic waves. By using
oscillator LC-circuits, he not only produced and detected
electromagnetic waves, but also demonstrated their properties of
reflection, refraction and interference and established beyond
doubt that light radiation has wave nature.
Keep in Memory
1.Th
e amplitudes of electric and magnetic fields in free space,
in electromagnetic waves are related by E
0
= cB
0
or
0
0
=
E
C
B
2.The velocity of electromagnetic wave does not depend on
amplitude of field vectors.
3.The electric vector of an electromagnetic wave is responsible
for the optical effects & is called a light vector.
4.In a plane electromagnetic wave, the average energy
densities of electric and magnetic fields are equal.
In vacuum, the average electric energy density is given by,
2 22
00
111
( / 2)
224
=e
=e =e
E ooo
uEEE
The average magnetic energy density is given by
222
00
000
( / 2)
224
===
mmm
B
BBB
u
Also,
2 2 22
000
111
()
444
=e =e =e
E
ooo
u E cB cB

22
00
00
1
44
e
= ==
emm
o
B
o
BB
u
\ Total average energy density
= u
E
+ u
B
= 2u
E
= 2u
B

2
2 0
0
0
1
22
=e=
m
o
B
E
5. Poyntin
g vector: When an electromagnetic wave advances,
the electromagnetic energy flows in the direction
BE´.
The total
energy flowing perpendicularly per second per
unit area into the space in free space is called a poynting
vector
S
ur
where
0o
/)BE()BE(cSm´=´e=.
00
1
| | sin 90=
= °=mm
uur EB
S S EB
Th
e SI unit of S is watt/m
2
6.The intensity of a sinusoidal plane electromagnetic wave is
defined as average value of poynting vector taken over one
cycle. Thus,
22
000000
0 000
1
22222
æöæö
== =
==
ç÷ç÷
èøèø
m mmm
av
E B E B E cB
IS
c
7. Radiatio
n pressure : The pressure exerted by electromagnetic
waves is called as radiation pressure (P). When an
electromagnetic wave with Poynting vector
r
S is incident on a
perfectly absorbing surface, then radiation pressure on surface
is=
r
S
P
c
.
ELECTROMAGNETIC SPECTRU
M
All the known radiations form a big family of electromagnetic
waves according to frequency or wavelength. We call this family
as the complete electromagnetic spectrum. It includes: Gamma
rays, X-rays, ultraviolet light, visible light, infrared light,
microwaves and radio waves.

616. 612 PHYSICS
1Gamma r
ays
6 × 10
–14
to 1 × 10
–11
5 × 10
22
to 3 × 10
19
Nuclear origin Photographic film,
Geiger tubes,
ionisation chamber
2X-rays
1 × 10
–11
to 3 × 10
–8
3 × 10
19
to 1 × 10
16
Sudden deceleration of
high energy electron
(collision of electrons
with target)
Photographic film,
Geiger tubes,
ionisation chamber
3Ultra-violet
6 × 10
–8
to 4 × 10
–7
5 × 10
17
to 8 × 10
14
Excitation of atom,
spark and arc lamp
Photocells,
photographic film
4Vis ible light
4 × 10
–7
to 7 × 10
–7
8 × 10
14
to 4 × 10
14
Excitation of valency
electron
The eye, photocells,
photographic film
5Infrared
7 × 10
–7
to 3 × 10
–5
4 × 10
14
to 1 × 10
13
Excitation of atoms and
molecules
Th
ermopiles, bolometer
6 Heat radiations
10
–5
to 10
–1
3 × 10
13
to 3 × 10
9
Hot bodies
7Micro-waves
10
–3
to 0.3 3 × 10
11
to 1 × 10
9
Oscillating current in special vacuum tube
Point contact diodes
8Ultra high frequency
1 × 10
–1
to 1 3 × 10
9
to 3 × 10
8
Oscillating circuit
9Very high radio frequency
1 to 10
3 × 10
8
to 3 × 10
7
Oscillating circuit
10Radio frequency
10 to 10
4
3 × 10
7
to 3 × 10
4
Oscillating circuitReceiver's aerial
11Power
frequency
5 × 106 to 6 × 10660 to 50 Weak radiations from
a.c. circuit
DetectionS.No. Name
Wavelength range
(in m)
Frequency range
(in Hz)
Source
Example 1.
In a plane electrom
agnetic wave the electric field varies
with time having an amplitude 1 Vm
–1
. The frequency of
wave is 0.5 × 10
15
Hz. The wave is propagating along
Z-axis. What is the average energy density of (i) electric
field (ii) magnetic field (iii) total field. (iv) What is amplitude
of magnetic field?
Solution :
(i) Average energy density of electric field
2
0
o
2
oE
2
E
2
1
E
2
1
U
÷
÷
ø
ö
ç
ç
è
æ
e=e= ú
û
ù
ê
ë
é
=
2
E
E
0
Q
=
3122122
0o Jm1021.21)10854.8(
4
1
E
4
1 ---
´=´´´=e.
(ii) Averag
e energy density of magnetic field,
U
B
= Average energy density of electric field
= 2.21 × 10
–12
Jm
–3
(iii) Total average energy density U = U
E
+ U
B
= 2 U
E
= 2 × 2.21 ×10
–12
= 4.42 × 10
–12
Jm
–3
(iv) Amplitude of magnetic field, B
0
= E
0
/c
= 1/(3 × 10
8
) = 3.33 × 10
–9
T
Example 2.
A plane electromagnetic wave propagating in the
x-direction has a wavelength of 6.0 mm. The electric field
is in the y-direction and its maximum magnitude is 33 Vm
–1
.
Write suitable equations for the electric and magnetic fields
as a function of x and t.
Solution :
Here, l = 6.00 mm = 6 × 10
–3
m; E
0
= 33 V/m.
.s/rad10
106
1032c2
2
11
3
8
´p=
´
´´p
=
l
p
=pn=w
-
The maximum magnetic fi
eld, T1011
103
33
c
E
B
8
8
0
0
-
´=
´
==
The equatio
n for the electric field, along y-axis in the
electromagnetic wave is )c/xt(10sin33
c
x
tsinEE
11
0y
-´p=÷
ø
ö
ç
è
æ
-w=
The equation for
the magnetic field along z-axis in the
electromagnetic wave is )c/xt(10sin1011
c
x
tsinBB
118
0z
-´p´=÷
ø
ö
ç
è
æ
-w=
-
Example 3.
If
o
ε and
m
0
represent the permittivity and permeability of
vacuum and
ε and m represent the permittivity and
permeability of medium, then refractive index of the medium
is given by
(a)
0o
με
με
(b)
0o
με
με
(c)
0o
ε
με
(d)
0o
με
μ
Solution : (b)
Q Velo
city of light in medium
1
=
me
c
\ Refractive
index =
me
em
=
/1
/1
c
c o00
=
o0
em
me

617. 613Electromagnetic Waves
Example 4.
The magnetic field in a plane electromagnetic wave is given
by B = 2 × 10
–7
sin (0.5× 10
3
x +1.5 × 10
11
t) T.
(a) What is the wavelength and frequency of the wave?
(b) Write an expression for the electric field.
Solution :
(a) Comparing the given equation with

0
xt
B Bsin2
T
éùæö
= p+ ç÷êú
èølëû
we get,
3
2
m 1.26 cm
0.5 10
p
l==
´
and
111
(1.5 10 )/ 2 23.9 GHz
T
=n= ´ p=
(b)
E
0
= B
0
c = 2 × 10
–7
T × 3 × 10
8
m/s = 6 × 10
1
V/m
The electric field component is perpendicular to the
direction of propagation and the direction of magneticfield. Therefore, the electric field component along the z-axis is obtained as
E
z
= 60 sin (0.5 × 10
3
x + 1.5 × 10
11
t) V/m
Example 5.
A plane electromagnetic wave of frequency 25 MHz travelsin free space along the x-direction. At a particular point in
space and time, ˆ
E 6.3jV/m=
r
. What is
B
r
at this point ?
Soluti
on :
The magnitude of
B
r
is
8
8
E 6.3V/m
B 2.11 0T
c3 10 m/s
-
=
= =´
´
To find the direction, we note that
E
r
is along y-direction
and the wave propagates along x-axis. Therefore, B should
be in a direction perpendicular to both x- and y-axes. Using
vector algebra, EB´
rr
should be along x-direction.
Since, ˆ ˆˆ
( j) ( k)i.B+´+=
r
, B
r
is along the z-direction.
Thus,
8ˆ
B 2.1 10 kT
-
=´
r
USES OF ELECTROMAGNETIC WAVES
The following are some of the uses of electromagnetic waves
1.Radio waves are used in radio and T.V. communication
systems.
2.Microwaves are used in microwave oven.
3.Infrared radiations are used (a) in revealing the secret
writings on the ancient walls (b) in green houses to keep the
plants warm (c) in warfare, for looking through haze, fog or
mist as these radiations can pass through them.
4.Ultraviolet radiations are used in the detection of invisible
writing, forged documents, finger prints in forensic
laboratory and to preserve the food stuffs.
5. The study of infrared, visible and ultraviolet radiations help
us to know through spectra, the structure of the molecules
and arrangement of electrons in the external shells.
6.X-rays can pass through flesh and blood but not through
bones. This property of X-rays is used in medical diagnosis,
after X-rays photographs are made.
The study of X-rays has revealed the atomic structure and
crystal structure.
7. The study of g-rays provides us valueable information
about the structure of the atomic nuclei
8.Super high frequency electromagnetic waves (3000 to
30,000 MHz) are used in radar and satellite communication.
9.Electromagnetic waves (frequency 50 to 60 Hz) are ued for
lighting. These are weak waves having wavelength
5 × 10
6
to 6 × 10
6
m and can be produced from A.C. circuits.
Keep in Memory
1. Green h
ouse effect : It is the phenomenon which keeps the
earth’s surface warm at night. The earth absorbs solar radiationand reflects back only infrared rays due to its low temperature.
These rays are reflected back by the clouds and the gas molecules
of the lower atmosphere. This keeps the earth’s surface warm at
night.
PROPA
GATION OF RADIO WAVES THROUGH THE
ATMOSPHERE
It takes place in three ways :
(i) Ground wave propagation,
(ii) Sky wave propagation and
(iii) Space wave propagation.
(i) Ground wave propagation :When the radio wave travel
directly from one point to another following the surface of
the earth, it is called ground or surface wave. This type of
transmission is possible only with waves of wavelengths
above 200 m or frequencies below 1500 k Hz.
(ii) Sky wave propagation : When a radiowave is directed
towards the sky and is reflected by the ionosphere towards
desired location on the earth, it is called sky wave. This
method is useful for the transmission of waves of
wavelengths less than 200 m or frequencies above 1500 k
Hz upto 30 MHz.
(iii) Space wave propagation : For the transmission of television
signals (frequencies in the range 100-200 M Hz), space
wave propagation method is used, in which the wave travels
directly from a high transmitting antenna to the receiving
antenna. The relation between the height ‘h’ of the
transmitting antenna above the ground level and the distance
‘d’ upto which TV signal can be received is
2=d hR
= range of a TV tower where R is the radius of the earth.
Example 6.
A TV tower has a height of 100 m. How much population
is covered by the TV broadcast if the average population
density around the tower is 1000 km
–2
?(radius of the
earth = 6.37 × 10
6
m)
Solution :
Height of twoer h = 100 m
Radius of the earth R = 6.37 × 10
6
m
hR2d= = 2 100 6.37 106´´´
Popula
tion covered = p d
2
× population density
Solving we get population covered = 40 lakh.

618. 614 PHYSICS
CONCEPT MAP
ELECTROMAGNETIC WAVES
Constituted by mutually perpendicular
time varying electric and magnetic fields
Conduction current
Arises due to flow of
electrons in a definite
closed path
D
i
s
p
l
a
c
e
m
e
n
t
c
u
r
r
e
n
t
(
I
)
D
u
e
t
o
t
i
m
e
v
a
r
y
i
n
g
e
le
c
t
r
i
c

f
i
e
l
d
D
E
D
0
d
I
d
t
f
=
e
C
h
a
r
a
c
te
r
i
s
t
i
c
s
o
f
e
l
e
c
t
r
o
m
a
g
n
e
ti
c
w
a
v
e
s

Different types of
electromagnetic waves
Radiation pressure exerted
by an electromagnetic wave
energy associated
with em waves (u)
P
speed of light in vacuum(c)
=
Energy associated with an electromagnetic wave
2
2
0
0
11B
u E
2 2
=e +
m
D
o
n
o
t
r
e
q
u
i
r
e
a
n
y

m
a
t
e
r
i
a
l
m
e
d
i
u
m
f
o
r

p
r
o
p
a
g
a
t
i
o
n
P
r
o
d
u
c
e
d
b
y
a
c
c
e
l
e
r
a
te
d
c
h
a
r
g
e
T
r
a
v
e
l
s
w
i
th

s
p
e
e
d

o
f
l
i
g
h
t
i
n
f
r
e
e
s
p
a
c
e
0
08
1
C
3
1
0
m
/
s
=
m
e
=
´
I
n
f
r
e
e
s
p
a
c
e
;
m
a
g
n
i
t
u
d
e
o
f
e
l
e
c
t
r
i
c

f
i
e
l
d
(
E
)
m
a
g
n
e
ti
c
f
i
e
l
d
(
B
)
=
C

(
s
p
e
e
d

o
f
l
i
g
h
t



i
n
v
a
c
u
u
m
)
Radio waves
Wavelength > 0.1 m
Uses: in telecommu-
nication
Microwave
Wavelength 0.1 m to1 mm
Uses : in microwave
oven, RADAR
Infra -red
Wavelength 1 mm to 700 mm
Uses : treat muscular strain
Visible Wavelength :
700 nm to 400 nm
Uses : to see objects
Ultra-violet
Wavelength 400 nm
to 1nm
Uses : Preserve food
purifying water
g-rays
Wavelength : 1 nm to 10nm
Uses : Medical diagnosis
–3
g-rays
Wavelength : < 10nm
Uses : in medical science
information on nuclear
structure
–3
T
r
a
n
s
v
e
r
s
e

in
n
a
t
u
r
e

O
s
c
i
ll
a
ti
n
g
e
l
e
c
t
r
i
c

a
n
d

m
a
g
n
e
ti
c
f
i
e
l
d
s
a
r
e
i
n
p
h
a
s
e
a
n
d
th
e
i
r
m
a
g
n
i
t
u
d
e
s
b
e
a
r
c
o
n
s
t
a
n
t
r
a
t
i
o
C
=
B
0
E
0

619. 615Electromagnetic Waves
1.The electromagnetic radiation used in food processing
sterilizing agent is
(a) microwaves (b) UV rays
(c) gamma rays (d) radio waves
2.An electromagnetic wave is made up of joint electric (E) and
magnetic (B) fields. Its direction of propagation is
(a) parallel to E
(b) perpendicular to E but parallel to B
(c) parallel to B
(d) perpendicular to both E and B
3.Intensity of electromagnetic wave will be
(a)
2
00
I c B /2=m (b)
2
00
I c B /2=e
(c)
2
00
I B /c=m (d)
2
00
I E / 2c=e
4.The frequency modulated waves are
(a) reflected by atmosphere
(b) absorbed by atmosphere
(c) bend by atmosphere
(d) radiowaves
5.If
4
1 10?l=´ then it co rresponds to
(a) infrared (b) microwaves
(c) ultraviolet (d) X-rays
6.10 cm is a wavelength corresponding to the spectrum of(a) infrared rays(b) ultra-violet rays
(c) microwaves (d)
g-rays
7.In an elec
tromagnetic wave
(a) power is transmitted along the magnetic field
(b) power is transmitted along the electric field
(c) power is equally transferred along the electric and
magnetic fields
(d) power is transmitted in a direction perpendicular to
both the fields
8.Electromagnetic waves are transverse in nature is evident
by
(a) polarization (b) interference
(c) reflection (d) diffraction
9.If Å10=l then it corresponds to
(a) infrared (b) microwaves
(c) ultraviolet (d) X-rays
10.Intensity of electromagnetic wave will be
(a)
2
00
I c B /2=m (b)
2
00
I c B /2=e
(c)
2
00
I B /c=m (d)
2
00
I E / 2c=e
11.An electromagnetic wave passes through space and its
equation is given by E = E
0
sin (wt – kx) where E is electric
field. Energy density of electromagnetic wave in space is
(a)
2
00
1
E
2
e (b)
2
00
1
E
4
e (c)
2
00
Ee (d)
2
00
2Ee
12.The iono
sphere
(a) reflects back radiowaves in the AM band
(b) reflects back radiowaves in the FM band
(c) absorbs radiowaves in the AM band
(d) absorbs radiowaves in the FM band
13.The frequency 2 MHz belongs to
(a) visible light(b) X-rays
(c) microwaves (d) radiowaves
14.Radio waves and visible light in vacuum have
(a) same velocity but different wavelength
(b) continuous emission spectrum
(c) band absorption spectrum
(d) line emission spectrum
15.10 cm is a wavelength corresponding to the spectrum of
(a) infrared rays(b) ultra-violet rays
(c) microwaves (d)
g-rays
16.Th
e ozone layer absorbs radiation of wavelengths
(a) less than 3 × 10
–7
m
(b) more than 3 × 10
–7
m
(c) less than 3 × 10
–5
m
(d) more than 3 × 10
–5
m
17.The wave impendance of free space is
(a) zero (b) 376.6
W
(c) 33.66W (d) 3.76 W
18.The electromagnetic waves travel with a velocity
(a) equal to velocity of sound
(b) equal to velocity of light
(c) less than velocity of light
(d) None of these
19.The speed of electromagnetic wave in vacuum depends
upon the source of radiation. It
(a) increases as we move from g-rays to radio waves
(b) decreases as we move from g-rays to radio waves
(c) is same for all of them
(d) None of these
20.Maxwell in his famous equation of electromagnetism
introduced the concept of
(a) a.c. current
(b) d.c. current
(c) displacement current
(d) impedance
21.Which of the following shows green house effect ?
(a) Ultraviolet rays (b) Infrared rays
(c) X-rays (d) None of these

620. 616 PHYSICS
22.Radio wave
s diffract around building although light waves
do not. The reason is that radio waves
(a) travel with speed larger than c
(b) have much larger wavelength than light
(c) are not electromagnetic waves
(d) None of these
23.All components of the electromagnetic spectrum in vacuum
have the same
(a) energy (b) velocity
(c) wavelength (d) frequency
1.If a source is transmitting electromagnetic wave of frequency
8.2 × 10
6
Hz, then wavelength of the electromagnetic waves
transmitted from the source will be
(a) 36.6 m (b) 40.5 m
(c) 42.3 m (d) 50.9 m
2.In an apparatus, the electric field was found to oscillate with
an amplitude of 18 V/m. The magnitude of the oscillating
magnetic field will be
(a) 4 × 10
–6
T (b) 6 × 10
–8
T
(c) 9 × 10
–9
T (d) 11 × 10
–11
T
3.A TV tower has a height of 100m. How much population is
covered by the TV broadcast if the average population
density around the tower is 100 km
–2
(radius of the earth
= 6.37 × 10
6
m)
(a) 4 lakh (b) 4 billion
(c) 40,000 (d) 40 lakh
4.In a plane electromagnetic wave propagating in space has
an electric field of amplitude 9 × 10
3
V/m, then the amplitude
of the magnetic field is
(a) 2.7 × 10
12
T (b) 9.0 × 10
–3
T
(c) 3.0 × 10
–4
T (d) 3.0 × 10
–5
T
5.The ratio of electric field vector E and magnetic field vector
H i.e.,
÷
÷
ø
ö
ç
ç
è
æ
H
E
has the dime
nsions of
(a) resistance
(b) inductance
(c) capacitance
(d) product of inductance and capacitance
6.The energy of electromagnetic wave in vacuum is given by
the relation
(a)
0
2
0
2
µ2
B
2
E
+
e
(b)
2
0
2
0
Bµ
2
1
E
2
1
+e
(c)
c
BE
22
+
(d)
0
2
2
0
µ2
B
E
2
1
+e
7.An electroma
gnetic wave is propagating along Y-axis. Then
(a) oscillating electric field is along X-axis and
oscillating magnetic field is along Y-axis
(b) oscillating electric field is along Z-axis and
oscillating magnetic field is along X-axis
(c) both oscillating electric and magnetic fields are
along Y-axis, but phase difference between them
is 90°
(d) both oscillating electric and magnetic fields are
mutually perpendicular in arbitrary direction
8.The frequency of electromagnetic wave, which best suited
to observe a particle of radii 3 × 10
–4
cm is of the order of
(a) 10
15
(b) 10
14
(c) 10
13
(d) 10
12
9.If e
0
and m
0
are the electric permittivity and magnetic
permeability in vacuum, e and m are corresponding quantities
in medium, then refractive index of the medium is
(a)
0e
e
(b)
0
0
em
me
(c)
em
me
00
(d)
00me
em
10.A wave is propagating in a medium o
f electric dielectric
constant 2 and relative magnetic permeability 50. The wave
impedance of such a medium is
(a)5
W(b) 376.6 W(c) 1883W(d) 3776 W
11.The frequency of electromagnetic wave, which best suited
to observe a particle of radii 3 × 10
–4
cm is of the order of
(a) 10
15
(b) 10
14
(c) 10
13
(d) 10
12
12.The relation between electric field E and magnetic field H in
an electromagnetic wave is
(a) E = H (b)H
µ
E
0
0
e
=
(c) H
µ
E
0
0
e
= (d) H
µ
E
0
0
e
=
24.Approxima
te height of ozone layer above the ground is
(a) 60 to 70 km (b) 59 km to 80 km
(c) 70 km to 100 km (d) 100 km to 200 km
25.Biological importance of ozone layer is
(a) it stops ultraviolet rays
(b) ozone layer reduce green house effect
(c) ozone layer reflects radio waves
(d) ozone layer controls O
2
/H
2
radio in atmosphere

621. 617Electromagnetic Waves
13.To double the covering range of a T.V. transmitter tower, its
height should be made
(a) 2 times (b) 4 times
(c) 2 times (d) 8 time s
14.The transmitting antenna of a radiostation is mounted
vertically. At a point 10 km due north of the transmitter the
peak electric field is 10
–3
Vm
–1
. The magnitude of the radiated
magnetic field is
(a) 3.33 × 10
–10
T(b) 3.33 × 10
–12
T
(c) 10
–3
T (d) 3 × 10
5
T
15.It is possible to take pictures of those objects which are not
fully visible to the eye using camera films sensitive to
(a) ultraviolet rays(b) infrared rays
(c) microwaves (d) radiowaves
16.A plane electromagnetic wave is incident on a material
surface. If the wave delivers momentum p and energy E,
then
(a) p = 0, E = 0(b)
p0,E0¹¹
(c)p0,E0¹= (d)p0,E0=¹
17.An electromagnetic wave, going through vacuum is
described by
0
E E sin(kx t)= -w . Which of the following
is independent of wavelength ?
(a)k (b)w
(c) k/w (d) kw
18.In an electromagnetic wave, the electric and magnetic fields
are 100 V m
–1
and 0.265 A m
–1
. The maximum energy flow is
(a) 26.5 W/m
2
(b) 36.5 W/m
2
(c) 46.7 W/m
2
(d) 765 W/m
2
19.In which one of the following regions of the electromagnetic
spectrum will the vibrational motion of molecules give rise
to absorption ?
(a) Ultraviolet (b) Microwaves
(c) Infrared (d) Radio waves
20.The oscillating electric and magnetic vectors of an
electromagnetic wave are oriented along
(a) the same direction but differ in phase by 90°
(b) the same direction and are in phase
(c) mutually perpendicular directions and are in phase
(d) mutually perpendicular directions and differ in phase
by 90°
21.An electromagnetic wave going through vacuum is
described by E = E
0
sin(kx – wt); B = B
0
sin (kx – wt). Which
of the following equations is true?
(a)
00
EkB=w (b)
00
E Bkw=
(c)
00
EBk=w (d) None of these
22.A plane electromagnetic wave travels in free space along
x-axis. At a particular point in space, the electric field along
y-axis is 9.3V m
–1
. The magnetic induction (B) along z-axis is
(a) 3.1 × 10
–8
T (b) 3 × 10
–5
T
(c) 3 × 10
–6
T (d) 9.3 × 10
–6
T
23.Which of the following has/have zero average value in a
plane electromagnetic wave ?
(a) Both magnetic and electric field
(b) Electric field only
(c) Magnetic energy
(d) Electric energy
24.For sky wave propagation of a 10 MHz signal, what should
be the maximum electron density in ionosphere?
(a) ~1.2 × 10
12
m
–3
(b) ~10
6
m
–3
(c) ~10
14
m
–3
(d) ~10
22
m
–3
25.A new system of unit is evolved in which the values of µ
0
and
0
Î are 2 and 8 respectively. Then the speed of light in
this system will be(a) 0.25(b) 0.5 (c) 0.75 (d) 1
26.Light wave is travelling along y-direction. If the
corresponding
E
r
vector at any time is along the x-axis, the
direction of B
r
vector at that time is along
(a) y-axis
(b) x-axis
(c) + z-axis
x
y
z(d) – z-axis
27.An e
lectromagnetic wave of frequency
3.0 MHzn=
passes from vacuum into a dielectric medium with
permittivity 4.0Î= . Then
(a
) wavelength is halved and frequency remains
unchanged
(b) wavelength is doubled and frequency becomes half(c) wavelength is doubled and the frequency remains
unchanged
(d) wavelength and frequency both remain unchanged
28.In an electromagnetic wave
(a) power is transmitted along the magnetic field
(b) power is transmitted along the electric field
(c) power is equally transferred along the electric and
magnetic fields
(d) power is transmitted in a direction perpendicular to
both the fields
29.A plane electromagnetic wave is incident on a plane surface
of area A, normally and is perfectly reflected. If energy E
strikes the surface in time t then average pressure exerted
on the surface is (c = speed of light)
(a) zero (b) E/Atc
(c) 2E/Atc (d) E/c
30.If
E
r
and B
r
represent electric and magnetic field vectors of
the electromagnetic waves, then the direction of
propagation of the waves will be along

622. 618 PHYSICS
(a) EB
rr
´ (b)E
r
(c)B
r
(d) BE
rr
´
31.Which of the following statement is false for the properties
of electromagnetic waves?
(a) Both electric and magnetic field vectors attain the
maxima and minima at the same place and same time.
(b) The energy in electromagnetic wave is divided equally
between electric and magnetic vectors
(c) Both electric and magnetic field vectors are parallel to
each other and perpendicular to the direction of
propagation of wave
(d) These waves do not require any material medium for
propagation.
32.The electric and the magnetic field associated with an E.M.
wave, propagating along the +z-axis, can be represented by
(a)
00
ˆˆ
E Ei,B Bjéù==
ëû
rr
(b) 00
ˆ
E Ek,B Biéù==
ëû
rrr
(c) 00
ˆˆ
E E j,B Biéù==
ëû
rr
(d) 00
ˆˆ
E E j,B Bkéù==
ëû
rr
33.The electric field associated with an e.m. wave in vacuum is
given byˆEi=
r
40 cos (kz – 6 × 10
8
t), where E, z and t are in
volt/m, meter and seconds respectively. The value of wave
vector k is
(a) 2 m
–1
(b) 0.5 m
–1
(c) 6 m
–1
(d) 3 m
–1
34.The ratio of amplitude of magnetic field to the amplitude of
electric field for an electromagnetic wave propagating in
vacuum is equal to
(a) the speed of light in vacuum
(b) reciprocal of speed of light in vacuum
(c) the ratio of magnetic permeability to the electric
susceptibility of vacuum
(d) unity
35.The condition under which a microwave oven heats up a
food item containing water molecules most efficiently is
(a) the frequency of the microwaves has no relation with
natural frequency of water molecules.
(b) microwaves are heat waves, so always produce heating.
(c) infra-red waves produce heating in a microwave oven.
(d) the frequency of the microwaves must match the
resonant frequency of the water molecules.
36.The frequencies of X-rays, g-rays and ultraviolet rays are
respectively a, b and c then
(a) a < b, b > c (b) a > b, b > c
(c) a > b, b < c (d) a < b, b < c
37.If
vr
,ll and
m
l represent the wavelength of visible light
x-rays and microwaves respectively, then
(a)
mxv
l >l >l (b)vmy
l >l >l
(c)
mvx
l >l >l (d)
vxm
l >l >l
38.The electric and magnetic field of an electromagnetic wave
are in
(a) phase and parallel to each other
(b) opposite phase and perpendicular to each other
(c) opposite phase and parallel to each other
(d) phase and perpendicular to each other
39.The decreasing order of wavelength of infrared, microwave,
ultraviolet and gamma rays is
(a) microwave, infrared, ultraviolet, gamma rays
(b) gamma rays, ultraviolet, infrared, micro-waves
(c) microwaves, gamma rays, infrared, ultraviolet
(d) infrared, microwave, ultraviolet, gamma rays
40.Which of the following are not electromagnetic waves?
(a) cosmic rays (b)g-rays
(c)b-rays (d) X-rays.
41.Which of the following radiations has the least wavelength ?
(a)
g-rays (b)b-rays
(c)a-rays (d) X -ra ys
42.The frequency of electromagnetic wave, which is best suited
to observe a particle of radius 3 × 10
–4
cm is of the order of
(a) 10
15
(b) 10
14
(c) 10
13
(d) 10
12
43.Pick out the longest wavelength from the following types of
radiation.
(a) blue light (b) gamma rays
(c)X-rays (d) red light
DIRECTIONS for Qs. (44 to 50) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following-
(a) Statement -1 is false, Statement-2 is true
(b) Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c) Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d) Statement -1 is true, Statement-2 is false
44. Statement 1 : When variable frequency a.c. source is
connected to a capacitor, displacement current increases
with increase in frequency.
Statement 2 : As frequency increases conduction current
also increases.
45. Statement 1 : Short wave bands are used for transmission
of radio waves to a large distance.
Statement 2 : Short waves re reflected by ionosphere.
46. Statement 1 : Television signals are received through sky-
wave propagation.
Statement 2 : The ionosphere reflects electromagnetic waves
of frequencies greater than a certain critical frequency.
47. Statement 1 : Ultraviolet radiations of higher frequency
waves are dangerous to human being.
Statement 2 : Ultraviolet radiation are absorbed by the
atmosphere.
48. Statement 1 : Environmental damage has increased the
amount of ozone in the atmosphere.
Statement 2 : Increase of ozone increases the amount of
ultraviolet radiation on earth.
49. Statement 1 : The earth without atmosphere would be
inhospitably cold.
Statement 2 : All heat would escape in the absence of
atmosphere.
50. Statement 1 : Radio waves can be polarised.
Statement 2 : Sound waves in air are longitudinal in nature.

623. 619Electromagnetic Waves
Exemplar Questions
1.One requires 11 eV of energy to dissociate a carbon monoxide
molecule into carbon and oxygen atoms. The minimum
frequency of the appropriate electromagnetic radiation to
achieve the dissociation lies in
(a) visible region
(b) infrared region
(c) ultraviolet region
(d) microwave region
2.A linearly polarised electromagnetic wave given as
( )
0
ˆ
E E i coskz t= -w is incident normally on a perfectly
reflecting infinite wall at z = a. Assuming that the material ofthe wall is optically inactive, the reflected wave will be given
as
(a) ( )
r0
ˆ
EEikzt= -w
(b) ( )
r0
ˆ
EEicoskzt= +w
(c) ( )
r0
ˆ
E E i cos kz t=- +w
(d) ( )
r0
ˆ
EEisinkzt= -w
3.Light with an energy flux of 20 W/cm
2
falls on a non-reflecting
surface at normal incidence. If the surface has an area of 30
cm
2
, the total momentum delivered (for complete absorption)
during 30 min is
(a) 36 × 10
–5
kg-m/s(b) 36 × 10
–4
kg-m/s
(c) 108 × 10
4
kg-m/s(d) 1.08 × 10
7
kg-m/s
4.The electric field intensity produced by the radiations coming
from 100 W bulb at a 3 m distance is E. The electric field
intensity produced by the radiations coming from 50 W bulb
at the same distance is
(a)
E
2
(b) 2E
(c)
E
2
(d)2E
5.If E and B represent electric and magnetic field vectors of
the electromagnetic wave, the direction of propagation of
electromagnetic wave is along
(a)E (b) B
(c) B × E (d) E × B
6.The ratio of contributions made by the electric field and
magnetic field components to the intensity of an EM wave is
(a) c : 1 (b) c
2
: 1
(c) 1 : 1 (d)
c :1
7.An EM wave radiates outwards from a dipole antenna, with
E
0
as the amplitude of its electric field vector. The electric
field E
0
which transports significant energy from the source
falls off as
(a)
3
1
r
(b)
2
1
r
(c)
1
r
(d) remains constant
8.An electromagnetic wave travels in vacuum along z-direction
( )( )
12
ˆˆ
E E i E jcos kz t= + -w . Choose the correct options
from the following
(a) The associated magnetic field is given as
( )( )
12
1
ˆˆ
B E i E j cos kz t
c
= - -w
(b) The associated magnetic field is given as
( )( )
12
1
ˆˆ
B E i E j cos kz t
c
= - -w
(c) The given electromagnetic field is circularly polarised
(d) The given electromagnetic wave is plane polarised
NEET/AIPMT (2013-2017) Questions
9.An electromagnetic wave of frequency n = 3.0 MHz passes
from vacuum into a dielectric medium with relativepermittivity e = 4.0. Then [NEET Kar. 2013]
(a) wavelength is doubled and frequency is unchanged(b) wavelength is doubled and frequency becomes half(c) wavelength is halved and frequency remains
unchanged
(d) wavelength and frequency both remain unchanged
10.Light with an energy flux of 25 × 10
4
Wm
– 2
falls on a
perfectly reflecting surface at normal incidence. If the surfacearea is 15 cm
2
, the average force exerted on the surface is :
[2014]
(a) 1.25 × 10
– 6
N (b) 2.50 × 10
– 6
N
(c) 1.20 × 10
– 6
N (d) 3.0 × 10
– 6
N
11.A radiation of energy ‘E’ falls normally on a perfectlyreflecting surface. The momentum transferred to the surfaceis (C = Velocity of light) [2015]
(a)
2E
C
(b)
2
2E
C
(c)
2
E
C
(d)
E
C
12.The energy of the em waves is of the oder of 15 keV. To
which part of the spectrum does it belong? [2015 RS]
(a) Infra-red rays(b) Ultraviolet rays
(c) g-rays (d) X-rays
13.Out of the following options which one can be used toproduce a propagating electromagnetic wave ? [2016]
(a) A charge moving at constant velocity(b) A stationary charge(c) A chargeless particle(d) An accelerating charge
14.In an electromagnetic wave in free space the root meansquare value of the electric field is E
rms
= 6V/m. The peak
value of the magnetic field is :- [2017]
(a) 2.83 × 10
–8
T (b) 0.70 × 10
–8
T
(c) 4.23 × 10
–8
T (d) 1.41 × 10
–8
T

624. 620 PHYSICS
EXERCISE - 1
1. (b) 2.(a) 3.(b) 4. (c)
5. (a) 6.(c) 7.(d) 8. (a)
9. (d)
–10 –9
10Å 10 10 m 10 ml==´=
X- ray wa
velength is of the order of IÅ
10. (b)
11. (a) Energy density (EM waves)
=
2
22 0
0rms 0 00
E 1
EE
22
æö
e =e =e
ç÷
èø
12. (a) 13. (d)
14.
(a) In vacuum velocity of all EM waves are same but their
wavelengths are different.
15. (c) Microwave region wavelength = 10
–3
m to 1m
16. (a)
17. (b) Wave impedance = Z =
0
0
376.6
m
=W
e
18. (b) Velocit
y of EM waves
8
00
1
3 10 m/s = velocity of light= =´
mÎ
19. (c) Sp
eed of EM waves in vacuum
00
1
constant==
mÎ
20. (c)
21. (b
) Infrared radiations reflected by low lying clouds and
keeps the earth warm.
22. (b)
23. (b) All components of electromagnetic spectrum travel in
vacuum with velocity 3 × 10
8
m/s.
24. (a) 25. (a)
EXERCISE - 2
1. (a) Here, .m6.36
102.8
103c
6
8
=
´
´
=
n
=l
2. (b) Here, E
0
= 1
8 V/m; B
0
= ? T106
103
18
c
E
B
8
8
0
0
-
´=
´
==
3. (d)d 2hR= .
Population cove
red =
p d
2
× population density
4
. (d)
.T103
103
109
c
E
B
5
8
3
0
0
-
´=
´
´
==
5. (a)
6. (d)
2
00
1
E
2
e is electric energy density..
0
2
2
B
m
is magnetic energy de
nsity..
So, total energy =
2
2 0
00
0
B1
E
22
e+
m
7. (b)
8. (a) Si
ze of particle = 2 ×(3 × 10
–4
) = 6 × 10
–4
m. To observe a
particle, the wavelength of electromagnetic waves must
be of the size of particle.
9. (d) We know that velocity of electromagnetic wave in
vacuum
0
00
1
(v)=
me
and velocity of electromagnetic
wave in medium is
1
(v)=
me
.
Therefore ref
ractive index of the medium
0
vel. of E.M.wavein vacuum(v )
()
vel. of E.M. wavein medium(v)
m=
=
00
1/
1/
me
me
=
00
me
me
10. (c)11. (a)12. (c)
13. (b) It
is in accordance with relation =
hR2
[where h is hei
ght of antenna]
So, it must be made 4 times, to make coverage distance 2
times.
14. (b)
0
0
E
B
c
=
E
0
- Electric field, c -
speed of light, B
0
- Magnetic Field.
T1033.3
103
10
B
12
8
3
0
-
-
´=
´
=
15. (b) It is giv
en by relation =
00
1
me
16. (b) EM waves carry momentum and hence can exert
pressure on surfaces. They also transfer energy to thesurface so p ¹ 0 and E ¹ 0.
17. (c) The angular wave number
2
k
p
=
l
; where l is the
wave length
. The angular frequency is
2w = pn.
The ratio
k2/ 11
constant
2c
pl
= = ==
w pn nl
18.
(a) Here, amplitude of electric field, E
0
= 100 V/m; amplitude
of magnetic field, H
0
= 0.265 A/m. We know that the
maximum rate of energy flow,
S = E
0
× H
0
= 100 × 0.265 = 26.5 W/m
2
.
19. (b) Molecular spectra due to vibrational motion lie in the
microwave region of EM-spectrum. Due to Kirchhoff’s
law in spectroscopy the same will be absorbed.
20. (c)
E
r
and B
r
are mutually perpendicular to each other and
are in phase i.e., they become a zero and minimum at
the same place and at the same time.
21. (a)
0
0
E
c
B
=. also
2
k
p
=
l
and 2vw=p
These relation gives
00
EkB=w
Hints & Solutions

625. 621Electromagnetic Waves
22. (a) Velocity of light
8
8
E E 9.3
C B 3.1 10 T
BC 3 10
-
=Þ==
=´
´
23. (a) Both magnetic and electric fields have zero average
value in a plane e.m. wave.
24. (a) If maximum electron density of the ionosphere is N
max
per m
3
then the critical frequency f
c
is given by
f
c
= 9(N
max
)
1/2
6 1/2 123
10 10 9(N) N 1.2 10 m
-
Þ´ = Þ=´
2
5. (a) The speed of light
00
1 11
C 0.25
428
= = ==
me´
26
. (c) Light wave is an electromagnetic wave in which
E
and B are at right angles to each other as well as at
right angles to the direction of wave propagation.
27. (a) Frequency remains constant during refraction
med
00
1
2µ4
c
v==
δ
med med
air air
/21
2
v c
vc
l
= ==
l
\ wave
length is halved and frequency remains
unchanged.
28. (d)
29. (c) Incident momentum, p =
E
c
For perfectly reflecting surface with normal incidence
Dp = 2p =
2E
c
;
p 2E
F
t ct
D
==
D
F 2E
P
A ctA
==
30. (d
) Direction of propagation of Electro-magnetic waves is
perpendicular to electric field and magnetic field. Hence
direction is given by pointing
vector
0
BE
HES
m
´
=´=.
31. (c)
Electromagnetic waves are the combination of mutually
perpendicular electric and magnetic fields. So, option(c) is false.
32. (a) E.M. wave always propagates in a direction
perpendicular to both electric and magnetic fields. So,
electric and magnetic fields should be along + X and +
Y- directions respectively. Therefore, option (a) is the
correct option.
33. (a) On comparing the given equation to
E
r
=0
ˆaicos (wt – kz) , w = 6 × 10
8z
,
k =
2
rc
pw
= ork =
8
1
8
6 10
2m
3 10c
-w´
==
´
34. (b) The average energy stored in the electric field
2
0
1
2
EUE=e
The average
energy stored in the magnetic field
= U
B
=
2
0
1
,
2m
B
According to conservation of energy U
E
= U
B
2
00
2
em=
B
E
00
1
=em=
B
EC
35. (d)Required condition : Frequency of microwaves =
Resonant frequency of water molecules.
36. (a) rays x rays UV rays
VVV
g---
>>
37. (c
)
mvx
l >l >l
38. (d)
39
. (a) The decreasing order of the wavelengths is as given
below :
microwave, infrared, ultraviolet, gamma rays.
40. (c)
b-rays are the beam of fast moving electrons.
41. (a)
decreasing
RMIVUXGC
l
¾¾¾¾¾¾®
wavesRadioR® w avesMicroM®
raysredInfraI® raysVisibleV®
raystUltravioleU® raysxX®
raysGg® raysCosmicC®
Þ g rays has least wavelength
42. (b) Size of particle
c
=l=
n
8
63 10
3 10
- ´
´=
n
n = 10
14
Hz
Howev
er, when frequency is higher than this, wavelength
is still smaller. Resolution becomes better.
43. (d) Wavelength of red light is longest.
44. (b)
45. (c) Short wave (wavelength 30 km to 30 cm). These waves
are used for radio transmission and for general
communication purpose to a longer distance from
ionosphere.
46. (a) In sky wave propagation , the radio waves having
frequency range 2 MHz to 30 MHz are reflected back
by the ionosphere. Radio waves having frequency
nearly greater than 30 MHz penetrates the ionosphere
and is not reflected back by the ionosphere. The TV
signal having frequency greater than 30 MHz therefore
cannot be propagated through sky wave propagation.
In case of sky wave propagation, critical frequency is
defined as the highest frequency is returned to the
earth by the considered layer of the ionosphere after
having sent straight to it. Above this frequency, a wave
will penetrate the ionosphere and not reflected by it.
47. (c) The wavelength of these wave ranges between 4000 Å
to 100 Å that is smaller wavelength and higher
frequency. They are absorbed by atmosphere and
convert oxygen into ozone. They cause skin diseases
and they are harmful to eye and cause permanent
blindness.

626. 622 PHYSICS
48. (a)
Ozone layer in the stratosphere helps in protecting life
of organism form ultraviolet radiation on earth. Ozone
layer is depleted due to of several factors like use of
chlorofluoro carbon (CFC) which is the cause of
environmental damages.
49. (b) In the absence of atmosphere, all the heat will escape
from earth’s surface which will make earth in hospitably
cold.
50. (c) Radio waves can be polarised because they are
transverse in nature. Sound waves in air are longitudinal
in nature.
EXERCISE - 3
Exemplar Questions
1. (c)As we know that,
E =hn
As given that h = 6.62 × 10
–34
J-s
E = 11 eV =11 × 1.6 × 10
–19
n =?
11 eV =hn
So,n =
19
11 1.6 10
J
h
-
´´
=
19
34
11 1.6 10
J
6.62 10
-
-
´´
´
= 2.65 × 1
0
15
Hz
So, that frequency radiation belongs to ultraviolet region.
2. (b) The type of wave doesn't change when a wave is
reflected from denser medium but only its phase changesby 180°. As E is along positive x-axis so reflected ray will bealong negative x-axis and its component will also be oppositeto earlier in (–z) direction and phase will change.
For the reflected wave
ˆˆˆˆzz,ii=- =- and additional phase
of p in the in
cident wave.
As given that the incident electromagnetic wave is,
E = ( )
0
ˆ
Eicoskzt-w
So, the reflected electromagnetic wave is
E
r
=() ()( )
0
ˆ
Eicoskzt- - -w +p
= ( )( )
0
ˆ
Eicoskzt- - +w +p
= ( )( )
0
ˆ
Eicos kzt- p- +w
= ( )
0
ˆ
E i cos kz t+w
3.(b) As we know that
the momentum of incident light
=
U(total energy)
c
As given that the energy
flux f = 20W/cm
2
Surface are A = 30 cm
2
Time for total momentum delivered
t = 30 min =30 × 60 sec
So, total energy falling in time t sec is
U = fAt =20 × 30 × (30 × 60) J
Momentum of the incident light
=
U
c
(Q c = 3 × 10
8
)
Mome
ntum of incident light
=
( )
8
20 30 30 60
3 10
´´´
´
= 36 × 10
–4
kg-ms
–1
As no
reflection from the surface and for comidete
absorption.
Momentum of the reflected light = 0
Hence, momentum delivered to the surface
= Charge in momentum. = (p
f
– f
i
)
= 36 × 10
–4
– 0 =36 × 10
–4
kg-ms
–1
4. (c) As we know that the electric field intensity on a surface
due to incident radiation is,
I
av
µE
0
2
av
P
A
µE
0
2
(A is constant)
H
ere, P
av
µE
0
2
So, E
0
µ
av
P
\
()
()
01
02
E
E
=
()
()
av1
av2
P
P
()
()
01
02
E
E
=
100
50
=
2
1
éù
êú
ëû
()()
0021
E E /2=
5. (d)The direction
of propagation of electromagnetic wave
is perpendicular to both electric field E and magnetic field B,
i.e., in the direction of E × B by right thumb rule.
The diagram given below
z
E(x) B(–y) E(x) B(–y)
By
E(–x)
B(y)(y)
E(–x)
x
So, electromagnetic wave is along the z-direction which is
give the cross product of E and B direction is perpendicular
to E and B from EtoB
rr
. i.e., (E × B) in z-direction.
6. (c)Intensity in terms of electric field
U
av
=
2
00
1
E
2
e
Intensity in terms of magnetic field
U
av
=
2
0
0
B1
2m
We also know that the relationship between E and B is
E
0
= cB
0

627. 623Electromagnetic Waves
So the average energy by electric field is
(U
av
) =
2
00
1
E
2
e

= ()
2
000
1
E cB
2
e
=
22
0
1
cB
2
e´
00
1
c
æö
=
ç÷
me
èø
Q
\(U
av
)
Electric field
=
2
00
00
11
B
2
e´
me
=
2
0
0
B1
2m
= (U
av
)
Magnetic
field
So, the energy in electromagnetic wave is divided equally
between electric field vector and magnetic field vector.
Then, the ratio of contributions by the electric field and
magnetic field components to the intensity of an
electromagnetic wave is
Ratio =
()
()
avelectricfield
avMagnetic field
U
U
= 1 : 1
7. (
c)As we know that, the electric field is inversly
proportional to r, so
0
1
E
r
æö
µç÷
èø
From a diode antenna, an electromagnetic waves are radiated
outwards from dipole antenna with the amplitude of electricfield vector (E
0
) which transports significant energy from
the source falls off intensity inversely as the distance (r)from the antenna, i.e.,
radiated energy
0
1
E
r
æö
µç÷
èø
8. (d)In electromagnetic wave,
the electric field vector is
( )( )
12
ˆˆ
E E i E jcos kz t= + -w
and the associated magnetic field vector,
( )
12
ˆˆ
Ei EjE
B cosk zt
cc
+
= = -w
So, E and B are perpendicular to each other and the
propagation of electromagnetic wave is perpendicular
to E as well as B, so the electromagnetic wave plane
polarised.
NEET/AIPMT (2013-2017) Questions
9. (c) Given: frequency f = 2MHz, relative permittivity Î
r
= 4
From formula,
velocity '
22
r
cc l
n= = Þl=
Î
[Since fr
equency remains unchanged]
10. (b) Average force F
av
=
p 2IA
tc
D
=
D
(QPower = F.V)
=
44
8
2 25 10 15 10
3 10
-
´´ ´´
´
= 2.50 ×
10
– 6
N
11. (a) Momentum of light falling on reflecting surface p =
E
C
As surface is perfectly reflecting so momentum reflect p
1
=
–
E
C
E
C
E
P
C
=
So, momentum transferred
= P – P
1
=
EE
––
CC
æö
ç÷
èø
=
2E
C
12. (d) Energy of x-ray is (100 ev to 100 kev)
Hence energy of the order of 15 kev belongs to x-rays.
13. (d) To generate electromagnetic waves we need
accelerating charge particle.
14. (a) Given, E
rms
= 6 V/m
rms
rms
E
c
B
=
rms
rms
E
B
c
Þ= ...(i)
0
rms 0 rms
B
B B 2B
2
= Þ=
rms
0
E
B2
C
=´ From equ ation (i)
=
8
26
3 10
´
´
= 2.83
× 10
–8
T

628. 624 PHYSICS
REFLE
CTION OF LIGHT
It is the turning back of light in the same medium.
r
i
Incident ray
Reflected ray
Normal
i = angle of incidence
r = angle of reflection
Reflecting
urfaceS
Laws of Reflection
(i
) Angle of incidence (i) = angle of reflection (r)
(ii) The incident ray, reflected ray and normal are always in
same plane.
REFLECTION FROM PLANE SURFACE
Plane mirror has infinitely large radius of curvature. It produces
virtual image of same size but laterally inverted. Image is as much
behind the mirror as much is the object in front of it.
(i) If the direction of the incident ray is kept constant and the
mirror is rotated through an angle q about an axis in the
plane mirror, then the reflected ray rotates through an
angle 2q .
M
2
M
1
q
q
90° – q
(ii) If an
object moves towards (or away from) a plane mirror at
a speed v, the image will also approach (or recede) at the
same speed v, i.e. the speed of image relative to object will
be v – (–v) = 2v.
(iii) If two plane mirrors are inclined to each other at 90º, the
emergent ray is always antiparallel to incident ray if it suffers
one reflection from each whatever be the angle of incidence.
The same is found to hold good for three-plane mirrors
forming the corner of a cube if the incident light suffers one
reflection from each of them.
(iv) If there are two plane mirrors inclined to each other at an
angle q, the no. of images of a point object formed are
determined as follows :
(a) If
360°
q
is even integer (say m), no. of images formed
= (m – 1), for all positions of object.
(b) If
360°æö
ç÷
èøq
is odd integer (say m), no. of images formed,
n = m, if the object is not on the bisector of mirrors and
n = (m – 1), if the object is on the bisector of mirrors.
(c) If
360°æö
ç÷
èøq
is a fraction, the no. of images formed will
be equal to its integral part.
A plane mirror always forms a virtual image if object is
real and forms a real image if the object is virtual.
MIRROR FORMULA
111
+=
vuf
where, u = distance of the object from the pole of mirror
v = distance of the image from the pole of mirrorf = focal length of the mirror.
Mirror formula is valid only when the following
conditions are satisfied :(a) Object is placed very near to the principal axis.
(b) Object is placed far from the mirror.
Magnification

-
===
-
vvff
m
u f uf
or –
éù
==
êú
ëû
Iv
m
Ou
where, I = siz e of the image and O = size of the object and negative
sign implies that image is inverted w.r.t the object.
The above formulae are applicable only for paraxial rays (the rays
which makes very small angle with the principal axis).
24
Ray Optics and
Optical Instruments

629. 625Ray Optics and Optical Instruments
1.Rays retrace their path when their direction is reversed.
2.Focal length of a mirror depends only on the curvature of
the mirror ÷
ø
ö
ç
è
æ
=
2
R
f
. It does not depend on the material of
the mirror or on the wavelength of incident light.
3.Focal length of concave mirror is always negative.
Focal length of convex mirror is always positive.
4.The graph of
1
v
versus
1
u
for a concave mirror, if real
image is formed.
1/f
1/f
1/v
1/u
5.The gr aph shows variation of v with change in u for a
mirror.
f
2f
f2f
v
u
For
spherical mirror
For plane mirror
6.A person needs a plane mirror of minimum half of his height
to see his full image.
7.A person standing in the middle of room can see complete
wall behind him if the mirror in front of him is
3
1
rd of height
o
f wall.
8.A convex mirror is used as a rear view mirror (called driver
mirror).
9.If two or more optical components produce magnification,
then overall magnification (m) is the product of magnification
due to each component,
i.e., m = m
1
× m
2
× . . .
• If m is negative, the image is inverted
• If m is positive, the image is erect.
10.When an object moves with constant speed towards a
concave mirror from infinity to focus, the image will move
away from the mirror slower in the beginning and with the
speed of the object when it is at centre of curvature C and
faster later on.
11.Concave mirrors are used as reflectors, as objective in
reflecting telescope and by doctors (ENT) to examine ears,
nose and throat. It is also used as shaving mirrors.
12.The inability of a spherical mirror (or lens) of large aperture
to focus the paraxial rays and marginal rays to the same
point on the principal axis is called spherical aberration.
Due to this defect the image formed is blurred. This defect
can be removed by using parabolic mirror.
13.Chromatic aberration is absent in mirrors but present in
Areal magnification : When a two dimensional object is placed
with its plane perpendicular to the principal axis, its magnification
called superficial magnification or aerial magnification and is given
by
2
2

æö
= == ç÷
èø
s
Area of image v
Mm
Area of object u
Sign Conventions for Mirror and Lenses
New cartesian sign conventions :1. All the distances are measured from pole of spherical mirror
and optical centre in case of lenses.
2. The distances measured in a direction opposite to the direction
of incident light is taken as negative and vice-versa.
3. The heights measured upward and perpendicular to the
principal axis of mirror are taken as positive and vice -versa.
4. Angle measured from the normal in the anticlockwise
direction is positive and vice-versa.
+ve– ve
– ve
Incident ray
Object +ve
P
(Pole)
Reflecting or refracting surface
IMAGE FORMED BY CONCAVE AND CONVEX MIRROR
Im
age Formed by Concave Mirror
Pos ition of
object
Pos ition of
image
Magnification Nature of image
Between P
and F
Behind the
mirror
+ve, m > 1 Virtual and erect
A t F A t infin ity
– ve, Highly
magnified
Real and inverted
Between F
and C
Beyond C– ve, Magnified Real and inverted
A t C A t C m = –1 Real and inverted
Beyond C
between F
and C
Dismished Real and inverted
A t in fin ityAt F
Highly
diminshed
Real and inverted
Image Formed by Convex Mirror
Position of
object
Position of
image
Magnification Nature of image
Infront of
mirror
Between P
and F
m < +1 Virual and erect
A t in fin ityAt F m < < +1Virtual and erect
Keep in Memory

630. 626 PHYSICS
Example 3.
An obje
ct of length 2.5 cm is placed at a distance of 1.5 f
from a concave mirror where f is the magnitude of the
focal length of the mirror. The length of the object is
perpendicular to the principal axis. Find the length of the
image. Is the image erect or inverted?
Solution :
The given situation is shown in figure.
F
f
1.5f
O
The focal len
gth f = –f and u = –1.5f, we have
111
uvf
+= or
f
1
v
1
f5.1
1
-=+-
or
f3
1
f
1
f5.1
1
v
1 -
=-= or v = –3f
Now 2
f5.1
f3
u
v
m -=
-
=-= or 2
h
h
1
2
-=
or cm0.5h2h
12
-=-=
The image is 5.0 cm long. The minus sign shows that it is
inverted.
REFRACTION
Whenever a wave is bounced back into same medium at an
interface reflection is said to have occurred. Transmission of a
wave into the second medium at an interface is called refraction.
When a ray of light is passing from denser to rarer medium, it
bends away from the normal and when passing from rarer to
denser medium, it bends towards the normal.
• When a ray of light passing from one medium to another
medium frequency and phase do not change while
wavelength and velocity changes.
• Twinkling of stars, appearance of sun before actual sunrise
and after actual sunset etc. are due to atmospheric refraction.
Laws of Refraction
(i)Snell's Law : When a light ray is incident on a surface
separating two transparent media, the ray bends at the time
of changing the medium.
i.e.
12
12
21
sin
sin
m
= = =m
m
vi
rv
,
where i = angle of in
cidence
r = angle of refraction
v
1
= vel. of light in 1
st
medium
v
2
= vel. of light in 2
nd
medium
m
1
2
or
1
m
2
= refractive i
ndex of 2
nd
medium w.r.t. the 1
st
medium.
m
1
= refractive index of 1
st
medium w.r.t vacuum (or air)
m
2
= refractive index of 2
nd
medium w.r.t vacuum (or air)
(ii) The incident ray, the normal and the refracted ray at the
interface all lie in the same plane.
lenses. This is because the focal length of mirror is
independent of wavelength of light ÷
ø
ö
ç
è
æ
=
2
R
f
but that of
lens is dependent on wavelength.
14.Different colour rays travel with different velocity in a
medium but velocity of all coloured rays is same in vacuum
(and air).
15.If a hole is formed in a mirror, then also we will get full image
with no hole in the image. The hole will only reduce the
intensity of rays forming the image.
Newton’s Formula
In case of spherical mirrors if object distance (x
1
) and image
distance (x
2
) are measured from focus instead of pole,
u = (f + x
1
) and v = (f + x
2
), the mirror formula
f
1
u
1
v
1
=+
reduces to
f
1
)xf(
1
)xf(
1
12
=
+
+
+
which on simplification gives x
1
x
2

= f
2
Example 1.
The focal length of a concave mirror is 30 cm. Find the
position of the object in front of the mirror, so that the image
is three times the size of the object.
Solution :
Here image can be real or virtual. If the image is real
f = –30, u = ?, m = –3
f
m
fu
=
-
Þ
30
3
30u
-
-=
--
Þ u = –40 cm.
If
the image is virtual f
m
fu
=
-
Þ
30
3
30u
-
=
--
Þ u = –20 cm.
Example 2.
A squ
are ABCD of side 1mm is kept at distance 15 cm infront
of the concave mirror as shown in the figure. The focallength of the mirror is 10 cm. Find the perimeter of its image.
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
//
/
//
///
///////////////////
B C
A D
15cm.
Solution :
v = –
30, m =
v
u
- = – 2
\ A B C D 2 1 2mm= = ´=¢¢ ¢¢
Now
2
2
BCADv
4 B C A D 4 mm
BC AD u
¢¢ ¢¢
= ==Þ== ¢¢
¢¢
\ Perimeter = 2 + 2 + 4 + 4 = 12 mm

631. 627Ray Optics and Optical Instruments
Refractive Index of the Medium
(a)Refractive index of second medium w.r.t. first medium
2
12
1
m
m=
m
21
12
/
/
==
cvv
cvv
=
Velocity of light in first medium
Velocity of light in second medium
(b)Absolute refractive index of medium (n or µ)

sin
µ
sin
Velocity of
light in vacuum c i
n
Velocity of light in medium v r
== ==
Refractive index is the relative property of two media. If the
first medium carrying the incident ray is a vacuum, then the
ratio
Sin
Sin
i
r
is called the 'absolute refractive index of the
second medium'. The relative refractive index of any two
media is equal to the ratio of their absolute refractive indices.
Therefore, if the absolute refractive index of medium 1 and 2
be n
1
and n
2
respectively, then the refractive index of medium
2 with respect to medium 1 is
2
1 2 12
1
Sin
Sin
n i
nn
nr
= ==
n
1
sin i = n
2
si
n r
According to cauchy’s formula2
B
Am=+
l
where, A an d B are cauchy’s constant.
l
red
> l
violet
so, m
red
< m
violet
(c)12
21
1
n
n
=
(d)For thr
ee mediums 1, 2 and 3 due to successive refraction.
1
n
2
×
2
n
3
×
3
n
1
= 1
321
123
nnn
1
nnn
´´=
(e)For two mediums, n
1
and n
2
are refractive indices with respect
to vacuum, the incident and emergent rays are parallel thenn
1
sinf
1
= n
2
sinf
2
.
f
1
f
1
f
2
f
2
f
1 Medium
1
Medium
2
Medium
1
Factors affecting refractive index :
(i) Nature of the medium(ii) Wavelength(iii) Temperature of the medium-with increase in temperature,
refractive index of medium decreases.
Transmission of Wave
(i) The equation of the wave refracted or transmitted to the
next medium is given by : y = A´´ sin
()xkt
'
0
-w. This is
i
ndependent of the nature (rarer/denser) of the medium.
The wave is not inverted.
(ii) The amplitude (A´´) of the transmitted wave is less than
that (A) of the incident wave.
(iii) The angular frequency remains unchanged. However the
wave number changes. Note that the phase of the
transmitted wave is
()
'
0
t kxw- and that of t he incident
wave is (wt – kx).
(iv) The compression or rarefaction are transmitted as such and
same is the case with the crest or trough.
The wave velocity (v
p
), the angular frequency (w) and the wave
number (k) are related as v
p
= w/k = nl. Let the wave velocity in
the medium to which the wave is transmitted be v'
p
= w/k´ = nl'.
(i) If second medium is denser, in comparison to first medium
(i.e. m
2
> m
1
), then from Snell’s law
22
11
vsini
sinrv
m
==
m
here m
2
>m
1

so
12
vv>
Þ k
1
< k
2
and l
1
> l
2
.
It me
ans that if ray goes from rarer medium to denser
medium (i.e. from first medium to second medium), then
wave number increases & wavelength decreases.
(ii) If second medium is rarer in comparison to first medium,
then from Snell’s law
21
12
vsini
sinrv
m
==
m
here m
2

< m
1
so
12
vv<
Þk
1
> k
2
and l
1
< l
2
.
It me
ans that when ray goes from denser to rarer medium,
then wave number decreases & wavelength increases.
(iii)No change in wave number k occurs on reflection.
Image due to refraction at a plane surface, Apparent shift
B
r
m
1

m
2
= 1
i
A
r
Medium 1
Medium 2
t
O
I
(Denser)
(Rarer)
(a) Here O = position of object
I = position of image
2
1
1m
==
mm
Apparent depth
Real depth
i.e.
1
=
m
AI
AO
Þ =
m
t
AI
(b) The
image shifts closer to eye by an amount
OI = AO – AI or
1
1
æö
D=-
ç÷
è mø
tt where t = thickness of
medium over the object and Dt = apparent shift in its
position towards the observer.When an object in denser medium is seen through rarermedium, then apparent depth is less than real depth. Butwhen an aeroplane or bird flying is seen by an observers indenser medium, the apparent height is more by (m – 1)t

632. 628 PHYSICS
24.1
Lateral shift b
y a slab of uniform thickness t, is
sin()
cos
=-
t
x ir
r
t
r
i
m
x
The apparent shift throu gh a glass slab is in the direction of light
t
m I I¢
1
1
æö
¢==-
ç÷
è mø
II St
TOTAL INTERNAL
REFLECTION (TIR)
When the object is placed in an optically denser medium and if
the incident angle is greater than the critical angle then the ray
of light gets reflected back to the originating medium. This
phenomenon is called total internal reflection.
Critical angle (i
c
) : When a ray passes from an optically denser
medium to an optically rarer medium, the angle of refraction r is
greater than the corresponding angle of incidence i. From Snell’s law
i i
c
r = 90°
i > i
c
1
2
rsin
isin
m
m
=
Let m
1
= m and m
2
= 1 and
let for i = i
c
, r = 90º then
c
sini 1/=m
11
sin
-
\=
m
c
i ; i
c
is called the critical angle
This phenomen
on takes place in shining of air bubble, sparkling
of diamond, mirage and looming, in optical communication usingoptical fibre.
Keep in Memory
1.On travell
ing through a series of parallel layers, light follows
the following formula
332211
sinsinsin constant sin qm=qm=qm==qm
q
3
q
2
q
1
l
1
l
2
l
3
• It is important to note that the above relationship is
valid only when boundaries undeviated.
2.In case of refraction, if i = 0 then r = 0. This means that the
ray which strikes to a boundary at 90° passes through the
boundary undeviated.
3.If an object moves towards a denser medium with a
velocity v then the image moves faster with speed of
vm
as seen by the observer in denser medium.
I
mv
O v
Denser medium
If an object moves towards a rarer medium with a velocity
v then the image moves slower with a speed v/m as seen
by the observer in rarer medium.

Denser medium
I v
O
v/m
4.Denser the me
dium, smaller is the wavelength.
5.When light travels from one medium to another the
wavelength and velocity changes proportionally but
frequency of rays remains the same
6.
1 2
2
1
m
m=
m
and
b
a
c
a
b
c
m
m
=m
(‘a’ for air/va
cuum)
7.When a parallel compound slab consists of two media of
equal thickness and refractive indices
1
m and
2
m then
the equivalent refractive index
21
212
m+m
mm
=m
COMMON DEFAULT
ûIncorrect : If
a mirror or a lens is painted black on one
half, then half of image will be formed.
üCorrect : If half of the mirror or lens is blackened, we getfull image but with half the intensity.
REFRACTION AT A SPHERICAL SURFACE
For any curved spherical surfaces. Relation between u and v in
terms of refractive indices of the mediums and the radius of
curvature of the curved spherical surface.
2 1 21
m m m -m
-=
vuR
r
g
a
b
i
(Image)
(Object)
P
O m
1 m
2
Q
CIP
Spherical surface separating two media
(i) The lateral magnification in case of refraction from curved
surfaces
1
2
mæö
=
ç÷
èøm
v
m
u
(ii)Longitudinal magnification
22
1
'
m
=
m
mm

633. 629Ray Optics and Optical Instruments
m
1
is refractive index of medium 1 through which light
passes first before meeting the interface and m
2
is the
refractive index of medium 2 to which light encounters after
it passes through the interface.
REFRACTION BY A LENS
The focus point of a lens is the point where image of an object
placed at infinity is formed. And its distance from optical centre
of the lens is called focal length.
Focal length of convex lens is +ve, and of concave lens is –ve.
(i)Lens formula or thin lens formula

111
-=
vuf
(ii)Lens maker 's formula,
2
1 12
1 11
1
æöæöm
=--
ç÷ç÷
mèøèøf RR

12
11
( 1)
éù
=m--
êú
ëû
RR
where
1
21
2
m
m
=m
In the a
bove formula m
2
is refractive index of lens whereas m
1
is
the refractive index of surrounding medium.
R
1
is the radius of curvature of the lens reached first by light and
R
2
is the radius of curvature of the other surface.
Magnification : m = v/u
This relation holds for both convex and concave lenses for real
as well as virtual images.
Power of a lens, P = reciprocal of focal length expressed in metres.
i.e.,
1
(in metre)
=P
f
. Its uni t : dioptre(D).
To solve numerical problems use sign conventions
while substituting values in above equations.
Equivalent focal length of two lenses separated by distance d
1 2 12
111
=+-
d
F f f ff

f
1
f
2
d
Equivalent focal length of lens - mirror combination :
In such a case, the ray of light suffers two refraction from the lens
and one reflection from the mirror. The combination behaves as a
mirror whose focal length is given by
l
121
=+
m
Fff
lens oflength focalf
l
= , f
m
= focal length of mirror
It is important that in the above formula, we cannot apply the
sign conventions of cartesian system rather following sign
conventions are followed.
Focal length of a converging lens / mirror is taken as positiveand focal length of diverging lens/mirror is taken as negative.
Focal Length by Displacement Method
22
4
-
=
Dd
f
D
where D =
distance between an object and screen
and d = distance between two positions of lens.
D
d
I
2
I
1
Aperture of a lens : With reference to a lens, aperture means the
effective diameter of its light transmitting area. So the brightness
i.e. intensity of image formed by a lens which depends on the
light passing through the lens will depend on the square of
aperture i.e. I µ (aperture)
2
COMBINATION OF LENSES
(i) If a lens of focal length f is cut in two equal parts as shown in
figure, each part will have focal length = f
L L¢
(ii) If the above parts of lens are put in contact as shown then
the resultant focal length will be,
1112
=+=
Ffff
i.e.
2
=
f
F
LL¢
(iii)If the two parts are put as shown, then L will behave as
convergent lens of focal length f while the other (L´)
divergent of same focal length,
111
\ = + =¥
+-
orF
Fff
\ P = 0
LL¢
(iv) If a lens of focal length f is divided into equal parts as shown,
then each part will have focal length f',
i.e.
111
´2
´´
=+= orff
fff
i.e., each par
t will have focal length 2f.
(v) If these parts are put as shown, then the resultant focal
length of the combination will be

634. 630 PHYSICS
or
111
22
=+
Fff
or F = fi.e. initial value.
REFRACTION TH
ROUGH A PRISM
Prism is a transparent medium whose refracting surfaces are
inclined to each other.
(i) The angle of deviation is given by d = i + i¢ – A
where A= angle of prism. For d to be minimum, i = i¢ and
r = r¢
d
m
= A(m – 1)
R
r´
r
A
B C
U
T
Q
S
i i´
d
P
Refractive index of pris
m,
sin
2
sin
2
+dæö
ç÷
èø
m=
æö
ç÷
èø
m
A
A
where d
m
= minimu
m angle of deviation
If angle of prism A is small, than d
m
is also small.
( )/2
/2
+d
\ m=
m
A
A
Plot of ang
le of deviation (d) versus angle of incidence (i)
for a triangular prism.
d
d
m
ii = i¢i¢i
Dispersion
It is the breaking
up of white ray of light into its constituents
colours VIBGYOR. The band of seven constituents colours is
called spectrum.
Angular dispersion : It is defined as the difference of deviations
suffered by the extreme colours.
i.e.,
()q=d -d = m -m
vr
vr
A [For thin prism]
D
ispersive power : It is defined as the ratio of angular dispersion
to the mean deviation produced by the prism.
i.e.,
d -d m -m
w==
dm
vr vr
[For thin prism]
Keep in Memory
1.A ray en
tering a prism of angle A will not emerge out of
prism if
c
2Aq> where q
c
= critical angle
2.Maximum deviation through a prism will occur when angle
of incidence is 90°.
For this prism A)1(-m=d
This shows that for a small angled prism, deviation is
independent of angle of incidence.
3.Angle of emergence of a prism is 90° (called grazing
emergence) when angle of incidence
]AcosAsin)1([sini
21
--m=
-
4.A single prism produces devi
ation and dispersion
simultaneously.
5. Dispersion without deviation : When white light is incident
on a combination of two prisms of different materials andof suitable angles placed opposite to each other, theemergent light may have only dispersion without anydeviation ( of mean colour yellow).
Crown
Flint
White light
A
A¢
v
R
Y
For this to happen the conditions is
)1(
)1(
A
A
-m
-¢m
-=
¢
( For thin lenses
)
The net angular dispersion produced
d¢w-w=q)( ( For thin lenses)
6. Deviation without dispersion
W
hite light R
v
For this to happen
)(
)(
A
A
rv
rv
¢m-¢m
m-m-
=
¢
.... (1)
Net deviation ÷
ø
ö
ç
è
æ
¢w
w
-d=d1
net
Equation (1)
is said to be the condition of achromatism for
combination of two prisms.
7.Variation of refractive index of a medium with wavelength
causing incident light to split into constituent colours is
dispersion.
Cauchy’s equation :
24
m=++
ll
bc
a , where a, b and c are
con
stants.
8.Rayleigh scattering law explains blue colour of sky. Intensity
of scattered light is proportional to 1/l
4
. Hence the red
light having highest value of l
R
scatters less.

635. 631Ray Optics and Optical Instruments
9.Rainbow can be observed if light source is behind and the
droplets are in front of the observer, i.e. when the back of a
person is towards the sun.
It is a consequence of dispersion of sunlight by water
droplets due to a combinations of refraction and total internal
reflection. If the rainbow is formed after one internal
reflection in the droplets, it is called a primary rainbow. In
this the violet ray emerges at an angle of 40.8º and red rays
at an angle of 42.8º. If the rainbow is formed after two internal
reflections, it is called a secondary rainbow. In this the violet
rays emerge at 54º and red at 51º, i.e. the order of colours is
reversed. The primary rainbow is brighter than the
secondary.
10.When a point source of light is placed at a depth h below
the surface of water of refractive index m , then radius of
bright circular patch on the surface of water is given by
1
h
R
2
-m
=
h
B
11.When a lens made up of glass is immersed in water, its
focal length changes.
( )
a
g
12
1 11
1
f RR
æö
=m--
ç÷
èø
( )
w g
12
1 11
1
f RR
æö
=m--
ç÷
¢ èø
ag
ag
g
ag
g
a
g
)(
)(
1/
1/
1
1
f
f
mm-m
mm-m
=
-mm
-mm
=
-m
-m
=
¢
w
w
w
w
12.For ach
romatic combination of these lenses in contact, the
necessary condition is
0
ff
2
2
1
1
=
w
+
w
13.For two le
nses separated by distance d, spherical aberration
is minimum when d = f
1
– f
2
.
14.A convex lens forms a real image when the object is placed
beyond focus. When the object is placed between optical
centre and focus, convex lens forms a virtual image.
15.A concave lens always form a virtual image for a real object.
16.A lens is called thin when the thickness of the lens is small
compared to the object distance, image distance, radii of
curvatures of the lens.
In the case of thick lens, the problem has to be solved
using formula for each interface one by one.
17. Real image (inverted)Virtual image (erect)
Real image is formed byVirtual image is formed by actual
intersection of rays. extending the rays in the
back direction.
This image can be broughtThis image cannot be
on a screen. brought on screen.
COMMON DEFAULT
ûIncorrec
t : Using thin lens formula while the lens given in
the numerical problem is thick.
ü
Correct : The lens formula
÷
ø
ö
ç
è
æ
=-
f
1
u
1
v
1
and lens maker’ss
formula
ú
ú
û
ù
ê
ê
ë
é
÷
÷
ø
ö
ç
ç
è
æ
--m=
21
R
1
R
1
)1(
f
1
are valid on
ly for thin
lenses.
Exampl
e 4.
Consider the situation shown in figure. Find maximum
angle for which the light suffers total internal reflection
at the vertical surface.
q
q’
q’’
m = 1.0
m = 1.25
Solution :
Th
e critical angle for this case is
5
4
''sinor
5
4
sin
25.1
1
sin''
11
=q==q
--
Since '''
2
p
q = -q, we have
3
sin ' cos ''
5
q= q=
From Snell's law,
25.1
'sin
sin
=
q
q
4
3
5
3
25.1'sin25.1sinor =´=q´=q
4
3
sinor
1-
=q
If q² is greater than
the critical angle, q will be smaller than
this value. Thus, the maximum value of q¢ for which total
internal reflection takes place at the vertical surface is
sin
–1
(3/4).
m=
IB
OB
1
or, m=
m+
m=
+
IB
xt
or
IB
AOAB
1
or
t
BIx=+
m
The net shift is
÷
÷
ø
ö
ç
ç
è
æ
m
-=
÷
÷
ø
ö
ç
ç
è
æ
m
+-+=-=
1
1t
t
x)tx(IBOBIO
which
is independent of x.
Example 5.
A tank is filled with water to a height of 12.5 cm. The
apparent depth of a needle lying at the bottom of full tank
is measured by a microscope to be 9.4 cm. What is the
refractive index of water? If water is replaced by a liquid
of refractive index 1.63 upto the same height, by what
distance would the microscope have to be moved to focus

636. 632 PHYSICS
separating
surface at a distance of 100 cm from it.
Example 8.
One end of a horizontal cylindrical glass rod (
m = 1.5) of
radius 5.0 cm is rounded in the shape of a hemisphere. An
object 0.5 mm high is placed perpendicular to the axis of
the rod at a distance of 20.0 cm from the rounded edge.
Locate the image of the object and find its height.
Solution :
Taking the origin at the vertex, u = –20.0 cm and R = 5.0 cm.
We have,
Ruv
1212m-m
=
m
-
m
or
cm20
1
cm0.5
5.0
cm0.20
1
v
5.1
=+
-
=
or v = 30 cm
20.0cm 25 cm
5.0cm
The image is for
med inside the rod at a distance of 30 cm
from the vertex.
The magnification is
1
cm205.1
cm30
u
v
m
2
1
-=
´-
=
m
m
=
Thus, the image w
ill be of same height (0.5 mm) as the
object but it will be inverted.
Example 9.
A convex lens focuses a distant object on a screen placed
10 cm away from it. A glass plate (
m = 1.5) of thickness 1.5
cm is inserted between the lens and the screen. Where
should the object be placed so that its image is again
focused on the screen.
Solution :
The focal length of the lens is 10 cm. The situation with the
glass plate inserted is shown in figure. The object is placedat O. The lens would form the image at I
1
but the glass plate
intercepts the rays and forms the image at I on the screen.The shift
.cm5.0
5.1
1
1)cm5.1(
1
1tII
1
=÷
ø
ö
ç
è
æ
-=
÷
÷
ø
ö
ç
ç
è
æ
m
-=
Thus, the lens f
orms the image at a distance of 9.5 cm from
itself. Using
cm10
1
cm5.9
1
f
1
v
1
u
1
f
1
u
1
v
1
-=-=Þ=-
cm190u
=Þ
the needle again?
Solutio
n :
Here real depth =12.5 cm : apparent depth = 9.4 cm; m = ?
depthApparent
depthreal
=
m

33.1
4.9
5.12
==
Now in second ca
se, m = 1.63, real depth = 12.5 cm , apparent
depth = ? .cm67.7
63.1
5.12
yor
y
5.12
63.1 ===
Distance through
which microscope has to be moved up
= 9.4 – 7.67 = 1.73 cm.
Example 6.
A converging lens has a focal length of 20 cm in air. It is
made of a material of refractive index 1.6. If the lens is
immersed in a liquid of refractive index 1.3. What will be
the new focal length of the lens?
Solution :
R
2
)1(
f
1
1
1
-m= m
1
= 1.6, f
1
= 20
R
2
)16.1(
f
1
1
-=
or cm24
10
2026.0
R
R
26.0
20
1
=
´´
=\
´
=
R
2
)1
3.1
6.1
(
f
1
R
2
)1(
f
1
2
1
2
=-=\´-m=
52
1
12
1
3.1
3.0
24
2
)
3.1
3.16.1
(
f
1
2
=´=
-
=
f
2
= 52 cm.
Example 7.
Locate the im
age formed by refraction in the situation
shown in figure. The point C is the centre of curvature.
25 cm 20 cm
m=1.5
m=1.0
C
Solution :
We have,
2 1 21
vuR
m m m -m
-= ....(1)
Here u = –2
5 cm, R = 20 cm, m
1
= 1.0 and m
2
= 1.5
Putting the values in (1),
cm20
0.15.1
cm25
0.1
v
5.1 -
=+
or,
cm25
1
cm40
1
v
5.1
-=
or, v = –100 cm. As v is n
egative, the image is formed to the left of the

637. 633Ray Optics and Optical Instruments
\ Distance of image from the fish is mH + H +H/2 = H
(m + 3/2) below the fish.
(b) Here we have to find the images of fish as seen by the
eye.
Let h
apparent
be the apparent distance of the fish from
the surface
app.
real
h 1
h
\=
m
; app
H
h
2
m
=
m
\ Image formed is (H + H/2m) below the, eye,
i.e,
1
H1
2
æö
+
ç÷
è mø
below the eye.
Also image of fish, formed by plane mirror is H/2 below
the mirror.
2/H32/HHh
real
=+=\
(Qh
real
is distance of fish image formed by the mirror
from the surface)
Now
app
real.
h 1
h
=
m

m
=Þ
2
H3
h
app
\ Image fo
rmed is
3H
H
2
æö
+
ç÷
è mø
below the eye i.e,
3
H1
2
æö
+
ç÷
è mø
below the eye.
Example 12.
A lens has a power of +5 diopter in air. What will be its
power if completely immersed in water ?
Given gw
34
;
23
m= m=
Solution :
Let f
a
a
nd f
w
be the focal lengths of the lens in air water
respectively, then
a
a
1
P
f
=
and
w
w
w
P
f
m
=
; f
a
= 0.2
m = 20 cm
Using lens maker’s formula
ag
a 12
1 11
P ( 1)
f RR
éù
==m--
êú
ëû
...(i)
g
w w 12
1 11
1
f RR
mæö éù
=--
êúç÷
mèø ëû
Þ
w
w gw
w 12
11
P ()
f RR
éùm
==m-m-
êú
ëû
...(ii)
Dividin
g equation (ii) by equation (i), we get,
gww
ag
()P 1
P ( 1)3
m -m
==
m-
or
P 510a
PDw
333
+
=+=
OPTICAL INSTRUM
ENTS
(i) Simple Microscope
It is known as simple magnifier & consist of a convergent
Thus, the object should be placed at a distance of 190 cm
from the lens.
Example 10.
A lens is cut into two equal pieces and they are placed as
shown in figure. An object is placed at a distance of 12 cm
left from the first half lens. The focal length of original
lens was 30 cm. Find the position of final image.
O
Solution :
F
ocal length of each lens is 30 cm
For first lens u = – 12, f = 30 cm
11
111 11 1 11
v u f v u f 12 30
-=Þ =+ =-+
or, v
1
= – 20 cm.
For the second half lens image formed at v
1
acts as object.
Therefore object distance from second lens is = 20 + 20 = 40
cm or v
1
' = –40 cm
f of this lens = 30 cm
11
1111111 11 1
v v' f v v' f v 40 30 120
\- =Þ= +Þ=-+=
or v = 120 cm.Final image is 120 cm right from second lens.
Example 11.
Consider the situation in figure. The bottom of the plot is
a reflecting plane mirror. S is a small fish and T is a human
eye, refractive index of water is
m (a) At what distance (s)
from itself, will the fish see the image (s) of the eye (b) At
what distance (s) from itself will the eye see the image (s)
of the fish.
S
H
H
H/2
Solution :
(a)
We have the formula
app
app
real
h
hH
h1
m
= = =m (from the surface of water)
Now distance of fish from surface is H/2
\ Image of eye as seen by fish is = H(m + ½) above the
fish.
also the apparent image of eye, again makes an image
with the plane mirror, the apparent distance of eye is
mH + H from the plane mirror
\ Now image formed is mH + H below the plane mirror.
Distance of fish from the mirror is H/2

638. 634 PHYSICS
lens with
object between its focus & optical centre & eye
close to it.
The magnifying power of a simple microscope (M.P.) is
0
visual angle with instrument
M.P.
max v
isual angle for unaided eye
q
==
q
Here
u
h
v
h
,
D
h
1
0
==q=q so
u
D
h
D
v
h
.P.M
0
=´=
q
q
=
(a) If image is at infinity
[far point] then from lens formula
f 1
u
11
f 1
u
1
v
1
=
-
-
a
Þ=- i.e., u = f
&
=
D
M.P.
f
In this case M.P. is minimum if eye is least strained.
(b) If image is at D [near point] then u = –D
and from lens formula
f 1
u
1
v
1
=-
we get
)
f
D
1(
u
D
+= so, .1
D
MP
f
æö
=+
ç÷
èø
In this case M.P. is maximum and as final image is
close to eye, eye is under maximum strain.
(ii) Compound Microscope
M.P. of compound microscope is defined as
0

q
==
q
visual angle with the instrument
M.P
max. visual angle for unaided eye
where
D
h
0
=q,
e
1
u
h
=q
so
÷
÷
ø
ö
ç
ç
è
æ
÷
ø
ö
ç
è
æ
=´=
e
1
e
1
u
D
h
h
h
D
u
h
.P.M
(since for objecti
ve
u
v
h
h
u
v
o
T
m
1
-=Þ==, as u is –ive)
so ..
æö-
=
ç÷
èø
e
vD
MP
uu
(a)When image is formed at least distance of distinct
vision,
0
0
0
1
æö
= + =´ç÷
èø
e
e
v D
M MM
uf
(b)When the final image is fo
rmed at infinity
0
0
æö-
= ç÷
èøe
vD
M
uf
(iii) Astronomical Telescope
(a) If the final image is formed at a distance D,
0
1
-æö
=+
ç÷
èø
e
e
ff
M
fD
and length of tube is
0
=+
+
e
e
fD
Lf
fD
(b) If the final image is formed at infinity then
0
=
e
f
M
f
and the length of tube is
0
=+l
e
ff
(iv) Galilean Telescope
0
=
e
f
M
f
; Length of tube L = f
o
– f
e
(v) T
errestrial Telescope
0
=
e
f
M
f
;
Length of tube L = f
o
+ f
e
+ 4f, wh
ere f is the focal length of
erecting lens, which is used in this telescope.
Resolving Power of Microscope and Telescope
(i) The resolving power of microscope is
2 sinmq
=
l
R
where m = refra
ctive index of medium between object and
objective of microscope and q = angle subtended by a
radius of the objective on one of the objects. (When both
objects are not self illuminous).
(ii) The resolving power of a telescope is
1.22
=
l
a
R where
a = diam
eter of objective of telescope.

639. 635Ray Optics and Optical Instruments
Keep in Memory
1. R
efracting type
telescope (use of lenses)
R efl ecting type telescope
(use of mi rrors)
1 It suffers from spherical
aberration and chromatic
aberration.
1 It is almost free from
spherical aberration and
asbolutelyfreefrom
chromatic aberration.
2 The lenses used have
smallaperatureand
therefore light gathering
power is small.
2 The aperature of mirrors
used is large and therefore
light gathering power is
large.
2.
3.
4.Some common ey e defects are myopia, hypermetropia,
astigmatism and presbyopia.
5.
M yopia (or short
sightedness)
Hypermetropia (or long -
sightedness)
1 Eye can see near objects
clearly but cannot see far
objects clearly because
the light from the for off
object arriving the eye
lens may get converged in
front of retina.
1 Eye can see far off objects
clearly but cannot see near
objects clearly because the
light from the near by
object arriving the eye lens
may get converged at a
point behind the retina.
2 It can be corrected by
concave lens (power of
concave lens is –ve).
2 It can be corrected by
convex lens (Power of
convex lens is +ve).
6. Astigmatism : It is due to different curvature of cornea in
horizontal and vertical plane. It is corrected by using
cylindrical lens.
7. Presbyopia : The eye with this defect cannot see near
objects as well as far off objects clearly.
Example 13.
A small telescope has an objective of focal length 140cm
and an eye piece of focal length 5.0cm. What is the
magnifying power of the telescope for viewing distant
objects when
(a) the telescope is in normal adjustment?
(b) the final image is formed at the least distance of
distinct vision?
Solution :
Here, f
0
= 140 cm, f
e
= 5.0 cm
(a) The magnifying power in normal adjustment is given
by
28
5
140
f
f
m
e
0
===
(b) Wh
en image is formed at least distance of distinct
vision
ú
û
ù
ê
ë
é
+=
D
f
1
f
f
m
e
e
0
6.33
5
6
25
25
5
128 =´=
ú
û
ù
ê
ë
é
+=
cm25Dwhere=
Exampl
e 14.
A compound microscope has an objective of focal length
1 cm and an eyepiece of focal length 2.5 cm. An object has
to be placed at a distance of 1.2 cm away from the objective,
for normal adjustment. Find (a) the angular magnification
and (b) the length of the microscope tube.
Solution :
(a) If the first image is formed at a distance v from the
objective, then we have
cm1
1
)cm2.1(
1
v
1
=
-
-
or v = 6
cm.
The angular magnification in normal adjustment is,
.50
cm5.2
cm25
.
cm2.1
cm6
f
D
u
v
m
e
-=-==
(b) Fo
r normal adjustment, the first image must be in the
focal plane of the eyepiece.
The length of the tube is, therefore,
cm5.8cm5.2cm6fvL
e
=+=+=
Example 15.
A person cannot see objects in nearer than 500 cm from
the eye. Determine the focal length and the power of
glasses which enable him to read a book 25 cm from his
eye.
Solution :
Let f be the focal length of the glass. Then for the glass,
u = 25 cm, v = –500 cm
111
f uv
=+Q
or,
1 1 1 201
f 25 500 500
-
=-=
or,
500
f 26.3 cm
19
== = 0.263 m
P(powe
r) =
11
3.8 dioptre
f 0.263
==

640. 636 PHYSICS
Example 1
6.
What is the power of the spectacles required (a) by a
hypermetropic eye whose near point is 125 cm (b) by a
myopic eye whose far point is 50 cm ?
Solution :
(a) u = 25 cm, v = –125 cm
111
f uv
=+
or,
1 1 1 51
f 25 125 125
-
=-=
or,
125
f
4
= = 31.25 cm = 0.3125 m
11
P 3.2 dioptre
f 0.3125
===
(b) u = ¥, v = –50 cm
111
f uv
=+Q
111
fv
=+
¥
or,
1111
f 50 50
= - =-
¥
Þ f = –50 c
m = – 0.5 m
11
P 2 dioptre
f 0.5
= =- =-
Example 17
.
A person with normal vision has a range ofaccommodation from 25 cm to infinity. Over what rangewould he be able to see objects distinctly when wearing
the spectacles of a friend whose correction is +4 dioptres.
Solution :
P = 4 dioptres, \
1
f m 25 cm
4
==
For near p
oint, v = –25 cm,
v f 25 25
u 12.5cm
v f 25 25
-´
===
- --
For far poin
t, v =
¥
111
uvf
+=Q
111
u 25
+=
¥
or u = 25 cm
H
ence, the range of distinct vision is from 12.5 cm to 25 cm.
Example 18.
Where is the near point of an eye for which a spectaclelens of +1 dioptre is prescribed ?
Solution :
P = +1 dioptre.
11
f 1 m 100 cm
P1
\====
For the spe
ctacle lens,
u = 25 cm, f = 100 cm, v = ?
111
uvf
+=Q
or
1 1 1 uf
v f u uf
-
=-=
or
u f 25 100
v 33.33cm
u f 25 100
´
= = =-
--
i.e.,
the near point is 33.33 cm from the eye.
Example 19.
A certain person can see clearly at distance between 20 cm
and 200 cm from his eye. What sepctacles are required to
enable him to see distant objects clearly and what will be
his least distance of distinct vision when he is wearing them?
Solution :
For seeing distant objects
u =
¥, v = –200 cm, f = ?
Q
111
f uv
=+
or,
1111
or f 200cm
f 200 200
= - =- =-
¥
Fo
r finding the least distance of distinct vision
u = ?, v = –20 cm, f = –200cm.
vf 20( 200) (20 200)
u 22.2cm
v f 20 ( 200) 180
--´
=
= ==
- - --
i.e., his least distance of distinct vision is 22.2 cm when he
is wearing spectacles.
Example 20.
An elderly person cannot see clearly, without the use of
spectacles, objects nearer than 200 cm. What spectacles
will he need to reduce this distance to 25 cm ? If his eyes
can focus rays which are converging to points not less
than 150 cm behind them, calculate his range of distinct
vision when using the spectacles.
Solution :
Here u = 25 cm, v = – 200 cm
111
f uv
=+Q
or,
1 1 1 81
f 25 200 200
-
=-=
Þ
200
f 28.6 cm
7
==
i.e., he should us
e the converging lens of focal length
28.6 cm.
Let x be the object distance for v = 150 cm, then
1117
x 150 f 200
+ ==
Þ
1 7 1 214
x 200 150 600
-
= -=
Þ
600
x 35.3cm
17
==
\ Range of distinc
t vision is 25 cm to 35.3 cm.

641. 637Ray Optics and Optical Instruments
Example 21.
An angular magnification (magnifying power) of 30 × is
desired using an objective of focal length 1.25 cm. and an
eyepiece of focal length 5 cm. in a compound microscope.
What is the separation between objective and the eyepiece ?
Solution :
Let final image be formed at lease distance of distinct vision
For eyepiece,
e
e
25 25
M1 16
f5
æö æö
=+=+=
ç÷ç÷
èøèø
Now, M = M
0
× M
e
F
or objective,
0
e
M 30
M5
M6
-
= = =- ,
For objecti
ve, if u
0
= – x cm, v
0
= 5x cm.
Again,
111
vuf
-= or
111
5x x 1.25
-=
-
On simplificatio
n, x = 1.5
\ u
0
= – 1.5 cm, v
0
= 7. 5 cm.
For eyepiece,
e
e
e
v
M
u
=
or
e
e
e
v 25
u 4.17 cm.
M6
-
= = =-
Dista
nce between objective and eyepiece
= objective eyepiece
v | u | 7.5 4.17 11.67 cm.+ =+=
Example 22.
A sm
all telescope has an objective lens of focal length
144 cm. and eye-piece of focal length 6.0 cm. What is the
magnifying power of the telescope ? What is the separation
between objective and eye-piece ? Assume normal
adjustment.
Solution :
0
e
f144
M 24
f6
===
L = f
0
+ f
e
= (144
+ 6) cm. = 150 cm.
PHOTOMETRY
Ray optics is based on the assumption that light travels along
straight line.
(i)Luminous flux (f) of a source of light = amount of visible
light energy emitted per second from the source.
The SI unit of luminous fulx (f) is lumen.
(ii)Luminous intensity (I) of a light source = luminous flux
emitted per unit solid angle in any direction.
Its SI unit is candela.
Luminous intensity,
Df
=
Dw
I
For isot
ropic point source,
Solid angle,
2
A(surfacearea)
r
D
w=
2
2
4r
r
p
= 4=p steradian
( wh
ere DA = 4pr
2
= total surface area of sphere of radius r )
w
r
Isotropic point source
DA
o
so
4
f
=
p
I or f = 4p. I
Solid angle : W
e know that arc of a circle subtends an angle q on
the centre of circle O
o
r
q s
i.e.,
S Arcof circle
r Radiusofcircle
q== ...( i)
(a) The unit of q (plane angle) is radian.
Similarly in the case of a sphere, the surface area of spheresubtends an angle on the centre of sphere O, which is calledsolid angle & is denoted by w.
Let radius of sphere is r and a small area DA on its surface
subtends a solid angle w at the centre then
constant
r
A
2
=
D
=w ..(ii)
(b)
The unit of solid angle is steradian. If in eq (ii)
DA = r
2
, then w = 1 steradian
If DA = 4pr
2
(total surface area of sphere)
then w = 4p steradian.
(iii)Illuminance (E) of a surface is the luminous flux incident
normally on unit area of the surface.
Its unit is lux.
E
A
Df
=
D
For point source, the total normal area will be 4pr
2
,
so
22
1
4
EE
A rr
ff
== Þµ
p
(iv)Lum
inance or brightness of a surface is the luminous flux
reflected into our eyes from unit area of the surface.The unit of Brightness is lambert.
(v)Inverse square law for illuminance : Let S is a unidirectional
point source, whose luminous intensity is I. It has some
surface DA at distance r from source S.

642. 638 PHYSICS
Let central ray of source s falls perpendicularly on surface
DA, then luminous flux Df is given by
Df = I×Dw ...(i)
where
2
r
AD
=wD
2
r
A
I
D
´=fDÞ ...(ii)
or
2
Df
=
D
I
Ar
or
2
=
I
E
r
...(iii)
where E is cal
led illuminance or intensity of illumination.
If in eq. (iii) I is constant for a given source then
2
I
E.
r
µ
So intensity o
f illumination of any source is inversely
proportional to square of the distance between light source
& surface. This is called inverse square law.
Lambert's Cosine Law for Illuminance
Let S is unidirectional point source & its luminous intensity is I.
There is a surface of area DA at distance r from S, which is kept in
such a way that light from S falls obliquely on it and central ray
makes an angle q with normal to DA.
Then by fig. DA¢ = DA cos q
According to definition of luminous intensity :
Df = I × Dw
where
22
A' Acos
rr
DDq
Dw==
2
r
cosAIqD´
=fDÞ
or
2
cosDfq
==
D
I
E
A r
For any given source (I constant) & at a fixed distance (r constant)
E µ cos q
i.e., the, intensity of illumination of a surface is proportional to the
cosine of angle of the inclination of the surface. This is called
Lambert’s cosine law. As q increases, cos q decreases &
consequently E decreases.
q is the angle between normal to the area and direction of light
propagation.
Example 23.
What is the effect on the intensity of illumination on a
table if a lamp hanging 2 m directly above it is lowered by
0.5 m?
Solution :
1222
12
andE
rr
==
II
E
÷
ø
ö
ç
è
æ
=÷
ø
ö
ç
è
æ
=
÷
÷
ø
ö
ç
ç
è
æ
=\
25.2
4
5.1
2
r
r
E
E
22
2
1
1
2
Fractional increas
e in the intensity
÷
÷
ø
ö
ç
ç
è
æ
-=
-
= 1
E
E
E
EE
1
2
1
12
1001
25.2
4
´÷
ø
ö
ç
è
æ
-= = 78%
Example
24.
A lamp of power P is suspended at the centre of a circular
table of radius r. What should be the height of the lamp
above the table so that maximum intensity is produced at
the edge?
Solution :
See figure, the intensity of illumination at edge
(i.e., at point A)

2
)LA(
cos
E
q
=
I

)rh(
cos
22
+
q
=
1
I

From figure.
)rh(
h
cos
22
+
=q
322
)rh(
h
E
+
=\
I
For maximum intensit
y, dE/dh = 0
Applying this condition, we get
2rh=
24.2

643. 639Ray Optics and Optical Instruments
CONCEPT MAP
The angle of incidence
(I)is always equal to
angle of reflection (r)
i.e., i r Ð =Ð
The incident ray
the normal and the
reflected ray all lie
in the same plane
Necessary conditions
for TIR (i) ray of light
must travel from denser
to rarer medium
(ii) for two media i c Ð >Ð
The incident ray
the normal and the
refracted ray
all lie in the same
plane
Snell’s
l
a
w
s
i
n
i
µ
s
i
n
r
=
Mirror formula
111
f uv
=+
Laws of reflection
Critical angle(c) Angle in
denser medium for which
angle of refraction in rarer
medium is 90°
i
1
µ
sinC
=
Total internal
Reflection Ray totally
reflected back to denser
medium
Laws of refra
c
t
i
o
n

Relation between
f and R
R
f
2
=
Reflection of
light Turning back
of light in the same
medium after striking
the reflecting surface
or mirror
Ray optics
Optics - branch of study of
light (EM waves wavelength
400 nm to 750 nm). The path
of light (always travel in straight
line is ray of light
R
e
f
r
a
c
t
i
o
n
o
f
lig
h
t
B
e
n
d
i
n
g

o
f
l
i
g
h
t
ra
y

w
h
i
l
e

p
a
s
s
i
n
g

fro
m

o
n
e

m
e
d
i
u
m

t
o
a
n
o
t
h
e
r

m
e
d
i
u
m
R
e
f
r
a
c
t
i
v
e
i
n
d
e
x
,
µ

=

cv
r
e
a
r
d
e
p
t
h
a
p
p
a
r
e
n
t
d
e
p
t
h
=
R
e
f
r
a
c
t
i
o
n
a
t
a

s
i
n
g
l
e
s
p
h
e
r
i
c
a
l
s
u
r
f
a
c
e
2
1
2
1
v
u
R
m
m
m
-
m
-
=
RAY OPTICS AND
OPTICAL INSTRUMENTS
R
e
f
r
a
c
t
i
o
n
b
y

l
e
n
s
L
e
n
s

f
o
r
m
u
l
a
1
1
1
f
v
u
=
-
1
P
f
(
i
n
m
e
t
r
e
)
=
P
o
w
e
r
o
f

a

l
e
n
s
F
o
c
a
l
l
e
n
g
t
h

o
f
l
e
n
s
-
l
e
n
s
m
a
k
e
r
’
s

f
o
r
m
u
l
a
1
2
1
1
1
(
1
)
f
R
R
æ
ö
=
m
-
-
ç
÷
è
ø
R
e
f
r
a
c
t
i
o
n
t
h
r
o
u
g
h
P
r
i
s
m
A
n
g
l
e

o
f
d
e
v
i
a
t
i
o
n

=
A
(
µ

–
1
)
d
P
r
i
s
m
F
o
r
m
u
l
a
A
m
s
i
n
2
µ
s
i
n
A
/
2
+
d
æ
ö
ç
÷
è
ø
=
D
i
s
p
e
r
s
i
v
e

p
o
w
e
r
v
r
–
1
m
m
æ
ö
w
=
ç
÷
m
-
è
ø
Magnificatio
n
v height of image
m
u height of object
==
ffv
m
fu f
-
= =
-
Optical Instruments
Magnification
produced by
simple microscope
Microscope
Forms large
image of tiny objects
T
e
l
e
s
c
o
p
e
pro
v
i
d
e
a
n
g
u
l
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i
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c
a
t
i
o
n
o
f
di
s
t
a
n
t

o
b
j
e
c
t
s
Image formed
at near point
D
M1
f
=+
Image formed at infinity
D
M
f
=
M
a
g
n
i
f
i
c
a
t
i
o
n

b
y
com
p
o
u
n
d
m
i
c
r
o
s
c
o
p
e

When image
at near point
0 0
e
v
D
M
1
f
æ
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=
+
ç
÷
m
è
ø
W
h
e
n
f
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l

i
m
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f
m
f
=
-

644. 640 PHYSICS
1.What will be the
colour of the sky as seen from the earth if
there were no atmosphere?
(a) Black (b) Blue
(c) Orange (d) Red
2.Monochromatic light of wavelength l
1
travelling in a medium
of refractive index m
1
enters a denser medium of refractive
index m
2
. The wavelength in the second medium is
(a)l
1
(m
1
/m
2
) (b)l
1
(m
2
/m
1
)
(c)l
1
(m
2
– m
1
)/m
2
(d)l
1
(m
2
– m
1
)/m
1
3.A vessel of depth 2d cm is half filled with a liquid of refractive
index m
1
and the upper half with a liquid of refractive index m
2
.
The apparent depth of the vessel seen perpendicularly is
(a)
d
21
21
÷
÷
ø
ö
ç
ç
è
æ
m+m
mm
(b)
d
11
21
÷
÷
ø
ö
ç
ç
è
æ
m
+
m
(c)
d2
11
21
÷
÷
ø
ö
ç
ç
è
æ
m
+
m
(d)
d2
1
21
÷
÷
ø
ö
ç
ç
è
æ
mm
4.In a room con
taining smoke particles, the intensity due to a
source of light will
(a) obey the inverse square law
(b) be constant at all distances
(c) increase with distance from the source than the
inverse fourth power law
(d) fall faster with distance from the source than the
inverse fourth power law
5.What causes chromatic aberration?
(a) Non - paraxial rays
(b) Paraxial rays
(c) Variation of focal length with colour
(d) Difference in radii of curvature of the bounding surfaces
of the lens
6.Which of the following is not the case with the image formed
by a concave lens?
(a) It may be erect or inverted
(b) It may be magnified or diminished
(c) It may be real or virtual
(d) Real image may be between the pole and focus or beyond
focus
7.Critical angle of light passing from glass to water is minimum
for
(a) red colour (b) green colour
(c) yellow colour (d) violet colour
8.A normal eye is not able to see objects closer than 25 cm
because
(a) the focal length of the eye is 25 cm
(b) the distance of the retina from the eye-lens is 25 cm
(c) the eye is not able to decrease the distance between
the eye-lens and the retina beyond a limit
(d) the eye is not able to decrease the focal length beyond a limit
9.The one parameter that determines the brightness of a light
source sensed by an eye is
(a) energy of light entering the eye per second
(b) wavelength of the light
(c) total radiant flux entering the eye
(d) total luminous flux entering the eye
10.In vacuum the speed of light depends upon
(a) frequency
(b) wavelength
(c) velocity of light sources
(d) None of these
11.The intensity produced by a long cylindrical light source at
a small distance r from the source is proportional to
(a)
2
r
1
(b)
3
r
1
(c)
r
1
(d) None of these
12.The
refractive index of a piece of transparent quartz is the
greatest for
(a) violet light (b) red light
(c) green light (d) yellow light
13.Light travels through a glass plate of thickness t and having
refractive index m. If c be the velocity of light in vacuum, the
time taken by the light to travel this thickness of glass is
(a)
t
cm
(b)tcm (c)
t
c
m
(d)
tc
m
14.A convex mirr or of focal length f produces an image
(1/n)th of the size of the object. The distance of the object
from the mirror is
(a) (n – 1) f (b) f/n
(c) (n + 1) f (d) nf
15.Amount of light entering into the camera depends upon.
(a) focal length of objective lens
(b) product of focal length and diameter of the objective lens
(c) distance of object from camera
(d) aperture setting of the camera
16.In optical fibres, propagation of light is due to
(a) diffraction (b) total internal reflection
(c) reflection (d) refraction

645. 641Ray Optics and Optical Instruments
1.A lamp of 250 candle power is hanging at a distance of 6 m
from a wall. The illuminance at a point on the wall at a
minimum distance from lamp will be
(a) 9.64 lux (b) 4.69 lux
(c) 6.94 lux (d) None of these
2.An electric bulb illuminates a plane surface. The intensity
of illumination on the surface at a point 2 m away from the
bulb is 5 × 10
–4
phot (lumen/cm
2
). The line joining the bulb
to the point makes an angle of 60º with the normal to the
surface. The luminous intensity of the bulb in candela
(candle power) is
(a)
403 (b) 40
(c)
20 (d) 40 × 10
–4
3.If two mirrors are kept at 60º to each other, then the number
of images formed by them is
(a)5 (b) 6 (c)7 (d) 8
4.Wavelength of light used in an optical instrument are
l
1
= 4000Å and l
2
= 5000 Å, then ratio of their respective
resolving powers (corresponding to l
1
and l
2
) is
(a) 16 : 25 (b) 9 : 1(c) 4 : 5(d) 5 : 4
17.Rectilinear motion of light in a medium is caused due to
(a) high frequency
(b) short wavelength
(c) velocity of light
(d) uniform refractive index of the medium
18.Resolving power of a telescope increases with
(a) increase in focal length of eye-piece
(b) increase in focal length of objective
(c) increase in aperture of eye piece
(d) increase in apeture of objective
19.The distance between an object and its real image formed
by a convex lens cannot be
(a) greater than 2 f(b) less than 2 f
(c) greater than 4 f(d) less than 4 f
20.An electromagnetic radiation of frequency n, wavelength l,
travelling with velocity v in air enters in a glass slab of
refractive index (m). The frequency, wavelength and velocity
of light in the glass slab will be respectively
(a)
mm
l v
and,n (b)
m
l
v
and2,n
(
c)
mm
l
m
v
and,
n
(d) vand,
2
m
l
m
p
21.A conv
ex lens is dipped in a liquid whose refractive index is
equal to the refractive index of the lens. Then its focal length will
(a) remain unchanged
(b) become zero
(c) become infinite
(d) become small, but non-zero
22.Two thin lenses of focal lengths f
1
and f
2
are in contact and
coaxial. Its power is same as power of a single lens given by
(a)
21
21
ff
ff+
(b)
÷
÷
ø
ö
ç
ç
è
æ
2
1
f
f
(c)
÷
÷
ø
ö
ç
ç
è
æ
1
2
f
f
(d)
2
ff
21+
23.For the angl
e of minimum deviation of a prism to be equal to
its refracting angle, the prism must be made of a material
whose refractive index
(a) lies between
2and 1
(b)
lies between 2 and
2
(c) is less than 1
(d) is greater than 2
24.Which of the following is not due to total internal reflection?
(a) Working of optical fibre
(b) Difference between apparent and real depth of pond
(c) Mirage on hot summer days
(d) Brilliance of diamond
25.An astronomical telescope has a large aperture to
(a) reduce spherical aberration
(b) have high resolution
(c) increase span of observation
(d) have low dispersion
5.The critical angle for light going from medium X into medium
Y is q. The speed of light in medium X is v, then speed of
light in medium Y is
(a) v(1 – cos q) (b) v/sin q
(c) v/cos q (d) v cos q
6.A man wants to see two poles, separately, situated at 11 km.
The minimum distance (approximately) between these poles
will be
(a) 5 m (b) 0.5 m
(c) 1 m (d) 3 m
7.The index of refraction of diamond is 2.0. The velocity of
light in diamond is approximately
(a) 1.5 × 10
10
cm/sec (b) 2 × 10
10
cm/sec
(c) 3.0 × 10
10
cm/sec (d) 6 × 10
10
cm/sec
8.The luminous intensity of 100 W unidirectional bulb is 100
candela. The total luminous flux emitted from bulb will be
(a) 100 p lumen (b) 200 p lumen
(c) 300 p lumen (d) 400 p lumen

646. 642 PHYSICS
18.An achroma
tic convergent doublet of two lenses in contact
has a power of + 2D. The convex lens has power + 5D. What
is the ratio of dispersive powers of convergent and divergent
lenses ?
(a) 2 : 5(b) 3 : 5(c) 5 : 2(d) 5 : 3
19.The dispersive power of material of a lens of focal length 20
cm is 0.08. What is the longitudinal chromatic aberration of
the lens ?
(a) 0.08 cm (b) 0.08/20 cm
(c) 1.6 cm (d) 0.16 cm
20.The magnifying power of a telescope is 9. When it is
adjusted for parallel rays, the distance between the objective
and the eye piece is found to be 20 cm. The focal length of
lenses are
(a) 18 cm, 2 cm (b) 11 cm, 9 cm
(c) 10 cm, 10 cm (d) 15 cm, 5 cm
21.The focal length of the objective of a telescope is 60 cm. To
obtain a magnification of 20, the focal length of the eye
piece should be
(a) 2 cm(b) 3 cm(c) 4 cm (d) 5 cm
22.The focal lengths of objective and eye lens of an
astronomical telelscope are respectively 2 meter and 5 cm.
Final image is formed at (i) least distance of distinct vision
(ii) infinity Magnifying power in two cases will be
(a) – 48, – 40 (b) – 40, – 48
(c) – 40, + 48 (d) – 48, + 40
23.We wish to see inside an atom. Assume the atom to have a
diameter of 100 pm. This means that one must be able to
resolve a width of say 10 pm. If an electron microscope is
used the energy required should be
(a) 1.5 keV (b) 50 keV
(c) 150 keV (d) 1.5 MeV
24.Which of the following is false ?
(a) Convex lens always forms image with m < 1
(b) A simple mirror produces virtual, erect and same-sized
image
(c) A concave mirror produces virtual, erect and magnified
image
(d) A convex lens can produce real and same-sized image
25.A plane mirror reflects a beam of light to form a real image.
The incident beam is
(a) parallel (b) convergent
(c) divergent (d) any one of the above
26.An object is placed at a distance 2f from the pole of a convex
mirror of focal length f. The linear magnification is
(a)
3
1
(b)
3 2
(c)
4
3
(d) 1
27.A beam of light c
onsisting of red, green and blue colours is
incident on a right-angled prism as shown. The refractive
index of the material of the prism for the above red, green
and blue wavelengths are 1.39, 1.44 and 1.47 respectively.
The prism will
9.The refractive index of water is 1.33. What will be speed of
light in water ?
(a) 3 × 10
8
m/s (b) 2.25 × 10
8
m/s
(c) 4 × 10
8
m/s (d) 1.33 × 10
8
m/s
10.A concave mirror of focal length f produces an image n
times the size of object. If image is real, then distance of
object from mirror, is
(a) (n – 1) f (b) { (n – 1)/n} f
(c) { (n + 1)/n} f (d) (n + 1) f
11.In a concave mirror, an object is placed at a distance x
1
from
focus, and image is formed at a distance x
2
from focus. Then
focal length of mirror is
(a)
12
xx (b)
12xx
2
-
(c)
2
xx
21+
(d)
2
1
x
x
12.A convex lens of focal
length f
1
and a concave lens of focal
length f
2
are placed in contact. The focal length of the
combination is
(a) (f
1
+ f
2
) (b) (f
1
– f
2
)
(c)
12
21
ff
ff-
(d)
12
12
ff
ff+
13.A lens of power + 2 diopter is placed in contact with a lens
of power – 1 diopter. The combination will behave like(a) a convergent lens of focal length 50 cm(b) a divergent lens of focal length 100 cm(c) a convergent lens of focal length 100 cm(d) a convergent lens of focal length 200 cm
14.Light takes t
1
sec to travel a distance x in vacuum and the
same light takes t
2
sec to travel 10 cm in a medium. Critical
angle for corresponding medium will be
(a)
÷
÷
ø
ö
ç
ç
è
æ
-
xt
t10
sin
1
21
(b)
÷
÷
ø
ö
ç
ç
è
æ
-
1
21
t10
xt
sin
(c)
÷
÷
ø
ö
ç
ç
è
æ
-
xt
t10
sin
2
11
(d)
÷
÷
ø
ö
ç
ç
è
æ
-
2
11
t10
xt
sin
15.A double conve
x lens of focal length 6 cm is made of glass
of refractive index 1.5. The radius of curvature of one surface
is double that of other surface. The value of small radius of
curvature is
(a) 6 cm(b) 4.5 cm(c) 9 cm (d) 4 cm
16.A prism has a refracting angle of 60º. When placed in the
position of minimum deviation, it produces a deviation of
30º. The angle of incidence is
(a) 30º(b) 45º (c) 15º (d) 60º
17.A ray of light passes through an equilateral prism such that
the angle of incidence is equal to the angle of emergence
and the latter is equal to 3/4th of the angle of prism. The
angle of deviation is
(a) 45º(b) 39º (c) 20º (d) 30º

647. 643Ray Optics and Optical Instruments
90°
45° 45°
A
B
C
(a) separate part of the red colour from the green and
blue colours.
(b) separate part of the blue colour from the red and
green colours.
(c) separate all the three colours from one another.
(d) not separate even partially any colour from the other
two colours.
28.A concave mirror of focal length f. in vacuum is placed in a
medium of refractive index 2. Its focal length in the medium is
(a)
2
f
(b) f (c) 2 f (d) 4 f
29.Th
e maximum and minimum distance between a convex lens
and an object, for the magnification of a real image to be
greater than one are
(a) 2f and f (b) f and zero
(c)
¥ and 2f (d) 4f a nd 2f
30.A plane convex lens of focal length 16 cm, is to be made of
glass of refractive index 1.5. The radius of curvature of the
curved surface should be
(a) 8 cm(b) 12 cm (c) 16 cm (d) 24 cm
31.A real image is formed by a convex lens. If we put a concave
lens in contact with it, the combination again forms a real
image. The new image
(a) is closer to the lens system.
(b) is farther form the lens system.
(c) is at the original position.
(d) may be anywhere depending on the focal length of
the concave lens.
32.A plano-convex lens of focal length 30 cm has its plane
surface silvered. An object is placed 40 cm from the lens on
the convex side. The distance of the image from the lens is
(a) 18 cm (b) 24 cm(c) 30 cm (d) 40 cm
33.Two convex lenses of focal lengths f
1
and f
2
are mounted
coaxially separated by a distance. If the power of the
combination is zero, the distance between the lenses is
(a)
12
|f f|- (b)
12
ff+
(c)
12
12
ff
|f f|-
(d)
12
12
ff
ff+
34.If D is the deviation of a normally falling light beam on a thin
prism of angle A and d is the dispersive power of the same
prism then
(a) D is independent of A.
(b) D is independent of refractive Index.
(c)
d is independent of refractive index.
(d)d is independent of A.
35.Why is refractive index in a transparent medium greater than
one ?
(a) Because the speed of light in vaccum is always less
than speed in a transparent medium
(b) Because the speed of light in vaccum is always greater
than speed in a transparent medium
(c) Frequency of wave changes when it crosses medium
(d) None of these
36.Two convex lenses of focal lengths 0.3 m and 0.05 m are used
to make a telescope. The distance kept between the two in
order to obtain an image at infinity is
(a) 0.35 m (b) 0.25 m(c) 0.175 m (d) 0.15 m
37.The refractive indices of glass and water with respect to air
are 3/2 and 4/3 respectively. Then the refractive index of
glass with respect to water is
(a) 8/9(b) 9/8 (c) 7/6 (d) 2
38.The wavelength of a monochromatic light in vacuum is l. It
travels from vacuum to a medium of absolute refractive index
µ. The ratio of wavelength of the incident and refracted
wave is
(a)µ
2
: 1 (b) 1 : 1(c) µ : 1(d) 1 : µ
39.An object is placed at a distance of 40 cm in front of a
concave mirror of focal length 20 cm. The image produced is
(a) real, inverted and smaller in size
(b) real, inverted and of same size
(c) real and erect
(d) virtual and inverted
40.The frequency of a light wave in a material is 2 × 10
14
Hz and
wavelength is 5000 Å. The refractive index of material will
be
(a) 1.50 (b) 3.00 (c) 1.33 (d) 1.40
41.A ray incident at 15° on one refracting surface of a prism of
angle 60° suffers a deviation of 55°. What is the angle of
emergence ?
(a) 95° (b) 45°
(c) 30° (d) None of these
42.A man’s near point is 0.5 m and far point is 3 m. Power of
spectacle lenses required for (i) reading purposes, (ii) seeing
distant objects, respectively, are
(a) –2 D and + 3 D (b) +2 D and –3 D
(c) +2 D and –0.33 D(d) –2 D and + 0.33 D
43.Two light sources with equal luminous intensity are lying at
a distance of 1.2 m from each other. Where should a screen
be placed between them such that illuminance on one of its
faces is four times that on another face ?
(a) 0.2 m (b) 0.4 m (c) 0.8 m (d) 1.6 m
44.A lamp is hanging along the axis of a circular table of radius
r. At what height should the lamp be placed above the table,
so that the illuminance at the edge of the table is
1
8
of that
at its c
entre?
(a) r/2(b)
2/r(c) r/3 (d) 3/r

648. 644 PHYSICS
45.A rectangula
r block of glass is placed on a mark made on the
surface of the table and it is viewed from the vertical position
of eye. If refractive index of glass be m and its thickness d,
then the mark will appear to be raised up by
(a)
( 1)dm+
m
(b)
( 1)dm-
m
(c)
( 1)
d
m+
m
(d)
( 1)
d
m-m
46.Light passes through a glass plate of thickness d and
refractive index m. For small angle of incidence i, the lateral
displacement is
(a) id (b) id (m – 1)
(c)
i d ( 1)m-
m
(d)
id
–1
m
m
47.A glass slab of thic
kness 4 cm contains the same number of
waves as 5 cm of water when both are traversed by the same
monochromatic light. If the refractive index of water is 4/3,
what is that of glass?
(a) 5/3(b) 5/4 (c) 16/15(d) 1.5
48.An air bubble in glass slab (m = 1.5) from one side is 6 cm
and from other side is 4 cm. The thickness of glass slab is
(a) 10 cm (b) 6.67 cm
(c) 15 cm (d) None of these
49.A vessel is half filled with a liquid of refractive index m. The
other half of the vessel is filled with an immiscible liquid of
refrative index 1.5 m. The apparent depth of the vessel is
50% of the actual depth. Then m is
(a) 1.4(b) 1.5 (c) 1.6 (d) 1.67
50.A man 160 cm high stands in front of a plane mirror. His eyes are
at a height of 150 cm from the floor. Then the minimum length of
the plane mirror for him to see his full length image is
(a) 85 cm (b) 170 cm(c) 80 cm(d) 340 cm
51.It is desired to photograph the image of an object placed at
a distance of 3 m from plane mirror. The camera, which is at
a distance of 4.5 m from mirror should be focussed for a
distance of
(a) 3 m(b) 4.5 m(c) 6 m (d) 7.5 m
52.Two thin lenses are in contact and the focal length of the
combination is 80 cm. If the focal length of one lens is 20 cm,
then the power of the other lens will be
(a) 1.66 D (b) 4.00 D
(c) – 100 D (d) – 3.75 D
53.A thin convergent glass lens (m
g
= 1.5) has a power of
+ 5.0 D. When this lens is immersed in a liquid of refractive
index m, it acts as a divergent lens of focal length 100 cm.
The value of m must be
(a) 4/3(b) 5/3 (c) 5/4 (d) 6/5
54.A ray of light passes through an equilateral prism
(m = 1.5). The angle of minimum deviation is
(a) 45º(b) 37º 12' (c) 20º (d) 30º
55.Two lenses in contact form an achromatic lens. Their focal
lengths are in the ratio 2 : 3. Their dispersive powers must
be in the ratio of
(a) 1 : 3(b) 2 : 3(c) 3 : 2(d) 3 : 1
56.A combination is made of two lenses of focal length f and f'
in contact, the dispersive powers of the materials of the
lenses are w and w'. The combination is achromatic, when
(a)w = w
0
, w' = 2w
0
f¢ = 2f
(b)w = w
0
, w' = 2w
0
f¢ = f/2
(c)w = w
0
, w' = 2w
0
f¢ = –f/2
(d)w = w
0
, w' = 2w
0
f¢ = –2f
57.An achromatic convergent lens of focal length 20 cms is
made of two lenses (in contact) of materials having dispersive
powers in the ratio of 1 : 2 and having focal lengths f
1
and f
2
.
Which of the following is true ?
(a)f
1
= 10 cms, f
2
= –20 cms
(b) f
1
= 20 cms, f
2
= 10cms
(c)f
1
= –10 cms, f
2
= –20 cms
(d) f
1
= 20 cms, f
2
= –20 cms
58.A simple telescope, consisting of an objective of focal length
60 cm and a single eye lens of focal length 5 cm is focussed on
a distant object in such a way that parallel rays emerge from
the eye lens. If the object subtends an angle of 2º at the
objective, the angular width of the image is
(a) 10º(b) 24º (c) 50º (d) (1/6)º
59.An astronomical telescope has an angular magnification of
magnitude 5 for distant objects. The separation between
the objective and the eye-piece is 36 cms and the final image
is formed at infinity. The focal length f
0
of the objective and
f
e
of the eye piece are
(a)f
0
= 45 cm and f
2
= –9 cm
(b) f
0
= 50 cm and f
e
= 10 cm
(c)f
0
= 7.2 cm and f
e
= 5 cm
(d) f
0
= 30 cm and f
e
= 6 cm
60.Two lens of focal length f
1
and f
2
are kept in contact coaxially.
The resultant power of combination will be
(a)
21
21
ff
ff
-
(b)
21
21
ff
ff+
(c )
21
ff+ (d)
1
2
2
1
f
f
f
f
+
61.When white li
ght enters a prism, its gets split into its
constituent colours. This is due to
(a) high density of prism material
(b) because
m is different for different wavelength
(c) diffraction of light(d) velocity changes for different frequency
62.A pencil of light rays falls on a plane mirror and form a real
image, so the incident rays are
(a) parallel (b) diverging
(c) converging (d) statement is false

649. 645Ray Optics and Optical Instruments
63.Astronauts look down on earth surface from a space ship
parked at an altitude of 500 km. They can resolve objects of
the earth of the size (It can be assumed that the pupils
diameter is 5mm and wavelength of light is 500 nm)
(a) 0.5 m(b) 5 m (c) 50 m (d) 500 m
64.Spherical aberration in a thin lens can be reduced by
(a) using a monochromatic light
(b) using a doublet combination
(c) using a circular annular mark over the lens
(d) increasing the size of the lens
65.A lens produces an image of an object on a screen. If a slab
of refractive index n is placed in between lens and screen,
the screen has to be moved by distance d behind. The
thickness of slab is
(a) nd (b)
nd
1n-
(c)
n
d)1n(-
(d)
1n
nd
-
66.An object is m
oved along the principal axis of a converging
lens from a position 5 focal lengths from the lens to a position
that is 2 focal lengths from the lens. Which statement about
the resulting image is most accurate?
(a) The image increases in size and decreases in distance
from the lens
(b) The image increases in size and increases in distance
from the lens
(c) The image decreases in size and decreases in distance
from the lens
(d) The image decreases in size and increases in distance
from the lens
67.An object is placed upright on the axis of a thin convex lens
at a distance of four focal lengths (4f) from the center of the
lens. An inverted image appears at a distance of 4/3 f on the
other side of the lens. What is the ratio of the height of he
image of the height of the object?
(a) 1/3(b) 3/4 (c) 4/3 (d) 3/1
68.A paper, with two marks having separation d, is held normal
to the line of sight of an observer at a distance of 50m. The
diameter of the eye-lens of the observer is 2 mm. Which of
the following is the least value of d, so that the marks can be
seen as separate ? (The mean wavelength of visible light
may be taken as 5000 Å)
(a) 1.25 m (b) 12.5 cm (c) 1.25 cm(d) 2.5 mm
69.A diver inside water sees the setting sun at
(a) 41° to the horizon(b) 49° to the horizon
(c) 0° to the horizon(d) 45° to the horizon
70.A concave mirror forms the image of an object on a screen.
If the lower half of the mirror is covered with an opaque
card, the effect would be to make the
(a) image less bright.
(b) lower half of the image disappear.
(c) upper half of the image disappear.
(d) image blurred.
71.The layered lens as shown is made of two types of
transparent materials-one indicated by horizontal lines and
the other by vertical lines. The number of images formed of
an object will be
(a)1 (b) 2 (c )3 (d) 6
72.In the displacement method, a concave lens is placed in
between an object and a screen. If the magnification in the
two positions are m
1
and m
2
(m
1
> m
2
), and the distance
between the two positions of the lens is x, the focal length
of the lens is
(a)
21mm
x
+
(b)
21mm
x
-
(c)
2
21
)mm(
x
+
(d)
2
21)mm(
x
-
73.A thin lens has
focal length f, and its aperture has diameter
D. It forms an image of intensity I. If the central part of the
aperture, of diameter
2
D
, is blocked by
an opaque paper,,
the focal length of the lens and the intensity of image will
become
(a)2
I
,
2
f
(b) f,
4
I
(c)
2
I
,
4
f3
(d) f,
4
I3
74.The graph show
s the variation of magnification m produced
by a convex lens with the image distance v. The focal length
of the lens is
a
b
c
m
)b,a(
)b,ca(+
v
(a)
c
b
(b)
b
c
(c)b (d)
c
ab
75.A ray of light tr
aveling in water is incident on its surface
open to air. The angle of incidence is q, which is less than
the critical angle. Then there will be
(a) only a reflected ray and no refracted ray
(b) only a refracted ray and no reflected ray
(c) a reflected ray and a refracted ray and the angle between
them would be less than 180°–2q
(d) a reflected ray and a refracted ray and the angle between
them would be greater than 180°–2q

650. 646 PHYSICS
76.Air has refr
active index 1.0003. The thickness of air column,
which will have one more wavelength of yellow light (6000
Å) than in the same thickness of vacuum is
(a) 2 mm(b) 2 cm(c) 2 m (d) 2 km.
77.The position of final image formed by the given lens
combination from the third lens will be at a distance of
(f
1
= + 10 cm, f
2
= – 10 cm and f
3
= + 30 cm).
30 cm5 cm10 cm
(a) 15 cm (b) infinity (c) 45 cm(d) 30 cm
78.A ray of light is travelling from glass to air. (Refractiveindex of glass = 1.5). The angle of incidence is 50°.
The deviation of the ray is
(a) 0° (b) 80°
(c)
1sin 50
50 sin
1.5
- °éù
°-
êú
ëû
(d)
1sin 50
sin 50
1.5
- °éù
-°
êú
ëû
79.If f
V
and f
R
are the focal
lengths of a convex lens for violet
and red light respectively and F
V
and F
R
are the focal lengths
of concave lens for violet and red light respectively, then we
have
(a)f
V
< f
R
and F
V
> F
R
(b) f
V
< f
R
and F
V
< F
R
(c)f
V
> f
R
and F
V
> F
R
(d) f
V
> f
R
and F
V
< F
R
80.A ray is incident at an angle of incidence i on one surface of
a prism of small angle A and emerges normally from the
opposite surface. If the refractive index of the material of
prism is m, the angle of incidence i is nearly equal to
(a)
A
m
(b)
2
A
m
(c)m A (d)
2
Am
81.One face of a rectangular glass plate 6 cm thick is silvered. An
object held 8 cm in front of the first face, forms an image 12 cmbehind the silvered face. The refractive index of the glass is
(a) 0.4(b) 0.8 (c) 1.2 (d) 1.6
82.A convex lens of focal length 80 cm and a concave lens of
focal length 50 cm are combined together. What will be their
resulting power?
(a) + 6.5 D (b) –6.5 D(c) + 7.5 D(d) –0.75 D
83.A luminous object is placed at a distance of 30 cm from the
convex lens of focal length 20 cm. On the other side of the
lens, at what distance from the lens a convex mirror of radius
of curvature 10 cm be placed in order to have an upright
image of the object coincident with it?
(a) 12 cm (b) 30 cm(c) 50 cm (d) 60 cm
84.Light enters at an angle of incidence in a transparent rod of
refractive index n. For what value of the refractive index of
the material of the rod the light once entered into it will not
leave it through its lateral face whatsoever be the value of
angle of incidence?
(a)
2n>(b) n = 1 (c) n = 1.1(d) n = 1. 3
85.The radius of curvature of a thin plano-convex lens is 10 cm
(of curved surface) and the refractive index is 1.5. If the
plane surface is silvered, then it behaves like a concave
mirror of focal length
(a) 10 cm (b) 15 cm(c) 20 cm(d) 5 cm
86.An air bubble in a glass slab (m = 1.5) is 5 cm deep when
viewed from one face and 2 cm deep when viewed from the
opposite face. The thickness of the slab is
(a) 7.5 cm (b) 10.5 cm (c) 7 cm(d) 10 cm
87.A light ray falls on a rectangular glass slab as shown. The
index of refraction of the glass, if total internal reflection is
to occur at the vertical face, is
Glass
45º
(a) 2/3(b)
()
2
13+
(c)
()
2
12+
(d) 2/5
88.An equiconvex l
ens is cut into two halves along (i) XOX'
and (ii) YOY' as shown in the figure. Let f, f', f'' be the focallengths of the complete lens, of each half in case (i), and ofeach half in case (ii), respectively
Choose the correct statement from the following
(a) f ' = 2f, f '' = 2f(b) f ' = f, f '' = 2f
(c) f ' = 2f, f '' = f(d) f ' = f, f '' = f
89.A plano-convex lens is made of material of refractive index1.6. The radius of curvature of the curved surface is 60 cm.The focal length of the lens is
(a) 50 cm (b) 100 cm(c) 200 cm(d) 400 cm

651. 647Ray Optics and Optical Instruments
90.The refractive index of the material of a prism is Ö2 and its
refracting angle is 30º. One of the refracting surfaces of the
prism is made a mirror inwards. A beam of monochromatic
light enters the prism from the mirrored surface if its angle
of incidence of the prism is
(a) 30º(b) 45º (c) 60º (d) 0º
91.A telescope has an objective lens of 10 cm diameter and is
situated at a distance of one kilometer from two objects.
The minimum distance between these two objects, which
can be resolved by the telescope, when the mean
wavelength of light is 5000 Å, is of the order of
(a) 5 cm(b) 0.5 m(c) 5 m (d) 5 mm
92.The refractive index of the material of the prism is
3; then
the angle o
f minimum deviation of the prism is
(a) 30º(b) 45º (c) 60º (d) 75º
93.A ray of light travelling in a transparent medium of refractive
index
m, falls on a surface separating the medium from air at
an angle of incidence of 45°. For which of the followingvalue of
m the ray can undergo total internal reflection?
(a)m = 1.33 (b)m = 1.40 (c)m = 1.50 (d)m = 1.25
94.A biconv
ex lens has a radius of curvature of magnitude 20
cm. Which one of the following options best describe the
image formed of an object of height 2 cm placed 30 cm from
the lens?
(a) Virtual, upright, height = 1 cm
(b) Virtual, upright, height = 0.5 cm
(c) Real, inverted, height = 4 cm
(d) Real, inverted, height = 1cm
95.A thin prism of angle 15º made of glass of refractive index
µ
1
= 1.5 is combined with another prism of glass of refractive
index µ
2
= 1.75. The combination of the prism produces
dispersion without deviation. The angle of the second prism
should be
(a) 7° (b) 10° (c) 12° (d) 5°
96.A person is six feet tall. How tall must a vertical mirror be if
he is able to see his entire length?
(a) 3 ft(b) 4.5 ft (c) 7.5 ft(d) 6 ft
Directions for Qs. (97 to 100) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following-
(a)Statement -1 is false, Statement-2 is true
(b)Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c)Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d)Statement -1 is true, Statement-2 is false
97. Statement 1: Two convex lenses joined together cannot
produce an achromatic combination.
Statement 2 : The condition for achromatism is
12
12
0
ff
ww
+=
where symbols have their usual meaning.
98. Statement 1: Critical angle is minimum for violet colour.
Statement 2 : Because critical angle
1
c
1
sin
-æö
q=
ç÷
mèø
and
1
mµ
l
.
99. Sta
tement 1: Optical fibres are used to transmit light without
any appreciable loss in its intensity over distance of several
kilometers.
Statement 2 : Optical fibres are very thick and all the light is
passed through it without any loss.
100. Statement 1 : If P
1
and P
2
be the powers of two thin lenses
located coaxially in a medium of refractive index µ at a
distance d, then the power P of the combination is
P = P
1
+ P
2
– P
1
P
2
d/µ
Statement 2 : Because for above given system equivalent
focal length is given by
12
12
ff
F
f f d/
=
+-m
and
1
P
F
=.
Exemplar Questions
1.A ray of light incident at an angle q on a refracting face of a
prism emerges from the other face normally. If the angle of
the prism is 5° and the prism is made of a material of refractive
index 1.5, the angle of incidence is
(a) 7.5° (b) 5°
(c) 15° (d) 2.5°
2.A short pulse of white light is incident from air to a glass
slab at normal incidence. After travelling through the slab,
the first colour to emerge is
(a) blue (b) green
(c) violet (d) red
3.An object approaches a convergent lens from the left of the
lens with a uniform speed 5 m/s and stops at the focus. The
image
(a) moves away from the lens with an uniform speed 5 m/s
(b) moves away from the lens with an uniform acceleration
(c) moves away from the lens with a non-uniform acceleration
(d) moves towards the lens with a non-uniform acceleration

652. 648 PHYSICS
4.A passenge
r in an aeroplane shall
(a) never see a rainbow
(b) may see a primary and a secondary rainbow as
concentric circles
(c) may see a primary and a secondary rainbow as
concentric arcs
(d) shall never see a secondary rainbow
5.You are given four sources of light each one providing a
light of a single colour - red, blue, green and yellow. Suppose
the angle of refraction for a beam of yellow light
corresponding to a particular angle of incidence at the interface
of two media is 90°. Which of the following statements is
correct if the source of yellow light is replaced with that of
other lights without changing the angle of incidence?
(a) The beam of red light would undergo total internal
reflection
(b) The beam of red light would bend towards normal while
it gets refracted through the second medium
(c) The beam of blue light would undergo total internal
reflection
(d) The beam of green light would bend away from the
normal as it gets refracted through the second medium
6.The radius of curvature of the curved surface of a plano-
convex lens is 20 cm. If the refractive index of the material of
the lens be 1.5, it will
(a) act as a convex lens only for the objects that lie on its
curved side
(b) act as a concave lens for the objects that lie on its curved
side
(c) act as a convex lens irrespective of the side on which
the object lies
(d) act as a concave lens irrespective of side on which the
object lies
7.The phenomena involved in the reflection of radiowaves by
ionosphere is similar to
(a) reflection of light by a plane mirror
(b) total internal reflection of light in air during a mirage
(c) dispersion of light by water molecules during the
formation of a rainbow
(d) scattering of light by the particles of air
8.The direction of ray of light incident on a concave mirror is
shown by PQ while directions in which the ray would travel
after reflection is shown by four rays marked 1, 2, 3 and 4
(figure). Which of the four rays correctly shows the direction
of reflected ray?
1
Q
42
C F
3 P
(a)1 (b) 2 (c)3 (d) 4
9.The optic
al density of turpentine is higher than tnat of water
while its mass density is lower. Figure shows a layer of
turpentine floating over water in a container. For which one
of the four rays incident on turpentine in figure, the path
shown is correct?
(a)1 (b) 2 (c)3 (d) 4
12 34
Air
T
W
G
10.A car is mov
ing with at a constant speed of
60 km h
–1
on a straight road. Looking at the rear view mirror,
the driver finds that the car following him is at a distance of100 m and is approaching with a speed of 5 kmh
–1
.
In order to keep track of the car in the rear, the driver beginsto glance alternatively at the rear and side mirror of his carafter every 2 s till the other car overtakes. If the two carswere maintaining their speeds, which of the followingstatement (s) is/are correct?
(a) The speed of the car in the rear is 65 km h
–1
(b) In the side mirror, the car in the rear would appear to
approach with a speed of 5 kmh
–1
to the driver of the
leading car
(c) In the rear view mirror, the speed of the approaching car
would appear to decrease as the distance between the
cars decreases
(d) In the side mirror, the speed of the approaching car would
appear to increase as the distance between the cars decreases
11.There are certain material developed in laboratories which
have a negative refractive index figure. A ray incident from
air (Medium 1) into such a medium (Medium 2) shall follow a
path given by
i
r
1
2
(a)
2
1
i
r
(b)
(c)
ir
2
1
(d)
2
1

653. 649Ray Optics and Optical Instruments
NEET/AIPMT (2013-2017) Questions
12.A plano convex lens fits exactly into a plano concave lens.
Their plane surfaces are parallel to each other. If lenses are
made of different materials of refractive indices m
1
and m
2
and R is the radius of curvature of the curved surface of the
lenses, then the focal length of the combination is [2013]
(a)
( )
12
2
R
m -m
(b)
( )12
R
m -m
(c)
( )21
2R
m -m
(d)
( )122m +m
R
13.For a norm
al eye, the cornea of eye provides a converging
power of 40D and the least converging power of the eyelens behind the cornea is 20D. Using this information, thedistance between the retina and the eye lens of the eye canbe estimated to be [2013]
(a) 2.5 cm (b) 1.67 cm
(c) 1.5 cm (d) 5 cm
14.Two plane mirrors are inclined at 70°. A ray incident on onemirror at angle q after reflection falls on second mirror and is
reflected from there parallel to first mirror. The value of q is
[NEET Kar. 2013]
(a) 50° (b) 45°
(c) 30° (d) 55°
15.The reddish appearance of the sun at sunrise and sunset isdue to [NEET Kar. 2013]
(a) the colour of the sky
(b) the scattering of light
(c) the polarisation of light
(d) the colour of the sun
16.If the focal length of objective lens is increased then
magnifying power of : [2014]
(a) microscope will increase but that of telescope decrease.
(b) microscope and telescope both will increase.
(c) microscope and telescope both will decrease
(d) microscope will decrease but that of telescope increase.
17.The angle of a prism is ‘A’. One of its refracting surfaces is
silvered. Light rays falling at an angle of incidence 2A on
the first surface returns back through the same path after
suffering reflection at the silvered surface. The refractive
index m, of the prism is : [2014]
(a) 2 sin A (b) 2 cos A
(c)
1
cosA
2
(d) tan A
18.T
he refracting angle of a prism is ‘A’, and refractive index of
the material of the prism is cot(A/2). The angle of minimumdeviation is : [2015]
(a) 180° – 2A (b) 90° – A
(c) 180° + 2A (d) 180° – 3A
19.Two identical thin plano-convex glass lenses (refractive
index 1.5) each having radius of curvature of 20 cm are placed
with their convex surfaces in contact at the centre. The
intervening space is filled with oil of refractive index 1.7.
The focal length of the combination is [2015]
(a) –25 cm (b) –50 cm
(c) 50 cm (d) –20 cm
20.A beam of light consisting of red, green and blue colours is
incident on a right angled prism. The refractive index of the
material of the prism for the above red, green and blue
wavelengths are 1.39, 1.44 and 1.47, respectively. [2015 RS]
45°
Blue
Green
Red
The prism will:
(
a) separate all the three colours from one another
(b) not separate the three colours at all(c) separate the red colour part from the green and blue
colours
(d) separate the blue colour part from the red and green
colours
21.In an astronomical telescope in normal adjustment a straight
black line of lenght L is drawn on inside part of objective
lens. The eye-piece forms a real image of this line. The length
of this image is l. The magnification of the telescope is :
(a)
L
1
I
- (b)
LI
LI
+
-
[2015 RS]
(c)
L
I
(d)
L
1
I
+
22.The angle of incidence for a ray of light at a refracting surface
of a prism is 45° . The angle of prism is 60°. If the ray suffersminimum deviation through the prism, the angle of minimum
deviation and refractive index of the material of the prism
respectively, are : [2016]
(a)
1
45,
2
° (b)30,2°
(c)45,2° (d)
1
30,
2
°
23.A astronom
ical telescope has objective and eyepiece of focal
lengths 40 cm and 4 cm respectively. To view an object 200cm away from the objective, the lenses must be separatedby a distance : [2016]
(a) 37.3 cm (b) 46.0 cm
(c) 50.0 cm (d) 54.0 cm
24.Match the corresponding entries of column-1 with column-2(Where m is the magnification produced by the mirror):
Column-1 Column-2 [2016]
(A) m = –2 (a) Convex mirror
(B) m =
1
2
- (b) Concave mirror
(C) m = +2 (c) Real image
(D) m = +
1
2
(d) Virtual image

654. 650 PHYSICS
(a) A ® b
and c, B ®b and c, C ® b and d,
D ® a and d.
(b) A ® a and c, B ® a and d, C ® a and b,
D ® c and d
(c) A ® a and d, B ® b and c, C ® b and d,
D ® b and c
(d) A ® c and d, B ® b and d, C ® b and c,
D ® a and d
25.A beam of light from a source L is incident normally on a
plane mirror fixed at a certain distance x from the source.
The beam is reflected back as a spot on a scale placed just
above the source I. When the mirror is rotated through a
small angle q, the spot of the light is found to move through
a distance y on the scale. The angle q is given by [2017]
(a)
y
x
(b)
x
2y
(c)
x
y
(d)
y
2x
26.A thin prism having refracting angle 10° is made of glass of
refractive index 1.42. This prism is combined with anotherthin prism of glass of refractive index 1.7. This combinationproduces dispersion without deviation. The refracting angleof second prism should be [2017]
(a) 6° (b) 8°
(c) 10° (d) 4°

655. 651Ray Optics and Optical Instruments
EXERCISE - 1
1. (a) As no scattering of light occurs, sky appears dark.
2. (a)
a
c
l=
n
or
mm
m
vcc
()
v
l= = m=
n mn
Q
\
nm
=l
1
1
c
and
nm
=l
2
2
c
or
2211
ml=ml or )/(
2
112
mml=l
3. (b) Apparent depth = d/m
1
+ d/m
2
4. (d) 5.(c) 6.(d) 7.(d)
8. (d) Because, the focal length of eye lens can not decrease
beyond a certain limit.
9. (d) It is the total luminous flux.
10. (d)
11. (c) The intensity of cylindrical source at small distance r is
inversely proportional to r.
1
I
r
µ (since
21
A &IA
r
µµ )
12. (a)
1
,mµ
l
red violet
l >l
13. (c)
plateglassinlightofvelocity
vacuuminlightofvelocity
=m
or m =
c
c
¢
or
m
=¢
c
c
Time taken = distan
ce/velocity =
c
t
)/c/(t
m
=m
14. (a)
)u(
v
n
1
m
-
-
==
u
v
n
Þ=
As
f
1
u
1
v
1
=+ \
f
1
u
1
u
n
=-
u = (n – 1) f
15. (
d) Amount of light entering into the camera depends upon
aperture setting the camera.
16. (b) Optical fibre is a device which transmits light introduced
at one end to the opposite end, with little loss of thelight through the sides of the fibre. It is possible with thehelp of total internal reflection.

17. (d) If the medium is heterogeneous having a gradient of
refractive index. Then light rays will not follow arectilinear (straight line path).
18. (d) Resolving power
l
l
=
d
plane transmission grating
Resolving power for telescope
resolution oflimit
1
=
l
=
22.1
d
l
0
d
d
=
by incre
asing the aperture of objective resolving power
can be increased.
19. (d) Least distance is 4f when object is at radius of curvature,
and greatest is infinity.
20. (a) When electromagnetic wave enters in other medium,
frequency reamains unchanged while wavelength and
velocity become
m
1
times.
so, e
.m. entering from air to glass slab (m ), frequency
remains n, wavelength
m
l
=l'
veloc
ity of light in medium
m
=
v
'v
21. (c)( )
÷
÷
ø
ö
ç
ç
è
æ
--m=
2
g
R
1
R
1
1
f
1
l
where 1
g
=m
l is gi
ven, we get
Þ 0
R
1
R
1
)11(
f 1
21
=
÷
÷
ø
ö
ç
ç
è
æ
--= or
Þ ¥=f
22. (a)
21
12
21 ff
ff
f
1
f
1
F
1 +
=+=
;
21
21
ff
ff
F 1
P
+
==
23. (b)
A
A
Prism angle
S
min
Angle of
minimum
deviation
e Angle of
emergence
B C
Incident
angle
C
r
1
r
2
The angle of minimum deviation is given as
min
d= i + e–A
for mi
nimum deviation
min
d= A then
2A = i +
e
in case of
min
d i = e
2A = 2i r
1
= r
2
=
2
A
i = A = 90°
from s
mell’s law
1 sin i = n sin r
1
sin A = n sin
2
A
Hints & Solutions

656. 652 PHYSICS
2 sin cos = sin
222
AAA
n
2 cos =
2
A
n
when A = 90° = i
min
then n
m
in
=
2
i = A = 0 n
max
= 2
24. (b
) Difference between apparent and real depth of a pond
is due to the refraction of light, not due to the total
internal reflection. Other three phenomena are due to
the total internal reflection.
25. (b) Large aperture increases the amount of light gathered
by the telescope increasing the resolution.
EXERCISE - 2
1. (c) At minimum distance, incidence is normal. Therefore,
22
I 250
E 6.94 lux
r6
===
2. (b) r = 2 m = 200
cm
E = 5 × 10
–4
phot, q = 60º
From,
2
r
cosI
E
q
= ,
2 42
E r 5 10 (200)
I
cos cos 60
-
´
==
q
= 40 C .P
3. (a) Number of images 1
º360
)n(
1
-
q
=
where q = ang
le between mirrors
Here, q = 60º
So, number of images
1
º60
º360
n
1
-= = 5
4. (d) Re
solving power of an optical instrument
l
µ
1
1
2
2
1
atpower Resolving
λatpower Resolving
l
l
=
l
÷
÷
ø
ö
ç ç
è
æ
µ
powerresolving
1
resolutionofLimit
\ Ra
tio of resolving power =
4000
5000
= 4:5
4
5
=
5. (b)
6 (d) 7.(a)
8. (d)f=p
=p = p
4 I 4 (100) 400 lumen .
9. (b)
33.1
1
v
v
2
1
1
2
=
m
m
= or
81
2
v
v 2.25 10 m /s
1.33
==´
10. (c) Gi
ven v = nu As
f
1
u
1
v
1
=+
\
f
1
u 1
nu
1
=+ or
n
f)1n(
u
+
=
11. (a) Her
e, u = f + x
1
, n = f + x
2
use
n+
n
=
u
u
f and solve to get
21
xxf=
12. (c)
21
12
21 ff
ff
f
1
f
1
f
1 -
=
-
+= ;
12
21
ff
ff
f
-
=
13. (c) P = P
1
+ P
2
=
+ 2 – 1 = + 1 dioptre, lens behaves as
convergent
cm100m1
1
1
P
1
F ====
14. (c)
21 t
10
v,
t
x
c ==
x
t
t
10
c v1
isin
1
2
c ´==
m
=
;
÷
÷
ø
ö
ç
ç
è
æ
=
-
xt
t10
sini
2
11
c
15. (b) If
R
1
= R, R
2
= –2 R
÷
÷
ø
ö
ç
ç
è
æ
--m=
21R
1
R
1
)1(
f
1
R2
35.0
R2
1
R
1
)15.1(
6
1 ´
=
÷
÷
ø
ö
ç
ç
è
æ
+-=
R = 4.5 cm
16
. (b)
º45
2
3060
2
A
i
m
=
+
=
d+
=
17. (d) A
4
3
ii
21==
As
21
iiA +=d+
\ AA
4
3
A
4
3
Aii
21 -+=-+=d = º30
2
º60
2
A
==
18. (b) Here, P
1
=
5 D
P
2
= P – P
1
= 2 – 5 = –3 D
5
3
P
P
f
f
1
2
2
1
2
1
=
-
=-=
w
w
19. (c) Longitudinal chromatic aberration = w f
= 0.08 × 20 = 1.6 cm
20. (a) 9
f
f
e
0
=, \
0e
f 9f=
Also f
0
+ f
e
= 20 (Qfinal image is at infinity)
9 f
e
+ f
e
= 20, f
e
= 2 cm,\ f
0
= 18 cm
21. (b) In normal adjustment,
20
f
f
M
e
0
==, cm3
20
60
20
f
f
0
e ===
22. (a) (i) ÷
ø
ö
ç è
æ
+-=
d
f
1
f
f
M
e
e
0
= 48
25
5
1
5
200
-=÷
ø
ö
ç
è
æ
+-
(since le
ast distance d = 25cm)
(ii)
40
5
200
f
f
M
e
0
-=-=-=
23. (b)
24. (a
) Convex lens can form image with m < 1, m > 1 and
m = 1 depending upon the position of the object.
Convex lens forms magnified image (m > 1) when the

657. 653Ray Optics and Optical Instruments
object is pole and 2f, same size as the object (m = 1)
when the object is at 2f and smaller image (m < 1), when
the object is beyond 2f.
25. (b) Virtual object forms real image on a plane mirror, so
rays convergent.
26. (a)
f
3
2
v
f2
3
v
1
f
1
f2
1
v
1
=Þ=Þ=-
3
1
f2
f
3
2
u
v
m ===\
27. (a)
Difference in refractive indices of blue and green colour
are less so they are seen together and red is seen separate
because deviation depends on refractive index.
28. (b) Medium doesn't effect focal length of a mirror.
29. (a)
30. (a)
÷
÷
ø
ö
ç
ç
è
æ
--m=
21R
1
R
1
)1(
f
1
÷
ø
ö
ç
è
æ
¥
--=
1
R
1
)15.1(
16
1
cm8R
R
1
5.0
16
1
=Þ´=Þ
31. (b)
When we bring in contact a concave lens the effective
focal length of the combination decreases.
f
1
u
1
v
1
f
1
u
1
v
1
+=Þ=-
accordin
g to above relation as f reduces, v increases.
32. (b) 33. (b) 34. (d) 35. (b)
36. (a) In a telescope, to obtain an image at infinity or in normal
adjustment, the distance between two convex lenses
one called objective (greater focal length) and the other
called eye piece (shorter focal length) is L.
L = f
0
+ f
e
= 0.3 + 0.05 = 0.35 m.
37. (b) given :
aa
gw
34
,
23
m= m=

awa
w gg
m ´ m=mQ
\
a
gw
g
a
w
3/29
.
4/38
m
m= ==
m
38. (c)
39
. (b) Object distance u = – 40 cm
Focal length f = – 20 cm
According to mirror formula
111 111
or
uvf vfu
+= =-
or
()
11 1 11
v 20 40 20 40
+- =+
---
1211
or v 40cm.
v 40 40
-+
= =- =-
N
egative sign shows that image is infront of concave
mirror. The image is real.
Magnification,
()
()
40v
m1
u 40
--
= =- =-
-
The image is of
the same size and inverted.
40. (b)
velocity of light in vacuum (c)
velocity of light in medium (v)
m=
But v = ul = 2
× 10
14
× 5000 × 10
–10
In the medium, v = 10
8
m/s.
8
vac
8
med
v 3 10
3.
v 10
´
\m===
41. (d) Here, i
1
= 15°, A = 60°, d = 55°, i
2
= e = ?
As i
1
+ i
2
= A + d
i
2
= A + d – i
1
= 60° + 55° – 15° = 100°.
42. (c) For reading purposes :
u = – 25 cm, v = – 50 cm, f = ?
50
1
25
1
50
1
u
1
v
1
f
1
=+-=-= ; D2
f
100
P +==
For distant vis
ion, f' = distance of far point = –3 m
D33.0D
3
1
f
1
P -=-=
¢
=
43. (c) E
2
=
4 E
1
. If x is distance from 1st source,
then,
22
Ι 4Ι
(1.2x)x
=
-
or
x
2
x2.1
1
=
-
3x = 2.
4, x = 0.8 m
44. (d)
12
E
8 1
E= or,
2
22
22
h
1
8
1
hr
h
)hr(
1
=
+
´
+
(by Lambert's co
sine law)
or,
32/322
)h2()hr(=+ or, h2)hr(
2/122
=+
or, r
2
+ h
2
= 4 h
2 3/rh=
45. (b)
Since
Apparent depth 1
Realdepth
=
m
Þ Apparent depth = d/m
So mark raised up = Real depth – Apparent depth
= ÷
÷
ø
ö
ç
ç
è
æ
m
-=
m
-
1
1d
d
d

d
1
÷
÷
ø
ö
ç
ç
è
æ
m
-m
=
46. (c
) From figure
d
m
r
i
(i–r)
B
x
D
C
)ri()ri(sin
BC
x
-»-= ...(1)
Furthe
r,
1rcos
BC
d
»=
(When i i
s small r is small)
\ dBC»
From eq. (1),
x
(i r)
d
»- or x d(i r)»-

658. 654 PHYSICS
or ÷
ø
ö
ç
è
æ
-»
i
r
1dix
Now
m==
r
i
rsin
isin
\
÷
÷
ø
ö
ç
ç
è
æ
m
-»
1
1dix
47. (a) Giv
en that
4
5
gw
=m and
3
4
wa
=m
\
3 5
3 4
4 5
wagwga
=´=m´m=m
48. (c) We
know that
depthapparent
depthreal
=m
Let the
thickness of the slab be t and real depth of the
bubble from one side be x. Then
4
)xt(
6
x -
==m or
4
xt
6
x
5.1
-
== .
This gives x = 9 and
4
9t
5.1
-
= or t = 15 cm
4
9. (d) Let d be the depth of two liquids.
Then apparant depth
2
d
5.1
)2/d()2/d(
=
m
+
m
or 1
3
21
=
m
+
m
Solving we get m = 1.67
1
50. (c) The minimum length of the mirror is half the length of
the man. This can be proved from the fact that Ði = Ðr.
51. (d) Distance of image from plane mirror = 3 m at the back.
To photograph the image, camera must be focussed
for a distance of 4.5 + 3 = 7.5 m.
52. (d)
D75.3
20
100
80
100
PPP
12 -=-=-=
53. (b) 5
100/100
5
1
1
P
P
1
g
a
g
1
a
-=
-
+
=
÷
÷
ø
ö
ç
ç
è
æ
-
m
m
÷
÷
ø
ö
ç
ç
è
æ
-
m
m
=
115
a
g
1
g
-
m
m
=
÷
÷
ø
ö
ç
ç
è
æ
-
m
m
-
1.0)15.1(
5
1
1
5.1
1
-=-
-
=-
m
;
3
5
9.0
5.1
1
==m
54. (b)
2/Asin
2/)A(sin
md
+
=m
`2/Asin
2
)A(sin
m
m=
d+
Þ = 1.5 × sin 30º = 0.75
63º48)75.0(sin
2
A 1m
¢==
d+ -
\ d
m
= 37º12'
55. (b) Co
ndition for achromatism is
0
ff
2
2
1
1
=
w
+
w
\
3
2
f
f
2
1
2
1
=
w
w
-=
(leaving sign)
5
6. (d) The necessary condition is
f
f
¢
-=
w¢
w
which is satisfied
by (d)
57. (a)
2
1
f
f
2
1
2
1
-=
w
w
-= \
21
f 2f=-
As
21f
1
f
1
F 1
+=
\
111f2
1
f2
1
f
1
20
1
=-=\ f
1
= 10 cm
f
2
= –20 cm
5
8. (b)
0
e
f
M
f
b
==
µ
\
0
e
f 60
2 24
f5
b= µ= ´°=°
59. (d) 5
f
f
M
e
0
==, L = f
0
+ f
e
= 36
\ f
e
= 6
cm, f
0
= 30 cm
60. (b) When two lenses are placed coaxially then their
equivalent focal length F is given as
21
21
211 ff
ff
F
f
1
f
1
F
1
+
=Þ+=
f
1
f
2
Now power
21
21
ff
ff
lengthfocal
1 +
==
61. (b) Refract
ive index of medium is given by
( )
2
B
A whereA and Ba reconstant .m=+
l
Light has seven different colour, so its each colour has
different wavelength and so different refractive index.
Due to difference in refractive index different refractive
angle .
rsin
isin
÷
ø
ö
ç
è
æ
=m
So this is due to dep
endence on wavelength of
refractive index.
62. (c) When a object is real plane mirror form a virtual image
and when object is virtual image will be real. Thus in this
question object is virtual. Virtual object means object is
at infinity. So rays (incident) converge on the mirror.

659. 655Ray Optics and Optical Instruments
63. (c) Resolving power of eyea/l=
radians10
105
10500 4
3
9
-
-
-
=
´
´
=
Now, radiusanglearc ´=\ = 10
–4
× (500 × 10
3
)m = 50m
64. (c) Spherical abberation occurs due to the inability of a
lens to coverage marginal rays of the same wavelength
to the focus as it converges the paraxial rays, This defect
can be removed by blocking marginal rays. This can be
done by using a circular annular mask over the lens.
65. (d) Shift in the position of image after introducing slab,
(d) = t
÷
ø
ö
ç
è
æ
-
n
1
1
nd = t (n – 1) Þ t =
)1n(
nd
-
66. (b) The easiest way to answer this question is with a fast
sketch. For a given object’s position, draw two rays
from the top of the object. One ray is parallel to the
principal axis and passes through the focal point on
the opposite side of the lens. The other ray passes
through the center of the lens. The top of the image
appears where these two rays intersect.
Object AObject B
Image A Image B
F
F
Change the object’s position and repeat the process. You will observe that as the object approaches the lens while remaining beyond the focal length, the image produced on the opposite side of the lens moves away from the lens and increases in size. As an aside, the
image is real and inverted.
67. (a) The ratio of object to image distance equals the ratio of
object to image height. The ratio of image to object height
is found by rearranging the ratios to give 4f /(4/3)f = 1/3.
The image is demagnified by a factor of 3. Thus, answer
choice A is the best answer.
68. (b) Angular limit of resolution of eye,
d
l
q=, where d is
diameter of eye lens. Also if y is the minimum just resolution separation between two objects at distance D from eye then
y
D
q=
d
D
y
dD
y l
=Þ
l
=Þ ...............(1)
Here : wavelength m105Å5000
7-
´==l
D = 50m
Diameter of eye lens = 2 mm = m102
3-
´
From eq. (1) minimum separation is
cm5.12m105.12
102
50105
y
3
3
7
=´=
´
´´
=
-
-
-
69. (a)
70. (a) Due to covering the reflection from lower part is not
there so it makes the image less bright.
71. (b) Due to difference in refractive indices images obtained
will be two. Two mediums will form images at two
different points due to difference in focal lengths.
72. (b) 73. (d)
74. (b)
u
v
m=
f
1
u
1
v
1
=-
Multiply the equation by v
f v
1
u
v
f v
u
v
1 -=Þ=-
f
v
1m-=\
Slope =
b
c
f
c
b
f 1
=Þ=-
75. (c) The ray is partly reflected and partly refracted.
ÐMOB = 180 – 2q
But the angle between refracted and reflected ray is
ÐPOB. Clearly ÐPOB is less than ÐMOB.
76. (a)
77. (d) For 1
st
lens, u
1
= – 30, f
1
= + 10cm,
Formula of lens,
1
111
v 30 10
+=
or, v
1
= 15 cm at I
1
behind the lens.
The images I
1
serves as virtual object for concave lens.
For second lens, which is concave, u
2
= (15 – 5) = 10
cm. I
1
acts as object. f
2
= – 10 cm.
The rays will emerge parallel to axis as the virtual object
is at focus of concave lens, as shown in the figure. Image
of I
1
will be at infinity. These parallel rays are incident on
the third lens viz the convex lens, f
3
= + 30 cm. These
parallel rays will be brought to convergence at the focus
of the third lens.
\ Image distance from third lens = f
3
= 30 cm

660. 656 PHYSICS
78. (b)
a
µ
g
= 1.5
\ 1.5 = co
secC or C = 42°. Critical angle for glass
= 42°. Hence a ray of light incident at 50° in glass
medium undergoes total internal reflection. d denotes
the deviation of the ray.
\ d = 180° – (50° + 50°) or d = 80°.
79. (b)
÷
÷
ø
ö
ç
ç
è
æ
--m=
21R
1
R
1
)1(
f
1
According to
Cauchy relation
......
CB
A
42
l
+
l
+=m
Hence lµf .
Hence
red light having maximum wavelength has
maximum focal length.
VR
ff<and also F
V
< F
R
80. (
c) As refracted ray emerges normally from oppostite
surface, r
2
= 0
As A = r
1
+ r
2
\ r
1
= A
Now,
11
11
sin
;
sin
ii i
iA
rrA
m= = = =m
(where t = thickness of glass plate)
81. (c) Thickness of glass plate (t) = 6 cm;
Distance of the object (u) = 8 cm. and
distance of the image (v) = 12 cm.
Let x = Apparent position of the silvered surface in cm.
Since the image is formed due to relfection at the silvered
face and by the property of mirror image distance of object
from the mirror = Distance of image from the mirror
or x + 8 = 12 + 6 – x or x = 5 cm.
Therefore refractive index of glass
=
2.1
5
6
depthApparent
depthalRe
==
.
82. (d) We kn
ow that
å
=
=
n
1i
if
1
f
1
21f
1
f
1
f 1
+=
f
1
= 80 cm, f
2
= –
50 cm
100
50
1
100
80
1
f
1
-=
1
P
f
Þ= = 1.25 – 2 = – 0.75D
83. (c)
f = 20 cm
30 cm
10 cm
60 cm
O
l
f
1
u
1
v
1
=-;
20
1
30
1
v 1
=
-
-
Þ v = 60 cm
Coincide
nce is possible when the image is formed at
the centre of curvature of the mirror. Only then the rays
refracting through the lens will fall normally on the
convex mirror and retrace their path to form the image
at O. So the distance between lens and mirror = 60 – 10
= 50 cm.
84. (a) Let a ray of light enter at A and refracted beam is AB.
This is incident at an angle q. For no refraction at the
lateral face q > C
sin q > sin CBut q + r = (90°)
B
A
i
r
q
\ sin (90° – r
) > sin C or cos r > sin C..(1)
From Snell’s law
n
isin
rsin
rsin
isin
n =Þ=
2
2
2
sini
cosr1sinr1
n
æö
\ =- =- ç÷
èø
\ equation (1) gives, Csin
n
isin
–1
2
2
>
Csin
n
isin
–1
2
2
2
>Þ
Also sin
C
n
1
=
2
n
2
2
22
2
1
n
isin
1or
n
1
n
isin
–1 +>>\
or 1)1i(sin
n
1 2
2
<+ or 1isinn
22
+>
Max
imum value of sin i = 1
2n2n
2
>Þ>\
85. (c) The si
lvered plano convex lens behaves as a concave
mirror; whose focal length is given by
m1f
1
f
2
F
1
+=
If plane surface i
s silvered
¥=
¥
==
22
R
f
2
m
÷
÷
ø
ö
ç
ç
è
æ
m=\
211
R
1
–
R
1
)1–(
f
1

R
1–1
–
R
1
)1–(
m
=÷
ø
ö
ç
è
æ
¥
m=

661. 657Ray Optics and Optical Instruments
R
)1–(21
R
)1–(2
F
1 m
=
¥
+
m
=\

)1–(2
R
F
m
=
Here R =
20 cm, m = 1.5
cm20
)1–5.1(2
20
F ==\
86. (b)
)cm5(depthApparent
)R(depthalRe
5.1
1
=
\ R
1
= 1.5 × 5 = 7.5 cm
air bubble
5 cm
2 cm
R
1
R
2
For oppos
ite face,
2
R
5.1
2
= Þ R
2
= 3.0 cm
\ Thickness of the slab = R
1
+ R
2
= 7.5 + 3
= 10.5 cm
87. (a) For point A,
rsin
45sin
ga=m
Þ
ga2
1
rsin
m
=
for poin
t B, sin (90 – r) =
g
m
a
(90 – r) is critical angle.
Glass
B
90 – r
90 – r
45º
A
r
Air
\
ga
ag
1
rcos
m
=m=
Þ
rcos
1
ga=m
=
2
ga
2
2
1
1
1
rsin1
1
m
-
=
-
Þ
12
2
2
1
1
1
2
ga
2
ga
2
ga
2
ga
-m
m
=
m
-
=m
Þ 212
2 ga=-m
Þ
2
3
ga
=m
88. (b) ÷
÷
ø
ö
ç
ç
è
æ
--m=
21
R
1
R
1
)1(
f
1

in this case,
R
1
a
nd R
2
are unchanged
So, f will remain unchanged for both pieces of the
lens
\ f = f '
21f
1
f
1
f
1
+=

This is combination of two lenses
of equal focal length
\ ú
û
ù
ê
ë
é
¢
=
¢
+
¢
=
f
2
f
1
f
1
f
1
Þ f '' = 2f
89. (b) R
1
= 60 cm, R
2
= µ, m = 1.6
÷
÷
ø
ö
ç
ç
è
æ
--m=
21
R
1
R
1
)1(
f
1
÷
ø
ö
ç
è
æ
-=
60
1
)16.1(
f
1
Þ f = 100 cm.
9
0. (b) The angle must be equal to the critical angle,
º45
2
1
sin
1
sinC
11
=
÷
÷
ø
ö
ç
ç
è
æ
=
÷
÷
ø
ö
ç
ç
è
æ
m
=
--
91
. (d) Here
D
22.1
1000
x l
= or
2
3103
1010
101010522.1
x
-
-
´
´´´´
=
or x = 1.22 × 5 ×
10
–3
m = 6.1 m
x is of the order of 5 mm.
92. (c) Angle of minimum deviation
2
2
m
A
sin
A
sin
+dæö
ç÷
èø
m=Þ
æö
ç÷
èø
60
2
3
60
2
m
sin
sin
°+dæö
ç÷
èø
=
°æö
ç÷
èø
3
30
22
m
sin
dæö
Þ °+=ç÷
èø
30 60
2
m
d
Þ°+ =°
Þ d
m
= 60°.
93.
(c) For total internal reflection,
1
sinC
m³
2³>1.414
Þµ = 1.50
94. (
c) R = 20 cm, h
0
= 2, u = –30 cm
We have,
12
1 11
( 1)
f RR
æö
=m--
ç÷
èø
=
311
1
2 20 20
éùæö æö
- --ç÷ ç÷êú
èø èø
ëû
132
1
f 2 20
æö
Þ= -´
ç÷
èø
\ f = 2
0 cm
111
f vu
=-
111
20 v 30
Þ =+
111
v 20 30
=- =
10
600
v = 60 cm

662. 658 PHYSICS
m =
i
0
hv
hu
= i0
v
hh
u
Þ =´ =
60
2 – 4 cm
30
´=
So, image is inve
rted.
95. (b) Deviation = zero
So, d = d
1
+ d
2
= 0 Þ (m
1
– 1)A
1
+ (m
2
– 1) A
2
= 0
Þ A
2
(1.75 – 1) = – (1.5 – 1) 15°
Þ A
2
=
0.5
15
0.75
- ´° or A
2
= – 10°.
Negative
sign shows that the second prism is inverted
with respect to the first.
96. (a) To see his full image in a plane mirror a person requires
a mirror of at least half of his height.
M
H
E
L
2
M'
H
H
97. (b) 98. (b) 99. (d) 100. (b)
EXERCISE - 3
Exemplar Questions
1. (a) As we know that the deviation
d = (µ – 1) A..... (i)
By geometry, the angle of refraction by first surface is
5° and given µ = 1.5
So,d = (1.5 – 1) × 5°
= 2.5°
also,d = q – r, ..... (ii)
By putting the value of d and r in equation (ii)
2.5° = q – 5°
So, q = 5 + 2.5 = 7.5°
2. (d) As we know that when light ray goes from one medium
to other medium, the frequency of light remains
unchanged.
And, c = nl
So, c µ l the light of red colour is of highest wavelength
and therefore of highest speed. Thus, after travelling
through the slab, the red colour emerge first,
3. (c) According to the question, when object is at different
position, and if an object approaches towards a
convergent lens from the left of the lens with a uniform
speed of 5 m/s, the image move away from the lens to
infinity with a non-uniform acceleration.
4. (b) When a passenger in an aeroplane then he may see
primary and secondary rainbow such as

concentric
circles.
5. (c) Among all given sources of light, the bule light have
smallest wavelength.According to Cauchy
relationship, smaller the wavelength higher the
refractive index and consequently smaller the critical
angle as
1
µ
sinc
= .
Hence, correspo
nding to blue colour, the critical angle
is least which facilitates total internal reflection for the
beam of blue light and the beam of green light would
also undergo total internal reflection.
6. (c) Using lens maker’s formula for plano-convex lens, so
focal length is
1
f
=( )
21
12
11
RR
æö
m-m-
ç÷
èø
If object on curved suface
so,
2
R=¥
then, f =
( )
1
21
R
m -m
Lens placed in air, µ
1
= 1.
(As gi ven that, R = 20cm, µ
2
= 1.5, on substituting the
values in)
f =
1
R
1m-
=
20
1.51-
= 40 cm
So,
f is converging nature, as f > 0. Hence, lens will
always act as a convex lens irrespective of the side on
which the object lies.
7. (b) The reflection of radiowaves by ionosphere is similar
to total internal reflection of light in air during a mirage
because angle of incidence is greater than critical angle
so that internal reflection of radio wave, take place.
8. (b) The incident PQ ray of light passes through focus F on
the concave mirror, after reflection should become
parallel to the principal axis, i.e., ray-2.
9. (b) As we know, when the ray goes from rarer medium air
to optically denser turpentine, then it bends towards
the normal i.e., i > r whereas when it goes from optically
denser medium turpentine to rarer medium water, then
it bends away from normal i.e., i < r.
So, the path of ray 2 is correct.
10. (d) As we know that, the image formed by convex mirror
does not depend on the relative position of object w.r.t.
mirror.
So, when the car approaches in the rear side, initially it
appear at rest as images is formed at focus. Hence the
speed of the image of the car would appear to increase
as the distance between the cars decreases.

663. 659Ray Optics and Optical Instruments
11. (a) When the negative refractive index materials are those
in which incident ray from air (Medium 1) to them refract
or bend differently to that of positive refractive index
medium.
r
r
i
Air
glass
1
2
NEET/AIPMT (2013-2017) Questions
12. (b)
Plano-convexPlano-concave
Combination
1
f
=
1
1
f
+
2
1
f
= (m
1
– 1)
11
R
æö
-
ç÷
¥-èø
+ (m
2
– 1)
11
R
æö
-
ç÷
¥èø
=
()1
1
R
m-
–
()2
1
R
m-
Þ
1
f
=
12
R
m -m
Þ f =
12
R
m -m
Hence, focal length of the combination is
12
R
m -m
.
13. (b) P
cornea
= + 40 D
P
e
= + 20 D
Total power of combination = 40 + 20 = 60 D
Focal length of combination
1
100 cm
60
=´
5
cm
3
=
For minimum converging state of eye lens,
u v?f
3
5
=-¥==
From lens formula,
1
f
=
1
v
–
1
u
Þ v =
5
cm
3
Distance between retina and cornea-eye lens
1.67 m
3
5
==
14. (a)
70°
20°
20°
40°
qq
Incident ray
From fig. 40° + q = 90° \ q = 90° – 40° = 50°
15. (b) It is due to scattering of light. Scattering
4
1
µ
l
. Hence
the light reaches us is rich in red.
16. (d) Magnifying power of microscope
=
0e0
LD1
fff
µ
Hence with increase f
0
magnifyig power of microscope
decreases.
Magnifying power of telescope =
0
0
e
f
f
f
µ
Hence with increase f
0
magnifying power of telescope
increases.
17. (b)
According to Snell’s law m =
sini
sinr
Þ(1) sin 2A = (m) sin AÞm = 2 cos A
18. (a) As we know, the refractive index of the material of the
prism
mA
sin
2
sin (A/ 2)
d+æö
ç÷
èø
m=
cot A/2 =
m
A
sin
cos (A/ 2)2
sinA/2 sin(A/2)
+dæö
ç÷
èø
=
[Q µ = cot (A/2)]
ÞSin
m
A
2
d+æö
ç÷
èø
= sin(90° + A/2)
Þd
min
= 180° – 2A

664. 660 PHYSICS
19. (b)
Using lens maker’s formula,
12
1 11
(–1)–
f RR
æö
=m ç÷
èø

n = 1.5
n = 1.7
n = 1.5
1
11.5 11
–1–
f 1 –20
æöæö
=
ç÷ç÷
¥èøèø
Þ f
1
= 40cm
2
11.7 11
–1–
f 1 –20 20
æöæö
=
ç÷ç÷
+èøèø
2
100
f – cm
7
Þ=
and
3
11.5 11
–1–
f 1 –20
æöæö
=
ç÷ç÷
¥èøèø
Þ f
3
= 40 cm
eq123
1111
f fff
=++
eq
1111
f 40 –100 / 7 40
Þ=++
\ f
eq
= –50 c
m
Therefore, the focal length of the combination is – 50
cm.
20. (c) For total internal reflection, incident angle (i) > critical
angle (i
c
)
i=45°
45°
So, sin i > sin i
c
sin 45° >
1
m
Þ m > 2 Þ 1.414
Since refractiv
e index m of green and voilet are greater
than 1.414 so they will total internal reflected. But red
colour will be refracted.
21. (c) L
d=f +f
e0
Objective lens
Eye-piece
Magnification by eye piece
m =
f
fu+
ee
e0e0
ffI
Lf[(ff)]f
- = =-
+-+
or,
e
0
fI
Lf
=
Magnific
ation, M =
0
e
fL
fI
=
22. (b) Given: Angle of incidence
angle of prism,
i = 45°;
A = 60°;
Angle of minimum deviation,
d
m
= 2i – A = 30°
Refractive index of material of prism.
m =
m
A
sin
2
sinA/2
+dæö
ç÷
èø
=
sin45 12
·2
sin3012
°
==
°
23. (d)G
iven: Focal length of objective, f
0
= 40cm
Focal length of eye – piece f
e
= 4 cm
image distance, v
0
= 200 cm
Using lens formula for objective lens
000
111
vuf
-=
Þ
000
1 11
vfu
=+
Þ
0
1 1 1 51
v 40 200 200
+-
=+=
-
Þ v
0
= 50 cm
Tube l
ength l = |v
0
| + f
e
= 50 + 4 = 54 cm.
24. (a) Magnitude m = +veÞvirtual image
m = –veÞreal image
magnitude of magnification,
| m | > 1Þmagnified image
| m | < 1Þdiminished image
25. (d) When mirror is rotated by angle q reflected ray will be
rotated by 2q.
y
2
x
=q
light
spot
y
source
(L) x
2q
q
Mirror
Þ q =
y
2x
26. (a) For dispersion without deviation
12
(1)A('1) A0m-+m-=
12
( 1)A ( ' 1)Am- = m-
(1.42–1) × 10° = (1.7–1)A
2
4.2 =
0.7A
2
A
2
= 6°

665. WAVEFRONT
The locus of all particl
es of the medium vibrating in the same
phase at a given instant is called a wavefront. Depending on
the shape of source of light, wavefront can be of three types.
(i) Spherical wavefront: A spherical wavefront is produced by a
point source of light. This is because the locus of all such
points which are equidistant from the point source will be a
sphere. Spherical wavefronts are further divided into two
headings: (i) converging spherical and (ii) diverging spherical
wavefront.
Spherical
wave
front
Convergingspherical wavefront
Diverging
spherical wavefront
O
OO
(ii) Cylindrical wavefront: When the
source of light is linear in shapesuch as a slit, the cylindricalwavefront is produced. This isbecause all the points equidistantfrom a line source lie on the surface
Cylindrical wavefront
S
O¢
Oof a cylinder.
(iii) Plane wavefront: A small part
of a spherical or cylindricalwavefront due to a distantsource will appear plane andhence it is called plane wave-front. The wavefront of parallelrays is a plane wavefront.
Plane wavefront
HUYGENS WAVE THEORY
(Geometrical method to find the secondary wavefront)
(i) Each point source of light is a centre of disturbance from
which waves spread in all directions.
(ii) Each point on primary wavelets acts as a new source of
distrubance and produces secondary wavelets which travel
in space with the speed of light.
(iii) The forward envelope of the secondary wavelets at any
instant gives the new wavefront.
A
B
A¢
B¢
Primary
wav
efront
Secondary wave
lets
Secondary
wavefront
B²
A²

AA¢A²
BB¢B²
(iv) In a homogeneous medium the wavefront is always
perpendicular to the direction of wave propagation.
With the help of Huygen’s wave theory, law of
reflection and refraction, total internal reflection and dispersion
can be explained easily. This theory also explain interference,
diffraction and polarization successfully.
Drawbacks of Huygens Wave Theory
(a) This theory cannot explain photo-electric effect, compton,
and Raman effect.
(b) Hypothetical medium in vacuum is not true imagination.
(c) The theory predicted the presence of back wave, which
proved to be failure.
REFLECTION AND REFRACTION OF PLANE WAVES
USING HUYGENS PRINCIPLE
Reflection on the Basis of Wave Theory
According to Huygens principle, every point on AB is a source
of secondary wavelets. Let the secondary wavelets from B strike
reflecting surface M
1
M
2
at A¢ in t seconds.
\ '
BA ct=´ … (i)
where c
is the velocity of light in the medium.
N
1
2
3
3¢
A
D
B
P A¢
D¢
B¢
2¢
1¢
ri
i r
M
2
M
1
25
Wave Optics

666. 662 PHYSICS
The second
ary wavelets from A will travel the same distance c ×
t in the same time. Therefore, with A as centre and c × t as radius,
draw an arc B¢, so that
AB¢ = c × t … (ii)
A¢B¢ is the true reflected wavefront.
angle of incidence, i =
'BAAÐ
and angle of reflection, r = ''BAAÐ
In Ds AA¢B and AA¢B¢,
AA¢ is com
mon,
'',BA AB c t= =´ and ' 90BBÐ=Ð=°
\ Ds are congruent \ ' ' ' , . .,BAA B A A i e i rÐ =Ð Ð =Ð … (iii)
Which is th
e first law of reflection.
Further, the incident wavefront AB, the reflecting surface M
1
M
2
and the reflected wavefront A¢B¢ are all perpendicular to the plane
of the paper. Therefore, incident ray, normal to the mirror M
1
M
2
and reflected ray all lie in the plane of the paper. This is secondlaw of reflection.Refraction on the Basis of Wave Theory
XY is a plane surface that separates a denser medium of refractive
index µ from a rarer medium. If c
1
is velocity of light in rarer
medium and c
2
is velocity of light in denser medium, then by
definition.
µ = 1
2
c
c
… (iv)
A
P A¢
1
2
3
r
1¢
B
D
2¢
3¢
X Y
Rarer-C
1
Denser-C
2
r
i
D¢
i
N
B¢
AB is a plane wave front incident on XY at 'BAAÐ = iÐ. 1, 2, 3
are th
e corresponding incident rays normal to AB.
According to Huygens principle, every point on AB is a sourceof secondary wavelets. Let the secondary wavelets from B strikeXY at A¢ in t seconds.
\BA¢ = c
1
× t … (v)
The secondary wavelets from A travel in the denser medium with
a velocity c
2
and would cover a distance (c
2
× t) in t seconds.
A¢B¢ is the true refracted wavefront. Let r be the angle of refraction.
As angle of refraction is equal to the angle which the refracted
plane wavefront A¢B¢ makes with the refracting surface AA¢,
therefore,
''AABrÐ= .
Let ''AABrÐ= , angle of refraction.
In DAA¢B,
1'
sin
''
ctBA
i
AA AA
´
==
In D AA¢B¢,
2'
sin
''
ctAB
r
AA AA
´
==
\
1
2
sin
sin
ci
µ
rc
==
[using (iv)]
Hence
sin
sin
i
r
m= … (vi)
which proves Snell’s law of refraction.
It is clear from fig. that the incident rays, normal to the interface
XY and refracted rays, all lie in the same plane (i.e., in the plane of
the paper). This is the second law of refraction.
Hence laws of refraction are established on the basis of wave
theory.
Keep in Memory
1.In 1873, Maxwell sho
wed that light is an electromegnetic
wave i.e. it propagates as transverse non-mechnical waveat speed c in free space given by
81
00
1
c 3 10 ms
-
= =´
me
2.There are so
me phenomenon of light like photoelectric
effect, Compton effect, Raman effect etc. which can beexplained only on the basis of particle nature of light.
3.Light shows the dual nature i.e. particle as well as wavenature of light. But the wave nature and particle natureboth cannot be possible simultaneously.
4.Interference and diffraction are the two phenomena that can
be explained only on the basis of wave nature of light.
INTERFERENCE OF LIGHT WAVES AND YOUNG'S
DOUBLE SLIT EXPERIMENT
The phenomenon of redistribution of light energy in a medium
on account of superposition of light waves from two coherent
sources is called interference of light waves.
Young performed the experiment by taking two coherent sources
of light. Two source of light waves are said to be coherent if the
initial phase difference between the waves emitted by the source
remains constant with time.
(i) The rays of light from two coherent sources S
1
and S
2
superpose each other on the screen forming alternately
maxima and minima (constructive and destructive
interference).
D
y
P
d
Screen
S
1
S
2
M
2
M
1
S
O
(ii) Let th
e equation of waves travelling from
21
S&S are
tSinAy
11
w= ...(1)
tSinAy
22
w= ...(2)
where A
1
& A
2
are amplitudes of waves starting from S
1
& S
2
respectively. These two waves arrive at P by traversing
different distances S
2
P & S
1
P. Hence they are superimposed
with a phase difference (at point P) given by

667. 663Wave Optics
(B) Position of fringe:
(i)If D = S
2
P – S
1
P = nl, then we obtain bright fringes at
point P on the screen and it corresponds to
constructive interference. So from equation (4) the
position of n
th
bright fringe
D
yd
nPSPS
12
=l=-=D
or
æö
=l
ç÷
èø
nD
y
d
...(7)
(Posi
tion of n
th
bright fringe)
(ii) If
2
)1n2(PSPS
12
l
+=-=D, then we obtain dark
fringe at point P on the screen and corresponds todestructive interference. So from equation(4), the
position of, n
th
dark fringe is
D
yd
2
)1n2(PSPS
12 =
l
+=-=D
or
(2 1)
2
+l
=
nD
y
d
...(8)
(Posi
tion of n
th
dark fringe)
(C) Spacing or fringe width :
Let y
n
and y
n+1
are the distance of n
th
and (n+1)
th
bright
fringe from point O then
d
Dn
y
n
l
= &
d
)1n(D
y
1n
l+
=
+
So spacing b between n
th
an
d (n+1)
th
bright fringe is
d
D
yy
n1n
l
=-=b
+ ...(9)
Since
it is independent of n, so fringe width or spacing
between any two consecutive bright fringes is same.
Similarly the fringe width between any two consecutive
dark fringe is
l
b=
D
d
...(10)
(D) Cond
itions for sustained interference:
(i) The two sources should be coherent i.e they should
have a constant phase difference between them.
(ii) The two sources should give light of same frequency
(or wavelength).
(iii) If the interfering waves are polarized, then they must
be in same state of polarization.
(E) Conditions for good observation of fringe:
(i) The distance between two sources i.e. d should be
small.
(ii) The distance of screen D from the sources should be
quite large.
(iii) The two interfering wavefronts must intersect at a very
small angle.
(F) Conditions for good contrast of fringe :
(i) Sources must be monochromatic i.e they emit waves
of single wavelength.
(ii) The amplitude of two interfering waves should be
equal or nearly equal.
( )
( )
21
2
phase difference path difference
2
(S P S P
).
p
d = ´D
l
p
=-
l
....(3)
whe
re PS
2
(from fig) =
2
2 d
Dy
2
æö
++ç÷
èø
)]dy(D[
D2
)2/dy(
2
1
D
2
+>>\
+
+»
Similarly,
2
1
(y – d /2)
SPD
4D
»+
so,
D
yd
PSPS
12
=- ....(4)
(A) Co
nditions for maximum & minimum intensity :
(i)Conditions for maximum intensity or constructiveinterference : If phase difference
d = 0, 2p, 4p – – – 2np
or, path difference
21
SPSP0 ,,nD= - = l 2l--- l
then resultant intensity at point P due two waves
emanating from
21
S&Sis
2 22
1 2 12
2 cos==++dI A A A AA
2
( I A)\µ
or
2
12
()=+I AA
or
1 2 12
2=++I I I II ....(5)
It means th
at resultant intensity is greater than the
sum of individual intensity ( where A is the amplitudeof resultant wave at point P).
(ii)Conditions for minimum intensity or destructiveinterference : If phase difference,
,3 ,5 (2n 1)d=p p p---- + p
or, path difference
21
3
S P S P , (2n 1)
222
lll
D= -= - --+
then resultant intensity at point P is
2 22
1 2 12
– 2 cos== +dI A A A AA
or
2
12
()=-I AA
or
12 12
2=+-III II ...(6)
It means that res
ultant intensity I is less than the sum
of individual intensities. Now as the position of pointP on the screen changes, then the path difference at
point P due to these two waves also changes &
intensity alternately becomes maximum or minimum.
These bright fringes ( max. intensity) & dark fringes
(min. intensity) make an interference pattern.
It must be clear that there is no loss of energy
( at dark fringe) & no gain of energy ( at bright fringe),
but, only there is a redistribution of energy.
The shape of fringe obtained on the screen is
approximately linear.

668. 664 PHYSICS
(iii)
Both sources must be narrow.
(iv) As Intensity I is directly proportional to the square of
amplitude, hence Intensity of resultant wave at P,12 12 120
I I I 2 II co s ; ifI I I.=++ f == , then
()
2
0
4 cos
2
f
I=I
(v)()
2
max 12
I I I.=+
If I
1
= I
2
= I
0
, then
I
max
= 4I
0()
2
min 12
I II=- , if I
1
= I
2
= I
0
, then I
min
= 0
(vi)
2
12max
min 12
III
.
I II
æö+
=ç÷
-èø
(vii) Angular fringe-width
0
Dd
bl
q==
(viii) The width of all interference fringes are same. Since
fringe width b is proportional to l, hence fringes with
red light are wider than those for blue light.
(ix) If the interference experiment is performed in a medium
of refractive index m instead of air, the wavelength of
light will change from l to
m
l
.
i.e. ´
æölb
b==ç÷
mmèø
D
d
(x) If a trans
parent sheet of refractive index m and
thickness t is introduced in one of the paths of
interfering waves, then due to its presence optical path
will become mt instead of t. Due to this a given fringe
from present position shifts to a new position. So the
lateral shift of the fringe,
0
( 1) ( 1)
b
= m- = m-
l
D
y tt
d
(xi) In You
ng’s double slit experiment (coherent sources
in phase): Central fringe is a bright fringe. It is on the
prerpendicular bisector of coherent sources. Central
fringe position is at a place where two waves having
equal phase superpose.
(xii) Young’s experiment with the white light will give white
central fringe flanked on either side by coloured bands.
COHERENCE
The phase relationship between two light waves can very from
time to time and from point to point in space. The property of
definite phase relationship is called coherence.
1. Temporal coherence : A light wave (photon) is produced
when an excited atom goes to the ground state and emits
light.
(i) The duration of this transition is about 10
–9
to 10
–10
sec. Thus the emitted wave remains sinusoidal for this
much time. This time is known as coherence time (
c
t).
(ii) Definite phase relationship is maintained for a length
c
cLt= called coherence length.
For neon
6238 Ål= ,
10
c
10
-
t» sec. and L = 0.0 3 m.
For cadmium 6238 Ål= ,
9
c
10
-
t= and L = 0.3 m
For Laser
5
c
10
-
t= sec and L = 3 km.
(iii) The
spectral lines width
lD is related to coherence
length L and coherence time t
c
.
2
l
Dl»
t
c
c
or
2
l
Dl»
L
2. Spatial coherence : Two points in space are said to be
spatially coherence if the waves reaching there maintains a
constant phase difference. Points P and Q are at the same
distance from S, they will always be having the same phase.
Points P and P¢ will be spatially coherent if the distance
between P and P¢ is much less than the coherence length i.e.
c
PPc<<t¢
Monochromatic
source of light
P¢
P
Q
Methods of Obtaining Coherent Sources
T
wo coherent sources are produced from a single source of light
by two methods :
(i) By division of wavefront and (ii) By division of amplitude.
(i) Division of wavefront : The wavefront emitted by a narrow
source is divided in two parts by reflection, refraction or
diffraction. The coherent sources so obtained are imaginary.
Example : Fresnel’s biprism, Llyod’s mirror, Young’s double
slit, etc.
(ii) Division of amplitude : In this arrangement light wave is
partly reflected (50%) and partly transmitted (50%) to
produced two light rays. The amplitude of wave emitted by
an extended source of light is divided in two parts by partial
reflection and partial refraction. The coherent sources
obtained are real and are obtained in Newton’s rings,
Michelson’s interferometer, etc.
Incoherence of Two Conventional Light Sources
Let two conventional light sources L
1
and L
2
(like two sodium
lamps or two monochromatic bulbs) illuminate two pin holes S
1
and S
2
. Then we will find that no interference pattern is seen on
the screen.
The reason is as follows : In conventional light source, light
comes from a large number of independent atoms, each atom
emitting light for about 10
–9
seconds i.e., light emitted by an
atom is essentially a pulse lasting for only 10
–9
seconds.

669. 665Wave Optics
Screen
S
2
S
1L
1
L
2
Even if all the ato
ms were emitting light pulses under similar
conditions, waves from different atoms would differ in their initial
phases. Consequently light coming out from the holes S
1
and S
2
will have a fixed phase relationship only for 10
–9
sec. Hence any
interference pattern formed on the screen would last only for
10
–9
sec. (a billionth of a second), and then the pattern will
change. The human eye can notice intensity changes which last
at least for a tenth of a second and hence we will not be able to
see any interference pattern. Instead due to rapid changes in the
pattern, we will only observe a uniform intensity over the screen.
LIoyd’s Mirror
The two sources are slit S ( parallel to mirror ) and its virtual image
S'.
Flat black
glass
D
Super-position
occurs in this region
O
Screen
S'
d
S
Lloyd mirror arrangement
(i) If screen is moved so that, point O touches the edge of
glass plate, the geometrical path difference for two wave
trains is zero. The phase change of p radian on reflection at
denser medium causes a dark fringe to be formed.
·The fring width remains unchanged on introduction
of transparent film.
·If the film is placed in front of upper slit S
1
, the fringe
pattern will shift upwards. On the other hand if the
film is placed in front of lower slit S
2
, the fringe pattern
shifts downwards.
(ii) This interference pattern is frequently seen in a ripple tank
when one uses a wave train to demonstrate the law of
reflection.
(iii) In this case, fringe width
l
b=
D
d
Optical path : (Equivalent path in vacuum or air) In case of medium
of refractive index m and thickness t, the optical path = mt.
Interference in Thin Films
We are familiar with the colours produced by a thin film of oil on
the surface of water and also by the thin film of a soap bubble.
Hooke observed such colours in thin films of mica and similar
thin transparent plates. Young was able to explain the phenomenon
on the basis of interference between light reflected from the top
and bottom surface of a thin film. It has been observed that
interference in the case of thin films takes place due to
(i) reflected light and (ii) transmitted light.
Interference due to reflected light
From the figure, the optical path difference between the reflected
ray (AT) from the top surface and the reflected ray (CQ) from the
bottom surface can be calculated. Let it be x, then
Air
Air
C
QN
S
t
i
i
T
M
F
P
B
r
r
A
m
x (AB BC) AN=m+-
On simplification, we get
x = 2mt cos r
1. If 2 cos (2 1)
2
l
m =+trn , where n = 0,1,2, ..............then
constructive interference takes place and the film appears
bright.
2. If 2tcosrnm =l , where n = 0, 1, 2, 3,............ then destructive
interference takes place and the film appears dark.
Interference due to transmitted light
Air
Air
C
Q
R
S
t
i
i
P
M
D
N
B
r
rr
r
r
A
m
The optical path difference between the reflected ray (DQ) and
the transmitted ray (NR) is given by
x (BC CD) BN=m+-
On sim
plification, we getx 2 tcosr=m
1. If 2tcosrnm =l , where n = 0, 1, 2, 3, .............then
constructive interference takes place and the film appears
bright.
2. If 2 t cos r(2n 1)
2
l
m =+ , n = 0, 1, 2 , ....... then destructive
interference takes place and the film appears dark.

670. 666 PHYSICS
Newton's Rings
Newton obser
ved the formation of interference rings when a
plano-convex lens is placed on a plane glass plate. When viewed
with white light, the fringes are coloured while with monochromatic
light, the fringes are bright and dark. These fringes are produced
due to interference between the light reflected from the lower
surface of the lens and the upper surface of the glass plate.
Interference can also take place due to transmitted light.
Air film
Newton's rings by reflected light :
Here, interference takes place due to reflected light. Therefore,for bright rings,
2 tcos (2n 1) where n 1,2,3,
2
l
mq =-= ......
And for dar
k rings,
2 tcos n , n 1,2, 3,m q=l= ......
Air film
L
G
Proceeding further, we get the radius of rings as follows:
For bright rings,
-l
=
(2 1)
2
nR
r
For dark rings, =l ,r nR where R = radius of curvature of
lens.
(i) The centre is dark and alternately dark and bright rings
are produced.
(ii) While counting the order of the dark rings 1, 2, 3, etc.
the central ring is not counted. Therefore,
for 1st dark ring, n = 1 and
1rR=l
for 2nd dark ring, n = 2 and
2r 2R=l
Newton's rings by transmitted light
Here, interference takes place due to transmitted light.
L
G
Therefore,
For bright rings, 2 t cos n ,n 0,1,2,....m q=l=
For dark rings, 2 t cos (2n 1), n 0,1, 2,......
2
l
mq=-=
Procee
ding further, we get Radius of bright ring,
(2n 1)R
r
2
-l
=
(i) The centre is bright and alternately bright and dark
rings are obtained.
(ii) The ring pattern due to reflected light is just opposite
to that of transmitted light.
Keep in Memory
1. If D
n
an
d D
n + m
be the diameters of n th and (n + m)th dark
rings then the wavelength of light used is given by
22
nmn(D ) (D )
4mR
+-
l=
where, R is th
e radius of curvature of the lens.
2. IfD
n
= diameter of nth dark ring when air is present betwen
the glass plate and the lens
D
n+m
= diameter of (n+m)th dark ring when air is present
between the glass plate and the lens
D¢
n
= diameter of n th dark ring when a liquid is poured
between the plate and the lens
D¢
n+m
=diameter of (n+m)th dark ring when a liquid is
poured between the plate and the lens
Then the refractive index of the liquid is given by
m =
22
22
nmn
(D ) (D )
nmn
(D ) (D )
+
-
+
¢¢-
o r ,,
m =
22
nmn
4mR
(D ) (D )
+
l
¢¢-
Example 1.
In Young’s exp
t., two coherent sources are placed 0.90 mm
apart and fringes are observed one metre away. If it
produces second dark fringe at a distance of 1 mm from
central fringe, what would be the wavelength of
monochromatic light used?
Solution :
For dark fringes,
d2
D
)1n2(x
l
-=
D)1n2(
dx2
-
=l\
1)122(
109.0102
33
´-´
´´´
=
--
or, .cm106m106.
0
56 --
´=´=l

671. 667Wave Optics
Ex
ample 2.
Two beam of light having intensities I and 4 I interfere to
produce a fringe pattern on a screen. The phase difference
between the beams is
p/2 at point A and p at point B. Then
find the difference between the resultant intensities at A
and B.
Solution :
Here, p=qp=q==
2121
,2/;4III;I
12121A
cos2q++=IIIII
IIIII 52/cos424=p´++=
22121B
cos2q++=IIIII
= IIIIIII=-=p++ 45cos424 ;
IIIII45
BA
=-=-\
Example 3.
In a biprism ex
periment, 5th dark fringe is obtained at a
point. If a thin transparent film is placed in the path of oneof waves, then 7th bright fringe is obtained at the samepoint. Determine the thickness of the film in terms ofwavelength
l and refractive index m .
Solution :
For 5th dark fringe,
d2
D9
d
D
2
)1n2(x
1
l
=
l
-=
For 7th bright fri
nge,
d
D7
d
D
nx
2
l
=l=
but
d
D
t)1(xx
12-m=-;
d
D
t)1(
2
9
7
d
D
-m=
ú
û
ù
ê
ë
é
-
l
\ Thickness,
)1(
5.2
t
-m
l
=
Example 4.
In Young’
s experiment, the interference pattern is found to
have an intensity ratio between the bright and dark fringes
as 9. What is the ratio of (a) intensities (b) amplitudes of
the two interfering waves ?
Solution :
In case of interference, ()1 2 12
I I I 2 I I cos=++f
(a) For I to be maximum and minimum fcos is 1 and –1
respectivel
y, i.e., 2
max12 1 2 12
I I I 2 II (I I)=++ =+ and
2
min12 1 2 12
I I I 2 II (I~ I)=+-=
Acco
rding to given problem,
()
( )
2
12
max
2
min
12
II
I 9
I1
II
+
==
-
,
i.e.,
12max
min 12
III 3
I1 II
+
==
-
By comonendo a
nd dividendo,
1
2
I31
31I
+
=
-
i.e.,
1
2
I4
4
I1
==
(b) Now as for a wave
2
IAµ,
22
1 1 11
2 2 22
IAAA
, 4,i.e .,2
IAAA
éùéù
= ==êúêú
ëûëû
Example 5
.
In a Young’s double slit experiment the angular width of a
fringe formed on a distant screen is 1°. The wavelength of
the light used is 6280 Å. What is the distance between the
two coherent sources ?
Solution :
The angular fringe width is given by
d
l
a=
where l is wavelength and d is the distance between two
coherent sources. Thus d
l
=
a
Given, 6280 Å, 1 radian
180
p
l= a= °=
T
hus
10
56280 10
d 180 3.6 10 m 0.036 mm
3.14
-
-´
= ´=´=
DIFFRACTION
W
hen a wave is obstructed by an obstacle, the rays bend round
the corner. This phenomenon is known as diffraction.
Fraunhoffer Diffraction by Single Slit
In Fraunhoffer diffraction experiment, the source and the screen
are effectively at infinite distance from the diffracting element.
In single slit diffraction, imagine aperature to be divided into two
equal halves. Secondary sources in these two halves give first
minima at b sin q = l
D
q
q
q
b
P
P
0
In general, b sin q = nl for minima and, ( )sin 21
2
l
q=+bn for
maxima. (i
) The points of the maximum intensity lie nearly midway
between the successive minima. The amplitude E
0
' of the
electric field at a general point P is
00
sin
E'E
b
=
b
where
l
qp
=b
sinb
and
E
0
= amplitude at the po
int P
0
i.e. at q = 0
The intensity at a general point P is given as
2
2
0
sin
b
b
=II

672. 668 PHYSICS
(ii) Th
e graph for the variation of intensity as a function of
sinq is as follows :
I
I
0
I
0 / 22I
0 / 62.5
O
b
3l-
b
2l-
b
l-
b
l
b
2l
b
3l
sinq
(iii) The width of the central maxima is
2D
b
læö
ç÷
èø
and angular
width of centr
al maxima is
2
b
læö
ç÷
èø
.
Fraunh offer Diffraction by a Ci
rcular Aperture
(i) The 1
st
dark ring is formed by the light diffracted from the
circular aperture at an angle q with the axis where
b
22.1
sin
l
»q where l = wavele
ngth of light used,
b = diameter of circular aperture
q
b
D
Screen
Circular aperture
(ii) If the screen is at a distance D (D >> b) from the circular
aperture, the radius of the 1
st
dark ring is,
b
D22.1
R
l
»
(iii) If the lig
ht transmitted by the hole is converged by a
converging lens at the screen placed at the focal plane of
the lens, the radius of the 1
st
dark ring is
b
f22.1
R
l
=
This radius is also calle
d the radius of diffraction disc.
For plane transmission diffraction grating
(a + b) sin q
n
= nl for maxima, where a = width of transparent
portion, b = width of opaque portion.
Difference between Interference and Diffraction of light
Interference Diffraction 1. Interference is due to the
superposition of two
wavefronts originating
from two coherent
sources.
1. Diffraction is due to the
superposition of two
secondary wavelets
originating from the
different points of the
same wavefront.
2. In Interference pattern,
all the maxima i.e. bright
fringes are of the same
intensity.
2. In diffraction pattern, the
bright fringes are of
varying intensity.
3. In Interference pattern, the
dark fringes are usually
almost perfectly dark.
3. In diffraction pattern, the
dark fringes are not
perfectly dark.
4. In Interference pattern, the
width of fringes (bright
and dark) is equal.
4. In diffraction pattern, the
widths of fringes are not
equal.
5. In Interference, bands are
large in number.
5. In diffraction, bands are
a few in number.
6. In Interference, bands are
equally spaced.
6. In diffraction, bands are
unequally spaced.

Example 6.
In a single sli
t diffraction experiment, the angular position
of the first (secondary) maximum is found to be 5.2°, whenthe slit width is 0.01 mm. If sin 52° = 0.0906, then find thewavelength of light used.
Solution :
For single-slit diffraction, the angular position of the firstmaximum is determined from the relation
1
3
a sin
2
l
q=¢
It is given that a = 0.01 mm
=
1 × 10
–5
,
11
5.2 , sin 0.0906q= ° q=¢¢ . Therefore,

5
1
22
a sin 10 0.0906 6040 Å
33
-
l=q=´´=¢
Exam
ple 7.
In Fraunhaufer diffraction from a single slit of width 0.3mm the diffraction pattern is formed in the focal plane of alens of focal length 1m. If the distance of third minimumfrom the central maximum is 5mm, then find the wavelengthof light used.
Solution :
The distance of n
th
minimum from the central maximum is
given by
n
nf
X
a
l
=
where it is giv
en that
a = 0.3 × 10
–3
m, n = 3, f = 1m, X
n
= 5 × 10
–3
m
Therefore,
33
7n
aX0.3 10 5 10
5 10 m 500 nm
nf 31
--
-´ ´´
l== =´=
´

673. 669Wave Optics
POLARISATION
An ordinary source such as bulb consists of a large number of
waves emitted by atoms or molecules in all directions
symmetrically. Such light is called unpolarized light (see fig - a)
Source
Direction of
wave motion
Z
Y
X
Fig (a) Unpolarised light
Fig (b) Polarised light
If we confine the direction of wave vibration of electric vector inone direction perpendicular to direction of wave propagation,then such type of light is called plane polarised or linearly
polarised (with the help of polaroids or Nicol prism). The
phenomenon by which, we restrict the vibrations of wave in aparticular direction (see fig-b) ^ to direction of wave
propagation is called polarization.
The plane of vibration is that which contains the vibrations of
electric vector
E
r
and plane of polarisation is perpendicular to
the plane of vibration
·Tourmaline and calcite polarizes an e.m. wave passing
through it.
Polarization by Reflection (Brewster’s Law)
During reflection of a wave, we obtain a particular angle called
angle of polarisation, for which the reflected light is completely
plane polarised.
ip
Reflected light
is polarised.
Rarer
Denser
90º
i +r =90º
pp
r
p
m = tan (
i
p
)
where, i
p
= angle of incidence, such that the reflected and refracted
waves are perpendicular to each other.
Law of Malus : If the electric vector is at angle q with the
transmission axis, light is partially transmitted. The intensity of
transmitted light is
I = I
0
cos
2
q where I
0
is the intensity when the incident electric
vector is parallel to the transmission axis.
·Polarization can also be achieved by scattering of light
·(a) Plane polarized : oscillating
E
r
field is in a single plane.
(b) Circularly polarized : tip of oscillating E
r
field describes
a circle.
(c) Elliptically polarized : tip of oscillating E
r
field
descr
ibes an ellipse.
Example 8 :
The intensity of the polarised light becomes 1/20th of its
initial intensity after passing through the analyser. What
is the angle between the axis of the analyser and the initial
amplitude of the light beam ?
Solution :
Here
00
1
I I 0.05 I
20
==
Using
2
0
I I cos=q , we get
2
00
0.05 I I cos=q
Þ
2
cos 0.05q= or cos 0.05 0 .2236q==
1
cos (0.2236) 76 9
-
\q= =° ¢
Example 9 :
A beam of polarised light makes an angle of 60°
with the
axis of the polaroid sheet. How much is the intensity oflight transmitted through the sheet ?
Solution :
Here q = 60°,
Using
2
0
I I cos=q , we get
2
00
1
I I (cos60) I
4
= °=
1
cos 60
2
æö
°=ç÷
èø
Q
\ Intensity o
f transmitted light =
1
100 25%
4
´=
Thus, the in
tensity of the transmitted light is 25% of the
intensity of incident light.
Example 10:
A ray of light strikes a glass plate at an angle of 60° withthe glass surface. If the reflected and refracted rays are at
right angles to each other, find the refractive index of the
glass.
Solution.
When the reflected and refracted rays are at right angle to
each other, the angle of incident is known as angle of
polarisation (i
p
).
Here, q = 60°, Using m = tan i
p
, we get
tan 60 3 1.732m= °==

674. 670 PHYSICS
Example 1
1 :
A beam of light AO is incident on a glass slab (
m = 1.54) in
a direction as shown in fig. The reflected ray OB is passed
through nicol prism. On viewing through a nicol prism,
we find on rotating the prism that
A BN
O
33? 33?
57?
(a) the inten
sity is reduced down to zero and remains
zero
(b) the intensity reduces down somewhat and rises again
(c) there is no change in intensity
(d) the intensity gradually reduces to zero and then again
increases
Solution : (d)
For complete polarisation of reflected light
m = tan f (f = Brewster’s angle)
\ f = tan
–1
m = tan
–1
(1.54) = 57º
From fig, angle of incidence = 90° – 33º = 57º
Hence the reflected light is completely polarised. When the
plane polarised light is viewed through a rotating nicol prism,
the intensity gradually reduces to zero and then again
increases.
RESOLVING POWER OF AN OPTICAL INSTRUMENT
The resolving power of an optical instrument, is its ability to
distinguish between two closely spaced objects.
Diffraction occurs when light passes through the circular, or
nearly circular, openings that admit light into cameras, telescopes,
microscopes, and human eyes. The resulting diffraction pattern
places a natural limit on the resolving power of these instruments.
For example, for normal vision, the limit of resolution of normal
human eye is ~0.1 mm from 25 cm. (i.e., distances less than 0.1 mm
cannot be resolved). For optical microscope the limit of resolution
~ 10
–5
cm and for electron microscope ~5 Å or less.
The limit of resolution of a microscope
0.61l
=x
a
where a is the
a
perture of the microscope.
DOPPLER’S EFFECT FOR LIGHT WAVES(a) When the source moves towards the stationary observer
or the observer moves towards the source, the apparentfrequency.
)shiftBlue(
c
v
1´÷
ø
ö
ç
è
æ
+n=n
(b) When the source moves away from the stationary observer
or vice-versa,
v
´1
c
æö
n=n-ç÷
èø
(Red shift)
where n´
= apparent frequency, n = active frequency
v = velocity of source, c = velocity of light
But in both cases, the relative velocity v is small.
Example 12 :
The time period of rotation of the sun is 25 days and its
radius is 7 × 10
8
m. What will be the Doppler shift for the
light of wavelength 6000 Å emitted from the surface of the
sun?
Solution :
Doppler’s shift ÷
ø
ö
ç
è
æl
÷
ø
ö
ç
è
æp
=÷
ø
ö
ç
è
æl
w=l´=l
cT
2
R
c
R
c
v
d
Å
103
6000
760602425
222107
8
8
´
´
´´´´
´´´
= Å04.0=
Example 13 :
How far in advanc
e can one detect two headlights of a car
if they are separated by a distance of 1.57 m ?
Solution :
The human eye can resolve two objects when the angle
between them is 1 minute of arc. Thus, we have
x
D=
q
Here x = 1.57 m,
1
1 rad
60 180
p
q==´¢ ,
Thus
1.57 10800 1.57
D 5400 m 5.4 km
1 3.14
60 180
´
= = ==
p
´
Examp
le 14 :
The numerical aperture of a microscope is 0.12, and the
wavelength of light used is 600 nm. Then find its limit of
resolution.
Solution :
The limit of resolution of a microscope is given by
0.61
x
sin
l
=
mq
It is given that
7
610m
-
l=´ , and the numerical apertur
e
sin 0.12m q= .
Therefore,
7
60.61 6 10
x 3.05 10 m 3 µm
0.12
-
-´´
= =´»
Example 1
5 :
A person wants to resolve two thin poles standing neareach other at a distance of 1 km. What should be theminimum separation between them?
Solution :
Angular limit of resolution of eye q = 1 minute of arc
= 1/60 degree.Therefore, the minimum separation should be such that
x = Dq
with D = 1 km. = 10
3
m and
1
60 180
p
q=´ radian
Thus
3
10 3.14 31.4
x 0.29m
60 180 108
´
= ==
´
orx 30 cm.»

675. 671Wave Optics
CONCEPT MAP
Cylindrical wavefront
Linear light source
Effective distance -finite
1
Intensity I
r
µ
1
Amplitude A
r
µ
Huygens’ principle Each point on the
primary wavefront
is the source of a
secondary wavelets
Constructive interference
Phase diff. = 2n
Path diff. x =2n (/2)
Resultant amplitude A = a + a
Resultant intensity I =
d p
D l
1 2
2
1 2 (II) +
Destructive interference
Phase diff. = (2n – 1)
Path diff. x = (2n – 1)/2
Resultant amplitude A =
a
–

a
Resultant intensity I =
d p
D l
1
2
2
1
2
(
I
–
I
)
Plane wavefront
Light source at large
distance: Effective
distance infinite
Intensity and
amplitude independent
of distance
Forms of
wavefront
Wavefront
Locus of all particles
vibrating in same phase
Doppler effect
in light
radial
v v
v c
D
=
Interference
of light Redistr
i
b
u
t
i
o
n
of energy due t
o
s
u
p
e
r
position of wav
e
s
Spherical wavefront
Point light source
Effective distance-finite
2
1
Intensity I
r
µ1
Amplitude A
r
µ
WAVE OPTICS
Describes the connection
between waves and
rays of light
Super
p
o
s
i
t
i
o
n
o
f

w
a
v
e
s
.
When
t
w
o

s
i
m
i
l
a
r

w
a
v
e
s
propa
g
a
t
e

s
i
m
u
l
t
a
n
e
o
u
s
l
y
then r
e
s
u
l
t
a
n
t

displa
c
e
m
e
n
t

y
=
y

+

y
1
2
C
o
h
e
r
e
n
t

s
o
u
r
c
e
s
o
f

l
i
g
h
t
S
o
u
r
c
e
s

o
f
l
i
g
h
t
,
e
m
i
t
t
i
n
g
l
i
g
h
t
o
f
s
a
m
e
w
a
v
e
l
e
n
g
t
h
s
a
m
e
f
r
e
q
u
e
n
c
y

h
a
v
i
n
g
a

z
e
r
o
o
r

c
o
n
s
t
a
n
t
p
h
a
s
e
d
i
f
f
e
r
e
n
c
e
.
Y
o
u
n
g
’
s
d
o
u
b
l
e

s
l
i
t
E
x
p
e
r
i
m
e
n
t
Y
D
S
E
F
r
i
n
g
e

w
i
d
t
h

(
)
D
i
s
t
a
n
c
e

b
e
t
w
e
e
n
c
e
n
t
r
e
s

o
f
t
w
o

c
o
n
s
e
c
u
t
i
v
e
b
r
i
g
h
t

o
r
d
a
r
k

f
r
i
n
g
e
sb
Dd
b
=
l
D
i
s
t
a
n
c
e

b
e
t
w
e
e
n
c
e
n
t
r
a
l
f
r
i
n
g
e
a
n
d

n

d
a
r
k
f
r
i
n
g
e
t
h
n
(
2
n
1
)
D
X
2
d
-
l
=
D
i
s
t
a
n
c
e

b
e
t
w
e
e
n
c
e
n
t
r
a
l
f
r
i
n
g
e
a
n
d
n

b
r
i
g
h
t

f
r
i
n
g
e
D
=

d
i
s
t
a
n
c
e

b
e
t
w
e
e
n




s
o
u
r
c
e
a
n
d
s
c
r
e
e
n
d

=
d
i
s
t
a
n
c
e
b
e
t
w
e
e
n




t
w
o
s
l
i
t
s

t
h
n
n
D
X
dl
=
D
i
f
f
r
a
c
t
i
o
n
B
e
n
d
i
n
g
o
f

li
g
h
t
w
a
v
e
s
a
r
o
u
n
d

t
h
e
c
o
r
n
e
r
s

o
f
a
n
o
b
s
t
a
c
l
e
Linear width o
f
central maximu
m
2
D
2
a
a
l
l
=
+
For secondary maximum
Path diff. =
(2n
1
)
2+
l
Linear distance
(
2
n
1
)
D
2
a
+
l
=
F
o
r
s
e
c
o
n
d
a
r
y
m
i
n
i
m
u
m
p
a
t
h
d
i
f
f
e
r
e
n
c
e
=
n
l
L
i
n
e
a
r

d
i
s
t
a
n
c
e

=
n
Da
l
Polarisation
Restricting the vibration
o
f

light in a particular dire
c
t
i
o
n
perpendicular to the dire
c
t
i
o
n
of propagation of wave
Brewster’s law
µ = tan
= angle of polarisation
q
q
p
p
Law of Malus
I = Icos
I = intensity of
transmitted light
from analyser
0
2
q

676. 672 PHYSICS
1.Which one of t
he following phenomena is not explained by
Huygens construction of wavefront?
(a) Refraction (b) Reflection
(c) Diffraction (d) Origin of spectra
2.Which of the following phenomena is not common to sound
and light waves ?
(a) Interference (b) Diffraction
(c) Coherence (d) Polarisation
3.Interference is possible in
(a) light waves only
(b) sound waves only
(c) both light and sound waves
(d) neither light nor sound waves
4.A single slit diffraction pattern is obtained using a beam of
red light. If the red light is replaced by the blue light, then
the diffraction pattern
(a) remains unchanged(b) becomes narrower
(c) becomes broader (d) will disappear
5.In Young's double slit experiment, if the slit widths are in the
ratio 1 : 2, the ratio of the intensities at minima and maxima
will be
(a) 1 : 2(b) 1 : 3(c) 1 : 4(d) 1 : 9
6.If a wave can be polarized, it must be
(a) a transverse wave(b) a longitudinal wave
(c) a sound wave (d) a stationary wave
7.To demonstrate the phenomenon of interference, we require
two sources which emit radiation of
(a) nearly the same frequency
(b) the same frequency
(c) different wavelengths
(d) the same frequency and having a definite phase
relationship
8.Angular width (B) of central maxima of a diffraction pattern
of a single slit does not depend upon
(a) distance between slit and source
(b) wavelength of the light used
(c) width of slit
(d) frequency of light used
9.The phenomenon by which stars recedes from each other
is explained by
(a) black hole theory(b) neutron star theory
(c) white dwarf (d) red shift
10.Which of the following does not support the wave nature
of light?
(a) Interference (b) Diffraction
(c) Polarisation (d) Photoelectric effect.
11.The colours seen in the reflected white light from a thin oil
film are due to
(a) diffraction (b) interference
(c) polarisation (d) dispersion
12.Which of the following cannot be polarised ?
(a) Radio waves (b)
b rays
(c) Infr
ared rays (d)
g rays
13.The phenom
enon of interference is shown by
(a) longitudinal mechanical waves only
(b) transverse mechanical waves only
(c) non-mechanical transverse waves only
(d) All of the above
14.The transverse nature of light is shown by
(a) interference of light (b) refraction of light
(c) polarization of light (d) dispersion of light
15.If the intensities of the two interfering beams in Young’s
double-slit experiment are I
1
and I
2
, then the contrast
between the maximum and minimum intensities is good
when
(a) | I
1
and I
2
| is large(b) | I
1
and I
2
| is small
(c) either I
1
or I
2
is zero (d) I
1
= I
2
16.The idea of the quantum nature of light has emerged in an
attempt to explain
(a) interference
(b) diffraction
(b) polarization
(d) radiation spectrum of a black body
17.. The contrast in the fringes in an interference pattern
depends on
(a) fringe width
(b) wavelength
(c) intensity ratio of the sources
(d) distance between the slits
18.Polarisation of light establishes
(a) corpuscular theory of light
(b) quantum nature of light
(c) transverse nature of light
(d) all of the three
19.Huygens concept of wavelets is useful in
(a) explaining polarisation
(b) determining focal length of the lenses
(c) determining chromatic aberration
(d) geometrical reconstruction of a wavefront
20.When a compact disc is illuminated by small source of white
light, coloured bands are observed. This is due to
(a) dispersion (b) diffraction
(c) interference (d) reflection
21.A nicol prism is based on the action of
(a) refraction (b) double refraction
(c) dichroism (d) both (b) and (c)
22.The deflection of light in a gravitational field was predicted
first by
(a) Einstein (b) Newton
(c) Max Planck (d) Maxwell
23.When light passing through rotating nicol is observed,
no change in intensity is seen. What inference can be
drawn ?
(a) The incident light is unpolarized.
(b) The incident light is circularly polarized.
(c) The incident light is unpolarized or circularly
polarized.
(d) The incident light is unpolarized or circularly polarized
or combination of both.

677. 673Wave Optics
1.The width of a slit is 0.012 mm. Monochromatic light is
incident on it. The angular position of first bright line is
5.2º. The wavelength of incident light is
[sin 5.2º = 0.0906].
(a) 6040 Å (b) 4026 Å
(c) 5890 Å (d) 7248 Å
2.A ray of light is incident on the surface of a glass plate at an
angle of incidence equal to Brewster’s angel f. If m represents
the refractive index of glass with respect to air, then the angle
between the reflected and the refracted rays is
(a) 90° + f (b) sin
–1
(mcosf)
(c) 90º (d)
÷
÷
ø
ö
ç
ç
è
æ
m
f
-
-sin
sinº90
1
3.Light of
wavelength 6.5 × 10
–7
m is made incident on two
slits 1 mm apart. The distance between third dark fringe and
fifth bright fringe on a screen distant 1 m from the slits will
be
(a) 0.325 mm (b) 0.65 mm
(c) 1.625 mm (d) 3.25 mm
4.The max. intensity produced by two coherent sources of
intensity I
1
and I
2
will be
(a)
21
I+I (b)
2
2
2
1
I+I
(c)
2121
2II+I+I(d) zero
5.The pa
th difference between two wavefronts emitted by
coherent sources of wavelength 5460Å is 2.1 micron. The
phase difference between the wavefronts at that point is
(a) 7.692 (b) 7.692 p
(c)
p
692.7
(d)
p3
692.7
6.In Young’s expt., the distance between two slits is
d
3
and
the di
stance between the screen and the slits is 3 D. The
number of fringes in
1
m
3
on the screen, formed by
monochromatic light of wavelength 3l, will be
(a)
lD9
d
(b)
lD27
d
(c)
lD81
d
(d)
lD
d
7.In Young’
s double slit expt. the distance between two
sources is 0.1 mm. The distance of the screen from thesource is 20 cm. Wavelength of light used is 5460 Å. Theangular position of the first dark fringe is(a) 0.08º (b) 0.16º
(c) 0.20º (d) 0.32º
8.The separation between successive fringes in a double slitarrangement is x. If the whole arrangement is dipped underwater what will be the new fringe separation? [Thewavelenght of light being used is 5000 Å](a) 1.5 x (b) x
(c) 0.75 x (d) 2 x
9.Light of wavelength 6328 Å is incident normally on a slithaving a width of 0.2 mm. The angular width of the centralmaximum measured from minimum to minimum of diffractionpattern on a screen 9.0 metres away will be about(a) 0.36 degree (b) 0.18 degree(c) 0.72 degree (d) 0.09 degree
10.A plane wave of wavelength 6250 Å is incident normally ona slit of width 2 × 10
–2
cm. The width of the principal maximum
on a screen distant 50 cm will be(a) 312.5 × 10
–3
cm (b) 312.5 × 10
–3
m
(c) 312.5 × 10
–3
m (d) 312 m
11.A ray of light strikes a glass plate at an angle of 60º. If thereflected and refracted rays are perpendicular to each other,the index of refraction of glass is
(a)
2
1
(b)
2
3
(c)
2
3
(d) 1.732
12.T
he wavelength of H
a
line in hydrogen spectrum was found
to be 6563 Å in the laboratory. If the wavelength of same
line in the spectrum of a milky way is observed to be 6568Å,
then recession velocity of milky way will be
(a) 105 m/s (b) 1.05 × 10
6
m/s
(c) 10.5 × 10
6
m/s (d) 0.105 × 10
6
m/s
13.A star is receding away from earth with a velocity of
10
5
m/s. If the wavelength of its spectral line is 5700 Å, then
Doppler shift will be
(a) 200 Å (b) 1.9 Å
(c) 20 Å (d) 0.2 Å
14.A slit of width a is illuminated by red light of wavelength
6500 Å. If the first minimum falls at q = 30°, the value of a is
(a) 6.5 × 10
–4
mm (b) 1.3 micron
(c) 3250 Å (d) 2.6 × 10
–4
cm
24.In refraction, light waves are bent on passing from one
medium to the second medium, because, in the second
medium
(a) the frequency is different
(b) the coefficient of elasticity is different
(c) the speed is different
(d) the amplitude is smaller
25.Interference was observed in an interference chamber when
air was present. Now, the chamber is evacuated and if the
same light is used, a careful observation will show
(a) no interference
(b) interference with bright bands
(c) interference with dark bands
(d) interference in which breadth of the fringe will be
slightly increased

678. 674 PHYSICS
15.Two beam
s of light of intensity I
1
and I
2
interfere to give
an interference pattern. If the ratio of maximum intensity to
that of minimum intensity is 25/9, then I
1
/I
2
is
(a) 5/3 (b) 4
(c) 81/625 (d) 16
16.The condition for obtaining secondary maxima in the
diffraction pattern due to single slit is
(a)
asinnq=l (b) ( )asin 2n1
2
l
q=-
(c) ()asin 2n1q= -l (d)
n
a sin
2
l
q=
17.Light from two coh
erent sources of the same amplitude A
and wavelength l illuminates the screen. The intensity of
the central maximum is I
0
. If the sources were incoherent,
the intensity at the same point will be
(a) 4I
0
(b) 1I
0
(c)I
0
(d) I
0
/2
18.In Young's double slit experiment with sodium vapour lamp
of wavelength 589 nm and the slits 0.589 mm apart, the half
angular width of the central maximum is
(a) sin
–1
(0.01) (b) sin
–1
(0.0001)
(c) sin
–1
(0.001) (d) sin
–1
(0.1)
19.In Young's double slit experiment with sodium vapour lamp
of wavelength 589 nm and the slits 0.589 mm apart, the half
angular width of the central maximum is
(a) sin
–1
0.01 (b) sin
–1
0.0001
(c) sin
–1
0.001 (d) sin
–1
0.1
20.When the light is incident at the polarizing angle on the
transparent medium, then the completely polarized light is
(a) refracted light
(b) reflected light
(c) refracted and reflected light
(d) neither reflected nor refracted light
21.In the phenomena of diffraction of light, when blue light is
used in the experiment in spite of red light, then
(a) fringes will become narrower
(b) fringes will become broader
(c) no change in fringe width
(d) None of these
22.The wavefronts of a light wave travelling in vacuum are
given by x + y + z = c. The angle made by the direction of
propagation of light with the X-axis is
(a) 0º (b) 45º
(c) 90º (d)
–1
cos (1/ 3)
23.In Fresnel’s biprism expt., a mica sheet of refractive index
1.5 and thickness 6 × 10
–6
m is placed in the path of one of
interfering beams as a result of which the central fringegets shifted through 5 fringe widths. The wavelength oflight used is
(a) 6000 Å (b) 8000 Å
(c) 4000 Å (d) 2000 Å
24.Two sources of light of wavelengths 2500 Å and 3500 Å are
used in Young’s double slit expt. simultaneously. Which
orders of fringes of two wavelength patterns coincide?
(a) 3rd order of 1st source and 5th of the 2nd
(b) 7th order of 1st and 5th order of 2nd
(c) 5th order of 1st and 3rd order of 2nd
(d) 5th order of 1st and 7th order of 2nd
25.A radar sends radiowaves of frequency v towards an
aeroplane moving with velocity v
a
. A change Dn is observed
in the frequency of reflected waves which is higher than
original frequency. The velocity of aeroplane is (v
a
<< c)
(a)
cDn
n
(b)
2cDn
Dn
(c)
c
2
Dn
n
(d)
2c
Dn
n
26.In Young’s double slit experiment, we get 10 fringes in the
field of view of monochromatic light of wavelength 4000Å.If we use monochromatic light of wavelength 5000Å, thenthe number of fringes obtained in the same field of view is(a)8 (b) 10 (c) 40 (d) 50
27.With a monochromatic light, the fringe-width obtained in aYoung’s double slit experiment is 0.133 cm. The whole set-up is immersed in water of refractive index 1.33, then thenew fringe-width is(a) 0.133 cm (b) 0.1 cm
(c) 1.33 × 1.33 cm (d)
cm
2
33.1
28.A slit of width a is illumin
ated by white light. The first
minimum for red light (l = 6500 Å) will fall at q = 30º when a
will be(a) 3250 Å (b) 6.5 × 10
–4
cm
(c) 1.3 micron (d) 2.6 × 10
–4
cm
29.The Fraunhoffer ‘diffraction’ pattern of a single slit is formedin the focal plane of a lens of focal length 1 m. The width ofslit is 0.3 mm. If third minimum is formed at a distance of 5mm from central maximum, then wavelength of light will be(a) 5000 Å (b) 2500 Å
(c) 7500 Å (d) 8500 Å
30.Two points separated by a distance of 0.1 mm can just beinspected in a microscope, when light of wavelength 600Åis used. If the light of wavelength 4800 Å is used, the limitof resolution will become(a) 0.80 mm (b) 0.12 mm (c) 0.10 mm(d) 0.08 mm
31.Unpolarised light of intensity 32 W m
–2
passes through
three polarizers such that the transmission axis of the lastpolarizer is crossed with that of the first. The intensity offinal emerging light is 3 W m
–2
. The intensity of light
transmitted by first polarizer will be(a) 32 W m
–2
(b) 16 W m
–2
(c) 8 W m
–2
(d) 4 W m
–2
32.A parallel beam of monochromatic unpolarised light isincident on a transparent dielectric plate of refractive index
3
1
. The reflected beam i
s completely polarised. Then the
angle of incidence is(a) 30º (b) 60º
(c) 45º (d) 75º

679. 675Wave Optics
33.Two nicols are oriented with their principal planes making
an angle of 60º. Then the percentage of incident unpolarised
light which passes through the system is
(a) 100(b) 50 (c) 12.5 (d) 37.5
34.A beam of unpolarised light passes through a tourmaline
crystal A and then through another such crystal B oriented
so that the principal plane is parallel to A. The intensity of
emergent light is I
0
. Now B is rotated by 45º about the ray.
The emergent light will have intensity
(a) 2/
0
I (b)
2/
0
I (c) 2
0
I (d)
0
2I
35.A rocket is r
eceding away from earth with velocity 0.2 c.
The rocket emits signal of frequency 4 × 10
7
Hz. The
apparent frequency observed by the observer on earth will
be
(a) 4 × 10
6
Hz (b) 3.2 × 10
7
Hz
(c) 3 × 10
6
Hz (d) 5 × 10
7
Hz
36.The heavenly body is receding from earth, such that the
fractional change in l is 1, then its velocity is
(a)c (b)
5
c3
(c)
5
c
(d)
5
c2
37.Fluores
cent tubes give more light than a filament bulb of
same power because(a) the tube contains gas at low temperature(b) ultraviolet light is converted into visible light by
fluorescence
(c) light is diffused through the walls of the tube(d) it produces more heat than bulb
38.In young’s double-slit experiment, the intensity of light at a
point on the screen where the path difference is
l is I, l
being the wavelength of light used. The intensity at a point
where the path difference is
4
l
will be
(a)
4
I
(b)
2
I
(c)I (d) zero
39.A
perture of the human eye is 2 mm. Assuming the mean
wavelength of light to be 5000 Å, the angular resolution
limit of the eye is nearly
(a) 2 minute (b) 1 minute
(c) 0.5 minute (d) 1.5 minute
40.If the polarizing angle of a piece of glass for green light is
54.74°, then the angle of minimum deviation for an
equilateral prism made of same glass is
[Given, tan 54.74° = 1.414]
(a) 45° (b) 54.74°
(c) 60° (d) 30°
41.In Young's double slit experiment, the fringes are displaced
by a distance x when a glass plate of refractive index 1.5 is
introduced in the path of one of the beams. When this
plate is replaced by another plate of the same thickness,
the shift of fringes is (3/2) x. The refractive index of the
second plate is
(a) 1.75 (b) 1.50
(c) 1.25 (d) 1.00
42.A single slit Fraunhoffer diffraction pattern is formed with
white light. For what wavelength of light the third
secondary maximum in the diffraction pattern coincides
with the second secondary maximum in the pattern for red
light of wavelength 6500 Å?
(a) 4400 Å (b) 4100 Å
(c) 4642.8 Å (d) 9100 Å
43.When the angle of incidence is 60° on the surface of a
glass slab, it is found that the reflected ray is completely
polarised. The velocity of light in glass is
(a)
81
2 10 ms
-
´ (b)
81
3 10 ms
-
´
(c)
81
2 10 ms
-
´ (d)
81
3 10 ms
-
´
44.A lens havi
ng focal length f and aperture of diameter d
forms an image of intensity I. Aperture of diameter
2
d
in
central
region of lens is covered by a black paper. Focal
length of lens and intensity of image now will be
respectively
(a)f and
4
I
(b)
3
4
f
and
2
I
(c)f and
3
4
I
(d)
2
f
and
2
I
45.In Young’s double slit experiment, the slits are 2 mm apart and
are illuminated by photons of two wavelengths l
1
= 12000Å
and l
2
= 10000Å. At what minimum distance from the
common central bright fringe on the screen 2 m from the slit
will a bright fringe from one interference pattern coincide
with a bright fringe from the other ?
(a) 6 mm (b) 4 mm (c) 3 mm (d) 8 mm
46.A parallel beam of fast moving electrons is incident normally
on a narrow slit. A fluorescent screen is placed at a large
distance from the slit. If the speed of the electrons is
increased, which of the following statements is correct ?
(a) The angular width of the central maximum of the
diffraction pattern will increase.
(b) The angular width of the central maximum will decrease.
(c) The angular width of the central maximum will be
unaffected.
(d) Diffraction pattern is not observed on the screen in
case of electrons.
DIRECTIONS for Qs. (47 to 50) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following-
(a)Statement -1 is false, Statement-2 is true
(b)Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(c)Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(d)Statement -1 is true, Statement-2 is false
47. Statement 1 : In YDSE, if a thin film is introduced in front of
the upper slit, then the fringe pattern shifts in the downward
direction.
Statement 2 : In YDSE if the slit widths are unequal, the
minima will be completely dark.
48. Statement 1 : In YDSE, if I
1
= 9I
0
and I
2
= 4I
0
then
max
min
I
25.
I
=
Statem
ent 2 : In YDSE
2
max 12
I ( I I)=+ and
2
min 12
I (I I)=- .

680. 676 PHYSICS
49. Stat
ement 1 : In Young’s double slit experiment if
wavelength of incident monochromatic light is just doubled,
number of bright fringe on the screen will increase.
Statement 2: Maximum number of bright lunge on the screen
is inversely proportional to the wavelength of light used
Exemplar Questions
1.Consider a light beam incident from air to a glass slab atBrewster’s angle as shown in figure.A polaroid is placed in the path of the emergent ray at pointP and rotated about an axis passing through the centre andperpendicular to the plane of the polaroid.
P
(a) For a particular orientation, there shall be darkness as
observed through the polaroid
(b) The intensity of light as seen through the polaroid shall
be independent of the rotation
(c) The intensity of light as seen through the polaroid shall
go through a minimum but not zero for two orientations
of the polaroid
(d) The intensity of light as seen through the polaroid shall
go through a minimum for four orientations of the
polaroid
2.Consider sunlight incident on a slit of width 10
4
Å. The
image seen through the slit shall
(a) be a fine sharp slit white in colour at the centre
(b) a bright slit white at the centre diffusing to zero
intensities at the edges
(c) a bright slit white at the centre diffusing to regions of
different colours
(d) only be a diffused slit white in colour
3.Consider a ray of light incident from air onto a slab of glass
(refractive index n) of width d, at an angle q. The phase
difference between the ray reflected bv the top surface of
the glass and the bottom surface is
(a)
12
2
2
2d1
1 sin
n
pæö
- q +pç÷
lèø
(b)
12
2
2
4d1
1 sin
n
pæö
-qç÷
lèø
(c)
12
2
2
4d1
1 sin
2n
ppæö
- q+ç÷
lèø
(d)
12
2
2
4d1
1sin2
n
pæö
- q +pç÷
lèø
4.In a Young’s double
-slit experiment, the source is white
light. One of the holes is covered by a red filter and another
by a blue filter. In this case,
(a) there shall be alternate interference patterns of red and
blue
(b) there shall be an interference pattern for red distinct
from that for blue
(c) there shall be no interference fringes
(d) there shall be an interference pattern for red mixing with
one for blue
5.Figure shows a standard two slit arrangement witn slits S
1
,
S
2
, P
1
, P
2
are the two minima points on either side of P (figure).
Screen
Second
screen
S
S
1
S
2
S
3
S
4
P
2
P
P
1
At P
2
on the
screen, there is a hole and behind P
2
is a
second 2-slit arrangement with slits S
3
, S
4
and a second
screen behind them.(a) There would be no interference pattern on the second
screen but it would be lighted
(b) The second screen would be totally dark(c) There would be a single bright point on the second
screen
(d) There would be a regular two slit pattern on the second
screen
NEET/AIPMT (2013-2017) Questions
6.In Young’s double slit experiment, the slits are 2 mm apart andare illuminated by photons of two wavelengths l
1
= 12000Å
and l
2
= 10000Å. At what minimum distance from the com-
mon central bright fringe on the screen 2 m from the slit willa bright fringe from one interference pattern coincide with abright fringe from the other ? [2013]
(a) 6 mm (b) 4 mm
(c) 3 mm (d) 8 mm
50. Statement 1 : In YDSE number of bright fringe or dark fringe
can not be unlimitedStatement 2 : In YDSE path difference between the
superposing waves can not be more than the distancebetween the slits.

681. 677Wave Optics
7.A parallel beam of fast moving electrons is incident nor-
mally on a narrow slit. A fluorescent screen is placed at a
large distance from the slit. If the speed of the electrons is
increased, which of the following statements is correct ?
[2013]
(a) The angular width of the central maximum of the dif-
fraction pattern will increase.
(b) The angular width of the central maximum will decrease.
(c) The angular width of the central maximum will be un-
affected.
(d) Diffraction pattern is not observed on the screen in
case of electrons.
8.In Young’s double slit experiment the distance between the
slits and the screen is doubled. The separation between
the slits is reduced to half. As a result the fringe width
(a) is doubled [NEET Kar. 2013]
(b) is halved
(c) becomes four times
(d) remains unchanged
9.A parallel beam of light of wavelength l is incident normally
on a narrow slit. A diffraction pattern is formed on a screen
placed perpendicular to the direction of the incident beam.
At the second minimum of the diffraction pattern, the phase
difference between the rays coming from the two edges of
slit is [NEET Kar. 2013]
(a)pl (b) 2p
(c)3p (d) 4p
10.A beam of light of l = 600 nm from a distant source falls on
a single slit 1 mm wide and the resulting diffraction pattern
is observed on a screen 2 m away. The distance between
first dark fringes on either side of the central bright fringe
is: [2014]
(a) 1.2 cm (b) 1.2 mm
(c) 2.4 cm (d) 2.4 mm
11.In the Young’s double-slit experiment, the intensity of light
at a point on the screen where the path difference is l is K,
(l being the wave length of light used). The intensity at a
point where the path difference is l/4, will be: [2014]
(a)K (b) K/4
(c) K/2 (d) Zero
12.In a double slit experiment, the two slits are 1 mm apart and
the screen is placed 1 m away. A monochromatic light
wavelength 500 nm is used. What will be the width of each
slit for obtaining ten maxima of double slit within the central
maxima of single slit pattern ? [2015]
(a) 0.1 mm (b) 0.5 mm
(c) 0.02 mm (d) 0.2 mm
13.For a parallel beam of monochromatic light of wavelength
'l', diffraction is produced by a single slit whose width 'a' is
of the wavelength of the light. If 'D' is the distance of the
screen from the slit, the width of the central maxima will be:
(a)
D
a
l
(b)
Da
l
[2015]
(c)
2Da
l
(d)
2D
a
l
14.At the first minimum adjacent to the central maximum of a
single-slit diffraction pattern, the phase difference between
the Huygen's wavelet from the edge of the slit and the
wavelet from the midpoint of the slit is : [2015 RS]
(a)
radian
2
p
(b)p radian
(c)radian
8
p
(d)radian
4
p
15.Two slits in Young’s experiment have widths in the ratio 1 :
25. The ratio of intensity at the maxima and minima in the
interference pattern,
max
min
I
I
is: [2015
RS]
(a)
121
49
(b)
49
121
(c)
4
9
(d)
9
4
16.In a diffraction pattern due to a single slit of width 'a', the
first minimum is observed at an angle 30° when light of
wavelength 5000 Å is incident on the slit. The first
secondary maximum is observed at an angle of : [2016]
(a)
11
sin
4
-æö
ç÷
èø
(b)
12
sin
3
-æö
ç÷
èø
(c)
11
sin
2
-æö
ç÷
èø
(d)
13
sin
4
-æö
ç÷
èø
17.The intensity at the maximum in a Young's double slit
experiment is I
0
. Distance between two slits is d = 5l, where
l is the wavelength of light used in the experiment. What
will be the intensity in front of one of the slits on the screen
placed at a distance D = 10 d ? [2016]
(a)I
0
(b)
0
I
4
(c) 0
3
I
4
(d)
0
I
2
18.The ratio of resolving powers of an optical microscope for
two wavelengths l
1
= 4000 Å and l
2
= 6000 Å is [2017]
(a) 9 : 4 (b) 3 : 2
(c) 16 : 81 (d) 8 : 27
19.Young's double slit experment is first performed in air and
then in a medium other than air. It is found that 8
th
bright
fringe in the medium lies where 5
th
dark fringe lies in air.
The refractive index of the medium is nearly [2017]
(a) 1.59 (b) 1.69
(c) 1.78 (d) 1.25
20.Two Polaroids P
1
and P
2
are placed with their axis
perpendicular to each other. Unpolarised light I
0
is incident
on P
1
. A third polaroid P
3
is kept in between P
1
and P
2
such
that its axis makes an angle 45° with that of P
1
. The intensity
of transmitted light through P
2
is [2017]
(a)
0
I
4
(b)
0
I
8
(c)
0
I
16
(d)
0
I
2

682. 678 PHYSICS
EXERCISE - 1
1. (d) 2. (d) 3
. (c) 4. (b)
5. (d) 6. (a) 7. (d)
8. (a) For single slit diffraction pattern l=qsine Angular
width,
e= slit width
÷
ø
ö
ç
è
æl
=q\
-
e
sin
1
It is indepen
dent of D, distance between screen and
slit.
9. (d) Doppler effect in light explains the phenomenon of
receding of stars and approaching of star by red shift
and blue shift respectively.
10. (d) Photoelectric effect does not support the wave nature
of light.
11. (b)
12. (b) Longitudinal waves cannot be polarised.
13. (d)
14. (c) 15. (d) 16. (d)
17. (c) 18. (c) 19. (d)
20. (b) The line rulings, each of 0.5
mm width,on a compact
disc function as a diffraction grating. When a small
source of light illuminates a disc, diffracted light forms
coloured ‘lanes’ that are the composite of the
diffraction patterns from the ruling.
21. (d) When a ray of light enters nicol prism, it splits into
two plane polarised light in mutually perpendicular
direction. One of this light undergoes total reflection
and absorption whereas other comes out as a plane
polarised light.
22. (b) Newton first predicted deflection of light by
gravitational field.
23. (c)
24. (c) Speed of light is different in different media and different
medium has different refractive index.
1
m
2
=
Speed of light in medium 1
Speed of light inmedium 2
25. (d) As fringe width b =
D
d
l
m =
24
BC
A ...+++
ll
m
Vacuum
< m
Air
so l
Vac
uum
> l
Air
Therefore when chamber is evacuated fringe width b
slightly increases.
EXERCISE - 2
1. (d) It is
a one of Fraunhoffer diffraction from single slit.
so for bright fringe where a is the width of slit.
2
)1n2(sina
l
+=q
112
0906.0102.12
1n2
sina2
5
+´
´´´
=
+
q
=l
-
.
Å7248m107248
10
=´=
-
2. (c)Q i
p
= f, therefore, angle between reflected and
refracted rays is 90º.
3. (c)
3
7
5
10
1105.65
d
D
nx
-
-
´´´
=l= m105.32
4-
´=
3
7
3
102
1105.65
d
D
2
1
)1n2(x
-
-
´
´´´
=
l
-=
m1025.16
4-
´=
4
53
x x 16.25 10 m 1.625 mm .
-
-=´=
4. (c) As R
2
= a
2
+
b
2
+ 2 ab cos f º0cos2
2121max
II+I+I=I\
2121
2II+I+I=
5. (b) Phase
diff.
x
2
l
p
=
Path difference
10
6
105460
101.22
-
-
´
´´p
= = 7.692 p radian.
6. (
c)
.
d
D
27
3/d
D33
d
D l
=
l
=
¢
¢l¢
=b
No. of fringes .
D81
d3/1
l
=
b
=
7. (b) The posi
tion of n
th
dark fringe. So position of first
dark fringe in d2/Dx
1
l= .
d = 20 cm, D = 0.1mm, l = 5460 Å, x
1
= 0.16
8. (c) When the arrangement is dipped in water;
x3
/ x 0.75x
4/34
¢b=bm= ==
9. (
a) The angular width of central maxi. is
.radian
102
1063282
a
22
4
10
-
-
´
´´
=
l
=q

6180
6328 10 degree 0.36º
-
=´´=
p
Hints & Solutions

683. 679Wave Optics
10. (a) Width of central maximum
4
10
102
5.01062502
a
D2
-
-
´
´´´
=
l
=
63
3125 10 m 312.510 cm.
--
=´ =´
q
o
D
Intensity
a
a
2l
– 2l
=
=
a
y sin
2
q
y sin
2
q
y sin
1
q = /al
sinq = 0
y sin
1
q = – /al
Screen p
osition of various minima for Fraunhoffer
diffraction pattern of a single slit of width a.
11. (d) As reflected and refracted rays are perpendicular to
each other, therefore, i
p
= i = 60º, where i
p
is called
angle of polarisation.
m = tan i
p
= tan 60º =
3= 1.732.
12.
(b)
8
103
6563
)62636586(
c ´´
-
=´
l
lD
=u
.s/m1005.1
6
´=
13. (b) Å9.15700
103
10
c
8
5
=´
´
=l´
u
=lD .
14. (b) A
ccording to principle of diffraction, a sin q = nl
where, n = order of secondary minimum
or, a sin 30° = 1 × (6500 × 10
–-10
)
or, a = 1.3 × 10
–6
m, or, a = 1.3 micron.
15. (d)
2
max 12
min 12
I aa25 25
or
I9aa9
æö+
==
ç÷
-èø
where a
denotes amplitude.
12
12 12
12
aa 5
or 5a 5a 3a 3a
aa3
+
= - =+
-
or, 5
a
1
– 5a
2
= 3a
1
+ 3a
2
or 2a
1
= 8a
2
or,
2
11
22
aa
4 or 16
aa
æö
==
ç÷
èø
1
2
I
I
=.
16. (b)
17
. (d) For two coherent sources, I
1
= I
2
I
max
= (A
1
+ A
2
)
2
=
()
2
12
II+
This is gi
ven as I
0
for maximum and zero for minimum.
If there are two noncoherent sources, there will be nomaximum and minimum intensities. Instead of all theintensity I
0
at maximum and zero for minimum, it will
be just I
0
/2.
18. (c)
sin
d
l
q=
9
3
3
589 10 1
10 0.0001
10000.589 10
-
-
-
´
= ===
´
1
9. (c) In Young's double slit experiment, half angular width
(q) is given by
sin
d
l
q=
–9
–3
–3
589 10
10
0.589 10
´
==
´
\ q = sin
–1
0.
001
20. (b) When the light is incident at the polarising angle on
the transparent medium, the reflected light is
completely polarised.
21. (a)
22. (d) Let any
R
r
its components are
zRyRxRR
rrrr
++=
with ==R
|R|
rr
–RzRyRxR|R|
22
++==
rr
& x yz
Rx Ry Rz
cos , cos , cos
RRR
q= q q=
th
ere cosq
x
, cosq
y
and cosq
z
one called direction
cosines.
Hence )R(czyx
r
==++
So, magn
itude of
222
cIII3= ++=
and
3
1
cos
x=q
23. (a)
Where n is equivalent number of fringe by which the
centre fringe is shifted due to mica sheet
5
106)15.1(
n
t)1(
6-
´-
=
-m
=l
Å6000m106
7
=´=
-
24. (b
) Let nth fringe of 2500 Å coincide with (n – 2)th fringe
of 3500Å.
\ 3500 (n – 2) = 2500 × n
1000 n = 7000, n = 7
\ 7th order fringe of 1st source will coincide with 5th
order fringe of 2nd source.
25. (c) In Radar, the source & receiver are together, the
receiver being turned for frequencies other than the
source or radar frequency.
To measure the speed of helicopter
(i) The moving object of speed v
a
receive a frequency
÷
ø
ö
ç
è
æ
+n=n
c
v
1'
a

684. 680 PHYSICS
(ii) T
he object which receives n’ frequency now acts as a
moving source. The detector observes a frequency
n
0
÷
÷
÷
÷
ø
ö
ç
ç
ç
ç
è
æ
-
n=n
c
v
1
1
'
a
0
÷
÷
÷
÷
ø
ö
ç
ç
ç
ç
è
æ
-
+
=
c
v
1
c
v
1
a
a
c
v
)()(
a
00
n+n=nD=n-nÞ
or a
0
2
0
c
v
2
\n »n
Þn+n»n
Dn æö
= ç÷
n èø
26. (a) As lµb
\ fringe width becomes
4
5
times,
No, of frin
ges
810
5
4
=´=
27. (b) cm1.0
33.1
133.0
==
m
b
=b¢
28. (c) Th
e position of n
th
dark fringe in Fraunhoffer
Diffraction from a single slit is
a sin q = n l
º30sin
105.61
sin
n
a
7-
´´
=
q
l
= ,
(for first min. n = 1)
.m3.1m1013
2/1
105.6 7
7
m=´=
´
=
-
-
29. (a) l=qnsina
l=3
f
xa
(since q is very
small so f/xtansin =q»q»q)
or
13
105103.0
f3
xa
33
´
´´´
==l
--
.Å5000m105
7
=´=
-
30. (d) Limit of r
esolutionlµR.L
.mm08.0
6000
4800
1.0R.LR.L=´=
l
l¢
´=¢\
\ Resolution imp
roves when light of lower wavelength
is used.
31. (b) Intensity of polarised light transmitted from 1st
polariser,
I
1
= I
0
cos
2
q
but (cos
2
q)
av

2
1
=
So
2
10
1 32
I I 16Wm
22
-
= ==
32. (a) Wh
en angle of incidence i is equal to angle of
polarisation i.e, then reflected light is completely plane-
polarised whose vibration is perpendicular to plane of
incidence.
33. (c) Suppose intensity of unpolarised light = 100.
\ Intensity of polarised light from first nicol prism
0
Ι1
100 50
22
==´=
According to law of Malus,
22
0
Ι Ιcos 50 (cos 60º)= q= 5.12
2
1
50
2
=÷
ø
ö
ç
è
æ
´=
34. (a) Ac
cording to law of Malus
22
1
)º45(coscos
0
2
0
2
0
2
0
Ι
ΙΙΙΙ =
÷
÷
ø
ö
ç
ç
è
æ
==q=
35. (b) Hz108.0104
c
c2.0
c
77
´=´´=n´
u
=nD
77
108.0104´-´=nD-n=n¢ .Hz102.3
7
´=
36. (a) When p
lanet or stars are receding away from earth. If
f is frequency of vibration.
Then ,
l
=
c
f
If =vvelocity of body moving away
=l'apparent wavelength to an observer on the earth
()
f
vc
'
+
=l (c and v are in opposi
te to each other)
l÷
ø
ö
ç
è
æ+
=
c
vc
l÷
ø
ö
ç
è
æ
+=l
c
v
1'
c
v'
=
l
l-l
Fractional ch
ange in wavelength
cv
c
v
1 =Þ=\
37. (b) The
fluorescent material present in the tube converts
u.v. rays into visible light.
38. (b) For path difference l, phase
difference = 2p
22
Q x .2
ppæö
= = l=p
ç÷
llèø
Þ I = I
0
+ I
0
+ 2I
0
cos 2p
Þ I = 4I
0
( cos 2 1)\ p=

685. 681Wave Optics
For
4
x
l
=, phase difference =
2
p
2
cosII2II'I
2121
p
++=\
If I
1
= I
2
= I
0
then
0
II
I' 2I 2.
42
===
39. (b) If the angular limit of resolution of human eye is R
then
rad
102
10522.1
a
22.1
R
3
7
-
-
´
´´
=
l
=
minute1nutemi60
180
102
10522.1
3
7
=´
p
´
´
´´
=
-
-
40. (d) By principle of polarization, µ = tanq
p
or µ = tan 54.74° or µ = 1.414
For an equilateral prism, ÐA = 60°
( ) ( )
A 60
sin sin
22
sin A/2 sin 60 /2
+d °+dæö æö
ç÷ ç÷
èø èø
\m==
°
or,
1.141 1 60
sin 1.414 2
22
´ °+dæö
éù==ç÷ ëûèø
Q
or,
2 60 1 60
sin or sin
222 2
°+d °+dæö æö
==
ç÷ ç÷
èø èø
or, sin 45° = sin
60 60
or 45
22
°+d °+dæö æö
°=ç÷ ç÷
èø èø
41. (a)
42. (c)
()2n1D
x
2a
+l
=
For red light,
()4 1D
x 6500Å
2a
+
=´
For other light,
()6 1D
xÅ
2a
+
= ´l
x is same for each.
5
5 6500 7 6500 4642.8Å.
7
\´=´lÞl=´=
43. (b)
a
µ
g
= tan q
P
where q
P
= polarising angle.
or,
a
µ
g
= tan 60°
or,
g
c
3
v
=
or,
8
81
g
c 3 10
v 3 10 ms
33
-´
= = =´
44. (c) By covering aperture, focal length does not change.
But intensity is reduced by
1
4
times, as aperture
diameter
2
d
is covered.
\I' =
3
44
II
I-=
\New focal length = f and intensity =
3
4
I
.
45. (a)Q y =
nD
d
l
\ n
1
l
1
= n
2
l
2
Þ n
1
× 12000 × 10
–10
= n
2
× 10000 × 10
–10
or, n (12000 × 10
–10
) = (n + 1) (10000 × 10
–10
)
Þ n = 5
( )
10 10
12
12000 10 m; 10000 10 m
--
l=´ l=´Q
Hence, y
common

1nD
d
l
=
( )
10
3
5 12000 10 2
2 10
-
-
´´
=
´
( )d 2 mm and D 2m==Q
= 5 × 12 × 10
–4
m
= 60 × 10
–4
m
= 6 × 10
–3
m = 6 mm
46. (b)
q
S
1
S
2
d
D
Y
Angular width, q =
Y
D
=
nD
dD
l

D
Y
d
léù
=
êú
ëû
Q
so, q =
d
l
, v ­l ¯ q ¯[For central maxima n = 1]
Hence, with increase in speed of electrons angular
width of central maximum decreases.
47. (a) 48. (b) 49.(a) 50. (b)

686. 682 PHYSICS
EXERCISE - 3
Exemplar Questions
1. (c) Let us consider the diagram shown below the light beam
incident from air to the glass slab at Brewster’s angle
(i
p
). The angle between reflected ray BE and BC is 90°.
Then only reflected ray is plane polarised represented
by arrows.
As the emergent and incident ray are unpolarised, so,
polaroid rotated in the way of CD then the intensity
cannot be zero but varies in one complete rotation.
DPolaroid
(P)
Polarised
E
B
C
90?
r
A
N
i
P
2. (a) As giv
en that the width of the slit
=10
4
Å = 10000 Å
= 10
4
× 10
–10
m = l0
–6
m = 1 µm
Wavelength of visible sunlight varies from 4000 Å to8000 Å.Thus the width of slit is 10000 Å comparable to that ofwavelength visible light i.e., 8000 Å. So diffraction occurswith maxima at centre. Hence at the centre all coloursappear i.e., mixing of colours form white patch at thecentre.
3. (a) Let, us consider the diagram, the ray (P) is incident at
an angle q and gets reflected in the direction P' and
refracted in the direction P' through O'. Due to reflectionfrom the glass medium there is a phase change of p.
The time difference between two refracted ray OP' andO'P'' is equal to the time taken by ray to travel alongOO'.
Dt =
g
OO
V
¢
=
d cosr
cn
=
nd
ccosr
From Snell's law,n =
sin
sinr
q
sin r =
sin
n
q
As we know that,
co
s r =
2
1 sinr- ,
so by putting s
in r value in that relation.
So, cos r =
2
1 sinr-
cos r =
2
2
sin
1
n
q
-
drr
r¢
O
N¢
q
q
P¢
N
P
P¢
q
P
\ Dt =
12
2
2
nd
sin
c1
n
æö q
-ç÷
èø
=
12
2
2
nd sin
1
cn
-
æö q
-ç÷
èø
Phase difference=Df =
2
t
T
p
´D
=
12
2
2
2 d sin
1
1 n
-
æöpq
-ç÷
èø
× nl
n
Df =
12
2
2
2 d sin
1
n
-
éùpq
-
êú
lëû
\Hence the net phase difference = Df + p
=
12
2
2
2d1
1 sin
n
-
pæö
- q +pç÷
lèø
4. (c) Fo
r sustained interference pattern to be formed on the
screen, the sources must be coherent and emits lights
of same frequency and wavelength.
In a Young’s double-slit experiment, when one of the
holes is covered by a red filter and another by a blue
filter. In this case due to filteration only red and blue
lights are present which has different frequency. In this
monochromatic light is used for the formation of fringes
on the screen. So, in that case there shall be no
interference fringes.
5. (d) Consider the given figure there is a hole at point P
2
. By
Huygen’s principle, wave will propagates from the
sources S
1
and S
2
. Each point on the screen will acts as
sources of secondary wavelets.
Wavefront starting from P
2
reaches at S
3
and S
4
which
will again act as two monochromatic or coherent sources.
Hence, there will be always a regular two slit pattern on
the second screen.
Past Years (2013-2017) NEET/AIPMT
6
. (a)Q y =
nD
d
l
\n
1
l
1
= n
2
l
2
Þn
1
× 12000 × 10
–10
= n
2
× 10000
× 10
–10
or, n (12000 × 10
–10
) = (n + 1) (10000 × 10
–10
)
Þn = 5

687. 683Wave Optics
( )
10 10
12
12000 10 m; 10000 10 m
--
l=´ l=´Q
Hence, y
common

1nD
d
l
=
( )
10
3
5 12000 10 2
2 10
-
-
´´
=
´
( )d 2 mm and D 2m==Q
= 5 × 12 × 10
–4
m
= 60 × 10
–4
m
= 6 × 10
–3
m = 6 mm
7. (b)
Y
d
Angular width, q =
Y
D
=
nD
dD
l

D
Y
d
léù
=
êú
ëû
Q
so, q =
d
l
, v ­l ¯ q ¯
[For central maxima n = 1]
Hence, with increase in speed of electrons angular
width of central maximum decreases.
8. (c) Fringe width ;
D
d
l
b=
From question D¢ = 2D and
2
d
d=¢
1
1
'4
l
\b= =b
D
d
9. (d) Conditions for diffraction minima are
Path diff. xnD=l and Phase diff. df = 2np
Path diff. = 2nl=l
Phase diff. = 24np=p ( 2)n=Q
10. (d) Given: D = 2m; d = 1 mm = 1 × 10
– 3
m
l = 600 nm = 600 × 10
– 6
m
Width of central bright fringe (= 2b)
=
6
3
2 D 2 600 10 2
m
d 1 10
-
-
l ´´´
=
´
= 2.4 × 10
– 3
m = 2.4 mm
11. (c) For path difference l, phase difference = 2p rad.
For path difference
4
l
, phase difference =
2
p
rad.
As K = 4I
0
so intensity at given point where path
difference is
4
l
K¢ =
2
0
4I cos cos cos 45º
44
ppæöæö
=ç÷ç÷
èøèø
= 0
K
2I
2
=
12. (d) Here, distance between two slits,
d = 1mm = 10
–3
m
distance of screen from slits, D = 1 m
wavelength of monochromatic light used,
l = 500nm = 500 × 10
–9
m
width of each slit a = ?
Width of central maxima in single slit pattern =
2D
a
l
Fringe width in double slit experiment
D
d
l
b=
So, required condition
10D 2D
da
ll
=
Þ
–3d1
a 10
5D5
= =´ m = 0.2 mm
13. (d) Linear width of central maxima y
= D(2q) = 2Dq =
2D
a
l
\ q
a
l
=
D
q
q
y
14. (b) For first minima at P
AP – BP = l
AP – MP =
2
l

688. 684 PHYSICS
A
M
B
P
f
f
O
So phase difference, f =
2
2
pl
´
l
= p radian
15. (d) The ratio of slits width =
1
25
(given)
\
1
2
I
I
=
25
1
I µ A
2
Þ
2
11
2
2 2
IA 25
I1A
== or
1
2
A5
A1
=
max 12
min 12
A AA 5163
A A A 5142
+ +
= = ==
--
\
22
max max
2
min min
IA 39
I 24A
æö
= ==
ç÷
èø
16. (d) For the first minima,
q =
a
hl
Þsin30° =
1
a2
l
=
First secondary maxima will be at
sin q =
3 31
2a 22
læö
=ç÷
èø
Þq = sin
–1
3
4
æö
ç÷
èø
17. (d) Let P is a point infront of one slit at which intensity is
to be calculated. From figure,
S
1
O
S
2
P
B
D
d
Path difference = S
2
P – S
1
P
=
22
DdD+- =
2
2
1d
D1D
2D
æö
+-ç÷
èø
=
22
2
dd
D11
2D2D
éù
+ -=êú
ëû
Dx =
2
d d5
2 10d 20 20 4
ll
===
´
Phase difference,
Df =
2
42
plp
´=
l
So, resultant intensity at the desired point 'p' is
I = I
0
cos
2
2
f
= I
0
cos
2
0
I
42
p
=
18. (b) Resolving power of a microscope =
2 sinmq
l
i.e.,
1
Rµ
l
or,
12
21
R
R
l
=
l
Given that the two wavelengths,
1
4000Ål=
2
6000Ål=
\
1
2
R
R
=
6000Å 3
4000Å 2
=
19. (c) According to question
8
th
bright fringe in medium = 5th dark fring in air
Y
8th bright
=
D
8
d
l
m
Y
5th dark
= (2 × 5 – 1)
D 9D
2d 2d
ll
=
Þ
9DD
8
2dd
ll
=
m
or, refractive index m =
16
1.78
9
=
20. (b)
According to malus law, I = I
0
cos
2
q
I
1
=
0I
2
I
2
=
20I
cos 45
2
° =
00II1
224
´=
I
3
=
20I
cos 45
4
°
I
3
=
0
I
8

689. CATHODE RAYS
Discharge Tube Experiments
When a very strong potential difference is applied across the two
electrodes in a discharge tube and the pressure of the air is lowered
gradually, then a stage is reached at which the current begins to
flow through the air with cracking noise. The potential at which
this happens is called sparking potential.
+
–
Negative glow
Aston’s
dark space
Cathode glow
Faraday’s dark space
Positive striation
·As pressure is lowered to 0.1 m.m. Hg – cathode glow,
Crooke’s dark space, negative glow, Faraday’s dark spaceand striations are observed.
·At a pressure 0.01 m.m. Hg entire tube is dark (Crooke’s darkspace) except the glass wall behind anode. Colour isyellowish-green for soda glass and greyish-blue for lead
glass.
·The luminous streaks travelling from cathode to anode,
below the pressure 0.01 m.m. Hg, are called cathode rays.
Properties of Cathode Ray :
(i) Emitted perpendicularly to cathode, (ii) Travel in straight lines
(iii) Carry energy (iv) Possesses momentum (v) Deflected by
electric and magnetic fields (vi) Excite fluroescence (vii) Ionise
gas (viii) Produce highly penetrating secondary radiation when
suddenly stopped (ix) Effect photographic plate etc.
J.J. Thomson’s e/m value of electron :
2
11 –1
2
/ 1.76 10 kg
2
= =´
E
emC
VB
. Thi
s value of
÷
ø
ö
ç
è
æ
m
e
is for
electr
on.
Millikan’s oil drop method for e/ m :
In fig.(a) we consider a single drop of mass m carrying a negative
charge –q in the absence of electric field. Then
mg
mg
F = 6 rv
2viscous 2
ph
–qE
n
2n
1 E
F =6r
viscous1 1
(a) (b)
F
viscous 1
= m g [from Stoke’s law F
viscous
= 6phrv]
or6ph rv
1
= mg …(1)
wher
e
h is coefficient of viscocity of air, r is radius of drop and v
1
is the terminal velocity of drop.
In fig. (b) we consider a single drop of mass m, radius r carrying a
negative charge –q in the presence of electric field acting
downward. Then by free body diagram (fig. (b)), we get
2viscous
FmgE)q(+=- 2
rv6mgph+= …(2)
where v
2
is the terminal speed in this case. Then from eqn (1), we
have.
E
)vv(r6
)q(
21+ph
=- …(3)
and radius o
f drop from equation (1)gr
3
4
mgrv6
3
1 rp==ph or
g2
v9
r
1
r
h
= …(4)
where r is de
nsity of drop.
Millikan repeated these measurements on thousands of drops
and he found that the charge q calculated for each drop was some
integral multiple of an elementary charge e. (e = 1.6 × 10
–19
C).
Hence, q = ne, n = 0, ±1, ±2 ...(5)
This experiment gives the evidence of quantisation of charge.
EMISSION OF ELECTRON
Electrons from the metal surface are emitted by anyone of the
following physical processes :
(i)Thermionic emission : The emission of electrons by suitably
heating the metal surface.
(ii)Field emission : The emission of electrons by applying very
strong field of the order of 10
8
Vm
–1
to a metal.
26
Dual Nature of
Radiation and Matter

690. 686 PHYSICS
(1) No ph
otoelectrons are emitted, if the frequency of incident
light is less than some cut-off frequency (i.e., threshold
frequency) n
0
. It is inconsistent with the wave theory of
light, which predicts that photoelectric effect occurs at any
frequency provided intensity of incident light is sufficiently
high.
(2) The maximum kinetic energy of the photoelectrons is
independent of light intensiy, but increases with increasing
the frequency of incident light.
(3) Electrons are emitted from surface almost instantaneously
(less than 10
–9
sec after the surface illumination), even at
low intensity of incident light(classicaly we assume that the
electrons would require some time to absorb the incident
light before they acquire enough kinetic energy to escape
from metal).
These above points were explained by Einstein in 1905 by treating
the light as stream of particles.
Taking Max Planck assumptions, Einstein postulated that a beam
of light consists of small packets of energy called photons or
quanta. The energy E of a photon is equal to a constant h times
its frequency n
i.e.,
l
=n=
hc
hE ...(2)
where h is a
universal constant called Planck’s constant &
numerical value of h = 6.62607 × 10
–34
J.s
When a photon arrives at surface, it is absorbed by an electron.
This energy transfer is an All-or-None process, in contrast to
continuous transfer of energy in classical theory; the electrons
get all photon’s energy or none at all. If this energy is greater than
the work function (f) of the metal (f is the minimum energy
required to remove the electron from metal surface), the electron
may escape from the surface. Greater intensity at a particular
frequency means greater number of photons per second absorbed
& consequently greater number of electrons emitted per second
& so greater current.
To obtain maximum kinetic energy
2
max max
1
2
K mv=
for an emitted ele
ctron, applying law of conservation of energy.
According to it

2
max max
1
2
K mvh= = n-f
0
[]hf=n

max0
()Kh= n-n ...(3)
or
0m ax0
()eVKh= = n-n ...(4)
or,
2
max0
0
1 11
()
2
æö
=-= -=
ç÷
llèø
s
mv h v v hc ev
This is the Einste in’s photoelectric equation.
where V
0
= cut-off potential
n
max
= maximum velocity obtained by photoelectrons
n = frequency of incident light i.e., photon
n
0
= cut off frequency or threshold frequency.
(iii)Photo-electric emission : The emission of electrons when
light of suitable frequency illuminates metal surface.
PHOTOELECTRIC EFFECT (EINSTEIN’S PHOTOELECTRIC
EQUATION
In 19
th
century, experiments showed that when light is incident
on certain metallic surfaces, electrons are emitted from the
surfaces. This phenomenon is known as the photoelectric effect
& emitted electrons are called photoelectrons. The first
discovery of this phenomenon was made by Hertz.
G
V
C
Anode A
Light
Battery
C
a t hode
C
When light strikes the cathode C (metallic surface), photo
electrons are ejected. Electrons are collected at anode A,
constituting a current in the circuit. (Photoelectric effect)
Fig. shows, when light strikes the cathode C, electrons are emitted
& they are collected on anode A due to potential difference provided
by battery and constitutes the current in the circuit (observed by
Galvanometer G.)
A plot of photoelecric current versus the potential difference V between
cathode & anode is shown in fig below.
Photoelectric current versus voltage for two light intensities.
At a voltage less than –V
0
the current is zero.
It is clear from fig. that photoelectric current increases as weincrease the intensity of light & obtain saturation value at largervalue of potential difference V between cathode & anode. If V isnegative then, photoelectrons are repelled by negative cathodeand only those electrons reaches anode, who have energy equalto or greater than eV. But if V is equal to V
0
, called stopping
potential (i.e., cut off. potential), no electrons will reach the anode
i.e., Maximum kinetic energy of electron = eV
0
or K
max
= eV
0
...(1)
where e is charge of electron (e = 1.6 × 10
–19
coulomb).
But some features of photoelectric effect cannot be explained byclassical physics & the wave theory of light.

691. 687Dual Nature of Radiation and Matter
n
0
is different for different metallic surfaces. For most metals the
threshold frequency is in ultarviolet region of spectrum.
(Corresponding to l between 200 & 300 nm), but for potassium &
cesium oxides, it is in the visible spectrum (l between 400 & 700 nm).
Work Function (f) of Some Elements Given in Brackets :
Al (4.3eV) Ni (5.1 eV)
C (5.0 eV) Si (4 .8 eV)
Cu (4.7 eV) Ag (4.3 eV)
Au (5.1eV) Na (2 .7 eV)
where 1 eV = 1.602 × 10
–19
joule.
Within the framework of photon theory of light (Quantum theory
of light) we can explain above failures of classical physics.(1) It is clear from eq. (3) that if energy of photon is less than
the work function of metallic surface, the electrons will neverbe ejected from the surface regardless of intensity of incidentlight.
(2)K
max
is independent of intensity of incident light, but it
depends on the frequency of incident light i.e., K
max
µ n
(frequency of light).
(3) Electrons are emitted almost instantaneously consistent with
particle view of light in which incident energy is concentrated
in small packets (called photons) rather than over a large
area (as in wave theory).
Various Graphs Related to Photoelectric Effect
1. Photocurrent versus intensity of light graph
I
Intensity
(W/m)
2
(m
A)
nn >
0
2. Photocurrent versus potential graph
I
1
I
2
–V
0 V
(mA)
I n n n
1 0= >
2 I > I
21
I
(mA)
–V
0
2
–V
0
1
nnn
210
> >
I = I
12
V
3. Maximum kinetic energy versus potential graph.
n
0
2
n
0
1
1
2
mv
max
2
n
0
1
n
0
2
for metal 1 for metal 2
n
Keep in Memory
1.
Mass spectrograph is an appratus used to determine the
mass or the specific charge (e/m) of positive ions. Examples
are (a) Thomson mass spectrograph (b) Bain bridge mass
spectrograph (c) Aston mass spectrograph (d) Dempster
mass spectrograph etc.
2.In photoelectric effect all photoelectrons do not have same
kinetic energy. Their KE ranges from zero to E
max
which
depends on frequency of incident radiation and nature of
cathode.
3.The photoelectric effect takes place only when photons
strike bound electrons because for free electrons energy
and momentum conservations do not hold together.
4.Cesium is the best photo sensitive material.
5.Efficiency of photoemission,
e
p
Number of photoelectrons emitted
nper unitarea per unit time
Number of photons incident n
per unit area per unit time
h==
e
p
IIntensity of emitted electrons
Intensityof incident radiation I
h==
Therefore, h==
ee
pp
nI
nI
6.Maximum ve locity of emitted electrons
00
max
0
2( )2(
)2
s
h v v hc ev
v
mmm
- l -l
===
ll
7.Sto
pping potential
00
0
()()- l-l
==
ll
s
h v v hc
V
ee
de-B
ROGLIE EQUATION (DUAL NATURE OF MATTER)
In 1924, Louis de Broglie, wrote a doctoral dissertation in whichhe proposed that since photons have wave and particlecharacterstics, perhaps all forms of matter have wave as well asparticle properties.This is called dual nature of matter. According to which A matter
particle moving with a velocity v can be treated as a wave ofwavelength l. This l is called de-Broglie wavelength & it is defined
as :
hh
P mv
l== ...(1)
whe
re m is the mass of matter particle & these waves are called
matter waves.Further with the analogy of photon, the frequency of matter wavesis
E
h
n= ...(2)
The dual
nature of matter is quite apparent in these two equations
(equations (1) & (2)). i.e., each equation contains both particleconcepts (mv & E) & wave concepts (l & n). It is clear from next
topic that Compton effect confirm the validity of p = h/l for
photons, and the photoelectric effect confirms the validity andE = hn for photons.

692. 688 PHYSICS
de-Brog
lie wavelength associated with electron accelerated under
a potential difference V volt is given by
12.27
Å
V
l=
de-Broglie wave
is not an electromagnetic wave but the matter
wave.
Wavelength of matter waves associated with accelerated charged
particles :
If V is the accelerating voltage applied then :
(a)For the charged particle :
Energy E = qV ;
Velocity v =
22qVE
mm
=
Momentum p = 22mE mqV= ;
Wavelengt
h l
=
22
hh
mqV mE
=
(b)For electronl
e
=
12.27
V
Å
(c)For proton l
p
=
0.286
V
Å
(d)For alpha particlel
a
=
0.101
V
Å
(e)For deuteronl
d
=
0.202
V
Å
For neutral particles (neut
ron, atom or molecule) :
(a) If E is the energy of the particle, then,
2
hh
p mE
=
(b) If T is the t
emperature, then,
3
h
mkT
l=
DAVISSON-GERMER EXPERIMENT
Idea of
de-Broglie wave was tested beautifully in 1926 in an
experiment performed by C. Davission (1881-1958) and L.H. Jermer
(1896-1971). They directed a beam of electrons at a crystal and
observed that the electrons scattered in various directions for a
given crystal orientation.
In this experiment the pattern
formed by the electrons
reflected from the crystal
lattice of aluminium is almost
identical to that produced by
X-rays. This strongly
suggests that the electrons
have a wavelength l
associated with them and that the Bragg condition for X-ray
diffraction holds true for electron also :
Bragg’s equation
nl = D sin q or n l = 2d sin f.
Diffraction maximum of electrons accelerated with 54 volt is
obtained at q = 50º for the Nickel crystal.
Explanation of Bohr’s quantum condition :
(a)According to Bohr’s quantum conditions :
Angular momentum, mvr
n
=
h
n
2p
(b)Matter waves associated with the electrons moving in an
orbit are stationary waves.
(c)For the production of stationary waves in the orbit the
circumference of the orbit should be integral multiple of
wavelength of waves associated with the electron,
i.e., 2pr
n
= nl, where
h
mv
l=
\mvr
n
=
2
nh
p
COMPTON EFFECT
Further experime
ntal proof for photon concept(i.e., particle nature
of light) was discovered in 1923 by American Physicist, A.H.
Compton. According to which, when a monochromatic beam of
X-rays (wavelength l
0
) strikes the electron in a carbon target, two
types of X-rays are scattered. The first type of scattered wave has
same wavelength (l
0
) as the incoming X-rays, while second type
has a longer wavelength (l) than incident rays (First type of X-
rays are called unmodified x-rays, while second type of X-rays are
called modified X-rays.) This change in wavelength i.e. Dl = l –
l
0
is called Compton shift & this effect is called Compton effect”.
Incident x-rays
Carbon target
which consists
of free electron
Recoiling
electron
q
f
Scattered X
-ray
Ehc/l
p=h/l
m
E =hc/
0
l
p =h/
0
l
0
cosq
sinq
q
f
cosf
sinf
Diagram shows c
ompton scattering of an x-rays by free electron in
a carbon target. The scattered x-rays has less energy than the
incident x-rays. The excess energy is taken by recoiling electrons.
This effect cannot be explained by classical theory (by wave nature
of light). According to classical model, when X-rays of frequency
n
0
is incident on the material containing electrons, then electrons
do oscillate & reradiate electromagnetic waves of same frequency
n
0
. Hence scattered X-rays has same frequency n
0
& same
wavelength as that of incident X-rays.
Compton treated this processes as a collision between a photon

693. 689Dual Nature of Radiation and Matter
& an electron. In this treatment, the photon is assumed as a particle
of energy
E = hn
0
= hc/l
0
...(1)
Further, the rest mass of photon is zero (because photon travels
with the speed of light) hence the momentum of photon can be
written as
Ehch
p
cc
===
ll
...(2)
To der
vie the compton shift. Dl, we apply both conservation of
energy & momentum.
Conservation of energy :
0
e
hc hc
K=+
ll
...(3)
Where hc/l i
s energy of scattered X-rays, K
e
is kinetic energy of
recoling electron & hc/l
0
is the energy of incoming X-rays. Since
the electron may travel with the speed of light, so we must use
relativistic expression of K
e
in equation (3), and we obtain
22
0
mcmc
hchc
-g+
l
=
l
...(4)
wher
e m is rest mass of electron and mc
2
is the rest mass energy
of the electron
where
22
1
1 v /c
g=
-
Conser
vation of momentum :
0
hh
cos mvcos=q+gf
ll
x - com
ponent ...(5)
h
0 sin mv sin=q-gf
l
y - com
ponent ...(6)
where p
e
= gmv is the relativistic expression for momentum of
recoiling electron.
By eliminating v & f from equation (4) to (6), we obtain

)cos1(
mc
h
0
q-=l-l=lD ...(7)
or 0.0243(1 cos )ÅDl= -q ...(8)
It is clear
from expression (7) that compton shift Dl depends on
scattering angle q & not on the wavelength. Keep in Memory
1.The
wave nature of light shows up in the phenomena of
interference, diffraction and polarisation whereas
photoelectric effect and compton effect shows particle nature
of light.
2.The maximum kinetic energy of the photoelectrons varies
linearly with the frequency of incident radiation but is
independent of its intensity.
X-RAYS
·The X-rays were discvoered by Prof. Roentgen, a German
scientist in 1885. He was awarded Nobel Prize for this
discovery in 1901. X-rays are electromagnetic waves.
·The modern apparatus for the production of X-rays was
developed by Dr. Coolidge in 1913.
·X-rays are produced when fast moving electrons are
suddenly stopped on a metal of high atomic number.
Properties of X-rays
(i)They are not deflected by electric or magnetic field.
(ii)They travel with the speed of light.
(iii)There is no charge on X-rays.
(iv)X-rays show both particle and wave nature.
(v)They are invisible.
Continuous and Characteristic X-rays
Experimental observation and studies of spectra of X-rays reveal
that X-rays are of two types and so are their respective spectras.
Characteristic X-rays and Continuous X-rays.
Characteristic X-rays: The spectra of this group consists of
several radiations with specific sharp wavelengths and frequency
similar to the spectrum (line) of atoms like hydrogen. The
wavelengths of this group show characteristic discrete radiations
emitted by the atoms of the target material. The characteristic
X-rays spectra helps us to identify the element of target material.
When the atoms of the target material are bombard with high
energy electrons (or hard X-rays), which posses enough energy
to penetrate into the atom, knock out the electron of inner shell
(say K shell, n = 1). When an electron is missing in the ‘K’ shell, an
electron from next upper shell makes a quantum jump to fill the
vacancy in ‘K’ shell. In the transition process the electron radiates
energy whose frequency lies in the X-rays region. The frequency
of emitted radiation (i.e. of photon) is given by
2
22
12
11
e
v RZ
nn
æö
=-
ç÷
èø
; where R is constant and Z
e
is effective
atomic number. Generally Z
e
is taken to be equal to Z – s, where Z
is proton number or atomic number of the element and s is called
the screening constant. Due to the presence of the other electrons.The charge of the nucleus as seen by the electron will be different
in different shells.
KL
k
e
i
e
i
e
KL
hv
(x-ray)
Knocking out e
-
of K shell by incident electron e
i
emission of X-ray photon (K
a
- series)
Another vacancy is now created in the ‘L’ shell which is againfilled up by another electron jump from one of the upper shell (M)which results in the emission of another photon, but of different
X-rays frequency. This transition continues till outer shells are
reached. Thus resulting in the emission of series of spectral line.
The transitions of electrons from various outer shells to the inner
most ‘K’ shell produces a group of X-rays lines called as
K-series. These radiations are most energetic and most
penetrating. K-series is further divided into
K , K ,K
a bg ….
dependin
g upon the outer shell from which the transition is made.
The jump of electrons from outer shells to ‘L ’ shell results in
L-Series X-rays

694. 690 PHYSICS
K-series
L-series
M-series
n=2
n=3
n=4
n=1
K
a
K
b
K
gK
d
L
a
L
bL
g
M
aM
b
N
a
N
L
M
K
n=a
and so on.
Continuo
us X-rays : In addition to characteristic X-rays tubes
emit a continuous spectrum also. The characteristic line spectra is
superimposed on a continuous X-rays spectra of varying
intensities. The wavelength of the continuous X-rays spectra are
independent of material. One important feature of continuous X-
rays is that they end abruptly at a certain lower wavelength for a
given voltage. If an electron beam of energy eV (electron volts) is
incident on the target material; the electrons are suddenly stopped.
If the whole of the energy is converted to continuous radiation,
then l
min
(corresponding to energy maximum) = hc/Ve where V is
the voltage applied.
X-ray photon
hv
Target atom
K K¢
The class
ical theory of electromagnetism states that the suddenly
accelerated or decelerated electrons emit radiations ofelectromagnetic nature called as bremsstrahlung (braking
radiation) and wavelength of such radiation is continuous because
the loss in energy is statistical. At the peak, the probability ofmaximum number of electrons producing radiation.The wavelength of X-rays emitted is minimum corresponding to
the electron which hits the target with maximum speed. This
electron is completely stopped and will emit the photon of highest
energy.
As the electrons lose energy by collision, longer wavelengths are
produced the shape of the curve is statistical.
I
n
t
e
n
s
i
t
y
Wavelength (nm)
0.01 0.1 1.0
minl
K
g
K
b
K
a
L
a
L
b
M
a
50 kV
40 kV
30 kV
20 kV
00.020.040.060.080.10
Wavelength (nm)
R
e
l
a
t
i
v
e

i
n
t
e
n
s
i
t
y
Wavelength of X-Rays (Daume Hunt Law)
(i) When an electron is accelerated through a potential
difference V then
the energy accquired by electron
21
2
eV mv== .
(ii) When th
ese high energy electrons fall on target T of high
atomic number, then X-rays are produced, whose wavelength
is given by
;\ = n= l=
l
hc hc
eVh
eV
.
(iii) The en
ergy of X-rays of wavelength l is
hc
Eh= n=
l
(iv) The shortest wavelength of X-rays emitted is
1240hc
nm
eVV
l== i.e.
1
V
lµ.
It is called Daume Hunt law.
Ty
pes of X-Rays
1.Hard X- rays : The X-rays of high frequency or low
wavelength are said to be hard X-rays. They have higher
penetrating power.
2.Soft X-rays : The X-rays of longer wavelength are called
soft X-rays.
Moseley’s Law
Moseley used different elements as target in the X-ray tube. He
found that K
a
radiation of different elements were different
Mathematically,
()aZbn=-
where a and b depend on the particular line of the radiation
For Ka,
3
a Rc
4
= and b = 1
wher
e R = Rydberg constant and c = speed of light
In general the wavelength of K - lines are given as
2
2
11
( 1)1RZ
n
éù
=--
êú
l ëû
where n = 2, 3, .....
Absorption of X-rays
·X-rays are absorbed by the materials according to the
relation I = I
0
e
–mx
, where m is absorption coefficient and x
is the thickness of the mateiral. Here I is the intensity after
penetrating the material through distance x and I
0
is the
initial intensity of the X-rays.
·The coefficient of absorption (m) of the material is given by
1/2
0.6931
x
m= where x
1/2
is the distanc
e after traversing which
the intensity of X-rays is reduced to half.

695. 691Dual Nature of Radiation and Matter
·Absorption coefficient depends on the nature of material
and wavelength of X-rays i.e. m = cZ
4
l
3
.
It means that (a) m µz
4
(b) mµl
3
(c) mµn
–3
.
Fluorescen
ce : Certain substances (like quinine sulphate,
fluoricine, barium platinocyanide, uranium oxide etc.), whenilluminated with light of high frequency (ultraviolet, violet, etc.)
emit light of lower frequency. The phenomenon is called
fluorescence.
·When quinine sulphate is illuminated with ultraviolet or
violet light it gives out blue light. The fluroescence of barium
sulphate as well as uranium oxide gives out green light when
illuminated with ultraviolet or violet light.
·The house hold tubes are painted from inside with
magnesium tungstate or zinc-beryllium silicate. They are
fluorescent materials. The ultraviolet light generated inside
the tube falls on the walls, where magnesium tungstate gives
blue light and zinc beryllium silicate gives yellow orange
light. The mixture of the two produces white light. If the
inner side of the tube is painted with cadmium borate it
gives fluorescence of pink light and when painted with zinc
silicate, it gives fluorescence of green light.
·The fluorescence occurs as long as the material is
illuminated.
Phosphorescence : Fluorescent materials emit light only so long
as light is incident on them. There are certain susbstances which
continue emitting light for some time after the light incident on
them is stopped. This phenomenon is called phosphorescence.
For example, if we make blue light incident on a zinc-sulphide
(ZnS) screen, then it produces phosphorescence of green colour.
Calcium sulpide and barium sulphide, after absorbing sunlight,
produce blue phosphorescence for some time. Time of
phosphorescence is different for different materials.
Keep in Memory
1.The
stopping potential (and hence the maximum kinetic
energy of emitted electrons) is independent of the intensity
of light but that the saturation current (and hence the number
of emitted photoelectrons) is proportional to the intensity.
2.Photoelctric effect doesn't take place below the threshold
frequency for the photometal used.
3.In compton effect, the change in wavelength is independent
of incident photon as well as of the nature of the scatterer,
but depends only on the angle of scattering (q).
4.The quantity
h
( 0.02426Å)
mc
= is called Co
mpton
wavelength.
5.The maximum wavelength change possible in comptoneffect is 0.05Å.
6.Compton effect can't be observed for visible light rays.
7.In compton effect, the direction of recoil electron is given
by tan
sin
.
' cos
lq
f=
l-lq
8.The kinetic energy of recoil electron is given by
2
x(1cos)h
T h ,whe rex
1 x(1 cos)mc
éù-qn
=n=
êú
+-qëû
9.de-Br
oglie wavelength of a particle of K.E., E
k
is given by
k
h
.
2mE
l=
10.de-Broglie w
avelength for a charged particle with charge q
and accelerated through a potental difference V is given by
h
.
2mqV
l=
11.de-Brog
lie wavelength of a material particle at temperature
T is given by
h
.
2mkT
l= , where k is Boltzmann’s constant.
Application of X-rays
Following are some important and useful applications of X-rays.
1. Scientific applications: The diffraction of X-rays at crystals
opened new dimension to X-rays crystallography. Various
diffraction patterns are used to determine internal structure
of crystals. The spacing and dispositions of atoms of a crystal
can be precisely determined by using Bragg’s law :
nl = 2d sin q.
2. Industrial applications: Since X-rays can penetrate through
various materials, they are used in industry to detect defects
in metallic structures in big machines, railway tracks and
bridges. X-rays are used to analyse the composition of alloys
and pearls.
3. In radio therapy: X-rays can cause damage to the tissues of
body (cells are ionized and molecules are broken). So X-rays
damages the malignant growths like cancer and tumors which
are dangerous to life, when is used in proper and controlled
intensities.
4. In medicine and surgery: X-rays are absorbed more in heavy
elements than the lighter ones. Since bones (containing
calcium and phosphorus) absorb more X-rays than the
surrounding tissues (containing light elements like
,,HCO),
their shadow is
casted on the photographic plate. So the
cracks or fracture in bones can be easily located. Similarly
intestine and digestive system abnormalities are also detectedby X-rays.
Example 1.
An electron is moving with velocity 10
7
m/s on a circular
path of radius 0.57 cm in a magnetic field of 10
–2
Wb/m
2
.
Find the value of e/m for electron.
Solution :
)r/mv(Bve
2
= or
Br
v
m
e
=
\
0057.010
10
m
e
2
7
´
=
-
= 1.76
× 10
11
coulomb/kg.
Example 2.
Find the ratio of specific charge e/m of a proton to that ofan
a particle.

696. 692 PHYSICS
Solution :
÷
÷
ø
ö
ç
ç
è
æ
=÷
ø
ö
ç
è
æ
pproton
m
e
m
e
and
p
m4
e2
m
e
=÷
ø
ö
ç
è
æ
a
\
1
2
)m4/e2(
)m/e(
)m/e(
)m/e(
p
pproton
==
a
or 2 : 1
Exam
ple 3.
Determine the ratio of momentum of an electron and an
alpha particle which are accelerated from rest by a
potential difference of 100 V.
Solution :
Q
Vevm
2
1 2
= or
2/1
)m/Ve2
(v=
\
2/1
)Vem2(vmp==
Now
2/1
ee
)100em2(p´´=
and
2/1
)100em4(p´´=
aa
\
)m2/m(p/p
ee aa
=
Example 4.
The work function of cesium is
1.8 eV. Light of 5000 Å is
incident on it. Calculate (a) threshold frequency andthreshold wavelength. (b) maximum K.E. of emittedelectrons. (c) maximum velocity of emitted electrons(d) if the intensity of the incident light be doubled, thenwhat will be the maximum K.E. of the emittedelectrons? (h = 6.6 × 10
–34
joule second, mass of electron
m = 9.0×10
–31
kg and speed of light c = 3 x 10
8
m/s).
Solution :
(a)
00
vhW= or
00
W/hv=
)106.1(8.1W
19
0
-
´´= = 2.9 × 10
–19
joule
\
19
0
0
34
W 2.9 10 joule
h6.6 10 joule second
-
-
´
n==
´
\ Threshold fre
quency = 4.4 × 10
14
sec
–1
Threshold wavelength
114
8
0
0
s104.4
s/m100.3c
-
´
´
=
n
=l
= 6.8 × 10
–7
metre =
6800 Å
(b)
00k
W
hc
WhE -
l
=-n=
=
348
19
10
(6.63 10 )(3 10 )
(2.9 10 joule)
5000 10
-
-
-
´´
-´
´
= (4.0 – 2
.9) ×10
–19
= 1.1 × 10
–19
joule.
(c)
2
maxk
mv
2
1
E=
\
ú
ú
û
ù
ê
ê
ë
é
´
´´
=÷
ø
ö
ç
è
æ
=
-
-
31
19
k
max
100.9
)101.1(2
m
E2
v
= 5.0 × 10
5
m/sec.
(d
) The K.E. of emitted electrons does not depend upon
the intensity of light. Hence if the intensity of incident
light be doubled, the energy will remain unchanged.
Example 5.
What will be the ratio of the de-Broglie wavelength of
proton and
a particle of same energy?
Solution :
mE2
h
mv
h
==l
\
2
mv
2
1
E= or 2 m E = m
2
v
2
or mE2mv=
so, 2
1
4
m
m
p
p
==
÷
÷
ø
ö
ç
ç
è
æ
=
la
l
a
or 1:2:
p
=ll
a
Exam
ple 6.
Calculate the energy and momentum of a photon of
wavelength 6600Å.
Solution :
Energy of photon
348
19
10
hc 6.62 10 3 10
E 310J
6600 10
-
-
-
´ ´´
= = =´
l ´
Momentum
of photon 34
27
10
h 6.6 10
p 10 kg m / sec
6600 10
-
-
-
´
===
l ´
Example 7.
Find
the number of photons emitted per second by a
25 watt source of monochromatic light of wavelength 6000Å.
Solution :
Energy of one photon
348
19
10
hc 6.62 10 3 10
E h 3.315 10 J
6000 10
-
-
-
´ ´´
=n== =´
l ´
No. of ph
otons emitted per second
=
Total energy emitted per second P
Energy of the photon E
=
19 1925
10 7.54 10
3.315
-
= ´ =´
Example 8.
In a photoelect
ric experiment, with light of wavelength
l,
the fastest electron has speed v. If the exciting wavelength is changed to 3
l/4, then find the speed of the fastest emitted
electron.
Solution :
21 hc
mv
2
= -f
l
1 hc 4hc
m
2(3/4)3
= -f= -f¢
ll
Clearly,
4
vv
3
>¢

697. 693Dual Nature of Radiation and Matter
Example 9.
What is the de-Broglie wavelength associated with (a) an
electron moving with a speed of 5.4 × 10
6
m/s, and (b) a
ball of mass 150 g travelling at 30.0 m/s?
Solution :
(a)For the electron
Mass m = 9.11 ×10
–31
kg, speed v = 5.4 × 10
6
m/s. Then,
momentum p = m v
= 9.11 ×10
–31
(kg) × 5.4 × 10
6
(m/s)
p = 4.92 × 10
–24
kg m/s
de-Broglie wavelength, l = h/p =
34
24
6.63 10 Js
4.92 10 kg m/ s
-
-
´
´
= 0.1
35 nm
(b)For the ball: Mass m' = 0.150 kg,
Speed v' = 30.0 m/s.
Then momentum p' = m'v'
= 0.150 (kg) × 30.0 (m/s)
p' = 4.50 kg m/s
de Broglie wavelength l' = h/p'
=
34
6.63 10 Js
4.50 kg m / s
-
´
= 1.47
× 10
–34
m
The de-Broglie wavelength of electron is comparable
with X-ray wavelengths. However, for the ball it is about
10
–19
times the size of the proton, quite beyond
experimental measurement.
Example 10.
What is the de Broglie wavelength associated with an
electron, accelerated through a potential difference of 100
volt?
Solution :
Accelerating potential V = 100 V. The de Broglie wavelength,
h 1.227
nm
p V
l==

1.227
nm 0.123nm
100
l==
The de-B
roglie wavelength associated with an electron in
this case is of the order of X-ray wavelengths.
Example 11.
If the K
a
radiation of Mo (Z = 42) has a wavelength of
0.71 Å, calculate wavelength of the corresponding radiation
of Cu, i.e., K
a
for Cu (Z = 29) assuming s = 1.
Solution :
According to Moseley’s law :
v a(Z 1)=-
Þ
2
(Z1)v-µ or
2
(Z 1) 1/- µl
Þ
2
M0 Cu
2
MoCu
( Z 1)
( Z 1)
-l
=
l-
Þ

2
M0
Cu Mo
2
Cu
(Z 1)
(Z 1)
-
l =l
-

2
41
0.71 Å 1.52Å
28
æö
=´=
ç÷
èø
Example 12.
The wave
length of K
a
x-rays of two metals ‘A’ and ‘B’ are
4
1875R
and
1
675R
respectively, where ‘R’ is Rydbergg
constant. Find the number of elements lying between A and
B according to their atomic numbers.
Solution :
Using
2
22
21
1 11
R(Z 1)
nn
éù
=-- êú
l êúëû
For a particle, n
1
= 2, n
2
= 1
For metal A :
2
11
1875 R 3
R (Z 1) Z 26
44
æö
= - Þ= ç÷
èø
For
metal B :
2
22
3
675 R R (Z1) Z 31
4
æö
= - Þ= ç÷ èø
Therefor
e, 4 elements lie between A and B.

698. 694 PHYSICS
Thermionic emission
Emission of electrons
by suitably heating of
metal surface.
Field emission
Emission of electrons by
applying a very strong
electric field to a metal
Photoelectric emissi
o
n
Emission of free ele
c
t
r
o
n
s
from the surface of m
e
t
a
l
s

when light radiation
o
f
suitable frequency fa
l
l

o
n
i
t
Work function
Minimum energy required
to just escape electron from
metal surface
0 0hv f=
Methods of electron
emission
Photon Tiny packets of light energy
Energy of a photon
E = h n
2m
a
x
0
0
1
m
v
h
(
v
v
)
h
v
h
v
2
=
-
=
-
0
1
1
h
c
æ
ö
=
-
ç
÷
l
l
è
ø
Electron emission
Emission of electrons
from the surface of metal
De-Broglie wavelength
m
h h h
p mv2ev
l== =
Wave nature of
particles De-Broglie
hypothesis
Davisson and Germer
experiment Confirms
the wave nature
of electrons
Uses of photocell
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entering an auditorium
burglar alarm, in
motion picture and
television
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0
CONCEPT MAP

699. 695Dual Nature of Radiation and Matter
1.In which of the following, emission of electrons does not
take place?
(a) Thermionic emission (b) X-rays emission
(c) Photoelectric emission (d) Secondary emission
2.In Davison-Germer experiment, an electron beam is incident
on a crystal. The reflected beam consists of
(a) photons (b) protons
(c) x-rays (d) electrons
3.The de-Broglie wavelength of an electron moving in the nth
Bohr orbit of radius r is given by
(a)
2r
n
p
(b)nrp (c)
nr
2p
(d)
nr
p
4.The kinetic energy of electron (in electron volt) moving with
a velocity of 4 × 10
6
m/s will be
(a) 60 eV (b) 50 eV(c) 30 eV(d) 45.5 eV
5.Photoelectric effect is the phenomenon in which
(a) photons come out of a metal when it is hit by a beam of
electrons.
(b) photons come out of the nucleus of an atom under the
action of an electric field.
(c) electrons come out of a metal with a constant velocity
which depends on the frequency and intensity of
incident light wave.
(d) electons come out of a metal with different velocities
not greater than a certain value which depends only
on the frequency of the incident light wave and not on
its intensity.
6.The momentum of a photon of wavelength l is
(a)hl (b) h/l (c)l/h (d) h/cl
7.A photo sensitive metal is not emitting photo-electrons when
irradiated. It will do so when threshold is crossed. To cross
the threshold we need to increase
(a) intensity (b) frequency
(c) wavelength (d) None of these
8.A photoelectric cell is a device which
(a) converts light into electricity
(b) converts electricity into light
(c) stores light
(d) stores electricity
9.A particle with rest mass m
0
is moving with speed of light c.
The de-Broglie wavelength associated with it will be
(a)µ (b) zero(c)m
0
c/h(d) hn/m
0
c
10.Einstein’s work on photoelectric effect provided support
for the equation
(a) E = hn (b) E = mc
2
(c)
2
n
Rhc
E
-
= (d)
2
mv
2
1
.E.K=
11.The maximum ve
locity of an electron emitted by light of
wavelength l incident on the surface of a metal of work-
function f is
(a)
l
lf+
m
)hc(2
(b)
l
lf+
m
)hc(2
(c)
l
lf-
m
)hc(2
(d)
m
)h(2f-l
12.Einstein’s pho
toelectric equation is
k
Eh
= n-f. In this
equatio
n E
k
refers to
(a) kinetic energy of all the emitted electrons
(b) mean kinetic energy of emitted electrons
(c) maximum kinetic energy of emitted electrons
(d) minimum kinetic energy of emitted electrons
13.If E
1
, E
2
, E
3
are the respective kinetic energies of an electron,
an alpha-particle and a proton, each having the same
de-Broglie wavelength, then
(a)E
1
> E
3
> E
2
(b) E
2
> E
3
> E
1
(c)E
1
> E
2
> E
3
(d) E
1
= E
2
= E
3
14.When the speed of electrons increase, then the value of its
specific charge
(a) increases
(b) decreases
(c) ramains unchanged
(d) increases upto some velocity and then begins to
decrease
15.In an electron gun the control grid is given a negative
potential relative to cathode in order to
(a) decelerate electrons
(b) repel electrons and thus to control the number of
electrons passing through it
(c) to select electrons of same velocity and to converge
them along the axis.
(d) to decrease the kinetic energy of electrons
16.X-rays are produced in X-ray tube operating at a given
accelerating voltage. The wavelength of the continuous
X-rays has values from
(a) 0 to ¥
(b)l
min
to ¥, where l
min
> 0
(c) 0 to l
max
, where l
max
< ¥
(d)l
min
to l
max
, where 0 < l
min
< l
max
< ¥
17.As intensity of incident light increases
(a) photoelectric current increase
(b) K.E. of emitted photoelectron increases
(c) photoelectric current decreases
(d) K.E. of emitted photoelectrons decreases
18.Which of the following shows par.ticle nature of light?
(a) Refraction (b) Interference
(c) Polarization (d) Photoelectric effect
19.Which of the following when falls on a metal will emit
photoelectrons ?
(a) UV radiations (b) Infrared radiation
(c ) Radio waves (d) Microwaves
20.Light of frequency v falls on a material of threshold
frequency v
0
. Maximum kinetic energy of emitted electron
is proportional to
(a) v–v
0
(b) v
(c)
0
v–v (d) v
0

700. 696 PHYSICS
1.In a photoele
ctric experiment the stopping potential for the
incident light of wavelength 4000Å is 2 volt. If the wavelength
be changed to 3000 Å, the stopping potential will be
(a) 2 V (b) zero
(c) less than 2 V (d) more than 2 V
2.A proton and a-particle are accelerated through the same
potential difference. The ratio of their de-Broglie wavelength
will be
(a) 1 : 1(b) 1 : 2(c) 2 : 1 (d)
1:22
3.An electron is a
ccelerated by a p.d. of 1000 V. Its velocity
will be
(a) 3.78 × 10
7
m/s (b) 1.89 × 10
7
m/s
(c) 5.67 × 10
7
m/s (d) 0.95 × 10
7
m/s
4.The photoelectric work function for a metal surface is
4.125 eV. The cut-off wavelength for this surface is
(a) 4125 Å (b) 3000 Å(c) 6000 Å (d) 2062 Å
5.The energy of a photon of green light of wavelength 5000Å
is
(a) 3.459 × 10
–19
joule(b) 3.973 × 10
–19
joule
(c) 4.132 × 10
–19
joule(d) 8453 × 10
–19
joule
6.4eV is the energy of incident photon and the work function
is 2eV. The stopping potential will be
(a ) 2V(b) 4V ( c ) 6 V(d)
V22
7.The velocity of
a body of rest mass
o
m is
c
2
3
(where c i
s
the velocity of light in vacuum). The mass of this body is:
(a)
o
m
2
3
÷
÷
ø
ö
ç
ç
è
æ
(b) om
2
1
÷
ø
ö
ç
è
æ
(c)
o
m3 (d) 2m
0
8.If the en
ergy of a photon is 10 eV, then its momentum is
(a) 5.33 × 10
–23
kg m/s(b) 5.33 × 10
–25
kg m/s
(c) 5.33 × 10
–29
kg m/s(d) 5.33 × 10
–27
kg m/s
9.The photoelectric threshold of Tungsten is 2300Å. The energy
of the electrons ejected from the surface by ultraviolet light
of wavelength 1800Å is
(a) 0.15 eV (b) 1.5 eV(c) 15 eV(d) 150 eV
10.The maximum kinetic energy (E
max
) of photoelectrons
emitted in a photoelectric cell varies with frequency (n) as
shown in the graph. The slope of the graph is equal to
(a) charge of the electron
(b)
m
e
of the electron
max
E
n
n
0
(c) work function of the emitter
(d) Plank’s constant
11.Ultraviolet radiation of 6.2 eV falls on an aluminium surface
(workfunction 4.2 eV). The kinetic energy in joule of the
faster electron emitted is approximately
(a) 3 × 10
–21
(b) 3 × 10
–19
(c) 3 × 10
–17
(d) 3 × 10
–15
12.A light having wavelength 300 nm fall on a metal surface.
The work function of metal is 2.54 eV, what is stopping
potential ?
(a) 2.3 V(b) 2.59 V (c) 1.59 V (d) 1.29 V
13.A and B are two metals with threshold frequencies
1.8 × 10
14
Hz and 2.2 × 10
14
Hz. Two identical photons of
energy 0.825 eV each are incident on them. Then
photoelectrons are emitted in (Take h = 6.6 × 10
–34
Js)
(a) B alone (b) A alone
(c) neither A nor B(d) both A and B.
21.The ratio of de-Broglie wavelengths of proton and
a-particle having same kinetic energy is
(a) 1:2(b) 1:22(c) 2 : 1(d) 4 : 1
22.W
hite X-rays are called white due to the fact that
(a) they are electromagnetic radiations having nature
same as that of white light.
(b) they are produced most abundantly in X ray tubes.(c) they have a continuous wavelength range.(d) they can be converted to visible light using coated
screens and photographic plates are affected by
them just like light.
23.In case of electrons and photons having the same
wavelength. What is same for them?
(a) Energy (b) Velocity
(c) Momentum (d) Angular momentum
24.The energy of a photon of wavelength l is
(a)hc l (b)
hc
l
(c)
hc
l
(d)
h
c
l
25.In the Davisson and Germer experiment, the velocity of
electrons emitted from the electron gun can be increased by
(a) increasing the potential difference between the anode
and filament
(b) increasing the filament current
(c) decreasing the filament current
(d) decreasing the potential difference between the anode
and filament

701. 697Dual Nature of Radiation and Matter
14.The de-Broglie wavelength of a proton (mass = 1.6 ×
10
–27
kg) accelerated through a potential difference of 1 kV
is
(a) 600 A (b) 0.9 × 10
–12
m
(c) 7 Å (d) 0.9 nm.
15.An ionisation chamber, with parallel conducting plates as
anode and cathodes has singly charged positive ions per
cm
3
. The electrons are moving toward the anode with
velocity 0.4 m/s. The current density from anode to cathode
is X10
4
m A/m
2
. The velocity of positive ions moving
towards cathode is
(a) 0.1 m/s (b) 0.4 m/s (c) zero (d) 1.6 m/s
16.A parallel beam of light is incident normally on a plane surface
absorbing 40% of the light and reflecting the rest. If the
incident beam carries 60 watt of power, the force exerted by it on
the surface is
(a) 3.2 × 10
–8
N (b) 3.2 × 10
–7
N
(c) 5.12 × 10
–7
N (d) 5.12 × 10
–8
N
17.Radiations of two photon’s energy, twice and ten times the
work function of metal are incident on the metal surface
successsively. The ratio of maximum velocities of
photoelectrons emitted in two cases is
(a) 1 : 2(b) 1 : 3(c) 1 : 4(d) 1 : 1
18.A photocell is illuminated by a small bright source placed 1 m
away. When the same source of light is placed 2 m away, the
number of electrons emitted by photocathode are reduced by
a factor of
(a) 1/8(b) 1/16(c) 1/2 (d) 1/4
19.In a photoelectric effect measurement, the stopping potential
for a given metal is found to be V
0
volt when radiation of
wavelength l
0
is used. If radiation of wavelength 2 l
0
is
used with the same metal then the stopping potential (in
volt) will be
(a)
2
V
0
(b) 2 V
0
(c)
0
0
e2
hc
V
l
+ (d)
0
0
e2
hc
V
l
-
20.An electro
n of mass m and charge e initially at rest gets
accelerated by a constant electric field E. The rate of change
of de-Broglie wavelength of this electron at time t ignoring
relativistic effects is
(a)
2
Ete
h-
(b)
E
eht-
(c) 2
Ete
mh-
(d)
Ee
h-
21.The wave
length of K
a
-line characteristic X-rays emitted by
an element is 0.32 Å. The wavelength of K
b
-line emitted by
the same element will be
(a) 0.32 Å (b) 0.39 Å (c) 0.49 Å (d) 0.27 Å
22.The X-rays of wavelength 0.5 Å are scattered by a target.
What will be the energy of incident X-rays, if these are
scattered at an angle of 72º ?
(a) 12.41 keV (b) 6.2 keV
(c) 18.6 keV (d) 24.82 keV
23.In a photo-emissive cell with exciting wavelength l, the
fastest electron has speed V. If the exciting wavelength is
changed to 3l/4, the speed of the fastest emitted electron
will be :
(a)
2/1
(¾)n (b)
2/1
)3/4(v
(c) less the
2/1
)3/4(v(d) greater than
2/1
)3/4(v
24.The momentum of photon whose frequency f is
(a)
c
hf
(b)
f
hc
(c )
f
h
(d)
hf
c
25.A small photoce
ll is placed at a distance of 4 m from a
photosensitive surface. When light falls on the surface the
current is 5 mA. If the distance of cell is decreased to 1 m,
the current will become
(a) 1.25 mA (b)
÷
ø
ö
ç
è
æ
16
5
mA
(c) 20 mA (d
) 80 mA
26.If 5% of the energy supplied to a bulb is radiated as visible
light, the number of visible quanta emitted per second by a
100 W bulb, assuming the wavelength of visible light to be
5.6 × 10
–5
cm, is
(a) 1.4 × 10
19
(b) 1.4 × 10
20
(c) 2 × 10
19
(d) 2 × 10
20
27.A material particle with a rest mass m
0
is moving with speed
of light c. The de-Broglie wavelength associated is given
by
(a)
cm
h
0
(b)
h
cm
0
(c) zero (d)¥
28.Which on e of the following graphs represents the variation
of maximum kinetic energy (E
K
) of the emitted electrons
with frequency u in photoelectric effect correctly ?
(a)
E
K
u
(b)
E
K
u
(c)
E
K
u
(d)
E
K
u
u
0
29.In a photoelec
tric effect experiment, for radiation with
frequency u
0
with hu
0
= 8eV, electrons are emitted with
energy 2 eV. What is the energy of the electrons emitted for
incoming radiation of frequency 1.25 u
0
?
(a) 1 eV (b) 3.25 eV
(c) 4 eV (d) 9.25 eV.

702. 698 PHYSICS
30.Two ins
ulating plates are both uniformly charged in
such a way that the potential difference between them is
V
2
– V
1
= 20 V. (i.e., plate 2 is at a higher potential). The
plates are separated by d = 0.1 m and can be treated as
infinitely large. An electron is released from rest on the inner
surface of plate 1. What is its speed when it hits plate 2?
(e = 1.6 × 10
–19
C, m
e
= 9.11 × 10
–31
kg)
1 2
0.1 m
Y
X
(a) 2.65 × 10
6
m/s (b) 7.02 × 10
12
m/s
(c)
1.87 × 10
6
m/s (d) 32 × 10
–19
m/s
31.A steel ball of mass m is moving with a kinetic energy K. Thede-Broglie wavelength associated with the ball is
(a)
h
2mK
(b)
h
2mK
(c)
h
2mK
(d) meaningless
32.If the X-ray tube is working at 20 kV then the minimum
wavelength of X-rays will be
(a) 0.31 Å (b) 0.62 Å (c) 0.93 Å (d) 0.47 Å
33.In an electron gun, the potential difference between the
filament and plate is 3000 V. What will be the velocity of
electron emitting from the gun?
(a) 3 × 10
8
m/s (b) 3.18 × 10
7
m/s
(c) 3.52 × 10
7
m/s (d) 3.26 × 10
7
m/s
34.Which metal will be suitable for a photoelectric cell using
light of wavelength 4000Å. The work functions of sodium
and copper are respectively 2.0 eV and 4.0 eV.
(a) Sodium (b) Copper
(c) Both (d) None of these
35.What is the energy of k
a
X-ray photon of copper (Z = 29) ?
(a) 7.99 keV (b) 8.29 keV (c) 8.25 keV (d) 7.19 keV
36.When X-rays of wavelength 0.5 Å would be transmitted by
an aluminium tube of thickness 7 mm, its intensity remains
one-fourth. The attenuation coefficient of aluminium for
these X-rays is
(a) 0.188 mm
–1
(b) 0.189 mm
–1
(c) 0.198 mm
–1
(d) None of these
37.An X-ray tube with Cu target is operated at 25 kV. The
glancing angle for a NaCl. Crystal for the Cu k
a
line is 15.8°.
Find the wavelength of this line.
(d for NaCl = 2.82 Å, h = 6.62 × 10
–27
erg-sec)
(a) 3.06 Å (b) 1.53 Å
(c) 0.75 Å (d) None of these
38.The de-Broglie wavelength of a neutron at 927°C is l.
What will be its wavelength at 27 °C ?
(a)
2
l
(b)l (c) 2 l (d) 4 l
39.The maximu
m distance between interatomic lattice planes is
15 Å. The maximum wavelength of X-rays which are
diffracted by this crystal will be
(a) 15 Å(b) 20 Å(c) 30 Å (d) 45 Å
40.A monochromatic source of light operating at 200 W emits
4 × 10
20
photons per second. Find the wavelength of light.
(a) 400 mm (b) 200 nm
(c) 4 × 10
–10
Å (d) None of these
41.An X-ray tube is operated at 15 kV. Calculate the upper limit
of the speed of the electrons striking the target.
(a) 7.26 × 10
7
m/s (b) 7.62 × 10
7
m/s
(c) 7.62 × 10
7
cm/s (d) 7.26 × 10
9
m/s
42.All electrons ejected from a surface by incident light of
wavelength 200nm can be stopped before travelling 1m in
the direction of uniform electric field of 4N/C. The work
function of the surface is
(a) 4 eV(b) 6.2 eV(c) 2 eV (d) 2.2 eV
43.The stopping potential (V
0
) versus frequency (v) plot of a
substance is shown in figure, the threshold wavelength is
2
1
45678
V
0
v × 10 Hz
14
(a) 5 × 10
14
m
(b) 6000 Å
(c) 5000 Å
(
d) Cannot be estimated from given data
44.A photon of 1.7 × 10
–13
joule is absorbed by a material
under special circumstances. The correct statement is
(a) Electrons of the atom of absorbed material will go the
higher energy states
(b) Electron and positron pair will be created
(c) Only positron will be produced
(d) Photoelectric effect will occur and electron will be
produced
45.Light from a hydrogen discharge tube is incident on
the cathode of a photoelectric cell, the work function of
the cathode surface is 4.2 eV. In order to reduce the
photocurrent to zero the voltage of the anode relative to the
cathode must be made
(a) – 4.2 V (b) – 9.4 V
(c) – 17.8 V (d) + 9.4 V
46.The glancing angle in a X-ray diffraction is 30º and the
wavelength of X-rays used is 20 nm. The interplanar spacing
of the crystal dffracting these X-rays will be
(a) 40 nm (b) 20 nm(c) 15 nm(d) 10 nm

703. 699Dual Nature of Radiation and Matter
47.The frequency and work function of an incident photon are
v and f
0
. If v
0
is the threshold frequency then necessary
condition for the emission of photoelectron is
(a) v < v
0
(b)
0
v
v
2
=
(c) v ³ v
0
(d) Non
e of these
48.The work function of aluminium is 4.2 eV. If two photons,
each of energy 3.5 eV strike an electron of aluminium, then
emission of electrons
(a) will be possible
(b) will not be possible
(c) Data is incomplete
(d) Depends upon the density of the surface
49.For intensity I of a light of wavelenght 5000Å the
photoelectron saturation current is 0.40 µA and stopping
potential is 1.36 V, the work function of metal is
(a) 2.47 eV (b) 1.36 eV
(c) 1.10 eV (d) 0.43 eV
50.A source S
1
is producing, 10
15
photons per second of
wavelength 5000 Å. Another source S
2
is producing
1.02×10
15
photons per second of wavelength 5100Å Then,
(power of S
2
) to the (power of S
1
) is equal to :
(a) 1.00 (b) 1.02 (c) 1.04 (d) 0.98
51.The potential difference that must be applied to stop the
fastest photoelectrons emitted by a nickel surface, having
work function 5.01 eV, when ultraviolet light of 200 nm falls
on it, must be
(a) 2.4 V(b) – 1.2 V (c) – 2.4 V (d) 1.2 V
52.Photoelectric emmision occurs only when the incident light
has more than a certain minimum
(a) power (b) wavelength
(c) intensity (d) frequency
53.In photoelectric emission process from a metal of work
function 1.8 eV, the kinetic energy of most energetic
electrons is 0.5 eV. The corresponding stopping potential is
(a) 1.8 V (b) 1.2 V (c) 0.5 V (d) 2.3 V
54.The threshold frequency for a photosensitive metal is
3.3 × 10
14
Hz. If light of frequency 8.2 × 10
14
Hz is incident
on this metal, the cut-off voltage for the photoelectric
emission is nearly
(a) 2 V (b) 3 V(c) 5 V (d) 1 V
55.If the momentum of electron is changed by P, then the de
Broglie wavelength associated with it changes by 0.5%.
The initial momentum of electron will be
(a) 200 P(b) 400 P(c)
200
P
(d) 100 P
56.Two
radiations of photons energies 1 eV and 2.5 eV,
successively illuminate a photosensitive metallic surface of
work function 0.5 eV. The ratio of the maximum speeds of
the emitted electrons is
(a) 1 : 4(b) 1 : 2(c) 1 : 1(d) 1 : 5
57.A 200 W sodium street lamp emits yellow light of wavelength
0.6 µm. Assuming it to be 25% efficient in converting
electrical energy to light, the number of photons of yellow
light it emits per second is
(a) 1.5 × 10
20
(b) 6 × 10
18
(c) 62 × 10
20
(d) 3 × 10
19
58.Monochromatic radiation emitted when electron on
hydrogen atom jumps from first excited to the ground state
irradiates a photosensitive material. The stopping potential
is measured to be 3.57 V. The threshold frequency of the
materials is
(a) 4 × 10
15
Hz (b) 5 × 10
15
Hz
(c) 1.6 × 10
15
Hz (d) 2.5 × 10
15
Hz
59.The magnitude of the de-Broglie wavelength (l) of electron
(e), proton (p), neutron (n) and a-particle (a) all having the
same energy of 1 MeV, in the increasing order will follow the
sequence
(a)l
e
, l
p
, l
n
, l
a
(b)l
e
, l
n
, l
p
, l
a
(c)l
a
, l
n
, l
p
, l
e
(d)l
p
, l
e
, l
a
, l
n
60.For a given photosensitive material and frequency (>
threshold frequency) of incident radiation, the photoelectric
current varies with the intensity of incident light as
(a)
Current
Intensity
(b)
Current
Intensity
(c)
Current
Intensity
(d)
Current
Intensity
61.The anode vo ltage of a photocell is kept fixed. The
wavelength l of the light falling on the cathode is gradually
changed. The plate current I of the photocell varies as
follows
(a)
I
O l
(b)
I
O l
(c)
I
O l
(d)
I
O l

704. 700 PHYSICS
62.Radiations o
f intensity 0.5 W/m
2

are striking a metal plate.
The pressure on the plate is
(a) 0.166 × 10
–8
N/m
2
(b) 0.332 × 10
–8
N/m
2
(c) 0.111 × 10
–8
N/m
2
(d) 0.083 × 10
–8
N/m
2
63.Hard X-rays for the study of fractures in bones should have
a minimum wavelength of 10
–11
m. The accelerating voltage
for electrons in X-ray machine should be
(a) < 124 kV
(b) > 124 kV
(c) between 60 kV and 70 kV
(d) = 100 kV
64.The wavelength of a 1 keV photon is 1.24 × 10
–9
m. What is
the frequency of 1 MeV photon ?
(a) 1.24 × 10
15
(b) 2.4 × 10
20
(c) 1.24 × 10
18
(d) 2 × 4 × 10
23
65.Photoelectric work function of a metal is 1eV. Light of
wavelength l = 3000 Å falls on it. The photo electrons come
out with velocity
(a) 10 metres/sec (b) 10
2
metres/sec
(c) 10
4
metres/sec (d) 10
6
metres/sec
66.Energy levels A, B, C of a certain atom correspond to
increasing values of energy i.e., E
A
< E
B
< E
C
. If l
1
, l
2
, l
3
are
the wavelengths of radiation corresponding to the
transitions C to B, B to A and C to A respectively, which of
the following relation is correct?
(a)
213
l+l=l (b)
21
21
3
l+l
ll
=l
(c) 0
321
=l+l+
l (d)
2
2
2
1
2
3
l+l=l
67.A radio transmitter operates at a freqency 880 kHz and a
power of 10 kW. The number of photons emitted per second
is
(a) 1.72 × 10
31
(b) 1.327 × 10
25
(c) 1.327 × 10
37
(d) 1.327 × 10
45
68.The momentum of a photon of an electromagnetic radiation
is 3.3 × 10
–29

kgms
–1
. What is the frequency of the associated
waves ?
[h = 6.6 × 10
–34
Js; c = 3 × 10
8
ms
–1
)
(a) 1.5 × 10
13
Hz (b) 7.5 × 10
12
Hz
(c) 6.0 × 10
3
Hz (d) 3.0 × 10
3
Hz
69.The quantity
ch2
e
0
2
e
has a value
(a)
1
ms
137
1 -
(b)
1
ms
137
2 -
(c)
137
1
(d)
137
2
70.The threshold frequency for photoelectric effect on sodium
corresponds to a wavelength of 5000 Å. Its work function is
(a) 4 × 10
–19
J (b) 1 J
(c) 2 × 10
–19
J (d) 3 × 10
–19
J
DIRECTIONS (Qs. 71 to 75) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following-
(a)Statement-1 is false, Statement-2 is true
(b)Statement-1 is true, Statement-2 is true; Statement-2 is a
correct explanation for Statement-1
(c)Statement-1 is true, Statement-2 is true; Statement-2 is not
a correct explanation for Statement-1
(d)Statement-1 is true, Statement-2 is false
71. Statement-1 : In process of photoelectric emission, all
emitted electrons do not have same kinetic energy.
Statement-2 : If radiation falling on photosensitive surface
of a metal consists of different wavelength then energy
acquired by electrons absorbing photons of different
wavelengths shall be different.
72. Statement-1 : Though light of a single frequency
(monochromatic) is incident on a metal, the energies of
emitted photoelectrons are different.
Statement-2 : The energy of electrons emitted from inside
the metal surface, is lost in collision with the other atoms in
the metal.
73. Statement-1 : The de-Broglie wavelength of a molecule (in a
sample of ideal gas) varies inversely as the square root of
absolute temperature.
Statement-2 : The rms velocity of a molecule (in a sample of
ideal gas) depends on temperature.
74. Statement-1 : Photoelectric saturation current increases
with the increase in frequency of incident light.
Statement-2 : Energy of incident photons increases with
increase in frequency and as a result photoelectric current
increases.
75. Statement-1 : Photosensitivity of a metal is high if its work
function is small.
Statement-2 : Work function = hf
0
where f
0
is the threshold
frequency.

705. 701Dual Nature of Radiation and Matter
Exemplar Questions
1.A particle is dropped from a height H. The de-Broglie
wavelength of the particle as a function of height is
proportional to
(a)H (b) H
1/2
(c)H
0
(d) H
–1/2
2.The wavelength of a photon needed to remove a proton
from a nucleus which is bound to the nuclear with 1 MeV
energy is nearly
(a) 1.2 nm (b) 1.2 × 10
–3
nm
(c) 1.2 × 10
–6
nm (d) 1.2 × 10 nm
3.Consider a beam of electrons (each electron with energy E
0
)
incident on a metal surface kept in an evacuated chamber.
Then,
(a) no electrons will be emitted as only photons can emit
electrons
(b) electrons can be emitted but all with an energy, E
0
(c) electrons can be emitted with any energy, with a
maximum of E
0
– f (f is the work function)
(d) electrons can be emitted with any energy, with a
maximum of E
0
4.Consider figure given below. Suppose the voltage applied
to A is increased. The diffracted beam will have the maximum
at value of q that
(a) will be larger than the earlier value
(b) will be the same as the earlier value
(c) will be less than the earlier value
(d) will depend on the target
El ectron bea m
Nickel
target
Vacc
um
chamber
Diffracted
el ec tron
beam
Movable
collector
To galvanometer
Electron
gun
A
HT
LT
– +
5.A proton, a neutron, an electron and an a-particle have
same energy. Then, their de-Broglie wavelengths compare
as
(a)l
p
= l
n
> l
e
> l
a
(b)l
a
< l
p
= l
n
> l
e
(c)l
e
< l
p
= l
n
> l
a
(d)l
e
= l
p
= l
n
= l
a
6.An electron is moving with an initial velocity v = v
0
i
r
and
is in a m
agnetic field B = B
0
j
r
. Then, it's de-Broglie
wavelength(a) remains constant(b) increases with time
(c) decreases with me
(d) increases and decreases periodically
7.An electron (mass m) with an initial velocity v = v
0
i (v
0
> 0) is
in an electric field E = – E
0

ˆ
i (E
0
= constant > 0). It's
de-Broglie wavelength at time t is given by
(a)
0
0
0
eEt
1
mv
l
æö
+
ç÷
èø
(b)
0
0
eEt
1
mv
æö
+
ç÷
èø
(c)l
0
(d)l
0
t
8.An ele
ctron (mass m) with an initial velocity
0
ˆ
v vi= is in
an elect
ric field
0
ˆ
E Ej=. If l
0
= h /mv, it' s de-Broglie
wavelnegth at time t is given by
(a)l
0
(b)
2 22
0
0
22
0
eEt
1
mv
l+
(c)
0
2 22
0
22
0
eEt
1
mv
l
+ (d)
0
2 22
0
22
0
eEt
1
mv
l
æö
+ç÷
ç÷
èø
NEET/AIPMT (2013-2017) Questions
9.For photoelectric emission from certain metal the cut-off
frequency is n. If radiation of frequency 2n impinges on the
metal plate, the maximum possible velocity of the emitted
electron will be (m is the electron mass) [2013]
(a)
/hmn (b)2/hmn
(c)2/hmn (d)()/2hmn
10.The wav
elength l
e
of an electron and l
p
of a photon are of
same energy E are related by [2013]
(a)
pe
l µl (b)
pelµl
(c)
1
p
e
lµ
l
(d)
2
pe
l µl
11.A sou rce of light is placed at a distance of 50 cm from a
photocell and the stopping potential is found to be V
0
. If
the distance between the light source and photocell is made25 cm, the new stopping potential will be(a) 2V
0
(b) V
0
/2 [NEET Kar. 2013]
(c)V
0
(d) 4V
0

706. 702 PHYSICS
12.The de-B
roglie wavelength of neutron in thermal equilibrium
at temperature T is [NEET Kar. 2013]
(a)
30.8
Å
T
(b)
3.08
Å
T
(c)
0.308
Å
T
(d)
0.0308
Å
T
13.When the energy
of the incident radiation is incredased by
20%, the kinetic energy of the photoelectrons emitted from
a metal surface increased from 0.5 eV to 0.8 eV. The work
function of the metal is : [2014]
(a) 0.65 eV (b) 1.0 eV
(c) 1.3 eV (d) 1.5 eV
14.If the kinetic energy of the particle is increased to 16 times
its previous value, the percentage change in the de-Broglie
wavelength of the particle is : [2014]
(a) 25 (b) 75
(c) 60 (d) 50
15.Which of the following figures represents the variation of
particle momentum and the associated de-Broglie
wavelength? [2015]
(a)
p
l
(b)
p
l
(c)
p
l
(d)
p
l
16.A certain metallic surface is illuminated with monochromatic
light of wavelength l. The stopping potential for photo-
electric current for this light is 3V
0
. If the same surface is
illuminated with light of wavelength 2l, the stopping
potential is V
0
. The threshold wavelength for this surface
for photo-electric effect is [2015]
(a)4l (b)
4
l
(c)
6
l
(d) 6l
17.Light of wavele
ngth 500 nm is incident on a metal with work
function 2.28 eV. The de Broglie wavelength of the emittedelectron is: [2015 RS]
(a) < 2.8 × 10
-9
m(b)³. 2.8 × 10
-9
m
(c)£ 2.8 × 10
-12
m(d) < 2.8 × 10
-10
m
18.A photoelectric surface is illuminated successively by
monochromatic light of wavelength l and 2
l
. If the maximum
k
inetic energy of the emitted photoelectrons in the second
case is 3 times that in the first case, the work function of the
surface of the material is :
(h = Planck's constant, c = speed of light) [2015 RS]
(a)
hc
l
(b)
2hc
l
(c)
hc
3l
(d)
hc
2l
19.An electron of mass m and a photon have same energy E.
The ratio of de-Broglie wavelengths associated with them
is : [2016]
(a)
1
21E
c 2m
æö
ç÷
èø
(b)
1
2E
2m
æö
ç÷
èø
(c)
1
2c(2mE)
(d)
1
21 2m
xcE
æö
ç÷
èø
20.When a metallic surface is illuminated with radiation of
wavelength l, the stopping potential is V. If the same surface
is illuminated with radiation of wavelength 2l, the stopping
potential is
V
4
. The threshold wavelength for the metallic
surface is : [2016]
(a)4l (b) 5l
(c)
5
2
l (d) 3l
21.The de-Br
oglie wavelength of a neutron in thermal
equilibrium with heavy water at a temperature T (Kelvin)and mass m, is :- [2017]
(a)
h
3mkT
(b)
2h
3mkT
(c)
2h
mkT
(d)
h
mkT
22.The photoelectric threshold wavelength of silver is 3250 ×
10
–10
m. The velocity of the electron ejected from a silver
surface by ultraviolet light of wavelength 2536 × 10
–10
m is
(Given h = 4.14 × 10
–15
eVs and c = 3 × 10
8
ms
–1
) [2017]
(a)» 0.6 × 10
6
ms
–1
(b) »61 × 10
3
ms
–1
(c) »0.3 × 10
6
ms
–1
(d) »6 × 10
5
ms
–1

707. 703Dual Nature of Radiation and Matter
EXERCISE - 1
1. (b) 2. (d) 3. (a) 4. (d) 5. (d)
6. (b) 7. (b) 8. (a)
9. (b) 0
vm
c/v1h
vm
h
0
22
=
-
==l ( v c)=Q
10. (a) Einstein’s photo electric effect & compton effect
established particle nature of light. These effects can
be explained only, when we assume that the light has
particle nature (To explain, Interference & Diffraction
the light must have wave nature. It means that light
has both particle and wave nature, so it is called dual
nature of light)
11. (c)
f-
l
=
hc
mv
2
1 2
m
)hc(2
v
l
lf-
=Þ
12. (c)
13
. (a) According to relation, E =
21
mv
2
2E
v
m
=
mE2
h
=l
Because m
1
< m
3
< m
2
So f
or same l,
231
EEE>>.
14. (b) Here the velocity of electron increases, so as per
Einstein’s equation mass of the electron increases,
hence the specific charge
e
m
decreases.
15
. (b) 16. (b)
17. (a) Because when intensity of incident light increases, it
means that number of photons increases in incident
light. If number of incident photons increases, then
number of emitted photo electrons also increases,
consequently the photo electric current increases.
18. (d)
19. (a) Emission of electron from a substance under the action
of light is photoelectric effect. Light must be at a
sufficiently high frequency. It may be visible light, U.V,
X-rays. So U.V. cause electron emission.
20. (a) By photoelectric equation:
hv = hv
0
+ K.E
electron
K.E
electron
= hv – hv
0
K.E
electron
µ (v – v
0
) [Q h is plank's constant]
21. (c) de-Broglie wavelength,
l =
E.KmE2
h
\
p
p
p
p
m
m4
m
m
==
l
l
a
a
[Q
)p(E.K)(E.K
EE=
a
]
\
al
l
p
=
1
2
22. (c)
23
. (c) Momentum
hh
p
p
= Þl=
l
(photon)
elect
ronphoton
If l=l
p(momentum)will become same.
24. (b) Energy of a photon E = hn =
hc
l
25. (a) In the Davisson and Germer experiment, the velocity
of electrons emitted from the electron gun can be
increased by increasing the potential difference
between the anode and filament.
EXERCISE - 2
1.
(d)
0s W
hc
eV -
l
= . If l decreases, V
s
i ncreases
2. (d)
2
vm
2
1
Vq= or mVq2vm= ;
So
mVq2
h
vm
h
==l i.e.
1
qm
lµ ;
so 2242
mq
mq
pp
p
=´==
l
l
aa
a
3. (b)
2
vm
2
1
eV= or m/eV2v=
4. (b) Si
nce work function for a metal surface is
0
hc
W
l
=
where l
0
i
s threshold wavelength or cut-off wavelength
for a metal surface.
here W = 4.125 eV = 4.125 × 1.6 × 10
–19
Joule
so
348
6.6 10 3 10
3000Å
0 19
4.125 1.6 10
-
´ ´´
l==
-
´´
5. (b) E
= hc/l = 6.6 × 10
–34
× 3 × 10
8
/5000 × 10
–10
= 3.973 × 10
–19
J
6. (a) Einstein equation E.KhE
o
+u=
where E = energy of incident photon.
o
hu = work function of metal
K.E = max. kinetic energy of e
–
4 eV 2 eV K.E\=+ or K.E 2 eV=
Stopping potential is the potential difference which
may stop this e
–
.
Let it be V, then eV 2e V 2 volt.= Þ=
7. (d)
2/1
2
2
o
c
v
1
m
m
÷
÷
ø
ö
ç ç
è
æ
-
=
( )
o
2/1
o
m2
431
m
=
-
=
Hints & Solutions

708. 704 PHYSICS
8. (d) Mom
entum of a photon
E
c
= =
8
19
103
106.110
´
´´
-
127
mskg1033.5
--
´=
9. (a)
k
0
34 8 10 10
19
hc11
E (in
eV)
c
6.6 10 3 10 10 10
0.15eV
1800 23001.6 10
-
-
æö
=-
ç÷
llèø
æö´ ´´
= -=
ç÷
´ èø
10. (d)
1
1. (b) E
k
= E – W
0
= 6.2 – 4.2 = 2.0 eV
= 2.0 × 1.6 × 10
–19
= 3.2 × 10
–19
J
12. (c)
13. (b) Photoelectrons are emitted in A alone. Energy of
electron needed if emitted from
h
A eV
e
u
=
()()
34 14
A
19
6.6 10 1.8 10
E 0.74eV
1.6 10
-
-
´ ´´
\ ==
´
()()
34 14
B
19
6.6 10 2.2 10
E 0.91 eV
1.6 10
-
-
´ ´´
==
´
Incident ener
gy 0.825 eV is greater than E
A
(0.74 eV)
but less than E
B
(0.91 eV).
14. (b)
hhh
p2mE 2mqV
l===
()( )
34
27 19
6.6 10
2 1.6 10 1.6 10 1000
-
--
´
\l=
´´ ´´ ´
or
34 22
126.6 10 10
0.9 10 m
1.6 20
-
-´´
l= =´
15. (a) H
ere, No. of electrons
n
e
= 5 × 10
7
/cm
3
= 5 × 10
7
×10
6
/m
3
No. of positive ions,
n
p
= 5 × 10
7
× 10
6
= 5 × 10
13
/m
3
v = 0.4 m/s ; J = 4 × 10
–6
A/m
2
; v
p
= ?
Use the relation J = n
e
e v
e
+ n
p
e v
p
and solve it
for v
p
16. (b) Momentum of incident light per second
7
8
1 102
103
60
c
E
P
-
´=
´
==
Momentum of re
flected light per sec
7
8
2 102.1
103
60
100
60
c
E
100
60
P
-
´=
´
´=´=
Force on the surfa
ce = change in momentum per sec
= P
2
– (–P
1
) = P
2
+ P
1
= (2 + 1.2) × 10
–7
= 3.2 × 10
–7
N
17. (b)
000
2
1
WWW2m
2
1
=-=n and
000
2
2 W9WW10m
2
1
=-=n
\
3
1
W9
W
0
0
2
1
==
n
n
18. (d) Inte
nsity
2
1/ (distance)µ ; No. of photoelectr ons
emitted is proportional to intensity of incident light.
19. (d)
0
0
0
W
hc
eV -
l
= and
0
0
W
2
hc
Ve -
l
=¢
Subtracting the
m we have
00
0
2
hc
2
1
1
hc
)VV(e
l
=
ú
û
ù
ê
ë
é
-
l
=¢-
or
0
0
e2
hc
VV
l
-=¢
20. (a) Here,
u = 0 ;
m
eE
a= ; v = ? ; t = t
\ t
m
eE
0atuv+=+=
de-Broglie w
avelength,
eEt
h
)m/eEt(m
h
mv
h
===l
Rate of change of
de-Broglie wavelength
22
tEe
h
t
1
eE
h
dt
d -
=
÷
÷
ø
ö
ç ç
è
æ
-=
l
21. (d)
2
22
2
)Z(
4
R3
2
1
1
1
)Z(R
1
a-=
ú
û
ù
ê
ë
é
-a-=
l
a
2
22
2
)Z(R
9
8
3
1
1
1
)Z(R
1
a-=
ú
û
ù
ê
ë
é
-a-=
l
b
\
32
27
=
l
l
a
b
or Å27.032.0
32 27
32 27
=´=l=l
ab
22. (d)
J
105.0
103106.6hc
Energy
10
834
-
-
´
´´´
=
l
=
= eV
106.1105
103106.6
1911
834
--
-
´´´
´´´
= 24.82 keV
23. (d)
0
2 hchc
mv
2
1
l
-
l
= Where l is excit
ing wavelength, l
0
is threshold wavelength & v is speed of fastest electron.
24. (a) Moment of photon
l
==
h
p 2
mcE=Q
But, p = mc
c.mcE=\
So, E = pcor
l
=
hc
E
\
pc
hc
=
l
or
l
=
h
p and
f
c
=l
\
c
hf
p=

709. 705Dual Nature of Radiation and Matter
25. (d)
2
d
1
Iµ
26. (a)
Energy associated with wave.
Let number of quanta = x
5% of 100 =
l
nhc
348
7
5 n 6.26 10 3 10
100
100 5.6 10
--
-
´ ´ ´´
Þ´=
´
19
5
n
3.35 10
-
Þ=
´

19
n 1.4 10Þ=´
27. (c)
0
2
mh
,v
mv
v
1
c
l==
æö
-ç÷
èø
, v c,m® ®¥
hence, 0®l .
28. (d) hu –
hu
0
= E
K
, according to photoelectric equation,
when u = u
0
, E
K
= 0.
Graph (d) represents E
K
– u relationship.
29. (c) hu = W
ex
+ maximum kinetic energy
hu
0
= 8 eV = W
ex
+ 2 eV
ÞW
ex
= 6 eV
For
incoming radiation, energy is
h × 1.25 u
0
= 10 eV, W
ex
= 6 eV
\ Kinetic energy (maximum) = 4 eV.
30. (a)
2
mv
2
1
eV=
31
19
101.9
20106.12
m
eV2
v
-
-
´
´´´
==Þ
s/m1065.2
6
´=
31. (c)
de-Broglie’s relation,
p
h
=l
momentum mE2p=
h
2mE
Þl=
h
=
2mK
( )E=KQ
32. (b) Å62.0
1020
12400
V
12400
eV
hc
3
.min =
´
===l
33. (d) V
= 3000 volt. 21
mv eV
2
= Þ
m
eV2
v=
\
19
31
2 1.6 10 3000
v
9.1 10
-
-
´´´
=
´
= 32
.6 × 10
6
= 3.26 × 10
7
m/s.
34. (a)Q l
0
=
hc
f
\ (l
0
)
sodium
=
348
19
6.6 10 3 10
2 1.6 10
-
-
´ ´´
´´
= 6
188 Å
Q l
0
µ 1
f
Þ
copper0 sodium
0 copper so dium
()()
( ) ()
fl
=
lf
Þ (l
0
)
copper
=
2
4
× 6188 = 3094 Å
To eje
ct photo-electrons from sodium the longest
wavelength is 6188 Å and that for copper is 3094 Å.
Hence for light of wavelength 4000 Å, sodium is
suitable.
35. (a) E(k
a
) = 10.2 (Z – 1)
2
eV
= 10.2 × 28
2
= 10.2 × 784
= 7.997 keV = 8 KeV
36. (c) 0
I
I
= 4, x = 7mm (given)
Þ
0
10
I
2.303 log
I
x
m=
10
2.303 log 42.303 0.6023
77
´
m==
m = 0.198 mm
–1
3
7. (b) According to Bragg’s law, Þ 2d sin q = nl,
n = 1 for first orderÞ 2 × 2.82 sin 15.8 = l Þ l = 5.64 × 0.2723 = 1.53 Å
38. (c) de-Broglie wavelength of a material particle at
temperature T is given by
h
.
2mkT
l= , wher e k is Boltzmann's constant.
Þ
1
T
lµ
\
21
12
T
T
l
=
l
or
2
1
1200
2
300
l
==
l
\ 21
22l=l=l
39. (c)
Å30
1
º90sin152
n
sind2
.min
.max =
´´
=
q
=l
40. (a)
The energy of each photon =
20
200
4 10´
= 5 × 10
–19
J
Waveleng
th = l =
hc
E
=
348
19
(6.63 10) (3 10 )
5 10
-
-
´ ´´
´
Þ l = 4.
0 × 10
–7
= 400 nm
41. (a) The maximum kinetic energy of an electron accelerated
through a potential difference of V volt is
1
2
mv
2
= eV

710. 706 PHYSICS
\ maximu
m velocity v =
2eV
m
v =
19
31
2 1.6 10 15000
9.1 10
-
-
´´´
´
= 7.26 × 10
7
m/s
42
. (d) The electron ejected with maximum speed v
max
are
stopped by electric field E =4N/C after travelling a
distance d =1m
2
max
1
mv eEd 4eV
2
==
The energy of
incident photon =
1240
6.2 eV
200
=
From equation
of photo electric effect
2
max0
1
mvh
2
= n-f
\
0
6.2 4 2.2 eVf= -=
43. (b)
8
7
0 14
0
c 3 10
6 10 m 6000Å
v5 10
-´
l== =´=
´
44.
(b) For electron and positron pair production, minimum
energy is 1.02 MeV.
Energy of photon is given
13
3
19
1.7 10
1.7 10 J
1.6 10
-
-
-
´
´=
´
= 1.0
6 MeV.
Since energy of photon is greater than 1.02 MeV.
so electron positron pair will be created.
45. (b) E = W
0
+ eV
0
For hydrogen atom, E = +13.6 eV
\ + 13.6 = 4.2 + eV
0
0
(13.6 4.2)eV
V 9.4V
e
-
Þ==
Potential at ano
de = – 9.4 V
46. (b) 2 d sin q = nl or
nm20
º30sin2
201
sin2
n
d =
´
´
=
q
l
=
47. (c) 48. (b)
For emission of electrons incident energy of each
photon must be greater than work function (thresholdenergy).
49. (c) By using E = W
0
+ K
max12375
E
5000
= =2.475 eV and K
max
= eV
0
= 1.36 eV
So 2.475 = W
0
+ 1.36 Þ W
0
= 1.1 eV.
50. (a) Energy emitted/sec by 111
1
,
hc
SPn=
l
Energy emitted/sec by
222
2
,
hc
SPn=
l
\
2 21
1 12
Pn
Pn
l
=×
l
15
15
1.02 10 5000
1.0
510010
´
= ×=
51. (d) K
max

=
5.01
hc hc
W-=-
ll
=
12375
5.01
(in Å)
-
l
12375
2000
= –5.01 = 6.1875 – 5.01 = 1.17775 ;1.2 V
52. (d) For
occurence of photoelectric effect, the incident light
should have frequency more than a certain minimumwhich is called the threshold frequency (v
0
).
We have,
2
0
1
mv hh
2
vv=-
For photoelec
tric effect emission n > n
0
where n is the frequency of the incident light.
53. (c) The stopping potential is equal to maximum kinetic
energy.
54. (a) K.E. = hn – hn
th
= eV
0
(V
0
= cut off voltage)
14 14
0
h
V (8.2 10 3.3 10 )
e
Þ= ´ -´
=
34 14
19
6.6 10 4.9 10
2V.
1.6 10
-
-
´ ´´
»
´
55. (a) The de-Broglie’s wavelength associated with the
moving electron
h
P
l=
Now, according to problem

d dp
P
l
=-
l

0.5
100 '
P
P
=
P¢ = 200 P
56. (b) Accord
ing to Einsten’s photoelectric effect, the K.E.
of the radiated electrons
K.E
max
= E – W

1
2
mv
1
2
= (1 – 0.5) eV = 0.5 eV

1
2
mv
2
2
= (2.5 – 0.5) eV = 2 eV

1
2
0.511
224
= ==
v
v
57. (a) Give th
at, only 25% of 200W converter electrical energy
into light of yellow colour
25
200
100
æö
´=´ç÷
èøl
hc
N
Where N is the No. of
photons emitted per second,
h = plank’s constant, c, speed of light.
200 25
100
´l
=´N
hc

6
348
200 25 0.6 10
100 6.2 10 3 10
-
-
´´´
=
´ ´ ´´
= 1.5 ×
10
20
58. (c)n ® 2 – 1
E = 10.2 eV
kE = E – f
Q = 10.20 – 3.57
h u
0
= 6.63 eV

711. 707Dual Nature of Radiation and Matter
19
0
34
6.63 1.6
10
6.67 10
-
-
´´
u=
´
= 1.6 × 10
15
Hz
59. (
c)
h
2mE
l=
So
1
h
m
µ
since
m
a
> m
n
> m
p
> m
e
so de-Broglie wave length in increasing order will be
l
d
, l
m
, l
p
, l
e
60. (a) For a given photosensitive material and µ frequency
> threshold frequency photoelectric current µ
intensity.
61. (d) As l is increased, there will be a value of l above which
photoelectrons will be cease to come out so
photocurrent will become zero. Hence (d) is correct
answer.
62. (a) We know that P = Fv or F = P/v
28
8
m/N10166.0
103
5.0
F
-
´=
´
=
63. (a)
Accelerating voltage =
hc
e´
l
=
34 8 19
11
6.6 10 3 10
1.6 10
10
--
-
´ ´´ ´´
= 123.
75 KV < 124 KV
64. (b) Here,
3
10
hc
=
l
eV and hn = 10
6
eV
Hen
ce,
9–
833
1024.1
10310c10
´
´´
=
l
=n
= 2.4 ×
10
20
Hz
65. (d)
21
2
h W mvn=+ or
21
2
hc
W mv=+
l
Here l = 3000 Å = 3000 × 10
–10
m
an
d W = 1 eV = 1.6 × 10
–19
joule
348
10
(6.6 10 )(3 10)
3000 10
-
-
´´
\
´
19 31 21
(1.6 10) (9.1 10 )v
2
--
=´ +´´
So
lving we get,
6
10 m/sv@
66. (b)
21
()
hc
EEh- = n=
l
\ ( )
1
,
CB
hc
EE=-
l
( )
2
BA
hc
EE=-
l
and ( )
3
CA
hc
EE=-
l
Now,
()()()
CA CB BA
EE EE EE-=-+-
or,
3 12
hc hc hc
=+
l ll
or
3 12
1 11
=+
l ll
\
21
21
3
1
ll
l+l
=
l
or
21
21
3
l+l
ll
=l
67. (a)
No. of photons emitted per sec,
Power
Energy of photon
n=

343
10000
6.6 10 880 10
P
h
-
==
n ´ ´´
= 1.7
2 × 10
31
68. (a) As
h
p
l= and ;
c
l=
n
so
c cP
h
n==
l

34
29
8
106.6
103.3
103
-
-
´
´
´´=
= 1.5 ×
10
13
Hz
69. (c)
2 192
34 128
0
e (1.6
10)
2hc
2 6.26 10 8.857 10 3 10
-
--
´
=
e ´ ´ ´ ´ ´´

137
1
= it is unit less
quantity..
70. (a)
0
0
hc
W=
l
10
834
105000
1031063.6
-
-
´
´´´
= = 4 × 10
–
19
J
71. (b) Both statement I and II are true; but even it radiation of
single wavelength is incident on photosensitivesurface, electrons of different KE will be emitted.
72. (a) When a light of single frequency falls on the electrons
of inner layer of metal, then this electron comes out ofthe metal surface after a large number of collisions with
atom of it's upper layer.
73. (a) de-Broglie wavelength associated with gas molecules
varies as 1
1
T
lµ
74. (d
) Photoelectric saturation current is independent of
frequency. It only depends on intensity of light.
75. (b) Less work function means less energy is required for
ejecting out the electrons.EXERCISE - 3
Exemplar Questions
1. (d) Velocity of a body freely falling from a height H is
v 2gH=
So,
hhh
mvm 2gH m 2g H
l= = Þ=
(h
, m and g are constant)
Here,
h
is
m 2g
also constant
So,
1/21
h orH
H
-
µ Þ lµ

712. 708 PHYSICS
2. (b) En
ergy of a photon is
hc
E=
l
Where l is the minimum wa velength of the photon
required to eject the proton from nucleus.
Energy of photon must be equal to the binding energy
of proton.
So, energy of a photon, E = 1 MeV Þ 10
6
eV
(given)
Now,
hc
E
æö
=ç÷
èøl
So,
348
6
hc 6.63 10 3 10
E 10 eV
-æö´ ´´
l== ç÷
èø
Þ So,
348
6 19
hc 6.63 10 3 10
E 10 1.6 10 J
-
-
´ ´´
l==
´´
= 1.24 × 10
–9
×10
–3
= 1.24 × 10
–3
nm
3. (d) Wh
en a beam of electrons of energy E
0
is incident on
a metal surface kept in vacuum or evacuated chamber
so electrons can be emitted with maximum energy E
0
(due to elastic collision) and with any energy less than
E
0
, when part of incident energy of electron is used in
liberating the electrons from the surface of metal. So
maximum energy of emitted electrons can be E
0
.
4. (c) We know that,
In Davisson – Germer experiment, the de-Broglie
wavelength associated with electron is
d
12.27
Å
V
l= ...(i)
where V is th
e applied voltage. If there is a maxima of
the diffracted electrons at an angle q, than
2d sinq = l ...(ii)
From equation (i) if V is inversely proportional to thewavelength l
d
, then the applied voltage V will increase
with the decrease in the wavelengh l
d
.
From equation (ii) if wavelength l
d
is directly
proportional to sin q and hence q.
So, with the decrease in l
d
sin q, q will also decrease.
Hence, when the voltage applied to A is increased.The diffracted beam will have the maximum at a valueof q that will be less than the earlier value.
5. (b) The relation between l and K is given by
h
2mK
l=
So, for the given
value of kinetic energy K,

h
2K
is a constant.
Thus,
1
m
lµ
\Þ pne
:::
a
llll
Þ
pne
1111
:::
mmmm
a
=
if (m
p
= m
n
), then l
p

= l
n
if (m
a
> m
p
), then l
a
< l
p
if (m
e
< m
n
), then l
e
> l
n
Hence, l
a
< l
p
= l
n
< l
e
.
6. (a) As given that, v = v
0

i
r
and B = B
0
j
r
Force on moving electron due to perpendicular
magnetic field B is, F = –e (v × B)
00
ˆˆ
F e vi Biéù=-´
ëû
= –e v
0
B
0
(ˆˆ
ij´)
Þ
00
ˆ
evBk=- ( )
ˆ ˆˆ
k ij=´Q
So, the force is perpendicular to v and B, both as the
force is ^ to the velocity so the magnitude of v will
not change, so momentum is (= mv) will remain sameor constant in magnitude. Hence,
de-Broglie wavelength
h
mv
l= remians constant.
7
. (a) de-Brogile wavelength of electron,
0
0
h
mv
l=
...(i)
Force on electron
Þ ()()
00
ˆˆ
F eE e Ei eEi=-=--=
Accelera
tion of electron
0
ˆ
eEiF
a
mm
== ( )F ma=Q
Velocity of electron after time t, is v = (v
0
+ at)
00
00
ˆ
eE i eE
ˆˆ
v vi t v ti
mm
æö æö
=+ =+ç÷ ç÷
èøèø
0
0
eE
ˆ
v v 1 ti
m
æö
=+ç÷
èø
Now for new de-Broglie wavelength associated with
electron at time t is
h
mv
l=
Þ
0
0
0
0
0 0
h
eEeEt
ˆ
vm1i 1t
mv mv
l
l==
é ùéùæö
+ +ê úêúç÷
èø
ëû ëû

[ ]
0
0
h
ˆ
i1
mv
éù
l==
êú
ëû
Q
0
0
0
eE
1t
mv
l
l=
éùæö
+
êúç÷
èøëû
8. (c)We know tha t,
de-Broglie wavelength of electron,
0
0
h
mv
l=
Force on mo
ving electron due to eletric field,
0
ˆ
EFeFeEj= =- =-

713. 709Dual Nature of Radiation and Matter
Acceleration in electron due to force
F = ma or
0
ˆ
eEjF
a
mm
-
==
It is acting along
negative y – axis direction and the
initial velocity of electron along x – axis,
0
x0
ˆ
v vi=
Initial velocity
of electron along y – axis,
0
y
v0=.
So, veloc
ity of electron after time t along y – axis,
v = u + at (
Q u = 0)
00
y
eE eE
ˆˆ
v 0 j t tj
mm
æö
= + =-ç÷
èø
Magnitude of velocity of electron after time t is
2
222 0
xy0
eE
vvvvt
m
-æö
=+=+ ç÷
èø
So,
2 22
0
0
22
0
eEt
vv1
mv
=+
de-Brogli
e wavelength,
h
'
mv
l=
So,
( )
2 22 22
000
h
mv 1 eEt /mv
=
+
0
0
h
mv
éù
l=
êú
ëû
Q
0
2 22 22
00
'
1 eEt /mv
l
l=
+
Hence, opt
ion (c) is verified.
NEET/AIPMT (2013-2017) Questions
9. (b) From photo-electric equation,
hn' = hn + K
max
...(i)
h.2n = hn +
1
2
mV
2
max
[\ n¢ = 2 n]
Þhn =
1
2
mV
2
max
Þ V
max
=
2h
m
n
10. (d) As P =
E
c
l
p
=
hc
E
...(i)
2
2
ee
hh
2mE2mE
l= Þl= ...
(ii)
From equations (i) and (ii)
l
p
µ l
e
2
11. (c) Since, stopping potential is independent of distance
hence new stopping potential will remain unchanged
i.e., new stopping potential = V
0
.
12. (a) From formula
l =
2
h
mKT
=
34
27 23
6.63 10
2 1.
67 10 1.38 10
m
T
-
--
´
´´ ´´
[By pla
cing value of h, m and k)
=
30.8
Å
T
13. (b
) According to Einstein’s photoelectric equation,
hn = f
0
+ K
max
We have
hn = f
0
+ 0.5 ...(i)
and 1.2hn = f
0
+ 0.8 ...(ii)
Therefore, from above two equations f
0
= 1 . 0 eV.
14. (b) As we know
l =
h
P
=
h
2mK
( P 2mK E)=Q
or
12
21
K 16K 4
K K1
l
===
l
Therefor
e the percentage change in de-Broglie
wavelength =
14
100 75%
4
-
´ =-
15. (
a) According to De-broglie
h
p=
l
or
1
Pµ
l
where
P = par
ticle momentum;
l = de-Broglie wavelength
h = Plank’s constant
1
Pµ
l
represents rectangular hyperbola.
16. (a) As we know,
s
hc
eV–=Y
l
o
hc
3eV–=Y
l
...(1)
o
hc
eV–
2
=Y
l
...(2)
o
3hc
3eV –3
2
=Y
l
...(3)
Multiply
ing eqn. (2) by (3) and subtracting it from eqn (1)
hc
4
Y=
l
So, threshold wavelength,
th
hc hc
4
hc/4
l = = =l
Yl
17. (b)Gi
ven : Work function f of metal = 2.28 eV
Wavelength of light l = 500 nm = 500 × 10
–9
m

714. 710 PHYSICS
KE
max
=
hc
l
–f
KE
max
=
348
7 19
6.6 10 3 10
5 10 1.6 10
-
--
´ ´´
´ ´´
– 2.28
KE
max
= 2
.48 – 2.28 = 0.2 ev
l
min
=
h
p
=
()
max
h
2m KE
=
34
31 19
20
10
3
2 9 10 0.2 1.6 10
-
--
´
´´
´´´
l
min
=
25
9
× 10
–9
= 2.80 × 10
–9
nm\ l ³ 2.8
× 10
–9
m
18. (d) Photoelectric equations
Ek
1max
=
hc
-f
l
...(i)
andEk
2max
=
hc
/2
-f
l
EK
2max
=
2hc
λ
-f ...(ii)
From question
, Ek
2max
= 3Ek
1max
Multiplying equation (i) by 3
3Ek
1max
=
hc
3
æö
-f
ç÷
lèø
...(iii)
From equ
ation (ii) and (iii)
3hc 2hc
3- f= -f
ll
\ f (work functi
on) =
hc
2l
19. (a) For electron De-Broglie wavelength,
e
h
2mE
l=
For photon E =
pc
ÞDe-Broglie wavelength,
Ph
hc
E
l=
\
1/2
e
Ph
hEE1
hc2mc2mE
l æö
= ´= ç÷
l èø
20. (d)
According to Einstein's photoelectric effect,
eV =
0
hc hc
-
ll
…(i)
eV/4 =
0
hc hc
2
-
ll
…(ii)
Dividing equation (i)
by (ii) by
Þ4 =
0
0
11
11
2
-
ll
-
ll
on solving we get,
l
0

= 3l
21. (a) We know that,
de-Broglie wavelength l =
P
h
=
h
2m(KE)
K.E. of thermal neutron =
3
kT
2
=
h
3
2m kT
2
æö
ç÷
èø
l =
h
3mkT
22. (a, d) Both answers are correct
Given,
l
0
= 3250 × 10
–10
m
l = 2536 × 10
–10
m
f =
158
10
0
hc 4.14 10 3 10
3.82eV
3250 10
-
-
´ ´´
==
l ´
hv =
158
10
hc 4.14 10 3 10
4.89eV
2536 10
-
-
´ ´´
==
l ´
Accor
ding to Einstein's photoelectric equation,
K
max
= hv – f
KE
max
= (4.89–3.82)eV=1.077 eV
21
mv
2
= 1.077 × 1.6 × 10
–19
Þ v =
19
31
2 1.077 1.6 10
9.1 10
-
-
´ ´´
´
or, v
= 0.6 × 10
6
m/s or 6 × 10
5
m/s

715. THOMSON'S ATOMI C MODEL
This model suggests an atom to be a tiny sphere of radius
10
10m
-
» , containing
the positive charge. The atom is electrically
neutral. It contains an equal negative charge in the form of
electrons, which are embedded randomly in this sphere, like seeds
in a watermelon.
This model failed to explain (i) large scattering angle of
a-particle and (ii) origin of spectral lines observed in the spectrum
of hydrogen atom.
ALPHA-PARTICLE SCATTERING AND RUTHER-FORD
NUCLEAR MODEL OF ATOM
In Rutherford a- particle scattering experiment a very fine beam
of a-particle passes through a small hole in the lead screen. This
well collimated beam is then allowed to fall on a thin gold foil.
While passing through the gold foil, a-particles are scattered
through different angles. A zinc sulphide screen is placed out the
other side of the gold foil, this screen is movable, so as to receive
the a-particles, scattered from the gold foil at angles varying
from 0 to 180°. When an a-particle strikes the screen, it produces
a flash of light.
It was found that :
•Most of the a-particles went straight through the gold foil
and produced flashes on the screen as if there were nothinginside gold foil. This suggests that the most part of the
atom is empty.
•Few particles collided with the atoms of the foil which havescattered or deflected through considerable large angles.
Very few particles even turned back towards source itself.
Conclusions :
•The entire positive charge and almost whole mass of the
atom is concentrated in small centre called a nucleus.
•The electrons revolving round the nucleus could not
deflected the path of a-particles. This suggests that
electrons are very light.
In 1911 Rutherford , proposed a new type of model of the atom.
According to this model, the positive charge of the atom, instead
of being uniformly distributed throughout a sphere of atomic
dimension is concentrated in a very small volume at its centre.
This central core, called nucleus, is surrounded by clouds of
electrons makes the entire atom electrically neutral.
According to Rutherford scattering formula, the number of
a-particles scattered at angle q by a target,
N µ cosec
4
(q/2)
+
b
Nucleus
a
Impact parameter b =
2
2
00
2 cot( / 2)
4
Ze
mv
q
pe
Distance of closest approach
0
22
00 00
2 2( )(2)
44
Zeq Zee
r
mv mv
==
pe pe
Result o
f Rutherford scattering experiment : Nucleus- is central,
massive, positively charged core, its size of the order of 10
–15
m,
number of electrons surrounding nucleus is such that atom iselectrically neutral.
Unit for nuclear dimension measurement : 1 fermi = 10
–15
m.
BOHR’S ATOMIC (HYDROGEN ATOM) MODEL
In 1913 Bohr gave his atomic theory primarily to explain, the
spectra of hydrogen and hydrogen-like atoms. His theory,
contained a combination of views from Plank’s quantum theory,
Einstein’s photon concept and Rutherford model of atom. The
Bohr theory can explain, the atomic spectra of hydrogen atom
and hydrogen-like ions such as He
+
, Li
2+
, Be
3+
.......(one electron
ions). But his theory failed to explain, the spectra of more complex
atom and ions.
27
Atoms

716. 712 PHYSICS
The basic postulates of Bohr’
s model are :
(i) The electron moves in circular orbits around the nucleus
under the influence of coulombic force of attraction between
the electron and the positively charged nucleus (as shown
in figure below).
+
r
F
c
e
–
v
Bohr’s model of hydrogen atom
(ii) The electron rotates about the nucleus in certain stationary
circular orbits, for which the angular momentum of electron
about the nucleus is an integral multiple of =
p2
h
h, where
h is plank
’s constant
i.e., Angular Momentum,
2
nh
mvrn==
p
h ...(1)
(where n = 1, 2, 3....
..... principal quantum number)
(iii) When the electron is in one of its stationary orbits, it does
not radiate energy, hence the atom is stable.These stationary
orbits are called allowed orbits.
(iv) The atom radiates energy when the electron “jumps” from
one allowed stationery state to another. The frequency of
radiation follows the condition
hn = E
i
–

E
f
...(2)
Where E
i
and E
f
are total energies of initial and final stationary
states. This difference in energy (E
i
-E
f
) between two
allowed stationory states is radiated/absorbed in the form
of a packet of electromagnetic energy (hn - one photon of
frequency n) called a photon.
Now we calculate the allowed energies of hydrogen atom,
For moving an electron in a circular orbit the required
centripetal force is provided by the coulomb force of attraction
which acts between nucleus [Ze
+
, here Z = 1 (atomic number)
for hydrogen atom] & electron (e
–
),
i.e.,
22
2
mv ke
r r
= ...(3)
where
o4
1
k
pe
= is electrost atic constant & e
o
is
permittivity of free space.Eliminating v from eqn. (1) and (3) we obtain radius of n
th
orbit
2222
22
o
nhnh
r
mke me
e
==
p
(where n = 1, 2, 3 .....)...(4)
Equatio
n (4) gives the radii of various orbits (have discrete
values).The smallest radius (also called Bohr radius ) corresponds
to n = 1 is
2
0
2
0.527
h
rÅ
mke
=@ ...(5)
Þr = 0.529 n
2
Å for h
ydrogen atom and
r = 0.529 ×
2
n
Z
for hydrogen like ions.
From equation (4) & (1) we obtain,
Velocity of electron in n
th
state
222 42
2
22 22
n k e ke
vv
nmrn
= = Þ=
h
hh
or
2
2
o
e
v
nh
=
e
1
137
C
n
æö
=´
ç÷
èø
(for hydroge n atom ) ...(6)
137
CZ
v
n
æö
=ç÷
èø
for hydrogen like ions
16
ms1019.2
137
C -
´=
The total energy of el
ectron is given by
E = K.E. + P.E. = Kinetic energy + Potential energy

r
)e(ke
mv
2
1 2 -
+=
22
o
42
hn8
me
r2
ke
E
e
-
=
-
= ...(7)
(Allowed energ
y state)
After substituting numerical values in eqn.(7), we obtain
2
13.6
E eV
n
-
= (for hydrogen atom) ...(8)
2
2
13.6
eV/atom
Z
E
n
-
= for hydrogen like ions.
The lowest ene
rgy state, or ground state, corresponds to n = 1 is
4
0
2
o
me
E 13.6eV
8h
-
= =-
e
The next state co
rresponds to n = 2 i.e., first excited state has an
energy, E = –3.4 eV
Limitations of Bohr's Model
1.It could not explain the spectra of atoms containing more
than one electron.
2.There was no theoretical basis for selecting mvr to be an
integral multiple of p2/h.
3.It involved the orbit concept which could not be checked
experimentally.
4.It could not explain Zeeman & Stark effect and fine lines of
spectra.
5.It was against de-Broglie concept and uncertainty principle.
Keep in Memory
1.Total energ
y of electron = – Kinetic energy
=
2
energy Potential
2.The reference level for potential energy has been taken as
infinity
3.The energy gap between two successive levels decreases
as the value of n increases
4.The radius difference between the successive orbit (or
shells) increases as the value of n increases
5.The velocity of electrons around the nucleus goes on
decreasing as n increases
6.The time period of the electron in an orbit
32
rTµ
7.Maximum number of spectral lines that can be emitted when
an electron jumps from n
th
orbit is
2
)1n(n-

717. 713Atoms
ENERGY LEVELS AND THE L
INES SPECTRA OF
HYDROGEN ATOM
An energy level diagram of the hydrogen atom is shown in figure.
The upper most level corresponding to n®
¥, represents the
state for which the electron is completely removed from the atom.
¥
–0.54 eV
0 eV
–0.85 eV
Paschen
s
eries–1.51 eV
Balmer
series
–3.4 eV
–13.6 eV
(Ground state)
Some transitions for Lyman, Balmer & Paschen series are shown.
The quantum numbers are at left & energies of levels are at right.
E = 0 for r = ¥ (Since n = ¥)
If the ele
ctron jumps from allowed state n
i
to allowed state n
f
,
then frequency of emitted photon is given by
4
322
11
8
fi
o if
EE me
h hnn
æö-
n==- ç÷
ç÷eèø
...(1)
and the wavele
ngth of emitted photon is
4
322
1 11
8
o if
me
c chnn
æö
n
==- ç÷
ç÷l e èø
22
1 11
if
R
nn
æö
=-ç÷
ç÷l
èø
for hydrogen atom ...(2)
a
nd
2
22
12
1 11
RZ
nn
éù
n==- êú
l
êúëû
( for H-like atoms)
w
here R = 1.096776 × 10
7
m
–1
is known as Rydberg constant. By
using this expression we can calculate the wavelengths for various
series (Lyman, Balmer...) in hydrogen spectrum, i.e.
(i) Lyman series n
i
= 1 & n
f
= 2, 3, 4...............
(ii) Balmer series n
i
= 2, & n
f
= 3, 4, 5...............
(iii) Paschen series n
i
= 3 & n
f
= 4, 5, 6..............
(iv) Bracett series n
i
= 4 & n
f
= 5, 6, 7...............
(v) P fund series n
i
= 5 & n
f
= 6, 7, 8...............
First three series of hydrogen atom are shown in figure.
But in practice, the value of Rydberg constant varies between
2
R
and R
This
is because in above calculations we assumed that electron
revolves around a massive fixed nucleus of mass M. But in reality,
the electron and nucleus each revolve round their common center
of mass i.e., the motion of nucleus cannot be ignored. The
correction for nuclear motion amounts to replacing electronic
mass m by reduced mass m which is defined as
mM
mM
m=
+
...(3)
So total e
nergy by taking this correction is
4
22
8
o
e
E
hn
-m
=
e
...(4)
If we
are dealing with hydrogen like ions such as – He
+
, Li
2+
,
Be
3+
, Be
4+
(one electron ions), each can be considered as a system
of two charges, the electron of mass m & charge –e & nucleus of
mass M and charge +Ze, where Z is atomic number. The radii of
circular orbits for these one electron ions can be written as
22 22
22
o
nh n
r
Ze kZe
e
==
pmm
h
(n = 1, 2, 3............) ...(5)
and the allowed energies are given by
24
22
8
o
Ze
E
hn
-m
@
e
(n = 1, 2, 3.........)...(6)
Wave
length Limits in Various Spectral Series of Hydrogen Atom
(i) For Lyman series (lies in ultraviolet region)
max min
1216 Å, 91 2 Ål = l=
Here ...............3,2n,1n
fi ==
(ii) For Balmer series (lies in visible region)
max min
6563Å and 36 46 Ål = l=
Here ...............5,4,3n,2n
fi ==
(iii) For Pasc hen series (lies in infrared region)
max min
18751Å and 8107 Ål = l=
Here ...............6,5,4n,3n
fi==
(iv) For B rackett series (lies in infrared region)
max min
40477Å and 145 72Ål = l=
Here ...............7,6,5n,4n
fi
==
(v) For p-fu nd series (lies in infrared region)
max min
74515 Å and 2276 8 Ål = l=
Here ...............8,7,6n,5n
fi
==
Keep in Memory
1.
The first line of Lyman series is when electron jumps from
2 ® 1, It is also called a – line
The second line of lyman series is when electron jumpsfrom 3 ® 1, It is also called b – line
The limiting line of lyman series is when electron jumpsfrom ¥ ® 1
2. Energy of electrons in different orbits in an atom varies
inversely with the square of the number of orbits. So, energyof electrons increases (decreases in negative) as the orbitbecomes higher.
3. If energy of a particular orbit is E for H-atom then its value
for a H-like atom with atomic number Z is given byE' = E × Z
2
.
4. If the radius of a particular orbit of H-atom is R then its valu
for a H-like atom is given by
R' =
R
Z
.
5. If ve
locity of an electron in a particular orbit of H-atom be v
then its value for a H-like atom is given by
v'= v × Z.
6. If kinetic energy and potential energy of an electron in a
particular orbit of H-atom be T and V respectively then
their corresponding values for H-like atom are given by
T' = T × Z
2
and V' = V× Z
2
.

718. 714 PHYSICS
COMMON DEFAULT
ûInco
rrect. Bohr's formula for spectral lines does not
differentiate between isotopes. For example the first line of
Lyman series in hydrogen and deuterium will have same
wavelength because
ú
û
ù
ê
ë
é
-=
l
22
2
2
1
1
1
)1(R
1
üCorrect. The val
ue of R will be different for hydrogen and
deuterium and therefore
λ will be different for the two cases.
In fact
HD
λλ< )RR(
HD
>Q
Example 1.
A hydrogen at
om in the ground state is excited by
radiations of wavelength 975Å. Find (a) the energy state
to which the atom is excited. (b) how many lines will be
possible in emission spectrum ?
Solution :
(a)
10
975Å 975 10 m
-
l= =´
22
1 11
R
1n
éù
=-
êú
l ëû
\
7
10 22
1 11
1.1 10
975 10 1 n
-
-
éù
=´-
êú
´ ëû
or n = 4
(b)Q Numbe
r of spectral lines (N) =
n (n 1)
2
-
\
4 (4 1)
N6
2
´-
==
Possible transition
4 ® 3, 4 ® 2, 4 ® 1, 3 ® 2, 3 ® 1, 2 ® 1.
Example 2.
Find the longest and shortest wavelength when a hydrogen
atom in the ground state is excited by radiations of
wavelength 975Å.
Solution :
hc 12400
Å
eE E (eV)
l=»
348
6.6 10 ; 3 10 m/sh JsC
-éù= ´ =´
ëû
Q
\For longest w
avelengthmax
43
12400 12400
E 0.66
®
l==
= 18787.8 Å
For smalles
t wavelength
min
41
hc 12400 12400
973 Å
eE E 12.75
®
l===»
Example
3.
If the wavelength of the first line of the Lyman series for the
hydrogen atom is 1210 Å, then what will be the wavelength
of the first line of the Balmer series of the hydrogen
spectrum?
Solution :
We know that,
ú
ú
û
ù
ê
ê
ë
é
-=
l
2
2
2
1n
1
n
1
R
1
For first lin
e of Lyman series, n
1
=1 and n
2
= 2
\
4
R3
4
1
1
1
R
1
1
=
ú
û
ù
ê
ë
é
-=
l
or
13
4
R
l
=
For first lin
e of Balmer series n
1
=2 and n
2
= 3
\
ú
û
ù
ê
ë
é
l
==
ú
û
ù
ê
ë
é
-=
l
12
3
4
36
5
36
R5
9
1
4
1
R
1
or,
1210
1
3
4
36
51
2
´´=
l
or, Å6434
45
1210336
2
=
´
´´
=l
Example4.
If the electron in hydroge
n atom jumps from the third orbit
to second orbit, then find the wavelength of the emitted
radiation.
Solution :
ú
ú
û
ù
ê
ê
ë
é
-=
l
2
2
2
1n
1
n
1
R
1
or
36
R5
9
1
4
1
R
1
=
ú
û
ù
ê
ë
é
-=
l
\
R5
36
=l
Example 5 :
How many different
wavelengths may be observed in the
spectrum from a hydrogen sample if the atoms are excited
to states with principal quantum number n ?
Solution :
From the nth state, the atom may go to (n – 1)th state, ….,
2
nd
state or 1
st
state. So there are (n – 1)th possible transi-
tions starting from the
nth sate. The atoms reaching (n –
1)th state may make (n – 2) different transitions. Similarly forother lower states, the total number of possible transitions
are
(n 1) (n 2) (n 3) .....2 1-+-+-++
n(n 1)
2
-
=
Example 6.
In the hydrogen atom, an elect
ron makes a transition
from n = 2 to n = 1. The magnetic field produced by the
circulating electron at the nucleus
(a) decreases 16 times (b) increases 4 times
(c) decreases 4 times(d) increases 32 times
Solution :(d)
Q
0
I
B
2r
m
= and
e
I
T
=

0
e
B
2rT
m
= Þ
5
1
B
n
µ [
25
rn,TnµµQ ] ;

719. 715Atoms
CONCEPT MAP
Bohr model of Hydrogen
atom Electron can revolve
only in those orbits in which
angular momentum about the
nucleus is an integral multiple
hnh
ofi.e., mvr
22
=
pp
Drawbacks of Bohr model
unable to explain the fine
structure of spectral lines
Valid only for single electron
system
Various parameters
Radius of n-th
or bit
22
n
2 2
2
nh
r
4 KZem
n
0.53Å
Z
=
p
=
Velocity of electron
in n-th orbit
2
n
6
2 KZe
v
nh
Z
2.2 10m/s
n
p
=
= ´
Potential energy
(Un) in n-th orbit
2
2
n
2
n
KZe27.2
UZ ev
rn
- -
= =
Kinetic energy
2
2
k
2
n
KZe 1
3
.6
Z
E
e
v
2r
n
= =
J
.J
T
h
o
m
s
o
n

m
o
d
e
l
o
f

a
t
o
m

F
i
r
s
t
m
o
d
e
l

o
f

a
t
o
m
p
l
u
m

p
u
d
d
i
n
g

m
o
d
e
l
-
p
o
s
i
t
i
v
e

c
h
a
r
g
e
i
s

u
n
i
f
o
r
m
l
y
d
i
s
t
r
i
b
u
t
e
d
a
n
d

n
e
g
a
t
i
v
e
l
y
c
h
a
r
g
e
d

e
l
e
c
t
r
o
n
s

a
r
e
e
m
b
e
d
d
e
d
i
n

i
t
l
i
k
e
s
e
e
d
s

i
n
a
w
a
t
e
r
m
e
l
o
n
Model of atom
ATOMS
Consists of three
elementary particles
electrons, protons and neutrons
Various series of line
spectra of hydrogen
atom
Lyman series
2
2
1
1
1
v R
1
n
é
ù
==
-
ê
ú
l
ë
û
n = 2, 3, 4
.
..
..
..
..
.
in uv-regio
n
Balmer series
2
2
1
1
1
v
R
2
n
é
ù
=
=
-
ê
ú
l
ë
û
n

=
3
,
4
,
5
i
n

v
i
s
i
b
l
e
r
e
g
i
o
n
P
a
s
c
h
e
m
s
e
r
i
e
s
2
2
1
1
1
v
R
3
n
é
ù
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Rutherford - particle scattering experim
e
n
t
Most of the -particles passed through th
e

gold foil i.e., atom has lot of empty spa
c
e
Only about 0.14% of the -particles
scatter by more than 1° and one -particl
e

i
n

every 8000 -particles deflected by > 90°
positively charged particles protons
confined to core called nucleus, size abo
u
t

10to 10m
a
a
a
a
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–15–14
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r

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scatter
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720. 716 PHYSICS
1.In Millik
an’s oil drop experiment, an oil drop is observed to
move vertically upward. The upward motion of the drop is
due to
(a) gravity (b) viscosity
(c) buoyancy (d) electric field
2.The following statements are all true. Which one did
Rutherford consider to be supported by the results of
experiments in which a-particles were scattered by gold
foil?
(a) The nucleus of an atom is held together by forces
which are much stronger than electrical or
gravitational forces.
(b) The force of repulsion between an atomic nucleus
and an a-particle varies with distance according to
inverse square law.
(c)a-particles are nuclei of Helium atoms.
(d) Atoms can exist with a series of discrete energy levels.
3.The proof of quantization of energy states in an atom is
obtained by the experiment performed by
(a) Thomson (b) Millikan
(c) Rutherford (d) Franck and Hertz
4.The angular speed of the electron in the n
th
orbit of Bohr
hydrogen atom is
(a) directly proportional to n
(b) inversely proportional to n
(c) inversely proportional to n
2
(d) inversely proportional to n
3
5.According to Bohr’s model of hydrogen atom
(a) the linear velocity of the electron is quantised.
(b) the angular velocity of the electron is quantised.
(c) the linear momentum of the electron is quantised.
(d) the angular momentum of the electron is quantised.
6.As the quantum number increases, the difference of energy
between consecutive energy levels
(a) remain the same
(b) increases
(c) decreases
(d) sometimes increases and sometimes decreases.
7.Line spectrum is obtained whenever the incandescent
vapours at low pressure of the excited substance are in
their
(a) atomic state (b) molecular state
(c) nuclear state (d) None of these
8.According to the Bohr theory of H-atom, the speed of the
electron, its energy and the radius of its orbit varies with
the principal quantum number n, respectively, as
(a)
2
2
11
,n,
n n
(b)
2
2
1
n, ,n
n
(c)
22
11
n,,
nn
(d)
2
2
11
, ,n
nn
9.In a hydrogen atom follo
wing the Bohr’s postulates the
product of linear momentum and angular momentum is
proportional to (n)
x
where ‘n’ is the orbit number. Then ‘x’
is
(a)0 (b) 2
(c) –2 (d) 1
10.In an atom, the two electrons move round the nucleus in
circular orbits of radii R and 4R. The ratio of the time taken
by them to complete one revolution is
(a) 1/4 (b) 4/1 (c) 8/1 (d) 1/8
11.The ratio of the energies of the hydrogen atom in its first to
second excited states is
(a) 1/4 (b) 4/9 (c) 9/4 (d) 4
12.For the principal quantum number n = 4, the total number of
different possible values of the azimuthal quantum number
l, is
(a)2 (b) 8 (c)4 (d) 3
13.With increasing quantum number, the energy difference
between adjacent energy levels in atoms
(a) decreases
(b) increases
(c) remains constant
(d) decreases for low Z and increases for high Z
14.The angular momentum of the electron in hydrogen atom in
the ground state is
(a) 2h (b)
2
h
(c)
p2
h
(d)
p4
h
15.If e is electron
ic charge, the equivalent current for an electron
revolving with a frequency n in the ground state of
hydrogen atom is
(a)ne2
p (b)
n
e
p
(c)
n2
e
p
(d) en
16.The pri
nciple of LASER action involves
(a) amplification of particular frequency emitted by the
system
(b) population inversion
(c) stimulated emissin
(d) All of the above.
17.In which of the following series, does the 121.5 nm line of
the spectrum of the hydrogen atom lie ?
(a) Lyman series (b) Balmer series
(c) Paschen series (d) Brackett series.
18.For the azimuthal quantum number l = 3, the total number of
different possible values of the magnetic azimuthal quantum
number m
1
, is
(a)3 (b) 4 (c)5 (d) 7
19.The ratio between Bohr radii is
(a) 1 : 2 : 3 (b) 2 : 4 : 6 (c) 1 : 4 : 9 (d) 1 : 3 : 5

721. 717Atoms
20.The kinetic en
ergy of the electorn in an orbit of radius r in
hydrogen atom is (e = electronic charge)
(a)
2
2
e
r
(b)
2
e
2r
(c)
2
e
r
(d)
2
2
e
2r
21.Accor ding to the Rutherford’s atomic model, the electrons
inside the atom are
(a) stationary (b) not stationary
(c) centralized (d) None of these
22.According to classical theory, the circular path of an
electron in Rutherford atom is
(a) spiral (b) circular
(c) parabolic (d) straight line
23.Rutherford’s a-particle experiment showed that the atoms
have
(a) Proton (b) Nucleus
(c) Neutron (d) Electrons
24.Electrons in the atom are held to the nucleus by
(a) coulomb’s force(b) nuclear force
(c) vander waal’s force(d) gravitational force
25.The angular momentum of electron in n
th
orbit is given by
(a) nh (b)
h
2np
(c)
h
n
2p
(d)
2h
n
2p
1.In a hypothetical Bohr hydrogen atom, the mass of the electron
is doubled. The energy
'
0
E and radius
'
0
r of the first orbit will
be (r
0
is the Bohr radius)
(a) –11.2 eV (b) –6.8 eV
(c) –13.6 eV (d) –27.2 eV
2.A 15.0 eV photon collides with and ionizes a hydrogen
atom. If the atom was originally in the ground state
(ionization potential =13.6 eV), what is the kinetic energy of
the ejected electron?
(a) 1.4 eV (b) 13.6 eV (c) 15.0 eV (d) 28.6 eV
3.The ratio of areas between the electron orbits for the first
excited state to the ground state for the hydrogen atom is
(a) 2 : 1(b) 4 : 1(c) 8 : 1(d) 16 : 1
4.The potential energy associated with an electron in the
orbit
(a) increases with the increases in radii of the orbit
(b) decreases with the increase in the radii of the orbit
(c) remains the same with the change in the radii of the orbit
(d) None of these
5.Energy levels, A, B, C of a certain atom correspond to
increasing values of energy i.e. E
A
< E
B
< E
C
. If l
1
, l
2
, l
3
are
the wavelengths of radiations corresponding to the
transitions C to B, B to A and C to A respectively, which of
the following relations is correct ?
(a)
213
l+l=l (b)
21
21
3
l+l
ll
=l
(c) 0
321=l+l+l (d)
2
2
2
1
3
3
l+l=l
6.Consider
the spectral line resulting from the transition
1n2n
=®= in the atoms and ions given below. The
shortest wavelength is produced by
(a) hydrogen atom (b) deuterium atom
(c) singly ionized Helium (d) doubly ionised Lithium
7.A hydrogen atom emits a photon corresponding to an
electron transition from n = 5 to n = 1. The recoil speed of
hydrogen atom is almost
(a) 10
–4
ms
–1
(b) 2 × 10
–2
ms
–1
(c) 4 ms
–1
(d) 8 × 10
2
ms
–1
8.In Bohr’s model of the hydrogen atom, let R, V, T and E
represent the radius of the orbit, speed of the electron, time
period of revolution of electron and the total energy of the
electron respectively. The quantity proportional to the
quantum number n is
( a ) VR(b) E (c)r (d) T
9.Let n
1
be the frequency of the series limit of the Lyman series,
n
2
be the frequency of the first line of the Lyman series, and
n
3
be the frequency of the series limit of the Balmer series.
(a)n
1
– n
2
= n
3
(b)n
2
– n
1
= n
3
(c)n
3
=

2
1
( n
1
+ n
2
) (d)n
1
+ n
2
= n
3
10.The approximate v
alue of quantum number n for the circular
orbit of hydrogen 0.0001 mm in diameter is
(a) 1000(b) 60 (c) 10000(d) 31
11.The ionization energy of Li
++
is equal to
(a) 9hcR(b) 6hcR (c) 2hcR (d) hcR.
12.The Rutherford a-particle experiment shows that most of
the a-particles pass through almost unscattered while some
are scattered through large angles. What information does
it give about the structure of the atom?
(a) Atom is hollow.
(b) The whole mass of the atom is concentrated in a small
centre called nucleus
(c) Nucleus is positively charged
(d) All the above
13.Which of the following series in the spectrum of hydrogen
atom lies in the visible region of the electromagnetic
spectrum?
(a) Paschen series (b) Balmer series
(c) Lyman series (d) Brackett series
14.The ratio of maximum to minimum wavelength in Balmer
series is
(a) 3 : 4(b) 1 : 4(c) 5 : 36(d) 5 : 9
15.What element has k
a
line of wavelength 1.785 Å?
R = 109737 cm
-1
.
(a) Platinum (b) Zinc(c) Iron (d) Cobalt

722. 718 PHYSICS
16.If the k
a
radiation of Mo (Z = 42) has a wavelength of
0.71Å. Calculate the wavelength of the corresponding
radiation of Cu (Z = 29).
(a) 1.52Å(b) 2.52Å(c) 0.52Å(d) 4.52Å
17.The extreme wavelengths of Paschen series are
(a) 0.365 mm and 0.565 mm (b) 0.818 mm and 1.89 mm
(c) 1.45 mm and 4.04 mm(d) 2.27 mm and 7.43 mm
18.The third line of Balmer series of an ion equivalnet to
hydrogen atom has wavelength of 108.5 nm. The ground
state energy of an electron of this ion will be
(a) 3.4 eV (b) 13.6 eV
(c) 54.4 eV (d) 122.4 eV
19.The first line of Balmer series has wavelength 6563 Å. What
will be the wavelength of the first member of Lyman series
(a) 1215.4 Å (b) 2500 Å
(c) 7500 Å (d) 600 Å
20.Hydrogen atom excites energy level from fundamental state
to n = 3. Number of spectral lines according to Bohr, is
(a)4 (b) 3 (c)1 (d) 2
21.The ratio of the longest to shortest wavelengths in Brackett
series of hydrogen spectra is
(a)
25
9
(b)
17
6
(c)
9
5
(d)
4
3
22.In a Rutherford experiment, the number of particles scattered
at 90° angle are 28 per minute then number of scattered
particles at an angle 60° and 120° will be
(a) 117 per minute, 25 per minute
(b) 50 per minute, 12.5 per minute
(c) 100 per minute, 200 per minute
(d) 112 per minute, 12.4 per minute
23.In Hydrogen spectrum, the wavelength of H
a
line is 656
nm, whereas in the spectrum of a distant galaxy, H
a
line
wavelength is 706 nm. Estimated speed of the galaxy with
respect to earth is
(a) 2 × 10
8
m/s (b) 2 × 10
7
m/s
(c) 2 × 10
6
m/s (d) 2 × 10
5
m/s
24.The energy levels of the hydrogen spectrum is shown in
figure. There are some transitions A, B, C, D and E.
Transition A, B and C respectively represent
n = 5
n = 4
n = 3
n = 2
n = 6
n = 1
n = ¥ – 0.00 eV
– 0.36 eV
– 0.54 eV
– 0.85 eV
– 1.51 eV
– 3.39 eV
– 13.5 eV
A
B
C
D
E
(a) first member of Lyman series, third spectral line of
Balmer series and the second spectral line of Paschen
series
(b) ionization potential of hydrogen, second spectral line
of Balmer series, third spectral line of Paschen series
(c) series limit of Lyman series, third spectral line of Balmer
series and second spectral line of Paschen series
(d) series limit of Lyman series, second spectral line of
Balmer series and third spectral line of Paschen series
25.Which of the following statements are true regarding Bohr’s
model of hydrogen atom?
(I) Orbiting speed of electron decreases as it shifts to
discrete orbits away from the nucleus
(II) Radii of allowed orbits of electron are proportional to
the principal quantum number
(III) Frequency with which electrons orbit around the
nucleus in discrete orbits is inversely proportional to
the cube of principal quantum number
(IV) Binding force with which the electron is bound to the
nucleus increases as it shifts to outer orbits
Select correct answer using the codes given below.
Codes :
(a) I and II (b) II and IV
(c) I, II and III (d) II, III and IV
26.If in hydrogen atom, radius of n
th
Bohr orbit is r
n
, frequency
of revolution of electron in n
th
orbit is f
n
, choose the correct
option.
(a)
r
n
O n
(b)
O
log n
n
1
r
log
r
æö
ç÷
èø
(c)
O
log n
n
1
f
log
f
æö
ç÷
èø
(d) Both (a) and (b)
27.Excitation energy of a hydrogen like ion in its excitation
state is 40.8 eV. Energy needed to remove the electron from
the ion in ground state is
(a) 54.4 eV (b) 13.6 eV (c) 40.8 eV(d) 27.2 eV
28.The ionisation potential of H-atom is 13.6 V. When it is
excited from ground state by monochromatic radiations of
970.6 Å, the number of emission lines will be (according to
Bohr’s theory)
(a) 10 (b) 8 (c)6 (d) 4
29.Electrons in a certain energy level n = n
1
, can emit 3 spectral
lines. When they are in another energy level, n = n
2
. They
can emit 6 spectral lines. The orbital speed of the electrons
in the two orbits are in the ratio of
(a) 4 : 3(b) 3 : 4(c) 2 : 1(d) 1 : 2
30.The energy of electron in the nth orbit of hydrogen atom is
expressed as
n 2
13.6
E eV.
n
-
= The shortest and longest
wavelength of Lyman series will be
(a) 910 Å, 1213 Å (b) 5463 Å, 7858 Å
(c) 1315 Å, 1530 Å (d) None of these

723. 719Atoms
31.A hydr
ogen atom in its ground state absorbs 10.2 eV of
energy. The orbital angular momentum is increased by
(a) 1.05 × 10
–34
J-s (b) 3.16 × 10
–34
J-s
(c) 2.11 × 10
–34
J-s (d) 4.22 × 10
–34
J-s
32.Taking Rydberg’s constant R
H
= 1.097 × 10
7
m, first and
second wavelength of Balmer series in hydrogen spectrum
is
(a) 2000 Å, 3000 Å (b) 1575 Å, 2960 Å
(c) 6529 Å, 4280 Å (d) 6552 Å, 4863 Å
33.If the atom
100
Fm
257
follows the Bohr model and the radius
of
100
Fm
257
is n times the Bohr radius, then find n.
(a) 100 (b) 200 (c)4 (d) 1/4
34.In Rutherford scattering experiment, the number of
a-particles scattered at 60° is 5 × 10
6
. The number of
a-particles scattered at 120° will be
(a) 15 × 10
6
(b) 3
5
× 10
6
(c)
5
9
× 10
6
(d) None of th ese
35.The energy of He
+
in the ground state is – 54.4 eV, then the
energy of Li
++
in the first excited state will be
(a) – 30.6 eV (b) 27.2 eV
(c) – 13.6 eV (d) – 27.2 eV
36.The wavelength of radiation is l
0
when an electron jumps
from third to second orbit of hydrogen atom. For the electronto jump from the fourth to the second orbit of the hydrogenatom, the wavelength of radiation emitted will be
(a)
0
16
25
l(b) 0
20
27
l(c) 0
27
20
l (d) 0
25
16
l
37.If the frequency of K
a
X-rays emitted from the element with
atomic number 31 is n, then the frequency of K
a
X-rays
emitted from the element with atomic number 51 would be(a)
5
3
n (b)
51
31
n(c)
25
9
n (d)
9
25
n
38.A hydrogen atom is in ground state. Then to get six lines in
emission spectrum, wavelength of incident radiation should be
(a) 800 Å(b) 825 Å (c) 975 Å (d) 1025 Å
39.In the hydrogen atom, an electron makes a transition from
n = 2 to n = 1. The magnetic field produced by the circulating
electron at the nucleus
(a) decreases 16 times(b) increases 4 times
(c) decreases 4 times(d) increases 32 times
40.One of the lines in the emission spectrum of Li
2+
has the
same wavelength as that of the 2
nd
line of Balmer series in
hydrogen spectrum. The electronic transition corresponding
to this line is n = 12 ® n = x. Find the value of x.
(a)8 (b) 6 (c)7 (c)5
41.The wavelength K
a
of X-rays for two metals ‘A’ and ‘B’ are
4
1875R
and
1
675R
respectively, where ‘R’ is Rydbergg
constant. Find the number of elements lying between Aand B according to their atomic numbers
(a)3 (b) 1 (c)4 (d) 5
42.Doubly ionised helium atom and hydrogen ions are
accelerated, from rest, through the same potential difference.
The ratio of final velocities of helium and hydrogen is
(a)
2:1 (b) 1:2
(c) 1 : 2 (d)
2 : 1
43.Ionization potential of hydrogen atom is 13.6 eV. Hydrogen
atoms in the ground state are excited by monochromatic
radiation of photon energy 12.1 eV. The spectral lines
emitted by hydrogen atom according to Bohr’s theory will
be
(a) one(b) two (c) three(d) four
44.Which of the following transitions in hydrogen atoms emit
photons of highest frequency?
(a) n = 1 to n = 2 (b) n = 2 to n = 6
(c) n = 6 to n = 2 (d) n = 2 to n = 1
45.The energy of a hydrogen atom in the ground state is
– 13.6 eV. The energy of a He
+
ion in the first excited state
will be
(a) –13.6 eV (b) – 27.2 eV
(c) – 54.4 eV (d) – 6.8 eV
46.An alpha nucleus of energy
21
2
mv bombards a heavy
nuclear target of charge Ze. Then the distance of closest
approach for the alpha nucleus will be proportional to
(a)
1
Ze
(b)
2
v (c)
1
m
(d)
4
1
v
Directions for Qs. (47 to 50) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a
correct explanation for Statement -1
(b) Statement-1 is True, Statement -2 is True; Statement-2 is
NOT a correct explanation for Statement - 1
(c) Statement-1 is True, Statement- 2 is False
(d) Statement-1 is False, Statement -2 is True
47. Statement-1 : Balmer series lies in the visible region of
electromagnetic spectrum.
Statement-2 :
22
1 11
R,
2n
éù
=-
êú
l ëû
where n = 3, 4, 5.
48. Statement-1 : The force of repulsion between atomic nucleus
and a-particle varies with distance according to inverse
square law.
Statement-2 : Rutherford did a-particle scattering experiment.
49. Statement-1 : Bohr had to postulate that the electrons in
stationary orbits around the nucleus do not radiate.
Statement-2 : According to classical physics all moving
electrons radiate.
50. Statement 1 : In Lyman series, the ratio of minimum and
maximum wavelength is
3
4
.
Statemen
t 2 : Lyman series constitute spectral lines
corresponding to transition from higher energy to ground
state of hydrogen atom.

724. 720 PHYSICS
Exemplar Questions
1.Taking the Bohr radius as a
0
= 53 pm, the radius of Li
++
ion
in its ground state, on the basis of Bohr's model, will be
about
(a) 53 pm (b) 27 pm
(c) 18 pm (d) 13 pm
2.The binding energy of a H – atom, considering an electron
moving around a fixed nuclei (proton), is
( )
4
222
0
me
B m electron mass
8nh
=-=
e
If one decid
es to work in a frame of reference where the
electron is at rest, the proton would be moving around it.
By similar arguments, the binding energy would be
( )
4
222
0
me
B – m proton mass
8nh
==
e
This last
expression is not correct, because
(a) n would not be integral
(b) Bohr – quantisation applies only two electron
(c) the frame in which the electron is at rest is not inertial
(d) the motion of the proton would not be in circular orbits,
even approximately.
3.The simple Bohr model cannot be directly applied to
calculate the energy levels of an atom with many electrons.
This is because
(a) of the electrons not being subject to a central force
(b) of the electrons colliding with each other
(c) of screening effects
(d) the force between the nucleus and an electron will no
longer be given by Coulomb's law
4.For the ground state, the electron in the H – atom has an
angular momentum = h, according to the simple Bohr model.
Angular momentum is a vector and hence there will be
infinitely many orbits with the vector pointing in all possible
directions. In actuality, this is not true,
(a) because Bohr model gives in correct values of angular
momentum
(b) because only one of these would have a minimum
energy
(c) angular momentum must be in the direction of spin of
electron
(d) because electrons go around only in horizontal orbits
5.O
2
molecule consists of two oxygen atoms. In the molecule,
nuclear force between the nuclei of the two atoms
(a) is not important because nuclear forces are short –
ranged
(b) is as important as electrostatic force for binding the
two atoms
(c) cancels the repulsive electrostatic force between the
nuclei
(d) is not important because oxygen nucleus have equal
number of neutrons and protons
6.Two H atoms in the ground state collide inelastically. The
maximum amount by which their combined kinetic energy
is reduced, is
(a) 10.20 eV (b) 20.40 eV
(c) 13.6 eV (d) 27.2 eV
NEET/AIPMT (2013-2017) Questions
7.An electron in hydrogen atom makes a transition n
1
® n
2
where n
1
and n
2
are principal quantum numbers of the two
states. Assuming Bohr’s model to be valid the time period
of the electron in the initial state is eight times that in the
final state. The possible values of n
1
and n
2
are
(a)n
1
= 4 and n
2
= 2 [NEET Kar. 2013]
(b)n
1
= 6 and n
2
= 2
(c)n
1
= 8 and n
2
= 1
(d)n
1
= 8 and n
2
= 2
8.Hydrogen atom in ground state is excited by a
monochromatic radiation of l = 975 Å. Number of spectral
lines in the resulting spectrum emitted will be [2014]
(a)3 (b) 2
(c)6 (d) 10
9.Consider 3
rd
orbit of He
+
(Helium), using non-relativistic
approach, the speed of electron in this orbit will be [given
K = 9 × 10
9
constant, Z = 2 and h (Plank's Constant)
= 6.6 × 10
–34
J s] [2015]
(a) 1.46 × 10
6
m/s(b) 0.73 × 10
6
m/s
(c) 3.0 × 10
8
m/s (d) 2.92 × 10
6
m/s
10.Two particles of masses m
1
, m
2
move with initial velocities
u
1
and u
2
. On collision, one of the particles get excited to

725. 721Atoms
higher
level, after absorbing energy e. If final velocities of
particles be v
1
and v
2
then we must have [2015]
(a)
2222
11 22 1
1 22
1111
mu mu mv mv –
22 22
+=+e
(b)
22 22
1122 112 2
11 11
mu mu – m v mv
22 22
+ e=+
(c)
22 22 22 22
11 22 11 22
11 11
mu mu mv mv
22 22
++e=+
(d)
22 22
11 22 11 22
mu mu – mv mv+ e=+
11.In t
he spectrum of hydrogen, the ratio of the longest
wavelength in the Lyman series to the longest wavelength
in the Balmer series is [2015 RS]
(a)
9
4
(b)
27
5
(c)
5
27
(d)
4
9
12.When an a-p article of mass 'm' moving with velocity 'v'
bombards on a heavy nucleus of charge 'Ze', its distance of
closest approach from the nucleus depends on m as : [2016]
(a)
1
m
(b)
1
m
(c)
2
1
m
(d) m
13.Given th
e value of Rydberg constant is 10
7
m
–1
, the wave
number of the last line of the Balmer series in hydrogen
spectrum will be : [2016]
(a) 0.025 × 10
4
m
–1
(b) 0.5 × 10
7
m
–1
(c) 0.25 × 10
7
m
–1
(d) 2.5 × 10
7
m
–1
14.The ratio of wavelengths of the last line of Balmer series
and the last line of Lyman series is :- [2017]
(a) 1 (b) 4
(c) 0.5 (d) 2

726. 722 PHYSICS
EXERCISE - 1
1. (d) 2. (b)
3. (d) 4. (d) 5. (d)
6. (c) 7. (a) 8. (d) 9. (a)
10. (d)
4
1
n
n
R
R
2
2
2
1
2
1
==\
2
1
n
n
2
1
=
8
1
2
1
n
n
T
T
3
3
2
1
2
1
=÷
ø
ö
ç
è
æ
=
÷
÷
ø
ö
ç ç
è
æ
=
11. (c) Ist e
xcited state corresponds to n = 2
2nd excited state corresponds to n = 3
\
4
9
2
3
n
n
E
E
2
2
2
1
2
2
2
1
===
12. (c)l = 0 to (n – 1) i.e. 0, 1, 2, 3,. In al
l 4 values.
13. (a) As n increases, energy difference between adjacent
energy levels decreases.
14. (c) According to Bohr's theory,
Angular momentum,
nh
mvr
2
=
p
So in ground state, angular momentum =
h
2p
.
15. (d) Charge = e
, frequency = n, time period =
n
1
So, equivalent current =
e
en
1/n
=
16. (d) Laser action involves all of the following.
(i) Amplification of particular frequency.
(ii) Population inversion.
(iii) Stimulated emission
17. (a) Since 121.5 nm line of spectrum of hydrogen atom lies
in ultraviolet region, therefore it is Lyman series.
18. (d) Different values of magnetic azimuthals quantum
number are m
1
= 2 l + 1 = 2(3) + 1 = 7
19. (c)
20. (b) Potential energy of electron in n
th
orbit of radius r in
H-atom
2
e
U
r
=- (in CGS)
Q
2
1e
K.E. | P.E.| K
2 2r
= Þ=
21. (b) 22. (a)2
3. (b)24. (a)
25. (c) According to Bohr’s second postulate.EXERCISE - 2
1. (d) As
m
1
rµ \ 00r
2
1
r=¢
As mEµ \
'
E
0
= 2(–13.6) = –27.2 eV
2. (a) C
onservation of energy requires that the 15.0 eV
photon energy first provides the ionization energy to
unbind the electron, and then allows any excess energy
to become the electron’s kinetic energy. The kinetic
energy in this case is 15.0 eV – 13.6 eV = 1.4 eV.
3. (d)
422
nrnr µpÞµ
So area Þµ
4
n Ratio is 2
4
: 1 Þ 16 : 1.
4. (b)
r4
Ze
.E.P
0
2
pe
-
= . Negative sign indic
ates that revolving
electron is bound to the positive nucleus.
So, it decreases with increase in radii of orbit.
5. (b) As energy emitted varies inversely as wavelength,
therefore
21
12
213
111
ll
l+l
=
l
+
l
=
l
Þ
21
21
3
l+l
ll
=l
6. (d) We know tha
t,
22
22
21
1 111
RZZ
nn
éù
= - Þµêú
ll
êúëû
l is short
est when
l
1
is largest i.e., when Z is big . Z is
highest for lithium.
7. (c)
h
p1
h~p
D
=
l
Þl´D
We know that
÷
÷
ø
ö
ç ç
è
æ
-=
l
2
2
2
1n
1
n
1
R
1
÷
÷
ø
ö
ç ç
è
æ
-=Þ
2
2
2
1
HHH
n
1
n
1
hRvm
341
H
24
v 6.26 10 R 4 m s
25
-- æö
=´´= ç÷
èø
8. (a)
2
Rnµ
1
V
n
µ so VR is nµ
2
3
n
1
EnTµÞµ
9. (a) Ser
ies limit of Lyman series is 1
®¥ n
1
First line of Lyman
series is 12
®n
2
Series limit
for Balmer series 2
®¥
321
vvv=-\
10. (d)
102
10n528.
0r
-
´=
7
2 101 10
0.528n 10
2
-
-
æö´
=´
ç÷
èø
8
2
10
5 10
n
0.528 10
-
-
´
Þ=
´
31n»Þ
Hints & Solutions

727. 723Atoms
11. (a) I
onisation energy of Li
++
= 9hcR
Ionisation energy = RchZ
2
= Rch(3)
2
(as Z = 3 for Li
++
)
= 9hcR.
12. (d)
13. (b) Transition from higher states to n = 2 lead to emission
of radiation with wavelengths 656.3 nm and 365.0 nm.
These wavelengths fall in the visible region and
constitute the Balmer series.
14. (d)
2
22
12
1 11
RZ
nn
æö
=-ç÷
ç÷l
èø
For Balmer series n = 2
2
22
max
1 11
RZ
2
æö
=-
ç÷
l è ¥ø
and
2
22
min
1 11
RZ
23
æö
=-
ç÷
l èø
or
22
min
2
max
1/ (1/ 2 1/3 )
1/ (1/2)
l -
=
l
=
11
449
1 5/9
1/49
-
=-=
Þ
max
min
5
9
l
=
l
15. (d) For
k
a
line
k
1
a
l
= R (Z – 1
)
2

22
11
12
éù
-
êú
ëû
(Z – 1)
2
=
k
411
3R
a
l
=
8
411
3 1097371.785 10
-
´´
´
Þ (Z – 1)
2
= 6
80.6 Þ Z – 1 = 26 Þ Z = 27
Thus, the element is cobalt.
16. (a) From Mosley's law, we have,
(Z – 1)
2
µ n \ (Z – 1)
2
= A
k
c
a
l
where A
is some constant,
\
2
MO Cu
2
MOCu
(Z 1)
(Z 1)
-l
=
l-
or
2
Cu41
28 0.71
læö
=
ç÷
èø
\
Cu
l = 0.71 ×
2
41
28
æö
ç÷
èø
= 1.52Å
17. (b
) In Paschen series
22
max
1 11
R
(3) (4)
éù
=-
êú
l
ëû
6
max
7
144 144
1.89 10 m
7R7 1.1 10
-
Þl=
= =´
´´
= 1.89 mm
Similarly
min
7
99
0.818 m
R1.1 10
l== =m
´
18. (c) Fo
r third line of Balmer series n
1
= 2, n
2
= 5 22
22 12
2 2 22
1 2 21
nn1 11
RZ g
ivesZ n n (n n)R
éù
\=-=
êú
l -l
ëû
On putt
ing values Z = 2
From
22
22
13.6Z 13.6(2)
E 54.4 eV
n (1)
-
=- = =-
19. (
a)
22
Balmer
1 1 1 5R
R,
3623
éù
= -=
êú
l ëû
Lyman
1
l

22
1 1 3R
R
412
éù
= -=
êú
ëû
Lyman Balmer
5
1215.4Å
27
\l =l ´=
20
. (b) No. of lines
E
n(n 1) 3
(3 1)
N3
22
--
= ==
21. (a) For Bracket series 22
max
1 119
RR
25 1645
éù
= -=
êú
l´ ëû
and
max
22
min min
1 1 1 R 25
R
1694
léù
= -=Þ=
êú
ll ë ¥û
22
. (d) No. of particles scattered through an angle
q = N(q)
2
42
kZ
sin (K.E.)
2
=
qæö
ç÷
èø
\
2
2
4kcz
28 for 90
(K.E.)
= q=° \
2
2
kz 28
7
4(K.E.)
==
\
4
7
N(60)
60
sin
2
°=
°æö
ç÷
èø
= 16 × 7 = 112 /min.
4
7
N(120 ) 12.4 / min
120
sin
2
°==
°æö
ç÷
èø
23.
(b)
1 1 c–v
' cv
=
ll+
Here, 'l = 706 nm,l = 656 nm
\
22
c – v 656
0.86
c v ' 706
læö
æö
===
ç÷ç÷
+lèøèø
Þ
v 0.14
c 1.86
= Þ v = 0.075 × 3 × 10
8
= 2.25 × 10
7
m/s
24. (c) Transition A (n = ¥ to 1) : Series lime of Lyman series
Transition B (n = 5 to n = 2) : Third spectral lien of
Balmer series
Transition C (n = 5 to n = 3) : Second spectral line of
Paschen series
25. (a) Orbital speed varies inversely as the radius of the orbit.
1
v
n
µ

728. 724 PHYSICS
26. (d) Ra
dius of n
th
orbit r
n
µ n
2
, graph between r
n
and n is
a parabola. Also,
2
nn
ee
11
rrn
log 2log (n)
r1r
æöæö
=Þ=ç÷ ç÷èø èø
Comparing
this equation with y = mx + c,
Graph between
n
e
1
r
log
r
æö
ç÷
èø
and log
e
(n) will be a straight
line, passing from origin.
Similarly it can be proved that graph between
n
e
1
f
log
f
æö
ç÷
èø
and log
e
n is a straigh t line. But with
negative slops.
27. (a) Excitation energy DE = E
2
–E
1
= 13.6 Z
2

22
11
12
éù
-
êú
ëû
23
40.8 13.6 Z Z 2.
4
Þ = ´´ Þ=
Now required energy to remove the electron from
ground state
2
2
2
13.6Z
13.6(Z) 54.4 eV.
(1)
+
===
28. (c) 22
12
1 11
R
nn
éù
=-
êú
l
ëû
7
2
10 22
2
1 11
1.097 10 n 4
970.6 10 1 n
-
éù
Þ = ´ - Þ=
êú
´
ëû
\ Number of e
mission line
n(n1) 43
N6
22
-´
= ==
29. (a) Number of emission spectral lines,
n(n 1)
N
2
-
=
11n (n 1)
3,
2
-
\= in first case.
O
r
2
11 11
n n 6 0 or (n 3 )(n 2) 0--= - +=
Take positive root.
\ n
1
=
3
Again,
22n (n 1)
6,
2
-
= in second case.
Or
2
22 22
n n 12 0 or (n 4 )(n 3) 0.- -= - +=
Take positive root, or n
2
= 4
Now velocity of electron
2
2 KZe
nh
p
u=
12
21
n4
.
n3
u
\==
u
30. (a)
max
22
max
1114
R 1213Å
3R(1) (2)
éù
= - Þl
=»
êú
l
ëû
and
min
2
min
1111
R 910Å.
R(1)
éù
= - Þl =»
êú
l¥
ëû
31.
(a) Electron after absorbing 10.2 eV energy goes to its
first excited state (n = 2) from ground state (n = 1).
\ Increase in momentum
h
2
=
p
34
346.6 10
1.05 10 J-s.
6.28
-
-´
= =´
32. (d) 22
12
1 11
R.
nn
éù
=-
êú
l
ëû
For first wavelength, n
1
= 2, n
2
= 3
Þ l
1
= 6563 Å. For second wavelength, n
1
= 2, n
2
= 4
Þ l
2
= 4861 Å
33. (d) For an atom following Bohr’s model, the radius is given
by
2
0
m
rm
r
Z
= where r
0
= Bo hr’s radius and m = orbit
number.
For Fm, m = 5 (Fifth orbit in which the outermost
electron is present)
\r
m
2
0
0
5
100
r
nr== (given) Þ n =
1
4
34. (c)
4
1
N
sin /2
µ
q
;
4
21
4
1 2
N sin ( / 2)
Nsin ( / 2)
q
=
q
or
4
2
64
N sin (60 / 2)
5 10 sin (120 / 2)
°
=
´°
or
4
2
64
N sin 30
5 10 sin 60
°
=
´°
or
44
66
2
125
N 5 10 10
293
æöæö
=´ ´ =´ç÷ç÷
èøèø
35. (a)
Energy of electron in n
th
orbit is 2
n
2
Z
E (Rch) 54.4 eV
n
=- =-
For He
+
is ground state
2
1
2
(2)
E (Rch) 54.4 Rch 13. 6
(1)
=- =- Þ =
\ For
Li
++
in first excited state (n = 2)
E' = – 13.6 ×
2
2
(3)
30.6 eV
(2)
=-
36. (b) 22
12
1 11
R
nn
æö
=-ç÷
l èø
22
0
1 1 1 1 1 5R
RR
4 9 3623
æö æö
= - = -=
ç÷ç÷
èøèøl

729. 725Atoms
22
1 1 1 1 1 3R
RR
4 16 1624
æö æö
= - = -=
ç÷ç÷
èøèøl
0
51620
36 3 27
l
=´=
l
37. (c) (Z 1)nµ- for K
a
X-ray
Þ
2 2
22
2
11
Z1 51 1 25
Z1 3119
æön- -æö
= Þn= n=n
ç÷ç÷
èøn--èø
38. (c
) Number of possible spectral lines emitted when
an electron jumps back to ground state from n
th
orbit =
n (n 1)
2
-
Here,
n (n 1)
6 n4
2
-
=Þ=
Wavelength l from transition from n = 1 to n = 4 is
given by,
2
1 1 1 16
R 975Å
1 15R4
æö
=-Þl==
ç÷
èøl
39
. (d)
Q
0
I
B
2r
m
= and
e
I
T
=

0
e
B
2rT
m
= [
25
rn,Tnµµ ] ;
5
1
B
n
µ
40. (b) For
2
nd
line of Balmer series in hydrogen spectrum
22
1 113
R(1)R
1624
æö
= -=
ç÷ èøl
For Li
2+

22
1 1 1 3R
R9
16x 12
éù æö
=´ -=êú ç÷ èøl
ëû
which is satisfied by n = 12 ® n = 6.
41. (c) Using
2
22
21
1 11
R(Z 1)
nn
éù
=-- êú
l êúëû
For a particle, n
1
= 2, n
2
= 1
For metal A :
2
11
1875 R 3
R (Z 1) Z 26
44
æö
= - Þ= ç÷
èø
For me
tal B :
2
22
3
675 R R (Z 1) Z 31
4
æö
= - Þ= ç÷ èø
Therefore, 4
elements lie between A and B.
= 220 MeV.
42. (a)
2
vm
2
1
Vq= or
m
Vq2
v= i.e.
m
q
vµ
\
He
H
H
He
H
He
m
m
q
q
v
v
´=
=
2
1
m4
m
e
e2
=´
43. (c) T
otal energy of electron in excited state = –13.6 + 12.1
= –1.5 eV, which corresponds to third orbit. The
possible spectral lines are when electron jumps from
orbit 3rd to 2nd ; 3rd to 1st and 2nd to 1st.
44. (d) We have to find the frequency of emitted photons.
For emission of photons the transition must take place
from a higher energy level to a lower energy level which
are given only in options (c) and (d).
Frequency is given by
22
21
11
h 13.6
nn
æö
n=--ç÷
ç÷
èø
For transition from n = 6 to n = 2,
1
22
13.6 1 1 2 1 3.6
h 9h62
- æö æö
n= - =´
ç÷ ç÷
èøèø
For transition from n = 2 to n = 1,
2
22
13.6 1 1 3 1 3.6
h 4h21
-æö æö
n= - =´
ç÷ç÷
èøèø
.
\12n <n
45. (a)Energy of a H-like atom in it's n
th
state is given by
E
n
=
2
2
13.6
Z eV
n
-´
For, first
excited state of He
+
, n = 2, Z = 2
\ 2
4
13.6 13.6
2
He
E eV
+=- ´ =-
46.
(c) Kinetic energy of alpha nucleus is equall to
electrostatic potential energy of the system of the alphaparticle and the heavy nucleus. That is,
21
2
mv =
00
1
4
q Ze
r
a
pe
where
0
ris the distance of closest approach
0
2
0
2
4
q Ze
r
mv
a
=
pe
Þ
0
r Zeq
a
µµ
1
m
µ
2
1
v
µ
Hence, correct opt
ion is (c).
47. (a) The wavelength in Balmer series is given by
22
1 11
R , n 3,4, 5,...
2n
éù
=--
êú
l ëû
22
max
1 11
R
23
éù
=-
êú
l ëû
max 7
36 36
6563 Å
5R5 1.097 10
l===
´´
and 22
min
1 11
R
2
éù
=-
êú
l ë ¥û
min
7
44
3646 Å
R1.097 10
l===
´
The
wavelength 6563 Å and 3646 Å lie in visible region.
Therefore, Balmer series lies in visible region.

730. 726 PHYSICS
48. (b) Rut
herford confirmed that the repulsive force on
a-particle due to nucleus varies with distance according
to inverse square law and that the positive charges
are concentrated at the centre and not distributed
throughout the atom.
49. (b) Bohr postulated that electrons in stationary orbits
around the nucleus do not radiate.
This is the one of Bohr’s postulate. According to this
the moving electrons radiates only when they go from
one orbit to the next lower orbit.
50. (b)
EXERCISE - 3
Exemplar Questions
1. (c) According to Bohr's model of atom rading of an atom
in ground state is
0r
r
z
=where r
0
is Bohr 's radius and
z is a atomic number. Given r
0
= 53 pm
The atomic number of lithium is 3, therefore, the radiusof Li
++
ion in its ground state, on the basis of Bohr's
model, will be about
1
3
times to that of Bohr radius.
So, the radius of lithium ion is
0
r53
18 pm
z3
==» .
2. (c) If one d
ecides to work in a frame of reference where
the electron is at rest, the given expression is not trueas it forms the non – inertial frame of reference.
3. (a) The simple Bohr model cannot be directly applied to
calculate the energy levels of an atom with manyelectrons. So the nuclear the electrons not beingsubject to a central force.
4. (a) Accroding Bohr's second postulate states that the
electron evolves around the nucleus only in thoseorbits for which the angular momentum is some
integral multiple of
h
2p
where h is the Planck's
c
onstant (= 6.6 × 10
–34
J-s). So, the magnitude of
angular momentum is kept equal to some integral
multiple of
h
2p
, where, h is Planck's constant and thus,
the Bohr model does not gives correct value of
angular momentum.
5. (a) As we know that, the nuclear force is too much
stronger than the Coulomb force acting between
charges or the gravitational forces between masses.
The nuclear binding force has to dominate over the
Coulomb repulsive force between protons inside the
nucleus.
The nuclear force between two nucleons falls rapidly
to zero as their distance is increase than a few
femtometres. So in case of oxygen molecule, the
distance between atoms of oxygen is larger as
compared to the distance between nuclears in a
necleus. So, nuclear force between the nuclei of the
two oxygen atoms is not important because nuclear
forces are short – ranged.
6. (a) We know that, electron on the lowest state of the
atom, called the ground state have the lowest energy
and the electron revolving in the orbit of smallest
radius, the Bohr radius, r. The energy of this state (n
= 1), E
1
is – 13.6 eV.
Total energy of two H – atoms in the ground state
collide in elastically = 2 × (–13.6 eV) = –27.2 eV.
The maximum amount by which their combined kinetic
energy is reduced when any one H–atom goes into
first excited state after the inelastic collision. So that
the total energy of the two H – atoms after the
inelastic collision
( )
2
13.6
13.6 17.0eV
2
æö
= +=
ç÷
èø
( )for excited state n 2éù =ëûQ
.
So, maximum loss of
their combined kinetic energy.
Due to inelastic collision
= 27.2 –17.0 = 10.2 eVNEET/AIPMT (2013-2017) Questions
7. (a)Q
3
Tnµ
3
121
222
8TTn
TTn
æö
==
ç÷
èø
Hence, n
1
= 2n
2
8. (c) For t
he l = 975 Å
22
12
1 11
R
nn
æö
=-ç÷
ç÷l
èø
where R is the Rydberg constant
Solving we get n
2
= n = 4
(Q n
1
= 1 ground state)
Therefore number of spectral lines
=
n(n 1) 4(4 1)
6
22
--
==
9. (a) Speed of electron in nth orbit
V
n
=
2
2 KZe
nh
p
V = (2.19 × 10
6
m/s)
Z
n
V = (2.19 × 10
6
)
2
3
(Z = 2 & n = 3)
V =
1.46 × 10
6
m/s

731. 727Atoms
10. (b) By la
w of conservation of energy,
K.E
f
= K.E
i
– excitation energy (e)
or
2222
1 2 11
22
1111
mv mv mu mu222 2
+ = + -e
11. (d) For Lyman series (2 ® 1)
L
1
l
= R
2
1
1
2
éù
-
êú
ëû
=
3R
4
For Balmer series (3 ® 2)
B
1
l
= R
11
49
éù
-
êú
ëû
=
5R
36
Þ
L
B
l
l
=
4
3R
36
5R
=
4
36

5
3
æö
ç÷
èø
=
5
27
12. (a) At closest distance of approach, the kinetic energy of
the particle will convert completely into electrostatic
potential energy.
Kinetic energy K.E. =
21
mv
2
Potential energy P.E. =
KQq
r
21 KQq
mv
2r
= Þr µ
1
m
13. (c) According to Bohr's theory, the wave number of the
last line of the Balmer series in hydrogen spectrum,
For hydrogen atom z = 1
2
22
21
1 11
RZ
nn
æö
=-
ç÷
l
èø
= 10
7
× 1
2

22
11
2
æö
-
ç÷
è ¥ø
Þwave number
1
l
= 0.25 × 10
7
m
–1
14. (b
) For last line of Balmer series : n
1
= 2 and n
2
= ¥22
1
2 2 22
B 12
1 1 1 11
RZR
nn2
éùéù
= -=-
êúêú
l ë ¥ûëû
B
4
R
l= ...(1)
For last lin
e of Lyman series : n
1
= 1 and n
2
= ¥22
22 22
L 12
1 11 11
RZ RZ
nn1
éùéù
= -=-
êúêú
l ë ¥û
ëû
L
1
R
l= ...(ii)
Dividing equ
ation (i) by (ii)
B
L
4
R
1
R
l
=
l
Ratio of wav
elengths is
B
L
4
l
=
l

732. 728 PHYSICS
SOME IMPORT
ANT FACTS ABOUT ATOMIC MASS, SIZE
AND COMPOSITION OF NUCLEUS
1.Proton was discovered by Goldstein
2. Atomic mass unit, 1.a.m.u = 1/12
th
of mass of C-12 isotope,
1 a.m.u = 1.660565 × 10
–27
kg
3.Mass of a proton, m
p
= 1.0073 a.m.u = 1.6726 × 10
–27
kg
4. Chadwick’s experiment : Neutrons were detected.
4 9 121
24 60
HeBe Cnéù+ ®+
ëû
5.Mass of neutron, m
n
= 1. 00866 a.m.u = 1.6749 × 10
–27
kg
6.Mass of electron = 9.1 ×10
–31
kg
7.Mass number, A = total number of nucleons (neutrons +
protons present in the nucleus of an atom)
8.Atomic number, Z = number of protons = number of
electrons
9. Types of nuclei :
(i)Isotopes : The atoms of the element which have the
same atomic number but different atomic mass
numbers. e.g.,
1
H
1
,
1
H
2
,
1
H
3 ;
8
O
16
,
8
O
17
,
8
O
18
(ii)Isobars : The atoms of differents element which have
the same atomic mass number but different atomic
numbers. e.g.,
6
C
14
,
7
N
14
,
18
Ar
40
,
20
Ca
40
etc.
(iii)Isotones : The nuclides which contain the same number
of neutrons e.g.,
3 4 9 10
222 22555
H , He ; Be , Be etc.
(iv)Isomers : hav
ing same mass number, same atomic
number but different radioactive properties.
10.Rest mass of nucleus is less than sum of rest masses ofconstituent nucleons, the difference is called mass defect.
11. Size of the nucleus : Radius of nucleus, R = R
0
A
1/ 3
where
R
0
= 1.1 × 10
–15
m.
Nuclear density of all elements ~ 10
17
kg m
–3
.
MASS ENERGY AND NUCLEAR BINDING ENERGY
Einstein’s mass energy equivalence i.e., E = mc
2
gives :
.J106.1eV1where,V.Me5.931u.m.a1
19-
´=º
and 1Mev = 1.6 × 10
–13
J
Binding Energy
Binding energy of a nucleus is the energy with which nucleons
are bound in the nucleus. It is measured by the work required to
be done to separate the nucleons an infinite distance apart from
the nucleus, so that they may not interact with one another.
Total
[ ]
2
()
P nN
BE Zm AZm mc=+--
m
p
= mass of
proton; m
n
= mass of neutron
m
N
= mass of nucleons (protons + neutrons); Z = Atomic number;
A = atomic mass
Binding Energy per Nucleon
The binding energy per nucleon of a nucleus is the average
energy required to extract a nucleon from the nucleus.
Binding energy per nucleon 2
Total bindingenergy BE mc
B
Total number of nucleons AA
D
= ==
=
2
[ () ()]
A
p nZ
c
Zm AZmMX
A
+--
The graph of b
inding energy per nucleon with mass number A is
as shown below.
Mass number A
B
i
n
d
i
n
g
e
n
e
r
g
y
p
e
r
n
u
c
l
e
o
n
(MeV)
8.0
6.0
4.0
2.0
0
50 100 150 200 250
56
2
4
He
6
12
C
8
16
O
Binding energy per nucleon gives a measure of stability of
nucleus. More is binding energy per nucleon more is the stability
of nucleus. Binding energy per nucleon is small for lighter nuclei
i.e.
1
H
1
,
1
H
2
etc.
For A < 28 at A = 4n the curve shows some peaks at
2
He
4
,
4
Be
8
,
6
C
16
,
8
O
16
,
10
Ne
20
,
12
Mg
24
.
This represents extra stability of these elements with respect to
their neighbours.
28
Nuclei

733. 729Nuclei
Keep in Memory
1.A nu
clide is a specific nucleus of an atom characterised as
Z
X
A
where A = mass number and Z = atomic number.
2.Binding energy per nucleon is nearly 8.4 MeV for nuclei in
the range of mass number 40 to 120.
3.Binding energy is highest in Fe
56
.( 8.8 MeV)
4. Binding energy curve predicts :
(a)Fission : Breaking up of a heavy nucleus
(A > 200) into two nuclei of approximately equal size,
and release of energy.
(b)Fusion : Lighter nuclei ( A < 20) combine together to
form heavier nucleus and release of energy.
(c) BE/ A varies by less than 10% above A = 10 suggests
that each nucleon interacts with its neighbouring
nucleon only.
(d) For A > 56, BE/A decreases because of the
destabilising effect of long-range coloumbic force.
NUCLEAR FORCE
It is the force acts in the nucleus between the nucleons and is
responsible for binding the nucleon.
Characteristics of Nuclear Force :
1.It is a short range force effective only in range 10
–15
m
2.It is charge independent. It acts between proton-proton,
proton-neutron and neutron – neutron.
3.It is not a central force.
4.It is spin dependent.
5.It is 10
38
times stronger than gravitational force and 10
2
times stronger than electric force.
6.The main cause of nuclear force is the exchange of
p- mesons between nucleus
+
p+®np,
°p+®pp,
–
pnp+®
RADIOACTIVITY
It is the spontaneous disinte
gration of the heavy nucleus of an
atom (It occurs without external provocation).
There are three main types of radioactive radiations.
(i)a-rays (i.e., Helium nuclei or a – particles)
(ii)b-rays (i.e., electron or positron or b – particles)
(iii)g-rays (photons or gamma radiations)
It is a process by which an unstable nuclei achieves stability.
This process is not affected by
(a)chemical combination
(b)changing physical environment other than nuclear
bombardment.
Features of Radioactivity :
(i) It is a statistical process.
(ii) When a nucleus undergoes alpha or beta decay, its atomic
number and mass number changes (in b-decay only atomic
number changes) & it transforms into a new element.
(a)
4
2
4A
2Z
A
Z
HeYX+=
-
-
(a-particle), it means that
by emission of alpha particle (
a-particle), it loses
2 units of charge and 4 units of mass.
(b)
+
-
b+=
A
1Z
A
Z
YX (positron). It means that by
emission of beta particle (
b
+-particle), nucleus loses
one unit of charge. It is surprising to note that a
nucleus does not contain b
+
then how is it emitted.
Reason : During a b
+
particle(i.e., positron) decay, a
protron converts into a neutron
u+b+=
++ 0
np (neutrino).
A b
–
particle (i.e., electron) decays, when inside the
nucleus a neutron converts into a proton i.e.,
u+b+=
-+
pn
0
(antineutrino)
Since b
-
particle is an electron(or positron), so the loss
of mass in this decay is negligible.
In b
+
decay the daughter element is one place forward
in the periodic table.
(iii)When a nucleus emits a gamma ray, neither the mass nor
the charge of the nucleus changes
i.e.,
AA
XX
Excited state
Ground state
Z Z (gamm a ray or photon)= +g
The gamma ray
(g-ray) is photon & it carries away some
energy from the nucleus & nucleus returns from excited
state(unstable state) to ground state (stable state)
a and b-particles are not emitted simultaneously.
g rays are emitted after the emission of a and b-particle. a,
b and g-rays are known as Bequerel rays
The energy spectrum in the case of b-particles is continous
but that of a and g-rays is a line spectrum. This means that
b particles are emitted with any amount of kinetic energy.
Properties of a, b & g-rays
(A)Properties of a-rays
(a) It is a positively charged particle & contains a
charge of 3.2 × 10
–19
coulomb(exactly double the
charge of electron).
(b) The mass of a-particles is 6.645 × 10
–27
kg(It is
equal to mass of a helium nucleus). Actually a-
particle is nucleus of helium, hence it is called
doubly ionised helium.
(c) They (a-particles) get deflected in both electric
& magnetic fields.
(d) The velocity of a-particle is very less than the
velocity of light i.e.,
c
V
10
a
», wher
e c is velocity
of light.
(e) The range of a-particle in air depends on
radioactive substance.
(f) The ionisation power of a-particle is higher than
both b (100 times of b & 10,000 times of g) and g
particle.

734. 730 PHYSICS
Rutherford and S
oddy law for radioactive decay
It states that “at any time the rate at which particular
decay occurs in a radioactive substance is proportional
to number of radioactive nuclei present.
If N is the number of nuclei at any time t & at t + dt time, it
decrease to N-dN then the rate of decay of these nuclei is
dt
dN-
(negative si
gn comes because N decreases as t
increases). So according to Rutherford & Sodi,
-
µ
dN
N
dt
or =l
dN
N
dt
...(1)
Where l is decay con
stant (i.e., probabilily per unit time for
a nucleus to decay) and it is constant for a particular nuclei,but different for different nuclei. By integration of equation(1) w.r. to time we get
N = N
o
e
–lt
...(2)
where N
o
is the number of nuclei at t = 0.
Activity : The number of decays per unit time or decay rate
is called activity(R)
t
o
t
o eReN
dt
dN
]R[
l-l-
=l=
ú
û
ù
ê
ë
é
= where R
o
= N
o
l.
The S.I. unit of
R is bequerel,
1 bequerel = 1 Bq = 1 decay/sec
and 1 curie = 1Ci

= 3.7 × 10
10
decay/sec
The other unit of radioactivity is rutherford.
1 rutherford = 10
6
dacay/sec
Þ
N
N
log
303.2
t
0
l
=
00
Nm
and
Nm
éù
=
êú
ëû
m
0
= mass at t =0 and m = mass at t = t
COMMON DEFAULT
OIncorrect. Since b-particles (electrons) are emitted from the
nucleus shows that electrons exist in nucleus
PCorrect.b-particle cannot exists in nucleus. It is created
and ejected at once at the time of b-decay. b-particle cannot
exist in the nucleus because its wavelength is greater than
the size of nucleus
Half Life of a Radioactive Substance
Half life of a substance is the time, it takes for half of a given
number of radioactive nuclei to dacay
Let at = T
1/2

2
N
N
o
= then by eq. (2)
1
T
o 2
o
N
Ne
2
-l
= or
1/2
log20.693
e
T==
ll
Also
n
0
2
N
N= for n half-li
ves
Þ
n
0
2
m
m= where m
0
is mass of
radioacive substance at t = 0
and m is mass at t = t.
(g) The penetrating power of a particle is lowest (in
comparison to b & g particles). It is 1/100 times of
b-particles & 1/10,000 times of g-rays.
(h) The a-particles can produce fluorescence in
barium platinocynide and zinc sulphide.
(i) They show little effect on photographic plate.
(j) They show heating effect on stopping.
(B)Properties of b rays or b-particles :
(a) The beta particles (i.e., b
–
or b
+
) may be positive
& negative particle & contain C106.1
19-
´± of
charge. Actually b
–
is electron & b
+
is positron.
(b) They get deflected in both electric & magnetic
field.
(c) The velocity of b-particle varies between 0.01c
to .99c, where c is velocity of light.
(d) The mass of b particle is relativistic, because its
velocity is comparable to velocity of light
(e) They have both ionisation & penetration power.
Ionisation power less than a-particle and
penetration power more than a-particle.
(f) They produce fluorescence on barium
platinocynide & zinc sulphide.
(C)Properties of g-rays (or gamma radiation):
(a) They are electromagnetic waves as x-rays.
(b) They are not deflected in electric & magnetic field,
it means that they are chargeless.
(c) The velocity of g-particle is equal to velocity of
light.
(d) The ionisation power of gamma rays is less than
b & a rays but penetration power more than b
and a-rays.
(e) The g-particles are emitted from the nucleus, while
X-rays are obtained, when electron goes from
one state to another in an atom.
(f) When g-rays photon strikes nucleus in a
substance, then it gives rise to a phenomenon of
pair production i.e.,
h
-+
g-
n ¾¾®b+b
( rays or photon)
(Pair production)
The minimum energy of g-rays required for this
phenomena is 1.02 MeV, because the rest mass
energy of
±
b particle is 0 .51 MeV..
.
»
a -raysb -rays g -raysProperty Mass
Charge
Velo
city
Penetrating
power
Ionisation
power
Spectrum Line LineCon
tinuous
1
1000100
100
1
10,000
Zerom
e
c
e+2e
4m
p
Neutral
0.33 to
0.99c
2 × 10 m/s
2

735. 731Nuclei
N
t=o
t
N
0
N
0
N
0
N
0
/8
/4
/2
N
u
m
b
e
r
o
f

r
e
m
a
in
in
g

n
u
c
le
i
time
Mean Life of a Radioactive Substance:
Mean life (average life)
t is defined as the averge time the nucleus
survive before it decays.
It is given as :
1/2
T1
0.693
t==
l
The equ
ivalent
l and t for two nuclei A and B,
and
AB
AB
AB
tt
l=l +l t=
t +t
RADIOACTIVE SERIES
The heavy nuclides change their mass number by a decay and
atomic number by a and b decay.
They can decay to stable end products by four paths. The four
paths have mass numbers given as 4n, 4n + 1, 4n + 2, 4n + 3 where
n is integer.
Last element of series is stable and has a decay constant zero.
There are four radioactive series :
Uranium
92
U
238
®
82
Pb
206
(Half life years1047.4T
9
2/1
´= )
Actinium (natural)
92
U
235
®
82
Pb
207
(Half life years1004.7T
8
2/1
´= )
Thorium
92
Th
232
®
82
Pb
208
(Half life years1041.1T
10
2/1
´= )
Neptunium
93
Nb
237
®
83
Bi
209
(Half life years1014.2T
6
2/1
´= )
But only first three series occur in nature & fourth one is artificial.
Series Mass St artingStable end Natural /
numberis otopeproductArtificial
Thorium 4n
90
Th
232
82
Pb
208
Natural
Neptunium4n+1
93
Np
237
83
Bi
209
Artificial
Uranium 4n+2
92
U
238
82
Pb
206
Natural
Actinium4n+3
92
U
235
82
Pb
2
07
Natural
Radioactive EquilibriumWhen the rate of formation of daughter nuclei becomes equal torate of its decay then this is called as state of radioactiveequilibrium
N
A
l
A
= N
B
l
B
= ............or
AB
AB
NN
TT
= = ............
Carbon Dating
Carbon dating is the process of determination of time interval
which has passed by making use of radioactive decay of a sample
containing radioactive substance (
6
C
14
). It helps in calculating
age of geological specimens like rocks, biological specimens likes
bones of animals or trunk of trees and age of earth. The isotope
of carbon
6
C
14
is radioactive. It is formed in atmosphere by
bombardment of nitrogen atoms with cosmic rays

7
N
14
+
0
n
1
®
6
C
14
+
1
H
1
The
6
C
14
combines with oxygen to form carbondioxide which is
absorbed by plants so concentration of
6
C
14
is constant with
time. The living plants and animals have a fixed ratio of
6
C
14
to
ordinary carbon
6
C
12
. When a plant or animal dies the content of
6
C
14
decreases while that of
6
C
12
remains constant. The ratio of
two indicates the time that has passed since death of plant or
animal. The time interval is calculated from the laws of radioac-
tive disintegration
00
10
1 2.303
log log
e
NN
t
NN
==
ll

00NA
NA
æö
=
ç÷
èø
where N
o
is number of
6
C
14
nuclei at time of death, l is decay
constant of
6
C
14
and N is number of
6
C
14
nuclei currently present
in sample.
Keep in Memory
1.Spe
cific activity is the activity of 1 gram of material.
2. Geiger Muller Counter is used for detecting a and b
particles.
3. Cloud chamber is used for detecting radioactive radiations
and for determining their paths, range and energy.
4. Baryon number B = 1, for a neutron and a proton.
5. Lepton number (L) :L = 1 for electron, and neutrino
L = –1 for positron and antineutrino.
6. Radioactive isotopeUses
Iodine-131 For detecting the activity of
thyroid gland
Chromium-51 To locate the exact position of
haemorrhage
Phosphorous-32 In agriculture
C – 14 Carbon dating,
Photosynthesis in plants
Co
60
Cancer treatment
Na
24
For circu lation of blood
NUCLEAR REACTION
Nuclear reaction obeys following conservation laws :
(1)Charge conservation
(2)Conservation of linear momentum
(3)Conservation of angular momentum
(4)Conservation of energy (Rest mass energy + K.E.)
Standard way of representing a nuclear reaction
For a nuclear reaction
1
A 4 A3
1z 2 z1
XHe HY
+
+
+ ¾¾®+
Standar
d way
()
A A3
z z1
X,pY
+
+
a

736. 732 PHYSICS
Nuclear Fis
sion (By Otto Hans and Fstrassmann)
Nuclear fission is the disintegration of a heavy nucleus upon
bombardment by a projectile, such that the heavy nucleus splits
up into two or more segments of comparable masses with an
enormous release of energy.
()
235 1 141 92 1
92
0 53 36 0
U Ba Kr 3 200 MeVnn+®+++
The most of t
he energy released is by the mode of kinetic energy
of fission segment.
Uncontrolled Chain Reaction : It is the principle of atom bomb
(destructive use). The number of fission in this case goes on
increasing at a tremendous rate leading to the creation of a huge
amount of energy in a very small time.
The number of fissions in this case is maintained constant. Nuclear
reactor has beed devised for this purpose.
The main parts of nuclear reactor are
(a)Nuclear fuel : U
233
, U
235
, Pu
239
etc.
(b)Moderator : Graphite, heavy water (D
2
O). To slow down
the neutrons (or slow down the nuclear reaction).
(c)Control rods : (Cadmium, boron). To absorb excess
neutrons. It controls the chain reaction.
(d)Coolant : (water etc). To remove the heat produced in the
core to heat exchanger for production of electricity.
The reaction of controlled chain reactor is also called
critical reaction.
Critical mass : It is the minimum amount of fissionable material
required to carry out fission reaction. It is 10 kg for U
236
Reproduction factor K Rate of neutron production
Rate of neutron absorption
=
k = 1 for controlled r
eaction
1k³ for uncontrolled reaction
Breeder reactor :It converts U
238
non-fissionable to a fissionable
material
239
Pu or U
235.
Nuclear Fusion
:
Nuclear fusion is the fusion of two or more light nuclei to form aheavy nucleus with a release of huge amount of energy.
For a nuclear fusion to take place, very high temperature is
required to overcome the coulombic repulsive forces acting
between the nuclei. It is the principle of hydrogen bomb.
The nuclear fusion reaction, which is the source of the energy of
sun/ star are proton-proton cycle.
2
110
1111
H H H e 0.42MeV+ ¾¾® + +n+
MeV5
.5HeHH
3
2
1
1
2
1 +g+¾®¾+
MeV8.12HHHeHeHe
1
1
1
1
4
2
3
2
3
2 +++¾®¾+
MeV6.2422e2HeH4
0
1
4
2
1
1 +g+n++¾®¾
Stars with mass 0.4 to 2.5 solar mass produce energy by carbon-
nitrogen cycle. Stars with lower mas produce energy by proton-
proton cycle.
Nuclear holocaust : It is the name given to large scale destruction
which will be created upon the use of piled up nuclear weapons.
It is believed that if the existing nuclear weapons are used, then
the radioactive waste will hang like a cloud in the earth's
atmosphere. This clould will be capable to absorb solar radiation
due to which these radiation will not reach earth. This would
result to a long nuclear winter.
Radiation hazards : The g-radiations are highly energetic and
causes pathological and genetic damage.
Example 1.
The radius of a nucleus with nucleon number 16 is 3 × 10
–15
m.
What would be the radius of other nucleus with nucleon
number 128?
Solution :
We know that
3/1
ARµ
\
3/1
1
2
1
2
A
A
R
R
÷
÷
ø
ö
ç
ç
è
æ
= or
3/1
1
2
12
A
A
RR
÷
÷
ø
ö
ç
ç
è
æ
=
so, R
2
= 3 × 10
–15
× (128/1
6)
1/3
= (3 × 10
–15
) × 2 = 6 × 10
–15
m
Example 2.
The activity of a radioactive element decreased to one
third of the original activity
0
Ιin a period of nine years.
What will be i
ts activity after a further lapse of nine years?
Solution : T/t
0
2
1
÷
ø
ö
ç
è
æ
=
÷
÷
ø
ö
ç
ç
è
æ
I
I
or
T/9
2
1
3
1
÷
ø
ö
ç
è
æ
=
T/18
0
2
1
÷
ø
ö
ç
è
æ
=
÷
÷
ø
ö
ç
ç
è
æ
I
I¢
=
2
2
T/9
3
1
2
1
÷
ø
ö
ç
è
æ
=
ú
ú
û
ù
ê
ê
ë
é
÷
ø
ö
ç
è
æ
\ 9/
0
ΙΙ=¢
Example 3.
At any instant, t
he ratio of the amount of radioactive
substances is 2 : 1. If their half lives be respectively 12 and
16 hours, then after two days, what will be the ratio of
amount of the substances?
Solution :
For first substance
4
12
48
n ==
\ 16/N
2
1
NN
01
4
011 =÷
ø
ö
ç
è
æ
= ...(1)
For second substan
ce,
3
16
48
n ==
8/N
2
1
NN
02
3
022 =÷
ø
ö
ç
è
æ
= ...(2)
Now, 1
2
1
2
2
1
N
N
N
N
02
01
2
1
=´=´=(Q N
01
= 2 N
02
)
\ N
1
: N
2
= 1 : 1

737. 733Nuclei
Example4.
The ha
lf life of radium is 1620 years and its atomic weight
is 226. Find the number of atoms that will decay from its
1 gm sample per second.
Solution :
According to Avogadro’s hypothesis,
21
23
0
1066.2
226
1002.6
N ´=
´
=
Half life = years1620
6931.0
T =
l
=
\
111
7
s1035.1
1016.31620
6931.0 --
´=
´´
=l
Becaus
e half life is very much large as compared to its time
interval, hence N
» N
0
dtNdNorNN
dt
dN
00
l=l=l=
)1()1066.2()1035.1(dN
2111
´´=\
-

10
1061.3´=.
Example 5.
A nuc
leus breaks into two parts whose velocity is in ratio
of 2:1. Find the ratio of their radius.
Solution :
As per conservation of momentum m
1
v
1
+ m
2
v
2
= 0
so
12
21
mv
mv
=
Ratio of radi
i
1/3 1/3 1/3
111
222
RAm1
R
Am2
æö æö
æö
===
ç÷ç÷ ç÷
èøèø èø
\ R
1
: R
2
= 1
: 2
1/3
Example 6.
Calculate the binding energy per nucleon for
17
C
35
. Given
M (Cl
35
) = 34.9800 amu. m
n
= 1.008665 amu and
m
P
= 1.007825 amu.
Solution :
BE = Zm
P
+ (A – Z) m
n
– M (Cl
35
) = 17 × 1.007825 + 18 ×
1.008665 – 34.9800 = 0.308995 amu
BE = 0.308995 × 931.5 = 287.83 MeVBE 287.75
B 8.22
A 35
=== MeV
Exa
mple 7.
A star initially has 10
40
deutrons. It produces energy by
processes
1
H
2
+
1
H
2
¾®¾
1
H
3
+ p
and
1
H
2
+
1
H
3
¾®¾
2
He
4
+ n
If average power radiated by star is 10
16
W then find time
in which deutron is exhausted.M(
1
H
2
) = 2.01471 amu M(
2
He
4
) = 4.00388 amu
m
P
= 1.00783 amu and m
n
= 1.00866 amu
Solution :
Adding the two equation 3
1
H
2
¾®¾
2
He
4
+ p + n
mass defect = 3 × 2.01471 – 4.00388 – 1.00783 – 1.00866
= 0.02376 amu = 0.02376 × 931.5 MeV= 22.13 MeV
Power of star = 10
16
W = 10
16
J/s
Number of deutrons used per second
= 16
27
196
10
2.82 10
1.6 10 22.13 10
-
=´
´´´
Time in which deutrons will be used
=
Number of deutrons
Number of deutrons used per second
=
40
12
27
10
3.5 10
2.82 10
=´
´
sec
Exam
ple 8.
One gram of Radium emits 3.7 × 10
10

a particles per sec-
ond. Calculate half life and mean life of Radium. Given:
Atomic mass of Radium = 226.
Solution :
Rate of decay of Radium = rate of emission of a particles
or
dN
dt
-
= lN = 3.7 × 10
10
per s
econd
Number of active atoms
23
6.023 101
N
226
´´
=
\lN =
23
100.693 6.023 10
3.7 10
T 226
´
´ =´
or h
alf life T = 1583 years
Mean life t = 1.44 T = 1.44 × 1580 = 2279 years
Example 9.
Determine the disintegration energy of the process
210 206 4
828402
P Pb HeQ¾¾® ++ if an a particle of en-
ergy 5.3 MeV is emitted in it.
Solution :
Kinetic energy of a particle
A4
EQ
A
a
-
=
so Q =
A 210
E 5.3 5.40 MeV
A 4 210 4
a= ´=
--
Example 10.
The activit
y of a radioactive substance drops to 1/32 of its
initial value in 7.5 h. Find the half life.
Solution :
Using
t/T
0
A1
A2
æö
=
ç÷
èø
or
7.5/T
11
322
æö
=
ç÷
èø
or
5 7.5/T
11
22
æö æö
=
ç÷ ç÷
èø èø
or
7.5
5
T
= i.e. T = 1.5 hour

738. 734 PHYSICS
CONCEPT MAP
Isotopes
Nuclides with same
Z but different A or N
Isobars
Nuclides with same
A or N but different Z
Isotones
Nuclides with same
number of neutrons (N)
Nuclide
A
Z
X
Atomic mass A = Protons + neutrons
Atomic no. Z = no. of protons
Mass defect
(m) = M – m=[Zm+ (A – Z)m – m] D
pn N
Binding energy
E=(m)c
b D
2
Packing fraction
Exact nuclear mass – Mass number
P
Mass number
=
Atomic masses and
composition of Nucleus
NUCLEI
Nucleus consists of
protons and neutrons
Nuclear Force
Acting inside the
nucleus or acting
between the nucleons
due to continuous exchange
of meson particles
Nuclear energy
Radioactivity
Disintegration of heavy elem
e
n
ts
into comparatively lighte
r
elements by emission of , and

radiation
ab
g
Nuclear Fission Splitting of a heavy
nucleus into two or more lighter
nuclei
235 1 141 92
92 0 5636
1
0
U nBaKr
3 n energy
+ ® +
+ +
Nuclear fusion Combining two
lighter nuclei to form one heavy
nucleus
2 2 2
1 1 1
4 1 1
2 1 0
H H H
He Hn 21.6Mev
+ + ®
+ + +
Rate of decay law
dN
d
N
N
–
N
dt
d
t
-æ ö
µ
Þ
=
l
ç ÷
è ø
No. of u
n
d
e
c
a
y
e
d
a
t
o
m
s
at any in
s
t
a
n
t
N
=
N
eo
–
tl
H
a
l
f
l
i
f
e
1
/
2
0
.6
9
3
t
=
l
M
e
a
n

li
f
e
1
t
=
l
A
c
t
i
v
i
t
y

o
f
r
a
d
i
o
a
c
t
i
v
e

e
l
e
m
e
n
t
d
N
R
d
t
æ
ö
=
-
ç
÷
è
ø
A
c
t
i
v
i
t
y

a
f
t
e
r
t
i
m
e
,
R
=
R
e0
–
tl
A
t
o
m
i
c

m
a
s
s
u
n
i
t
(
a
m
u
)
1

a
m
u

=
1
th of mass
o
f
1
2
C
a
to
m
12
1 u = 1.660539 × 10
k
g
1amu = 931 Mev
–
2
7
Composition of Nucleus
Size of nucleus R = RA
0
1/3
R = 1.1 × 10m
0
– 15
1
7
3
30
3
m
D
e
n
s
i
t
y
2
.
3
1
0
k
g
/
m
4
R
r
=
=
´
p
Mass of
p
r
o
t
o
n
m
=

1
.0
0
7
2
7
u

=

1
.6
7
×
1
0
p
–
2
7
k
g
Mass of n
e
u
tr
o
n

m
=
1
.
0
0
8
6
6
u
=
1
.
6
7
×
1
0
k
g
n
–
2
7
Mass of e
le
c
tr
o
n

m
=
0
.0
0
0
5
5
u
=
9
.1
×

1
0
k
g
e
–
3
1
a
a
- decay
i
.
e
.
,
d
o
u
b
ly

i
o
n
is
e
d
helium io
n
A
f
te
r

e
m
i
s
s
i
o
n
o
f
one -part
i
c
l
e
a
to
m
i
c
n
o
.
d
e
c
r
e
a
s
e
s
by 2 and
m
a
s
s
n
u
m
b
e
r

b
y
4
b
b
-
d
e
c
a
y
i
.e
.,
f
a
s
t
m
o
v
i
n
g
e
l
e
c
t
r
o
n
s
A
f
t
e
r
e
m
is
s
i
o
n
o
f
o
n
e
-
p
a
r
t
i
c
l
e
a
to
m
i
c

n
u
m
b
e
r
i
n
c
r
e
a
s
e
s

b
y
1
a
n
d
m
a
s
s

n
u
m
b
e
r
r
e
m
a
i
n
s
u
n
c
h
a
n
g
e
d
g
-
d
e
c
a
y
A
f
t
e
r
e
m
i
s
s
i
o
n
t
h
e
r
e
i
s

n
o
c
h
a
n
g
e
i
n
a
t
o
m
i
c
n
u
m
b
e
r
a
n
d
m
a
s
s

n
u
m
b
e
r
Moderator Slow down
fast moving neutrons
e.g.: heavy water, graphite
Coolant Remove heat
eg: cold water, liquid oxygen
Control rods Absorb
neutrons eg., boron,
cadmium etc.

739. 735Nuclei
1.In gamma ray
emission from a nucleus
(a) only the proton number changes
(b) both the neutron number and the proton number
change
(c) there is no change in the proton number and the
neutron number
(d) only the neutron number changes
2.Fusion reaction occurs at temperatures of the order of
(a) 10
3
K (b) 10
7
K(c) 10 K (d) 10
4
K
3.An element A decays into an element C by a two step process
4
A B 2He®+ and B C 2e
-
®+ . Then,
(a) A
and C are isotopes (b) A and C are isobars
(c) B and C are isotopes (d) A and B are isobars
4.Nuclear forces are(a) spin dependent and have no non-central part(b) spin dependent and have a non-central part(c) spin independent and have no non-central part(d) spin independent and have a non-central part
5.A radioactive substance has a half life of four months. Threefourth of the substance will decay in(a) three months (b) four months
(c) eight months (d) twelve months
6.In which sequence the radioactive radiations are emitted inthe following nuclear reaction?
Z
X
A

¾ ¾®
Z + 1
Y
A
¾ ¾®
Z–1
K
A–4

¾ ¾®
Z – 1

K
A–4
(a)g, a, b (b)a, b, g
(
c)b, g, a (d)b, a, g
7.Radioactive substance emits
(a)a-rays (b)b-rays
(c)g-rays (d) All of the above
8.The curve of binding energy per nucleon as a function of
atomic mass number has a sharp peak for helium nucleus.
This implies that helium
(a) can easily be broken up
(b) is very stable
(c) can be used as fissionable meterial
(d) is radioactive
9.If the end A of a wire is irradiated with
a-rays and the other
end B is irradiated with b-rays. Then
(a) a current will flow from A to B(b) a current will flow from B to A(c) there will be no current in the wire(d) a current will flow from each end to the mid-point
of the wire
10.The electrons cannot exist inside the nucleus because
(a) de-Broglie wavelength associated with electron
in
b-decay is much less than the size of nucleus
(b) de-Broglie wavelength associated with electron
in b-decay is much greater than the size of
nucleus
(c) de-Broglie wavelength associated with electron
in b-decay is equal to the size of nucleus
(d) negative charge cannot exist in the nucleus
11.For a nuclear fusion process, suitable nuclei are
(a) any Nuclei
(b) heavy Nuclei
(c) light Nuclei
(d) nuclei lying in the middle of periodic table
12.A nucleus
m
n
X emits one a-par ticle and two b-particles.
The resulting nucleus is
(a)
m6
n4
Z
-
-
(b)
m6
n
Z
-
(c)
m4
n
X
-
(d)
m4
n2
Y
-
-
13.Which of
the following is best nuclear fuel
(a) thorium 236 (b) plutonium 239
(c) uranium 236 (d) neptunium 239
14.In the process of fission, the binding energy per nucleon
(a) increases
(b) decreases
(c) remains unchanged
(d) increases for mass number A < 56 nuclei but
decreases for mass number A > 56 nuclei
15One curie is equal to
(a) 3.7 × 10
10
disintegration/sec
(b) 3.2 × 10
8
disintegration/sec
(c) 2.8 × 10
10
disintegration/sec
(d) None of these
16.The volume of a nucleus is directly proportional to
(a)A (b) A
3
(c)
A (d) A
1/3
17.An ele
ctron is
(a) hadron (b) baryon
(c) a nucleon (d) a lepton.
18.Which of the following has the mass closest in value to
that of the positron? (1 a.m.u. = 931 MeV)
(a) Proton (b) Electron
(c) Photon (d) Neutrino
19.Nucleus of an atom whose atomic mass is 24 consists of
(a) 11 electrons, 11 protons and 13 neutrons
(b) 11 electrons, 13 protons and 11 neutrons
(c) 11 protons and 13 neutrons
(d) 11 protons and 13 electrons
20.Outside a nucleus
(a) neutron is stable
(b) proton and neutron both are stable
(c) neutron is unstable
(d) neither neutron nor proton is stable

740. 736 PHYSICS
21.Atomi
c number of a nucleus is Z and atomic mass is M. The
number of neutron is
(a) M – Z (b) M (c)Z (d) M + Z
22.Which of the following nuclear reactions is not possible?
(a)
12 12 20 4
66 102
C C Ne He+¾¾®+
(b)
9 1 64
4 1 32Be H Li He+¾¾®+
(c)
111 94
51
42
Be H Be He+¾¾®+
(d)
7 4 1 10
32 14
LiHe HB+ ¾¾®+
23.The mass d
efect per nucleon is called
1.A gamma ray photon creates an electron-positron pair. If
the rest mass energy of an electron is 0.5 MeV and the total
kinetic energy of the electron-positron pair is 0.78 MeV,
then the energy of the gamma ray photon must be
(a) 0.78 MeV (b) 1.78 MeV
(c) 1.28 MeV (d) 0.28 MeV
2.Actinium 231,
231
AC
89
, emit in succession two b particles,
four a-particles, one b and one a plus several g rays. What
is the resultant isotope?
(a)
221
Au
79
(b)
211
Au
79
(c)
221
Pb
82
(d)
211
Pb
82
3.If 1 mg of U
235
is completely annihilated, the energy liberated
is
(a) 9 × 10
10
J (b) 9 × 10
19
J
(c) 9 × 10
18
J (d) 9 × 10
17
J
4.If u denotes 1 atomic mass unit. One atom of an element
has mass exactly equal to Au, where A is mass number of
element.
(a) A = 1
(b) A = 12
(c) A = 16
(d) A can take up any integral value from 1 to 110
5.A radioactive substance contains 10000 nuclei and its half-
life period is 20 days. The number of nuclei present at the
end of 10 days is
(a) 7070 (b) 9000
(c) 8000 (d) 7500
6.If Avogadro number is 6 × 10
23
, then number of protons,
neutrons and electrons in 14 gms of
6
C
14
are respecitvely
(a) 36 × 10
23
, 48 × 10
23
, 36 × 10
23
(b) 36 × 10
23
, 36 × 10
23
, 36 × 10
23
(c) 48 × 10
23
, 36 × 10
23
, 48 × 10
23
(d) 48 × 10
23
, 48 × 10
23
, 36 × 10
23
7.If the distance between nuclei is 2 × 10
–13
cm, the density of
nuclear material is
(a) 3.21 × 10
–12
kg/m
3
(b) 1.6 × 10
–3
kg/m
3
(c) 2 × 10
9
kg/m
3
(d) 1 × 10
17
kg/m
3
8.The half-life of radioactive Radon is 3.8 days. The time at
the end of which (1/20)th of the Radon sample will remain
undecayed is (given log
10
e = 0.4343)
(a) 13.8 days (b) 16.5 days
(c) 33 days (d) 76 days
9.A freshly prepared radioactive source of half life 2 hr emits
radiation of intensity which is 64 times the permissible safe
level. The minimum time after which it would be possible to
work safely with this source is
(a) 6 hr (b) 12 hr
(c) 24 hr (d) 128 hr
10.The intensity of gamma radiation from a given source is I
0
.
On passing through 37.5 mm of lead it is reduced to I
0
/8.
The thickness of lead which will reduce it to I
0
/2 is
(a) (37.5)
1/3
mm (b) (37.5)
1/4
mm
(c) 37.5/3 mm (d) (37.5/4) mm
11.In the uranium radioactive series, the initial nucleus is
92
U
238
and that the final nucleus is
82
Pb
206
. When uranium nucleus
decays to lead, the number of a particles and b particles
emitted are
(a)8a, 6 b (b) 6a, 7b
(c)6a, 8b (d) 4a, 3b
12.Radioactive element decays to form a stable nuclide, then
the rate of decay of reactant is
(a)
t
N
(b)
N
t
(c)
N
t
(d)
t
N
(a) binding energy (b) packing fraction
(c) ionisation energy(d) excitation energy
24.Half life of radioactive element depends upon
(a) amount of element present
(b) temperature
(c) pressure
(d) nature of element
25.Fusion reaction takes place at high temperature because
(a) nuclei break up at high temperature
(b) atoms get ionised at high temperature
(c) kinetic energy is high enough to overcome the
coulomb repulsion between nuclei
(d) molecules break up at high temperature

741. 737Nuclei
13.A ra
dioactive nucleus undergoes
a-emission to form a
stable element. What will be the recoil velocity of the
daughter nucleus if v is the velocity of a emission?
(a)
4A
v4
-
(b)
4A
v2
-
(c )
4A
v4
+
(d)
4A
v2
+
14.If an el
ectron and positron annihilate, then the energy
released is
(a) J102.3
13-
´ (b) J106.1
13-
´
(c) J108.4
13-
´ (d) J104.6
13-
´
15.Radium
226
Ra, spontaneously decays to radon with the
emission of an a-particle and a g ray. If the speed of the a
particle upon emission from an initially stationary radium
nucleus is 1.5 ×10
7
m/s, what is the recoil speed of the
resultant radon nucleus? Assume the momentum of g ray is
negligible compared to that of a particle.
(a) 2..0 × 10
5
m/s (b) 2.7 × 10
5
m/s
(c) 3.5 × 10
5
m/s (d) 1.5 × 10
7
m/s
16.The energy released per fission of a
92
U
235
nucleus is nearly
(a) 200 eV (b) 20 eV
(c) 200 MeV (d) 2000 eV
17.At time t = 0, N
1
nuclei of decay constant
1
l and N
2
nuclei
of decay constant
2
l are mixed. The decay rate of mixture
is
(a)
t)(
21
21
eNN
l+l-
-
(b)
t)(
2
1 21
e
N
N l+l-
÷
÷
ø
ö
ç
ç
è
æ
-
(c) )eNe
N(
t
2
t
1
2111
l-l-
l+l-
(d)
t)(
2211
21
eNN
l+l-
ll-
18.A radioactive nuclide is produced at the constant rate of n
per second (say, by bombarding a target with neutrons).
The expected number N of nuclei in existence t seconds
after the number is N
0
is given by
(a)
t
0
eNN
l-
=
(b)
t
0eN
n
N
l-
+
l
=
(c)
t
0 e
n
N
n
N
l-
÷
ø
ö
ç
è
æ
l
-+
l
=
(d)
t
0 e
n
N
n
N
l-
÷
ø
ö
ç
è
æ
l
++
l
=
Where l is the dec
ay constant of the sample
19.One milligram of matter convert into energy will give
(a) 90 joule (b) 9 × 10
3
joule
(c) 9 × 10
5
joule (d) 9 × 10
10
joule
20.A radioactive element forms its own isotope after 3
consecutive disintegrations. The particles emitted are
(a) 3 b–particles
(b) 2 b–particles and 1 a–particle
(c) 3 b–particles and 1 a–particle
(d) 2 a–particles and 1 b–particle.
21.If the radius of a nucleus
256
X is 8 fermi, then the radius of
4
He nucleus will be
(a) 16 fermi (b) 2 fermi (c) 32 fermi (d) 4 fermi
22.Order of magnitude of density of uranium nucleus is
(m
p
= 1.67 × 10
–27
kg)
(a) 10
20
kg / m
3
(b) 10
17
kg / m
3
(c) 10
14
kg / m
3
(d) 10
11
kg / m
3
23.The binding energy per nucleon for
2
1
H and
4
2
He
respectively are 1.1 MeV and 7.1 MeV. The energy released
in MeV when two
2
1
H nuclei fuse to form
42
He is
(a) 4.4(b) 8.
2 (c) 24 (d) 28.4
24.Which one is correct about fission?
(a) Approx. 0.1% mass converts into energy
(b) Most of energy of fission is in the form of heat
(c) In a fission of U
235
about 200 eV energy is released
(d) On an average, one neutron is released per fission of
U
235
25.The average binding energy per nucleon is maximum for
the nucleus
(a)
2
He
4
(b)
8
O
16
(c)
26
Fe
56
(d)
92
U
238
26.The rest energy of an electron is
(a) 510 KeV (b) 931 KeV
(c) 510 MeV (d) 931 MeV
27.The mass and energy equivalent to 1 amu are respectively
(a) 1.67 × 10
–27
gm, 9.30 MeV
(b) 1.67 × 10
–27
kg, 930 MeV
(c) 1.67 × 10
–27
kg, 1 MeV
(d) 1.67 × 10
–34
kg, 1 MeV
28.From the following equations, pick out hte possible nuclear
reactions.
(a)
6
C
13
+
1
H
1
®
6
C
14
+ 4.3 MeV
(b)
6
C
12
+
1
H
1
®
7
N
13
+ 2 MeV
(c)
7
N
14
+
1
H
1
®
8
O
15
+ 7.3 MeV
(d)
92
U
235
+
0
n
1
®
54
X
140
+
38
Si
94
+
20
n
1
+ g + 200 MeV
29.Which of the following statements is true?
(a)
78
Pt
192
has 78 neutrons
(b)
84
Po
214
®
82
Pb
210
+ b
–
(c)
92
U
238
®
90
Th
234
+
2
He
4
(d)
90
Th
234
®
91
Pa
234
+
2
He
4
30.Consider a radioactive material of half-life 1.0 minute. If one
of the nuclei decays now, the next one will decay
(a) after 1 minute
(b) after
e
1
minute
log2
(c) a
fter
1
N
minute, where N is the number of nuclei
present at that moment
(d) after any time

742. 738 PHYSICS
31.If 200
MeV energy is released in the fission of a single U
235
nucleus, the number of fissions required per second to
produce 1 kilowatt power shall be (Given 1eV = 1.6 × 10
–19
J)
(a) 3.125 × 10
13
(b) 3.125 × 10
14
(c) 3.125 × 10
15
(d) 3.125 × 10
16
32.The fossil bone has a
14
C :
12
C ratio, which is
1
16
éù
êú
ëû
of that
in a living a
nimal bone. If the half-life of
14
C is 5730 years,
then the age of the fossil bone is
(a) 11460 years (b) 17190 years
(c) 22920 years (d) 45840 years
33.The ratio of half-life times of two elements A and B is
A
B
T
T
.
The ratio of r
espective decay constant
A
B
,
l
l
is
(a)T
B
/ T
A
(b) T
A
/ T
B
(c)
AB
A
TT
T
+
(d)
AB
A
TT
T
-
34.An archaeolo
gist analyses the wood in a prehistoric
structure and finds that C
14
(Half life = 5700 years) to C
12
is
only one-fourth of that found in the cells of buried plants.
The age of the wood is about
(a) 5700 years (b) 2850 years
(c) 11,400 years (d) 22,800 years
35.A radioactive nucleus undergoes a series of decay
according to the scheme
a
1234AAAAA
bag
¾¾® ¾¾® ¾¾® ¾¾®
If
the mass number and atomic number of ‘A’ are 180 and 72
respectively, then what are these numbers for A
4
(a) 172 and 69 (b) 174 and 70
(c) 176 and 69 (d) 176 and 70
36.1 g of hydrogen is converted into 0.993 g of helium in athermonuclear reaction. The energy released is
(a) 63 × 10
7
J (b) 63 × 10
10
J
(c) 63 × 10
14
J (d) 63 × 10
20
J
37.The mass defect in a particular nuclear reaction is 0.3 grams.
The amount of energy liberated in kilowatt hour is
(Velocity of light = 3 × 10
8
m/s)
(a) 1.5 × 10
6
(b) 2.5 × 10
6
(c) 3 × 10
6
(d) 7.5 × 10
6
38.The binding energies per nucleon for a deuteron and an
a-particle are x
1
and x
2
respectively. What will be the energy
Q released in the reaction
1
H
2
+
1
H
2
®
2
He
4
+ Q
(a) 4(x
1
+ x
2
) (b) 4(x
2
– x
1
)
(c) 2(x
1
+ x
2
) (d) 2(x
2
– x
1
)
39.Atomic weight of boron is 10.81 and it has two isotopes
5
B
10
and
5
B
11
. Then ratio of
5
B
10
:
5
B
11
in nature would
be
(a) 19 : 81 (b) 10 : 11
(c) 15 : 16 (d) 81 : 19
40.The masses of neutron and proton are 1.0087 a.m.u. and
1.0073 a.m.u. respectively. If the neutrons and protons
combine to form a helium nucleus (alpha particles) of mass
4.0015 a.m.u the binding energy fo the helium nucleus will
be (1 a.m.u. = 931 MeV)
(a) 28.4 MeV (b) 20.8 MeV
(c) 27.3 MeV (d) 14.2 MeV
41.The radioactivity of a sample is R
1
at a time T
1
and R
2
at a
time T
2
. If the half-life of the specimen is T, the number of
atoms that have disintegrated in the time (T
1
– T
2
) is
proportional to
(a) (R
1
T
1
– R
2
T
2
) (b) (R
1
– R
2
)
(c) (R
1
– R
2
)/T (d) (R
1
– R
2
) T
42.A radioactive element X converts into another stable
element Y. Half life of X is 2 hrs. Initially only X is present.
After time t, the ratio of atoms of X and Y is found to be 1 :
4, then t in hours is
(a)2 (b) 4
(c) between 4 and 6(d) 6
43.After 150 days, the activity of a radioactive sample is 5000
dps. The activity becomes 2500 dps after another 75 days.
The initial activity of the sample is
(a) 20000 dps (b) 40000 dps
(c) 7500 dps (d) 10000 dps
44.If the total binding energies of
2 4 56 235
1 2 26 92
H,He,Fe&U
nuclei are 2.22, 28.3, 492 and 1786 MeV respectively, identify
the most stable nucleus of the following.
(a)
56
26
Fe(b)
2
1
H (c)
235
92
U (d)
4
2
He
45.A neutron travelling with a velocity v and kinetic energy E
has a perfectly elastic head-on collision with a nucleus of
an atom of mass number A at rest. The fraction of total
energy retained by the neutron is approximately
(a) [(A – 1)/(A + 1)]
2
(b) [(A + 1)/(A – 1)]
2
(c) [(A – 1)/A]
2
(d) [(A + 1)/A]
2
46.At any instant, the ratio of the amount of radioactive
substances is 2 : 1. If their half lives be respectively 12 and
16 hours, then after two days, what will be the ratio of the
substances ?
(a) 1 : 1(b) 2 : 1(c) 1 : 2(d) 1 : 4
47.Half lives for a and b emission of a radioactive material are
16 years and 48 years respectively. When material decays
giving a and b emission simultaneously, time in which 3/
4
th
material decays is
(a) 29 years (b) 24 years
(c) 64 years (d) 12 years
48.In an a-decay the kinetic energy of a-particle is 48 MeV
and Q-value of the reaction is 50 MeV. The mass number of
the mother nucleus is X. Find value of X/25.
(Assume that daughter nucleus is in ground state)
(a)2 (b) 4 (c)6 (d) 8

743. 739Nuclei
49.A heav
y nucleus having mass number 200 gets
disintegrated into two small fragments of mass number 80
and 120. If binding energy per nucleon for parent atom is
6.5 MeV and for daughter nuclei is 7 MeV and 8 MeV
respectively, then the energy released in the decay is
X × 110 MeV. Find the value of X.
(a)3 (b) 4 (c)2 (d) 1
50.The activity of a radioactive sample is measured as 9750
counts per minute at t = 0 & 975 counts per minute at T = 5
minutes. The decay constant is approximately.
(a) 0.922 per minutes (b) 0.691 per minutes
(c) 0.461 per minutes (d) 0.230 per minutes
51.In a fission reaction
nnYXU
117117236
92
+++®
the binding energy per nucleon of X & Y is 8.5 MeV. Whereas
of
236
U is 7.6 MeV. The total energy liberated will be about
(a) 2000 MeV (b) 200 MeV
(c) 2 MwV (d) 200 KeV
52.Half lives of two radio active substance A & B are
respectively 20 minutes & 40 minutes. Initially the samples
of A & B have equal numbers of nulcei. After 80 minutes the
ratio of remaining numbers of A & B nuclei is
(a) 1 : 16 (b) 4 : 1
(c) 1 : 4 (d) 1 : 1
53.M
p
denotes the mass of a proton and M
n
that of a neutron.
A given nucleus, of binding energy B, contains Z protons
and N neutrons. The mass M(N, Z) of the nucleus is given
by (c is the velocity of light)
(a) M(N, Z) = NM
n
+ ZM
p
+ B/c
2
(b) M(N, Z) = NM
n
+ ZM
p
– Bc
2
(c) M(N, Z) = NM
n
+ ZM
p
+ Bc
2
(d) M(N, Z) = NM
n
+ ZM
p
– B/c
2
54.The mass of a
7
3
Li nucleus is 0.042 u less than the sum of
the masses of all its nucleons. The binding energy per
nucleon of
7
3
Li nucleus is nearly
(a) 46 MeV (b) 5.6 MeV
(c) 3.9 MeV (d) 23 MeV
55.The activity of a radioactive sample is measured as
N
0
counts per minute at t = 0 and N
0
/e counts per minute at
t = 5 minutes. The time (in minutes) at which the activity
reduces to half its value is
(a)
log 2/5
e
(b)
5
log2
e
(c) 5 log
10
2 (d) 5 log
e
2
56.The half life of a radioactive isotope 'X' is 50 years. It decays
to another element 'Y' which is stable. The two elements 'X'
and 'Y' were found to be in the ratio of 1 : 15 in a sample of
a given rock. The age of the rock was estimated to be
(a) 150 years (b) 200 years
(c) 250 years (d) 100 years
57.A radioactive nucleus of mass M emits a photon of
frequency n and the nucleus recoils. The recoil energy will
be
(a) Mc
2
– hn (b) h
2
n
2
/ 2Mc
2
(c) zero (d) hn
58.Two radioactive nuclei P and Q, in a given sample decay
into a stable nucleus R. At time t = 0, number of P species
are 4 N
0
and that of Q are N
0
. Half-life of P (for conversion
to R) is 1 minute where as that of Q is 2 minutes. Initially
there are no nuclei of R present in the sample. When number
of nuclei of P and Q are equal, the number of nuclei of R
present in the sample would be
(a) 3N
0
(b)
0
9N
2
(c)
0
5N
2
(d) 2N
0
59.The half life o
f a radioactive nucleus is 50 days. The time
interval (t
2
– t
1
) between the time t
2
when
2
3
of it has
decayed and the time t1

when
1
3
of it had decayed is
(a) 30 days (b) 50 days
(c) 60 days (d) 15 days
60.If the nuclear radius of
27
Al is 3.6 fermi, the approximate
nuclear radius of
64
Cu in fermi is
(a) 2.4 (b) 1.2
(c) 4.8 (d) 3.6
61.A mixture consists of two radioactive materials A
1
and A
2
with half lives of 20 s and 10 s respectively. Initially the
mixture has 40 g of A
1
and 160 g of A
2
. The amount of the
two in the mixture will become equal after
(a) 60 s (b) 80 s
(c) 20 s (d) 40 s
62.A certain mass of Hydrogen is changed to Helium by the
process of fusion. The mass defect in fusion reaction is
0.02866 a.m.u. The energy liberated per a.m.u. is
(Given : 1 a.m.u = 931 MeV)
(a) 26.7 MeV (b) 6.675 MeV
(c) 13.35 MeV (d) 2.67 MeV
63.The half life of a radioactive isotope ‘X’ is 20 years. It decays
to another element ‘Y’ which is stable. The two elements
‘X’ and ‘Y’ were found to be in the ratio of 1 : 7 in a sample
of a the given rock. The age of the rock is estimated to be
(a) 60 years (b) 80 years
(c) 100 years (d) 40 years
64.Fraction of tritium left after 150 years (half life of tritium is
12.5 years) is
(a) 1/ 1024 (b) 1/2048
(c) 1/4096 (d) 1/8192

744. 740 PHYSICS
65.Activity o
f a radioactive sample decreases to (1/3) rd of its
original value in 3 days. Then, in 9 days its activity will
become
(a) (1/27) of the orginal value
(b) (1/9) of the original value
(c) (1/18) of the original value
(d) (1/3) of the orignal value
66.A duetron strikes
16
8
Onucleus with subsequent emission
of alpha particle. Identify the nucleus so produced
(a)
7
3
Li (b)
10
5
B
(c)
13
7
N (d)
14
7
N
67.The more readily fissionable isotope of uranium has an
atomic mass of
(a) 234 (b) 235
(c) 236 (d) 238
68.A nuclear reaction is given by
n++®
-+
0
1
A
1Z
A
Z
eYX , represents
(a)
fission (b) b-decay
(c)
s-decay (d) fusion
69.Th
e half life of the radioactive substance is 40 days. The
substance will disintegrate completely in(a) 40 days (b) 400 days
(c) 4000 days (d) infinite time
70.The decay constant of radium is 4.28 × 10
–4
per year. Its
half life will be(a) 2000 years (b) 1240 years
(c) 63 years (d) 1620 years
Exemplar MCQs
1.Sup
pose we consider a large number of containers each
containing initially 10000 atoms of a radioactive material
with a half life of 1 yr. After 1 year,
(a) all the containers will have 5000 atoms of the material
(b) all the contains will conatin the same number of atoms
of the material but that number will only be
approximately 5000
(c) the containers will in general have different numbers
of the atoms of the material but their average will be
close to 5000
(d) none of the containers can have more than 5000 atoms.
2.The gravitational force between a H-atom and another
particle of mass m will be given by Newton's law
2
M.m
FG,
r
= where r is in km and
(a) M = m
proton
+ m
electron
(b) M = m
proton
+ m
electron
( )
2
B
B 13.6 eV
c
-=
(c) M is no
t related to the mass of the hydrogen atom
(d) M = m
proton
+ m
electron

(
2
V
V
c
-= magnitude of the potential energy of
electron in the H-atom.
3.When a nucleus in an atom undergoes a radioactive decay,the electronic energy levels of the atom(a) do not change for any type of radioactivity(b) change for a and b-radioactivity but not for g-
radioactivity
(c) change for a-radioactivity but not for others
(d) change for b-radioactivity but not for others
71.A radioactive sample with a half life of 1 month has the
label : ‘Activity = 2 micro curies on 1–8–1991. What would
be its activity two months earlier ?
(a) 1.0 micro curie(b) 0.5 micro curie
(c) 4 micro curie (d) 8 micro curie
72.The nucleus Cd
115
48
, after two successive b– decay will
give
(a)Pa
115
46
(b)In
114
49
(c)Sn
113
50
(d)Sn
115
50
DIRECTIONS (Qs.73 to 75) : Each of these question contains
two statements: Statement-1 (Assertion) and Statement-2
(Reason). Answer these questions from the following four
options.
(a) Statement-1 is True, Statement-2 is True; Statement-2 is a
correct explanation for Statement -1
(b) Statement-1 is True, Statement -2 is True; Statement-2 is
NOT a correct explanation for Statement - 1
(c) Statement-1 is True, Statement- 2 is False
(d) Statement-1 is False, Statement -2 is True
73. Statement- 1: The binding energy per nucleon, for nuclei
with atomic mass number A > 100, decrease with A.
Statement- 2 : The forces are weak for heavier nuclei.
74. Statement- 1 : Radioactivity of 10
8
undecayed radioactive
nuclei of half life of 50 days is equal to that of 1.2 × 10
8
number of undecayed nuclei of some other material with
half life of 60 days.
Statement- 2 : Radioactivity is proportional to half-life.
75. Statement- 1 : The ionising power of b-particle is less
compared to a-particles but their penetrating power is more.
Statement- 2 : The mass of b-particle is less than the mass
of a-particle.

745. 741Nuclei
4.M
x
and M
y
den
ote the atomic masses of the parent and the
daughter nuclei respectively in radioactive decay. The Q –
value for a b
–
decay is Q
1
and that for a b
+
decay is Q
2
. If m
e
denotes the mass of an electron, then which of the following
statements is correct?
(a)Q
1
= (M
x
– M
y
) c
2
and Q
2
= [M
x
– M
y
– 2m
e
] c
2
(b) Q
1
= (M
x
– M
y
) c
2
and Q
2
= (M
x
– M
y
) c
2
(c)Q
1
= (M
x
– M
y
– 2m
e
) c
2
and Q
2
= (M
x
– M
y
+ 2c
e
) c
2
(d) Q
1
= (M
x
– M
y
+ 2m
e
) c
2
and Q
2
= (M
x
– M
y
+ 2 m
e
)c
2
5.Tritium is an isotope of hydrogen whose nucleus triton
contains neutrons and 1 proton. Free neutrons decay into
p +
e + n. If one of the neutrons is Triton decays, it would
transform into He
3
nucleus. This does not happen. This is
because
(a) Triton energy is less than that of a He
3
nucleus
(b) The electron created in the beta decay process cannot
remain in the nucleus
(c) both the neutrons in triton have to decay simultaneously
resulting in a nucleus with 3 protons, which is not a
He
3
nucleus.
(d) free neutrons decay due to external perturbations
which is absent in triton nucleus
6.Heavy stable nuclei have more neutrons than protons. this
is because of the fact that
(a) neutrons are heavier than protons
(b) electrostatic force between protons are repulsive
(c) neutrons decay into protons through beta decay
(d) nuclear force between neutrons are weaker than that
between protons
7.In a nuclear reactor, moderators slow down the neutrons
which come out in a fission process. the moderator used
have light nuclei. Heavy nuclei will not serve the purpose,
because
(a) they will break up
(b) elastic collision of neutrons with heavy nuclei will
not slow them down
(c) the net weight of the reactor would be unbearably high
(d) substances with heavy nuclei do not occur in liquid
or gaseous state at room temperature
NEET/AIPMT (2013-2017) Questions
8.A certain mass of Hydrogen is changed to Helium by theprocess of fusion. The mass defect in fusion reaction is0.02866 a.m.u. The energy liberated per a.m.u. is [2013]
(Given : 1 a.m.u = 931 MeV)
(a) 26.7 MeV (b) 6.675 MeV
(c) 13.35 MeV (d) 2.67 MeV
9.The half life of a radioactive isotope ‘X’ is
20 years. It decays to another element ‘Y’ which is stable.
The two elements ‘X’ and ‘Y’ were found to be in the ratio
of 1 : 7 in a sample of a the given rock. The age of the rock
is estimated to be [2013]
(a) 60 years (b) 80 years
(c) 100 years (d) 40 years
10.a-particles, b-particles and g-rays are all having same energy.
Their penetrating power in a given medium in increasing
order will be [NEET Kar. 2013]
(a)b, g, a (b)g, a, b
(c)a, b, g (d)b, a, g
11.How does the binding energy per nucleon vary with the
increase in the number of nucleons? [NEET Kar. 2013]
(a) Increases continuously with mass number
(b) Decreases continuously with mass number
(c) First decreases and then increases with increase in
mass number
(d) First increases and then decreases with increase in
mass number
12.The Binding energy per nucleon of
7
3
Liand
4
2
Henuclei
ar
e 5.60 MeV and 7.06 MeV, respectively.
In the nuclear reaction
714
312
Li H HeQ+®+ , the value of
energy Q released is : [2014]
(a) 19.6 MeV (b) – 2.4 MeV
(c) 8.4 MeV (d) 17.3 MeV
13.A radio isotope ‘X’ with a half life 1.4 × 10
9
years decays to
‘Y’ which is stable. A sample of the rock from a cave was
found to contain ‘X’ and ‘Y’ in the ratio 1 : 7. The age of the
rock is : [2014]
(a) 1.96 × 10
9
years (b) 3.92 × 10
9
years
(c) 4.20 × 10
9
years (d) 8.40 × 10
9
years
14.If radius of the
27
12
Al nucleus is taken to be R
Al
, then the
radius of
125
53
Te nucleus is nearly: [2015]
(a) Al
5
R
3
(b) Al
3
R
5
(c)
1/3
Al
13
R
53
æö
ç÷
èø
(d)
1/3
Al
53
R
13
æö
ç÷
èø
15.A nucleus o
f uranium decays at rest into nuclei of thorium
and helium. Then : [2015 RS]
(a) the helium nucleus has less momentum than the
thorium nucleus.
(b) the helium nucleus has more momentum than the
thorium nucleus.
(c) the helium nucleus has less kinetic energy than the
thorium nucleus.
(d) the helium nucleus has more kinetic energy than the
thorium nucleus.
16.Radioactive material 'A' has decay constant '8 l' and material
'B' has decay constant 'l'. Initially they have same number
of nuclei. After what time, the ratio of number of nuclei of
material 'B' to that 'A' will be
1
e
? [2017]
(a)
1
7l
(b)
1
8l
(c)
1
9l
(d)
1
l

746. 742 PHYSICS
EXERCISE - 1
1. (c) 2. (b) 3
. (a) 4. (b)
2. (c) Substance left undecayed,
000
N
4
1
N
4
3
N =-
n
0 2
1
4
1
N
N
÷
ø
ö
ç
è
æ
==
\ Number of atoms l
eft undecayed,
n = 2 i.e. in two half lives
\ t = nT = 2 × 4 = 8 months
6. (d) 7. (d)8. (b)
9. (a) Due to irradiation of a-rays on end A will make it
(positive) and irradiation of b-rays on end B will make it
(negative) hence current will flow from A to B (or from
positive to negative).
10. (b)
11. (c)
12. (c) When
m
n
X emits one a-partic le then its atomic mass
decreases by 4 units and atomic number by 2. Therefore,
the new nucleus becomes
m4
n2
Y
-
-
. But as it emits
two b
–
particles, its atomic number increases by 2. Thus theresulting nucleus is
m4
n
X.
-
13. (b) Pluto
nium 239 is processed by breeder mechanism to
be used as nuclear feul.
14. (a) 15. (a)
16. (a) Radius of nucleus R = R
0
A
1/3
where A is the mass
number of nucleus.
\ Volume of nucleus
33
0
44
R RA
33
æö
=p=p
ç÷
èø
\ Volume is proportional to A.
17. (d) An electron is a lepton.
18. (b) Positron is the antiparticle of electron.
19. (c) Nucleus does not contain electron.
20. (c) Out side the nucleus, neutron is unstable (life » 932 s).
21. (a) N = M – Z = Total no. of nucleons – no. of protons.
22. (c) In this reaction mass is not conserved.
23. (b)
24. (d) Half life of a substance doesn’t depends upon amount,
temperature and pressure. It depends upon the nature
of the substance.
25. (c) When the coulomb repulsion between the nuclei is
overcome then nuclear fusion reaction takes place. This
is possible when temperature is too high.
EXERCISE - 2
1. (b) Energ
y of g-ray photon = Rest mass energy + K.E.
= 2 (0.5) + 0.78 = 1.78 MeV
2. (d)
3. (a) E = mc
2
= 10
–6
× (3 × 10
8
)
2
= 10
–6
× 9 × 10
16
= 9 × 10
10
J
4. (b)
5. (a)
t 10
T 20
0
N1 N1
or
N 2 10000 2
æö æö
==
ç÷ ç÷
èø èø
or
10000 1 10000
N 7070.
1.4142
´
= ==
6. (a) Each
atom of
6
C
14
contains 6 p, 6 e and 8 n
\ In 14 gram of
6
C
14
p = 6 × 6 × 10
23
= 36 × 10
23
n = 8 × 6 × 10
23
= 48 × 10
23
e = p = 36 × 10
23
7. (d) Density of nuclear material = mass/volume27 27
173
153
3
10 3 10
10 kg/m
4
4(210)
r
3
--
-
´
===
p´
p
8. (b)
20
1
N
N
2
1
0
n
==÷
ø
ö
ç
è
æ
gives n = 4.32
t = nT = 4.
3 × 3.8 = 16.5 days
9. (b) To work safely, intensity must reduce by 1/64
\
T/t
0 2
1
64
1
N
N
÷
ø
ö
ç
è
æ
== i.e.
T/t6
2
1
2
1
÷
ø
ö
ç
è
æ
=÷
ø
ö
ç
è
æ
or
6
T
t
= or t = 6 T = 1
2 hrs
10. (c) As
x
0
e
I
I m-
=
\
5.37
e
8
1 m-
= .......
(i) and
x
e
2
1 m-
= .......(ii)
Put (ii) in (i), e
–3 m x
= e
– m (37.5)
mm5.12
3
5.37
x ==
11. (a) Let no.
of a-particles emitted be x and no. of b particles
emitted be y.
Diff. in mass no. 4x = 238 – 206 = 32 Þ x = 8
Diff. in charge no. 2x – 1y = 92 – 82 = 10
16 – y = 10, y = 6
Hints & Solutions

747. 743Nuclei
12.
(c) No. of nuclide at time t is given by
t
o
eNN
l-
=
Where =
o
N initial nuclide
thus this equation is equivalent to
kx
y ae
-
=
thus cor
rect graph is
N
t
13. (a) We assume that mass number of nucleus when it was at
rest = A
Q mass number of a-particle = 4
\ mass number of remaining nucleus = A - 4
As there is no external force, so momentum of the system
will remain conserved
()v4v4A0+¢-=Þ
( )4A
v4
'v
-
-=Þ
negative sign rep
resents that direction is opposite to
the direction of motion of
a-particle.
14. (b) Energy of electron
2
e
cm=
Energy of positron
2
pcm=
pe
mm=, c = speed of light.
Thus according to conservation of energy released
2
e
cm2=
()
2
831
103101.92´´´=
- .Joules106.1
13-
´
=
15. (b) Conservation of linear momentum requires:
m
radon
v
radon
= m
helium
v
helium
with helium identified as the
alpha particle. The nuclear masses can be approximated
by their mass numbers (222 and 4). Thus, the recoil
speed of the radon is (4/222) × 1.5 × 10
7
m/s = 2.7 x 10
5
m/s.
16. (c) Energy released per fission is MeV200
»
17. (d)
18. (c
)
Nn
dt
dN
l-=
dt)Nn(dNl-=
òò
=
l-
t
0
N
N
dt
Nn
dN
0

t
Nn
dN1
N
N
0
=
l-
l-
l
-Þ
ò
[ ]t)Nn(log
1 N
Ne
0
=l-
l
-Þ
e
0
1 nN
logt
nN
éùæö-l
Þ-=êúç÷
l -lèøêúëû
ú
û
ù
ê
ë
é
÷
ø
ö
ç
è
æ
l-
l-
=lÞ
Nn
Nn
logt
0
e
Nn
Nn
e
0t
l-
l-
=
l
t
0
e)Nn(Nn
l-
l-=l-
NeN
nn t
0
=÷
ø
ö
ç
è
æ
-
l
-
l
l-
19. (d)
28
332
)103(1010mcE ´´==
--
= 9 × 10
10
J
20. (b) A nucleus is denoted by
Z
X
A
An isotope should have same Z.
a–particle =
2
He
4
; b–particle =
–1
b
0
The emission of one a particle and the emission of
two b particles maintain the Z same.
Hence, for isotope formation 2b particles and one a
particle are emitted.
21. (b)
1/3
0
R R (A)=
1/3 1/3
11
22
RA 256
4
RA4
æö æö
\=== ç÷ç÷
èøèø
1
2
R
R2
4
== fermi
22.
(b) The order of magnitude of mass and volume of uranium
nucleus will be
m ; A (1.67 × 10
–27
kg) (A is atomic number)
3 15 1/ 3344
V r [(1.25 10 m)A ]
33
-
=p p´;
; (8.2 × 10
–45
m
3
)A
H
ence,
27
453
m A(1.6710 kg)
V(8.2 10 m )A
-
-
´
r==
´
; 2.0 × 10
17
k
g / m
3
.
23. (c) The chemical reaction of process is HeH2
4
2
2
1
®
Energy released
4 (7.1) 4(1.1)=´- = 24 eV
24. (
a)
25. (c) Binding energy per nucleon increases with atomic
number. The greater the binding energy per nucleon
the more stable is the nucleus.
For
26
Fe
56
number of nucleons is 56.
This is most stable nucleus, since maximum energy is
needed to pull a nucleon away from it.

748. 744 PHYSICS
26. (a)
Rest energy of an electron = m
e
c
2
Here m
e
= 9.1 × 10
–31
kg and C = velocity of light
\ Rest energy = 9.1 × 10
–31
× (3 × 10
8
)
2
joule31 82
19
9.1 10 (3 10 )
eV 510 KeV
1.6 10
-
-
´ ´´
=»
´
27. (b) 28. (b
, c) 29. (c)
30. (d) Because radioactivity is a spontaneous phenomenon.
31. (a)
6 19
E n 200 10 1.6 10
P n
1000
tt
-
´ ´ ´´æö
= Þ=ç÷
èø
13n
3.125 10 .
t
Þ=´
32. (c)
14
12
C
C
=
0
1N
16N
=
Q
0
N
N
=
n
1
2
æö
ç÷
èø
Þ
1
16
=
n
1
2
æö
ç÷
èø
Þ
4
1
2
æö
ç÷
èø
=
n
1
2
æö
ç÷
èø
or, n = 4
or
t
T
= 4
or t = 4 × T =
4 × 5730 = 22920 years
33. (a)
1/2
1/2
ln2 ln2
T
T
= \l=
l
AB
AB
A B BA
TIn2 ln2
,.
TTT
l
Þl= l=Þ=
l
34. (c
)
t / 5700
14
12
C 11t
2 t 11400 years
C 4 2 5700
æö
= = Þ =
Þ=ç÷
èø
35. (a)
180 176 176
70 7172 12AAA
ab
¾ ¾® ¾ ¾®
172 172
69 6934
AA
g
a
¾¾® ¾¾®
36. (b)Dm = 1 –
0.993 = 0.007 gm
\ E = (Dm)c
2
= (0.007 ×10
–3
) (3× 10
8
)
2
= 63 × 10
10
J.
37. (d)
2 8 2 130.3
E m.c E (3 10 ) 2.7 10 J
1000
=D Þ= ´´ =´
13
6
6
2.7 10
7.5 10 kWh.
3.6 10
´
= =´
´
38. (b) Q = 4 (
x
2
– x
1
)
39. (a) Let the percentage of B
10
atoms be x, then average
atomic weight
10x 11(100 x)
10.81 x 19
100
+-
= = Þ=
10
B
11
B
N
19
N 81
\=
40. (a) B.E. = Dmc
2

= D × 931 MeV
= [2(1.0087 + 1.0073) – 4.0015] × 931 = 28.4 MeV
41.(d) Radioactivity at T
1
, R
1
= l N
1
Radioactivity at T
2
, R
2
= l N
2
\ Number of atoms decayed in time
(T
1
– T
2
) = (N
1
–N
2
)
=
693.0
T)RR()RR(
2121-
=
l
-
T)RR(
21
-µ
42. (c) Let N
0
be the
number of atoms of X at time t = 0.
Then at t = 4 hrs (two half lives)
0
x
N
N
4
= and
0
y
3N
N
4
=
\ N
x
/N
y
= 1/3
and at t = 6 hr
s (three half lives)
0
x
N
N
8
= and
0
y
7N
N
8
=
or
x
y
N 1
N7
=
The given ratio
1
4
lies between
1
3
and
1
7
.
Therefore, t l
ies between 4 hrs and 6 hrs.
43. (a) Activity of sample becomes 2500 from 5000 in 75 days
therefore its half life is 75 days, so
0
0
150
75
R
R 5000 R 5000 4 20, 000
2
= = Þ = ´=
44. (a) H
2.22
B.E 1.11
2
==
He
28.3
B.E 7.08
4
==
Fe
492
B.E 8.78
56
== = maximum
U
1786
B.E 7.6
235
==
56
26
Fe is most stable as it has maximum binding energy
per nucleon.

749. 745Nuclei
45. (a
)
1
v¢ =
1 2 1 22
12
( )2m m v mv
mm
-+
+
As v
2
is ze
ro, m
2
> m
1
, v '
1
is in the opposite direction.
m
1
= 1, m
2
= A.
\
11
( 1)
||
( 1)
A
vv
A
-
¢=
+
The frac
tion of total energy retained is2 2
1
22
1
1/2 ( 1)
1 / 2 ( 1)
¢ -
=
+
mv A
mvA
46. (
a) For substance A :
48/12
00
0A0
3
NN1
2N N 2N
28 2
æö
®= ==
ç÷
èø
For substa
nce B : 48/16
00
0 B0
3
NN1
N NN
28 2
æö
®= ==
ç÷ èø
N
A
: N
B
=
1 : 1
47. (b) Effective half life is calculated as
12
111
TTT
=+
111
T 12 years
T 16 48
= + Þ=
Tim
e in which
3
4
will decay is 2 half lives = 24 years
48. (b) We have
y
y
m
K .Q
mm
a
a
=
+
Þ
A4 A4
K .Q 48 .50 A 100
AA
a
--
= Þ = Þ=
49
. (c) Energy released = (80 × 7 + 120 × 8 – 200 × 6.5)
50. (c)
RactivityN
dt
dN
=l=
R
0
= lN
0
at t =
0, R
1
= lN
1
at t = 5 minutes
where
1
0
N
N
log
t
303.2
=l
5
1. (b) Liberated energy Q = 117 × 8.5 + 117 × 8.5 – 236 × 7.6 =
200 MeV. Thus, in fission of one Uranium nuclei nearly
200 MeV energy is liberated.
52. (c)
4
1
)2/1(
)2/1(
)2/1(
)2/1(
N
N
2
4
n
n
B
A
B
A
=== , where n
A
&
n
B
are
number of half lives of samples A & B respectively. N
A
& N
B
are the remaining numbers of A & B after 80 minutes
in this case.
53. (d) Mass defect
2
B.E
c
=
Mass of nuc
leus = Mass of proton
+ mass of neutron – mass defect
54. (b) B.E. = 0.042 × 931
;42 MeV
Nu
mber of nucleons in
7
3
Li is 7.
\B.E./ nucleon =
42
7
= 6 MeV ; 5.6 MeV
55. (d)N =
0
t
Ne
-l
Here, t = 5 minutes
50
0
N
Ne
e
-l
=×
Þ51l=, or
1
,
5
l=
Now, T
1/2
=
2n
l
l
= 5 2nl
56. (b) Let number of atoms in X = N
x
Number of atoms in Y = N
y
By question
x
y
N 1
N 15
=
\ Part of
N
x
=
xy
1
(N N)
16
+
= xy
4
1
(N N)
2
+
So, tota
l 4 half lives are passed, so, age of rock is 4 ×
50 = 200 years
57. (b) Momentum
Mu =
Eh
cc
n
=
Recoil ener
gy
222
21 1Mu 1h
Mu
2
2M 2Mc
næö
==
ç÷
èø
=
22
2
h
2Mc
n
58. (b) Initially P ® 4N
0
Q ® N
0
Half life T
P
= 1 min.
T
Q
= 2 min.
Let after time t number of nuclei of P and Q are equal,
that is
00
t/1 t/2
4NN
22
=
t/1 t/2
4N1
22
Þ= Þ 2
t/1
= 4.2
t/2
2
2
.2
t/2
= 2
(2
+ t/2)
tt
2
12
Þ=+
t
2
2
Þ= Þ t = 4 min
00
P
4/1
(4N)N
N
42
==

750. 746 PHYSICS
at t = 4 min.
00
0
NN
N
44
==
or populati
on of R00
00
NN
4NN
44
æ öæö
- +-
ç ÷ç÷
è øèø
=
0
9N
2
59. (b)N
1
= N
0
e
–lt 10
1
3
NN=
20
0
3
tN
Ne
-l
=
Þ
1
3
t
e
2
-l
= ...(i)
20
2
3
NN=

1
00
2
3
t
N Ne
-l
=
Þ
1
2
3
t
e
-l
= ...(ii)
Dividing equation
(i) by equation (ii)

21
()1
2
tt
e
-l-
=

21
( ) ln2ttl-=
21
ln2
tt-=
l
= T
1/2
= 50 days
60. (c
) The radius of the nuclears is directly proportional to
cube root of atomic number i.e. R µ A
1/3
ÞR = R
0
A
1/3
, where R
0
is a constant of
proportionality
1/3 1/3
22
11
644
273
æö æö
==ç÷ ç÷
èøèø
RA
RA
where R
1
= the r
adius of
27
Al, and
A
1
= Atomic mass number of Al
R
2
= the radius of
64
Cu and A
2
= Atomic mass number
of C4
2
4
3.6 4.8m
3
R= ´=
61. (d) L
et, the amount of the two in the mixture will become
equal after t years.
The amount of A
1
, which remains after t years
01
1
/20
(2)
t
N
N=
The amount
of A
2
, which remains, after t years02
2
/10
(2)
t
N
N=
According to the pr
oblem
N
1
= N
2
/20 /10
40 160
(2) (2)
tt
=
2
/20 10
22
t
t
æö
-ç÷
èø
=
2
20 10
tt
=-
2
20 10
tt
-=
2
20
t
=
t = 40 s
62. (b) Ma
ss defect Dm = 0.02866 a.m.u.
Energy = 0.02866 × 931 = 26.7 MeV
As
1
H
2
+
1
H
2
¾®
2
He
4
Energy liberated per a.m.u = 13.35/2 MeV
= 6.675 MeV
63. (a) The value of x is
1
8
=
0
x
8
=
0
3
x
2
Þt = 3T = 3 × 20 = 60 years
H
ence the estimated age of the rock is 60 years
ALTERNATE : X ® Y
0
at t = 0 N
0
0
at t = t N N
0
– N
0
N
NN-
=
1
7
=
0
N
N
=
1
8
t = 3T = 3 × 20 = 60 yea
rs
64. (c)
150
n 12
12.5
==
12
0
N1
N2
æö
=
ç÷
èø
65. (a) By Radioactive decay law,
N = N
0
e
–lt
Taking log on both sides, we get
lt =
0
e
N
log
N
AsN
a
= 0
1
N
3
and, t = 3 × 24 × 60 ×
60 sec
\l =
0.477
3 24 60 60´´´
= 1.8 × 10
–6
/s ec

751. 747Nuclei
Now
in 9 days, activity becomes,
N =
–6
–(1.8 10 )(9 24 60 60)
0
Ne
´ ´´´´
on
solving we get,
0
N
N
27
=
Hence,
in 9 days activity will become
1
27
of the
o
riginal value.
66. (d)
4
2
14
7
2
1
16
8
HeNHO+®+
67. (b)
68. (b)
0
1
e
-
is known as b
–
particle &
n is known as
antineutrino. Since in this reaction n is emitted with
0
1
e
-
(b
–
particle or electron), so it is known as b-
decay.
69. (d) Time taken to disintegrate completely by a substance is
infinity as t
N
N
log
0
l=
t
0
N
log
0
l=Þ
tlogl=¥
henc
e when
¥®®t,0N.
70. (d)
4
0.6931 1 0.6931
T year 1620 y ears
4.28 10
-
´
===
l ´
71.
(d) In two half lives, the activity becomes one fourth.
Activity on 1–8–91 was 2 micro–curie
\ Activity before two months,
4 × 2 micro-curie = 8 micro curie
72. (d) Two successive b decay increases the atomic number
by 2. Therefore, (d) is correct.
73. (c) Nuclear force is nearly same for all nucleus.
74. (c) Radioactivity
1/2
dN 0.693N
N
dtT
=- =l =
88
60.693 10 0.693 1.210
0.693 2 10 .
50 60
´ ´´
= = = ´´
Rad
ioactivity is proportional to 1/T
1/2,
and not to T
1/2.
75. (b)b-particles, being emitted with very high speed
compared to a-particles, pass for very little time near
the atoms of the medium. So the probability of the
atoms being ionised is comparatively less. But due to
this reason, their loss of energy is very slow and they
can penetrate the medium through a sufficient depth.
EXERCISE - 3
Exemplar Questions
1. (c) Radioactivity is a process due to which a radiactive
material spontaneously decays. Half-life time for a
radioactive remain average half of its or of radioactive
atoms will decay. So, the containers will in general
have different number of atoms of the material,after
one year means one half life i.e., average atom of
radioactive subtance remain after 1 year, in each
container is equal to
1
2
of 10000 = 5000 atoms.
2. (b) Due my formation of H-atom some mass of necleons
convert into energy by E = mc
2
, this energy is used to
bind the nucleons along with nucleus, so mass of
atom becomes slightly less than sum of actual masses
of electron and nucleons.
2
GMm
Given,F
r
=
M = effecti
ve mass of hydrogen atom
Actual mass of H-atom
2
B
is binding energy
c
æö
ç÷
èø
Q
2
(Me) (Mp)
B
mass of electron + mass of proton
c
=-
where
B is BE of hydrogen atom = 13.6 eV.
3. (b)b-particle carries one unit of negative charge and a-
particle carries 2 units of positive charge and g-particle
carries no charge, so electronic energy levels of the
atom charges in emission of a and b particle, but not
in g-decay.
4. (a) Let the nucleus is
Z
X
A
radiate b
+
decay is represented
as
Z
X
A
®
Z–1
Y
A
+
+1
e
0
+ v + Q
2
where Z and A have usual meaning : –
\Q
2
= [m
n
(
Z
X
A
) – m
n
(
Z –1
Y
A
) – 2m
e
] c
2
= [m
n
(
Z
X
A
) + Zm
e
– m
n
(
Z – 1
Y
A
) – (Z–1) m
e
– 2m
e
]c
2
= [m (
Z
X
A
) – m (
Z–1
Y
A
) – 2m
e
] c
2
= (M
x
– M
y
– 2m
e
) c
2
Let the parent nuclei
Z
X
A
is radioactive atom and b
–
decay as under is represented as
Þ
Z
X
A
®
Z + 1
A
Y
+
–1
e
0
+
v + Q
1
Q
1
= [m
n
(
Z
X
A
) – m
n
(
z+1
Y
A
) – m
e
] c
2
= [m
n
(
Z
X
A
)
+ Zm
e
– m
n
(
Z + 1
Y
A
) – (Z + 1) m
e
] c
2
= [m (
Z
X
A
) – m (
Z + 1
Y
A
)] c
2
= (M
x
– M
y
) c
2
5. (a) Tritium (
1
H
3
) has proton and 2 neutrons. If a neutron
decays as n ® P +
ev+, then the nucleus may have 2
protons and neutron, i.e., tritium atom will convert
into
2
He
3
(2 protons and 1 neutron).

752. 748 PHYSICS
Binding e
nergy of
1
H
3
is less than that of
2
He
3
nucleus, So, transformation is not possible
energetically.
6. (b) Stable heavy nuclei have more neutrons than protons
because electrostatic force between protons is
repulsive, which causes unstability of nucleus.
7. (b) The moderator used have light nuclei (like proton).
When protons undergo perfectly elastic collision with
the neutron emitted their velocities are exchanged,
i.e., neutrons come to rest and protons move with the
velocity of neutrons. To slowdown the speed of
neutrons substance should be made up of proton for
perfectly elastic i.e., we need light nuclei not heavy
nuclei because heavy nuclei will not serve the purpose
because elastic collisions of neutrons with heavy
nuclei will not slow them down or speed but only
direction will change.
NEET/AIPMT (2013-2017) Questions
8. (b) Mass defect Dm = 0.02866 a.m.u.
Energy = 0.02866 × 931 = 26.7 MeV
As
1
H
2
+
1
H
2
¾®
2
He
4
Energy liberated per a.m.u = 13.35/2 MeV
= 6.675 MeV
9. (a) The value of x is
1
8
=
0
x
8
=
0
3
x
2
Þt = 3T = 3 × 20 = 60 years
H
ence the estimated age of the rock is 60 years
ALTERNATE : X®Y
0
at t = 0 N
0
0
at t = t N N
0
– N
0
N
NN-
=
1
7
=
0
N
N
=
1
8
t = 3T
= 3 × 20 = 60 yea
rs
10. (c) Increasing order of penetrating power :
a < b < g.
For same energy, lighter particle has higher
penentrating power.
11. (d)
BE/A
A
From the graph of BE/A versus mass number A it is
clear that, BE/A first increases and then decreases with
increase in mass number.
12. (d) BE of
2
He
4
= 4 × 7.06 = 28.24 MeV
BE of
7
3
Li = 7 × 5.60 = 39.20 MeV
71 44 3122
Li
H He He Q39.20 28.24 2( 56.48 MeV)
+®++
´=
Therefo
re, Q = 56.48 – 39.20 = 17.28 MeV.
13. (c) As
x
y
N1
(Given)
N7
=
Þ
3
x
xy
N 11
NN82
æö
==
ç÷
+ èø
Therefore, age of the rock
t = 3T
1/2
= 3 × 1.4 × 10
9
yrs = 4.2 × 10
9
yrs.
14. (a) As we know, R = R
0
(A)
1/3
where A = mass number
R
AI
= R
0
(27)
1/3
= 3R
0
R
Te
= R
0
(125)
1/3
= 5R
0
=
5
3
R
AI
15. (d) In an
explosion a body breaks up into two pieces of
unequal masses both part will have numerically equal
momentum and lighter part will have more velocity.
U ® Th + He
KE
Th
=
2
Th
P
2m
, KE
He
=
2
He
P
2m
since m
He
is less so KE
He
will be mo
re.
16. (a) Given,
AB
8,
l = l l =l
N
B
=
A
N
e
Þ
A
B
t
t
oo
e
NeN
e
-l
-l
=
t 8t1
e ee
-l -l-
=
t 8t1
ee
-l - l-
=
Comparing both side
powerst 8t1-l=-l-
–1 = 7lt
t = –
1
7l
The best possible answer is t =
1
7l

753. )c,b,a( along the x, y and z axis and the angles (a, b and g) in
between the translational vectors are called lattice parameters.
Z
X
Y
a
b
g
a
b
c
Unit Cell
A unit cell is the s
mallest geometrical unit in three dimensions.
The repetition of which will give the entire crystal. The crystalline
solid is said to be consisting of a large no. of unit cells, each one
in contact with immediate neighbours.
Z
X
Y
a
b
g
a
b
c
Crystal System
B
ased on the different combination of lattice parameters (a, b, c)
and the angles (a, b and g) we can define crystal system as
follows.
SOLIDSAll solids are made up of atoms and molecules but due to differentarrangement of the molecules inside them, they are divided intotwo classes (a) crystalline and (b) amorphous.
Properties of Crystalline Solids(i) The atoms and molecules are arranged in a definite or regular
order.
(ii) Crystalline solids are bounded by flat surfaces.(iii) They posses uniform chemical composition.(iv) They have sharp melting point.(v) They are anisotropic.(vi) They have symmetry.Properties of Amorphous Solids(i) The atoms and molecules are arranged in an irregular
manner.
(ii) These are isotropic i.e. they have same physical properties
in all directions.
(iii) They do not have a sharp melting point.(iv) They do not have any symmetry.
CRYSTAL LATTICE
It is defined as the infinite array of atoms and molecules in
space (three dimensions) such that at every point, an atom or
the molecule has got the identical surroundings.
Lattice Vectors
Let a and b be the distance between the two successive atoms
(points) of the lattice along the x-axis and y-axis, then the lattice
vector
lis given by
12
na nb=+
ur ur ur
l where a and b are called translational vectors
and n
1
and n
2
are the integers.
For three dimensional lattice,
123
na nb nc=++
ur ur ur ur
l
Lattice Parameters
In a three dimensional crystal structure, the translational vectors
S. Crystal
Lattice type
U nit cell
Examples
No. system characteristics
1. Cubic (i) Simple cubic (sc) a = b = c (i) CsCl, ZnS, NaCl
(ii) Face centered cubic (fcc)s = b = g = 90° (ii) Silver, copper, gold, lead, nickel and platinum
(iii)Body centered cubic (bcc) (iii) Sodium, barium, iron, tungsten and uranium
2. Tetragonal (i) Simple cubic (sc) a = b ¹ c NiSO
4
, SnO
2
and white tin
(ii) Body centered cubic (bcc)s = b = g = 90°
29
Semiconductor Electronics:
Materials, Devices and
Simple Circuits

754. 750 PHYSICS
Character
istics of the unit cell of a cubic system
(i)Volume of a unit cell (V) = (a × b × c) or V = a × b × c
(ii)Mass and density of a unit cell :
The mass of unit cell = number of atoms or molecules in the
cell × mass of each atom
or, m = p × m
a
or
a
M
mp
N
=´ where M = mo lecular wt. of
the substance
\ density ==
a
a
M
p
N pM
d
V VN
If edge a, is given in picometre then
3
30
a
3
cm/g
10Na
pM
d
-
´
=
(iii) Numbe
r of atoms (N) in a unit cell or the total number of
atoms per unit cell is given by
28
f c
i
N N
NN=++
where N
i
= no. of bod
y centred or interior atoms
N
f
= no. of face centered atoms
N
c
= no. of corner atoms.
Coordination Number
The number of nearest neighbours of a given atom in the crystal
is called the coordination number.
Half of the distance between the centres of two neighbouring
atoms is called atomic radius . It is also called lattice constant or
lattice parameter.
Atomic Packing Factor or Density of Packing
It is defined as ratio of the volume occupied by all the atoms in
a unit cell to the total volume of the unit cell.
B.C.CS.C. F.C.C
( . .)=
Volume of atoms per unit cell
Packing factor P F
Volume
of unit cell
Example 1.
In a cubic unit cell of
bcc structure, the lattice points
i.e. number of atoms are
(a) 2 (b) 6
(c) 8 (d) 12
Solution : (a)
No. of lattice points in a crystal structure will be
C Fi
N NN
n
8 21
= ++
In bcc crys
tal N
C
= 8 and N
F
= 0 and 1N
i
= ;
801
n2
821
\ =++=
Example 2.
Sodium has b
ody-centred packing. Distance between two
nearest atoms is 3.7 Å. Determine the lattice parameter.
Solution :
Atomic radius for bcc structure is
4
3a
r= or Å3.4
732.13
75.14
3
)2/7.3(4
3
r4
a =
´
´
===
i.e., Lattic
e parameter = 4.3 Å
Example 3.
Determine the atomic packing factor for a face centred
cubic cell.
3. Hexagonal Simple cubic (sc) a = b ¹ c Zn, ZnO, Cd, Ni, graphite and quartz
s = b = 90°
g = 120°
4. Rhombo
hedral Simple cubic (sc) a = b = c Sb, Bi, calcite
(Trigonal) s = b = g ¹ 90° < 120°
5.
Orthorhombic(i) Simple cubic (sc) a ¹ b ¹ c KNO
3
, gallanium, mercury chloride, Rhombic
(Rhombic) (ii) Base centred cubics = b = g = 90° sulphur
(iii) Body centered cubic (bcc)
(iv) Face centered cubic (fcc)
6. Monoclinic (i) Simple cubic (sc) a ¹ b ¹ c Na
2
SO
4
, KClO
3
, FeSO
4
(ii) Base centered or s = g = 90° ¹ b
end centered cubic (ecc)
7. Triclinic Simple cubic (sc) a ¹ b ¹ c CuSO
4
, K
2
Cr
2
O
7
s ¹ b ¹ g ¹ 90°

755. 751Semiconductor Electronics : Materials, Devices and Simple Circuits
Solution :
Ato
mic packing factor
=
Volume occupiedby the atoms in a unit cell
Volume of the unit cell
For fcc sturcture, no. of atoms in a unit cell = 4
radius of each atom
22
a
r= ,
where a is
the lattice parameter.
Volume of 4 atoms 3
r4
4
3
p
´=
6
a2
3
p
=
so, atomic pac
king factor
3
3
a
6/a2p
=
6
2p
=
ENERGY BAND
THEORY OF SOLIDS
When two identical atoms, are far apart, the electronic levels (or
electronic wave functions) in one are not disturbed by the
presence of other atom fig 1(a). For example, the 3s electron of
each sodium atom has a single energy with respect to its nucleus.
But when the two atoms come closer, these electron’s wave
function starts to overlap and each electron revolves round both
nuclei.
Y
A
Y
Y
B
r
Fig. 1(a)

Y
A
+ Y
B
Y
r
Fig. 1(b)
P
robability of
finding the electron
is maximum
(symmetric wave
function)
Y
AB
– Y
Y
r
Fig. 1(c)
Probability of
finding the
electron is zero
(Antisymmetric
wave function)
Fig. 1 : Wave functions of two sodium atoms (each in G-state) :
(a) : At wide separation(b) : Small separation : Isolated wave function combined with
same sign (i.e., lower energy)
(c) : Small separation : Isolated wave function combined with
opposite sign i.e., antisymmetric state of joint wavefunctions Y
A
& Y
B
(higher energy state)
Splitting of 3s levels when two atoms are brought together.
3s
Energy
E
Fig.2(a)
Atomic
separation
Splitting of 3s levels when four atoms are brought together.
3s
Energy
E
Atomic
separation
Fig.2(b)
Formation of 3s band by a larger number of atoms.(G-state)
3s
Energy
E
Atomic
separationFig.2(c)
The wave function of each electron is roughly, a combination of
the wave functions of isolated atoms : Fig 1(b) & (c) show two
possible wave functions. In the first, the two isolated wave
functions are combined with (make symmetric state) same sign,
in the second with opposite signs(antisymmetic state). The
symmetric state has lower energy as compared to antisymmetric
state, because the electron is more likely to be found midway
between the nuclei, which leads to lower coulomb energy. This is
the effect, responsible for molecular bonding. Representation of
energy level (when two sodium atoms come closer) is shown in
fig 2(a).
As we bring together a large number of atoms to form a solid, the
same sort of effect occurs. When the sodium atoms are far apart,
all 3s electrons have same energy & as we begin to move them
together, the energy levels begin to “split”. The situation for four
sodium atoms is shown in fig 2(b). As the number of atoms is
increased (may be 10
20
atoms), the levels become so numerous
& so close that we can no longer distinguish the individual levels
as shown in fig 2(c). We can regard the N atoms as forming an
almost continuous band of energy. Since those levels were
identified with 3s atomic levels of sodium, we refer to the 3s band.
N
6N
2N
1N
3p
3s
2s
1s
2p
Fig. 3- Energy bands in sodium metal (G-state)

756. 752 PHYSICS
Each band has a to
tal of N individual levels. Each level can hold
2 × (2l + 1) electrons (l is azimuthal quantum number), so the
capacity of each band is 2(2l + 1)N electrons.
Fig - 3 shows a complete representation of energy bands in
sodium metal. The 1s, 2s & 2p bands are each full, 3s band is half
& 3p band is completely empty. The 1s & 2s bands each contain
2N electrons & 2p band contain 6N electrons. The 3s band contain
N electron, so it is half filled. The band, which can hold 6N
electrons is completely empty.
When we add energy to a system i.e., to sodium metal, the electron
can move from filled state to empty state. In this case, the electron
can move from partially full states of 3s band to empty states of
3s band by absorbing small amount of energy or move to 3p
band by absorbing larger amount of energy.
In a solid at zero temperature, the electron settle into the available
states of lowest energy. The lower energy bands will therefore
be completely filled & the upper most energy band will be either
completely filled or partially filled, depending on the number of
electrons & on the number of available states. The diffence
between conductor & insulator arises from a partially filled or a
completely filled upper most energy band.
CONDUCTORS, INSLULATORS AND
SEMICONDUCTORS
Conductors
A conductor such as sodium, has a partially filled band (in sodium,
the upper most band 3s is half filled). In these substance, electrons
are free to move by applying an electric field, because un-
occupied states are available in upper most band (in 3s band of
sodium only half states are completely filled & half states are
empty or un-occupied). Therefore these electrons are mobile &
can contribute to electrical & thermal conductivity. For example
the energy bands of sodium metal are shown in fig. 4.
Energy
E
Actual separation
of atom in crystal
Fig.4
3s
3p
R
R
0
O
2p
Origin of energy bands in sodium metals, as atoms move together
the energy level spreads into bands.In an isolated sodium atom, the six 3p lowest excited state for
valence electrons are about 2.1 eV above two 3s G-states. But in
solid sodium the 3s & 3p bands are spread out enough so that
they actually overlap, forming a single band.
Because each sodium atom has only one valence electron but
eight 3s & 3p states, that single band is only 1/8 filled & other
states are unoccupied.
Insulators
In these substances (such as diamond), the upper most level is
compeletely filled i.e., no unoccupied state is available for electron
to move. The nearest unoccupied states are in next band(called
C.B.), but this is separated from filled band (called V.B.) by an
energy gap (Eg) of about 6eV. Hence electron refuses to carry an
electric current.
Semiconductors
A semiconductor, has a completely filled valence band i.e., it
resembles an insulator at zero temperature. However, the gap
between this filled valence band & next band (C.B.) is small, about
1eV or less. Hence electrons can easily make the transitions from
one band to another at room temperature & then carry an electric
current (Silicon and germanium are semiconductors).
In a semiconductors, in thermal equilibrium, n
e
n
h
= ni
2
There are two types of semiconductor.
(i) Intrinsic (or pure) semiconductors : These semiconductors
are pure materials in which the thermal vibrations of the
lattice have liberated charge carriers (i.e., electrons & holes).
In intrinsic semiconductor, the number of electrons (n
e
) are
equal to the number of holes (n
h
).
(ii) Extrinsic (or impure or doped) semiconductors : They are
impure semiconductors in which minutes traces of impurity
introduces mobile charge carriers [which may be +ive (holes)
or -ive (electrons)] in addition to those liberated by thermal
vibration.
Again there are two types of extrinsic semiconductors.
(a) N-type semiconductor : When we add a pentavalent
impurity in intrinsic semiconductor, then we obtain N-
type semi-conductor. The pentavalent impurity
substances are P, As & Sb. In N-type semi-conductors,
the electrons are majority carriers & holes are
minority carriers.
i.e., n
e
>> n
h
(b) P-type semiconductor : When we add a trivalent
impurity in intrinsic semiconductor (such as B, Al, In),
we obtain p-type semiconductor.
In p-type semiconductors, the holes are majority
carriers & electrons are minority carriers.
i.e., n
h
>> n
e
In acceptor impurity (trivalent element) energy level is just above
the top of valence band (V.B.)
C.B.
V.B.
Acceptor energy level
*E
F
(*E
F
is called fermi leve
l & at T = 0 all the states below are
completely filled. In N-type semiconductor it lies betweenconduction band & donor energy level E
d
& in case of P-type, it
lies between acceptor level E
a
& valence band. In intrinsic
semiconductor the E
F
level lies in midway between C.B. & V.B.)

757. 753Semiconductor Electronics : Materials, Devices and Simple Circuits
COMMON DEFAULT
´Incorrect. N–type semiconductors are negatively and P–
type semiconductors are positively charged.
ÖCorrect. Both N–type and P–type semiconductors are
neutral because these are made up of neutral atoms.
´Incorrect. Mobility of holes is greater than mobility of
electrons
ÖCorrect. Mobility of holes is less than mobility of electrons
because mobility of holes takes place in valance band and
mobility of electrons takes place in conduction band.
Keep in Memory
1.Widt
h of forbidden energy gap
= Spacing between the top of valence band and the
bottom of conduction band.
2. At absolute zero temperature (0 K), there are no free
electrons in the conduction band of a semi-conductor.
3. A good number of semi-conductors (elemental and
compound) have been discovered but the most frequently
used are germanium (Ge) and silicon (Si) of group-IV of
periodic table because the forbidden energy gap of them is
small (of the order of 1 eV) so that at ordinary room
temperature covalent bonds break easily and free electrons
become available in the conduction band.
4. Important compound semi-conductors are gallium arsenide
(GaAs), lead sulphide (Pbs), cadmium sulphide (Cds),
indium phosphide (InP), etc.
5. Holes doesn’t exist in conductors but only in semi-
conductors. They are deflected by electric and magnetic
fields like electrons. Their movement contributes to electric
current similar to electrons.
Example 4.
Find the current produced at the room temperature in a
pure germanium plate of area 2 × 10
–4
m
2
and of thickness
1.2 × 10
–3
m when a potential of 5 V is applied across the
faces. Concentration of carriers in germanium at room
temperature is 1.6 × 10
6
per cubic metre. The mobility of
electrons and holes are 0.4 m
2
V
–1
s
–1
and 0.2 m
2
V
–1
s
–1
respectively. How much heat is generated in the plate in
100 second?
Solution :
Here, n
i
= 1.6 × 10
6
m
–3
m
e
= 0.4 m
2
V
–1
s
–1
m
h
= 0.2 m
2
V
2
s
–1
;
A = 2 × 10
–4
m
2
and d = 1.2 × 10
–3
m ; V = 5 V
As
)(en
hei
m+m=s
()
6 19
1.6 10 (1.610 )(0.4 0.2)
-
=´´´+
= 1.53
× 10
–13
Sm
–1
Current produced in germanium plate

V
ΙJAEAA
d
æö
= =s =s
ç÷
èø
= A1028.1102
)102.1(
5
1053.1
134
3
13 --
-
-
´=´´
´
´´
Heat gen
erated in plate,
H = VI t = 5 × 1.28 × 10
–13
×100 = 6.4 ×10
–11
J.
p-n JUNCTION DIODE
If we join a piece of n-type to a piece of p-type semiconductor by
appropriate method, then we obtain p-n junction diode. The two
ends of p-n junction is provided with metallic conductors for the
application of an external voltage. It is clear that p-type has more
Comparison Between Conductors, Semiconductors and Insulators
Property Conductors Semiconductors Insulators
Resistivity range 10
–6
– 10
–8
Wm 10
–4
– 0.5 Wm 10
7
– 10
16
Wm
Conductivity range 10
–6
– 10
–8
mho/m 10
4
– 0.5

mho/m 10
–7
– 10
–16
mho/m
Temp. coefficient Positive Negative Negative
of resistance (a)
Flow of current Due to free electrons Due to electrons and holesNo current flow
Energy band diagram
E
l
e
c
t
r
o
n

e
n
e
r
g
y
Conduction band
Valence band
No gap
Overlapping
region
(CB)
(VB)
Forbidden
gap
E 1eV
g
E
l
e
c
t
r
o
n

e
n
e
r
g
y
Conduction band
Valence band
(CB)
(VB)
D Forbidden
gap
DE 6eV
g
E
le
c
t
r
o
n
e
n
e
r
g
y
Valence band
Conduction band
(VB)
(CB)
Forb
idden energy gap
@ 0eV @ 0.1 – 3eV ³ 6eV
Example
s Pt, Al, Cu, Ag etc. Ge, Si, C, Ga, As Wood, plastic
GaF
2
etc. diamond, mica

758. 754 PHYSICS
holes con
centration but less concentration of electrons than
n-type semicondutor.
Due to this difference in concentration, density gradient is
established across the junction resulting in majority carriers
diffusion : Holes (+ive ions) diffuses from p-type to n-type &
electrons (–ive ions) from n-type to p-type (shown in fig (a)) &
terminating their existence by recombination.
Diffusion and Drift Current
Diffusion current : We know that due to concentration difference,
holes diffuse from p-side to n-side (in figure -1(a)) they move
from left to right) & electron diffuse from n-side to p-side (in fig
1(a) they move from right to left). But electric field at junction
repels the holes as they come to depletion layer & only those
holes which have high kinetic energy are able to cross the
potential barrier. Similarly electric field at junction repels electrons
& those having high kinetic energy cross the junction.
The electric potential of n-side is more +ive (or higher) than
p-side –ive (or lower). The variation of potential barrier V
B
is
shown in fig (c).
The potential difference created across the p-n junction due to
the diffusion of electrons and holes is called potential barrier.
So diffusion results in an electric current from p-side to n-side
known as diffusion current.
Drift current : Due to thermal collision, some times a covalent
bond is broken & electron-hole pair is created. But those electron-
hole pairs are destroyed easily due to recombination. This process
[generation of electron-hole (e-h) pair] occurs in the whole part
of material.
But, if e-h pair is created in depletion region, the electron is quickly
pushed by electric field towards n-side & holes towards p-side.
As e-h pairs are continuously created in depletion region, there
is a regular flow of electron towards n-side & holes towards
p-side & so current flows from n-side to p-side. This current is
drift current.
In steady state the magnitude of drift current is equal to diffusion
current & since they are in opposite direction, hence there is no
net transfer of charge at any cross section.
This recombination produces narrow region near junction called
depletion layer. Since this region is depleted of free or mobile
charge carrier and contains only immobile charge carriers, hence
it is called depletion region. Due to these immobile charge carriers,
a potential barrier V
B
is established & further diffusion is stopped
& equilibrium is reached (shown by fig. 1(b)). The sign of V
B
is
+ive towards the n-type & –ive towards p-type.
The schematic symbol of diode & corresponding equivalance in
terms of V
B
are shown below in fig. 2 (a) & (b).
Fig (2)
AnodeCathode
(p) (n)
(a)
(b)
V
B
V
B
Depletion
layer
xo
(c)
In fig-2(a) the triangle shows the direction of current. For Si
diode the value of V
B
is 0.7V & for Ge diode the value of V
B
is
0 . 3 V.FORWARD AND REVERSE BIAS OF p-n JUNCTIONDIODEForward BiasWhen p-side of the p-n junction is connected to the +ve terminalof a battery and n to –ve terminal, conduction takes place andthe diode is said to be forward biased.
+–
pn
Reverse Bias When
p-side of he p-n junction is connected to the -ve terminal
of a battery and n-side to the +ve terminal, there is no conductiontakes place and diode is said to be reverse biased.
–+
pn
Effect of biasing : In forwar d bias, the thickness of depletion
layer decreases and potential barrier reduces, while in reversebias, the thickness of depletion layer increases and potentialbarrier also increases. In forward bias the current increasesexponentially & in reverse bias, the current remains constant atvery small magnitude upto breakdown voltage V
o
& after that it
increases sharply without any further increase of reverse voltageshown in fig.1.V-I characteristics of a p-n junction diode in forward and reversebias.
Breakdown
voltage
V
0
Reverse
bias
knee
voltage
Forward
bi
as
I
V
Fig (1)

759. 755Semiconductor Electronics : Materials, Devices and Simple Circuits
The effect of forward biasing & reverse biasing on potential barrier
are shown in fig.(2).
V
B
(x) np
Depletion layer
Junction
No bia
sing
Deple
-tion
layer
x
Barrier voltage
I
diffusion
I
drift
I = 0
net
Fig.(2)(a
) p-n junction without biasing
V
B
'(x)
np
Depletion layer
Junction
Forward biasing
Deple
-tion
layer
x
Barrier voltage
I
diffusion
I
drift
I
net
Fig.
(2)(b) p-n junction with forward biasing
V
B
''(x)
np
Depletion l
ayer
Junction
Reverse biasing
Deple
-tion
layer
x
Barrier voltage
(c)
I
diffusion
I
drift
I
net= Indrift – Idrift
Fig.(2) (c) p-n junction with reverse biasing
In forward bias the expression of current is )1
TK
eV
(expII
B
o -=
where
V is applied voltage
I
o
is reverse saturation current
e is fundamental electronic chargeK
B
is Boltzmann constant and
T is temperature
Keep in Memory
1.Width o
f depletion layer
1
Doping
µ
2.Depletion
is directly proportional to temperature.
3.Resistance is forward bias R
forward
» 10W – 25W and
resistance in reverse bias R
reverse
» 10
5
W.
4.Under suitable reverse bias break down occurs and voltagegets stabilized at Zener voltage.
5.Junction diode (p-n junction) can be used to convert thea.c. current to d.c. current i.e. as a rectifier. The process of
conversion from a.c. to d.c. is called rectification.
6.p-n junction diode can be used as a half wave or full waverectifier. A device used to convert dc into ac is calledinverter.
7.
Avalanche or zener diode or breakdown
diode used in voltage stabilisation.
8. Photodiode, used in automatic switching
of light, fire alarm, burglar alarm
LED (Light emitting diode) indicator, and
optical fibre communication.
9.A zener diode operating in the breakdown region isequivalent to a battery. This property of zener diode makes
it suitable for voltage regulation (holding voltage constant).
Some Important Results of Half-wave and Full-wave Rectifier
Circuit and Quantity Half wave rectifier Full wave rectifier
1. Circuit R
L
V
0
I/P
diode R
L
D
1
D
2
V
0
I/P
2. Input volt
age Output voltage Output voltage
V
in
t
V
0
t
V
0
t
3. The value of I
rms
of
0
/2I
0
/2I
input alternating c
urrent
I = I
0
sin wt

760. 756 PHYSICS
Example 5.
The
diode used in the circuit shown has a constant voltage
drop at 0.5 V at all currents and a maximum power rating
of 100 milli-watt. What should be the value of the resistor
R, connected in series with diode to obtain maximum
current?
0.5VR
1.5V
Solution :
V
P
= 0.5 V, Maxi
mum power rating P = 100 mW = 100×10
–3
W
and source voltage V
s
= 1. 5 V.
Now diode resistance
W=
´
==
-
5.2
10100
)5.0(
P
V
R
3
22
p
D
Current in diode
D
D
D
V 0.5
Ι 0.2A
R 2.5
= ==
\ Total resistan
ce in the circuit
W===5.7
2.0
5.1V
R
D
s
T
Ι
So, resistanc
e R in circuit = W=-=-55.25.7RR
DT
Example 6.
In p-n junction diode the reverse saturation current is
10
–5
A at 27ºC. Find the forward current for a voltage of
0.2 V. [exp (7.62) = 2038.6, k = 1.4 × 10
–23
J/K]
Solution :
eV / kT
0
Ι = Ι(e 1)-Q
By solving we get,
5 7.62
I 10 [e 1]
-
=- = [ ]
53
10 2038.6 1 20.376 10 A
--
-=´
So the forw
ard current for voltage 0.2V is 20.376 × 10
–3
A
TRANSISTOR
Transistors are three terminal (solid state) devices just like triode.
It can be assumed to consist of two back to back p-n junctions.
In practice a junction transistor (p-n-p) consists of silicon (or
germanium) bar crystal in which a layer of n-type silicon (or Ge) is
sandwiched between two layers of p-type silicon & we get p-n-p
transistor. Alternatively it may consist of a layer of p-type between
two layers of n-type material & we get a n-p-n transister as shown
by fig. (1).
n pp np nC C
(Collector) (Collector)
B(Base) B(Base)
E(Emitter)
Fig. (1)
E
(
E
m
i
t
t
e
r
)
(n – P – n) transistor(P – n – P) transistor
The schematic symbols of transistor are
Fig. (2)
C
B
E
(p-n-p) transistor
(a)
p p
n
C
B
E
(n-p-n) transistor
(b)
n n
p
Components of Tra
nsistor :
Emitter (E) : It supplies charge carriers (electron in n-p-n
transistor and holes in p-n-p transistor) & it has high density ofimpurity concentration i.e., highly doped. It is always forwardbiased.Collector (C) : It is a region on other side of base. It has maximum
area out of other sections(emitter & base) of transistor to dissipatethe heat. It collects the charge carriers & it is always reversebiased.
4. Peak inverse voltage E
0
2E
0
(i.e. maximum voltage
across diode when it
is not conducting)
5. Output direct current
0
/pI
0
2/pI
6. Ripple factor
2
rms
dc
I
1
I
æö
g=-
ç÷
èø
1.21 0.48
7. Efficiency
Output d.c.power
Input a.c. power
h= 40.6% 81.2%
8. Form factor =
rms rms
dc dc
IE
IE
= 1.57 1.11
9. The va
lue of d.c. component Less More
in output voltage as compared
to input a.c. voltage

761. 757Semiconductor Electronics : Materials, Devices and Simple Circuits
Base (B) : It is middle region which forms the two junctions
between emitter & collector. It is very lightly doped.
Biasing of Transistor
In proper biasing of transistor the input (i.e., base emitter junction)
is always forward biased & output (i.e., collector base junction)
is always reverse biased as shown in fig.(3)
This scheme of biasing is same for all three transistor
configurations; Common base configuration (CB), common
emitter configuration (CE) & common collector configuration (CC)]
B B
I
E
I
E
E EC C
p-n-p n-p-n
(a) (b)
Fig. (3)
Working of Trans
istor
Fig. (4) shows a common base (C.B.) configuration of p-n-p
transistor. The forward biasing of emitter junction lowers the
emitter base potential barrier, whereas the reverse biasing of
collector junction increases the collector-base potential barrier.
Hence holes (majority carriers in p-type) flows through emitter to
base & constitutes an emitter current I
E
. Since emitter is heavily
doped in comparison to base, so approximately (only 5% holes
recombine with electrons in base region & constitute base current
I
B
) 95% holes reach to collector & constitute collector current I
C
.
B
V
CC
V
EE
I
C
I
B
I
E
E C
p-n-p
Fig. (4)
Fro
m Kirch off’s current law,
I
E
= I
C
+ I
B
...(1)
Eq. (1) holds true regardless of circuit configuration or transistor
type (p-n-p or n-p-n) that is used.
The current gain (B) of transistor is defined as ratio of collector
current I
C
to base current I
B
i.e.,
b=
C
B
I
I
...(2)
The va
lue of b lies between 10 and 100. Since
EC
II»
Current gain (a) is defined as the ratio of collector current I
C
to
emitter current I
E
. The value of a is always less than unity.
a=
C
E
I
I
...(3)
Relati
on between a and b
or
11
ba
a= b=
+b -a
...(4)
Configurat
ions of Transistor
There are three types of transistor configuration. We can take
either terminal as input terminal and other terminal as output.
(i)Common base configuration (CB) : Here base terminal is
common to both input & output terminals. The emitter
terminal is taken as input.
B
V
CC
V
EE
I
C
I
B
I
E
E C
p-n-p
Ground
Fig. (5)
(ii)C
ommon emitter configuration (CE) : Here emitter terminal
is common to both input and output terminals as shown infig. (6). The base terminal is taken as input and collector istaken as output.
E
V
CC
V
BB
I
C
I
E
I
B
B
C
Ground
Fig. (6)
(iii)Common c ollector configuration (CC) : Here the collector
terminal is common to both input as well as output terminalsas shown in fig. (7). The base terminal is input & emitter isoutput terminal.
C
V
EE
V
BB
I
E
I
C
I
B
B
E
Fig. (7)
Transistor as an Amplifier
Amplification is the process of linearly increasing the amplitudeof a signal. It is one of the major properties of a transistor.
This amplifing action was produced by transfering a current froma low (base emitter loop is forward biased & hence provide lowresistance path and collector base junction is reverse biased &hence gives high resistance path in common emitter configuration)to a high resistance circuit.
Common- Emitter (CE) Amplifier : Fig. 8 shows a basic common-
emitter amplifier circuit in which we connect a signal source V
s
.
The input voltage loop is V
s
, then from base to emitter through
transister to ground (common refrence point) & then through

762. 758 PHYSICS
V
BE
back to V
s
. The outp
ut voltage loop consists ground, then
from emitter to collector through R
C
. The emitter is connected to
ground. During a +ive half cycle of V
s
, the forward bias across
the emitter base junction will be increased, while during – ive half
cycle of V
s
(signal source), the forward bias across emitter base
junction will be decreased. Hence more electrons flow during
+ive half cycle and so we obtain more collector current and so
large voltage drop across R
C
and less value and V
CE
. In negative
half cycle collector current decreases.
Fig. (8)
R
C
V
outV
CE
V
CC
V
BE
+
–
ss
n
In the abse
nce of input signal v
s
, a D.C collector current I
C
flows
in collector circuit due to forward biased battery V
BE
. This is
called zero signal collector current. So total collector current is
current
collector.c.a
c
current
collector.c.d
Ccurrentcollectortotal III +=
v
s
+v
s
–v
s
V
BE
Curve &
input signal
t
O
(a) (b)
t
– =–v
s ce
v
VV
s ce
= Curve of
output
signal
V
CE
V
out
Fig. (9)
(a) input signal superimposed on V
(a) Input signal superimposed on V
BE
(b) Output amplifying signal wave from super imposed on V
CE
The voltage from the collector to ground determines the output
signal. When I
C
(during +ive half cycle) increases, V
CE
decreases
& when I
C
decreases (during – ive half cycle) V
CE
increase. So
output voltage is 180º out of phase with input signal in CE
amplifier:
The current gain A
i
is defined as,
C
i
B
IChange in load current/collector current
A
I Change in input current/base current
D
==
D
The voltage g
ain A
v
is defined as,
CE
v
BE
V Change in load voltage
A
V Change in input voltage
D
==
D
The power gain A
p
is de
fined as, A
p
= A
i
× A
v
For example : Let V
BE
=0 .7V, V
CC
= 20V
I
B
= 60mA, I
C
= 3m A, R
C
= 3 kW
Then
50
A60
mA3
I
I
B
C
c.d =
m
==b
The collector
current flowing through R
C
results in voltage
drop:
VBE
=0.7V
dc bias
v v beBE= +
bias + signal
VBE
DV
BE
v
BE s
(signal)= v
time
0.5v–
V
CE
=11V
(d.c value)
I
C C
= 9V
(d.c value)
R
v v
CE
= +
(d.c value + signal)
V
CE ce
DV
CE
v v
ce out
(signal)=
time
10V–
20V
V
CC
= 20V
I
B
=60A
(d.c value)
m
i
B
= I
+ i
(d.c value + signal)
Bb
DI
B
i
b
(signal
current)
t
40 mA
60mA I
B
80mA
Base
current
(a)
I
C
=3mA
(d.c value)
i
C
= I +
i
(d.c value + signal)
cc
DI
C
i
c
(signal
current)
t
2m A
I
C
4mA
(b)
Fig. (a) Base wave forms or input wave forms
(b) Collector wave froms or output wave forms.
I
C
R
C
= .003A × 3000W = 9.0V
So by K.V.L the voltage from collector to emitter is
V
CE
= V
CC
– I
C
R
C
= 20 – 9 = 11.0V
Now we introduce signal voltage v
s
, which has peak value
5 0m V.
During +ive half cycle of v
s
= + 50mV, I
B
rises to 80mA & during
–ive half cycle of v
s
= – 50mV. I
B
decreases to 40mA.
So change in base current is DI
B
(80 – 60) mA = 20mA
and change in collector current is DI
C
(b
ac
DI
B
) = 50 × 20 = 1mA
It means during +ive half cycle I
C
becomes 4mA hence
V
CE
= V
CC
– I
C
R
C
= 20 – 12 = 8.V

763. 759Semiconductor Electronics : Materials, Devices and Simple Circuits
During the negative half cycle I
C
becomes 2mA, hence
V
CE
= 20 – 6 = 14V.
so DV
CE
= 3V
so 50
A20
mA1
I
I
A
B
C
i =
m
=
D
D
=
60
mV50
V3
V V
A
BE
CE
v ==
D D
=
and A
p
= A
i
×
A
v
= 3000
V
CE
=11.0V n
CE
= + V
CE ce
v
v
ce
time
8.0V
11.0V
14.0V
Collector to
emitter voltage
Keep in Memory
1.Amp
lifier is a device used to increase the amplitude of input
signal.
2.Amplification factor of triode valve m = g
m
× R
p
where
g
m
= mutual or transconductance and R
p
= plate resistance.
3.Transistor is also used as an oscillator.
Frequency of transistor LC oscillator is given by,
LC2
1
f
p
=
4.In phase sh
ift oscillator (which consists of three RC section),
the frequency is given by RC62
1
f
p
=
5. Oscillat
or gives continuous, undamped oscillating output.
6. Inverter : An oscillator which converts d.c. into a.c.
7.An amplifier with feed back in proper phase and magnitude
behaves as an oscillator.
8. Berkhausen criteria : for amplifier to act as oscillator :
Gain with feed back, A´ = A/[1 - KA], K = feed back ratio.
Criteria to become oscillator, (1 - KA) = 0.
“Where A' is called closed loop gain & A is called open
loop gain.
9.The basic rules for normal operation of a transistor are
(i) the emitter-base junction is forward blased to offer a
low resistance to the flow of current.
(ii) the collector-base junction is reverse biased to offer a
high resistance to the flow of current.
Comparative study of CB, CE and CC Amplifier
Use and property
Used for
ImpedanceImpedance
Output
Impedance
Current gain
Voltage gain
Power gain
Phase shift
A < 1
()
i
a
A > 1
v
A > 1
p
Nil 180° Nil
A > 1
vA > 1
p
A > 1
v A < 1
v
A < 1
( ) ( 20 – 100)
i
b b »
A (B + 1)> 1
(20 – 200)
i
High High Low
Low High Low
Voltage
amplification
Common base
(CB) amplifier
Common emitter
(CE) amplifier
Power
amplification
Current
amplification
Common collector
(CC) amplifier
Exa
mple 7.
The correct relationship between two current gains
a
and b in a transistor is
(a)
a+
a
=b
1
(b)
a
a+
=b
1
(c)
b+
b
=a
1
(d)
b
b+
=a
1
Solution : (c)
a-
a
=b
1
or a=a-b)1( or a=ba-b
or )1(b+a=b; )1/(b+b=a\
Example 8.
The
current gain of a transistor in common base mode is
0.99. To change the emitter current by 5 mA, what will
be the necessary change in collector current?
Solution :a=
D
D
e
c
Ι
Ι
or
ec
ΙΙD a=D; 599.c´=DΙ = 4.95 mA
Exa
mple 9.
For a transistor, the current amplification factor is 0.8,
the transistor is connected in common emitter
configuration. Find the change in the collector current
when the base current changes by 6 mA.
Solution :
a = 0.8 ;
4
8.01
8.0
1
=
-
=
a-
a
=b ;
bc
/ΙΙDD=b or mA2464
bc
=´=Db=DΙΙ
Example 10.
The current gain of a transistor in
a common emitter
configuration is 40. If the emitter current is 8.2 mA, then
find the base current.

764. 760 PHYSICS
Solution :
1
b
e
b
be
b
c
-=
-
==b
Ι
Ι
Ι
ΙΙ
Ι
Ι
or b+=1
b
e
Ι
Ι
or mA20.
41
2.8
401
2.8
1
e
b ==
+
=
b+
=
Ι
Ι
Example 11.
In a common emitter trans
istor amplifier
b = 60,
R
0
= 5000
W and internal resistance of a trnasistor is
500
W. What will be the voltage amplification of amplifier?
Solution :
0
v
i
R 5000
A 60 600
R 500
=b=´=
Example 12.
A p-n-p transistor
is used in common-emitter mode in an
amplifier circuit. A change of 40
mA in the base current
brings a change of 2 mA in collector current and 0.04 V in
base emitter voltage. Find the (i) input resistance R
i
(ii)
the base current amplification factor (
b). (iii) If a load
resistance of 6 k
W is used, then also find the voltage gain
of the amplifier.
Solution :
Here, A1040
6
b
-
´=DΙ ; A102
3
c
-
´=DΙ ;
V04.0V
i
=D ; R
0
= 6000W
(i) Input resistance,
W=
´
=
DI
D
=
-
3
6
b
i
i 10
1040
04.0V
R
(ii) Base curren
t amplification factor,
50
1040
102
6
3
b
c
=
´
´
=
DI
DI
=b
-
-
(iii) Vol
tage gain,
0
3
i
R 6000
Av 50 300
R 10
=b=´= .
LOGIC GATES
A logic
gate is a digital circuit that follow certain logical
relationship between input and output voltages. They control
the flow of information.
Basic logic gates are : AND, OR and NOT gates.
AND gate : The output is high or 1 when all inputs are high.
A B Y
0 0 0
0 1 0
1 0 0
Truth table

Logic symbol
A
B
Y
Boolean expression : A . B = y
OR gate : Output is high even if one of the inputs is high.
A B Y
0 0 0
0 1 1
1 0 1
1 1 1
Truth table

Logic symbol
A
B
Boolean expression : A + B = y
NOT gate : Output is not the input.
A Y
0 1
1 0
Truth table

Logic symbol
A
B
Boolean expression
:
yA=
NAND gate : It is the combinatin of AND and NOT gate. The
output is high, even if all inputs are low (i.e., 0) or one input is
low.
A B Y
0 0 1
1 0 1
0 1 1
1 1 0
Truth table

Logic symbol
A
B
Y
Boolean expression : .ABY=
NAND gate is called the building block of all digital circuits.
NOR gate : It is the combination of NOT and OR gate. The output
is high, when all inputs are low.
A B Y
0 0 1
1 0 0
0 1 0
1 1 0
Truth table

Logic symbol
A
B
Y
Boolean expression : ABY+=
Exclusive OR (XOR) gate : The output of a two input exclusive
OR gate is at logical 1 if one and only one input accepts logical 0
(zero)
Logic symbol:
A
B
Y = A + B
Truth table
Input
A
0
0
1
1
B
0
1
0
1
Y = A B
0 1
1
0
Output
Å
Logic expression : =Å=+Y A B AB AB
Note : This circuit is also called inequality comparator or detector
because it produces an output only when the two inputs are
different.
BINARY NUMBERS AND DECIMAL NUMBERS
·In a binary system, a number is expressed by only two
digits 0 and 1. The base of the binary system is thus 2.
·0 and 1 represent the coefficients of powers of 2.
·A binary digit (0 or 1) is known as a bit.
·A group of 4 bit is called a nibble & a group of 8 bit is called
byte.

765. 761Semiconductor Electronics : Materials, Devices and Simple Circuits
Decimal to Binary Conversion
(A)When the decimal number is an integer –
Divide the decimal number progressively by 2 until the
quotient is zero. The remainders of the successive divisions,
written in the reverse order, give the binary number.
For example : We want to convert (9)
10
into binary system.
2 9 1 Fi rst remainder
4
0 Second remai nder
2
1
0 Third remainder
0
1 Fourth remainder
®
®
®
®
( ­ Read up)
So (9)
10
=
(1001)
2
(B)When the decimal number is a fraction –
The number is multiplied repeatedly by 2, and the carry in
the integer position is recorded each time. The process is
continued until the fractional part is zero or sufficient binary
bits have been obtained.
Let we want to convert (0.85)
10
into binary no. then
0.85 × 2 = 1.7 = 0.7 with a carry 1
0.7 × 2 = 1.4 = 0.4 with a carry 1
0.4 × 2 = .8 = 0.8 with a carry 0 (¯ Read down)
0.8 × 2 = 1.6 = 0.6 with a carry 1
0.6 × 2 = 1.2 =0 .2 with a carry 1
0.2 × 2 = .4 = 0.4 with a carry 0
So (0.85)
10
= (0.110110)
2
Binary to Decimal Conversion
(A)When the binary number is an integer –
Binary can be converted into its decimal equivalent by
noting that the successive digits from the extreme right of a
binary number are the coefficients of ascending power of 2,
beginning with the zeroth power of 2 at the extreme right.
(B)When the binary number is a fraction –
The decimal equivalent of the binary number is found by
multiplying each digit in fraction sucessively by
2
–1
, 2
–2
, 2
–3
...etc.
Let we want to convert (101.011)
2
into its decimal equivalent.
Then
(101.011)
2
= 1 × 2
2
+ 0 × 2
1
+ 1 × 2
0

+ 0 × 2
–1
+ 1 × 2
–2
+ 1 × 2
–3
= (5.375)
10
5.375 = 5 × 10
0
+ 3 × 10
–1
+ 7 × 10
–2
+ 5 × 10
–3
Demorgan Theorems
(i)
BAB.A+= (ii) B.ABA=+
These theor
ems are self proved.
Some basic boolean identies :
OR operationAND operation NOT operation
A+ 0 = A A.0 = 0 A+
A=1
A + 1 = 1 A.
1= A A
A= 0
A + A = A A.
A = A
A= A
A + A= 1 A.A= 0
A + B =
B + A A . B = B . A
A(B + C) = AB + ACA + BC = (A + B) (A + C)
A + AB = AA(A + B) = A
A +
AB = A + BA(A + B) = AB
These ab
ove theorems, identities & relations can be easily proved.Keep in Memory
1.Th
e basic concept of digital circuit has been provided by
George Boole.
2.Claude shanon established an analogy between function
of mechanical switches and Boolean algebra.
3.Series combination of switches is equivalent to AND logic
operation.
4.Parallel combination of switches is equivalent to OR logic
operation.
5.NOT logic operation is performed on a single variable. That’s
why it is called unary operation.
6.AND, OR and NOT logic operations follow closure property,
i.e., input as well as output are in either of the binary states.
7.NAND and NOR gates are universial gates
8.If the logic gate is changed from positive to negative or
vice-versa; AND changes into OR, OR changes into AND,
NAND changes into NOR and NOR charges into NAND.
Example 13.
Identify the gate represented by the block diagram shownin fig. Write the Boolean expression and truth table.
A
B
y
III
I
II
Solution :
He
re for the input, the two NOR gates have been used as
NOT gates (by joining the input terminals of NOR gate).Their outputs are jointly fed to the NOR gate. From the
NOR gate I, for the input A, the output is A. From the NOR
gate II, for the input B, the output is B. From NOR gate III,
the output is given by Y A B A.B=+=
Thus Bo
olean expression for this combination of gate is
Y=A+B=A. B
which is for AND gate. Thus the combination will work as
AND gate. The truth table of the combination of gates is

766. 762 PHYSICS
A B A B Y
0 0 1 1 0
1 0 0 1 0
0 1 1 0 0
1 1 0 0 1
Example 14.
The combinations of the ‘NAND’ gates shown here in fig.
are equival
ent to
A
A
B
B
C
1
A
B
C
2
(a) an ‘OR’ gate and an ‘AND’ gate respectively
(b) an ‘AND’ gate and a ‘NOT’ gate respectively
(c) an ‘AND’ gate and an ‘OR’ gate respectively
(d) an ‘OR’ gate and a ‘NOT’ gate respectively
Solution : (a)
For first case,
)BA(B.AC
1 +==(by Demorgan's
theorem)The truth table is shown below
A B
A B B.AB.A
100 1 0 1
011 0 0 1
001 1 1 0
110 0 0 1
This is truth table fo
r C
1
= A + B i.e. OR gate
For second case,
B.AB.AC
2
==i.e., AND gate.

767. 763Semiconductor Electronics : Materials, Devices and Simple Circuits
Semiconductors Resistivity ()
or conductivity () intermediate
to metals and insulators
= 10– 10 m; = 10 – 10sm
E for Ge = 0.72 eV ; Si = 1.1 eV
r
s
rW s
–5 65–6–1
g
Metals Resistivity = 10– 10m
conductivity = 10 – 10 Sm
E = 0 eV
rW
s
–2–8
2 8–1
g
Insulators Resistivity = 10 – 10 m
conductivity = 10– 10 Sm
Eg > 3 eV
rW
s
1119
– 11–19
–1
Extrinsic or impure
semiconductor
Due to desirable
addition of impurity
atoms or dopants
This is to improve
conductivity
Intrinsic or pure
semiconductors e.g.,
Intrinsic carrier concentration
n = n = n
Total current I = n + n
i e h
e h
N
-
t
y
p
e
s
e
m
i
c
o
n
d
u
c
t
o
r
S
i
o
r

G
e

d
o
p
e
d
w
i
t
h
p
e
n
t
a
v
a
l
e
n
t
A
s
,
S
b
, B
i
e
t
c
.
E
l
e
c
t
r
o
n
s

m
a
j
o
r
i
t
y
a
n
d

h
o
l
e
s

m
i
n
o
r
i
t
y

c
a
r
r
i
e
r
s
n

>
>

n
e
h
P-type semiconductor
Si or Ge doped with
trivalent, B, Al etc.
Electrons minority and
holes majority carriers
n > > n
h e
SEMICONDUCTOR ELECTRONICS;
MATERIALS, DEVICES AND
SIMPLE CIRCUITS
Classification
of metals, insulators
and semiconductors
P-n
j
u
n
c
t
i
o
n
An
a
r
r
a
n
g
e
m
e
n
t
m
a
d
e
b
y
a cl
o
s
e
c
o
n
t
a
c
t
o
f

n
-
t
y
p
e
sem
i
c
o
n
d
u
c
t
o
r
a
n
d
p
-
t
y
p
e
sem
i
c
o
n
d
u
c
t
o
r
P-n Junction transistor
A three terminal semi-
conductor device.
n-p-n and p-n-p transistors
CE, CC and CB transistors
In transistor I =I + I
e b c
Uses of transistor
O
s
c
i
l
l
a
t
o
r

f
r
e
q
u
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c
y
1
v
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L
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=
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m–
6
Parameters of
amplifier
Current gain
C
e
I
I
a=
C
DC
b
I
I 1
a
b= =
-a
c
AC
b
I
I
Dæ ö
b=
ç ÷
D
è ø
Transconductance
CAC
m
iin
I
g
V R
D b æ ö
= =
ç ÷
D
è ø
Logic gates Digital circuit follows certain
logical relationship
between the input and
output voltage
OR gate
A
B
Y = A + B
A
B
AND gate
Y = A.B
A
Y = A
NOT gate
NAND gate Combination
of NOT and AND gate
A
B
Y = A.B
NOR gate Combination
of NOT and OR gate
A
B
Y = A + B
C
O
N
C
E
P
T

M
A
P

768. 764 PHYSICS
1.Choose the only
false statement from the following.
(a) In conductors, the valence and conduction bands
may overlap.
(b) Substances with energy gap of the order of 10 eV are
insulators.
(c) The resistivity of a semiconductor increases with
increase in temperature.
(d) The conductivity of a semiconductor increases with
increase in temperature.
2.Application of a forward bias to a p–n junction
(a) widens the depletion zone.
(b) increases the potential difference across the depletion
zone
(c) increases the number of donors on the n side.
(d) increases the electric field in the depletion zone.
3.A transistor has three impurity regions. All the three regions
have different doping levels. In order of increasing doping
level, the regions are
(a) emitter, base and collector
(b) collector, base and emitter
(c) base, emitter and collector
(d) base, collector and emitter
4.In a semiconductor diode, the barrier potential offers
opposition to
(a) holes in P-region only
(b) free electrons in N-region only
(c) majority carriers in both regions
(d) majority as well as minority carriers in both regions
5.In Boolean algebra, Y = A + B implies that
(a) output Y exists when both inputs A and B exist
(b) output Y exists when either input A exists or input B
exists or both inputs A and B exist
(c) output Y exists when either input A exists or input B
exists but not when both inputs A and B exist
(d) output Y exists when both inputs A and B exists but
not when either input A or B exist
6.Which of the following is unipolar transistor?
(a)p – n – p transistor(b)n – p – n transistor
(c) Field effect transistor (d) Point contact transistor
7.In a junction diode, the holes are due to
(a) protons (b) extra electrons
(c) neutrons (d) missing electrons
8.By increasing the temperature, the specific resistance of a
conductor and a semiconductor
(a) increases for both(b) decreases for both
(c) increases, decreases (d) decreases, increases
9.The energy band gap is maximum in
(a) metals (b) superconductors
(c) insulators (d) semiconductors.
10.When n-P-n transistor is used as an amplifier, then
(a) electrons move from collector to emitter
(b) electrons move from emitter to collector
(c) electrons move from collector to base
(d) holes move from emitter to collector
11.An oscillator is nothing but an amplifier with
(a) positive feedback(b) large gain
(c) no feedback (d) negative feedback
12.Minority carriers in a p-type semiconductor are
(a) free electrons
(b) holes
(c) neither holes nor free electron
(d) both holes and free electrons.
13.In insulator
(a) valence band is partially filled with electrons
(b) conduction band is partially filled with electrons
(c) conduction band is filled with electrons and valence
band is empty
(d) conduction band is empty and valence band is filled
with electrons.
14.The intrinsic semi conductor becomes an insulator at
(a) 0ºC (b) 0 K
(c) 300 K (d) –100ºC
15.NAND and NOR gates are called universal gates primarily
because they
(a) are available universally
(b) can be combined to produce OR, AND and NOT gates
(c) are widely used in Integrated circuit packages
(d) are easiest to manufacture
16.One serious drawback of semi-conductor devices is
(a) they do not last for long time.
(b) they are costly
(c) they cannot be used with high voltage.
(d) they pollute the environment.
17.Radiowaves of constant amplitude can be generated with
(a) FET (b) filter
(c) rectifier (d) oscillator
18.Zener diode is used for
(a) amplification (b) rectification
(c) stabilisation (d) all of the above
19.Which one is the weakest type of bonding in solids ?
(a) Ionic (b) Covalent
(c) Metallic (d) Vander Wall’s
20.The band gap in germanium and silicon in ev respectively
is
(a) 1.1, 0 (b) 0, 1.1
(c) 1.1, 0.7 (d) 0.7, 1.1

769. 765Semiconductor Electronics : Materials, Devices and Simple Circuits
21.The transistor are usually made of
(a) metal oxides with high temperature coefficient of
resistivity
(b) metals with high temperature coefficient of resistivity
(c) metals with low temperature coefficient of resistivity
(d) semiconducting materials having low temperature
coefficient of resistivity
22.The device that can act as a complete electronic circuit is
(a) junction diode (b) integrated circuit
(c) junction transistor(d) zener diode
23.In forward biasing of the p–n junction
(a) the positive terminal of the battery is connected to
p–side and the depletion region becomes thick
(b) the positive terminal of the battery is connected to
n–side and the depletion region becomes thin
(c) the positive terminal of the battery is connected to
n–side and the depletion region becomes thick
(d) the positive terminal of the battery is connected to
p–side and the depletion region becomes thin
24.If a small amount of antimony is added to germanium crystal
(a) it becomes a p–type semiconductor
(b) the antimony becomes an acceptor atom
(c) there will be more free electrons than holes in the
semiconductor
(d) its resistance is increased
25.At absolute zero, Si acts as
(a) non-metal (b) metal
(c) insulator (d) none of these
1.On doping germanium with donor atoms of density
10
17
cm
–3
its conductivity in mho/cm will be
[Given : m
e
= 3800 cm
2
/V–s and n
i
= 2.5 × 10
13
cm
–13
]
(a) 30.4 (b) 60.8
(c) 91.2 (d) 121.6
2.The ratio of electron and hole currents in a semiconductor
is 7/4 and the ratio of drift velocities of electrons and holes
is 5/4, then the ratio of concentrations of electrons and
holes will be
(a) 5/7 (b) 7/5
(c) 25/49 (d) 49/25
3.In the half wave rectifier circuit operating from 50 Hz mains
frequency, the fundamental frequency in the ripple would
be
(a) 25 Hz (b) 50 Hz
(c) 70.7 Hz (d) 100 Hz
4.In a full wave rectifier circuit operating from 50 Hz mains
frequency, the fundamental frequency in the ripple would
be
(a) 25 Hz (b) 50 Hz
(c) 70.7 Hz (d) 100 Hz
5.Distance between body centred atom & a corner atom in
sodium(a = 4.225 Å) is
(a) 3.66 Å (b) 3.17 Å
(c) 2.99 Å (c) 2.54 Å
6.If the forward voltage in a semiconductor diode is changed
from 0.5V to 0.7 V, then the forward current changes by
1.0 mA. The forward resistance of diode junction will be
(a) 100 W (b) 120 W
(c) 200 W (d) 240 W
7.Current gain of a transistor in common base mode is 0.95.
Its value in common emitter mode is
(a) 0.95 (b) 1.5
(c) 19 (d) (19)
–1
8.A transistor has b = 40. A change in base current of 100 m A,
produces change in collector current
(a) 40 × 100 microampere (b) (100 – 40) microampere
(c) (100 + 40) microampere (d) 100/40 microampere
9.A transistor has a base current of 1 mA and emitter current
90 mA. The collector current will be
(a) 90 mA (b) 1 mA
(c) 89 mA (d) 91 mA
10.In a common emitter transistor amplifier b = 60, R
o
= 5000 W
and internal resistance of a transistor is 500 W. The voltage
amplification of amplifier will be
(a) 500 (b) 460
(c) 600 (d) 560
11.For a common base amplifier, the values of resistance gain
and voltage gain are 3000 and 2800 respectively. The current
gain will be
(a) 1.1 (b) 0.98
(c) 0.93 (d) 0.83
12.In a common base amplifier the phase difference between
the input signal voltage and the output voltage is
(a)0 (b)p/4
(c)p/2 (d)p
13.The current gain in transistor in common base mode is 0.99.
To change the emitter current by 5 mA, the necessary
change in collector will be
(a) 0.196 mA (b) 2.45 mA
(c) 4.95 mA (d) 5.1 mA

770. 766 PHYSICS
14.The electr
ical conductivity of a semiconductor increases
when electromagnetic radiation of wavelength shorter than
2480 nm is incident on it. The band gap (in eV) for the
semiconductor is
(a) 0.9 (b) 0.7
(c) 0.5 (d) 1.1
15.What is the voltage gain in a common emitter amplifier,
where input resistance is 3 W and load resistance 24 W, b =
0.6 ?
(a) 8 . 4 (b) 4 . 8
(c) 2 . 4 (d) 480
16.A half-wave rectifier is being used to rectify an alternating
voltage of frequency 50 Hz. The number of pulses of rectified
current obtained in one second is
(a) 50 (b) 25
(c) 100 (d) 2000
17.In a triode, g
m
= 2 × 10
–3
ohm
–1
; m = 42; resistance of load,
R = 50 kilo ohm. The voltage amplification obtained from
this triode will be
(a) 30.42 (b) 29.57
(c) 28.18 (d) 27.15
18.In a p-n junction having depletion layer of thickness
10
–6
m the potential across it is 0.1 V. The electric field is
(a) 10
7
V/m (b) 10
–6
V/m
(c) 10
5
V/m (d) 10
–5
V/m
19.The difference in the variation of resistance with temperature
in a metal and a semiconductor arises essentially due to the
difference in the
(a) crystal structure
(b) variation of the number of charge carriers with
temperature
(c) type of bonding
(d) variation of scattering mechanism with temperature
20.In common emitter amplifier the
e
c
I
I
is 0.98. The cu
rrent
gain will be
(a) 4.9 (b) 7.8
(c) 49 (d) 78
21.In the diagram, the input is across the terminals A and C
and the output is across B and D. Then the output is
A
B
C
D
(a) zero (b) s ame as the input
(c) full wave rectifier(d) half wave rectifier
22.The cause of the potential barrier in a p-n diode is
(a) depletion of positive charges near the junction
(b) concentration of positive charges near the junction
(c) depletion of negative charges near the junction
(d) concentration of positive and negative charges near
the junction
23.The intrinsic conductivity of germanium at 27° is 2.13 mho
m
–1
and mobilities of electrons and holes are 0.38 and 0.18
m
2
V
–1
s
–1
respectively. The density of charge carriers is
(a) 2.37 × 10
19
m
–3
(b) 3.28 × 10
19
m
–3
(c) 7.83 × 10
19
m
–3
(d) 8.47 × 10
19
m
–3
24.In a reverse biased diode when the applied voltage changes
by 1 V, the current is found to change by 0.5 µA. The reverse
bias resistance of the diode is
(a) 2 × 10
5
W (b) 2 × 10
6
W
(c) 200 W (d) 2 W.
25.When the forward bias voltage of a diode is changed from
0.6 V to 0.7 V, the current changes from 5 mA to 15 mA.
Then its forward bias resistance is
(a) 0.01 W (b) 0.1 W
(c) 10 W (d) 100 W
26.The current gain of a transistor in common base mode is
0.995. The current gain of the same transistor in common
emitter mode is
(a) 197 (b) 201
(c) 198 (d) 199
27.The gate for which output is high if atleast one input is
low?
(a) NAND (b) NOR
(c) AND (d) OR
28.What is the conductivity of a semiconductor if electron
density = 5 × 10
12
/cm
3
and hole density = 8 × 10
13
/cm
3
(µ
e
= 2.3 m
2
V
–1
s
–1
, µ
h
= 0.01 m
2
V
–1
s
–1
)
(a) 5.634 (b) 1.968
(c) 3.421 (d) 8.964.
29.In a p-type semiconductor the acceptor level is situated 60
meV above the valence band. The maximum wavelength of
light required to produce a hole will be
(a) 0.207 × 10
–5
m (b) 2.07 × 10
–5
m
(c) 20.7 × 10
–5
m (d) 2075 × 10
–5
m
30.In a npn transistor 10
10
electrons enter the emitter in
10
–6
s. 4% of the electrons are lost in the base. The current
transfer ratio will be
(a) 0.98 (b) 0.97
(c) 0.96 (d) 0.94
31.The grid voltage of any triode valve is changed from –1
volt to –3 volt and the mutual conductance is 3 × 10
–4
mho.
The change in plate circuit current will be
(a) 0.8 mA (b) 0.6 mA
(c) 0.4 mA (d) 1 mA

771. 767Semiconductor Electronics : Materials, Devices and Simple Circuits
32.The ratio of work function and temperature of two emitters
are 1 : 2, then the ratio of current densities obtained by
them will be
(a) 4 : 1 (b) 2 : 1
(c) 1 : 2 (d) 1 : 4
33.The transfer ratio b of transistor is 50. The input resistance
of a transistor when used in C.E. (Common Emitter)
configuration is 1kW. The peak value of the collector A.C
current for an A.C input voltage of 0.01V peak is
(a) 100 mA (b) .01 mA
(c) .25 mA (d) 500 mA
34.The manifestation of band structure in solids is due to
(a) Bohr’s correspondence principle
(b) Pauli’s exclusion principle
(c) Heisenberg’s uncertainty principle
(d) Boltzmann’s law
35.The frequency response curve of RC coupled amplifier is
shown in figure. The band with of the amplifier will be
max
A
m
a
x
A
7
0
7
.
0
1f2f
3
f
4f
(a)f
3
– f
2
(b) f
4
– f
1
(c)
2
ff
24-
(d) f
3
– f
1
36.A diode ha
ving potential difference 0.5 V across its junction
which does not depend on current, is connected in series
with resistance of
W20 across source. If 0.1 A current
passes through resistance then what is the voltage of the
source?
(a) 1.5 V (b) 2.0 V
(c) 2.5 V (d) 5 V
37.Copper has face centered cubic (fcc) lattice with interatomic
spacing equal to 2.54 Å. The value of lattice constant for
this lattice is
(a) 2.54 Å (b) 3.59 Å
(c) 1.27 Å (d) 5.08 Å
38.Carbon, Silicon and Germanium atoms have four valence
electrons each. Their valence and conduction bands are
separated by energy band gaps represented by (E
g
)
C
, (E
g
)
Si
and (E
g
)
Ge
respectively. Which one of the following
relationship is true in their case?
(a) (E
g
)
C
> (E
g
)
Si
(b) (E
g
)
C
< (E
g
)
Si
(c) (E
g
)
C
= (E
g
)
Si
(d) (E
g
)
C
< (E
g
)
Ge
39.Application of a forward bias to a p–n junction
(a) widens the depletion zone
(b) increases the potential difference across the depletion
zone
(c) increases the number of donors on the n side
(d) increases the electric field in the depletion zone.
40.A semi-conducting device is connected in a series circuit
with a battery and a resistance. A current is found to pass
through the circuit. If the polarity of the battery is reversed,
the current drops to almost zero. The device may be
(a) a p-n junction
(b) an intrinsic semi-conductor
(c) a p-type semi-conductor
(d) an n-type semi-conductor
41.Which of the following gates will have an output of 1?
1 0
00
1 1
01
(A) (B)
(C) (D)
(a)D (b) A
(c
)B (d) C
42.The following circut represents
Y
B
A
(a) OR gate (b) XOR gate
(c) AND gate (d) NAND gate
43.The diagram of a logic circuit is given below. The output F
of the circuit is represented by
X
W
Y
F
W
(a) W . (X + Y) (b) W . (X . Y)
(c) W + (X . Y) (d) W + (X + Y)

772. 768 PHYSICS
44.If A is the atomic mas
s number of an element, N is the
Avogadro number and a is the lattice parameter, then the
density of the element, if it has bcc crystal structure, is
(a)
3
Na
A
(b)
3
Na
A2
(c)
3
Na
A3
(d)
3
Na
A22
45.Metallic solids are alw
ays opaque because
(a) they reflect all the incident light.
(b) they scatter all the incident light.
(c) the incident light is readily absorbed by the free
electrons in a metal.
(d) the energy band traps the incident.
46.Assuming that the silicon diode having resistance of
20
W, the current through the diode is
(knee voltage 0.7 V)
0 V2 V
W=180R
(a) 0 mA (b) 10 mA
(c) 6.
5 mA (d) 13.5 mA
47.The following configuration of gate is equivalent to
OR
NAND
AND
Y
A
B
(a) NAND gate (b) XOR g
ate
(c) OR gate (d) NOR gate
48.Two junction diodes one of Germanium (Ge) and other of
silicon (Si) are connected as shown in figure to a battery of
emf 12 V and a load resistance 10 k W. The germanium diode
conducts at 0.3 V and silicon diode at 0.7 V. When a current
flows in the circuit, the potential of terminal Y will be
12 V
Ge
Si
Y
Wk10
(a) 12 V (b) 11 V
(c
) 11.3 V (d) 11.7 V
49.The current gain
b may be defined as
(a) the ratio of change in collector current to the change
in emitter current for a constant collector voltage in a
common base arrangement.
(b) the ratio of change in collector current to the change
in the base current at constant collector voltage in a
common emitter circuit
(c) the ratio of change in emitter current to the change in
base current for constant emitter voltage in common
emitter circuit.
(d) the ratio of change in base current to the change in
collector current at constant collector voltage in
common emitter circuit.
50.A solid that is not transparent to visible light and whose
electrical conductivity increases with temperature is formed
by
(a) ionic binding (b) covalent binding
(c) metallic binding(d) vander Waal’s binding
51.In a transistor, the change in base current from 100 µA to
125 µA causes a change in collector current from 5 mA to
7.5 mA, keeping collector-to-emitter voltage constant at 10
V. What is the current gain of the transistor?
(a) 200 (b) 100
(c) 50 (d) 25.
52.The truth table given below is for
(a) NOR
(b) AND
(c) XOR
ABY
001
011
101
110(d) NAND
53.In the fo
llowing circuit, the output Y for all possible
inputs A and B is expressed by the truth table.
A
B
B
A
Y
(a)ABY
001
011
101
110
(b)ABY
001
010
100
110
(c)ABY
000
011
101
111
(d)ABY
000
010
100
111

773. 769Semiconductor Electronics : Materials, Devices and Simple Circuits
54.In common emitter amplifier, the current gain is 62. The
collector resistance and input resistance are 5 kW an 500W
respectively. If the input voltage is 0.01 V, the output voltage
is
(a) 0.62 V (b) 6.2 V
(c) 62 V (d) 620 V
55.The real time variation of input signals A and B are as shown
below. If the inputs are fed into NAND gate, then select the
output signal from the following.
A
B
t (s)

A
B
Y
(a)
Y
t (s)
02468
(b)
Y
t (s)
02468
(c)Y
t (s)
02468
(d)Y
t (s)
02468
56.The time variations of signals are given as in A, B and C.
Point out the true statement from the following :
t
e
1.0
0
(A)

t
e
1.0
0
(B)

t
e
1.0
0
(C)
(a) A, B and C are analogue signals
(b) A and B are analogue, but C is digital signal
(c) A and C are digital, but B is analogue signal
(d) A and C are analogue, but B is digital signal
57.If the ratio of the concentration of electrons to that of holes
in a semiconductor is
5
7
and the ratio of currents is
4
7
,
then wh
at is the ratio of their drift velocities?
(a)
8
5
(b)
5
4
(c)
4
5
(d)
7
4
58.The circuit has two oppositively connected ideal diodes in
parallel. What is the current flowing in the circuit?
4W
D
1
D
2
2W3W
12V
(a) 1.71 A (b)
2.00 A
(c) 2.31 A (d) 1.33 A
59.In the energy band diagram of a material shown below, the
open circles and filled circles denote holes and electrons
respectively. The material is
E
g
E
v
E
c
(a) an in
sulator
(b) a metal
(c) an n-type semiconductor
(d) a p-type semiconductor
60.In a P -N junction
(a) the potential of P & N sides becomes higher
alternately
(b) the P side is at higher electrical potential than N side.
(c) the N side is at higher electric potential than P side.
(d) both P & N sides are at same potential.
61.In the case of a common emitter transistor amplifier the
ratio of the collector current to the emitter current I
c
/I
e
is
0.96. The current gain of the amplifier is
(a)6 (b) 48
(c) 24 (d) 12
62.Barrier potential of a P-N junction diode does not depend
on
(a) doping density (b) diode design
(c) temperature (d) forward bias

774. 770 PHYSICS
63.A n-p-n transist
or conducts when
(a) both collector and emitter are negative with respect
to the base
(b) both collector and emitter are positive with respect to
the base
(c) collector is positive and emitter is negative with
respect to the base
(d) collector is positive and emitter is at same potential
as the base
64.Following diagram performs the logic function of
Y
A
B
(a) XOR gate (b)
AND gate
(c) NAND gate (d) OR gate
65.Reverse bias applied to a junction diode
(a) increases the minority carrier current
(b) lowers the potential barrier
(c) raises the potential barrier
(d) increases the majority carrier current
66.In semiconductors at a room temperature
(a) the conduction band is completely empty
(b) the valence band is partially empty and the
conduction band is partially filled
(c) the valence band is completely filled and the
conduction band is partially filled
(d) the valence band is completely filled
67.The peak voltage in the output of a half-wave diode rectifier
fed with a sinusoidal signal without filter is 10V. The d.c.
component of the output voltage is
(a) 20/p V (b) 10/Ö2 V
(c) 10/p V (d) 10V
68.In a p-n junction photo cell, the value of the photo-
electromotive force produced by monochromatic light is
proportional to
(a) the voltage applied at the p-n junction
(b) the barrier voltage at the p-n junction
(c) the intensity of the light falling on the cell
(d) the frequency of the light falling on the cell
69.Of the diodes shown in the following diagrams, which one
is reverse biased ?
(a)
+5 V
+10 V
R
(b)
–12 V
–5 V
R
(c)
–10 V
R
(d)
+5 V
R
70.Choose the only false statement from the following.
(a) In conductors the valence and conduction bands may
overlap.
(b) Substances with energy gap of the order of 10 eV are
insulators.
(c) The resistivity of a semiconductor increases with
increase in temperature.
(d) The conductivity of a semiconductor increases with
increase in temperature.
71.Which one of the following statement is false ?
(a) Pure Si doped with trivalent impurities gives a p-type
semiconductor
(b) Majority carriers in a n-type semiconductor are holes
(c) Minority carriers in a p-type semiconductor are
electrons
(d) The resistance of intrinsic semiconductor decreases
with increase of temperature
72.A common emitter amplifier has a voltage gain of 50, an
input impedance of 100W and an output impedance of
200W. The power gain of the amplifier is
(a) 500 (b) 1000
(c) 1250 (d) 50
73.For transistor action
(1) Base, emitter and collector regions should have similar
size and doping concentrations.
(2) The base region must be very thin and lightly doped.
(3) The eimtter-base junction is forward biased and base-
collector junction is reverse based.
(4) Both the emitter-base junction as well as the base-
collector junction are forward biased.
(a) (3) and (4) (b) (4) and (1)
(c) (1) and (2) (d) (2) and (3)
74.The following figure shows a logic gate circuit with two
inputs A and B and the output Y. The voltage waveforms of
A, B and Y are given
Logic gate
circuit
A
B
Y
A
B
1
0
1
0
1
0Y
t
1
t
2
t
3
t
4
t
5
t
6
The logic gate is :
(a) NAND
gate (b) NOR gate
(c) OR gate (d) AND gate

775. 771Semiconductor Electronics : Materials, Devices and Simple Circuits
75.A transistor is operated in common emitter configuration at
V
C
= 2V such that a change in the base current from 100 mA
to 300 mA produces a change in the collector current from
10mA to 20 mA. The current gain is
(a) 50 (b) 75
(c) 100 (d) 25
76.Symbolic representation of four logic gate are shown as
(i) (ii)
(iii) (iv)
Pick out which ones are for AND, NAND and NOT gates,
respectively
(a) (ii), (iii) and (iv)(b) (iii), (ii) and (i)
(c) (iii), (iii) and (iv) (d) (ii), (iv) and (iii)
77.Pure Si at 500K has equal number of electron (n
e
) and hole
(n
h
) concentrations of 1.5 × 10
16
m
–3
. Doping by indium
increases n
h
to 4.5 × 10
22
m
–3
. The doped semiconductor is
of
(a) n–type with electron concentration
n
e
= 5 × 10
22
m
–3
(b) p–type with electron concentration
n
e
= 2.5 ×10
10
m
–3
(c) n–type with electron concentration
n
e
= 2.5 × 10
23
m
–3
(d) p–type having electron concentration
n
e
= 5 × 10
9
m
–3
78.A zener diode, having breakdown voltage equal to 15V, is
used in a voltage regulator circuit shown in figure. The
current through the diode is
250W
1kW15V20V
(a) 10 mA (b) 1
5 mA
(c) 20 mA (d) 5 mA
79.Two ideal diodes are connected to a battery as shown inthe circuit. The current supplied by the battery is
D
1
D
2
20W
5V
A
C
E
B
D
F
10W
(a) 0.75 A (b) zero
(c) 0.25 A (d) 0.5 A
80.In a CE transistor amplifier, the audio signal voltage across
the collector resistance of 2kW is 2V. If the base resistance
is 1kW and the current amplification of the transistor is
100, the input signal voltage is
(a) 0.1 V (b) 1.0 V
(c) 1 mV (d) 10 mV
81.Transfer characteristics [output voltage (V
0
) vs input
voltage (V
1
)] for a base biased transistor in CE configuration
is as shown in the figure. For using transistor as a switch, it
is used
V
0
III
II
I
V
i
(a) in region (III)
(b) both in region (I) and (III)
(c) in region (II)
(d) in region (I)
82.The figure shows a logic circuit with two inputs A and B
and the output C. The voltage wave forms across A, B and
C are as given. The logic gate circuit is:
A
B
C
t
1
t
2
t
3
t
4
t
5
t
6
(a) OR gate (b) NOR gate
(c) AND gate (d) NAND gate
83.The input resistance of a silicon transistor is
100 W. Base current is changed by 40 mA which results in a
change in collector current by 2 mA. This transistor is used
as a common emitter amplifier with a load resistance of 4
KW. The voltage gain of the amplifier is
(a) 2000 (b) 3000
(c) 4000 (d) 1000

776. 772 PHYSICS
84.In a n-type sem
iconductor, which of the following statement
is true?
(a) Electrons are minority carriers and pentavalent atoms
are dopants.
(b) Holes are minority carriers and pentavalent atoms are
dopants.
(c) Holes are majority carriers and trivalent atoms are
dopants.
(d) Electrons are majority carriers and trivalent atoms are
dopants.
85.In a common emitter (CE) amplifier having a voltage gain G,
the transistor used has transconductance 0.03 mho and
current gain 25. If the above transistor is replaced with
another one with transconductance 0.02 mho and current
gain 20, the voltage gain will be
(a) 1.5 G (b)
1
3
G
(c)
5
4
G (d)
2
3
G
86.In the study of tr
ansistor as amplifier, ifC
E
I
I
a= and
C
B
I
I
b= , where, I
C
, I
B
and I
E
are the collector,,
base and emitter currents, then
(a)
a
a+
=b
)1(
(b)
a
a-
=b
)1(
(c)
)1(a-
a
=b
(d)
)1(a+
a
=b
87.An oscillator i
s nothing but an amplifier with
(a) positive feedback(b) large gain
(c) no feedback (d) negative feedback
88.A TV tower has a height of 100 m. How much population is
covered by the TV broadcast if the average population
density around the tower is 1000 km
–2
? (radius of the earth
= 6.37 × 10
6
m)
(a) 4 lakh (b) 4 billion
(c) 40,000 (d) 40 lakh
89.The energy gap of silicon is 1.14 eV. The maximum
wavelength at which silicon starts energy absorption, will
be (h = 6.62 × 10
–34
Js ; c = 3 × 10
8
m/s)
(a) 10.888 Å (b) 108.88 Å
(c) 1088.8 Å (d) 10888 Å
90.The value of
b
(a) is always less than 1
(b) lies between 20 and 200
(c) is always greater than 200
(d) is always infinity
91.In a bridge rectifier, the number of diodes required is
(a)1 (b) 2
(c)3 (d) 4
92.The ratio of forward biased to reverse biased resistance for
pn junction diode is
(a) 10
–1
: 1 (b) 10
–2
: 1
(c) 10
4
: 1 (d) 10
–4
: 1
93.In germanium the energy gap is about 0.75 eV. The
wavelength of light which germanium starts absorbing is
(a) 5000 Å (b) 1650 Å
(c) 16500 Å (d) 165000 Å
Directions for Qs. (94 to 100) : Each question contains
STATEMENT-1 and STATEMENT-2. Choose the correct answer
(ONLY ONE option is correct ) from the following-
(a) Statement -1 is true, Statement-2 is true; Statement -2 is a
correct explanation for Statement-1
(b) Statement -1 is true, Statement-2 is true; Statement -2 is not
a correct explanation for Statement-1
(c) Statement -1 is true, Statement-2 is false
(d) Statement -1 is false, Statement-2 is true
94. Statement-1 : NAND or NOR gates are called digital building
blocks.
Statement-2 : The repeated use of NAND (or NOR) gates
can produce all the basis or complicated gates.
95. Statement-1 : When two semi conductor of p and n type
are brought in contact, they form p-n junction which act
like a rectifier.
Statement-2 : A rectifier is used to convent alternating
current into direct current.
96. Statement-1 : NOT gate is also called invertor circuit.
Statement-2 : NOT gate inverts the input order.
97. Statement 1 : Diode lasers are used as optical sources in
optical communication.
Statement 2 : Diode lasers consume less energy.
98. Statement 1 : A pure semiconductor has negative
temperature coefficient of resistance.
Statement 2 : In a semiconductor on raising the temperature,
more charge carriers are released, conductance increases
and resistance decreases.
99. Statement 1 : A transistor amplifier in common emitter
configuration has a low input impedence.
Statement 2 : The base to emitter region is forward biased.
100. Statement 1 : If the temperature of a semiconductor is
increased then it’s resistance decreases.
Statement 2 : The energy gap between conduction band
and valence band is very small.

777. 773Semiconductor Electronics : Materials, Devices and Simple Circuits
Exemplar Questions
1.The conductivity of a semiconductor increases with
increase in temperature because
(a) number density of free current carries increases
(b) relaxation time increases
(c) both number density of carries and relaxation time
increase
(d) number density of carries increases, relaxation time
decreases but effect of decrease in relaxation time is
much less than increase in number density
2.In figure given below V
0
is the potential barrier across a
p–n junction, when no battary is connected across the
junction
1
2
3
V
0
(a) 1 an
d 3 both correspond to forward bias of junction
(b) 3 corresponds to forward bias of junction and 1
corresponds to reverse bias of junctions
(c) 1 corresponds to forward bias and 3 corresponds to
reverse bias of junction
(d) 3 and 1 both correspond to reverse bias of junction
3.In figure given below, assuming the diodes to be ideal(a)D
1
is forward biased and D
2
is reverse biased and hence
current flows from A to B
(b) D
2
is forward biased and D
1
is reverse biased and
hence no current flows from B to A and vice–versa
(c)D
1
and D
2
are both forward biased and hence current
flows from A to B
(d) D
1
and D
2
are both reverse biased and hence no current
flows from A to B and vice – versa.
A
–10V
R
D
1
D
2
B.
4.A 220 V AC s
upply is connected between points A and B
(figure). What will be the potential difference V across thecapacitor? A
B
C V200 AC
(a) 220 V ( b) 110 V
(c) 0 V (d)220 2V
5.Hole in semiconductor is
(a) an anti – particle of electron
(b) a vacancy created when an electron leaves a covalent
bond
(c) absence of free electrons
(d) an artificially created particle
6.The output of the given circuit in figure given below,
m
vsintw
(a
) would be zero at all times
(b) would be like a half wave rectifier with positive cycles
in output
(c) would be like a half wave rectifier with negative cycles
in output
(d) would be like that of a full wave rectifier
7.In the circuit shown in figure given below, if the diode
forward voltage drop is 0.3 V, the voltage difference between
A and B is
(a) (b)
0.3V
r
1
r
25kW
A
0.2mA
5K
B
5K
A
0.2mA
5K
B
(a) 1.3 V (
b) 2.3 V
(c)0 (d) 0.5 V
8.Truth table for the given circuit is
..
..A
B
C
D
E
(a)ABE
001
010
101
110
(b)ABE
001
010
100
110
(c)ABE
000
011
100
111
(d)ABE
000
011
101
110

778. 774 PHYSICS
NEET/AIPMT (2013-2017) Questions
9.In a n-type semiconductor, which of the following
statements is true? [2013]
(a) Electrons are minority carriers and pentavalent atoms
are dopants.
(b) Holes are minority carriers and pentavalent atoms are
dopants.
(c) Holes are majority carriers and trivalent atoms are
dopants.
(d) Electrons are majority carriers and trivalent atoms are
dopants.
10.In a common emitter (CE) amplifier having a voltage gain G,
the transistor used has transconductance 0.03 mho and
current gain 25. If the above transistor is replaced with
another one with transconductance 0.02 mho and current
gain 20, the voltage gain will be 2013]
(a) 1.5 G (b)
1
3
G
(c)
5
4
G (d)
2
3
G
11.The output(X) of the logic circuit shown in figure will be
[2013]
(a) X = .AB (b) X = A.B
(c) X = AB+ (d) X = ,AB
12.In an unbiased p-n junction, holes diffuse from the p-region
to n-region because of [NEET Kar. 2013]
(a) the potential difference across the p-n junction
(b) the attraction of free electrons of n-region
(c) the higher hole concentration in p-region than that in
n-region
(d) the higher concentration of electrons in the n-region
than that in the p-region
13.One way in which the operation of a n-p-n transistor differs
from that of a p-n-p [NEET Kar. 2013]
(a) the emitter junction is reversed biased in
n-p-n
(b) the emitter junction injects minority carriers into the
base region of the p-n-p
(c) the emitter injects holes into the base of the p-n-p and
electrons into the base region of n-p-n
(d) the emitter injects holes into the base of
n-p-n
14.The output from a NAND gate is divided into two in parallel
and fed to another NAND gate. The resulting gate is a
[NEET Kar. 2013]
A
B
C
C¢
(a) NOT gate (b) AND gate
(c) NOR gate (d) OR gate
15.The given graph represents V - I characteristic for a semiconductor device.
A
B
I
V
Which of the following statement is correct ? [2014]
(a) It is V - I characteristic for solar cell where, point A
represents open circuit voltage and point B short
circuit current.
(b) It is a for a solar cell and point A and B represent open
circuit voltage and current, respectively.
(c) It is for a photodiode and points A and B represent
open circuit voltage and current, respectively.
(d) It is for a LED and points A and B represent open
circuit voltage and short circuit current, respectively.
16.The barrier potential of a p-n junction depends on: [2014]
(A) type of semi conductor material
(B) amount of doping
(C) temperature
Which one of the following is correct ?
(a) (A) and (B) only (b) (B) only
(c) (B) and (C) only (d) (A), (B) and (C)
17.Which logic gate is represented by the following
combination of logic gate ? [2015]
A
Y
1
B
Y
2
Y
(a) NAND (b) AND
(c) NOR (d) OR
18.If in a p-n junction, a square input signal of 10 V is applied
as shown, then the output across R
L
will be [2015]
+5V
–5V
R
L
(a)
10V
(b)
–5V
(c)
5V
(d)
–10V

779. 775Semiconductor Electronics : Materials, Devices and Simple Circuits
19.In the given figure, a diode D is connected to an external
resistance R = 100 W and an e.m.f. of 3.5 V. If the barrier
potential developed across the diode is 0.5 V, the current in
the circuit will be: [2015 RS]
R = 100W
D
3.5V
(a) 40 mA (b) 2 0 mA
(c) 35 mA (d) 30 mA
20.The input signal given to a CE amplifier having a voltage
gain of 150 is V
i
= 2 cos 15t
3
pæö
+ç÷
èø
. The corresponding
output signal will be : [2015 RS]
(a)
2
75cos 15t
3
pæö
+ç÷
èø
(b)
5
2cos 15t
6
pæö
+ç÷
èø
(c)
4
300cos 15t
3
pæö
+ç÷
èø
(d)300cos 15t
3
pæö
+ç÷
èø
21.Consider the junction diode as ideal. The value of current
flowing through AB is : [2016]
B
–6V
1kWA
+4V
(a) 0 A (b) 10
–2
A
(c)
10
–1
A (d) 10
–3
A
22.A npn transistor is connected in common emitter
configuration in a given amplifier. A load resistance of 800
W is connected in the collector circuit and the voltage drop
across it is 0.8 V. If the current amplification factor is 0.96
and the input resistance of the circuit is 192W, the voltage
gain and the power gain of the amplifier will respectively be:
(a) 4, 3.84 (b) 3.69, 3.84 [2016]
(c) 4, 4 (d) 4, 3.69
23.To get output 1 for the following circuit, the correct choice
for the input is [2016]
A
Y
B
C
(a) A = 0, B = 1, C = 0 (b) A = 1, B = 0, C = 0
(c) A = 1, B = 1, C = 0 (d) A = 1, B = 0, C = 1
24.In a common emitter transistor amplifier the audio signalvoltage across the collector is 3V. The resistance of collectoris 3 kW. If current gain is 100 and the base resistance is 2
kW, the voltage and power gain of the amplifier is [2017]
(a) 15 and 200 (b) 150 and 15000
(c) 20 and 2000 (d) 200 and 1000
25.The given electrical network is equivalent to : [2017]
A
B
Y
(a) OR gate (b) NOR gate
(c) NOT gate (d) AND gate
26.Which one of the following represents forward bias diode?
(a)
–4V R
–3V
[2017]
(b)
–2V R
+2V
(c)
3V R
5V
(d)
0V R
–2V

780. 776 PHYSICS
EXERCISE - 1
1. (c) 2. (c) 3
. (d) 4. (c) 5. (b)
6. (c) Field effect tansistor is unipolar in nature because in
this either holes or electrons are present. But in case
of BJT both holes and electrons are present due to
which they are bipolar in nature.
7. (d)
8. (c) The resistivity of conductor increases with increase
in temperature. The resistivity of semiconductor
decreases as the temperature increases
9. (c) Maximum in insulators and overlaping in metals
10. (b)
11. (a)
12. (a) Minority carriers in a p-type semiconductor are free
electrons.
13. (d)
14. (b) [Hint : At 0K (–273ºC) motion of free electron stop i.e.,
there is no electron in conduction band therefore at
0K intrinsic semiconductor becomes insulator.]
15. (b) Combination of NAND & NOR gates can produce OR,
AND & NOT gates
16. (c)
17. (d) Radiowaves of constant amplitude can be produced
by using oscillator with proper feedback.
18. (c) Zener diode is used as a voltage regulator i.e. for
stabilization purposes.
19. (d) 20. (d)
21. (a) Metal oxides with high temperature coefficient of
resistivity.
22. (b) Integrated circuit can act as a complete electronic
circuit.
23. (d) In forward biasing of the p-n junction, the positive
terminal of the battery is connected to p-side and the
negative terminal of the battery is connected to n-
side. The depletion region becomes thin.
24. (c) When small amount of antimony (pentavalent) is added
to germanium crystal then crystal becomes n-type semi
conductor. Therefore, there will be more free electrons
than holes in the semiconductor.
25. (c) Semiconductors are insulators at low temperature
EXERCISE - 2
1. (b) Cond
uctivity 3800)106.1(10en
1917
ei
´´´=m=s
-
= 60.8 mho/cm
2. (b)
d
I nA ev= or
d
I nvµ
\
e ee
h hh
I nv
I nv
=
or
ee h
hhe
nI v747
n I v 455
= ´ =´=
3. (b) In half wave rectifier, we get the output only in one half
cycle of input a.c. therefore, the frequency of the rippleof the output is same as that of input a.c. i.e. 50 Hz
4. (d) In full wave rectifier, we get the output for the positive
and negative cycle of input a.c. Hence the frequency ofthe ripple of the output is twice than that of input a.c.i.e. 100 Hz
5. (a) [Hint
Þ for B.C.C cell a
4
3
r=
so distance betw
een body centered atom & a corner
atom is
ºA66.32a
4
3
r2 =´´=]
6. (c) For
ward resistance =
3
V 0.7 0.5
200
I1.0 10
-
D-
= =W
D ´
7. (c) 19
05.0
95.0
95.01
95.0
1
==
-
=
a-
a
=b
8. (a)
C
B
I
I
D
b=
D
or
CB
I I 40 100 AD=bD= ´ m
9. (c)I
C
= I
E
–I
B
= 90 – 1 = 89 m A
10. (c) Voltage amplification
o
v
i
R 5000
A 60 600
R 500
=b=´=
11. (
c) Current gain,
V
R
A 2800
0.93
A 3000
a===
12. (
a) The phase difference between output voltage and input
signal voltage in common base transistor circuit is zero
13. (c)
CE
I I 0.99 5 4.95m AD=aD= ´=
14. (
c) Band gap,
eV
106.1102480
)103()106.6(hc
E
199
834
g
--
-
´´´
´´
=
l
= = 0.49 eV
15
. (b) Voltage gain,
L
v
i
R 24
A 0.6 4.8
R3
=b =´=
Hints & Solutions

781. 777Semiconductor Electronics : Materials, Devices and Simple Circuits
16. (b) In half wave rectifier only half of the wave is rectified
17. (b)
R)g/(
R
Rr
R
A
mp
+m
m
=
+
m
=
n
= 57.29
1050)102/(42
)1050(42
33
3
=
´+´
´´
-
18. (c) m/V10
10
1.0
d
V
E
5
6
===
-
19. (b)
20
. (c) We know that for common base
c
e
icollectorcurrent
0.98
i emiter current
a===
& for common emitter
currentbase
currentcollector
i
i
b
c
==b
.49
02.0
98.0
98.01
98.0
1
==
-
=
a-
a
=b
21. (c)
The given circuit is a circuit of full wave rectifier.
22. (d) During the formation of a junction diode, holes from p-
region diffuse into n-region and electrons from n-region
diffuse into p-region. In both cases, when an electrons
meets a hole, they cancel the effect at each other and as
a result, a thin layer at the junction becomes free from
any of charges carriers. This is called depletion layer.
There is a potential gradient in the depletion layer,
negative on the p-side, and positive on the n-side. The
potential difference thus developed across the junction
is called potential barrier.
23. (a) Conductivity,
ee hh
1
σ= =e(nμ+nμ)
ρ
ie, 2.13 = 1.6 × 10
–19
(0.38 + 0.18) n
i
(S
ince in intrinsic semi-conductor, n
e
= n
h
= n
i
)
\ density of charge carriers, n
i
193
19
2.13
2.37 10 m
1.6 10 0.56
-
-
= =´
´´
.
24. (b
) Reverse resistance
6
6
V1
2 10
I0.5 10
-
D
= = =´W
D ´
25. (c
) Forward bias resistance
V
I
D
=
D
( )
3
(0.7 0.6)V
0.1
10.
15 5 mA 10 10
-
-
= = =W
- ´
26. (d)
Current gain in common emitter mode0.995 0.995
199.
1 1 0.995 0. 005
a
== ==
-a-
27. (a)
28. (b) Given : µ
e
= 2.3 m
2
V
–1
s
–1
µ
h
= 0.01 m
2
V
–1
s
–1
, n
e
= 5 × 10
12
/ cm
3
= 5 × 10
18
/m
3
n
h
= 8 × 10
13
/cm
3
= 8 × 10
19
/m
3
.
Conductivity s = e[n
e
µ
e
+ n
h
µ
h
]
= 1.6 × 10
–19
[5 × 10
18
× 2.3 + 8 × 10
19
× 0.01]
= 1.6 × 10
–1
[11.5 + 0.8]
= 1.6 × 10
–1
× 12.3 = 1.968 W
–1
m
–1
.
29. (b)
E
hc
=l =
m1007.2
)106.11060(
1031062.6 5
193
834
-
--
-
´=
´´´
´´´
30. (c)
No. of electrons reaching the collector,
1010
C 1096.010
100
96
n ´=´=
Emitter curr
ent,
E
E
ne
I
t
´
=
Collecto
r current,
C
C
ne
I
t
´
=
\Current
transfer ratio,
CC
EE
In
In
a== = 96.0
10
1096.0
10
10
=
´
31. (b)
g
p
m
V
I
g
D
D
=
or
4
pmg
I g V 3 10 [ 3 ( 1)]
-
D = ´D = ´ ´---
= –0.6 × 10
–3
A = shortage of 0.6 × 10
–3
A
32. (d)
22
11
kT/W2
2
kT/W2
1
2
1
eAT
eAT
J
J
-
-
=
=
4
1
e
2
1
e
T
T
0
2
kT
W
kT
W
2
2
1 2
2
1
1
=÷
ø
ö
ç
è
æ
=
÷
÷
ø
ö
ç
ç
è
æ
+-
33. (d) [
Hint
Þ
5s
B
3
in
V0.01
i 110A
R 10
-
= = =´

782. 778 PHYSICS
Now b of tran
sistor is defined as
b
c
ac
i
i
=b
or A5001050i
5
c
m=´=
-
]
3
4. (b) Pauli’s exclusion principle.
35. (b)
36. (c)'V = V + IR = 0.5 + 0.1 × 20 = 2.5 V
W20
V
0.5V
0.1A
37. (b) Give
n interatomic spacing = 2r = 2.54 Å
4r =
2a, where a is lattice constant
2r =
2
2
a
=
a
2
a = 2r 2 = (2.54 Å)(1.414) = 3.59 Å
38. (a) Du
e to strong electronegativity of carbon.
39. (c)
P N

Number of donors is more because electrons from –
ve terminal of the cell pushes (enters) the n side and
decreases the number of uncompensated
pentavalent ion due to which potential barrier is
reduced. The neutralised pentavalent atom are again
in position to donate electrons.
40. (a) In reverse bias the current through the p-n junction is
almost zero.
41. (d) (A) is a NAND gate so output is
0111==´
(B) is a NOR
gate so output is
0110+==
(C) is a NAND gate so output is 1010==´
(D) is a XOR
gate so output is 000
=Å
0
1
1
Following is NAND Gate ABY=
42. (b) Output of upper AND gate =
BA
Output of lower AND gate =
AB
\ Output of OR gate, ABBAY +=
This is boolean expression for XOR gate.
43. (c) (W + X) . (W + Y) = W + (X . Y)
44. (b)
3
aN
AZ
d
´
´
=
For bcc, Z
= 2
3
Na
A2
d=
45. (c)
46. (c
) Here diode is forward biased with
voltage = 2 – 0 = 2 V.
V
B
= V
knee
+ IR
2 = 0.7 + I× 200
(\ Total resistance = 180 + 20 = 200W)
1.3
I 6.5mA
200
\==
47. (b)
Y
Y
1
Y
2
A
B
B.AY,BAY
21 =+=

783. 779Semiconductor Electronics : Materials, Devices and Simple Circuits
Y (A B) AB=+ g AA AB BA BB= +++gggg
0ABBA0ABB A=+++=+gg gg
This expression is for XOR
48. (d) 49. (b) 50. (b)
51. (b) Current gain
C
B
I
I
D
=
D
when V
CE
is con stant.
3
3
6
2.5 10
0.1 10 100
25 10
-
-
´
= =´=
´
[DI
B
= 1
25 µA – 100 µA = 25 µA
DI
C
= 7.5 mA – 5 mA = 2.5 mA]
52. (d) The given truth table is for NAND gate.
53. (c)
A
B
B
A
Y
Y'
Y ' A B. Y A B A B.=+ =+=+
Truth table of the given circuit is given by
ABY'Y
0010
0101
1001
1101
54. (b)
3
oo
in in
VR 5 10 62
10 62 620
V R 500
´´
= ´b=
=´=
V
o
= 620 × V
in
= 620 × 0.01 = 6.2 V
\ V
o
= 6.2 volt.
55. (b) From input signals, we have,
A B Output NAND gate
001
101
001
110
001
T
he output signal is shown at B.
56. (d) A and C are analogue but B is digital signal.
57. (c)
h
e
hh
ee
h
e
v
v
5
7
4
7
eAvn
eAvn
I
I
´=Þ=
4
5
v
v
h
e
=Þ
58. (b) D
2
is fo
rward biased whereas D
1
is reversed biased.
So effective resistance of the circuit
R = 4 + 2 = 6W
A2
6
12
i ==\
59. (d) For a p
-type semiconductor, the acceptor energy level,
as shown in the diagram, is slightly above the top E
v
of the valence band. With very small supply of energy
an electron from the valence band can jump to the
level E
A
and ionise acceptor negatively.
60. (b) [Hint : For easy flow of current the P side must be
connected to +ive terminal of battery i.e., it is
connected to higher potential in comparison to N. This
connection is called forward biased. In this case the
input resistance is very low.
In reverse-biased, the P-side is connected to –ive
terminal & N side to (+ive) terminal to battery. In this
case input resistance is very high.
V
I
forward
biased
reverse
biased
V
B
V
T
B
re a
k d
o w
n
v o lta g
e
61. (c)
96.0
I
I
e
c
=

ec
I96.0I=Þ
But
bebce
II96
.0III+=+=
Þ
eb
I04.0I=
\ Current gai
n
24
I04.0
I96.0
I
I
e
e
b
c
===b
62. (b)
[Hint : Barrier potential depends on, doping density,
temperature, forward/reverse bias but does not depend
on diode design.]
63. (c) When the collector is positive and emitter is negative
w.r.t. base it causes the forward biasing for each
junction, which causes conduction of current.
64. (b)
A
B
X
Y
ABX=
ABXY ==\
Y = AB by Demorgan theorem
\ This diagram performs the function of AND gate.
65. (c) In reverse biasing, the conduction across the p-n junction

784. 780 PHYSICS
does not take
place due to majority carriers, but takes
place due to minority carriers if the voltage of external
battery is large. The size of the depletion region increases
thereby increasing the potential barrier.
66. (c) In semiconductors, the conduction band is empty and
the valence band is completely filled at 0 K. No
electron from valence band can cross over to
conduction band at 0 K. But at room temperature some
electrons in the valence band jump over to the
conduction band due to the small forbidden gap, i.e.,
1 eV.
67. (c)
o
V10
VV==
pp
68. (c) Electromotive force depends upon intensity of light
falling, it does not depend on frequency of barriervoltage.
69. (d) Positive terminal is at lower potential (0V) and negative
terminal is at higher potential 5V.
70. (c) Semiconductors have –ve temperature coefficient of
resistivity.
71. (b) Majority carriers in an n-type semiconductor are
electrons.
72. (c) Power gain = voltage gain × current gain
=
00
GG
ii
VI
VI
VI
×=×
=
2
0
2
0
i
i
VR
RV
× =
100
50 50
200
´´
=
2500
1250
2
=
73. (d) For tr
ansistor action, the base region must be very
thin and lightly doped. Also, the emitter-base junctionis forward biased and base-collector junction is reversebiased.
74. (a) From the given waveforms, the truth table is as follows.
110
001
011
1 01
ABY
The above truth table is for NAND gate.
Therefore, the logic gate is NAND gate.
75. (a) The current gain
b =
3
C
B
I10mA 10 10
50
I 200 A 200
D ´
===
Dm
76. (d)
77. (d)n
i
2
= n
e
n
h
(
1.5 × 10
16
)
2
= n
e
(4.5 × 10
22
)
Þ n
e
= 0.5 × 10
10
or n
e
= 5 × 10
9
Given n
h
= 4.5 × 10
22
Þ n
h
>> n
e
\ Semiconductor is p-type and
n
e
= 5 × 10
9
m
–3
.
78. (d) Voltage across zener diode is constant.
250W
1kW
ii
5V
20v
15V
1kW
i–i
1kW
15V
Current in 1kW re sistor,
(i)
1kW
=
15volt
1kW
= 15 mA
Current in
250W resistor,
(i)
250W
=
(20 15)V 5V
250 250
-
=
WW

20
A 20 mA
1000
==
\ zener diode
(i) (20 15) 5mA.= -=
79. (d) Her
e D
1
is in forward bias and D
2
is in reverse bias so,
D
1
will conduct and D
2
will not conduct. Thus, no
current will flow through DC. 51
Amp.
102
===
V
I
R
80. (d)
InputR
B
R
C
I
C
OP/ = 2 Volt
The output voltage, across the load R
C

785. 781Semiconductor Electronics : Materials, Devices and Simple Circuits
V
0
= I
C
R
C
= 2
The collector current (I
C
)
3
3
2
10 Amp
2 10
-
==
´
C
I
Current
gain (b)
(b) current gain =
100=
C
B
I
I
3
510
10 Amp
100 100
-
-
===
C
B
I
I
In
put voltage (V
i
)
V
i
= R
B
I
B
= 1 × 10
3
× 10
–5
= 10
–2
Volt
V
i
= 10 mV
81. (b) I ® ON
II ® OFF
In II
nd
state it is used as a amplifier it is active region.
82. (a)
A 0 1 1 0
B 0 0 1 1
C 1 1 1 1
OR gate
83.
(a) Voltage gain (A
V
) =
out out out
in in in
=×
VIR
VIR
A
V

–33
6
2 10 4 10
10040 10
-
´´
=´
´
= 2 × 100 = 2000
8
4. (b) In a n-type semiconductor holes are minority carriers
and pentavalent atoms are dopants.
85. (d) Voltage gain D
Ú
= b
out
in
R
R
Þ G = 25
out
in
R
R
...(i)
Transcon
ductance g
m
=
in
R
b
Þ R
in
=
m
g
b
=
25
0.03
Putting this value of R
in
in eqn. (i)
G = 25
out
R
25
× 0.03 ...(ii)
\G' = 2
0
out
R
20
× 0.02 ...(iii)
From eqs. (i
i) and (iii)
Voltage gain of new transistor G' =
2
3
G
86. (c
) As we know that
e cb
I II=+
Divide both side by I
e
11
11
eb
cc
II
II
=+ Þ =+
ab
a-
a
=b
1
87. (a)
88
. (d)
2d hR=
Popula
tion covered
=
p d
2
× population density
= 3.14 × 2hR × 1000 × (10
–3
)
2
= 3.14 × 2 × 100 × 6.37 × 10
6
× 1000 × 10
–6
= 40 × 10
5
= 40 lakhs.
89. (a)
E
hc
=l =
Å888.10
)106.114.1(
1031062.6
19
834
=
´´
´´´
-
-
90. (b) 9
1. (d) 92. (d)
93. (c)
l
=
hc
E
Å16500
106.175.0
1031063.6
E
hc
19
834
=
´´
´´´
==lÞ
-
-
94. (a
) These gates are called digital building blocks because
using these gates only (either NAND or NOR) we can
compile all other gates also (like OR, AND, NOT, XOR)
95. (b) Study of junction diode characteristics shows that
the junction diode offers a low resistance path, when
forward biased and high resistance path when reverse
biased. This feature of the junction diode enables it to
be used as a rectifier.
96. (a) A NOT gate puts the input condition in the opposite
order, means for high input it give low output and for
low input it give high output. For this reason NOT
gate is known as invertor circuit.
97. (c) Statement - 1 is True, Statement- 2 is False
98. (a) In semiconductors, by increasing temperature,
covalent bond breaks and conduction hole and
electrons increase.
99. (a) Input impedance of common emitter configuration.
constant
BE
BV
CE
V
i
=
D
=
D
where DV
BE
= vo
ltage across base and emitter (base
emitter region is forward biased)

786. 782 PHYSICS
Di
B
= base curre
nt which is order of few microampere.
100. (a) In semiconductors the energy gap between conduction
band and valence band is small (»1 eV). Due to
temperature rise, electron in the valence band gain
thermal energy and may jumpy across the small energy
gap, (to the conduction band). Thus conductivity
increases and hence resistance decreases.
EXERCISE - 3
Exemplar Questions
1. (d) In semiconsuctor the density of change carriers
(electron hole) are very small, so its resistance is high
when the conductivity of a semiconductor increases
with increase in temperature, because the number
density of current carries increases then the speed of
free electron increase and relaxation time decreases
but effect of decrease in relaxation is much less than
increase in number density.
2. (b) When p–n junction is forward biased then the depleton
layer is compresses or decrease so it opposes the
potential junction resulting decrease in potential
barrier junction when p–n junction is reverese biased,
it supports the potential barrier junction, resulting
increase in potential across the junction.
3. (b) As the given circuit, p–side of p–n function D
1
is
connected to lower voltage and n –side of D
1
of higher
voltage.
So, D
1
is reverse biased.
In circuit A is at –10V and B is at 0 (zero) V. So B is
positive then A or
The p–side of p–n junction D
2
is at higher potential
and n–side of D
2
is at lower potential.
So, D
2
is forward biased.
Hence, No current flows through the junction B to A
and vice–versa.
4. (d) As the given figure p–n junction conducts during
positive half cycyle only, then diode connected here
will work is positive half cycle. Potential difference
across C will be peak voltage when diode is in forword
bias then the peak voltage of the given AC voltage
=
0 rms
V V 2 220 2V==
5. (b) Atom o
f semiconductor are bounded by covalent
bonds between the atoms of same or different type.The concept of hole describes the lack of an electronat a position where one could exist in an atom oratomic lattice. If an electron is excited into a higher
state, it leaves a hole in its old state. So, hole can be
defined as a vacancy created when an electron leaves
a covalent bond.
6. (c) When the diode will be in forward biased during
positive half cycle of input AC voltage, the resistance
of p–n junction is low. The current in the circuit is
maximum. So, a maximum potential difference will
appear across resistance connected in series of circuit.
So, potential across PN junction will be zero. When
the diode will be in reverse biase during negative half
cycle of AC voltage, the resistance of p–n juntion
becomes high which will be more than resistance in
series. So, there will be voltage across p–n junction
with negative cycle in output.
7. (b) Let the potential difference between A and B is V, Given
here r
1
= 5 kW and r
2
= 5 kW are resistance in series
connection.
So,
V
AB
– 0.3 = [(r
1
+ r
2
) 10
3
] × (0.2 × 10
– 3
)] [
QV = ir]
(V
AB
– 0.3) = 10 × 10
3
×
0.2 × 10
–3
= 2
So, V
AB
= 2 + 0.3 = 2.3 V
8. (c) As the given figure the output of C,D and E are :
C = A.B and D =
A.B
E = C + D = (A.B) + ()A.B
So, the truth table of given arrangement of gates can
be written as :
( )ABACA.BDA. BE CD
001 0 0 0
011 0 1 1
100 0
0 0
110 1 0 1
= = =+
NEET/AIPMT (2013-2017) Questions
9. (b) In a n-type semiconductor holes are minority carriers
and pentavalent atoms are dopants.
10. (d) Voltage gain D
Ú
= b
out
in
R
R
Þ G = 25
out
in
R
R
...(i)
Transconduct
ance g
m
=
inR
b
Þ R
in
=
m
g
b
=
25
0.03
Putting this value of R
in
in eqn. (i)
G = 25
out
R
25
× 0.03 ...(ii)
\G' = 20
out
R
20
× 0.02 ...(iii)

787. 783Semiconductor Electronics : Materials, Devices and Simple Circuits
From eqs. (ii) and (iii)
Voltage gain of new transistor G' =
2
3
G
11. (b)
i.e., output X = A.B
ALTERNATE :
ABX
000
100
010
111
12. (c) In p-region of p-n junction
holes concentration > electrons concentration and in
n-region
electrons concentration > holes concentration.
13. (c) In p-n-p transistor holes are injected into the base while
electrons are injected into the base of n-p-n transistor.
Emitter-base junction is forward biased.
14. (b)
C AB C AB AB=×Þ=×=×¢
Hence the resultant gate is AND gate.
15. (a) The given graph represents V-I characteristics of solar
cell.
16. (d) The barrier potential of a p-n junction depends on
amount of doping, type of semiconductor material and temperature.
17. (b) First two gates are NOT gates and the last gate is NOR
gate.
Thus, y
1
=
A, y
2
=
B
and y =
12
yy+
The truth table corresponding to this is as follows:
AB A . B
1
yA=
2
y B= 1 2yy+
1 2
yyy=+
00 1 1 1 0 0
01 1 0 1 0 0
10 0 1 1 0 0
11 0 0 0 1 1
Thus the combination of gate represents AND gate.
18. (c) Here P-N junction diode rectifies half of the ac wave
i.e., acts as half wave rectifier. During + ve half cycle
Diode ® forward biased output across will be
5V
During –ve half cycle Diode ® reverse biased output
will not obtained.
19. (d) Current I =
V
R
=
(3.5 0.5)
100
-
A
[Q Barrier potential V
B
= 0.5V]
=
3
100
= 30mA
20. (c)Given : Voltage gain A
V
= 150
V
i
= 2cos15t
3
pæö
+
ç÷
èø
; VV
0
= ?
For CE transistor phase difference between input and
output signal is p = 180°
Using formula, A
V
=
0
i
V
V
Þ V
0
= A
V
× V
i
= 150 × 2cos 15t
3
pæö
+
ç÷
èø

or V
0
= 300 cos15t
3
pæö
+ +p
ç÷
èø
V
0
= 300 cos
4
15t
3
æö
+p
ç÷
èø
21. (b) Since diode is in forward bias, so the value of current
flowing through AB
i =
()
33
V 4 6 10
R1 10 10
D --
==
´
= 10
–2
A
22. (a) Given: amplification factor a = 0.96
load resistance, R
L
= 800 W
input resistance, R
i
= 192W
So, b =
0.96
24
1 0.04
a
= Þ b=
-a
Voltage gain for common emitter configuration
A
v
= b
.

L
i
R
R
= 24 ×
800
192
= 100
Power gain for common emitter configuration

788. 784 PHYSICS
P
v
= bA
v
= 24 × 100 = 2
400
Voltage gain for common base configuration
A
v
= a,
L
P
R
R
= 0.96 ×
800
192
= 4
Power gain for
common base configuration
P
v
= A
v
a = 4 × 0.96 = 3.84
23. (d) The Boolen expression for the given combination is
output Y = (A + B) · C
Truth table
A B C Y = (A + B) · C
0 0 0 0
1 0 0 0
0 1 0 0
0 0 1 0
1 1 0 0
0 1 1 1
1 0 1 1
1 1 1 1
Hence, A = 1, B = 0, C = 1
24. (b) Given, current gain b = 100, R
c
= 3kW, R
b
= 2kW
Voltage gain (A
v
) =
c
b
R
R
b =
3
100
2
æö
ç÷
èø
= 150
Power gain =
A
v
b = 150 (100) = 15000
25. (b)
A
B
Y Y
2Y
1
y
1
=
AB+
y
2
=
11 1
y y y AB+ = =+
y =
2
y =
AB+
i.e. NOR gate
2
6. (d)
V
1
V
2
In forward bias, V
1
> V
2
i.e., in figure (d) p-type semi-
conductor is at higher potential w.r.t. n-type semicon-
ductor.