Jee Advance Tmh

JEE ADVANCED Comprehensive

Physics

2019

JEE ADVANCED

2019

Comprehensive

Physics

McGraw Hill Education (India) Private Limited CHENNAI

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McGraw Hill Education (India) Private Limited Published by McGraw Hill Education (India) Private Limited, 444/1, Sri Ekambara Naicker Industrial Estate, Alapakkam, Porur,Chennai - 600 116, Tamil Nadu, India Comprehensive Physics—JEE Advanced Copyright © 2018 by McGraw Hill Education (India) Private Limited No Part of this publication may be reproduced or distributed in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise or stored in a database or retrieval system without the prior written permission of the publishers. The program listings (if any) may be entered, stored and executed in a computer system, but they may not be reproduced for publication. This edition can be exported from India only by the publishers. McGraw Hill Education (India) Private Limited 1 2 3 4 5 6 7 8 9 D102542 22 21 20 19 18

ISBN (13) : 978-93-87572-57-7 ISBN (10) : 93-87572-57-9 Information contained in this work has been obtained McGraw Hill Education (India), from sources believed to be reliable. However, neither, McGraw Hill nor its authors guarantee the accuracy or completeness of any information published herein, and neither McGraw Hill Education (India) nor its authors shall be responsible for any errors, omissions, or damages arising out of use of this information. This work is published with the understanding that McGraw Hill Education (India) and its authors are supplying information but are not attempting to render engineering or other professional services. If such services are required, the assistance of an appropriate professional should be sought.

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A Word to the Reader Comprehensive Physics—JEE Advanced is highly useful for aspirants appearing for JEE Advanced. It is as per the syllabus of JEE Advanced. Divided into 29 core chapters, it covers the entire gamut of the subject through a combination of conceptual theory supported by solved problems and practice exercises. Each chapter opens with a review of the basic concepts, formulae, laws and definition that is linked to the concepts and problems of that chapter. There are plenty of fully solved questions of all types as per the latest pattern and syllabus. Questions in each chapter have been classified as follows: Section Section Section Section Section

I: II : III : IV : V:

Multiple Choice Questions with Only One correct choice. Multiple Choice Questions with One or More correct choices. Multiple Choice Questions based on passage. Matching of Column type questions. Assertion-Reason type questions.

Key Features

New Feature

THE PUBLISHERS

Syllabus General: Units and dimensions, dimensional analysis; least count, significant figures; Methods of measurement and error analysis for physical quantities pertaining to the following experiments: Experiments based on using Vernier

specific resistance of the material of a wire using meter bridge and post office box. Mechanics: tion; Relative velocity.

-

Law of gravitation; Gravitational potential and field; Acceleration due to gravity; Motion of planets and satellites in circular orbits; Escape velocity. Rigid body, moment of inertia, parallel and perpendicular axes theorems, moment of inertia of uniform bodies with simple geometrical shapes; Angular momentum; Torque; Conservation of angular momentum; Dynamics of rigid bodies with fixed axis of rotation; Rolling without slipping of rings, cylinders and spheres; Equilibrium of rigid bodies; Collision of point masses with rigid bodies. Linear and angular simple harmonic motions. theorem and its applications.

(in sound). Thermal physics: Thermal expansion of solids, liquids and gases; Calorimetry, latent heat; Heat conduction in one heats (Cv and Cp for monoatomic and diatomic gases); Isothermal and adiabatic processes, bulk modulus of gases;

Electricity and magnetism: of point charges and of electrical dipoles in a uniform electrostatic field; Electric field lines; Flux of electric field; charged infinite plane sheet and uniformly charged thin spherical shell. in a capacitor. applications; Heating effect of current.

viii Syllabus

circular coil and inside a long straight solenoid; Force on a moving charge and on a current-carrying wire in a uniform magnetic field. Magnetic moment of a current loop; Effect of a uniform magnetic field on a current loop; Moving coil galvanometer, voltmeter, ammeter and their conversions. d.c. and a.c. sources. Optics: Magnification. Modern physics: Atomic nucleus; Alpha, beta and gamma radiations; Law of radioactive decay; Decay constant; these processes.

Contents A Word to the Reader Syllabus

v vii

1. Units and Dimensions

1.1–1.24

2. Motion in One Dimension

2.1–2.36

3. Vectors

3.1–3.13

4. Motion in Two Dimensions

4.1–4.34

5. Laws of Motion and Friction

5.1–5.61

6. Work, Energy and Power

6.1–6.39

7. Conservation of Linear Momentum and Collisions

7.1–7.38

8. Rigid Body Rotation

8.1–8.68

9. Gravitation

9.1–9.36

10. Elasticity

10.1–10.24

11. Hydrostatics (Fluid Pressure and Buoyancy)

11.1–11.32

12. Hydrodynamics (Bernoulli’s Theorem and Viscosity)

12.1–12.25

13. Simple Harmonic Motion

13.1–13.47

14. Waves and Doppler’s Effect

14.1–14.48

15. Thermal Expansion

15.1–15.14

16. Measurement of Heat

16.1–16.11

17. Thermodynamics (Isothermal and Adiabatic Processes)

17.1–17.41

18. Kinetic Theory of Gases

18.1–18.16

19. Transmission of Heat

19.1–19.23

20. Electrostatic Field and Potential

20.1–20.50

21. Capacitance and Capacitors

21.1–21.35

22. Electric Current and D.C. Circuits

22.1–22.50

23. Heating Effect of Current

23.1–23.23

24. Magnetic Effect of Current and Magnetism

24.1–24.52

25. Electromagnetic Induction and A.C. Circuits

25.1–25.54

26. Ray Optics and Optical Instruments

26.1–26.52

27. Wave Optics

27.1–27.34

28. Atomic Physics

28.1–28.43

29. Nuclear Physics

29.1–29.27

x Contents

Model Test Paper I Model Test Paper II IIT-JEE 2012 Paper–I IIT-JEE 2012 Paper–II JEE JEE JEE JEE JEE JEE

Advanced Advanced Advanced Advanced Advanced Advanced

2013 Paper–I 2013 Paper–II 2014: Paper–I 2014: Paper–II 2015: Paper–I (Model Solutions) 2015: Paper–II (Model Solutions)

   JEE Advanced 2016: Paper-I (Model Solutions)    JEE Advanced 2016: Paper-II (Model Solutions)    JEE Advanced 2017: Paper-I    JEE Advanced 2017: Paper-II

MTPI.1–MTPI.12 MTPII.1–MTPII.12 IJI.1–IJI.8 IJII.1–IJII.10 JAI.1–JAI.8 JAII.1–JAII.9 JAI.1–JAI.11 JAII.1–JAII.10 JAI.1–JAI.11 JAII.1–JAII.12 P-I.1–P-I.15 P-II.1–P-II.19 P-I.1–P-I.9 P-II.1–P-II.11

1

Units and Dimensions

Chapter

REVIEW OF BASIC CONCEPTS 1.1

THE SI SYSTEM OF UNITS

The internationally accepted standard units of the fundamental physical quantities are given in Table 1.1. Table 1.1

Fundamental SI Units

Physical Quantity

Name of the Unit

Symbol

Mass

kilogram

kg

Time

second

s

Electric current

ampere

A

Temperature

kelvin

K

Luminous intensity

candela

cd

Amount of substance

mole

mol

Physical Quantity

Name of the Unit

Symbol

Angle in a plane

radian

rad

Length

metre

m

Solid angle

steradian

sr

Table 1.2 Physical Quantity Area

Dimensional Formulae of some Physical Quantities

Dimensional Formula

Physical Quantity

Dimensional Formula

0 2 0

Heat energy

ML2T–2

0 3 0

Entropy

ML2T–2K–1

ML T

Volume

ML T

Density

ML–3T 0

Velocity

M0LT –1 0

M0L2T –2K–1 Latent heat

–2

Acceleration

M LT

Momentum

MLT–1 2 –1

M0L2T–2 ML2T–2K–1 mol–1

Thermal conductivity

MLT–3K–1

Angular momentum

ML T

Wien’s constant

M0LT0K

Force

MLT–2

Stefan’s constant

ML0T–3K–4

Energy, work

ML T

Boltzmann’s constant

ML2T–2K–1

Power

ML2T–3

Torque, couple Impulse Frequency

2 –2

Molar gas constant

ML2T–2K–1 mol–1

2 –2

Electric charge

TA

–1

Electric current

A

Electric potential

ML2T–3 A–1

ML T MLT

0 0 –1

MLT

0 0 –1

Angular frequency

MLT

Angular acceleration

M0L0T–2

MLT–3 A–1 Capacitance

M –1L–2T4A2

Pressure

–1 –2

ML T

Inductance

ML2T–2A–2

Elastic modulii

ML–1T–2

Resistance

ML2T–3A–2

Stress

–1 –2

ML T

ML2 T–2A–1 (Contd.)

1.2 Comprehensive Physics—JEE Advanced

Physical Quantity

Dimensional Formula

Physical Quantity

Dimensional Formula

2 0

ML0T–2A–1

Moment of inertia

ML T

Surface tension

ML0T–2

Permeability

MLT –2A–2

Viscosity

ML–1T –1

Permittivity

M–1L–3T4A2

Planck’s constant

ML2T –1

Gravitational constant

1.2

–1 2 –2

M LT

PRINCIPLE OF HOMOGENEITY OF DIMENSIONS

Consider a simple equation, A + B = C. If this is an equation of physics, i.e. if A, B and C are physical quantities, then this equation says that one physical quantity A, when added to another physical quantity B, gives a third physical quantity C. This equation will have no meaning in physics if the nature (i.e. the dimensions) of the quantities on the left-hand side of the equation is not the same as the nature of the quantity on the right-hand A is a length, B must also be a length and the result of addition of A and B In other words, the dimensions of both sides of a physical equation must be identical. This is called the principle of homogeneity of dimensions.

1.3

Dimensions of at = dimensions of x [ x] [L] or [a] = [LT 1 ] [M 0 LT 1 ] [t ] [T] Dimensions of bt2 = dimensions of x L x or b= 2 = [LT–2] = [M0 LT–2] T2 t 1.2 The pressure P, volume V and temperature T of a gas are related as a P V b = cT V2 where a, b, and c are constants. Find the dimensions a of . b

USES OF DIMENSIONAL ANALYSIS

Dimensional equations provide a very simple method of deriving relations between physical quantities involved in any physical phenomenon. The analysis of any phenomenon carried out by using the method of dimensions is called dimensional analysis. This analysis is based on the There are four important uses of dimensional equations: 1. Checking the correctness of an equation. 2. Derivation of the relationship between the physical quantities involved in any phenomenon. 3. Finding the dimensions of constants or variables in an equation. 4. Conversion of units from one system to another. 1.1 The distance x travelled by a body varies with time t as x = at + bt2, where a and b are constants. Find the dimensions of a and b. SOLUTION The dimensions of each term on the right hand of the given equation must be the same as those of the left hand side. Hence

SOLUTION Dimensions of

a

= dimensions of P. V2 Dimensions of a = dimensions of PV2 Also dimensions of b = dimensions of V Dimensions of

a [ PV 2 ] = b [V ] = [PV] = [ML–1 T–2] = [ML2 T – 2]

[L3]

NOTE 1. Trigonometric function (sin, cos, tan, cot etc) are dimensionless. The arguments of these functions are also dimensionless nents are also dimensionless

1.3 When a plane wave travels in a medium, the displacement y of a particle located at x at time t is given by y = a sin(bt + cx) where a, b and c are constants. Find the dimensions b of . c

Units and Dimensions 1.3

this equation must be the same. Equating the powers of M, L and T, we have 1 and a = 0, b + c = 0 and – 2c = 1, which give b = 2 1 c= . 2 Hence t = k m0 l1/2 g(– 1/2)

SOLUTION Terms bt and cx must be dimensionless. Hence 1 [T 1] [b] = [t ] and

[c] =

1 [ x]

1 [L 1 ] [ L]

Thus t is independent of the mass of the bob and is directly proportional to proportional to

1.4 2

a e ax b P is pressure, x is a distance and a and b are constants. Find the dimensional formula for b. P=

SOLUTION a2 b

= [P]

Also ax is dimensionless. Hence [a] = [L – 1]. [b] =

[a 2 ] [L 2 ] = [ P] [ML 1 T 2 ]

= [M–1 L–1 T2] The principle of homogeneity of dimensions can also on other physical quantities.

SOLUTION Let

t

m al bg c

or t = k m al bg c, where k is a dimensionless constant. Writing the dimensions of each quantity, we have 2 c

[T] = [M ]a [L]b [LT or

[M 0 L0 T] = [M a Lb

c

T

2c

]

]

According to the principle of homogeneity of dimensions, the dimensions of all the terms on either side of

l and inversely

g.

The dimensional method can also be used to convert a physical quantity from one system to another. The method is based on the fact that the magnitude of a physical quantity X remains the same in every system of its measurement, i.e. (1) X = n1 u1 = n2 u2 where u1 and u2 are the two units of measurement of quantity X and n1 and n2 are their respective numerical values. Suppose M1, L1 and T1 are the fundamental units of mass, length and time in one system of measurement and M2, L2 and T2 in the second system of measurement. Let a, b and c be the dimensions of mass, length and time of quantity X, the units of measurement u1 and u2 will be u1 = M1a Lb1 T1c and

u2 = M a2 Lb2 T2c

Using these in Eq. (1), we have n1 M1a Lb1 T1c

1.5 The time period (t) of a simple pendulum may depend upon m the mass of the bob, l the length of the string and g the acceleration due to gravity. Find the dependence of t on m, l and g.

l g

t= k

b = [LT –1] = M0 LT –1] c Note that the dimensions of a are the same as those of y.

= n2 M a2 Lb2 T2c

n2 = n1

M1 M2

a

L1 L2

b

T1 T2

c

(2)

Knowing (M1, L1, T1), (M2, L2, T2), (a, b and c) and the value of n1 in system 1, we can calculate the value of n2 in system 2 from Eq. (2). 1.6 Dyne is the unit of force in the c.g.s. system and newton is the unit of force in the SI system. Convert 1 dyne into newton. SOLUTION The dimensional formula of force is [F] = [M1 L1 T–2] Therefore, a = 1, b = 1 and c = –2

1.4 Comprehensive Physics—JEE Advanced

System 1 (Given System)

System 2 (Required System)

n1 = 1 dyne

n2 = ? (number of newtons)

M1 = 1g

M2 = 1 kg

L1 = 1 cm

L2 = 1 m

T1 = 1 s

T2 = 1 s

n2 = n1 =1 =1

a

M1 M2

1

1g 1 kg

1 cm 1m 1

1g

=1

1

T1 T2

2

(2)

1

1

102 cm

1.4

1.7 Convert 72 kmh–1 into ms–1 by using the method of dimensions.

quantity on other quantities of a given system, it has its own limitations, some of which are listed as follows: 1. In more complicated situations, it is often not easy

2.

SOLUTION Given System

Required System

n1 = 72 units

n2 = ?

L1 = 1 km

L2 = 1 m

T1 = 1 h

T2 = 1 s

= 72 = –1

M1 M2

1

L1 L2

1000 m 1m

3.

1

60

60 s 1s

1

72 1000 = 20 60 60

Hence 72 km h = 20 ms

LIMITATIONS OF DIMENSIONAL ANALYSIS

Though the dimensional method is a simple and a very

10–5 newtons in 1 dyne, i.e. 1

n2 = n1

(3)

Using L = 10 m in Eq. (3), we get T = 2 s. Using T = 2 s and L = 10 m in either Eq. (1) or Eq. (2), we get M = 4 kg.

10–5

Hence there are 1 dyne = 10–5 N.

[ML2 T–2] = 100 J

Dividing Eq. (2) by Eq. (1), we get 100 J 100 Nm L= = = 10 m 10 N 10 N

c

1s 1s

1 cm

103 g

(1)

[LT–1] = 5 ms–1

b

L1 L2

SOLUTION [MLT–2] = 10 N

–1

4.

5.

NOTE

will depend. In such cases, one has to make a guess which may or may not work. This method gives no information about the dimensionless constant which has to be determined either vation. This method is used only if a physical quantity varies as the product of other physical quantities. It fails if a physical quantity depends on the sum or difference of two quantities. Try, for instance, to 1 2 at using the method obtain the relation S = ut + 2 of dimensions. This method will not work if a quantity depends on another quantity as sin or cos of an angle, i.e. if the dependence is by a trigonometric function. The method works only if the dependence is by power functions only. This method does not give a complete information in cases where a physical quantity depends on more than three quantities, because by equating the powers of M, L and T, we can obtain only three equa-

Sometimes it is more convenient to use units rather than 72 km h–1 =

72 km 72 1000 m = = 20 ms–1 1h 60 60 s

1.8 If the units of force, energy and velocity are 10 N, 100 J and 5 ms–1 time.

1.5

SIGNIFICANT FIGURES

indicates the degree of precision of that measurement. The degree of precision is determined by the least count of the measuring instrument. Suppose a length measured by a metre scale (of least count = 0.1 cm) is 1.5 cm, then

Units and Dimensions 1.5

with a vernier callipers (of least count = 0.01 cm) the 0.001 cm) the same length may be 1.536 cm which has

1.9

It must be clearly understood that we cannot increase the accuracy of a measurement of changing the unit. For 39.4 kg. It is understood that the measuring instrument has g or 39400000 mg, we cannot change the accuracy of measurement. Hence 39400 g or 39400000 mg still have

SOLUTION M

only the magnitude of measurement. M 2. We use the following rule to determine the number and division of various physical quantities. Do not worry about the number of digits after the decimal place. Round off the result so that it has the

The least accurate quantity determines the accuracy of the sum or product. The result must be rounded off to the appropriate digit.

accurate quantity.

(preceding) digit to be retained is left unchanged. 4 which is less than 5. 2. If the digit to be dropped is more than 5, the preced3. If the digit to be dropped happens to be 5, then (a) the preceding digit to be retained is increased by 1 if it odd, or (b) the preceding digit is retained unchanged if it is even. result of the indicated rounding-off is therefore, 5.3. off to 3.4. 1. For addition and subtraction, we use the following rule. Find the sum or difference of the given measured that it has the same number of digits after the decimal place as in the least accurate quantity (i.e., the

1.10 A man runs 100.2 m in 10.3 s. Find his average speed

SOLUTION Average speed (v) =

100.2 m 10.3s

–1

the time 10.3 s has only three. Hence the value of the The correct result is v = 9.73 ms–1

1.6

LEAST COUNTS OF SOME MEASURING INSTRUMENTS

1. Least count of metre scale = 1 mm = 0.1 cm 2. Vernier constant (or least count) of vernier callipers = value of 1 main scale division – value of 1 vernier scale division = 1 M.S.D. – 1 V.S.D Let the value of 1 M.S.D = a unit If n vernier scale divisions coincide with m main scale divisions, then value of m of 1 M.S.D 1.V.S.D = n ma = unit n

1.6 Comprehensive Physics—JEE Advanced

ma m a unit 1 n n 3. Least count of a micrometer screw is found by the formula Least count = Pitch of screw Total number of divisions on circular scale Least count = a –

where pitch = lateral distance moved in one complete rotation of the screw.

1.7

ORDER OF ACCURACY: PROPORTIONATE ERROR

The order of accuracy of the result of measurements is determined by the least counts of the measuring instruments used to make those measurements. Suppose a length x is measured with a metre scale, then the error in x is x, where x = least count of metre scale = 0.1 cm. If the same length is measured with vernier callipers of least count 0.01 cm, then x = 0.01 cm. x . x x 100. x 1. Suppose a quantity is given by a=x+y Then a = x + y a x = a (x 2.

mum error is

y y)

If a = x – y

-

a= x+ y We take the worst case in which errors add up. a x y = a ( x y) 3. Suppose we determine the value of a physical quantity u by measuring three quantities x, y and z whose true values are related to u by the equation u = x y z– quantities x, y and z be respectively ± x, ± y and ± z so that the error in u by using these observed quantities is ± u. The numerical values of x, y and z are given by the least count of the instruments used to measure them. Taking logarithm of both sides we have log u = log x + log y – log z Partial differentiation of the above equation gives

x y z u = x y z u The proportional or relative error in u is u/u. The values of x, y and z may be positive or negative and in some uses the terms on the right hand side may counteract each other. This effect cannot be relied upon and it is necessary to consider the worst case which is the case when all errors add up giving an error u given by the equation: x y z u = x y z u ma u, multiply the proportional errors in each factor (x, y and z) by the numerical value of the power to which each factor is raised and then add all the terms so obtained. proportional error in the result of u. When the proportional error of a quantity is multiplied by 100, we get the percentage error of that quantity. 1.11 ) of a rectangular metal block, a student makes the following measurements. Mass of block (m) = 39.3 g Length of block (x) = 5.12 cm Breadth of block (y) = 2.56 cm Thickness of block (z) = 0.37 cm The uncertainty in the measurement of m is ± 0.1 g and in the measurement of x, y and z is ± 0.01 cm. Find the value of (in g cm – 3) up to appropriate of . SOLUTION =

m = xyz 5.12

39.3 2.56 0.37 –3

y m x z + + + y m x z ma 0.1 0.01 0.01 0.01 = 39.3 5.12 2.56 0.37 = 0.0353 = 0.0353 = 0.0353 Round off error = 0.3 gcm – 3. =

–3

Units and Dimensions 1.7 –3 Hence is not accurate to the fourth decimal place. In fact, it is accurate only up to the much be

is written as

SOLUTION –3

1.12 The time period of a simple pendulum is given by T = 2 L/ g . The measured value of L is 20.0 cm using a scale of least count 1 mm and time t for 100 oscillations is found to be 90 s using a watch of least count 1 s. Find the value of g (in m s–2) up to appro-

A B 1 C D X +3 = 2 + + A B 3 C D X =2

1% + 2% +

= 17%

1 3

3% + 3

4%

NOTE 2. The quantity which is raised to the highest power conured to a high degree of accuracy.

the value of g. SOLUTION If t is the time for n oscillations, then T = T=2

of quantities A, B, C and D are 1%, 2%, 3% and 4% respectively. Find the percentage error in the measurement of X.

L . Squaring, we get g 4 2 Ln 2 g= t2

t . Given n

(1)

Putting L = 0.200 m, n = 100 and t = 90 s, g=

4

(3.14) 2

0.2

(100) 2

(90) 2

= 9.74 m s–2 From Eq. (1), the relative error in g is g L 2 t = g L t Note that there is no error in counting the number (n) of oscillations. Thus g 0.1 cm 2 1s = g 20.0 cm 90s = 0.005 + 0.022 = 0.027 g = 0.027 9.74 = 0.26 m s–2 Rounding off g g = 0.3 m s–2. Hence the value of g must be rounded off as g = 9.7 m s–2. Hence g = (9.7 ± 0.3) m s–2 1.13 In the measurement of a physical quantity X = A2 B , the percentage errors in the measurements C1/3 D3

1. For a simple pendulum T = 2 T 1 = T 2 2. For a sphere of radius r,

A 2 r = A r

Surface area A = 4 r2 Volume V =

g

4 3 r 3

3 r r

V V

3. Acceleration due to gravity g =

GM R

g 2 R M = g R M 4. For resistances connected in series Rs Rs

Rs = R1 + R2

R1 R1

R2 R2

5. For resistances connected in parallel 1 1 = R1 Rp

1 R2 Rp RP2

–

=

R1

R12

Rp R 2p

R1

R12

R2

R22

R2

R22

6. Kinetic energy K and linear momentum p are related as p2 K 2 p K = 2m K p

1.8 Comprehensive Physics—JEE Advanced

I Multiple Choice Questions with Only One Choice Correct 1. The pressure P is related to distance x, Boltzmann constant k and temperature as a –ax/k e P= b The dimensional formula of b is (a) [M –1 L–1T–1] (b) [MLT 2] (c) [M 0 L2T 0] (d) [M0 L0 T 0] 2. The magnitude of induced emf e in a conductor of length L B is given by 1 ( BL2) a The dimensional formula of a is (a) [M 0 L0T] (b) [ML2T –2] (c) [M 2 LT –1] (d) M 0 L0 T 0] 3. Two resistors R1 = [3.0 ± 0.1] and R2 = (6.0 ± 0.3) are connected in parallel. The resistance of the combination is (a) (2.0 ± 0.4) (c) (2.0 ± 0.2) (d) (2.00 ± 0.04) 4. If the resistances in Q. 3 above were connected in tance of the combination will be (a) 1.1% (b) 2.2% (c) 3.3% (d) 4.4% 5. Which of the following pairs of physical quantities do not have the same dimensions? (a) Pressure and Young’s modulus (b) Emf and electric potential (c) Heat and work e=

6. Which of the following pairs of physical quantities have different dimensions? (a) Impulse and linear momentum (b) Planck’s constant and angular momentum (c) Moment of inertia and moment of force (d) Torque and energy 7. A = A0 e–a/kT, k is Boltzmann constant and T is the absolute temperature. The dimensions of a are the same as those of (a) energy (b) time (c) acceleration (d) velocity 8. A cube has a side of 1.2 cm. The volume of the cube 3

(b) 1.73 cm3

9.

10.

11.

12.

13.

14.

(c) 1.7 cm3 (d) 17.3 cm3 The quantities L/R and RC (where L, C and R stand for inductance, capacitance and resistance respectively) have the same dimensions as those of (a) velocity (b) acceleration (c) time (d) force Which one of the following has the dimensions of ML–1T –2? (a) torque (b) surface tension (c) viscosity (d) stress The dimensions of angular momentum are (a) MLT –1 (b) ML2T–1 (c) ML –1T (d) ML0T –2 According to the quantum theory, the energy E of a photon of frequency is given by E=h where h is Planck’s constant. What is the dimensional formula for h? (b) M L2 T –1 (a) M L2 T –2 2 (c) M L T (d) M L2 T2 The dimensions of Planck’s constant are the same as those of (a) energy (b) power (c) angular frequency (d) angular momentum The volume V of water passing any point of a uniform tube during t seconds is related to the cross-sectional area A of the tube and velocity u of water by the relation V A u t which one of the following will be true? (a) = = (b) =

(c) = (d) 15. The frequency n of vibrations of uniform string of length l and stretched with a force F is given by p F n= 2l m where p is the number of segments of the vibrating string and m is a constant of the string. What are the dimensions of m? (b) M L–3 T0 (a) M L–1 T–1 –2 0 (c) M L T (d) M L–1 T0

Units and Dimensions 1.9

16. If velocity (V), acceleration (A) and force (F) are taken as fundamental quantities instead of mass (M), length (L) and time (T), the dimensions of Young’s modulus would be (a) FA2V –2

(b) FA2V–3

(c) FA2V–4 (d) FA2V –5 17. The dimensions of permittivity ( 0) of vacuum are (a) M–1 L–3 T4 A2 (b) ML–3 T2 A2 (c) M–1 L3 T4 A2 (d) ML3 T2 A2 18. What are the dimensions of permeability ( 0) of vacuum? (b) MLT–2 A–2 (a) MLT–2 A2 (c) ML–1 T–2 A2 (d) ML–1 T–2 A–2 19. The dimensions of 1/ 0 0 are the same as those of (a) velocity (b) acceleration (c) force (d) energy 20. (b) ML2 T–2 K–1 (a) MLT –2 K–1 (c) M0L2T–2 K–1 (d) M0LT–2 K–1 21. What are the dimensions of latent heat? (a) ML2 T –2 (b) ML–2 T–2 0 –2 (c) M LT (d) M 0 L2T –2 22. What are the dimensions of Boltzmann’s constant? (b) ML2T–2 K–1 (a) MLT –2 K–1 0 –2 –1 (c) M LT K (d) M0L2T–2 K–1 23. The dimensions of potential difference are (a) ML2T – 3 A – 1 (b) MLT–2 A–1 (c) ML2T – 2 A (d) MLT–2 A 24. What are the dimensions of electrical resistance? (b) ML2 T–3 A–2 (a) ML2T –2 A2 (c) ML2 T –3 A2 (d) ML2 T–2 A–2 25. (a) MLT –3 A–1 (b) MLT –2 A–1 –1 –1 (c) MLT A (d) MLT0 A–1 26. (b) M0L T –1 A–1 (a) ML0 T –1 A–1 –2 –1 (c) MLT A (d) ML 0 T–2 A–1 27. (b) ML2 T–2 A–2 (a) ML2 T –2 A–1 –2 –2 –1 (c) ML T A (d) ML–2 T–2 A–2 28. The dimensions of self inductance are (b) ML2 T–2 A–2 (a) ML2 T–2 A–1 (c) ML–2 T–2 A–1 (d) ML–2 T–2 A–2

29. The dimensions of capacitance are (a) M–1 L–2 TA2 (b) M–1 L–2 T2 A2 –1 –2 3 2 (c) M L T A (d) M–1 L–2 T4 A2 30. If velocity (V), force (F) and energy (E) are taken as fundamental units, then dimensional formula for mass will be (a) V –2F 0 E (b) V 0 FE 2 (c) VF –2E 0 (d) V–2 F 0E 31. Frequency (n) of a tuning fork depends upon length (l) of its prongs, density ( ) and Young’s modulus (Y) of its material. Then frequency and Young’s modulus will be related as (a) n Y (b) n Y 1 1 (d) n (c) n Y Y 1 2 32. The dimensions of E ( = permittivity of free 0 0 2 space and E (a) MLT –1 (b) ML2T –2 –1 –2 (c) ML T (d) ML2T–1 IIT, 2000 33. Of the following quantities, which one has dimensions different from the remaining three (a) Energy per unit volume (b) Force per unit area (c) Product of voltage and charge per unit volume (d) Angular momentum 34. If the time period t of a drop of liquid of density d, radius r, vibrating under surface tension s is given by the formula t = d a r b s c and if a = 1, c = – 1, then b is (a) 1 (b) 2 (c) 3 (d) 4 35. In the measurement of a physical quantity X = A2 B . The percentage errors introduced in the C 1/ 3 D 3 measurements of the quantities A, B, C and D are 2%, 2%, 4% and 5% respectively. Then the minimum amount of percentage of error in the measurement of X is contributed by: (a) A (b) B (c) C (d) D 36. Which of the following has the dimensions ML–1 T–1? (c) Bulk modulus (c) Angular momentum 37. Pressure gradient dp/dx is the rate of change of pressure with distance. What are the dimensions of dp/dx?

1.10 Comprehensive Physics—JEE Advanced

(a) ML–1 T–1 (b) ML–2 T–2 –1 –2 (c) ML T (d) ML–2 T–1 38. If E, M, J and G respectively denote energy, mass, angular momentum and gravitational constant, then EJ 2 has the dimensions of M 5G 2 (a) length (b) angle (c) mass (d) time IIT, 1990 39. If e, 0, h and c respectively represent electronic charge, permittivity of free space, Planck’s cone2 has the dimenstant and speed of light, then 0 hc sions of (a) current (b) pressure (c) angular momentum (d) angle 40. If L, R, C and V respectively represent inductance, resistance, capacitance and potential difference, L are the same as those then the dimensions of RCV of 1 (a) current (b) current (c) charge

(d)

1 charge

41. If E and B respe E has the B

dimensions of (a) displacement (b) velocity (c) acceleration (d) angle 42. If C and V respectively represent the capacitance of a capacitor and the potential difference between its plates, then the dimensions of CV2 are (a) ML2T–2 (b) ML3T–2 A–1

(c) ML2T–1A–1 (d) M0L0T0 43. If h and e respectively represent Planck’s constant h and electronic charge, then the dimensions of e are the same as those of

44. If energy E, velocity V and time T are chosen as the fundamental units, the dimensional formula for surface tension will be

45.

(a) E V2T–2

(b) E V–1T–2

(c) E V–2T–2

(d) E2V–1T–2

oscillates with a period proportional to Pa db Ec where P is the static pressure, d is the density of

and c respectively are 5 1 1 (a) , , 6 2 3 (c)

1 1 , , 3 2

5 6

a, b (b)

1 , 2

5 1 , 6 3

(d) 1, 1, 1

46. In a system of units in which the unit of mass is a kg, unit of length is b metre and the unit of time is c second, the magnitude of a calorie is 4.2c 4.2c 2 (b) (a) ab 2 ab 2 abc 4.2 (c) (d) 4.2 abc 47. The error in the measurement of the radius of a sphere is 1%. The error in the measurement of the volume is (a) 1% (b) 3% 48. If the error in the measurement of the volume of a sphere is 6%, then the error in the measurement of its surface area will be (a) 2% (b) 3% (c) 4% (d) 7.5% 49. The moment of inertia of a body rotating about a 2 in the SI system. What is the value of the moment of inertia in a system of units in which the unit of length is 5 cm and the unit of mass is 10 g? (a) 2.4 × 103 (b) 2.4 × 105 (c) 6.0 × 103 (d) 6.0 × 105 V 50. A quantity X is given by 0L where 0 is the t permittivity of free space, L is a length, V is a potential difference and t is a time interval. The dimensional formula for X is the same as that of (a) resistance (b) charge (c) voltage (d) current IIT, 2001 51. ) of a liquid by the the formula

R4 P l Q

where R = radius of the capillary tube, l = length of the tube, P = pressure difference between its ends, and

Units and Dimensions 1.11

Q Which quantity must be measured most accurately? (a) R (b) l (c) P (d) Q 52. The mass m of the heaviest stone that can be

dent also measures the diameter of the wire to be 0.4 mm with a uncertainty of ± 0.01 mm. Take g 2

from the reading is (a) (2.0 ± 0.3) 1011 N/m2 (b) (2.0 ± 0.2) 1011 N/m2 (c) (2.0 ± 0.1) 1011 N/m2 (d) (2.0 ± 0.05) 1011 N/m2

v, the speed of water, density (d) of water and the acceleration due to gravity (g). Then m is proportional to (b) v4 (a) v2 6 (c) v (d) v 53. The speed (v) of ripples depends upon their wavelength ( ), density ( ) and surface tension ( ) of water. Then v is proportional to (b)

(a) (c)

IIT, 2007 59. In a vernier callipers, one main scale division is x cm and n divisions of the vernier scale coincide with (n – 1) divisions of the main scale. The least count (in cm) of the callipers is n 1 nx x (b) (a) n (n 1) x x (c) (d) n (n 1)

1

(d)

1

54. The period of revolution (T) of a planet moving round the sun in a circular orbit depends upon the radius (r) of the orbit, mass (M) of the sun and the gravitation constant (G). Then T is proportional to (a) r1/2 (c) r3/2

(b) r (d) r2

55. If the velocity of light (c), gravitational constant (G) and planck’s constant (h) are chosen as fundamental units, the dimensions of time in the new system will be (a) c –5/2G2h–1/2 (b) c–3/2G–2h2 2

–2 1/2

–5/2

1/2 1/2

(c) c G h (d) c G h 56. The amplitude of a damped oscillator of mass m varies with time t as A = A0 e( at m ) The dimensions of a are (b) M0LT–1 (a) ML0T–1 –1 (c) MLT (d) ML–1T 57. A student measures the value of g with the help of a simple pendulum using the formula g=

58.

60.

4

2

L

2

T The errors in the measurements of L and T are L and T respectively. In which of the following cases is the error in the value of g the minimum? (a) L = 0.5 cm, T = 0.5 s (b) L = 0.2 cm, T = 0.2 s (c) L = 0.1 cm, T = 1.0 s (d) L = 0.1 cm, T = 0.1 s long, by Searle’s method. In a particular reading,

IIT, 2007 measuring the acceleration due to gravity (g) using a simple pendulum. They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown in the table. Least count for length = 0.1 cm Least count for time = 0.1 s

Student Length of the Number of Total time for pendulum oscillations (n) (cm) (n) oscillations (s)

Time period (s)

I

64.0

16.0

II

64.0

4

64.0

16.0

III

20.0

4

36.0

9.0

If EI, EII and EIII are the percentage errors in g. i.e. g 100 for student I, II and III, respectively, g (a) E1 = 0 (c) EI = EII

(b) EI is minimum (d) EII is minimum

61. The density of a solid ball is to be determined in an with a screw gauge, whose pitch is 0.5 mm and there are 50 divisions on the circular scale. The reading on the main scale is 20 divisions. If the measured mass of the ball has a relative error of 2%, the relative percentage error in the density is (a) 0.9% (b) 2.4% (c) 3.1% (d) 4.2% IIT, 2011

1.12 Comprehensive Physics—JEE Advanced

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61.

(c) (a) (d) (a) (a) (a) (b) (c) (b) (d) (c)

(a) (c) (b) (c) (d) (c) (b) (c) (d) (a)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56.

3. 9. 15. 21. 27. 33. 39. 45. 51. 57.

(b) (c) (d) (d) (a) (d) (d) (a) (a) (d)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58.

1. k x

1

JK

=

K

m

=Jm = [M L2T–2] = M L T –2 [a] Also [P] = [b] [a] = [ P]

3.

ML1 T 1

ML T

2 2

= [M 0 L2T 0]

BL2 [M 0 L0 T 0 ] [ML0 T 2 A 1 ] = e [ML2 T 3 A 1 ] = [M0 L0T1] 1 R2

1 1 = R1 R

R= R R

2

= =

R1 R2 3.0 6.0 = = 2.0 R1 R2 9.0 R1

R12

R2

R22

0.1

0.3

2

( 6) 2

(3)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60.

(c) (b) (b) (b) (d) (b) (a) (c) (c) (b)

L = 1.2 cm,

[L–1]

So the correct choice is (c). 2. a =

(d) (b) (a) (a) (d) (c) (b) (b) (d) (c)

3

–1

[b] =

5. 11. 17. 23. 29. 35. 41. 47. 53. 59.

5. The correct choice is (d). 6. The correct choice is (c). 7. Since kT has dimensions of energy, the correct choice is (a). 8. V = L3 = 1.2 cm 1.2 cm 1.2 cm

SOLUTIONS

[a] =

(d) (d) (c) (b) (b) (c) (b) (b) (c) (b)

= 0.019

R = 0.019 R2 = 0.019 (2)2 = 0.076 Hence the correct choice is (b). 4. R = R1 + R2 = 3.0 + 6.0 = 9.0 R = R1 + R2 = 0.1 + 0.3 = 0.4 R 100 R 0.4 = 100 9.0 = 4.4% So the correct choice is (d).

[L2 ]

Hence the correct choice is (c). 9. L/ R is the time constant of an L-R circuit and CR is the time constant of a C-R circuit. The dimension of the time constant is the same as that of time. Hence the correct choice is (c). 10. ML–1T –2 are the dimensions of force per unit area. Out of the four choices, stress is the only quantity that is force per unit area. Hence the correct choice is (d). 11. The angular momentum L of a particle with respect to point whose position vector is r is given by L=r p where p is the linear momentum of the moving particle. Dimensions of L = dimension of r dimensions of p = L MLT –1 = ML2T–1 Thus the correct choice is (b). dimension of E ML2 T 2 = 12. Dimensions of h = dimension of T 1 = ML2 T –1 Thus the correct choice is (b). 13. The correct choice is (d). 14. The dimensions of the two sides of proportionality are L 3 = L 2 (LT –1 ) T = L2 + T – Equating the powers of dimensions on both sides, we have 2 + =3 – =0

Units and Dimensions 1.13

1 (3 – ), i.e. = . 2 Thus the correct choice is (b). 15. Squaring both sides of the given relation, we get p2 F p2 F or m = n2 = 2 4l m 4 l 2 n2 dimensions of m dimensions of F = dimensions of l 2 dimensions of n 2 ( p is a dimensionless number) which give

=

= and

=

2

MLT L2

T

–1

= ML T

1 2

0

Hence the correct choice is (d). 16. Dimensions of Young’s modulus Y are ML–1 T–2. The dimension of V, A and F in terms of M, L and T are (V) = LT –1, (A) = LT –2 and (F) = MLT –2 Let (Y ) = (V a A b F c ) Putting dimensions of Y, V, A and F. We have (ML–1 T –2 ) = (LT –1) a ( LT – 2 ) b (MLT – 2) c or M1 L – 1 T – 2 = M c La + b + c T – a – 2b – 2c Equating powers of M, L and T we have c = 1, a + b + c = – 1 and – a – 2b – 2c = – 2 which give a = – 4, b = 2 and c = 1. 2 –4 Thus (Y) = (FA V ) Thus the correct choice is (c). 17. According to Coulomb’s law of electrostatics, force F between two charges q1 and q2 a distance r apart in vacuum, is given by q1q2 1 F= 4 0 r2 or

0

=

1 4

Dimensions of

q1q2 F 0

r2

Q2

=

MLT 2 L2 –1 –3 2 2 =M L T Q –1

–3

4

2

Q T

=M L T A A The correct choice is (a). 18. The force per unit length between two long wires carrying currents I1 and I2 a distance r apart in vacuum, is given by II 4 rf f = 0 1 2 or 0 = r 4 I1 I 2

Dimensions of

0

=

L MLT

2

L1

A2 = MLT A – 2 Therefore, the correct choice is (b). 19. Dimensions of 1 1 –2

0

0

MLT =

2

A2

1 1 2 2 2 T

M 1L 3 T 4 A

1 2 2

= LT – 1

L which are the dimensions of velocity. Hence the correct choice is (a). 20. The heat energy content H of a body of mass m at temperature is given by H = ms where s H s= m Dimensions of s dimensions of heat energy dimension of mass dimension of temperature ML2 T 2 = M 0 L 2T – 2 K – 1 M K Thus the correct choice is (c). 21. Latent heat L is the amount of heat energy H required to change the state of a unit mass without producing any change in temperature. Thus H L= m =

ML2 T 2 M 2 –2 = L T = M 0 L 2T – 2 Thus the correct choice is (d). 22. According to the law of equipartition of energy, the energy per degree of freedom of a gas atom or molecule at a temperature kelvin is given by 1 2E k or k = E= 2 where k is the Boltzmann’s constant. dimensions of E Dimensions of k = dimension of Dimensions of L =

ML2 T 2 ML2 T 2 K 1 K 23. The potential difference V between two points is the amount of work done in moving a unit charge from one point to the other. Thus,

V=

work done charge moved

W q

1.14 Comprehensive Physics—JEE Advanced

ML2 T 2 = ML2 T –2 Q –1 Q = ML2 T –3 A –1 ( Q = AT) Hence the correct choice is (a). 24. From Ohm’s law, resistance R is given by potential difference R= current Dimensions of V =

ML2 T 3 A 1 A 2 – 3 –2 = ML T A Thus the correct choice is (b). 25. Force F q in an electric E is given by F F = qE or E = q dimensions of F MLT 2 Dimensions of E = dimensions of Q AT Dimensions of R =

= MLT–3A–1.

26. The force F q moving with speed v perpendicular to the direction of a B is given by F F = qvB or B = qv MLT 2 Dimensions of B = = ML0 T –1Q–1 Q LT 1 = ML0T –2A–1 ( Q = AT) Hence the correct choice is (d) 27. linked with a circuit of area A B is given by = BA cos where vectors. Dimensions of = dimensions of BA ( cos is dimensionless) = ML0 T –2 A–1 L2 = ML2 T–2 A–1 Thus the correct choice is (a). 28. The self inductance L of a coil in which the current dI varies at a rate is given by dt dI e=–L dt where e is the e.m.f. induced in the coil. Now, the dimensions of e.m.f. are the same as those of potential difference, namely, ML2 T –3 A–1. e Now, L=– dI dt Dimensions of L dimensions of e = dimensions of I / dimensions of t

ML2 T 3 A 1 = ML2 T –2 A–2 A/T Thus the correct choice is (b). 29. When a capacitor of capacitance C is charged to a potential difference V, the charge Q on the capacitor plates is given by Q Q = CV or C = V dimensions of Q Dimensions of C = dimensions of V AT = 2 ML T 3 A 1 = M –1 L –2 T 4 A2 Hence the correct choice is (d). 30. Let (M) = V a F b E c Putting the dimensions of V, F and E, we have (M) = (LT –1 )a (MLT –2 )b (ML2T –2 )c =

M 1 = M b +c La+b+2c T –a–2b–2c

or

Equating the powers of dimensions, we have b+c=1 a + b + 2c = 0 – a – 2b – 2c = 0 which give a = – 2, b = 0 and c = 1. Therefore (M) = (V –2 F 0 E). Thus the correct choice is (d). 31. Let n l a b Yc Putting dimensions of all the quantities, we have (T –1) La (ML–3 ) b (ML–1 T –2)c Equating powers of M, L and T on both sides, we get b + c = 0, a – 3b – c = 0 and – 2c = – 1 1 1 and c = . Thus which give a = – 1, b = – 2 2 n l –1 1/2 Y 1/2 Hence the correct choice is (a). 32. We know that q1 q2

F= 0

Hence

1 2

0

=

E2 =

4

and E =

F q

q1 q2

4 F r2 q1 q2

F2

F r2

q2

q1 q2

F

= Dimensions of

r

0

2

1 2

q 0

2

r2

E 2 = dimensions of

F r2

=

Units and Dimensions 1.15

M LT

2

41. The force a velocity

= ML–1T –2

L2 Hence the correct choice is (c).

= q (E +

33. Energy per unit volume, force per unit area and product of voltage and charge density all have dimensions of ML2T–2, but the dimensions of angular momentum are ML2T–1. Hence the correct choice is (d). 34. Given t = da/2 rb/2 sc/2. Substituting dimensions, we have (T) = (ML–3)a/2 (L)b/2 (MT–2)c/2 = M(a + c)/2 · L(–3a/2 + b/2) T–c 3 a Equating powers of L, we have, 2 Given a = 1.

on a particle of charge q moving with in E and B

b = 0. 2

3 2

b = 0 or b = 3, which is choice(c). 2 A2 B 35. Given X = 1/ 3 3 C D Taking logarithm of both sides, we have 1 log X = 2 log A + log B – log C – 3 log D 3 Partially differentiating, we have A B 1 C D x = 2 3 A B 3 C D x A Percentage error in A = 2 = 2 2% A = 4% B Percentage error in B = = 2% B 1 C 1 = 4% Percentage error in C = 3 C 3 4 = % 3 D Percentage error in D = 3 = 3 5% D = 15% contributed by C. Hence the correct choice is (c). 36. The correct choice is (b) 37. The correct choice is (b) 38. Dimensions of J and G are ML2T–1 and M–1L3 T–2 respectively. 39. Dimensions of o and h are M–1L–3 T4 A2 and ML2T–1 respectively. 40. RC has the dimensions of time (T). V has the dimensions of emf which has the dimensions of dI . L dt

B)

Hence the dimensions of E are the same as those of vB. 42. Energy stored in a capacitor of capacitance C having a potential difference V between its plates is given by 1 U = CV2 2 Hence, the dimensions of CV 2 = dimensions of energy. Hence the correct choice is (a). (ML2 T 1 ) = ML2 T–2 A–1 AT

h e

43. Dimensions of

Dimensions of B = MT–2 A–1 B area 44. Let surface tension = Ea Vb Tc. Using the dimensions of , E, V and T and equating powers of M, a, b and c. The correct choice is (c). 45. The correct choice is (a). 46. Let n1 be the magnitude (i.e. numerical value) of a physical quantity when the fundamental units are (M1, L1, T1) and n2 the magnitude of the same physical quantity when the fundamental units are (M2, L2, T2), then, it is obvious that n1 (M1x L1y T1z ) = n2 (M 2x Ly2 T2z )

(i)

where x, y and z are the dimensions of the physical quantity in mass, length and time respectively. Now, we know that 1 calorie = 4.2 joule = 4.2 kg m2s–2 n1 = 4.2, x = 1, y = 2 and z = – 2. Hence, in the second system of units in which M2 = a kg, L2 = b m and T2 = c s, we have from (i) M1 M2

x

n2 = n1

L1 L2

1 kg a kg

x

= n1

1m bm

= 4.2

1 a

= 47. V = have

1

1 b

2

y

T1 T2 y

1 c

1s cs

z

z

2

4.2c 2 ab 2

4 3 r . Taking logarithm of both sides, we 3 log V = log 4 + log

+ 3 log r – log 3

1.16 Comprehensive Physics—JEE Advanced

Differentiating, we get V r =3 = 3 × 1% = 3% V r V r r r 48. =3 or 6% = 3 or = 2%. V r r r Now surface area s = 4 r2 or log s = log 4 + 2 log r s r =2 = 2 × 2% = 4%. s r 49. The dimensions of moment of inertia are (ML2). We have n1(u1) = n2(u2) or

n1(M1 L21 ) = n2 (M 2 L22 )

2

n1 (M1L22 )

M1 L1 n2 = = n1 2 M2 L2 (M 2 L 2 ) Given n1 = 6.0, M1 = 1 kg, L1 = 1 m, M2 = 10 g and L2 = 5 cm. Therefore, 2 1 kg 1m n2 = 6.0 × × 10 g 5 cm 2

1000 g 100 cm = 6.0 × × 10 g 5 cm = 6.0 × 100 × (20)2 = 2.4 × 105

50. The capacitance of a parallel plate capacitor is given by C = 0A/d. Hence the dimensions of 0L are the same as those of capacitance. V Dimension of 0L t dimension of C dimensions of V time dimension of Q = ( Q = CV) time Hence the correct choice is (d). =

51. The correct choice is (a). relation

is given by the

R l P Q R l P Q It is clear that the error in the measurement of R is = 4

of R4 in the formula. Hence the radius (R) of the capillary tube must be measured most accurately. Thus the quantity which is raised to the highest power needs the most accurate measurement. 52. Take m 53. Take v 54. Take T

v a d b g c and show that a = 6.

1 . 2 3 r a Mb G c and show that a = – . 2 a

b

c

and show that a = –

55. The correct choice is (d). 56. at/m is dimensionless. Therefore, dimension of m M = Dimension of a = dimension of t T = M L0 T–1 57. The proportionate error in the measurement of g is T g L = + 2 T g L Hence g will be minimum if L and T are minimum. Thus the correct choice is (d). FL 4MgL = (1) 58. Y= Al d 2l where –2 M g L l 10–3 m l = ± 0.05 mm, d = 0.4 mm = 0.4 10–3 m d = ± 0.01 mm Substituting the values of M, g, L, d and l in Eq. (1) we get Y = 2.0 1011 Nm–2 From Eq. (1) the proportionate uncertainty in Y is given by Y M = Y M

g g

L L

2 d d

Since the values of M, g and L g = 0 and L = 0. Hence Y 2 d = Y d

l l

=

2 0.01 mm 0.4 mm

l l M = 0, 0.05 mm . mm

= 0.05 + 0.0625 = 0.1125 Y = 0.1125 Y = 0.1125 2.0 1011 = 0.225 1011 Nm–2 Since the value of Y decimal place, the value of Y must be rounded Y = 0.2 1011 –2 Nm Y + Y = (2.0 0.2) 1011 Nm–2 Hence the correct choice is (b). 59. Vernier constant = value of 1 M.S.D – value of 1 V.S.D. Now n V.S.D = (n – 1) M.S.D = (n – 1) x cm n 1 x cm 1 V.S.D = n n 1 x V.C. = x cm – x cm = cm n n Hence the correct choice is (c).

Units and Dimensions 1.17

60. T = 2

l g

g=

4

2

T2

l

Thus the percentage error is minimum for student I. 61. Least count of screw gauge

Therefore,

=

2 T T g For student I, EI = 100 g g l = g l

pitch number of divisions on circular scale

0.5 mm = 0.01 mm 50 diameter of ball = 2.5 mm + 20 = 2.7 mm M Density = 4 3 r 3 is =

0.1 2 0.1 100 64.0 .0 5 = % 16 0.1 2 0.1 For student II, EII = 100 64 64.0 15 = % 32 0.1 2 0.1 For student III, EIII = 100 20.0 36.0 19 = % =

=

M M

0.01 mm

3 r r

0.01 100 2.7 = 2% + 1.1% = 3.1%

= 2% + 3

II Multiple Choice Questions with One or More Choices Correct 1. Which of the following are not a unit of time? (a) parsec (b) light year (c) micron (d) second 2. Choose the pair of physical quantities which have identical dimensions. (a) Impulse and linear momentum (b) Planck’s constant and angular momentum (c) Moment of inertia and moment of force (d) Young’s modulus and pressure 3. The dimensions of energy per unit volume are the same as those of (a) work (b) stress (c) pressure (d) modulus of elasticity 4. When a wave traverses a medium, the displacement of a particle located at x at time t is given by y = a sin (bt – cx) where a, b and c are constants of the wave. Which of the following are dimensionless quantities? (a) y/a (b) bt b (c) cx (d) c

5. In Q.4, the dimensions of b are the same as those of (a) wave velocity (b) wave frequency (c) particle frequency (d) wavelength b 6. In Q.4, the dimensions of are the same as those c of (a) wave velocity (b) angular frequency (c) particle velocity (d) wave frequency 7. The Van der Waal equation for n moles of a real gas is a P (V – b) = nRT V2 where P is the pressure, V is the volume, T is the absolute temperature, R is the molar gas constant and a, b are Van der Waal constants. The dimensions of (a) a are the same as those of PV2 (b) b are the same as those of V a (c) are the same as those of RT V (d) bP are the same as those of RT.

1.18 Comprehensive Physics—JEE Advanced

8. In Q.7, which of the following have the same dimensions as those of PV? a (a) nRT (b) V ab (c) Pb (d) V2 9. In Q.7, the dimensions of nRT are the same as those of (a) pressure (b) energy (c) work (d) force 10. Which of the following are dimensionless? (a) Boltzmann constant (b) Planck’s constant (c) Poisson’s ratio (d) relative density 11. For a body in uniformly accelerated motion, the distance x of the body from a reference point at time t is given by x = at + bt2 + c where a, b and c are constants of motion. (a) The dimensions of c are the same as those of x, at and bt2. (b) The dimensional formula of b is [M0 LT–2]. a (c) is dimensionless. b (d) The acceleration of the body is 2b. 12. The side of a cube is L = (1.2 ± 0.1) cm. The volume of the cube is 3 (b) (1.73 ± 0.02) cm3

0.3) 0.2) (c) The minimum resistance obtainable is (2.0 ± 0.3) (d) The minimum resistance obtainable is (2.0 ± 0.2) 14. A physical quantity P is given by P=

a 3b 2

d c The percentage errors in the measurements of a, b, c, and d are 1%, 3%, 4%, and 3% respectvely. P is 14% P is 10% measurement of b. (d) The minimum error is contributed by the measurement of c. 15. When a plane wave travels in a meduim, the diplacement y of a particle located at x at time t is given by y = a sin (bt – cx)

and R2 = (6.0 ±

where a, b, and c are constants. (a) The unit a is the same as that of y. (b) The SI unit of b is Hz. (c) The dimensional formule of c is [M0L–1T0] b are the same those of (d) The dimensions of c velocity.

1. Choices (a), (b) and (c) are units of length 2. The dimensions of moment of inertia are ML2T 0 and of moment of force are ML2T–2. All other pairs in (a), (b) and (d) have identical dimensions. 3. Dimensions of energy per unit volume are = dimensions of energy / dimensions of volume = ML2T –2 / L3 = ML–1T –2. Stress, pressure and modulus of elasticity all have the dimensions of ML–1T–2. The dimensions of work are ML2T–2. Hence the correct choices are (b), (c) and (d). 4. Since the sine function is dimensionless, sin (bt – cx) is dimensionless. Therefore, y and a must have the same dimensions, i.e. y/a is dimensionless. Since the argument of a sine function (or any trigonometric function) must be dimensionless, bt

and cx are also dimensionless. Hence the correct choice are (a), (b) and (c). 5. Since bt is dimensionless, the dimensions of b = dimensions of 1/t = T–1, which are the dimensions of angular frequency as well as wave frequency. Hence the correct choices are (b) and (d). 6. Dimensions of bt = dimensions of cx. Therefore b x = dimensions of = LT –1. Dimensions of c t Hence the correct choices are (a) and (c). 7. a ab = n RT PV – Pb + V V2 From the principle of homogeneity, it follows that all the four choices are correct.

(c) (1.7 ± 0.4) cm3

(d) (1.7 ± 0.3) cm3

13. Two resistances R1 = (3.0 ± 0.1) 0.2) are to be joined together. ANSWERS AND SOLUTIONS

Units and Dimensions 1.19

8. The correct choices are (a), (b), (c) and (d). 9. The dimensions of nRT = dimensions of PV = ML–1 T –2 L3 = ML2 T–2 which are dimensions of energy as well as work. Hence the correct choices are (b) and (c). 10. The correct choices are (c) and (d). 11. From the principle of homogeneity of dimensions, the dimensions of c must be the same as those of x at and bt2. Therefore, choice (a) is correct. Also dimension of bt2 = dimension of x. Hence [b]= [LT–2]. Hence choice (b) is also correct. Velocity of the body is dx d = [at + bt2 + c] = a + 2bt v= dt dt and acceleration is dv d = (a + 2 bt) = 2b, which is choice (d) dt dt a has dimension of choice (c) is wrong since b time [T] 12. Volume of cube (V = L3) = 1.2 cm 1.2 cm 3 . Now L = (1.2 ± 0.1) cm has V must

Rs ±

Rs = (9.0 ± 0.3)

Thus choice (a) is correct and choice (b) is wrong. The minimum value is obtained when the resistances are joined in parallel. 1 1 1 = Rp R1 R2 R1 R2 3.0 6.0 = = 2.0 R1 R2 3.0 6.0 RR Rp = 1 2 ( R 1 + R 2 = R s) Rs Rp Rs R1 R2 = R1 R2 Rs Rp Rp =

Now

0.1 0.2 0.3 3.0 6.0 9.0 = 0.033 + 0.033 + 0.033 = 0.099 0.1 Rp = 0.1 Rp = 0.1 2 = 0.2 =

Minimum value is (Rp ± Rp) = (2.0 ± 0.2)

.

Hence choice (c) is wrong and choice (d) is is correct. 1 14. log P = 3 log a + 2 log b – log d – log c 2 P 3 a 2 b d 1 c = P ma a b d 2 c 1 = 3 1% + 2 3% + 3% + % 2 = 3% + 6% + 3% + 2% = 14% Hence the correct choices are (a), (c) and (d). 15. The value of any trignomatric function is a dimensionless number. Hence choice (a) is correct. The argument of a trignometric function is also dimensionless. Hence (bt – cx) is dimensionless. Hence b has dimension [T–1] the same as that of frequency and c has dimension of [L–1]. Thus choices (b), (c) and (d) are all correct.

value of V = 1.7 cm3. Now V = L3 V 3 L 3 0.1 = = = 0.25 V L 1.2 V = 0.25 V = 0.25 1.7 cm3 = 0.425 cm3 The error in V is in the value of V should be rounded off as V = 0.4 cm3. Thus the correct result is V ± V = (1.7 ± 0.4) cm3, which is choice (c). 13. value is Rs= R1+ R2= 3.0 + 6.0 = 9.0 Error in Rs= Rs= R1+ R2= 0.2 + 0.1 = 0.3

III Multiple Choice Questions Based on Passage Passage-I The dimensional method is a very convenient way of

physical quantities of a given system. This method has its own limitations. In a complicated situation, it is often not easy to guess the factors on which a physical quantity will depend. Secondly, this method gives no information about the dimensionless proportionality constant. Thirdly, this

1.20 Comprehensive Physics—JEE Advanced

method is used only if a physical quantity depends on the product of other physical quantities. Fourthly, this method will not work if a physical quantity depends only on Finally, this method does not give complete information in cases where a physical quantity depends on more than three quantities in problems in mechanics. 1. The dimensional method cannot be used to obtain denpendence of (a) the height to which a liquid rises in a capillary tube on the angle of contact (b) speed of sound in an elastic medium on the modulus of elactricity. (c) height to which a body, projected upwards with a certain velocity, will rise on time t. (d) the decrease in energy of a damped oscillator on time t. 2. In dimensional method, the dimensionless proportionality constant is to be determined (b) by a detailed mathematical derivation (c) by using the principle of dimensional homogeneity. (d) by equating the powers of M, L and T.

least count of the measuring instrument used. The number degree of precision of that measurement. The importance calculation cannot increase the precision of a physical in the sum or product of a group of numbers cannot be greater than the number that has the least number of weakest link. 3. 0.0123 kg. What is the total mass supported by the ures? (c) 0.0124 kg

(d) 0.012 kg

4. The radius of a uniform wire is r = 0.021 cm. The value is given to be 3.142. What is the area of (a) 0.0014 cm2

2

(b) 0.00139 cm2

2

5. A man runs 100.5 m in 10.3 s. Find his average Passage-II In the study of physics, we often have to measure the physical quantities. The numerical value of a measured

(a) 9.71 ms–1

–1 –1

–1

ANSWERS AND SOLUTIONS

1. The correct choices are (a), (c) and (d). The height of a liquid in a capillary tube depends on cos , where is the angle of contact. The height 1 S to which a body rises is given by S = ut + at2, 2 1 which is a sum of two terms ut and at2. The 2 tially with time. 2. The correct choices are (a) and (b). 3. curate only upto the fourth decimal place. Hence

the sum must be rounded off to the fourth decimal place. Therefore the correct choice is (c). 2 . Now, 4. A = r2 = 3.142 (0.021)2 A = 0.0014 cm2, which is choice (a). 100.5 m –1 5. Average speed = 10.3 s the time has only three. Hence the result must 9.71 ms–1. Thus the correct choice is (a).

Units and Dimensions 1.21

IV Matrix Match Type 1. Match the physical quantities in column I with their SI units in column II Column I Column II (a) (b) (c) (d)

Stefan’s constant Universal gas constant Electrical permittivity Magnetic permeability

(p) (q) (r) (s)

JK–1 mol–1 Fm–1 Hm–1 Wm–2 K–4

ANSWER 1. (a) (c)

(s) (q)

(b) (d)

(p) (r)

2. Column I (b) 0.034 (c) 0.002504 (d) 1.25 1.07

Column II (q) 4 (r) 2 (s) 5

ANSWER 2. (a) (c)

(s) (q)

(b) (d)

(r) (p)

3. Match the quantities in column I with their order of magnitude given in column II Column I Column II (a) 2.6 (b) 3.9 (d) 4.2

104 104 10–24 10–24

(p) (q) (r) (s)

105 10–23 10–24 104

ANSWER 3.

x = 4.3 103, take its logarithm to the base 10. Log x = 3.633 and round if off as log x = 4 . So the order of magnitnde of x is 104. (a) (c)

4.

(s) (r)

(Q) given in column II. Column I (a) (b) (c) (d)

Angular momentum Latent heat Torque Capacitance

(b) (d)

(p) (q)

Column II (p) (q) (r) (s)

M L–2 T–2 M L2 Q–2 M L2 T–1 M L3 T–1 Q–2

1.22 Comprehensive Physics—JEE Advanced

(t) M–1 L–2 T2 Q2 (u) M0 L2 T–1

(e) Inductance (f) Resistivity

ANSWER 4. (a) (r) (c) (p) (e) (q)

(b) (d) (f)

(u) (t) (s)

SOLUTION (a) Angular momentum L = r p = r (mv) [L] = [L M L T–1] = [M L2 T–1] Q heat energy [M L2 T 2 ] = = [M0 L2 T–2] (b) Latent heat = m mass [M] (c) Torque

= r

[ ]= [L (d) Q = CV [C] = (e) |V | =

=

M L T–2] = [M L2 T–2] C= Q2

RA L

Q V

[M L2 T 2 ]

LdI dt

[L] = (f)

F

W Q

Q W /Q

Q2 W

= [M–1 L–2 T2 Q2] LQ T2

WT 2

[M L2 T 2 ] [T 2 ]

Q2

Q2

VA IL

WA QIL

= [M L2 Q–2]

[M L2 T 2 L2 ] = M L3 T–1 Q–2 Q Q L T

5. Column I gives three physical quantities. Select the appropriate units given in Column II. Column I Column II (a) Capacitance

(p) Ohm second

(b) Inductance

(q) (coulomb)2 (joule)–1 (r) coulomb (volt)–1 (s) newton (ampere metre)–1

(c) Magnetic induction or

–1

IIT, 1990

SOLUTION 5. (a) From Q = CV, unit of C = From U =

unit of Q = coulomb (volt)–1 unit of V

Q2 , unit of C = (coulomb)2 (volt)–1 2C

Units and Dimensions 1.23

(b) From |e| = L

dI , unit of L = volt second (ampere)–1 = ohm second dt

(c) From F = BIL, unit of B = newton (ampere metre)–1 Hence (a)

(q), (r)

(b)

(p), (t)

(c) (s) 6. Some physical quantities are given in Column I and some possible SI units in which these quantities may be indicate your answer by darkening appropriate bubbles in the 4 Column I Column II (p) (volt) (coulomb) (metre)

(a) GMeMs G – universal gravitational constant, Me – mass of the earth, Ms – mass of the Sun (b)

(c)

(d)

3RT M R – universal gas constant, T – absolute temperature, M – molar mass

(q) (kilogram) (metre)3 (second)–2

F2

(r) (metre)2 (second)–2

q2 B2 F – force, q – charge, B GM e Re

(s) (farad) (volt)2 (kg)–1

G – universal gravitational constant Me – mass of the earth, Re – radius of the earth IIT, 2007

SOLUTION 6. (a) F =

Gm1m2 . Therefore the SI unit of G is Nm2 kg–2. r SI unit of GMeMs = (Nm2kg–2) kg2 = Nm2 = kg ms–2

(b) SI unit of

3RT SI unit of PV SI unit of work = = = Nm kg–1 = kg ms–2 M SI unit of M kg

(c) From F = q v B (d) SI unit of

m2 = kg m3 s–2.

GM e Nm 2 kg 2 = Re m

kg

=

kgms

2

m

kg–1 = m2 s–2

F F2 = SI unit of v. Hence SI unit of 2 2 = (ms–1)2 = m2 s–2 qB q B m 2 kg 2 kg = m2 s–2 m

(p) Since volt coulomb = work, SI unit of (volt) (coulomb) (metre) = SI unit of work = Nm2 = kg ms–2 m2 = kg m3s–2

metre = Nm

coulomb and coulomb volt = work, the SI unit of (farad)(volt)2 (kg–1) = (coulomb) volt kg–1 = SI unit of work kg–1 = kg ms–2 m kg–1 = m2s–2

(s) Since farad =

m

(volt)

1.24 Comprehensive Physics—JEE Advanced

Hence the correct choices are as follows (a) (p), (q) (b) (r), (s)

(c)

(r), (s)

(d)

(r), (s)

V Assertion-Reason Type Questions In the following questions, Statement-1(Assertion) is followed by Statement-2 (Reason). Each question has the following four options out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and State(b) Statement-1 is true, Statement-2 is true but Statement-2 is not ment-1. (c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. 1. Statement-1 The order of accuracy of measurement depends on the least count of the measuring instrument.

Statement-2 The smaller the least count the greater is the num2. Statement-1 The dimensional method cannot be used to obtain the dependence of the work done by a force on the angle between force and displacement x. Statement-2 All trignometric functions are dimensionless. 3. Statement-1 The mass of an object is 13.2 kg. In this measureStatement-2

SOLUTIONS 1. The correct choice is (b). 2. Work done is W = F x cos . since cos is dimensionless, the dependence of W on cannot be determined by the dimensional method. Hence, the correct choice is (a)

3. The correct choice is (c). The degree of accuracy measurement cannot be increased by changing the unit.

2

Motion in One Dimension

Chapter

REVIEW OF BASIC CONCEPTS 2.1

SCALAR AND VECTOR QUANTITIES

A scalar quantity has only magnitude but no direction, such as distance, speed, mass, area, volume, time, work, energy, A vector quantity has both magnitude and direction, such as displacement, velocity, acceleration, force,

2.2

2.1 A particle moves from A to B along a circle of radius R placement in terms of R and

POSITION VECTOR AND DISPLACEMENT VECTOR

The position vector of a particle describes its instantaneous position with respect to the origin of the chosen frame of and is denoted by vector r For one-dimensional motion (say along x-axis), r = x i , where x is the distance of the particle from origin O For two-dimensional motion (say in the x–y plane), r = xi

Fig. 2.2

SOLUTION Path length = arc AB = R

y j , where (x, y) are the x and y coordinates of

For three-dimensional motion, r = x i

yj

zk

Displacement vector r1 is the position vector of a particle at time t1, and r2 at time t2, then the displacement vector is given by s = r2 – r1 Vector s is the resultant of vectors r2 and – r1 The displacement vector particle after a given interval

Fig. 2.3

Fig. 2.1

s = AB = AC + CB = 2 AC s = 2R sin ( /2)

2.2 Comprehensive Physics—JEE Advanced

2.3

The rate of change of displacement with time at a given instant is called instantaneous velocity and is given by dx v= dt vav =

total displacement time interval

2.2 The position of a particle moving along x-axis is given by x = 2t t2 + t , where x is in metre and t (a) Find the velocity of the particle at t (b) Find the average speed of the particle in the time interval from t = 2 s to t SOLUTION (a) x = 2t

t2 + t

dx = 2 – 6t dt v at (t = 2 s) = 2 – 6

t2

v=

(b) Position at t = 2 s is x1 = 2

(2)2 = 2 ms–1

(2)2 + (2) = 0

Position at t x2 = 2

Average velocity =

2

12 ms

1 2 at 2

and

1 2 at 2 v2 – u2 = 2 a(x – x0)

or

v2 – u2 = 2a s

or

s = ut +

in a straight horizontal direction, we will consider only the magnitudes of u, v, a and s and take care of their direction by assigning positive or negative sign a will mean acceleration – a Quantities directed vertically upwards are taken to be positive and those directed vertically downwards to gravity is directed downward for a body moving vertically up or falling vertically down, we take a –2 =–g nth second is given by sn = displacement in n seconds – displacement in (n – 1) seconds 1 1 a (n 1) 2 = un + a (n) 2 u (n 1) 2 2 sn = u +

a (2n 1) 2

Applications 1

INSTANTANEOUS ACCELERATION

The rate of change of velocity with time at a given instant is called instantaneous acceleration and is given by dv a= dt

2.5

x = x0 + ut +

2

x2 – x1

2.4

v = u + at

INSTANTANEOUS VELOCITY AND AVERAGE VELOCITY

EQUATIONS OF ONE DIMENSIONAL MOTION WITH CONSTANT ACCELERATION

Let x0 be the position of a particle at time t = 0 and let u be its velocity at t a for time t x and acquires a velocity v The particle suffers a displacement s = x – x0 in time t The equations of motion of the particle are

from A with initial velocity u and reaches B with a velocity v, then the velocity midway between A and B is u 2 v2 v = 2 (ii) A body starting from rest has an acceleration a for a time t1 and comes to rest under a retardation b for s1 and s2 are the distances travelled in a time t2 t1 and t2, (a)

s1 s2

b a

t1 t2

(b) Total distance travelled (s1 + s2) = where T = t1 + t2

1 ab T2, 2 a b

(c) Maximum velocity attained is vmax =

ab a

b

T

2.3

(d) Average velocity over the whole trip is v vav = max 2 (iii) At time t = 0 a body is thrown vertically upwards with a velocity u At time t = T, another body is thrown vertically upwards with the same velocity u T u t= 2 g (iv) A body is dropped from rest and at the same time another body is thrown downward with a velocity u from the same point, then (a) the acceleration of each body is g, (b) their relative velocity is always u, x (c) their separation will be x after a time t = u (v) From the top of a building, body A is thrown upwards with a certain speed, body B is thrown downwards with the same speed and body C is dropped t1, t2 and t are their respective times of reach the ground, then t =

t1 t2

(vi) A body of mass m is dropped from a height h on a x (a) the average retardation in sand is given by gh a= x because loss in PE (mgh) = work done against the resistive force of sand (max (b) total average force exerted by sand is F = mg + ma = m(g + a)

(f) the speed with which the body hits the ground is 2 g

v=

a h

u

g g

a a

NOTE Note that t1 is less than t2

Some useful tips …… are in the ratio 12 : 22

2

acceleration, the distance covered by it in the lst, air resistance is nelected, the speed with which it u, the maximum height attained is proportional to u2 and the time of ascent is proportional to u (v) For a freely falling body, (a) velocity time (b) distance fallen (time)2 (c) velocity distance fallen

2.6

GRAPHICAL REPRESENTATION Displacement – time (x – t) graphs (Fig. 2.4)

(vii) A body is thrown vertically upward with a velocity u a constant acceleration (or retardation) a (< g) (a) the net acceleration during upward motion = g + a, (b) the net acceleration during downward motion = g – a, (c) the maximum height attained is

ating

a > 0) a < 0) -

u2 2 g a (d) the time taken to reach the maximum height is h=

t1 =

2h g

u a

g

Fig. 2.4

a

NOTE

(e) the time of descent is t2 =

2h g

The slope of x – t graph gives velocity for uniform mo-

u a

g2

a2

1/ 2

2.4 Comprehensive Physics—JEE Advanced

v – t) graphs for uniformly acceler-

2.3 The displacement x (in metres) of a body varies with time t (in seconds) as 2 2 t 16t 2 x = How long does the body take to come to rest?

O

t a

O

o

u

o a

t o

O

t u

o a

o

SOLUTION v=

O

t

–u u

o a

O

o

t u

o a

O

O

u

t

–u u

o a

o

Fig. 2.5

0=

t

o

o a

o

Acceleration = slope of (v – t) graph v – t) graph

t 16

t 16

t = 12 s

2.4 vertically upwards with a velocity of 10 ms –1 (a) After how long will the ball hit the ground? (b) After how long will the ball pass through the point c will it hit the ground? Take g = 10 ms–2 SOLUTION u = + 10 ms –1, a = – 10 ms–2

(a) s

NOTE

dx = dt

Now

s = ut + t+

a–t

1 2 at 2 1 2

(– 10)t2

t2 – 2t (t + 2) (t t

Fig. 2.6

a – t), (v – t), (x – t

t is not

(b) s = 0, u = + 10 ms –1 and a = – 10 ms –2 0 = 10t t2 t=2s –1 (c) v = u + at = 10 – 10 The negative sign indicates the velocity v is di2.5

Fig. 2.7

elled by the body in 20 s, (b) the displacement of the body in 20 s and (c) the average velocity in the time interval t = 0 to t

2.5

S = u

2.7

29

29a 2

2 2

m

RELATIVE VELOCITY IN ONE DIMENSION

A and B are moving in a straight line with velocities vA and vB respectively, the relative velocity of A with respect to B vAB = vA – vB The relative velocity of B with respect to A will be vBA = vB – vA

Fig. 2.8

SOLUTION (a

2.7 A police van moving on a highway with a speed of 10 ms –1 –1

–1

Speed (ms–1) 6

, with what speed will

SOLUTION 5

10

15

20

( )

–1

Fig. 2.9

= area of OAB +

area of BCD

car is vC –1 respect to van is vBV vBV = vB – vV –1

vB be the velocvB = vBV + vV =

a speed =

1 2

(b

6 10

1 2

6 10

OAB – area of

(c) Average velocity =

displacement time

0

2.6 A body moving in a straight line covers a distance of

2.8 From the top of a tower 60 m tall, a body is thrown vertically down with a velocity of 10 ms –1 same time, another body is thrown vertically upward from the ground with a velocity of 20 ms –1 how long will the two bodies meet? (b) At what height above the ground do they meet? Take g = 10 ms –2

SOLUTION Sn = u S = u S = u

a 2n 1 2 9a 2 a 2 a = 2 ms –2

–1

vBC = vB – vC

60 m

BCD

–1

vV = 10 ms–1

Fig. 2.10

SOLUTION (a (1) (2) and

u=

C and let t be the

For body 1: s = – h1 , u1 = – 10 ms –1, a = – 10 ms –2 1 – h1 = – 10t + 10 t 2 2 h1 t (t + 2) (1)

2.6 Comprehensive Physics—JEE Advanced

For body 2: s = + h2, u2 = + 20 ms –1, a = – 10 ms –2 h2 = 20t t2 t (– t t or 60 = Adding (1) and (2), h1 + h2 t t (b) Using t h2 = 20 m Alternative method –1

u12 = u1 – u2

v

and obtain the expression for v(t x, use v = dx = v(t) dt

1 2 at 2

t

t h2 = 20t

t 2 = 20

(2)2 = 20 m

dx which gives dt

v and integrate both

sides t

x

dx =

a12 = a1 – a2 = – g – (– g s = ut

a dt 0

u

v dt 0

x0

s12 = – h1 – (h2) = – (h1 + h2) = – 60 m Using

t

dv =

2.10 A particle starts from rest at x –2 which varies with time as shown time t ticle and (b) its displacement in time interval from t = 0 to t –2

2.9 –1 The driver of train A sights another train B moving on the same track at a speed of 10 ms –1 ately applies brakes and achieves a uniform retardation of 2 ms –2 minimum distance between trains A and B when the driver of A sights B?

5ms

6s

Fig. 2.11

SOLUTION (a)

SOLUTION B is,

uAB

A = uA – uB

aAB

A B is = aA – aB = – 2 – 0 = – 2 ms –2

–1

m =

–2 c y = mx + c, the acceleration a varies with time as

a= dv = dt

A B vAB of A B is sAB which is found by using the relation v2 – u2 = 2 as which gives sAB = 100 m 0 – (20)2 = 2 (– 2) sAB

2.8

SOLVING PROBLEMS INVOLVING NONUNIFORM ACCELERATION

(a) Finding velocity and displacement if the dependence of acceleration on time is given. dv Use a = which gives dv = a dt dt given expression for a in terms of t and integrate

2

ms

6

v

6

t

6

t

t

dv =

0

6

0

v= zero at t Using t

12

t

dt

t2

(1)

t

v will be maximum at t –1 vmax

(b) From v =

dx , we have dx = v dt dt

x

t

dx = 0

v dt = 0

2 0

12

t2

t dt

2.7

t

x=

t 2

12

2

a=

0

2 2 = m 12 2 9 (c) Finding velocity and displacement if the dependence of acceleration on displacement is given dv dx d v dv Use a(x) = = =v dt dt dx dx dx v dt or v dv = a(x) dx

v

=

a(x) in terms

of x v

v dv =

t

dt 0

t

x

dx =

v dt 0

x0

Hence we get an expression for x 2.12

a x dx

dx = v (x)

ity u ates it at a rate a = neous velocity and k

t = 0) moving with a velock v where v is the instanta-

SOLUTION

t

dt (a) a =

0

where v 2.11 A particle is moving along the x-axis with an acceleration a = 2x where a is in ms –2 and x the particle starts from rest at x when it reaches the position x

dv dt

k v = v

dv v

u 1/ 2

2 v

1/ 2

u

dv dx v 0

2x = v

dv dx

v dv = 2x dx

v d v = 2 x dx

dv dt

2 1

–1 which gives v (c) Finding velocity and displacement if the dependence of acceleration on velocity is given

v

= – k dt

= – k dt 0

= – kt

Putting v = 0, we get t =

2 u k

dv dv = v dx dt dv k v = v dx = dx a=

1

v2 x2 = 2 2 2

dv

t

SOLUTION a =v

dt

Hence we obtain the expression for v(t dx v= dx = vdt dt

x0

Hence we get an expression for v(x) in terms of x dx v(x) = dt dx dt and integrate v x

x0

dv a

x

v

x

dv = a u

dv dt

x

dx = 0

x=

0

1 k 2u

v dv u /

k

v dv k

2.8 Comprehensive Physics—JEE Advanced

I Multiple Choice Questions with Only One Choice Correct 1.

last second of its fall, the ball covers a distance g = 10 ms–2, the height of the tower is

2. A ball is thrown vertically upwards from the foot of

(c) 100 m (d) 120 m 3. Two cars travelling on a straight road cross a kilometer stone A at the same time with velocities 20 ms–1 and 10 ms–1 with constant accelerations of 1 ms–2 and 2 ms–2 kilometer stone B at the same instant, the distance between A and B is (c) 1000 m (d) 1200 m 4. The acceleration a of a body moving with initial velocity u changes with distance x as a = k2 x , where k elled by the body when its velocity becomes 2u is u 2k

(c)

u 2k

/

(b)

u 2k

(d)

u 2k

/

2px ( p q)

(b)

k2 (d) 2 k2 2 7. The velocity of a particle at time t (in second) is related to its displacement x (in metre) as v= x (a) 1 ms–1 (b) 2 ms–1 –1

8. A car, starting from rest, has a constant accelera–2 for some time and then has a constant retardation of 2 ms–2 for some time and The maximum velocity of car during its motion is –1 (a) 12 ms–1 –1 (d) 21 ms–1 9. A freely falling body, falling from a tower of height h covers a distance h/2 in the last second of its moThe height of the tower is (take g = 10 ms–2) nearly 10.

/

2q x ( p q)

(d) – q x

6. The velocity of a particle moving along the x-axis is given by v = k x where k The acceleration of the particle is

k 2

A ms–1 towards a bigger ball B collision with ball B, ball A retraces the path and –1

instant is

(c)

(b)

(c)

/

5. A particle is moving along the x neous velocity when it is at a distance x from the origin is v = p q x 2 , where p and q are positive

(a) zero

k 2

–1

Take g = 10 ms–2

(a)

(a)

time interval 0 to 6 s? (a) zero –1

11.

A during the (b) 2 ms–1 –1 –1

A monkey is running northwards on the roof of the –1

of the monkey as observed by a person standing on the ground? –1 in the southward direction –1 in the northward direction –1 in the southward direction –1 in the northword direction 12. A police van moving on a highway with a speed –1 ing away in the same direction with a speed of –1

2.9 –1

, with what speed will the bullet hit the

(a) 120 ms–1

19.

(x-t)

–1

–1

–1

the velocity-time (v-t) graph of the motion of the

–1

13. Car A

on a B and C, each moving with –1 in opposite directions on the other lane are approaching car A stant when the distance AB = distance AC = 1 km, the driver of car B decides to overtake A before C car B so as to avoid an accident? (b) 2 ms–2 (a) 1 ms–2 –2

–2

14. The driver of a train A ms–1 sights another train B moving on the same track towards his train at a speed of 10 ms–1 immediately applies brakes and achieves a uniform

Fig. 2.12

–2

what must be the minimum distance between the trains? (a) 100 m (b) 200 m 15.

g = 10 ms–2, the

16. A body, starting from rest, moves in a straight line with a constant acceleration a for a time interval t during which it travels a distance s1 move with the same acceleration for the next time interval t during which it travels a distance s2 relation between s1 and s2 is (a) s2 = s1 (b) s2 = 2s1 (c) s2 s1 (d) s2 s1 17. v1 is the velocity of the body at the end v2 that at the end of the second time interval, the relation between v1 and v2 is (a) v2 = v1 (b) v2 = 2v1 (c) v2 v1 (d) v2 v1 18. A body dropped from the top of a tower hits the the tower? (a) 1 s (c) 2 2 s

Fig. 2.13

20.

g = 10 ms–2, what is the ratio of the distances travmotion? (a) 1 : 9

21.

–1

g = 10 ms–2, what is the ratio of the distances travelled by the bullet during the (a) 9 : 1

(b) 2 s (d)

s

(b) 2 : 9

(b) 9 : 2

22. A body moving in a straight line with constant acceleration of 10 ms–2

2.10 Comprehensive Physics—JEE Advanced th

in the 6 second? 23. A body, moving in a straight line with an initial –1 and a constant acceleration, covrd

distance will it cover in the next 2 seconds? (c) 90 m (d) 100 m 24. A body, moving in a straight line, with an initial velocity u and a constant acceleration a, covers a th second and a distance u and a of 60 m in the 6th respectively are –2 (b) 10 ms–1, 10 ms–2 (a) 10 ms–1 –1 –2 –1 , 10 ms–2 25. A car, starting from rest, has a constant acceleration a1 for a time interval t1 during which it covt2, the ers a distance s1 car has a constant retardation a2 and comes to rest after covering a distance s2 in time t2 following relations is correct? (a)

a1 a2

s1 s2

t1 t2

(b)

a1 a2

s2 s1

t1 t2

(c)

a1 a2

s1 s2

t2 t1

(d)

a1 a2

s2 s1

t2 t1

(a)

1 1 a1t1 = a2 t2 2 2

(b)

1 (a1 t1 2

27.

1

a1

a2 t1

t2

(a)

(c)

s a1a2 2 a1 a2

1 2

1 2

g = 10 ms–2

a2 t2 )

(d)

aa 2s 1 2 a1 a2 s a1a2 2 a1 a2

35.

(b)

v 2

–1

h1 during

s, 1 2

1 2

28. A car, starting from rest, is accelerated at a constant rate until it attains a speed v a constant rate (a) zero

(b) na

na

(c) n a (d) n a 34. Two balls are dropped from the same point after an

(d) zero

(b)

(b) s2 = 2s1 (a) s2 = s1 (c) s2 s1 (d) s2 s1 33. A car moving at a speed v is stopped in a certain distance when the brakes produce a deceleration a nv, what must be the deceleration of the car to stop it in the same distance and in the same time? 2

the maximum speed attained by it will be aa 2s 1 2 a1 a2

(d) v 2 2 29. The displacement of a body from a reference point, is given by x = 2t where x is in metres and t that the body is (a) at rest (b) accelerated (c) decelerated (d) in uniform motion 30. –1 (a) 2 ms–1 –1 (c) 6 ms (d) 12 ms–1 31. –2 (a) 2 ms–2 –2 –2 (c) 6 ms 32. A car, starting from rest, at a constant acceleration covers a distance s1 in a time interval t tance of s2 in the next time interval t at the same ac-

(a)

26.

(c)

v

(c)

th

h2 and h2 will be related as h2 (a) h1 (c) h1 = h2

g = 10 ms–2, h1

(b) h1 = 2h2 h2 (d) h1 =

36. A ball is thrown vertically downward with a velocity u u reach the ground is given by u 2u (b) (a) g g (c)

u g

(d)

u g

2.11

37. (a)

u2 g

(c)

u2 g

(b)

2u 2 g

(d)

u2 g -

38. –1 –1

train? 39.

and leaves it at a

–2

–2

–2

–2

platform?

(c) 10 s (d) 12 s 40. The motion of a body is given by the equation dV t dt

41.

V(t)

where V(t) is the speed (in ms–1) at time t (in sect = 0, the magnitude of the initial acceleration is –2 (b) 6 ms–2 –2 (c) 9 ms (d) zero (a) V(t) = (1 – e t ) (b) V(t) = 2 (1 – e t) (c) V(t) =

2

t

1 e

2

2t

(d) V(t) = 42.

2

1 e

tion is half the initial value is (b) 2 ms–1 (a) 1 ms–1 –1

-

–1

43. Two stones are thrown up simultaneously with initial speeds of u1 and u2 (u2 > u1 variation of

x = (x2 – x1), the relative position

t = 10 s? Assume that the stones do not rebound

Fig. 2.14

44. A body starts from rest at time t = 0 and undergoes

2.12 Comprehensive Physics—JEE Advanced

45. Acc (ms 2)

+3 0

1

2

3

3

4 Time (s)

46.

Fig. 2.15

the velocity-time (v–t) graph of the motion of the body from t = 0 s to t

47.

time t

–1

–1

–1

–1

from t = 0 to t (a) 6 m

(b) 9 m

represents the displacement-time (x – t) graph of the motion of the body from t = 0 s to t

Fig. 2.17

48. A body, moving in a straight line, covers half the distance with a speed V, the remaining part of the distance was covered with a speed V for half the time and with a speed V for the other half of the (a)

2V V V 2V V V

(b)

V V V 2V V V

(c)

2V V V V V

(d)

V V V V V

49. A particle moving in a straight line covers half the the distance is covered in two equal time interFig. 2.16

motion is

2.13

motion and air resistance, its velocity v varies with the height h v – t) graphs

50. (a) (b) (c) (d)

(i), (ii) and (iv) only (i), (ii) and (iii) only (ii) and (iv) only all

Fig. 2.19

54. A car, starting from rest, accelerates at a constant –2

constant rate of 10 ms–2 Fig. 2.18

51. A stone dropped from a building of height h reaches the ground after t ing if two stones are thrown (one upwards and the other downwards) with the same velocity u and they reach the ground after t1 and t2 seconds respectively, then the time interval t is

52.

(a) t = t1 – t2

(b) t =

(c) t =

(d) t =

t1 t2

t1

2 t12

t2 t22

x) of a particle is related to time (t) as x = at + bt2 – ct where a, b and c velocity of the particle when its acceleration is zero is given by 2 (a) a + b c

(b) a +

b2 2c

b2 c

(d) a +

b2 c

(c) a +

53. A ball is dropped vertically from a height h above vertically to a height h

attained by the car? –1

55.

(c) 20 ms

–1

(b) 10 ms–1 –1

car is 6 s?

(c) 100 m (d) 120 m 56. The distance x covered by a body moving in a straight line in time t is given by the relation 2x2 x=t v is the velocity of the body at a certain instant of time, its acceleration will be (a) – v (b) – 2v v v 57. The distance x covered by a body moving in a straight line in time t is given by x2 = t2 + 2t The acceleration of the body will vary as 1 1 (a) (b) 2 x x (c)

1 x

(d)

1 x

58. A body is thrown vertically up with a velocity u passes three points A, B and C

2.14 Comprehensive Physics—JEE Advanced

u u u with velocities , and 2 AB is BC 20 (a) (b) 2 (c)

10

average velocity of the particle is –1

–1 –1

(d) zero

(d) 1

59. A body is thrown vertically up with a velocity u passes a point at a height h above the ground at time t1 while going up and at time t2 Then the relation between u, t1 and t2 is 2u 2u (a) t1 + t2 = (b) t2 – t1 = g g (c) t1 + t2 =

u g

(d) t2 – t1 =

(a) t1t2 =

2h g

(c) (t1 + t2)2 =

(b) t1t2 = 2h g

h g

(d) (t1 + t2)2 =

t= x where x is in metres and t ment of the particle when its velocity is zero is 67.

the maximum time to reach the ground if H H (b) h = (a) h = 2 2 H H (d) h = 2 2 62. A body of density enters a tank of water of density after falling through a height h depth to which it sinks in water is (a) (c)

(

)

h

(b) (d)

h

(a) T =

2 u k

(b) T =

2u k

(c) T =

2u / k

(d) T =

2u 2 k

68.

–2

t

(

)

h

63. A body, falling freely under gravity, covers half the

will be

which varies with time as shown

ms–1 ms–1

ms–1 ms–1

for n seconds, then the value of n is (c) 2 – 64.

2

(d) 2 +

-

t = 0) moving with a velocity u result of which it decelerates at a rate a=–k v where v is the instantaneous velocity and k is a T taken by the particle to come to rest is given by

(c) h =

h

-

66. The displacement x of a particle moving in one dimension is related to time t by the equation

h g

61. A body dropped from a height H above the ground strikes an inclined plane at a height h above the

Fig. 2.20

zontal and with speed u travelled along the incline is (a) 2h (b) h h (d) h (c) 2

u g

t1, t2 and h is

60.

65. A body of mass m1 cally upwards with an initial velocity u reaches a maximum height h Another body of mass m2

2 A to point B, Fig. 2.21

2.15

69.

a body with position (x) from the origin O

v) of

a – t)

71.

v – t) of the motion of the body? Assume that x = 0 and v = 0 at t

variation of the acceleration (a) with position (x)? v v

Fig. 2.22

Fig. 2.25

v (ms–1) 0

v (ms–1)

2

4

0

( )

2

(a)

4

( )

(b)

Fig. 2.23

70. The velocity (v) of a body moving along the postive x-direction varies with displacement (x) from the origin as v = k x , where k the graphas show the displacement-time (x – t) graph of the motion?

v (ms–1) 0

v (ms–1)

2

4

0

( )

2

(c)

4

( )

(d)

Fig. 2.26

72.

-

val t = 2s to t (a) 2 m

(b) – 2 m

73. A particle of mass m moving with initial velocity u enters a medium at time t a resistive force F = kv wher k is a constant of the medium and v velocity of the particle varies with time t as kt kt (b) v = u – (a) v = u + m m (d) v = u ekt/m 74. x of the particle varies with time t as (assume that x = 0 at t = 0) (c) v = u e

Fig. 2.24

kt/m

2.16 Comprehensive Physics—JEE Advanced

mut k mu (b) x = t k (a) x =

75. 1 2 t 2

(c) x =

mu 1 e k

kt/m

(d) x =

mu 1 e 2k

kt/m

u at 2

time t given by m (a) t = ln(2) k (c) t =

(b) t =

m k

(d)

m ln(2) 2k

m 2k

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67. 73.

(c) (b) (a) (d) (d) (d) (d) (a) (d) (a) (c) (a) (c)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68. 74.

(a) (c) (b) (a) (a) (c) (c) (d) (c) (d) (a) (b) (c)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63. 69. 75.

(a) (a) (c) (a) (a) (c) (c) (c) (c) (c) (d) (d) (a)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64. 70.

(b) (b) (c) (b) (d) (c) (b) (c) (c) (a) (b) (c)

(d) (c) (b) (c) (b) (c) (b) (b) (a) (a) (a) (c)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65. 71.

6. 12. 18. 24 30. 36. 42. 48. 54. 60. 66. 72.

(c) (a) (c) (d) (b) (a) (a) (c) (a) (a) (d)

SOLUTIONS 1. Let h be the height of the tower and t be the time h= and

1 2 gt 2

h = ut +

(1)

2+ = 60 m

1 h = g (t – 1)2 2

Now h = h – 16 h

9h

=

=

1 2 gt 2

1 (– 10) (2)2 2

16 h

1 g(t – 1)2 2

(2) t

1 h= 2

Fig. 2.27

10

2

2. Let h = AB be the height of the tower and P be the by the ball to go from B to P time taken to go from A to P fore, time taken by the ball to go from A to B is t u –1 0 = u – 10 u

3. Let s be the distance between A and B 1 1 1 t2 = 20t + t2 2 2 For the second car, 1 (2) t2 = 10t + t2 s = 10t + 2 Equating (1) and (2), we get t value of t in either (1) or (2) gives s s = 20t +

(1)

(2)

2.17

a=

2u

2

1 1 1 h 1 gn2 = g g 2 2 h 2 2h 1 1 h2 1 = g = g g2 g

h=

v d v = a dx = k2 x dx v d v = k2

x dx

u

x= 5. a =

Hence h is given by

dv dv d v dx v = = dx dt dx dt

u 2k

2

1 (p – qx2)–1/2 2

10. The time taken by ball A to reach ball B =

12 m = 2 ms–1 which is 6s

11. –1

-

Therefore, the velocity of the monkey as observed

d (k x ) dx k k2 = 2 x 2

–1

The negative sign indicates that the direction of this

The car attains the maximum velocity at the end of t1 –1 vmax t1 1 1 h g (2n – 1) so that n = 2 2 g

–1

12.

The time during which the car decelerates is t2 = (t – t1) where t is the total time taken for the car t2, the initial velocity is v t1 t1 – 2t2 t1 – 2 (t – t1) t1 – 2t 2 2 t1 = t =

h 2

×2s

–1

7. Given v2 x v2 = u2 + 2ax, 2 u = 2 ms–1 we have u 8. Let t1 The velocity at the end of t1 is t1 t1 v = u + at1

9.

Average velocity =

v= k x

= k x

–1

(– 2 q x)

qx v = (p – qx2)1/2 v Hence a = – qx 6. The acceleration is dv dv d v dx = = v a= dx dt dx dt

a= k x

20

A covers a

distance of 20 m upto ball B

=

Given

g

g = 10 ms–2, we get h – 60h + 100 = 0 The positive root of this quadratic gives h

/

dv dv d v dx = = v dx dt dx dt dv d = p – qx2)1/2 dx dx =

2h

1

= 10 ms–1

moving, the net speed of the bullet = speed of the –1 –1

–1 –1 –1

and the

the bullet will hit the car with a speed which is the relative speed of the bullet with respect to the –1

13. Let us suppose that cars A and B are moving in the positive x C is moving in the –1 negative x vA –1 –1 –1 = + 10 ms , vB and vC –1 –1 B –1 with respect to A is vBA = vB – vA The relative velocity of C with respect to A is vCA –1 = vC – vA t = 0, the distance between A and B = distance between A and C C will cover a distance

2.18 Comprehensive Physics—JEE Advanced

A at a time t given

AC by t=

AC 1000 m = vCA ms 1

s2 = ut +

Car B will overtake car A C does and avoid an accident, if it acquires a minimum acceleration a such that it covers a distance s = AB = 1000 m in time t –1 speed u = vBA tion 1 2 s = ut + at 2 1 2 ×a 2 which gives a = 1 ms–2 14. The relative speed of train A with respect to train B –1

is given by

15.

for the next time interval t during which the body

s2

=2

18. Let h 1 g 2

which gives s

tance of s1 =

1 2 1 gt = 2 2

10

(2)2

h= 2

g

h The time t taken to fall through 2 h 1 2 1 2 = gt g= gt or t2 2 2 2

s

(u) of the bullet is ( –1 u = gt = 10 maximum height (h) attained by the bullet is h 1 2 1 2 gt = 10 = 2 2 time taken by the bullet to return to the point of

20.

t is

-

1 2 1 2 at = at 2 2 The velocity of the body at the end of this time interval is v = 0 + at = at s1 = 0 +

g is given by t= 2 2

total time taken by the stone to hit the ground is given by h=

1 2 gt or t = 2

2h g

2 10 h1

h1 =

1 g g(1)2 = 2 2

tance h given by

-

1 g 2 g 2 h2 through which the stone falls in the

h=

which gives t 16.

1 2 gt = 2

ment x increases linearly with time t velocity v is a positive constant between t = 0 and t t t v) is zero between t t t t = 20 s, displacement (x) decreases linearly with time (t v) is constant but negative between t t

given by

t seconds, where t is the time taken by the bullet to rise to a height h of t is given by 1 1 gt 2 t– 10 t2 h = ut – 2 2 or t2 – 10 t + 21 = 0 or (t t

u = at)

17. Here v1 = 0 + at = at and v2 = v1 + at = at + at = 2 at v2 = 2v1

19. 2

1 2 1 2 at = at2 + at = at2 ( 2 2 2 s1

Now h1/h2 21. The maximum height h attained by the bullet is given by u2 ( v = 0) v2 – u2 = – 2gh or h = 2g

2.19

or h =

2 10 en by the stone to attain this height is given u g 10 (t = 1 s), the stone covers a distance h1 given by

by t =

h1 = ut –

1 2 gt 2

1–

ers a height h given by 1 2

h

1 2 t

10

(1)2

s), the stone cov2

10

= 120 m

22. The distance covered in the nth second is given by 1 sn = u + a n 2 1 s =u+a =u+ a 2 2 u+

2

6

s6

23. Now, sn = u + a n a s and s

s2 =

2 2 1 a12 2 1 a2 t2 t1 = 2 a2 2 a2

or

s2 =

1 a 2 t22 2

Thus we have

26. Average speed =

1 2

which gives

1 which gives a = 10 ms–2 2 1 2 1 2

s =s +s a 11a and 60 = u + 24. u+ 2 2 –1 These equations give u and a = 10 ms–2

a 1t 1 = a 1t 1 time interval t2 have, from v = u + at, 0 = a1 t1 – a2 t2

a 1t 1 = a 2t 2)

a2 t22

a2 a12 t12 a1 a22 t22

=

a2 ( a1

a 1t 1 = a 2t 2)

t2 t1

total distance total time

s1 t1

s2 t2

1 1 a1 t12 + a 2 t22 2 2

1 a1t1 (t1 + t2) 2

(

a 1t 1 = a 2t 2)

1 a t t t 2 1 1 1 2 Average speed = t1 t2 1 1 = a 1t 1 = a 2t 2 2 2 27. The maximum speed v attained by the car = speed it attains at the end of time interval t1 during which v = a 1t 1 = a 2t 2 1 v2 Now s1 = a1 t 12 = 2a1 2 and s2 =

v2 1 a 2 t22 = 2a2 2

s = s1 + s2 = or

u = a 1 t 1)

a1 a2

=

t1 is v = 0 +

(

a1 t12

s1 s2

As shown above, s1 + s2 =

1 2

25.

s2 s1

(

(ii)

=

–1

10 or u

s6

or

From (i) and (ii) we get

h2

h1/h2

t1 is given by 2a1s1 = v2 – u2 = a12 t12 ( v = a 1t1 and u = 0) 1 a 1t 12 (i) or s1 = 2 The distance covered in the next time interval t 2 is given by – 2a2 s2 = 0 – a21 t21 ( v = 0 and u = a1t1 now)

v = 2s

v2 1 2 a1

a1 a2 a1 a2

1/ 2

(

v = a 1t 1)

(

v = a 2t 2) 1 a2

2.20 Comprehensive Physics—JEE Advanced

28. The distance s1 covered by the car during the time it is accelerated is given by 2 s1 = v2, which gives s2 covered during the time s1 = v2/2 the car is decelerated is, similarly given by s2 = v2/2 v2 1 1 s = s1 + s2 = (i) 2 t1 is the time of acceleration and t2 that of deceleration, then v = t1 = t2 or t1 = v/ and t2 = v/ Therefore, the total time taken is 1 1 (ii) t = t1 + t2 = v From (i) and (ii), the average speed of the car is given by total distance s v total time t 2

a* = (nv)2/2s = n2v2 34.

t1 1 1 falls a distance h1 = gt 12 = 10 2 2 The second ball falls for t2 1 1 it falls a distance h2 = gt 22 = 2 2 h1 – h2

2

2

10

h1 = ut –

1 2 gt 2

1–

1 2

10

(1)2 –1

2 seconds is v = u – gt x = 12t t2 + 9 x changes with time t, the body by

dx t dt v changes with time t, the body is not in uniform motion; it is accelerated because v increases with t 30. v t v = u + at u = 12 ms–1 v=

31. Now v given by

t a a=

or a

dv dt

d 12 dt

t

10

2

h1 = h2 36. From v = u + gt u = u + gt or 2u = gt or t = 2u/g u)2 – u2 u2 37. From 2gh = v2 – u2, we have 2gh 2 or h u /g 38. The total distance covered in order to cross the platform is s = length of train + length of platform = speed is u ms–1 2 = 2a 39.

–1

v

v2 – u2 = 2as a

2

– which is

–2

a

the relation, v = u + at

–2

t or t

–2

t is v = at distance covered during this time interval is s1 = 1 2 at v = at is the initial velocity for the 2 next time interval t elled in the next time interval t is s2 = a t2 + Thus s2

1 2

seconds is h2

v = u + at, we a is

–2

=

35. The total time taken by the bullet to reach the highest point (where its velocity becomes zero) is given by 0 = u – gt or t = u/g

29.

32.

a * = n 2a

1 2 at = at2 2 2

s1

33. The distance over which the car can be stopped is given by 2 as = v2 or a = v2/2s v becomes nv, the value a* of a to stop the car in the same distance is

40. Given t = 0 is

dV t V(t) (i) dt t = 0, V(0) = 0 (given), the acceleration at a(0) =

41.

dV t dt

–2 t 0

V(t) with respect to t given in each pression for dV(t)/dt dV t d 21 e = dt dt

t

= 2

d (1 – e dt

t

) = 6e

t

2.21

= 6 – 6 (1 – e

time interval = 6 ms–1 time interval from t = 2 s to t v = u + at t Therefore, velocity at t s after t = 2 s) – and velocity at t 2 s after t velocity time graph from t = 2 s to t -

V (t)

t

V(t) = 2 (1 – e t)

(ii)

42.

–1

at time t * given

V(t *) or V(t*) = 1 or

2(1 – e

t*

) = 1 or e

t*

t * ln e–1

The graph is again linear but its slope is negative BC in

t*

*

or t Putting this value of t V at t

)

e

–1

= 2(1 – e 43.

stones are given by 1 2 gt x 1 = u 1t – 2 and x2 = u2 t –

–1

46. The distance covered = area under the velocity time graph from t = 0 to t ABC in (

g is negative)

area of triangle ABD 1 BD AD = BD AD = 6 ms–1 2

2 s = 12 m

47.

x = x2 – x1 = (u2 – u1)t x2 – x1) varies linearly with t, the graph is a straight line upto t At t x1 = 0; so we have for time between 6 1 2 x = x 2 – x 1 = u 2t – gt 2 Thus (x2 – x1) versus t graph is not linear; it is a Now at t = 10 s, x = 0, therefore 1 g (10)2 0 = 10u2 – 2 –1 or u2 g 2 Hence x t t t = 0, u t = 0 to t = 2 s, acceleration a

=2 =2

1 2 gt 2

stones:

44.

45. As shown above, the velocity of the body in time interval t = 2 s to t v t At t v

x–t) t = 0 s and t t = 0 s to t = 2 s, the slope of the graph must increase with t AB t = 2 s to t BC of graph (c), the slope is increasing

48. Let the total distance be 2 s and let t1 and t2 be t1 = s/V –2

t = 0 to t = 2 s), the velocity of the body is v = 0 + at or v t –1 and at t = 2 s, v = 6 ms–1 at t = 1s, v from t = 0 to t = 2 s, the velocity–time graph is AB of For the next time interval from t = 2 s to t –2 acceleration a initial velocity u

t2 distance s1 covered in time with speed V is s1 2 t2 t2 =V and the distance s2 covered in time with 2 2 t2 , so that speed V is s 2 = V 2 s = s1 + s2 = V

t2 2

V

= (V + V )

t2 2 t2 , which gives 2

2.22 Comprehensive Physics—JEE Advanced

2s V V s Total time taken = t + t2 = V

t2 =

2s V V

s 2V V V V (V V ) Hence, average speed= =

2s V (V s 2V V

V ) V

total distance total time

2V V V 2V V V

49. Let the total distance be S S is 2 S /2 S t1 = = 6 Let t2 be the time taken to cover a distance S1 with t that to cover distance S 2 with Then Now

S1 S1 + S2 =

Therefore

t1 t12 t2 = t 2 which gives t = 52. Velocity is

t2 and S 2

t2 or

S 6

total distance = total time

S 2

S S S/

t22

v =

dx = a + 2bt dt

ct2

(i)

d2 x

= 2b – 6ct d t2 Acceleration is zero at time t given by 0 = 2b – 6ct b t or t = c b b2 b2 c a v = a + 2b c c 9c 2 Acceleration is

S = S1 + S 2 2 S t2 =

t2

t1 t2

t

S and t2 = t 2

Total time taken t = t1 + t2 =

1 2 gt1 (ii) 2 For the stone thrown downwards with velocity u, we have 1 2 gt (iii) h = ut 2 + 2 Notice that the displacement of the stone in the three case is the same, equal to h and (iii) we have 1 2 1 2 1 2 1 2 gt = – ut1 + g t 1 ut1 = g t 1 – gt 2 2 2 2 1 2 1 2 1 2 1 2 and gt = ut2 + g t2 ut2 = gt – g t2 2 2 2 2 h = – ut1 +

a =

50. Motion corresponding to graph (i) cannot be realised because if we draw a line parallel to the v-axis, the body will have two different velocities at a given value of t Graph (iii) is also not possible because at some values of t, the graph is parallel to the v-axis which responding to graphs (ii) and (iv) are possible in

53. The velocity at a height h is given by v 2 = u2 + 2gh For downward motion, u = 0 and the value of g is negative and h becomes more and more negav 2 increases with h vector is directed downwards, v becomes more and v2 h, the graph of v versus h is g is For upward motion, v 2 = u2 + 2gh directed downwards and h ly, v2 decreases with h velocity vector is positive, v becomes less and less v with h is parav becomes less and less positive, graph

51. For the stone dropped with zero initial velocity, we have 1 gt 2 (i) h= 2 For the stone thrown upwards with velocity u, we have

54. Let t1 be the time during which the car has an accel–2 and t2 be the time during which eration a1 the car has a deceleration a2 = – 10 ms–2 car starts from rest, the maximum speed attained t2 by the car v = 0 + a1t1 = a1t1 this v

Average speed =

2.23

Therefore, 0 = v – a2t2 or v = a2 t2 a1 t1 = t1 = 2t2 a2 t2 or t1/t2 = a2/a1 t1 + t2 = 6, we get t1 t2 55. The distance moved in t1 1 s1 = 0 × t1 + a1 t21 2 The distance moved in t2 = 2s is 1 a t2 s2 = vt2 – 2 2 2 56. x2 x = t with respect to t we have dx dx x =1 (i) dt dt dx =v xv v x v Now dt t, we have dx dt

2

x v2

d2x

d2x

dt 2 xa

dt 2 a=0

or

a=

where a =

=0 v2

(ii)

x

2h g 61. The time t1 taken by the body to strike the inclined plane is given by 2( H h) t1 = g The time t2 taken by the body to reach the ground after striking the plane is 2h t2 = g 2h 2( H h) Total time t = t1 + t2 = + g g 60. Product of roots is t1 t2 =

v

57. Given, x2 = t2 + 2t

dx dt or

2

or 62.

d x dt

2

x a = x2

(ii)

(iii)

x2 = t2 + 2t a = 2/x u

– 2g(BC) =

u

2

2

u u

H

h

=

d dh

2( H h ) g

1 (H 2

h)

1 h

1/ 2

2h g 1 h 2

or H – h = h

2

= 2

=

u2

u2

=–

u2

(i)

u2 u2 u2 – =– (ii) 16 9

= 0

1/ 2

= 0

or h = u=

Effective retardation in water is geff =

t2 + 2t + 1 + x a = x2

58. – 2g(AB) =

1

dt dh

2

d2x a= 2 we have dt (t + 1)2 + x a = x2 or

2 g

or

(i)

2

x

dx dt

x2

dx x =t+1 dt

1 2 gt 2

The roots of this quadratic equation are t1 and t2 2u The sum of the roots is t1 + t2 = g

v

x

dt 2

or

2h 2u t+ =0 g g

Time t will be maximum if

d x

dx 2x = 2t + 2 dt

t2 –

or

2

a

h = ut –

59.

g(

H 2

2gh )

v Find the maximum depth x from the relation v2– u2 = 2geff x 63. h is the total distance travelled in n seconds, then 1 gn2 h= 2 n – 1)th second, the h 1 = gn2 distance fallen is h = 2 1 1 Hence g(n – 1)2 = h = gn2 2 or n2 n+2=0 The positive root of this equation gives the required value of n

2.24 Comprehensive Physics—JEE Advanced

AB not displacement = t time 65. For the body of mass m1, we have 64. Average velocity =

becomes zero at t particle will be maximum at t

u2 2g For the body of mass m2, if S is the maximum distance travelled along the incline then v2 – u2 = 2aS Now, when S is maximum, v a = – g g sin = – g 2

vmax = –

h=

t=

66

x x = t2 – 6t + 9

velocity v = Find t

(i)

67. Given a = – k v v–1/2 dv = – k dt

or

–2

6

ms–2 per second

and its intercept is c y = mx + c, the acceleration a (in ms–2) as a function of time t is given by

or

x + v0

dv =– dt

v0 x0

dx dt

a=–

v0 x0

v

a=–

v0 x0

having a positive slope =

70. Given v = k x have

v=

or

v=–

t

6

0

12

t

t2

t+k

where k particle starts from rest, v = 0 at t (1) we get k v=–

12

t

2v

dt

2

t

(1)

(2)

v0 x x0

v0

2

dv =– t dt 6

or

-

v0 v02 or a= x x– x0 0 Thus the graph of a versus x is a straight line

t

6

(1)

where v0 and x0 tiating with respect to time t, we have

v02 intercept = – x0

a=–

v0 x0 v varies with x as

v0 x0

v=–

v1/2 = – kt + c given initial condition (v = u at t = 0), we get c= 2 u

68. The slope of the line is m = –

6 –1

and intercept is c = + v0

or

2 v1 2 u1 2 = – kt Now, use t = T and v

(6)2

12

69. The slope of the given v versus x graph is m = –

d 2 dx = (t – 6t + 9) = 2 t – 6 (ii) dt dt v t dv =–k v dt

t=6s

dv k2 = 2 dt

this in (2)

v=

2

and negative

v 2 = k 2x

dv dx = k2 = k 2v dt dt

dv =

v0 x0

k2 dt 2 k 2t 2

v

dx dt

2.25

dx k 2t = 2 dt dx = x= Thus x

k2 tdt 2

dv = dt

k v m

k2

dv = v

k dt m

t

2

t2

v

a – t graph is m =

71. line is

2

–2

c a=

2

2

ms–2

v

t

dt

kt m

v u

kt m

dv =

2

0

t

v=

t

v = ue

kt/m

dx = ue dt dx = u e

kt/m

x

2

t

t

t

(t

0

)

x=

t t2

t dt

x

dx =

t dt 2

t dt 2

2

t 2

or

2

1 2

e

dt

2

2

kt/m

dt

kt/m t

k /m

0

mu (1 e k

kt/m

)

u 2 u = ue 2

2

t

x= =

75. Putting v =

2

0

(1)

0

x= u

2

dx =

kt/m

dx = u e v–t

dx t = dt

=

dt

Thus v = 0 at t

72.

74.

t

0

k dt m0

|ln|vu = ln

t

2

t

dv = v u

dv = t 2 dt dv =

kv m

F m

73. Acceleration a =

kt/m

1 = e kt/m 2 kt ln(2) = m

2 = ekt/m t=

m ln(2) , k

2.26 Comprehensive Physics—JEE Advanced

II Multiple Choice Questions with One or More Choices Correct 1. At time t

–1

t

is neglected, which of the following statements will be true? (a) The two bullets will be at the same height above the ground at t (b) The two bullets will reach back their starting (c) The two bullets will have the same speed at t (d) The two bullets will attain the same maxi2. The ratios of the distances covered by a freely nth seconds of its motion (a) form an arithmetic progression (b) form the series corresponding to the squares n natural numbers (d) form a series corresponding to the differences of the squares of the successive natural numbers 3. The displacement x of a particle varies with time a bt according to the relation x = 1 e b (a) At t = 1/b, the displacement of the particle a/b) (b) The velocity and acceleration of the particle at t = 0 are a and – ab respectively (c) The particle cannot reach a point at a distance x from its starting position if x > a/b (d) The particle will come back to its starting point as t 4. of the following statements are correct? Take g = 10 ms–2 (a) The net displacement of the bullet in 10 s is zero (b) The total distance travelled by the bullet in (c) The rate of change of velocity with time is constant throughout the motion of the bullet

–1

5. Two bodies of masses m1 and m2 are dropped from heights h1 and h2 ground after time t1 and t2 and strike the ground with speeds v1 and v2 correct relation from the following: (a)

t1 t2

h1 h2

(b)

t1 t2

m2 h1 m1h2

(c)

v1 v2

h1 h2

v1 (d) v 2

m2 h1 m1h2

v-t) graphs shown in

6. motion of a particle?

Fig. 2.28

7. Two balls of different masses are dropped from the is neglected and the value of g remains constant, which of the following statements are true? (a) The heavier ball reaches the ground before time (c) The heavier ball hits the ground with a higher speed speed

2.27

8. A body is dropped from the top of a tower of height h h g = 10 ms–2, how long does it take to reach the ground? )s

(a) (2 +

(b) (2 –

(a) the stone passes the point from where it was (b) the speed with which it passes the point of –1

)s

6 )s 9. t = 0) located at x = 0 moves along the positive x-direction under the 6

with x as

(d) the stone hits the water with a speed of –1

13. A balloon is rising vertically upwards at a velocity of 10 ms–1

v=k x

opens his parachute and decelerates at a constant –2 g = 10 ms–2

where k (a) the displacement x varies with time t as x = k 2t 2 k 2t (b) the velocity v varies with time t as v = 2 k2 (c) the acceleration of the particle is 2 (d) the distance s travelled by the particle in time k 2T 2 T is s = 10. The motion of a body is given by dv v dt where v is the velocity (in ms–1) at time t (in sect (a) the velocity of the body when its acceleration is zero is 2 ms–1 (b) the initial acceleration of the body is 6 ms–2 (c) the velocity of body when the acceleration is half the initial value is 1 ms–1

11. A stone falls freely from rest and the total distance covered by it in the last second of its motion is

–1

(c) He hits the ground with a speed of 10 ms–1

14.

a constant reterdation a which varies with instantaneous velocity v as a = – kv where k of the particle is u at time t = 0, then (a) the velocity at time t is given by v = u – at (b) the velocity decreases exponentially with

1 u in time k 2 (d) the total distance covered by the particle before coming to rest is u/k 15. A body moves from point A to point B with a velocity v1 and returns to point A with a velocity v2 1 (a) the average speed is (v1 v2 ) 2 (c) the velocity will decrease to

g = 10 ms–2,

(b) the average speed is

2 v1v2 (v1 v2 )

(c) the stone hits the ground with a speed of –1

(d) the acceleration of the stone during the last -

12.

–1

g = 10 ms–2,

(d) the average velocity is 16.

1 (v1 2

v2 )

of a tower of height h with a speed u reaches the ground after time t1 cally downwards from the top of the tower with the same speed, it reaches the ground after time t2

2.28 Comprehensive Physics—JEE Advanced

Then

1 2 (a) h = g t1 2 (c) u =

17.

t22

1 g(t – t ) 2 1 2

where k of the particle is u at time t = 0, then (a) the velocity at time t is given by v = u – at (b) the velocity decreases exponentially with

1 (b) h = g t1t2 2 (d) u =

1 g(t + t2) 2 1

u 1 in time 2 k (d) the total distance covered by the particle before coming to rest is u/k (c) the velocity will decrease to

a constant ratardation a which varies with instantaneous velocity v as a = – kv

ANSWERS AND SOLUTIONS

1. The two bullets will attain the same height at time 1 t=n n– n2 n 2 1 2 n which gives n 2 t or t is 2 10 = 20 s while that of the second bullet is

placement x is maximum when t a a 1 e b b 4.

ms–1

–1

-

dx d a 1 e bt ae dt dt b Acceleration of the particle is given by

5.

a 1 e b

1

a 1 1 b

2a b

e

1

1

t = 0, the values v and respectively are v = ae–0 = a and = – abe–0 = – ab -

h1 = 0

t1 +

1 2 1 2 gt = gt 2 1 2 1

and h2 = 0 Therefore

t2 +

1 1 gt 22 = gt 2 2 2 2

t1 t2

bt

h1 h2

Also, we have v1 = 0 + gt1 = gt1 and v2 = gt 2

dv d ae bt abe bt = dt dt At t = 1/b, the displacement of the particle is x=

–1

For heights h << radius of the earth, the magnitude of g

3. Velocity of the particle is given by v=

u) of the bullet is

directed upwards which h) attained by 1 2 1 2 gt = 10 the bullet is h = 2 2 Therefore, the total distance travelled by the bullet

t

2. The distance covered in the nth second (with u = 0) is given by 1 1 g(2n – 1) sn = g n2 – (n – 1)2 2 2 1 s1 = g, s2 = g, s = g 2 2 2

tion, its net displacement is zero, which is choice ( u = gt = 10

-

xmax =

v1 Therefore v 2 6.

t1 t2

h1 h2 t

particle can have two different velocities at the

2.29

7.

speed with which a ball reaches the ground are in-

8. The total time t taken by the body to reach the 1 2 1 gt = 10 t2 = t2 ground is given by h1 = 2 2 The body moves for (t – 1) second before the beginan instant (t – 1) second is v = g(t – 1) = 10 (t This is the initial velocity for the motion in the last ond is 1 10 (1)2 = 10(t h2 = 10 (t – 1) 1 + 2 = 10t h1 h2 = t2

10t

or t2 – 6t The two roots of this quadratic equation are t= 9. Given v = k x

or

x–1/2 dx = k dt where c x = 0 at t Use v =

dx = kx1/2 or dt

x1/2 = kt + c

c 2 2

dx k t and x = dt

Acceleration a =

2

dv d k t = dt dt 2

dx k 2t = 2 dt S

O

which gives s =

v= =

k 2t 2

2

k = constant 2

k2 dx = tdt 2

or

k2 tdt dx = 2 O k 2T 2

v = 0 at -

(d) is wrong because acceleration a change with time because v 11. Let h1 – h1 = –

1 g 2

2

h1 =

9 g 2

Let t (t – 1) is g(t – 1) and the distance covered in the last one second is 1 g (1)2 – h2 = – g(t – 1) 1 – 2 g g h2 = g(t – 1) + = gt – 2 2 9 g g = gt – t 2 2 The height H from which the stone falls is 1 1 2 10 – H = – gt2 = – 2 2 H The speed with which the stone hits the ground is –1 –1 v – v = – gt = –10 Given h1= h2

(d) is wrong because for bodies not too far away from the earth, the acceleration due to gravity is –2

12. Let us assign a positive sign to quantities directed in the upward directon and a negative sign to those

s

T

dv v (1) dt (a) Velocity v0 when acceleration a = 0 is obta = 0, we get v0 = 0 or v0 = 2 ms–1

10. Given a =

(b) At t t v tion (a0) = 6 ms–2 a0 –2 (c) Putting a = 2 v v = 1 ms–1

ms–1

u = +10

g = – 10 ms–2 1 values in the relation s = ut + gt2, we get 2 1 t2 t + (–10)t2 = 10t 2 or t2 – 2t which gives t of t

t + 2)(t

2.30 Comprehensive Physics—JEE Advanced

The total time the parachutist takes (after his exit

coming back to the pointfrom where it was pros 0 = 10t +

1 (–10)t2 2

which gives t

t = 0 is not possible

– v = u + gt = 10 + (–10) –1 v 13.

14.

–1

dv = – kv dt v dv = v u

t

kdt 0

v = – kt u v = u e– kt

log e

(1) v=

city of the balloon and has an upward velocity of u = +10 ms–1 g = –10 ms–2 (act10 ms–1 t given by 1 2 gt s = ut + 2 1 2 = 10 (–10) 2

u in time t given by 2

u = ue– kt 2 1 = e– kt 2 = ekt 2 log e (2) 0 = t= k k

loge(2) = kt

dx = u e– kt dt x

the velocity of the balloon is 10 ms–1

t

dx =

v = u + gt = 10 + (–10)

ue

x – x0 = – –1

(directed downwards) At t u = –20 ms–1 and to hit the ground, his displacement s –2 (directed a 1 2 1 gt t + t2 2 2 or t2 t + 12 = 0 or (t – 6)(t – 2) = 0 which gives t t = 6 s, then the velocity with which he hits the ground is v = u + at = v is 6 = 10 ms–1 s = ut +

correct answer is t = 2 s, in which case, the velocity with which he hits the ground is v 2 = –10 ms–1

dt

u – kt e k

x = x0 –

u – kt e k

Now x = 0 at t = 0 which gives x0 = x=

by

kt

0

x0

u (1 – e– kt) k

u k (2)

v = 0 at t = distance travelled by the particle before coming to t = , which gives x = u/k 15. Let s = distance between points A and B t1 is the time taken to go from A to B and t2 the time taken to return from B to A, then t1 + t2 =

s v1

s v2

2.31

Average speed = =

Total distance travelled Total time taken s s v1

v=

u in time t given by 2

u = ue– kt 2

2v1v2 s = s (v1 v2 ) v2

1 = e– kt 2 = ekt 2 log e (2) 0 = t= k k

-

loge(2) = kt

16. The net displacement in both cases is –h (vertically) velocity is +u wards, the initial velocity is –u (since a = – g in both cases) 1 – h = ut1 – gt12 2 1 2 – h = ut2 – gt2 2

and

-

dx = u e– kt dt x

(1)

dv = – kv dt v dv = v u log e

x – x0 = –

(2)

kt

dt

u – kt e k

x = x0 –

u – kt e k

Now x = 0 at t = 0 which gives x0 = x=

t

kdt

u k

u (1 – e– kt) k

(2)

v = 0 at t = distance travelled by the particle before coming t= , which gives x = u/k

0

v = – kt u v = u e– kt

ue 0

x0

1 Equations (1) and (2) give u = g (t1 – t2) and 2 1 h = g t1t2 2 17.

t

dx =

(1)

III Multiple Choice Question Based on Passage Questions 1 to 4 are based on the following passage Passage I v=

total distance travelled total time taken

v =

net displacement time taken

by the total length of the path travelled by the body in a

2.32 Comprehensive Physics—JEE Advanced

The direction of the velocity vector is the same as that of 2.

1. A cyclist starts from centre O of a circular track of radius r = 1 km, reaches edge P of the track and then cycles along the circumference and stops at point Q the displacement of the cyclist is (a) r 1 +

6

(c) r 1 +

3.

4.

Fig. 2.29

(d)

r

km h–1of the cyclist is

approximately –1

is

) of the cyclist

(b) r

SOLUTION 1. Net displacement is OQ = shortest distance between the starting point O and end point Q = 1 km =r 2. Average velocity =

1km displacement = 1 = 6 km –1 time h 6

=r+

2 r =r 1 6

4. Average speed =

= 1 km 1 distance = time

km km 1 h 6

–1

OP + PQ (along the arc)

3.

Question 5 to 7 are based on the following passage. Passage II A is rolled along the positive x-direction with a –1 towards a bigger ball B collision with ball B, ball A retraces the path and reaches –1

5. The average velocity of ball A during the time ms–1 along positive x –1 along negative x (c) (d)

9

ms–1 along negative x ms–1 along positive x

6. The average velocity of ball A during the time interval 0 to 9 s is –1 along positive x (a) –1 along negative x ms–1 along negative x 9 (d) zero 7. The average velocity of ball A during the time interval 0 to 6 s is (a) 2 ms–1 along positive x (b) 2 ms–1 along negative x (c)

(c)

ms–1 along positive x

(d)

ms–1 along negative x

SOLUTIONS Therefore, the velocity of ball A –1 along positive x- direction, which is choice

5. Time taken by ball A to reach ball B is 20 t1 = ment of ball A = 20 m along positive x

-

6. Time taken by ball A to retrace its path and reach the starting point after collision with ball B is

2.33

t2 =

–1

20

Therefore, the net displacement in time interval 0 to 7.

A covers a distance of 20 m along positive x-direction up to

xNet displacement of A along positive xof A during 0 and 6 s is 12 –1 6s = 2 ms along positive x-direction

Questions 8 and 9 are based on the following passage Passage III t = 0) moving with a

velocity u which it decelerates at a rate

a= k v where v is the instantaneous velocity and k is a positive

(c) 2 k u (d) k u 9. The distance covered by the particle before coming to rest is (a)

8. The particle comes to rest in a time (a)

2 u k

(b)

(c)

u k

u

/

(b)

k u / 2k

(d)

2u / k 2u

/

k

SOLUTION a = – kv1/2

8. Given Thus

v

1 2

or

dv = – kv1/2 dt

1/ 2 v1/ 2 = u

kt 2

dv = – k dt v = u – ktu1/2 +

v or

1/ 2

d v = – k dt

2v1/ 2 = – kt + c

where c t = 0, v = u this value of c in (i), we have Let

(i) u1/2 = c

u1/ 2 ) = – kt

2(0 u

) =–k

or

2u1 / 2 = k

s = ut

ku 1 / 2t 2 2

or

s= u

1 1/ 2 ku 2

s=

2u / k

(ii)

(iii)

s covered in this time, we use

ds dt

k 2t 2 ds 1/ 2 = u kt u dt t = 0 to t = , we have

Therefore,

be the time taken by the particle to come to v = 0 at t = 1/ 2

9.

2(v1/ 2

v=

k 2t 2

–

u / 2k

k 2t 12 2

+

0

1 2 k (iv) 12 t from (iii) in (iv), we get u / 12k

or

s=

2u

/

k

2.34 Comprehensive Physics—JEE Advanced

IV Matching x–t

1.

Fig. 2.30

Column I

ANSWER 1. (a) (b)

Column II

(q)

(c)

(p)

(s)

(d)

(r)

Explanation: The slope of x – t AB CD, the slope is zero and in DE 2. The displacement x of a particle moving along the x-direction varies with time t x = a + bt – ct2 where a, b and c Column I (b) Acceleration when velocity = 0

b (q) –2c a a+

SOLUTION x = a + bt – ct2 x at (t = 0) = a Velocity

dx = b – 2ct dt b v = 0 at t = 2c v =

Column II c) b2 c

BC, the slope is

2.35

dv = –2c (which is constant) dt

Acceleration =

b 2c Hence the solution is (a) (s) (b) (q) Value of x at

t

b b –c 2c 2c

=a+b

(c) (d)

2

=a+

b2 c

(r) (p)

V Assertion-Reason Type Questions Statement-2 A body moving along a curve with a constant speed

following four options out of which only ONE choice is -

3. Statement-1 A wooden ball and a steel ball of the same mass, released from the same height in air, do not reach

-

Statement-2 The apparent weight of a body in a medium depends on the density of the body relative to that of the 4. Statement-1

1. Statement-1 A body moving in a straight line may have non-zero

body is a straight line parallel to the time axis, then Statement-2 Velocity is equal to the rate of change of displace-

Statement-2 the instant when it reaches the highest point is zero –2

5. Statement-1

2. Statement-1 A body moving in a straight line with a constant

a curve, then the body is either uniformly accelerStatement-2 The slope of the velocity-time graph gives the

SOLUTIONS 1. 2.

geff = g 1 moving along a curve continuously changes a body moving along a curve with a constant speed

3.

due to gravity in a medium is given by

where 4. 5.

= density of the medium and

= density

graph is parallel to the time axis, then, rate of

2.36 Comprehensive Physics—JEE Advanced

VI Integer Answer Type 1. The displacement x of a particle moving in one dimension, under the action of a constant force, is related to time t by the equation t=

x

where x is in metres and t placement (in metre) of the particle when its veloc-

2. The motion of a body is given by d v(t ) v(t) dt where v(t) is the velocity (in ms–1) of the body at time t t –1 its velocity (in ms ) when the acceleration is half

SOLUTIONS

1. Given

x =t x = t2 – 6t + 9

2. The acceleration of the body at time t is (i)

The instantaneous velocity of the particle is dx d 2 v= (t – 6t + 9) = 2t – 6 dt dt Now, v will be zero at time t given by 2t – 6 = 0 or t t obtained from relation (i) by putting t 2 –6 gives x (at t net displacement of the particle is zero when its

a(t) = Putting t a

d v(t ) dt

(1)

v(t)

v(0)

Hence a(0) = 6 ms–2 a(0) 6 a(t) = 2 2 v

t = 0, v

–2

, we have from

v = 1 ms–1

3

Vectors

Chapter

REVIEW OF BASIC CONCEPTS 3.1

EQUAL VECTORS

Two vectors A and B of the same physical quantity are equal, if and only if they have the same magnitude and the same direction. We can test the equality by shifting B parallel to itself until its tail coincides with the tail of A. If the tips of the two vectors also coincide, the vectors are equal. (Remember, the vector shifted parallel to itself is equal to the original vector since the magnitude and the direction of the vector do not change when it is shifted parallel to itself).

3.2

Fig. 3.1

UNIT VECTOR

A vector having a unit modulus is called unit vector. If A is a vector, then unit vector A along the direction of A A A = |A|

If the two vectors are represented in magnitude and direction by the two adjacent sides of a parallelogram drawn from a point, then their resultant is represented in magnitude and direction by the diagonal of the parallelogram passing through that point.

A A

Conventionally, unit vectors along x, y and z directions

Figure 3.1 show two vectors A and B of magnitudes A and B inclined at an angle . The magnitude R of the resultant vector R is given by R=

3.4

ADDITION OF VECTORS

vectors is known as the parallelogram law of vector addition and may be stated as follows.

B2

2 AB cos

The angle which the resultant vector R subtends with vector A is given by tan

are denoted by i , j and k respectively.

3.3 NULL VECTOR If two vectors A and B are equal, their difference (A – B) zero or null vector and is represented by the symbol 0. It has zero magnitude direction. A vector which is not null is called a proper vector. Thus, if A= B then A – B= 0

A2

=

B sin A B cos

B sin R In vector notation the resultant vector is written as R = A + B. or

sin

=

Special Cases (i) When the two vectors are in the same direction, i.e. = 0°, then R = A2 B 2 2 AB cos 0 = A + B. Therefore, the magnitude of the resultant is equal to the sum of the magnitudes of the two vectors. Also tan = 0 or = 0, i.e. the direction of the resultant is the same as that of either vector.

3.2 Comprehensive Physics—JEE Advanced

(ii) When the two vectors are in opposite directions, 2

2

B 2 AB cos180 i.e., = 180°, then R = A = A – B. Also = 0 (iii) When the two vectors are at right angles to each other, i.e. = 90°, then A2 B 2 2 AB cos 90 = A2 B tan = . A Note: Rmax = A + B and Rmin = A – B R

B 2 . Also

3.1 The following pairs of forces act on a particle at an angle which can have any value (a) 2 N and 3 N (b) 3 N and 3 N (c) 2 N and 6 N and (d) 3 N and 8 N The resultant of which pair cannot have a magnitude of 4 N? SOLUTION For pair (a) : Rmin = A – B = 3 – 2 = 1 N Rmax = A + B = 3 + 2 = 5 N For pair (b) : Rmin = 0, Rmax = 6 N For pair (c) : Rmin = 4 N, Rmax = 8 N For pair (d) : Rmin = 5 N, Rmax = 11 N Hence the correct answer is (d).

and

R = A or B. A2

B2

2 AB cos

A=

A2

A2

2 A2 cos

Suppose we wish to subtract a vector B from a vector A. Since A – B = A + (– B) the subtraction of vector B from vector A is equivalent to the addition of vector – B to vector A. Hence the A – B) is as follows:

Fig. 3.2

Choose a convenient scale and draw the vectors A and B as shown in Fig. 3.2 (a). If B is to be subtracted from A, draw the vector negative of B, i.e. draw the vector – B [see Fig. 3.2 (b)]. Now shift the vector – B parallel to itself so that the tail of– B is at the head of A. Vector C is the sum of vectors A and – B, i.e. [see Fig. 3.2 (c)] C = A + (– B) = A – B

3.7

RESOLUTION OF A VECTOR INTO RECTANGULAR COMPONENTS

2 A2 1 cos

A2 = 2A2(1 + cos )

cos

=

= 120°

3.5

SUBTRACTION OF VECTORS

Consider a vector A in the x–y plane making an angle with the x-axis. The x and y components of A are Ax and Ay. The magnitudes of Ax and Ay are (Fig. 3.3)

R=

=

3.6

and

3.2 The magnitude of the resultant of two vectors of the same magnitude is equal to the magnitude of either vector. Find the angle between the two vectors. SOLUTION Given A = B

If the two vectors are represented in magnitude and direction by the two sides of a triangle taken in the same order, then their resultant is represented in magnitude and direction by the third side of the triangle taken in the opposite order.

1 2 Fig. 3.3

TRIANGLE LAW OF VECTOR ADDITION

The parallelogram law of vector addition yields the triangle law of vector addition. In Fig. 3.1, vector QP = vector OS = B. In triangle OQP, vector OP = R. Hence, the triangle law of vector addition may be stated as follows:

and

Ax = A cos , along x-direction Ay = A sin , along y-direction

Thus

Ax = Ax i = (A cos ) i

Vectors

and

Ay = Ay j = (A sin ) j

From parallelogram law A = Ax + Ay = (A cos ) i + (A sin ) j The magnitude of A in terms of the magnitudes of its rectangular components is A= Also

tan

=

Ax2

Ay2

Ay Ax

3.3 Resolve A into x and y components. The magnitude of A is 4 units.

3.3

NOTE The magnitude of a component along + x direction or + y direction is taken to be positive while the magnitude of a component along – x direction or – y direction is taken to be negative.

3.8

ADDITION OF VECTORS USING COMPONENTS

Two or more vectors are added by using components as follows: (a) Resolve each vector into its rectangular components. (b) Add the magnitudes of all the x-components taking into account their signs. Rx = sum of x-components of all the vectors (c) Similarly Ry = sum of y-components of all the vectors (d) Magnitude of resultant is Rx2

R=

Ry2

(e) The angle which the resultant subtends with the x-axis is given by Ry tan = Rx

Fig. 3.4

SOLUTION

3.4 Find the resultant of two vectors A = 4 units and B = 3 units shown in Fig. 3.6 by using (a) parallelogram law and (b) components method.

Fig. 3.5

In Fig. 3.5(a): Ax = – A cos 60° = – 4 Ay = + A sin 60° = + 4 In Fig. 3.5(b): Ax = + A cos 30° = + 4 Ay = – A sin 30° = – 4

1 = – 2 units 2 3 = + 2 3 units 2 3 = 2 3 units 2 1 = – 2 units 2

Fig. 3.6

SOLUTION (a) Angle between the two vectors is (Fig. 3.7) = 90° R=

A2

B2

=

42

32

= 5 units

2 AB cos 2

4 3 cos 90

3.4 Comprehensive Physics—JEE Advanced

= 2 R2 = Rx2 (180° –

60°

)

sin

=

2 3

3 2

2

2

3 3 2

2

3 3 Ry 2 4.6 tan = 2.347 3 Rx 1.96 2 3 2 giving 67° above the x-axis. Therefore, the angle which the resultant subtends with A is 67° – 30° = 37°. 2

B sin

B sin R

Ry2

R = 5 units The angle subtended by the resultant with the x-axis is given by

Fig. 3.7

=

3 along + y direction

= 25

30°

R sin 180

3 2

3 sin 90 5

3 5

3 = sin 1 37° 5 (b) The x and y components of A and B are [see Fig. 3.8]

NOTE The components method is more useful if more than two vectors have to be added.

3.9

SCALAR OR DOT PRODUCT

The scalar (or dot) product of two vectors A and B is A and B and the cosine of the angle between them, i.e. A B = AB cos Fig. 3.8

3 2 = 2 3 units along + x direction

Ax = A cos 30° = 4

1 2 = 2 units along + y direction

Ay = A sin 30° = 4

1 2 = – 3/2 units along – x direction and By = B sin 60° and Bx = B cos 60° = – 3

3 units along + y direction 2 Rx = Ax + Bx = 3

= 2 3 Ry = Ay + By

3/ 2 along + x direction

The scalar product is represented by putting a dot between the two vectors. The scalar product of two vectors is a scalar quantity. Properties of the Scalar Product (i) The scalar product is commutative, i.e. A B = B A (ii) A A = A2 or A = A A . This is true because in this case = 0°. (iii) If two vectors A and B are perpendicular to each other, then = 90° and A B = AB cos 90° = 0. Note that A B can be zero when neither A nor B is zero. (iv) For unit vectors i , j and k along the three rectangular axes x, y and z, we have

and (v) A

i

i = j

j = k

k =1

i

j = j

k = k

i =0

B = (Ax i + Ay j + Az k ) = Ax Bx + Ay By + Az Bz

(Bx i + By j + Bz k )

Vectors

Some Examples of Scalar Product (i) Work done is W = F s where F is the force vector and s is the displacement vector. (ii) Power consumed is P = F v where F is the force vector and v is the velocity vector. (iii) Electric current is I = J A where J is the current density vector and A is the area vector. = B A where B is the magnetic induction vector and A is the area vector.

3.10

VECTOR OR CROSS PRODUCT

(vii) (A

i

j = – j

i = k,

j

k = – k

j = i

k

i = – i

k = j,

i

i = j

i B) = Ax

j Ay

k Az

Bx

By

Bz

If the smaller angle between two vectors A and B is , then the vector or cross product of vectors A and B is A

B = (AB sin

j = k

n is a unit vector perpendicular to the plane containing A and B. The vector product of two vectors A and B is equal to a vector C, i.e. A B =C The magnitude of vector C is given by C = AB sin The direction of C is perpendicular to the plane formed by A and B and is given by the right hand screw rule.

+ k (AxBy – AyBx) Some Examples of Vector Product (i) Torque = r F, where r is the position vector and F is the force vector. (ii) Linear velocity = r where is the angular frequency vector and r is the position vector. (iii) Angular momentum = r p where r is the position vector and p is the linear momentum vector. 3.5 Find the angle between vectors A = i + j and

Properties of a Vector Product (i) Vector product is anticommutative, i.e. (A B) = – (B A) (ii) A A = 0, i.e. the vector product of a vector by itself is zero. This is because, in this case, = 0, and hence sin = 0. Therefore A A = AA sin = 0 Hence, the condition for two vectors to be parallel ( = 0°) or antiparallel ( = 180°) is that their vector product should be zero. If A B = 0, it means either (i) A is zero or, (ii) B = 0 or (iii) the angle between them is 0° or 180°. (iii) The distributive law holds for both scalar and vector products, i.e. A (B + C) = A B + A C A (B + C) = A B + A C (iv) ( A) B = A ( B) = (A B); (v) |A B|2 = |A|2 |B|2 – (A B)2.

k =0

= i (AyBz – AzBy) + j (AzBx – AxBz)

) n

where A and B are the magnitudes of vectors A and B and

3.5

a real number.

(vi) i , j and k are the three mutually perpendicular unit vectors at the origin O and along OX, OY and OZ respectively; the right-hand rule gives:

B= i – j. SOLUTION A

B = AB cos

Magnitude of A is A =

12

Magnitude of B is B =

12

j) =

2

(i i i

j) ( i

12 =

2

12 =

2

2 cos

j j = 2 cos

(

1 – 1 = 2 cos gives cos

i j

j i

= 0 or

0)

= 90°

3.6 Two vectors A and B are inclined at an angle . Using of the resultant of vectors A and B is given by R=

A2

B2

2 AB cos

3.6 Comprehensive Physics—JEE Advanced

SOLUTION

SOLUTION R = A + B. Therefore, R = (A + B)

R

=A

A+A

R2 = A2 + 2 A

(A

(A + B) B+B

A+B

B) = (2 i = 10 i

B

B + B2

A2

B2

(5 i 8i

i ( i

2 AB cos

4 j)

j 15 j

i

j

= 0 8 k 15 k

0

= A2 + B2 + 2 AB cos R=

3 j)

j

i

12 j

0 and j

j k)

i

= 23 k

3.7 Given A = 2 i 3 j and B = 5 i nitude and direction of (A B).

Thus the magnitude of (A direction is along + z axis.

4 j . Find the mag-

B) is 23 units and its

I Multiple Choice Questions with Only One Choice Correct 1. The unit vector parallel to the resultant of vectors A= i (a) (c)

2j 1 14 1 17

3 k and B = 2 i

(3 i

j

(3 i

2j

2k)

(b)

2 k ) (d)

k is

j

1 (2 i 3 1 14

2. The value of n so that vectors A = 4 i B = 2i is

3j

j

(3 i nj

2k) j 2k) k and

2 k are perpendicular to each other

(a) – 1 (b) – 2 (c) – 3 (d) – 4 3. The angle between vectors A = i

5 j and

B = 2 i 10 j is (a) 0° (b) 90° (c) 120° (d) 150° 4. The angle between vectors A and B is 60°. The ratio A B is |A B| (a) (c)

1 2 1 3

(b)

2

(d)

3

5. The magnitude of the resultant of two equal vectors is equal to the magnitude of either vector. The angle between the two vectors is (a) 60° (b) 90° (c) 120° (d) 150° 6. Given P = A + B and Q = A – B. If the magnitudes of P and Q are equal, the angle between vectors A and B is (a) zero (b) 45° (c) 90° (d) 180° 7. A vector C when added to vectors A = 3 i

5j

k

and B = 2 i 3 j 4 k gives a unit vector along the y-axis as their resultant. Then C is (a)

5i

3j

3k

(b) 5 i

3j

3k

(c)

5i

2j

3k

(d) 5 i

2j

3k

8. If the sum A + B of two vectors A and B equals the difference A – B between them, then (a) A is a null vector (b) B is a null vector (c) both A and B are null vectors (d) neither A nor B is a null vector 9. Given A B 0 and A C = 0. What is the angle between B and C?

Vectors

10.

11.

12.

13.

14.

15.

16.

(a) 45° (b) 90° (c) 135° (d) 180° The resultant of two vectors A and B subtends an angle of 45° with either of them. The magnitude of the resultant is (a) zero (b) 2 A (c) A (d) 2 A A and B are two vectors in a plane at an angle of 60° with each other. C is another vector perpendicular to the plane containing vectors A and B. Which of the following relations is possible? (a) A + B = C (b) A + C = B (c) A B = C (d) A C = B Vector C is the sum of two vectors A and B and vector D is the cross product of vectors A and B. What is the angle between vectors C and D? (a) zero (b) 60° (c) 90° (d) 180° The resultant of two vectors of magnitudes 3 units and 4 units is 1 unit. What is the value of their dot product? (a) – 12 units (b) – 7 units (c) – 1 unit (d) zero The resultant of two vectors of magnitudes 3 units and 4 units is 1 unit. What is the magnitude of their cross product? (a) 12 units (b) 7 units (c) 1 unit (d) zero Three vectors A, B and C are related as A + C = B. If vector C is perpendicular to vector A and the magnitude of C is equal to the magnitude of A, what will be the angle between vectors A and B? (a) 45° (b) 90° (c) 135° (d) 180° The magnitude of the resultant of (A + B) and (A – B) is (a) 2A (b) 2B (c)

A2

B2

(d)

A2

B2

17. In Q.16, what is the angle between the resultant vector and vector A? A (a) zero (b) cos–1 B (c) cos–1

B A

(d) cos–1

A B A B

18. If i and j are unit vectors along x-axis and y-axis respectively, the magnitude of vector i + j will be (a) 1

(b)

2

(c)

3

3.7

(d) 2

19. In Q.18, the angle subtended by vector i + j with the x-axis is (a) 30° (b) 45° (c) 60° (d) 75° 20. If i , j and k are unit vectors along x, y and z-axes respectively, the angle between the vector i + j + k and vector i is given by 1 (a) = cos–1 (b) = sin –1 3 (c)

= cos –1

3 2

(d)

= sin–1

1 3 3 2

21. Given that 0.2 i + 0.6 j + a k is a unit vector. What is the value of a? (a) 0.3 (b) 0.4 (c)

0.6

(d)

0.8

22. Given A = i + j and B = i + k . What is the value of the scalar product of A and B? (a) 1 (b) 2 (c) 3 (d) 2 23. The cross product of vectors A and B in Q. 22 is (a) i + j + k

(b) i – j + k

(c) i + j – k

(d) i – j – k

24. Given A = 2 i + 3 j and B = i + j . The component of vector A along vector B is 1 3 (b) (a) i j i j 2 2 5 7 (c) (d) i j i j 2 2 25. In Q.24, above, what is the component of vector A perpendicular to vector B and in the same plane as B? 1 3 (a) (b) j i j i 2 2 (c)

5

i

(d)

7

j i 2 2 26. The magnitudes of vectors A, B and C are respectively 12, 5 and 13 units and A + B = C. The angle between A and B is (a) zero (b) (c)

j

(d) 2 4 27. The sum of two forces acting at a point is 16 N. If the resultant force is 8 N and its direction is perpen-

3.8 Comprehensive Physics—JEE Advanced

dicular to the smaller force, then the forces are (a) 6 N and 10 N (b) 8 N and 8 N (c) 4 N and 12 N (d) 2 N and 14 N 28. Vector A of magnitude 4 units is directed along the positive x-axis. Another vector B of magnitude 3 units lies in the x-y plane and is directed along 30º with the positive x-axis is as shown in Fig. 3.9. The A B is (a) 6 units

(b) 6 2 units

(c) 6 3 units

(d) 12 units

F = (4 i + j + 6 k ) newton. The work done by the force is (a) 60 J (b) 59 J (c) 58 J (d) 57 J 33. A vector A is along the positive z-axis and its vector product with another vector B is zero, then vector B could be

y

x^

A

x

Fig. 3.9

29. In Q. 28, the magnitude of A

(b) 6 2 units

(c) 6 3 units

(d) 12 units

B= i

(a)

between vectors A = i + 2 j + 3 k and

=

3 7

(b) cos

=

2 7

angle

32. A body moves from a position r1 = (2 i

3j

j

6k

j + 6k

2

(b) 3 i + j

6k

(d) 3 i + j

6k

with A, then B 2 (d) AB = 1

(a) A = 2B

=

metre to a position r2 = (3 i

7k

36. The angle between two vectors A and B is . Vector R is the resultant of the two vectors. If R makes an

1 (d) zero 7 31. Two vectors C = A + B and D = A – B are perpendicular to each other. Then (a) A is parallel to B (b) A is perpendicular to B (c) B is a null vector (d) A and B have equal magnitudes. (c) cos

3i

(c) 3 i

2 j + 3 k is given by

(a) cos

(d)

vectors (2 i 3 j + 4 k ) and ( i + 5 j + 2 k ) yields a unit vector along the y-axis. Then vector A is

B is

(a) 6 units 30. The angle

(c) j + k

F1 = i + k , F2 = 2 j + 3 k , F3 = 3 i and F4 = 3 j 4 i . In which plane will the body move? (a) x – y plane (b) x – z plane (c) y – z plane (b) none of these 35. A is a vector which when added to the resultant of

^ j 30°

(b) 4 i

34. A body, initially at rest, is acted upon by four forces

B

O

(a) i + j

(b) A =

(c) A = B

37. What is the torque of a force F = (2 i

newton acting at a point r = (3 i + 2 j + 3 k ) metre about the origin? (a) 6 i

4k)

(c)

4 j + 5 k ) metre under

6 j + 12 k

6 i + 6 j 12 k

(b) 17 i (d)

(d) (a) (a) (b) (a) (d) (b)

2. 8. 14. 20. 26. 32.

(b) (b) (d) (a) (c) (d)

3. 9. 15. 21. 27. 33.

(a) (b) (a) (c) (a) (d)

4. 10. 16. 22. 28. 34.

(c) (b) (a) (a) (c) (c)

5. 11. 17. 23. 29. 35.

(c) (c) (a) (d) (a) (a)

6. 12. 18. 24. 30. 36.

6 j 13 k

17 i + 6 j + 13 k

ANSWERS

1. 7. 13. 19. 25. 31. 37.

3 j + 4k)

(c) (c) (b) (c) (a) (c)

Vectors

3.9

SOLUTIONS 1. The resultant of A and B is R = A + B = (i

2j

= (3 i Magnitude of R is R =

3 k ) + (2 i

j

2k)

32

12

j

k)

( 2)2 = 14

10. As shown in Fig. 3.10, the angle between vectors A and B is 90°. Also A = B. Therefore, the magnitude of the resultant is given by R2 = A2 + B2 + 2 AB cos = A2 + A2 + 2A2 cos 90° = 2A2 or R = 2 A.

R 1 = (3 i j 2 k ) R 14 So the correct choice is (d). 2. A and B will be perpendicular to each other if A B = 0, i.e. R =

(4 i

nj

k ) (2 i

3j

8 + 3n – 2 = 0

2k) = 0

Fig. 3.10

n=–2

3. B = 2 ( i 5 j ) = 2 A. Hence the magnitude of B is twice that of A and the direction of B is the same as that of A. So the correct choice is (a). A B AB cos = = cot . 4. | A B | AB sin = cot 60° ( = 60°) 1

=

3 5. R = A + B + 2AB cos . It is given that R = A = B. Thus, we have A2 = A2 + A2 + 2A2 cos 1 cos = – = 120° 2 6. If is the angle between A and B, the magnitudes of P and Q are given by 2

and

2

2

P = A2

B2

2 AB cos

Q = A2

B2

2 AB cos

Given P = Q. If follows that cos

= 0 or

= 90°.

7. j = C + A + B = C + (3 i 5 j k ) (2 i 3 j 4 k ) = C +5i C=

5i

Hence the correct choice is (b). 11. Since C is perpendicular to both A and B, the sum of any two cannot yield the third vector. Hence choices (a) and (b) are not possible. Choice (d) is also not possible because B is not perpendicular to A. Choice (c) is possible. 12. Vector C lies in the plane containing vectors A and B, and vector D is perpendicular to both A and B. Hence D must be perpendicular to C. Hence the correct choice is (c). 13. Let be the angle between the two vectors. The resultant is given by R2 = A2 + B2 + 2AB cos Putting the values of R, A and B we get (1)2 = (3)2 + (4)2 + 2 3 4 cos or cos = – 1 or = 180° Now A · B = AB cos = 3 4 cos 180° = – 12 Hence the correct choice is (a). 14. The magnitude of A B = AB sin = 3 4 sin 180° = 0. Hence the correct choice is (d). 15. Since A + C = B, vector B is the resultant of vectors A and C. Using the triangle law of vector addition (see Fig. 3.11), we have = 45° ( A = C)

2 j 3k 3 j 3k

8. Given A + B = A – B which gives B = – B. This is possible only if B is a null vector. Hence the correct choice is (b). 9. Since A · B = 0, it follows that A is perpendicular to B. Also A C = 0. Therefore A is parallel to C. Hence B is perpendicular to C. Therefore, the correct choice is (b).

Fig. 3.11

Thus the correct choice is (a).

3.10 Comprehensive Physics—JEE Advanced

16. The resultant R of vectors (A + B) and (A – B) is R = (A + B) + (A – B) = 2A The magnitude of the resultant = 2A. Hence the correct choice is (a) 17. Since R = 2A, R is parallel to A. Hence the correct choice is (a). 18. Let i and j represent the magnitudes of vectors i

(A · B) = 2 i =2 i i

Also

B =

i = 1 and j = 1. Therefore, the magnitude of vector

tan

i

=

j or

i

2

j

2

k

=1 0 0 j i

2

1/ 2

j i

1 1 1 1/2

i

k i 1

j

i

k

i i

i k

j i

jk

=1+0+0+0=1 Thus the correct choice is (a). B= i = i =0 j

j i

j 1 2

1 5

j

j which is choice (c).

i

2

i

0 , vector i

j

j . Let i i

i

i

k

k

j

j = C. Now

j

C = 2i C

=

3j

i

j

1 2

i

1

j

2

i

j

i

j

j

i

i – j = 1 1 = 2 and (2 i 3 j) ( i j) = –1 Thus the correct choice is (a). 26. Since A + B = C, vector C is the resultant of vectors A and B. If the angle between A and B is , the magnitude of the resultant is given by C 2 = A2 + B2 + 2AB cos or

i

is per-

j

The required component is

1)

= 0.6 or a = 0.6 . So the correct choice is (c).

23. A

3j i 3 j j

3

Hence the correct choice is (a). 21. Here (0.2)2 + (0.6)2 + a2 = 1 or a2 = 1 – 0.04 – 0.36,

22. A · B = i

1

i

i

2

(A · C) = 2 i + 3 j

(A · C)

0 and i i

k i

i i

1

3

(

k .i

j

=

j

j

j

2

Thus the answer is

i

pendicular to vector i

= 45° which is choice (b). i

20. cos

i

=

25. Since i

1 =1 1

j

2i j

B B

2.

19. The angle subtended by vector i + j with the x-axis is given by

i

=2+0+0+3=5

and j respectively. Since i and j are unit vectors,

i + j = i 2+ j 2 12 12 Thus the correct choice is (b).

3j

j

k

k + i which is choice (d).

24. The component of vector A along vector B = (A · B) B B where B = where B is the magnitude of vector B B. Now

(13)2 = (12)2 + (5)2 + 2

12

5

cos

which gives cos = 0 or = /2. Hence the correct choice is (c). 27. F is the resultant of F1 and F2 and F1 is the smaller force. Now F 22 = F 21 + F 2 = F 21 + (8)2

and F1 + F2 = 16 or F2 = 16 – F1

(i) (ii)

Vectors

3.11

31. Since C and D are at right angles to each other, C D = 0 or (A + B) (A – B) = 0 or A2 – A B + B A – B2 = 0 or A = B ( A B = B A) 32. Displacement r = r2 – r1 = (3 i = i

Fig. 3.12

Using (ii) in (i) we have (16 – F1)2 = F 21 + 64, which gives F1 = 6 N. Therefore F2 = 16 – 6 = 10 N. Hence the correct choice is (a). 28. Let i and j be the unit vectors along positive x and y-axes respectively. Then A = 4 i and B = 3 cos 30º i + 3 sin 30º j =

29. A

3 3 3 i+ j 2 2

B=4i

(

i = 0; i

= (1 + 4 + 9)1/2 = 14

cos

=

6 14

14

33. The vector product of two non-zero vectors is zero if they are in the same direction. Hence, vector B must be parallel to vector A, i.e. along ± z-axis. 34. Resultant force = F1 + F2 + F3 + F4

6 14

36. The angle given by

given

3 7

1/ 2

j + 9 k ) newton metre

= (4 – 1 + 54) = 57 newton metre or 57 J.

which the resultant R makes with A is

= 0 + 6k

(3) 2

(i

14

B sin A B cos

tan

j =k )

A B = ( i + 2 j + 3k ) ( i – 2 j + 3k ) = (1)2 + 2 – 2 + (3)2 =1–4+9=6 A = {(1)2 + (2)2 + (3)2]1/2 ( 2) 2

= (4 i + j + 6 k )

35. Given A + (2 i – 3 j + 4 k ) + ( i + 5 j + 2 k ) = 1 j

= 6k Hence the correct choice is (a). 30. Angle between vectors A and B is given by A B cos = AB

B = (1)2

metre

= 5 j + 4k

i i = 1; i j = 0)

i

4k )

= ( i + k ) + (2 j + 3 k ) + 3 i + (3 j – 4 i )

=6 3

(

9k

j

3j

Work done W = F r

3 3 3 i+ j 2 2

3 3 3 i+ j 2 2

A B=4i

4 j + 5 k ) – (2 i

2 tan

or

sin cos

. Hence 2 2

B sin A B cos

=

=

2 B sin

2

cos 2 2 A B cos

which gives A = B.

37. Torque = r

i F= 3 2

j k 2 3 3 4

= i [8 – (– 9)] – j (12 – 6) + k (– 9 – 4) = 17 i – 6 j – 13 k

3.12 Comprehensive Physics—JEE Advanced

II Multiple Choice Questions with One or More Choices Correct 1. Two vectors of the same physical quantity are unequal if (a) they have the same magnitude and the same direction (b) they have different magnitudes but the same direction (c) they have the same magnitude but different directions (d) they have different magnitudes and different directions. 2. Given A = – B. This means that vectors A and B (a) have equal magnitudes (b) have unequal magnitudes (c) are in opposite directions (d) are in the same direction. 3. Which of the following is a null vector? (a) Velocity vector of a body moving in a circle with a uniform speed (b) Velocity vector of a body moving in a straight line with a uniform speed (c) Position vector of the origin of a rectangular coordinate system (d) Displacement vector of a stationary object 4. The magnitudes of four pairs of displacement vectors are given. Which pairs of displacement vectors cannot be added to give a resultant vector of magnitude 4 cm? (a) 1 cm, 1 cm (b) 1 cm, 3 cm

5.

6.

7.

8.

9.

(c) 1 cm, 5 cm (d) 1 cm, 7 cm The dot product of two vectors A and B is zero if (a) A is a null vector and B a proper vector (b) A is a proper vector and B is a null vector (c) A and B are both null vectors (d) A and B are proper vectors perpendicular to each other. The cross product of two vectors A and B is zero if (a) A is a null vector and B is a proper vector (b) A is a proper vector and B is a null vector (c) A and B are both null vectors (d) A and B are proper vectors parallel to each other. If A B = C, which of the following statements is/ are correct? (a) C is perpendicular to A (b) C is perpendicular to B (c) C is perpendicular to both A and B (d) C is parallel to both A and B Which of the following vector identities is/are false? (a) A B = B A (b) A B = – B A (c) A B = B A (d) A B = – B A A and B are two perpendicular vectors in a plane. C is another vector perpendicular to the plane containing vectors A and B. Which of the following relations is/are possible? (a) A + B = C (b) A + C = B (c) A B = C (d) A C = B

ANSWERS AND SOLUTIONS

1. The correct choices are (b), (c) and (d). 2. Given A = – B, i.e. A + B = 0. Two vectors add up to zero only if they have equal magnitudes and opposite directions. Hence the correct choices are (a) and (c). 3. The correct choices are (c) and (d) 4. The magnitude R of the resultant of two vectors A and B depends upon the magnitudes of A and B and the angle between them and is given by R2 = A2 + B2 + 2AB cos When = 0, R is maximum given by R2max = A2 + B2 + 2AB

Rmax = A + B = 180°, R is minimum given by R2min = A2 + B2 – 2AB or Rmin = A – B Thus, the magnitude of resultant will lie between A – B and A + B. Hence the correct choices are (a) and (d). 5. A · B = AB cos = 0 if A = 0 or B = 0 or = 90°. Hence all the four choices are correct. 6. A B = AB sin = 0 if A = 0 or B = 0 or = 0°. Hence all the four choices are correct. 7. The correct choices are (a), (b) and (c). or When

Vectors

choices (a) and (b) are not possible. Since A is perpendicular to B, the three vectors are mutually perpendicular. Hence choices (c) and (d) are possible.

8. The scalar product is commulative, i.e. A · B = B · A. Vector product is anti–commutative, i.e. A B = – B A. Hence choices (b) and (c) are false. 9. Since C is perpendicular to both A and B, the sum of any two cannot yield the third vector. Hence

III Matching 1. Match the following. Column I (a) A · B = B · A. (b) A B = B A (c) A · B = 0 (d) A B = 0 ANSWER 1. (a) (c)

(s) (q)

3.13

Column II (p) (q) (r) (s) (b) (d)

False A and B are perpendicular to each other A and B are parallel to each other True (p) (r)

4

Motion in Two Dimensions

Chapter

REVIEW OF BASIC CONCEPTS 4.1

PROJECTILE MOTION

Projectile is the name given to a body which, after having been given an initial velocity, is allowed to move under the .

Since y bolic.

x2, the trajectory of the body is para-

t f) Putting y = h and t = tf in Eq. (4.2), we get tf =

2h g

R) Putting t = tf and x = R in Eq. (4.1), we get 2h g Magnitude of resultant velocity at time t is R = u tf = u

2 v= u

Fig. 4.1

(i) A body projected horizontally with a velocity u from a height h. [Fig. 4.1] Horizontal and vertical distances covered in time t are x = ut (4.1) 1 2 (4.2) y = gt 2 Differentiating Eqs. (4.1) and (4.2) w.r.t time t, we get the horizontal and vertical velocities. dx (4.3) u vx = dt dy (4.4) gt dt Equation to trajectory Eliminating t from Eqs. (4.1) and (4.2), we get g y= x2 2 2u vy =

g2 t2

1/ 2

The angle which the resultant velocity vector subtends with the vertical is given by u tan = gt (ii) A body projected from the ground with a velocity u at an angle with the horizontal. (Fig. 4.2) The horizontal and vertical distances covered in time t are x = (u cos )t (4.5) 1 2 (4.6) and y = (u sin )t – gt 2 Horizontal and vertical components of the velocity at time t are vx = u cos (4.7) (4.8) and uy = u sin – gt Magnitude of resultant velocity at time t is v = v 2x

v 2y

1/ 2

4.2 Comprehensive Physics—JEE Advanced

The angle subtended by the resultant velocity vector with the horizontal is given by vy tan = vx

Equation of trajectory Eliminating t from Eqs. (4.5) and (4.6), we get

Putting y = 0 tf =

tf ) and

2h g

g2

1/ 2

The horizontal range is R = (u cos )T

(v) At the highest point, KE =

gx 2 and PE = total energy – KE =

2u 2 cos 2

= t = tf in Eq. (4.6), we get

2u sin g

h ) Put t = tf /2 and y = hmax in Eq. (4.6), we get hmax =

u 2 sin 2

u sin g

(i) The horizontal range is the same for angles and (90° – ). (ii) The horizontal range is maximum for = 45°. Rmax = u2/g R (iii) When horizontal range is maximum, hmax = max 4 1 2 (iv) At the point of projection, KE = mu , PE = 0. 2 1 2 Total energy E = mu . 2

Fig. 4.2

y = (tan )x –

which gives T =

1 mu 2 cos2 2 1 mu 2 2

1 mu 2 cos 2 2

1 2 2 mu sin . 2 R and hmax from the equation of trajectory y = ax – bx2

where a and b are constants, refer to Fig. 4.4.

u 2 sin 2 2g

H R) Putting t = tf and x = R in Eq. (4.5) we get R=

u 2 sin 2 g

(iii) A body projected from a height h with a velocity u at an angle with the horizontal. (Fig. 4.3)

Fig. 4.4

(a) At O and B, y = 0. Putting y = 0 in the above equation, we have 0 = ax – bx2 x = 0, a x = a/b. Therefore R . b a . Using these 2b a2 . values in y = ax – bx2, we get hmax = 4b (vii) If A and B are two points at the same horizontal level on a trajectory at a height h from the ground, (see Fig. 4.5), then (b) At A, y = hmax and x =

Fig. 4.3

If T h = uy T

1 g T 2 = u sin 2

T

1 gT2 2

R 2

Motion in Two Dimensions 4.3

and

Fig. 4.5

(a) tf =

2u sin g

t1

t2

1 gt1 t2 2 (c) Average velocity during time interval (t2 – t1) is vav = u cos ( during this interval, the vertical displacement is zero) (viii) Velocity and Direction of Motion of Projectile at any Height. Let P be a point on the trajectory of a projectile at a height h and let v be the velocity of the projectile at that height. If is the angle which the velocity vector makes with the horizontal, then the horizontal and vertical components of the velocity are given by vx = ux = constant or v cos = u cos (i) (b) h =

(v sin

)2 = (u sin )2 = – 2gh

vy = u sin ( –

)

The x and y components of acceleration due to gravity are – g sin and – g cos respectively, as shown in Fig. 4.6. Let Tf inclined plane. Since the net vertical displacement in time Tf is zero (i.e., h = 0), we have 1 0 = vy T f g cos T f2 2 1 or 0 = vy g cos T f 2 1 g cos T f or 0 = u sin ( – ) 2 2u sin or Tf = (iii) g cos

Fig. 4.6

(ii)

Squaring Eq. (i) and then adding to Eq. (ii), we get v2 = v 2x v 2y u2 2 gh or

)

v 2y = u 2y – 2gh

and or

vx = u cos ( –

v = (u2– 2gh)1/2

This gives the speed of the projectile at height h. The direction of the velocity vector (i.e., direction of motion) is obtained by taking the square root of Eq. (ii) and then dividing by Eq. (i). We get 1/ 2

u 2 sin 2 2 gh sin tan = = u cos cos (ix) Time of Flight and Range of a Projectile on an Inclined Plane Consider an inclined plane OAB making an angle with the horizontal (Fig. 4.6). Let a body be projected with a velocity u at an angle with the horizontal. Let us choose the x-axis along the plane OA and y-axis perpendicular to the plane OA. Let the body hit the inclined plane at point P so that R = OP is the range on the inclined plane. The x and y components of the velocity of the projectile are

During this time, the horizontal component of velocity u cos remains constant. Hence, horizontal distance OQ is OQ = (u cos )Tf Range of the projectile on the inclined plane is u cos OQ cos cos Using Eq. (iii) in Eq. (iv), we get R = OP =

R=

2 u 2 sin g cos

Tf

(iv)

cos 2

4.1 A body is projected horizontally with a velocity of 10 ms–1 from the top of building 20 m high. Find (a) horizontal distance from the bottom of the building at which the body will strike the ground. (b) the magnitude and direction of the velocity of the body 1 s after it is projected. Take g = 10 ms–2.

4.4 Comprehensive Physics—JEE Advanced

SOLUTION (a) Refer to Fig. 4.1 on page 4.1. Given h = – 20 m, u = 10 ms–1, and g = – 10 ms–2. Time taken by the body to go from A to B is 2h 2 20 =2s g 10 Horizontal distance OB = R = ut = 10 2 = 20 m (b) Refer to Fig. 4.1 again. Horizontal velocity at t = 1 s is vx = u = 10 ms–1. Vertical velocity at t = 1 s is (since initial vertical component of velocity uy = 0) vy = uy + at = 0 – 10 1 vy = – 10 ms–1 Magnitude of resultant velocity is t=

v=

v 2x

v 2y

102

Refer to Fig. 4.7. The horizontal and vertical components of initial velocity are (a) ux = 20 cos 30° = 10 3 ms–1 uy = 20 sin 30° = 10 ms–1 (vertically upwards) Horizontal acceleration ax = 0 and vertical acceleration ay = – 10 ms–2 (vertically downwards). Vertical displacement S = – H = – 15 m. Since the vertical and horizontal motions are independent of each other, we have, for vertical motion. 1 S = u yt + a yt 2 2 –15 = 10t +

( 10) 2

1 2

(– 10)t2

–15 = 10t – 5t2

200 10 2 ms 1 The angle which the resultant velocity vector subtends with the vertical is given by v 10 tan = x =1 = 45° v y 10

t2 – 2t – 3 = 0 t = – 1 s or 3 s.

=

4.2 A ball is thrown with a velocity of 20 ms–1 at an angle of 30° above the horizontal from the top of a building 15 m high. Find (take g = 10 ms–2) (a) the time after which the ball hits the ground. (b) the distance from the bottom of the building at which it hits the ground. (c) the velocity with which the ball hits the ground. (d) the maximum height attained by the ball above the ground. SOLUTION

Since t = –1 s is not possible, the ball with strike ground at point C after 3 seconds. (b) Horizontal range R = OC = uxt = 10 3 3 = 30 3 m (c) Horizontal velocity component at C is vx = ux = 10 3 ms–1 Vertical velocity component at C is vy = uy + ayt = 10 – 10 3 = – 20 ms–1 The negative sign shows that the ball is moving downwards. Resultant velocity v = v 2x v 2y = tan

=

(10 3 ) 2 vx vy

= tan

(20)2

10 3 20 1

3 2

10 7 ms

1

3 2 with the vertical.

(d) Maximum height attained above the ground is hmax = h + H =

u 2 sin 2 2g

=

(20) 2 sin 2 (30 ) + 15 2 10

+H

v v

Fig. 4.7

v

= 5 + 15 = 20 m

Motion in Two Dimensions 4.5

4.3 A stone thrown from the ground at an angle of 45° above the horizontal strikes a vertical wall at a point 10 m above the ground. If the wall is at a distance of 20 g = 10 ms–2) (a) the speed with which the stone is projected, (b) the magnitude and direction of the velocity of the stone when it strikes the wall. SOLUTION

(a) Let P be the point on the wall where the stone strikes it. Taking the point of projection O as the origin, the coordinates of P are (20 m, 10 m) [Fig. 4.8] x = (u cos )t (1) 1 2 (2) y = (u sin )t – gt 2 Putting x = 20 m, y = 10 m, g = 10 ms–2 and = 45° in Eqs. (1) and (2), we have u t 20 = (u cos 45°)t = 2 20 2 t= (3) u 1 and 10 = (u sin 45°)t – 10 t2 2 ut 2

– 5 t2

(4)

Hence point P is at the highest point on the trajectory where the velocity is only horizontal. Thus the stone strikes the wall at P with a velocity of 20 10 2 ms–1 in the horizontal direction. 2 4.4 A ball projected with a velocity of 10 ms–1 at an angle of 30° with the horizontal just clears two vertical poles, each of height 1.0 m. Find the separation between the poles. Take g = 10 ms–2.

Fig. 4.9

y-component of velocity is uy = u sin = 10 sin 30° = 5 ms–1. 1 h = uyt + gt 2 2 1 1.0 = 5t + ( 10)t 2 2 5t2 – 5t + 1.0 = 0 The two roots of this quadratic equations are t1 = 0.72 s and t2 = 2.76 s. Therefore OQ = uxt1 = 10 cos 30° 0.72 = 6.2 m and

Using (3) in (4) 10 =

u

20 2 u

2

10 = 20 –

5

4000

u = 20 ms

u

20 2 u

2

2

ms–1

uy = u sin = 20 sin 45° = v 2y =

2.76 = 23.9 m

4.5 A projectile has the same range R = 40 m for two angles of projection. If T1 and T2 are the times of T1 T2. Given g = 10 ms–2.

–1

20

OS = uxt2 = 10 cos 30° QS = 23.9 – 6.2 = 17.7 m

u2 = 400

2

(b) At point P, vx = 20 cos 45° =

v 2y = u 2y – 2g 10

vy = 0

SOLUTION Refer to Fig. 4.9. Let us calculate the two values of t at which the ball passes just above P and R. For each pole h = 1.0 m

Fig. 4.8

10 =

v 2y = 0

20 2

ms

–1

400 – 2 10 10 2

SOLUTION For a given speed u of projection, a projectile has the same range for angles of projection and (90° – ). Therefore

4.6 Comprehensive Physics—JEE Advanced

vBA = vB – vA = vB + (– vA )

2u sin g 2u sin (90 T2 = g T1 =

and

T 1T 2 = =

)

2u cos g

Magnitude of vector vBA is given by

4u 2 sin cos g2 2u 2 sin (2 ) 2

g 2 40 = = 8 s2 10

2R g

4.6 A ball is thrown from a point with a speed u at an angle with the horizontal. From the same point and at the same instant, a person starts running with a constant speed u/2 to catch the ball. Will he be able to catch the ball? If yes, what should be the value of ? SOLUTION The person will catch the ball if the horizontal rane i.e. if R= 2

u 2

u sin (2 ) u = g 2 1 cos = 2

4.2

Thus the magnitude and direction of vector vBA can be vB and –vA which is vector OC as shown in Fig. 4.10.

tf 2u sin g = 60°

vBA = v 2A

vB2

2v A vB cos

= v 2A

vB2

2v A vB cos

1/ 2

1/ 2

(

= 180° – )

The angle which the resultant vector OC subtends with vector OD is given by OC CD = sin sin sin

=

CD sin OC

vB sin vBA

(i) If vector vA and vB are in the same direction, = 0°, v 2A

then vBA =

vB2

2v A vB = vB – vA.

(ii) If vector vA and vB are in opposite direction, 180°, then vBA = vB + vA.

=

(i) To cross the river of width d along the shortest path which is PQ, the boat must move along PR making an angle (90° + ) with the direction of the stream such that the direction of the resultant velocity v is along PQ. Angle is given by (see Fig. 4.11)

RELATIVE VELOCITY IN TWO DIMENSIONS

The relative velocity of a body B with respect to body A vBA = vB – vA If vectors vA and vB are inclined to each other at an angle as shown in Fig. 4.10, the relative veliocity vBA is found as follows. Fig. 4.11

sin Also

=

v vb vb2

v=

v2

The time taken to cross the river along the shortest path is given by t= Fig. 4.10

d v

d vb2

v2

Motion in Two Dimensions 4.7

(b) To cross the river in the shortest time, the boat should move along PQ. The shortest time is given by d t= vb At this time, the boat will reach the point R on the opposite bank of the river at a distance x from the point Q (Fig. 4.12). From the Figure, we have

vm vr Thus, the man must hold the umbrella at an angle with the vertical towards north. or

4.3

= tan–1

UNIFORM CIRCULAR MOTION

(i) For a body moving in a horizontal circle The centripetal acceleration of a body of mass m moving in a circle of radius R with a constant speed v (or angular speed ) is ac =

v=

2

Centripetal force is

mv 2 R (ii) For a body moving in a vertical circle The minimum speed to complete the circle when the body is at the top of the circle is v = Rg . The minimum speed to complete the circle when the body is at the bottom of the circle is v = 5 Rg . fc = mac = m

Fig. 4.12

x = d tan , but tan

=

vw Therefore, vb

x= d

vw vb

v=

2

R=

NOTE

(ii) Holding an Umbrella to Project from Rain Let vr be the velocity of the rain falling vertically downward and vm the velocity of a man walking from north to south direction (Fig. 4.13). In order to protect himself from rain, he must hold his umbrella in the direction of the resultant velocity v, which is given by vr2

2 R= v R

vm2

Fig. 4.13

This is the speed with which the rain strikes the umbrella. If is the angle subtended by the resultant velocity v with the vertical, then from triangle ORM , we have vm RM tan = vr OR

The magnitude of velocity (v) and the magnitude of acceleration (v2/R) for a body in uniform circular motion are constant but the direction of velocity v (which is along the tangent and the direction of acceleration c (which is towards the centre keep on changing with time.

4.4

NON–UNIFORM CIRCULAR MOTION

If the speed of the body revolving in a circle changes with time, it is said to be in nonv uniform circular motion. In this case, the acceleration of the body is the resultant of two components (Fig. 4.14). (i) Radial (or centripetal) acceleration directed towards the centre O and has a magnitude ac = v2/R, where v is the Fig. 4.14 instantaneous speed and R is the radius of the circle. (ii) Tangential component directed along the tangent which causes the change in the magnitude of velocity. Its magnitude is given by dv at = dt The net acceleration of the body is a=

ac2

a t2

4.8 Comprehensive Physics—JEE Advanced

and it makes an angle tan

=

with the tangent given by

at2

ac2

=

k2

(4 kn)2

ac at

= k 1 16

4.7 A particle moves along a circle with a velocity v = kt, where k is a constant. Find the net acceleration of the particle at the instant when it has covered nth fraction of the circle after the beginning of motion.

v2 R Tangential acceleration at that instant is ac =

at =

dv dt

dv d x dt d x

dv dv d x dt dt d x at dx = v dv Integrating at =

(1)

vd v dx

0

v 2 (k x )2 2x 2x Centripetal acceleration is

v2 = at x = at (2 R)n 2 2at (2 R)n

k2 2

v2 k 2 x R R Angle between net acceleration and tangential acceleration is given by ac =

(3)

Using (3) in (1), we get ac =

v2 2

at =

0

v=

v dv 0

at x =

x

vd v dx

v

x

at dx =

v d v = at dx 0

n

SOLUTION Tangential acceleration is

vdv = at dx (2) Let x be the distance covered. Then x = (2 R)n. Integrating Eq. (2) we have v

2 2

4.8 A particle moves along a circle of radius R. Its speed v varies with distance x covered along the circle as v = k x , where k is a constant. If R = 1.0 m and x vector and the tangential acceleration vector.

SOLUTION Let v be the speed of the particle at the instant when it has covered nth fraction of the circle of radius R. Centripetal acceleration at that instant is

Now

Net acceleration a =

2at (2 R)n = 4 at n R

tan

dv d (kt ) = k dt dt ac = 4 kn

=

ac at

k 2 x/R 2

k /2

x 2R

Putting x = 0.5 m and R = 1.0 m. We get 2 0.5 tan = =1 = 45°. 1.0

at =

I Multiple Choice Questions with Only One Choice Correct 1. The maximum height attained by a projectile is

(a) 30°

(b) 45°

3 /4 times its horizontal range. The angle of projection of the projectile with the horizontal is

(c) 60°

(d) tan–1

3 2

Motion in Two Dimensions 4.9

2. A body is projected with a speed u at an angle with the horizontal. The speed of the body when it is at the highest point on its trajectory is 2 / 5 times its speed at half the maximum height. The value of is (a) 30° (b) 45° 3 (c) 60° (d) tan–1 2 3. A body of mass m is projected from the ground with linear momentum p such that it has the maximum horizontal range. The minimum kinetic energy of (a) zero

(b)

p2 m

p2 2m

(d)

p2 4m

(c)

4. Two bodies are projected simultaneously from the same point on the ground with speeds 10 ms–1 and 10/ 3 ms–1 at angles 30° and 60° respectively with the horizontal. The separation between them when they hit the ground is (take g = 10 ms–2) 10 (b) m (a) 10 3 m 3 (c) 5

3

1 3

m

(d) 5

3

1 3

m

5. A projectile is given an initial velocity = (2 i j) ms–1. The cartesian equation of its trajectory is (take g = 10 ms–2) (b) 2y = 2x – 5x2 (a) y = 2x – 5x2 2 (c) 4y = 2x – 5x (d) 4y = x – 5x2 6. A body is projected from the ground with a speed u at an angle with the horizontal. The magnitude of the average velocity of the body between the point of projection and the highest point of its trajectory is u (sin + cos ) (a) 2 u (b) (1 + 2 cos2 )1/2 2 u (c) (1 + 3 cos2 )1/2 2 u (d) 2 7. A body P is projected vertically upwards. Another body Q of the same mass is projected at an angle of 60° with the horizontal. If both attain the same

maximum height, the ratio of the initial kinetic energy of P to that of Q is (a)

3 4

3 2

1 2 2 8. With what minimum speed must a body be projected from the origin in the x-y plane so that it can pass through a point whose x and y coordinates are 30 m and 40 m respectively? Take g = 10 ms–2. (a) 10 ms–1 (b) 20 ms–1 (c) 30 ms–1 (d) 40 ms–1 9. It is possible to project a particle with a given velocity in two possible ways so as to make it pass through a point P at a distant r from the point of projection. The product of the times taken to reach this point in the two possible ways is then proportional to (c)

1

(b) (d)

(a) 1/r

(b) r

(c) r3

(d)

(a) 20 m

(b) 20 2 m

1

r2 10. A projectile has a maximum range of 200 m. What is the maximum height attained by it? (a) 25 m (b) 50 m (c) 75 m (d) 100 m 11. A body thrown along a frictionless inclined plane of angle of inclination 30° covers a distance of 40 m along the plane. If the body is projected with the same speed at angle of 30° with the ground, it will have a range of (take g = 10 ms–2) (c) 20 3 m (d) 40 m 12. Which of the following remains constant during the (a) kinetic energy (b) momentum (c) vertical component of velocity (d) horizontal component of velocity 13. A body is projected with kinetic energy K at an angle of 60° with the horizontal. Its kinetic energy at the highest point of its trajectory will be (a) 2 K (b) K (c) K/2 (d) K/4 14. A body, projected with a certain kinetic energy, has a horizontal range R. The kinetic energy will be minimum at a position of the projectile when its horizontal range is (a) R (b) 3R/4 (c) R/2 (d) R/4

4.10 Comprehensive Physics—JEE Advanced

15. Four projectiles are projected with the same speed at angles 20°, 35°, 60° and 75° with the horizontal. The range will be the longest for the projectile whose angle of projection is (a) 20° (b) 35° (c) 60° (d) 75° 16. A player throws a ball which reaches the other player in 4 seconds. If the height of each player is 1.8 m, what is the maximum height attained by the ball above the ground? (a) 19.4 m (b) 20.4 m (c) 21.4 m (d) 22.4 m 17. The maximum height attained by a projectile is increased by 1% by changing the angle of projection, without changing the speed of projection. The (a) 20% (b) 15% (c) 10% (d) 5% 18. A projectile has a range R T. If the range is doubled (by increasing the speed of projection, without changing the angle of projection), the (a)

T

(b)

2T

2 T (c) (d) 2 T 2 19. A projectile has the same range R when the maximum height attained by it is either h1 or h2.Then R, h1 and h2 will be related as (a) R =

h1h2

(c) R = 3 h1h2

(b) R = 2 h1h2 (d) R = 4 h1h2

20. A ball is projected vertically upwards with a certain initial speed. Another ball of the same mass is projected at an angle of 60° with the vertical with the same initial speed. At the highest point, the ratio of their potential energies will be (a) 4 : 1 (b) 3 : 2 (c) 2 : 3 (d) 2 : 1 IIT, 1989 21. A body of mass m1, projected vertically upwards with an initial velocity u reaches a maximum height h. Another body of mass m2 is projected along an inclined plane making an angle of 30° with the horizontal and with speed u. The maximum distance travelled along the incline is (a) 2h (b) h h h (d) (c) 2 4 22. Two balls A and B are projected from the same location simultaneously. Ball A is projected verti-

cally upwards and ball B at 30º to the vertical. They reach the ground simultaneously. The velocities of projection of A and B are in the ratio (a)

3 :1

(b) 1 :

3

(c)

3 :2

(d) 2 :

3

23. A body is projected with a velocity v = (3 i + 4 j ) ms–1. The maximum height attained by the body is (take g = 10 ms–2) (a) 0.8 m (b) 8 m (c) 80 m (d) 800 m 24. (a) 0.8 s (b) 1.0 s (c) 4.0 s (d) 8.0 s 25. A body is projected with a velocity at an angle with the horizontal. The velocity of the body will become perpendicular to the velocity of projection after a time t given by 2u sin u sin (a) (b) g g (c)

2u g sin

(d)

u g sin

26. A body is projected at an angle with the horizontal. When it is at the highest point, the ratio of the potential and kinetic energies of the body is (a) tan (b) tan2 (c) cot (d) cot2 27. At time t = 0 a body is projected horizontally from a certain height with a velocity u. The radius of curvature of its trajectory at time t is u2 1 g

gt u

(c)

u2 1 g

g 2t 2 u2

(b)

u2 1 g

g 2t 2

(d)

u2 1 g

g 2t 2

3/ 2

(a)

1/ 2

1/ 2

u2 3/ 2

u2

28. A body is projected from the bottom of an inclined plane which has an inclination of 10° with the horizontal. At what angle from the horizontal should the body be projected so that its range on the inclined plane is maximum for a given velocity of projection? (a) 45° (b) 50° (c) 55° (d) 60° 29. From the top of a tower, two balls are thrown horizontally with velocities u1 and u2 in opposite directions. If their velocities are perpendicular to each other just before they strike the ground, the height of the tower is

Motion in Two Dimensions 4.11

(a)

u1

u2 2g

u12

u22

2

(b)

u1 u2 2g

2

u1u2 2g 2g 30. A body is projected at time t = 0 with a velocity u at an angle with the horizontal. It hits the ground at time t = tf where tf velocity of the body during the time interval t = 0 to t = tf is (a) u cos (b) u sin u u (1 cos 2 )1 / 2 (d) (1 sin 2 )1 / 2 (c) 2 2 31. A body is projected at time t = 0 with a velocity u at an angle with the horizontal. The horizontal and vertical components of its velocity will become equal at time t 0, (c)

(a) if 0

(d)

4

(b) if

4

2

(c) if 0

(d) for no value of 6 32. A boy whirls a stone in a horizontal circle 2 m above the ground by means of a string 1.25 m long. tally, striking the ground 10 m away. What is the magnitude of the centripetal acceleration during circular motion? Take g = 10 ms–2. (a) 100 ms–2 (b) 200 ms–2 –2 (c) 300 ms (d) 400 ms–2 33. A body is moving in a circle with a uniform speed v. What is the magnitude of the change in velocity when the radius vector describes an angle ? (a) zero (c) 2v cos

2

(b) v (1 + cos )

36. A plumb line is hanging from the ceiling of a train. If the train moves along a horizontal track with a uniform acceleration a, the plumb line gets inclined to the vertical at a angle a g (b) tan–1 (a) tan–1 g a a g (d) cos–1 g a 37. A body moves along a circular track of radius 20 cm. It starts from one end of a diameter, moves along the circular track and reaches the other end of the diameter is 5 seconds. What is the angular speed of the body? (c) sin–1

(a)

2

rad s–1

(b)

3

rad s–1

rad s–1 (d) rad s–1 4 5 38. A cyclist is moving with a speed of 6 ms–1. As the approaches a circular turn on the road of radius 120 m, he applies brakes and reduces his speed at a constant rate of 0.4 ms–2. The magnitude of the net acceleration of the cyclist on the circular turn is (b) 1.0 ms–2 (a) 0.5 ms–2 (c) 2.0 ms–2 (d) 4.0 ms–2 39. A car is travelling at a velocity of 10 km/h on a straight road. The driver of the car throws a parcel with a velocity of 10 2 km/h when the car is passing by a man standing on the side of the road. If the parcel is to reach the man, the direction of throw makes the following angle with the direction of the car, (c)

(a) 135°

(b) 45°

1/2

(d) 2v sin 2 2 34. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follow that (a) its velocity is constant (b) its acceleration is constant (c) its kinetic energy is constant (d) it moves in a straight line. 35. A particle of mass M is moving in a horizontal circle or radius R with uniform speed V. When it moves from one point to a diametrically opposite point, its (a) kinetic energy changes by MV2/4 (b) momentum does not change (c) momentum changes by 2 MV (d) kinetic energy changes by MV2

(c) tan–1

(d) tan–1

2

1 2

40. Rain is falling vertically with a speed of 4 ms–1. After some time, wind starts blowing with a speed of 3 ms–1 in the north to south direction. In order to protect himself from rain, a man standing on the ground should hold his umbrella at an angle given by (a)

= tan–1

3 4

with the vertical towards south

(b)

= tan–1

3 4

with the vertical towards north

(c)

= cot–1

3 4

with the vertical towards south

(d)

= cot–1

3 4

with the vertical towards north

4.12 Comprehensive Physics—JEE Advanced

41. In Q.40 above, with what speed does the rain strike the umbrella? (a) 3 ms–1 (b) 4 ms–1 (c) 5 ms–1 (d) 6 ms–1 42. A swimmer can swim in still water with a speed of 5 ms–1. While crossing a river his average speed is 3 ms–1. If he crosses the river in the shortest (a) 2 ms–1 (c) 6 ms–1

46. A stone tied to a string of length L is whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at its lowest position and has a speed u. The magnitude of the change in its velocity as it reaches a position where the string is horizontal is

(b) 4 ms–1 (d) 8 ms–1 –1

speed of 2 ms . A person in a boat at a point P on the bank of the river wants to cross the river by the shortest path to reach a point Q directly opposite on the other bank. If he can row his boat with a speed of 4 ms–1 in still water, he show row his boat at an angle of (a) 30° upstream with the line PQ (b) 30° downstream with the line PQ (c) tan–1(0.5) upstream with the line PQ (d) tan–1(2) downstream with the line PQ. 44. In Q.43 above, the time taken by him to cross the river by the shortest path is 3 s

u2

2 gL

(b)

2gL

(c)

u2

gL

(d)

2(u 2

(b) 3 3 s

(c) 6 3 s (d) 18 3 s 45. A body moving in a circular path with a constant speed has a (a) constant velocity (b) constant momentum (c) constant kinetic energy (d) constant acceleration IIT, 1992

Fig. 4.15

IIT, 2008

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43.

(c) (a) (d) (d) (d) (b) (d) (a)

2. 8. 14. 20. 26. 32. 38. 44.

(c) (c) (c) (a) (b) (b) (a) (c)

3. 9. 15. 21. 27. 33. 39. 45.

(d) (b) (b) (a) (d) (d) (a) (c)

4. 10. 16. 22. 28. 34. 40. 46.

(d) (b) (c) (c) (c) (c) (b) (d)

5. 11. 17. 23. 29. 35. 41. 47.

6. 12. 18. 24. 30. 36. 42.

(c) (c) (d) (a) (d) (c) (c) (d)

(c) (d) (b) (a) (a) (a) (b)

SOLUTIONS 1. hmax =

u 2 sin 2 2g

gL)

IIT, 1998 47. A projectile is projected with a velocity u at an angle , which of the graphs shown in Fig. 4.15 shows the variation of range R versus u?

43.

(a)

(a)

u 2 sin (2 ) g

Given hmax =

2 u 2 sin cos = g

u 2 sin 2 2g

and R =

3 R. Hence. 4 =

3 4

2 u 2 sin cos g

Motion in Two Dimensions 4.13

5. Given u cos

tan = 3 = 60° 2. Maximum height attained is 2

2

2

u sin u sin gH = 2g 2 Speed at maximum height is ux = u cos Speed v at H/2 is given by H v2 = u2 – 2g 2 = u2 – gH

2

H=

= u2 – It is given that u x2 =

u 2 sin 2 2 2 2 v , i.e. 5 u 2 sin 2 2

u2(1 – sin2 ) =

2 2 u 5

u 2 sin 2 2

3 = 60° 2 3. For maximum horizontal range, = 45°. Also p = mu which gives u = p/m. The kinetic energy is minimum when the body is at the highest point of its trajectory. At this point the u . velocity is ux = u cos = u cos 45° = 2

y = x tan

= x

1 m 2

u 2

2

=

(mu ) 2 p2 = . 4m 4m

4. The vertical components of velocity of the two bodies are 10 sin 30° = 5 ms–1 and (10 / 3 ) sin 60° = 5 ms–1. Since their vertical velocity components the separation between them when they hit the ground is x = difference in their horizontal ranges = R1 – R2 (10) 2 sin 60 = 10

(10 / 3 ) 2 sin 120 – 10

5

= 5 3

3 1

m 3 So the correct choice is (d). 3

–

g x2 2

2 u cos 2 10 x 2

1 2

2 5

( 5 )2

2

2

6. Average velocity is net displacement vav = time R 2

H2 = where

2

(1)

T /2

=

Minimum K.E. =

= 5

These equations give u = 2 2 cos = . u 5 The equation of the trajectory is

x 5x2 2 4 2 4y = 2x – 5x , which is choice (c).

2 2 u 5

sin

= 1 ms–1. 1 –1 and 5 ms , tan = 2

=

=

u2 cos2

= 2 ms–1 and u sin

H = maximum height =

u 2 sin 2 2g

R = horizontal range =

u 2 sin (2 ) g

2u sin g

T Substituting in Eq. (1), we get u vav = (1 + 3 cos2 2 The correct choice is (c). 7. For body P, u 12 = 2 gh For body Q,

h= u 22 =

u22 sin 2 (60 ) 2g

)1/2

3 u22 8g

8 gh 3

1 mu12 u2 3 K.E. of P 2 = = 12 1 K.E. of Q u2 4 mu22 2 So the correct choice is (a). 8. In projectile motion, the equation of the trajectory is y = x tan

–

g x2 2 u 2 cos 2

4.14 Comprehensive Physics—JEE Advanced

= x tan

g x2

–

(1 + tan2 )

2u2 Substituting the given values, we have

or 900 tan2

–

10

(30)

The value of tan will be real if (6u2)2 4 900 (900 + 8u2) or u4 100 (900 + 8u2) or u4 – 800u2 90000 or (u2 – 400)2 – 160000 90000 or (u2 – 400)2 250000 u2 = 900 or u2 – 400 500 u = 30 ms–1 9. The range of a projectile is r = 2v 02 cos sin /g, the two possible angles of projection are and (90° – these two angles are

and

t1 =

2v0 sin g

t2 =

2v0 sin (90 g

so that t1t2 =

4v02 sin

cos 2

)

R=

2

(1 + tan2 ) 2u2 – (6u2 tan ) + (900 + 8u2) = 0

40 = 30 tan

projected at angle = 30° with this speed will have a range of

2v0 cos g

2r . g

g Thus t1t2 r. Hence the correct choice is (b). 10. For maximum range = 45°. Hence Rmax = v20/g and hmax = v02/4g. Thus hmax = Rmax/4 = 200/4 = 50 m. Hence the correct choice is (b). 11. Let u be the initial speed with which the body is thrown along the inclined plane. As shown in Fig. 4.16, the effective deceleration is given by a = g sin g = 5 ms –2 = g sin 30° = 2

u 2 sin 2 g

20 3 m

v02 sin 2 4g 2 = 2g 2g 2g = 2 9.8 = 19.6 m

hmax =

Now

Height above the ground = 19.6 + 1.8 = 21.4 m Hence the correct choice is (c). v 2 sin 2 17. Given h = 0 . Differentiating partially we 2g get ( h h

v0 = constant) h = 2cos sin

2 v0 sin g

The body stops after covering a distance, say, s along the plane, which is given by – 2as = 0 – u2 or u = 2 a s 2 5 40 = 20 ms –1. A projectile

20 sin 60 10

Hence the correct choice is (c). 12. Since the velocity of the projectile changes continuously, both kinetic energy and momentum undergo a change with time. Only the vertical component of velocity changes due to gravity; the horizontal component always remains constant. Hence the correct choice is (d). 13. At the highest point, the velocity has only the horizontal component vx = v cos = v cos 60° = v/2. 1 Now kinetic energy mv2 is proportional to v2. 2 Since the velocity is reduced to half, the kinetic energy becomes one–fourth, i.e. K/4. Hence the correct choice is (d). 14. Kinetic energy is minimum when the projectile is at the highest point of its trajectory. At the highest point, its range = half the horizontal range. Hence the correct choice is (c). 15. Range R = v20 sin 2 /g. For the same v0, R sin 2 . Since sin 2 is the largest for = 35°, the correct choice is (b). 16. t = 2v0 sin /g. Since t = 4 s, we have v0 sin = 2g.

T T

. Thus

= 0.005. We also have T =

which gives

cos sin

v02 2 sin cos 2g

= 0.01 (given). cos sin

Therefore,

Fig. 4.16

20

. But

T=

cos sin

T = 0.05 or T = 0.005 T. T

2 v0 cos g

. Thus

= 0.005. Therefore,

Motion in Two Dimensions 4.15

Hence T increases by 0.5%. Thus the correct choice is (d). 2 v02 sin cos 4 v02 sin 2 2 and T = . 18. Now R = g2 g2 From these two equations we have T2 = 2R tan or R . Hence the correct choice is (b). T 19. The range of a projectile is the same for two angles of projection and 90° – . For these two angles of projection, the maximum heights are

and

h1 =

v02 sin 2 2g

h2 =

v02 sin 2 90 2g

h 1h 2 = Also R2 =

20.

v02 cos 2 2g

v04 sin 2 cos 2

.

4g 2 4 v04 sin 2 g

cos 2

2

.

Which give R2 = 16 h1h2 or R = 4 h 1 h 2 . Hence the correct choice is (d). u2 2g where u is the initial speed of projection. The maximum height attained by the second ball is

22. The correct choice is (c). Use tA =

2u A g

= 90° – 60° = 30°)

u 2 sin 2 (30 ) u 2 h2 = 2g 8g Now, PE of ball 1 at height h1 = mgh1 and that of ball 2 at height h2 = mgh2. Therefore, the ratio of h u2 8 g = 4. Hence the potential energies = 1 h2 2 g u 2 correct choice is (a). 21. For the body of mass m1, we have u2 2g For the body of mass m2, if S is the maximum distance travelled along the incline then v2 – u2 = 2aS h=

2uB sin 60 g

23. The correct choice is (a). The magnitude of velocity is v = (3)2 (4) 2 = 5 ms–1 The angle subtended by the velocity vector with the horizontal (x-axis) is given by 4 4 tan = which gives sin = 3 5 Now, 4 2 2 ( 5 ) v 2 sin 2 5 = 0.8 m hmax = 2g 2 10 2v sin 24. tf = g

2 5 10

4 5 = 0.8 s,

which is choice (a). 25. Velocity of projection is v0 = (u cos ) i + (u sin ) j . At time t, the velocity of the body is v = (u cos ) i + (u sin

h1 =

(

tB =

and

– gt) j

The dot product of v0 and v is v0 v = u2 cos2 + u sin (u sin or v0 v = u2 – (u sin )gt

– gt) (i)

Since v is perpendicular to v0, v0 v = 0. Using this in (i), we have u 0 = u2 – (u sin )gt or t = g sin Hence the correct choice is (d). 26. The correct choice is (b). PE = mghmax = mg

u 2 sin 2 2g

mu 2 sin 2 2

1 m(u cos )2. 2 27. The horizontal and vertical distances travelled in time t are x = ut KE =

Now, when S is maximum, v = 0. Also a = – g sin g = – g sin 30° = – . Hence 2 g S 0 – u2 = 2 2

and

u2 = 2h, which is choice (a). or S = g

and

y=

1 2 gx 2 gt = 2 2u 2

dy gx = 2 dx u d2y dx

2

=

g u2

gt u

(

x = ut)

4.16 Comprehensive Physics—JEE Advanced

The radius of curvature of the trajectory at time t is given by 2 3/ 2

dy dx

1 R=

u2 1 g

d2y dx 2

g 2t 2

3/ 2

u2

Hence the correct choice is (d). 28. Refer to Fig. 4.6 on page 4.3. The range along the inclined plane is given by R=

=

2v02 sin(

) cos

g cos 2 v02

[sin(2 –

g cos 2

) – sin

]

R is maximum when sin(2 – ) = 1 or 2 – or

= 90°

1 1 (90° + ) = (90° + 10°) = 55°. 2 2 Hence the correct choice is (c). 29. Let h be the height of the tower. Let u1 be along positive x-direction u2 along negative x-direction. The two balls hit after at time t given by =

1 2 2h gt t= 2 g At time t, the respective vertical velocities (along the negative y directions) are gt each. Hence the velocities of the two balls at time t are h=

– v1 = u1 i – (gt) j

v1 = – u1 i + (gt) j

and – v2 = – u2 i – (gt) j

v2 = u2 i + (gt) j

Since v1 and v2 are perpendicular to each other, v1 v2 = 0 [– u1 i + (gt) j ]

[u2 i + (gt) j ] = 0

– u 1u 2 + g 2t 2 = 0 g 2t 2 = u 1 u 2 h=

g2

2h = u 1u 2 g

u1u2 . Hence the correct choice is (d). 2g

30. Average velocity =

Displacement time

2u 2 sin cos where R = g

R tf

2u sin g The correct choice is (a). 31. Horizontal component of velocity at a time t is vx = u cos vertical component of velocity at that time t is vy = u sin – gt They will become equal at time t = t* if u cos = u sin – gt* and tf =

which gives t* = u(sin – cos ) Now t* must be positive. Hence sin > cos or > /4. Hence the correct choice is (b). 32. Given, h = 2 m, R = 1.25 m and horizontal distance s = 10 m. When the string breaks, the stone is projected in the horizontal direction, which means that there is no initial vertical velocity. From s = ut + 1 gt 2, we have ( u = 0), 2 1 2 gt (i) 2 The horizontal distance travelled in time t is s = vt (ii) where v is the velocity of the stone in the horizontal direction which is the same as its velocity in circular motion. Eliminating t from (i) and (ii) we get h =

g s2 2h Now, centripetal acceleration is v2 =

ac =

v2 R

2

10 100 = 200 ms –2 2 1.25

Thus, the correct choice is (b). 33. Refer to Fig. 4.17. AB and CD represent the two velocity vectors. The change in velocity v = v2 – v1. which can written as v = v2 + (– v1). Thus, to v, we reverse the direction of vector AB as v2 and – v1 by triangle or parallelogram law. This is shown in Fig. (c). The magnitude of vector v is given by v = [(v1)2 + v22 + 2(v1)(v2)cos(180° – )]1/2 = [2v2(1 – cos )]1/2 = 2 v sin

is the horizontal range

g s2 2hR

2 Hence the correct choice is (d).

(

v1 = v2 = v)

Motion in Two Dimensions 4.17

Fig. 4.19

37. If r is the radius of the track, then distance moved in 5s = r = 20 cm. Therefore, speed along the 20 circle (v) = = 4 cms –1. Now, angular speed 5 v 4 rad s –1. Hence the correct choice = r 20 5 is (d). 2

–

1

Fig. 4.17

34. Since the force is always perpendicular to the velocity (i.e. the direction of motion) of the particle, no work is done by the force on the particle. Hence the kinetic energy of the particle remains constant. The particle will move in a circle in a plane. Thus the correct choice is (c). 35. As shown in Fig. 4.18, at diametrically opposite points A and B, the magnitude of the velocity is same (= V) but the directions of the velocity are opposite. Hence the change in momentum is MV – (– MV) = 2 MV. Thus the correct choice is (c).

38. Referring to Fig. 4.20, the cyclist is moving on a straight road from A to B with a velocity v = 6 ms–1. As he approaches the circular turn, he decelerates at rate at, represented by vector BD. The magnitude of deceleration is at = 0.4 ms –2. At point B, two accelerations t and c, the centripetal acceleration directed towards the centre C act on the cyclist. v 2 (6) 2 = 0.3 ms –2. Using the law of R 120 parallelogram of vector addition, vector BE gives the resultant acceleration a whose magnitude is Now ac =

(

DE =

a =

(a t2

c)

+ a c2)1/2 = {(0.4)2 + (0.3)2}1/2 = 0.5 ms –2

Hence the correct choice is (a).

Fig. 4.20 Fig. 4.18

36. When the train is at rest or moving with a uniform velocity, the plumb line hangs vertically along OB (Fig. 4.19). If the train moves with an acceleration a, the plumb line gets inclined along OC, the direction of the resultant of accelerations a and g. It is = a/g. Hence the correct choice is (a).

39. In Fig. 4.21 vc represents the velocity of the car and vP that of the parcel. M is the position of the man. From parallelogram law, the direction of the resultant velocity vr must be along the direction along which the man is standing. It follows from is given by vc 10 1 or = 45º sin = vp 10 2 2

4.18 Comprehensive Physics—JEE Advanced

Fig. 4.23 Fig. 4.21

Hence the correct choice is (a). 40. Velocity of rain (vr) = 4 ms –1 vertically downwards. Velocity of wind (vw ) = 3 ms–1 from north to south direction. A rain drop is acted upon by two velocities vr and vw as shown in Fig. 4.22. From the triangle law, the resultant velocity of the rain drop is v = OW. In order to protect himself from rain, he must hold his umbrella at an angle with the vertical (towards north) given by tan

=

RW OR

vw vr

43. Refer to Fig. 4.24. Let PR be the direction along which he should row his boat. The boat is acted upon by two velocities – boat velocity (vb) and water velocity (vw). The angle should be such that that the resultant velocity (v) is along PQ, i.e. sin

vw vb

=

2 = 0.5 4

= 30° (upstream).

3 4

Thus the correct choice is (b).

Fig. 4.24

Hence the correct choice is (a). 44. Time taken to cross the river by the shortest path PQ is

Fig. 4.22

41. The magnitude v of the resultant velocity gives the speed with which the rain strikes the umbrella, which is given by v=

[v2r

+

v2w]1/2

= [16 + 9]

1/2

= 5 ms

–1

Hence the correct choice is (c). 42. In order to cross the river in the shortest time, the resultant velocity v of the swimmer must be perpendicular to the velocity vw of water, as vs2 = v2 + v 2w or v 2w = v 2s – v2 = 25 – 9 = 16 –1 or vw = 4 ms which is choice (b).

t=

PQ v

PQ vb2

36 vw2

(4)

2

(2)2

6 3s

Hence the correct choice is (c). 45. Since the direction of the velocity changes from point to point on the circle, choices (a), (b) and (d) are incorrect. 46. From energy conservation, [see Fig. 4.25] 1 2 1 mu = mv 2 + mgL 2 2 v=

u2

2 gL

v=

v2

( u )2

2(u 2

gL)

Motion in Two Dimensions 4.19

47.

v O

,R

u2. So the correct graph is (d).

B v

L

A

v

–u

u

Fig. 4.25

II Multiple Choice Questions with One or More Choices Correct 1.

2.

3.

4.

5.

different angles of projection, say, and , that give it the same range. Then, and are such that (a) cosec = sec (b) tan ( + ) (c) sin2 – cos2 = sin2 – cos2 (d) cot = cos sec A ball is projected upwards at a certain angle with the horizontal. Which of the following statements are correct? At the highest point (a) the velocity of the projectile is zero (b) the acceleration of the projectile is zero (c) the velocity of the projectile is along the horizontal direction. (d) the acceleration of the projectile is vertically downwards. Choose the correct statements from the following. The range of a projectile depends upon (a) the angle of projection (b) the acceleration due to gravity (c) the magnitude of the velocity of projection (d) the mass of the projectile A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that: (a) its velocity is constant (b) its acceleration is constant (c) its kinetic energy is constant (d) it moves in a circular path A simple pendulum of length r and bob mass m swings in a vertical circle with angular frequency . When the string makes an angle with the

vertical, the speed of the bob is v. The radial acceleration of the bob at this instant is given by (a)

v2 r

(b)

v2 r

v2 (d) r 2 r 6. A body is moving in a circle of radius r with a uniform speed v, angular frequency , time period T and frequency . The centripetal acceleration is given by (c)

(a)

v2 r

(c)

v

(b) 4 (d)

4

r 2

2

r

2

T 7. Which of the following statements are true about a body moving in a circle with a uniform speed? (a) The speed of the body is constant but its velocity is changing (b) The acceleration is directed towards the centre (c) The velocity and acceleration vector are perpendicular to each other. (d) Elastic, frictional, gravitation and magnetic forces can cause a uniform circular motion. 8. from a place in the enemy country at a distance x half-way point. Then (a) the velocity with which the missile was projected is

gx

4.20 Comprehensive Physics—JEE Advanced

x 2g (c) the speed of the missile when it was detected is gx 2 (d) the maximum height attained by the missible x is . 4 (b) you have a warning time of

9. v. An armyman with an anti-aircraft gun on the ground sights the enemy plane when it is directly u. Then (a) the angle with the vertical at which the gun v u (b) the angle with the vertical at which the gun

(d) the initial speed of the projectile is (a2 + c2)1/2. 12. A projectle thrown at an angle of 30º with the horizontal has a range R1 and attains a maximum height h1. Another projectile, thrown with the same speed, at an angle of 30º with the vertical has a range R2 and attains a maximum height h2. Then (a) R2 = 2R1 (b) R2 = R1 (c) h2 = 2h1 (d) h2 = 3h1 13. The maximum height attained by a projectile is increased by 1% by increasing its speed of projection without changing the angle of projection. Then the percentage increase in the (a) horizontal range will be 2% (b) horizontal range will be 1%

cos–1

v . u (c) the maximum height at which the enemy sin–1

u2

v2

. 2g (d) the maximum height at which the enemy (u v) 2 . 2g 10. From the top of a tower of height 40 m, a ball is projected upwards with a speed of 20 ms–1 at an angle of elevation of 30º. The total time taken by the ball to hit the ground is T and the time taken to come back to the same elevation) is t. The horizontal distance covered by the ball is x. If g = 10 ms–2, then T T =2 (b) = 2 (a) t t (c) x = 40 2 m (d) x = 40 3 m 11. The horizontal distance x and the vertical height y of a projectile at time t are given by x = at

and

14. The speed of projection of a projectile is increased by 1% without changing the angle of projection. Then, the percentage increase in the (a) horizontal range will be 1%. (b) maximum height attained will be 2%

y = bt2 + ct

where, a, b and c are constants. Then (a) the speed of the projectile 1 second after it is a2 + b2 + c2)1/2 (b) the angle with the horizontal at which the –1 c a (c) the acceleration due to gravity is –2b.

15. A body is projected at time t = 0 from a certain point on a planet’s surface with a certain velocity at a certain angle with the planet’s surface (assumed horizontal). The horizontal and vertical displacements x and y (in meters) respectively vary with time t (in seconds) as x = 10 3 t y = 10 t – t2 (a) The acceleration due to gravity on the surface of the planet is 10 ms–2. (b) The maximum height attained by the body is 25 m. (d) The horizontal range is 100 m. 16. For a particle moving in a circle with a constant speed, (a) the velocity vector is always along the tangent to the circle. (b) the acceleration vector points towards the centre of the circle. (c) the velocity and acceleration vectors are perpendicular to each other. (d) the velocity and acceleration vectors are parallel to each other.

Motion in Two Dimensions 4.21

17. A particle is acted upon by a constant force which is always perpendicular to its velocity. The motion of the particle takes place in a plane. It follows that (a) its speed is constant (b) its acceleration is constant (c) its kinetic energy is constant (d) its momentum is constant. 18. A stone of mass 250 g is tied to the end of a string of length 1.0 m. It is whirled in a horizontal circle with a frequency of 30 rev./min. (a) The tension in the string changes as the stone moves in the circle. (b) The tension in the string is constant equal to

linear speeds of points A, B and C respectively at that instant, then (a) vA = vB = vC (b) vA > vB > vC (c) vA = 0, vB =

Fig. 4.26

(a) vA = vB = vC (b) vA = vB > vC (c) vA = vC > vB (d) vA < vB < vC 20. A uniform disc of radius R is rolling (without slipping) on a horizontal surface with an angular speed as shown in Fig. 4.27. O is the centre of the disc, points A and C are located on its rim and R from O. During rolling, point B is at a distance 2 the points A, B and C lie on the vertical diameter at a certain instant of time. If vA, vB and vC are the

(d)

vB vC

3 4

Fig. 4.27

2

newton. 4 (c) The speed of the stone is ms–1. (d) The maximum speed with which the stone can be whirled is 20 ms–1. 19. A uniform disc of radius R is rotating about its axis with angular speed . It is gently placed on a horizontal surface which is perfectly frictionless (Fig. 4.26). If vA, vB and vC are the linear speeds of points A, B and C respectively, then

3R 2

IIT, 2004 21. The trajectory of a projectile in a vertical plane is y = ax – bx2, where a and b are constants and x and y are respectively the horizontal and vertical distances of the projectile from the point of projection. a . (a) The horizontal range of the projectile is 2b (b) The maximum height attained by the projeca2 tile is . 4b 2a , where g is the bg acceleration due to gravity. (d) The angle of projection from the horizontal is = tan–1 (a – 2bx). IIT, 1990 22. The coordinates of a particle moving in a plane are given by x = a cos (pt) and

y = b sin (pt)

where a, b and p are positive constants and b < a. Then (a) the path of the particle is an ellipse (b) the velocity and acceleration of the particle are perpendicular to each other at t = /2p. (c) the acceleration of the particle is always directed towards a focus. (d) the distance travelled by the particle in time interval t = 0 to t = /2p is a. IIT, 1999

4.22 Comprehensive Physics—JEE Advanced ANSWERS AND SOLUTIONS

1. For the same range + = 90º or = 90º – . Choices (a), (b) and (d) satisfy this relation between and but choice (c) does not. 2. The correct choices are (c) and (d). u 2 sin 2 , is independent of the mass g of the projectile. Hence choices (a), (b) and (c) are correct.

9. Let G be the position of the gun and E that of the v, when u in a direction with the horizontal (Fig. 4.28). The horizontal component of u is

3. Range R =

vx

u cos

v

4. The magnitudes of velocity and acceleration remain constant but their directions are changing continuously. In uniform circular motion the force is radial (centripetal) and is always perpendicular to the velocity which is tangential. Thus, choice (c) and (d) are correct.

v

5. The radial component of acceleration is ar = centripetal acceleration v

v2 = =r 2 ( v=r ) r Hence, the correct choices are (c) and (d). 6. Since v = r = 2 correct.

r=

Fig. 4.28

Let the shell hit the plane at point P and let t be the time taken for the shell to hit the plane. It is clear that the shell will hit the plane, if the horizontal distance EP travelled by the plane in time t = the distance travelled by the shell in the horizontal direction in the same time, i.e. v t = vx v t or v = v x = u cos or cos = . u To avoid being hit, the plane should have a minimum altitude = maximum height attained by the shell which is

2 r , all the four choices are T

7. All the four choices are correct. 8. For maximum range = 45° for which Rmax = v20/g gx which is choice (a). Hence v0 = g Rmax The warni warning time is t=

tf 2

v0 sin g

gx sin 45 g

hmax =

x 2g

u2 1

Hence choice (b) is also correct. At half-way point, the missile is at its maximum height. Therefore, the vertical component of velocity is zero at this point. Hence the velocity is given only by the horizontal component which is vx gx = v0 cos = gx cos 45° = , which is choice 2 (c). The maximum height attained is hmax =

v02 sin 2 2g

gx sin 2 45 2g

Thus, all the four choices are correct.

u 2 sin 2 2g

x 4

=

2g

u 2 (1 cos 2 ) 2g v2 u2

(u 2

v2 ) 2g

Hence the correct choices are (a) and (c). 10.

t =

2u sin g

2 20 sin 30 =2s 10

Initial downward velocity = u sin = 20 sin 30° = 10 ms–1. The time taken to fall through a height of 40 m is given by 1 40 = 10t1 + 10 t21 which gives t1 = 2 s. Hence, 2 the total time taken to hit the ground is T = 2 + 2 = 4 s. Therefore T/t = 2. Also, the horizontal distance

Motion in Two Dimensions 4.23

travelled in 4 s = (u cos ) T = 20 cos 30° = 40 3 m. Hence the correct choices are (a) and (c). 11. The horizontal component of velocity is dx d at = a dt dt The vertical component of velocity is vx =

4

T= (i)

dy d (ii) b t 2 ct = 2 bt + c dt dt The value of vy at t = 1 s is (2b + c). Therefore, the magnitude of velocity at t = 1 s is vy =

v = (v 2x + v 2y) 1/2 = [a2 + (2b + c)2] 1/2 If a projectile is projected with an initial velocity v0 at an angle with the horizontal, the horizontal and vertical components of its velocity at time t are given by (iii) vx = v0 cos and vy = v0 sin – gt (iv) Comparing (iii) and (iv) with (i) and (ii) above we have v0 cos = a and v0 sin = c. Dividing, we get, tan = c/a. Compairing (iv) with (ii) we have g = – 2b We have seen above that v0 cos = a and v0 sin = c. Squaring and adding we get: v 20 = a2 + c2 or v0 = (a2 + c2) 1/2. Hence the correct choice are (b), (c) and (d). 12. The range is the same for and (90° – ). Hence R1 = R2 for = 30° or 60 °. v 2 sin 2 Since hmax = 0 2g h1 h2

and v0 is the same, we have 2

sin 30

1 3

sin 2 60

or

h h1 = 2 . 3

Thus the correct choice are (b) and (d). v02 sin 2 2g

13. We know that h =

. The increase

h in h

when v0 changes by v0 can be obtained by partially differenting this expression. Thus h=

2 v0 v0 sin 2 2g

v0 1 h = v0 2 h Now range R =

1 2

v02 sin 2 g

,

h h

0.01 = 0.005

2 v0 v0

2 v0 R = =2 v0 R

0.005 = 0.01 = 1 %

2v0 sin g

v0 T = = 0.005 = 0.5 % v0 T Hence the correct choices are (b) and (c). 14. The correct choices are (a) and (d). Use

R = R

v0 2 v0 h 2 v0 T , and . v0 v0 v0 h T 15. The horizontal and vertical displacements are given by x = (v0 cos )t (1) 1 2 (2) and y = (v0 sin )t – gt 2 Given x = 10 3 t (3) and y = 10 t – t2 (4) Comparing (3) with (1) and (4) with (2), we have (5) v0 cos = 10 3 v0 sin = 10 (6) 1 and g =1 g = 2 m s–2 2 Equations (5) and (6) give v0 = 20 ms–1 and = 30°. hmax =

v02 sin 2 2g

(20)2

sin 2 30 = 25 m 2 2

tf =

2v0 sin g

2 20 sin 30 = 10 s 2

R=

v02 sin 2 g

(20)2

sin 60 2

100 3 m

Hence the correct choices are (b) and (c). 16. The correct choices are (a), (b) and (c). 17. Since the force is always perpendicular to the velocity (i.e. the direction of motion) of the particle, no work is done by the force on a particle. The particle will move in a circle in a plane. Its speed and hence its kinetic energy will remain constant. Since the direction of the velocity is along the tangent to the circle, it will keep changing with time. Hence, the momentum will not remain constant. Since velocity is changing with time, the acceleration (which is perpendicular to velocity) will also keep changing. Hence the correct choices are (a) and (c). 18. Since the stone is whirled in a horizontal plane, the weight mg of the stone (which acts vertically down-

4.24 Comprehensive Physics—JEE Advanced

wards) is perpendicular to the plane of the circular motion and, therefore, has no component along this plane. Hence the tension in the string is constant. Given m = 250 g = 0.25 kg, R = 1.0 m, frequency ( ) 30 = 30 rev./min = = 0.5 Hz. Angular frequency 60 ( ) = 2 = 2 0.5 = rad s –1. The necessary cen-

dx dx dy 2bx = a dt dt dt vy = avx – 2bx vx = (a – 2bx)vx

At the maximum height vy = 0. Using this in Eq. (2), we get (a – 2bx) = 0 or x = a/2b. Putting this value of x in Eq. (1), we have (since y = hmax at this value of x)

tripetal force is provided by the tension in the string. Therefore, m v2 R

m

= 0.25

2

T=

R R

hmax = a

2

=m

2

R

4 Speed of the stone is v = R = 1.0 The maximum speed is given by

N =

ms–1.

100 R m = 20 ms –1

vmax =

100 1.0 0.25

vB vc

3R 2

1 2R

tf = 2t = a

2

a 2b

a2 4b

3 . 4

Hence the correct choices are (c) and (d). 21. Given y = ax – bx2 (1) (a) The value of y is zero at x = 0 and x = R (horizontal range). Putting y = 0 and x = R in Eq. (1), we get R = a/b. (b) Differentiating Eq. (1) with respect to time t, we have

2a 2 4bg

a 2bg

2 bg

dy d (ax – bx2) = a – 2bx dx dx Hence the correct choices are (b) and (d). 22. Given x = a cos (pt) (1) y = b sin (pt) (2) (d) tan

Hence the correct choices are (b), (c) and (d) 19. Since the surface is perfectly frictionless, the disc will not roll on the surface; it will simply keep on rotating at point A where it is placed. Now, linear speed = distance from centre angular speed. Therefore, R and vC = R vA = R, vB = 2 Hence the only correct choice is (c). 20. The disc is rolling about the point O. Thus the axis of rotation passes through the point A and is perpendicular to the plane of the disc. From the relation v = r where r is the distance of the point on the rim about the axis of rotation, we have 3R vA = 0, vB = (AB) = 2 and vC = (AC) = 2R Hence

b

2hmax g

t=

2 m vmax = 100 R

or

a 2b

(c) The time t to reach the maximum height is given by 1 hmax = gt 2 2

2

1.0 =

(2)

=

From (1) and (2)

x2

y2

2

2

= cos2 (pt) + sin2 (pt) = 1

a b Hence the path of the particle is an ellipse. Let the position vector of the particle at time t be r = xi Velocity v =

yj

dr dt

dx i dt

dy j dt

v = – a p sin (pt) i + bp cos (pt) j At

t = /2p, v = – ap sin

2

i + bp cos

2

= – ap i 2

j (3)

2

Acceleration a = – ap cos (pt) i – bp sin (pt) j At

t = /2p, a = – bp2 j At

t = /2p, v a = abp3 ( i j ) = 0.

Hence v a . It is easy to see that choices (c) and (d) are incorrect.

Motion in Two Dimensions 4.25

III Multiple Choice Questions Based on Passage

Two objects are projected from the same point with the same speed u at angles of projection and with the horizontal respectively. They strike the ground at the same point at a distance R from the point of projection. The respective maximum heights attained by the objects are h1 and h2 and t1 and t2 1. R, h1 and h2 are related as (a) R =

h1h2

(b) R =

(c) R = 2 2h1h2

(a) sin (b) (c) tan (d) 3. The ration h1/h2 is equal to (b) (a) sin2 (c) tan2 (d) 4. The sum (h1 + h2) is equal to

2h1h2

cos2 cot2

(a)

u2 sin2 g

(b)

u2 cos2 g

(c)

u2 g

(d)

u2 2g

(d) R = 4 2h1h2

t 2. The ratio 1 is equal to t2

cos cot

SOLUTIONS Since the horizontal range in the same Therefore = 90° – and we have

+

= 90°.

1. From Eqs. (1), (2) and (3), we have 2 gh1 2 gh2 u2 2 4 h1h2 g u2 u2 Hence the correct choice is (d) 2. From Eqs. (4) and (5) it follows that the correct choice is (c). 3. From Eqs. (1), (2), (4) and (5), we have R=

u2 sin2 h1 = 2g

(1)

h2 =

u2 u2 sin2 = cos2 2g 2g

(2)

R=

u2 2g

(3)

t1 =

2u sin g

t2 =

2u sin g

2 sin

cos

(4) 2u cos g

(5)

The position vector r with respect to the origin of a particle varies with time t as r = (at) i + (bt – ct2) j where a, b and c are constants. 5. The trajectory of the particle is a (a) straight line (b) circle (c) parabola (d) none of these

h1 = tan2 h2 Hence the correct choice is (c). 4. h1 + h2 =

u2 , which is choice (d). 2g

6. The magnitude of the initial velocity of the particle is (a)

a2

c2

(b)

b2

c2

(c) a 2 b 2 (d) (a + b – 2c) 7. The angle with the horizontal along which the particle is projected is given by

4.26 Comprehensive Physics—JEE Advanced

(a) sin (c) tan

= =

b c

(b) cos

b a

(d) tan

10. The maximum height to which the particle rises is

a c

b c

(b)

(a)

2c

= a

8. (a)

=

2

b2

ab c

2b 2 c

b2 2c

(b)

b2 4b 2 (d) 4c c 11. The horizontal range of the particle is (c)

c b (d) a ac 9. The acceleration due to gravity at that place is (a) 2 a (b) 2 b (c) 2 c (d) none of these

(a)

ab c

(b)

(c)

bc a

(d) abc

(c)

ac b

SOLUTIONS 5. Comparing Eq. r = (at) i + (bt – ct2) j with Eq. r = x i + y j , we get x = at (1) (2) and y = bt – ct2 b x– Eliminating t from (1) and (2) we get y = a c x2 which is the equation of a parabola. Hence 2 a the correct choice is (c). 6. From Eqs. (1) and (2), we have dx =a (3) dt dy = b – 2ct (4) vy = dt Putting t = 0 in Eqs. (3) and (4), the initial values of vx and vy are a and b respectively. The initial speed of the particle is vx =

2

1 2 gt (6) 2 Comparing Eqs. (5) and (6) with Eqs. (3) and (4) we have u cos = a and u sin = b which give tan = b/a, which is choice (c). 8. When t = tf, y = 0. Putting y = 0 and t = tf in Eq. (2) we get 0 = tf (b – ctf ) gives tf = 0 and tf = b/c. But tf = 0 is not possible. Hence the correct choice is (a). 9. Comparing Eq. (6) with Eq. (2), we get g = 2c, which is choice is (c). 1 b . Putting y = hmax 10. Now y = hmax when t = tf = 2 2c and t = b/2c in Eq. (2) we get y = (u sin )t –

and

11.

2

b2 4c

R = (u cos )tf a2

Now u =

2

u = a b , which is choice (c). 7. If the particle is projected with an initial velocity u at an angle with the horizontal, then the horizontal displacement x and vertical displacement y at time t are x = (u cos )t (5)

b b c 2c 2c

hmax = b

b 2 , tan

a cos

=

a

2

b

2

and tf =

(7) =

b which gives a

b . Putting these values c

ab . Hence the correct choice in Eq. (7), we get R = c is (a).

An object of mass m is whirled with a constant speed v in a vertical circle with centre O and radius R. T1, T2, T3 and T4 are the tensions in the string when the object is at A (top of the circle), B, C (the lowermost point of the circle) and D respectively (Fig. 4.29) Fig. 4.29

Motion in Two Dimensions 4.27

12. Tensions T1, T2 and T3 are related as (a) T1 = T2 = T3 (b) T1 < T2 < T3 (c) T1 > T2 > T3 (d) T1 = T3 < T2

(a)

Rg 2

(b)

13. Tension T4 is given by

(c)

2Rg

(d) 2 Rg

(a) T4 =

mv 2 + mg cos R

(b) T4 =

mv 2 – mg cos R

(c) T4 =

mv R

Rg

Here g is the acceleration due to gravity. 15. The minimum speed the object must have at the lowest point C to complete the circle is (a)

Rg

(b)

2Rg

(d)

5 Rg

2

+ mg sin

mv 2 – mg sin R 14. The minimum speed the object must have at the highest point A to complete the circle is (d) T4 =

SOLUTIONS 12. Refer to Fig. 4.30

(c) 2 2Rg

16. At the minimum speed of Q.15, the tension in the string is (a) 4 mg (b) 5 mg (c) 6 mg (d) mg

mv 2 + mg R Hence the correct choice is (b). 13. At point D, the force towards the centre is F4 = T4 – mg cos T3 =

mv 2 = T4 – mg cos R mv 2 + mg cos , which is choice (a). R 14. In order to keep a body of mass m in a circular path, the centripetal force, at the highest point A, must at least be equal to the weight of the body. Thus T4 =

Fig. 4.30

When the object is at A, since the weight mg acts vertically downwards, the force towards centre O is. F1 = T1 + mg mv 2 = T1 + mg R mv 2 – mg R At point B, weight mg has no component along BO. Hence the force towards O is T1 =

mv 2 = T2 R At point C, the weight mg acts in opposite direction to tension T3. Thus at C, the force towards centre O is F3 = T3 – mg

mv 2A = mg or vA = Rg R gives the minimum speed the body must have at the highest point so that it can complete the circle. Hence the correct choice is (b). 15. The minimum speed vC of the body must have at the lowest point C is given by v2C = v2A + 2 2 Rg where we have used v2 = u2 + 2gh, with h = 2 R. Thus

F2 = T2

mv 2 = T3 – mg R

v2C = Rg + 4 Rg = 5 Rg or vC =

5 Rg , which is choice (d).

16. The tension at this point is given by vC2 g = m(5 g + g) = 6 mg R Hence the correct choice is (c). T3 = m

4.28 Comprehensive Physics—JEE Advanced

The kinetic energy of a particle moving along a circle of radius R depends on distance (s) as K = as2 where a is a constant. 17. The centripetal force is given by (a)

as 2 2R

(b)

(a) mas (b) 2mas (b) as (d) 2as 20. The net force acting on the particle is

as 2 R

2as 2 4as 2 (d) R R 18. The speed of the particle around the circle is (c)

a (a) 2s m

1/ 2

a (b) s m

2a 1 / 2 a 1/ 2 (d) s m 2m 19. The tangential force acting on the particle is (c) s

s (a) 2as 1 R

(b) as 1

(c) 2as 1

1/ 2

R2

1/ 2

s2

1/ 2

s2

(d) zero

R2

SOLUTIONS 17. Given KE = force is

1 mv2 = as2. Therefore, the centripetal 2 1 2 mv 2 R

2 mv 2 R which is choice (c). fc =

2as 2 , R

or

2a v=s m

2a ds =v = s m dt

f = (f 2c + f 2t )1/2

dv dt

2a d s dt m

2a m

1/ 2

ds dt

A conical pendulum consists of a string of length L at one end carrying a body of mass m at the other end. The mass is revolved in a circle in the horizontal plane about-a

= 2 as 1

The angular frequency of revolution of the body is . The string makes an angle with the vertical axis. 21. The tension in the string is m 2 L (c) m 2L

L 2 m (d) m L2 (b)

1/ 2

2

(2 a s ) s2

2

1/ 2

R2 Thus the correct choice is (c).

22. The angle of inclination of the string with the vertical is given by (a) cos

(a)

2 a s2 R

=

Hence the correct choice is (c). 19. The tangential acceleration is at =

. Therefore,

2 a 1/ 2 2 a 1/ 2 2 a s s m m m Tangential force is ft = mat 2a s =m = 2as, which is choice (d). m 20. Net force acting on the particle is

1/ 2

1/ 2

1/ 2

at =

18. The speed v of the particle around the circle is given by 1 mv2 = as2 2

But

=

g 2 2

(c) cos

=

g

L L

(b) sin (d) sin

23. The linear speed of the body is (a) L (b) (c)

L cos

(d)

= =

L sin L tan

g 2

L

2

L

g

Motion in Two Dimensions 4.29

SOLUTIONS 21. Let T be tension in the string. Figure 4.31 shows the forces acting on the system. Tension T can be resolved into two mutually perpendicular components. The horizontal component T sin provides the centripetal force for circular motion and the vertical component T cos balances the weight mg.

Thus T cos T sin

and But T sin

=m

= mg

(1)

m v2 = m 2r r r = L sin . Therefore, =

2

L sin

or T = m

2

L

(2)

Hence the correct choice is (c). mg (3) 22. From (1), we have cos = T Using (2) in (3), we get mg g cos = , which is choice (a). 2 2 m L L 23. Linear velocity is v = r = L sin , which is choice (b).

Fig. 4.31

IV Matching 1. For a body projected at time t = 0 horizontally with velocity u from a height h. Column I Column II 2h g

(a) Horizontal displacement

(p)

(b) Vertical displacement

(q) u

(c) Time taken to hit the ground (d) Horizontal range

2h g (r) is proportional to t (d) is proportional to t2

ANSWER (a) (r) (c) (p)

(b) (d)

(s) (q)

2. A body is projected from the ground with velocity u such that its range is maximum. Column I Column II (a) Maximum height attained

(p)

(b) Horizontal range

(q)

2u g u g 2 u2 g

(d) Time to reach maximum height

(s)

u2 4g

4.30 Comprehensive Physics—JEE Advanced

ANSWER For maximum range (a) (c)

= 45°.

(s) (p)

(b) (d)

3. A body is projected with velocity u at angle Column I

(r) (q)

= 30° with the horizontal. Column II

(a) Velocity at maximum height

(p)

13u 4

(b) Velocity at half the maximum height

(q)

2 7 u 3

(c) Average velocity between the point of projection and highest point 2 (d) Velocity at t = 3

(r)

3u 2 7 u 8

ANSWER

Maximum height h =

u 2 sin 2 30 2g

u2 8g

(a) At maximum height, vx = ucos

and vy = 0. Therefore, v = ucos h . 2 vy2 = (u sin )2 – 2 gh = u2 sin2 2

= u2 sin2 30° – g vx = u cos 30° = v= Average velocity =

OA = Now

Now

v 2x

v 2y

2

u 8g

u 4

– gh u2 8

u2 8

3u 2 3u 2 4

u2 8

7 u 8

net displcemrnt OA = (see Fig. 4.32) 1 time taken tf 2 R2 4

h2

u2 R= g OA =

3u . 2

h =

(b) For half the maximum height,

(c)

= ucos 30° =

sin 60° = u4

3u 4

64 g 2

16 g 2

1 u sin tf = 2 g

u 2g

3u 2 2g 1/ 2

13u 2 8g Fig. 4.32

Motion in Two Dimensions 4.31

Average velocity =

At t =

(d)

OA 1 t 2 f

13u 2 2 g 8g u

2 t f, 3

vy = u sin

–g

2 tf = u sin 30º – g 3 u 2

2u 3

= u cos 60° =

3u 2

= vx = u cos v=

13u 4

v 2x

Thus the answer is as follows: (a) (r) (c) (p)

v 2y

3u 2 4 (b) (d)

u2 36

1/ 2

2 3

u g

u 6

2 7 u 3

(s) (q)

4. Match objects in circular motion listed in column I with the sources that provide the necessary centripetal force listed in column II. Column I Column II (a) A boy whirling a stone tied to a string in a circle (p) Frictional force (b) The moon revolving around the earth (q) Muscular force (c) The electron revolving around the proton (r) Gravitational force in a hydrogen atom (d) A car negotiating a curved road (s) Electrostatic force

ANSWER (a) (c)

(q) (s)

(b) (d)

(r) (p)

V Assertion-Reason Type Questions In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each questions has the following four choices out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Statement-1.

(c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. A body is projected horizontally with a velocity u from the top of a building of height h. It hits the ground after a time t = 2h / g . The vertical and horizontal motions can be treated independently.

4.32 Comprehensive Physics—JEE Advanced

A body is projected from the ground with kinetic energy K at an angle of 60° with the horizontal. If air resistance is neglected, its kinetic energy when it is at the highest point of its trajectory will be K/4.

The velocity vector at a point is always along the tangent to the trajectory at that point.

At the highest point of the trajectory, the directions of the velocity and acceleration of the body are perpendular to each other.

Then the force does no work on the body and its kinetic energy remains constant.

One end of a string of length R is tied to stone of mass m and the other end to a small pivot on a frictionless vertical board. The stone is whirled in a vertical circle with the pivot as the centre. The minimum speed the stone must have, when it is at the topmost point on the circle, so that the string does not slack is gR . At the topmost point on the circle, the centripetal force is provided partly by tension in the string and partly by the weight of the stone.

In a uniform circular motion, the centripetal force is always perpendicular to the velocity vector.

In a non-uniform circular motion, the particle has two acceleration-one along the tangent to the circle and the other towards the centre of the circle. In a non-uniform circular motion, the magnitude and the direction of the velocity vector both change with time. In a non-uniform circular motion, the acceleration of the particle is equal to sum of the tangential acceleration and the centripetal acceleration. The two accelerations are perpendicular to each other.

The maximum range on an inclined plane when a body is projected upwards from the base of the plane is less than that when it is projected downwards from the top of the same plane with the same speed. The maximum range along an inclined plane is independent of the angle of inclination of the plane. In projectile motion, the velocity of the body at a point on it trajectory is equal to the slope at that point.

In a uniform circular motion, the kinetic energy of the body remains constant. The momentum of the body does not change with time. In a uniform circular motion, the acceleration is always directed towards the centre of the circle. Otherwise the speed of the body moving along the circle will change with time.

SOLUTIONS 1. The correct choice is (a). The time taken by the body to hit the ground is the same as if it was dropped from that height and fell freely under gravity. 2. The correct choice is (b). If m is the mass of the body and u its velocity of projection, the initial kinetic energy is 1 K = mu2 2 At the highest point, the horizontal velocity is (u cos 60°) and vertical velocity is zero. Hence the kinetic energy at the highest point is K =

1 1 m (u cos 60°)2 = 2 4

1 K mu2 = 2 4

At the highest point of the trajectory, the velocity of the body is horizontal (parallel to the ground) but its acceleration is g directed vertically downwards. 3. The correct choice is (a). When the stone is at the topmost point A on the circle, the centripetal force is provided by (mg + T) as shown in Fig. 4.33. mv 2 = mg + T R When the stone is at A, the string will not slack if Thus

tension T = 0, which gives

mv 2 = mg R

v=

Rg

Motion in Two Dimensions 4.33

R max =

5.

. Fig. 4.33

4. The range along the inclined plane when a body is projected with velocity u at an angle with the horizontal is given by R =

=

2u 2 sin(

) cos

g cos u

6. 7. 8.

g cos

2

ac2

a=

2

g cos 2

u2

Thus R max > Rmax. Since the range R depends on angle , Statement is false. Hence the correct choice is (c). The correct choice is (d). At the highest point on the trajectory, the slope is zero but velocity is u cos . The correct choice is (a). The correct choice is (a). The correct choice is (d). The acceleration of the particle is given by

2

[1 – sin ] =

at2

where ac = centripetal acceleration and at = tangential acceleration.

[sin (2 – ) – sin ]

9. The correct choice is (c). The speed of the body remains constant but the momentum changes with time because the direction of the velocity vector changes with time.

where is the angle of inclination of the plane. Range R will be maximum of sin(2 – ) = 1 or 2 – = 90° in which case Rmax =

u2 g (1 sin )

u2 g (1 sin )

10. The correct choices is (a). If the acceleration vector is directed towards the centre of the circle, it will have a component along the tangent, as a result the speed of the body will change and the motion no longer remains uniform.

If the body is projected downwards from the top of the same inclined plane, the maximum range will be

VI Integer Answer Type 1. A body falling freely from a given height H hits an inclined plane in its path at a height h. As a result of this impact, the direction of the velocity of the body becomes horizontal. For what value of H/h, will the body take the maximum time to reach the ground? IIT, 1986 2. On a frictionless horizontal surface, assumed to be the x-y plane, a small trolley A is moving along a straight line parallel to the y-axis with a constant velocity vT = ( 3 1) ms–1 as shown Fig. 4.34. At a particular instant when OA makes an angle of 45° with the x-axis, a ball is thrown from origin O. Its velocity makes an angle of 60° with the x-axis and its velocity is such that it hits the trolley. Find the magnitude of the veolcity of the ball with respect to the horizontal surface. IIT, 2002

v

Fig. 4.34

3. A train is moving along a straight line with a constant acceleration ‘a’. A boy standing in the train throws a ball forward with a speed of 10 m/s, at an angle of 60° to the horizontal. The boy has to move forward by 1.15 m inside the train to catch the ball back at the initial height. The acceleration of the train, in m/s2, is IIT, 2011

4.34 Comprehensive Physics—JEE Advanced

SOLUTIONS 1. A body falling freely from point P at a height H hits the inclined plane at point Q at a height h. As a result, the velocity becomes horizontal and the the ground at point R. The motion of the body from P to R via Q is shown in Fig. 4.35. P

H d 2t It is easy to check that when = 2, is negah dh 2 tive. 2. Refer to Fig. 4.36. It follows that v B is the resultant of v and vT . In triangle OAB vB vT = sin135 sin15 vB =

H–h Q H

=

Q h

=

vT sin (90 45 ) sin (60 45 ) ( 3 1) cos 45 sin 60 cos 45 cos 60 sin 45 ( 3 1) 1 / 2 3 / 2 2 1/ 2 2

= 2 ms–1

R

Fig. 4.35

v

If t is the time taken by the body to fall from P to Q through a height (H – h), then we have (H – h) =

1 gt 2

2

2( H h) g

or t =

1/2

v v

The time t taken by the body to fall from Q to R is given by ( the initial velocity at O is zero) h=

1 gt 2

2

1/2

2h g

or t =

The total time taken by the body to reach the ground is 1/2

2( H h ) g

t =t + t =

2h g

(1) t=

dt d t = 0 and is dh dh 2 negative. Therefore, differenting Eq. (1) w.r.t. h and dt = 0, we have setting dh

Time t will be maximum if

= = which gives

2( H h) g

2 1 (H g 2 1 g

1 (H 1

(H

1/2

h)

1/2

h)

1/2

( 1)

2h g 1 h 2

1/2

1/2

1 1/2

h) =

1 1/2

h

3. u = 10 ms ,

= 60°

1/2

2

dt d =0 = dh dh

Fig. 4.36 –1

h1/2 H =2 h

2u sin g

2 10 sin 60 10

3s

Let v be the velocity of the train. The horizontal velocity of ball at the instant it is thrown = (v + ux) = (v + u cos ). Therefore, the horizontal range of the ball with respect to the ground is R = (v + ucos )t, where t =

3s

It is clear that Distance travelled by ball in time t + 1.15 = R 1 i.e. vt + at 2 + 1.15 = (v + u cos )t 2 1 2 at + 1.15 = (u cos )t 2 1 a ( 3 )2 + 1.15 = (10 cos 60°) 3 2 a = 5 ms–2

5

Laws of Motion and Friction

Chapter

REVIEW OF BASIC CONCEPTS 5.1

SOLUTION

NEWTON’S FIRST LAW OF MOTION -

every body continues in its state of rest or of uniform motion in a straight line unless it is compelled to change that state by an external unbalanced force.”

5.2

NEWTON’S SECOND LAW OF MOTION

5 2

v=

–1

–2 Using u v –2 we get a Force F = ma = 1.5

the rate of change of linear momentum of a body is directly proportional to the applied force and the change takes place in the direction in which the force acts.

t = 0.5 s in v = u + at 5 = 7.5 N.

5.2

Linear Momentum p = mv dv dp = m dt dt

SOLUTION F1 = F2 (

F=

m

= ma a=

=

dv dt

= 120 F12

F22

2 F1 F2 cos

102 102

2 10 10

1 2

= 10 N

NOTE

a= p–t

(F – t

F 10 = m 5

–2

5.3 5.1 v1

(2 i

3j

k)

1

to v 2

( 3i

2j

3k)

1

5.2 Comprehensive Physics—JEE Advanced

SOLUTION

2 10

v=

= m[ ( 3 i

2j

3 k ) (2 i

3j

v= 2

k

p = 2 ( 5)2 = 2 p t

F

=

( 1)2

(4)2

42 12 96

s

3

1 x

= 1

–1

v

12.96 = 4.32 ~ 4.3 N 3

5.3

1/ 2

1 2

m = 10–3

1/ 2

1

v= 2

x

= m ( 5 i j 4k)

1 x

k = 10–3

[

p = mv2 – mv1

or

10

3

1 1 0.5 –1

2 2

1/ 2

1 x

1/ 2

x=

1

NEWTON’S THIRD LAW OF MOTION

whenever one body exerts a force on a second body, the second body

EXAMPLE 5.4 x

action there is an equal and opposite reaction k

F= k = 10–3 v

x2 2

5.4

x x –1

law of conservation of linear momentum when no net external force acts on a system consisting of several particles, the total linear momentum of the system is conserved, the total linear momentum being the vector sum of the linear momentum of each particle in the system’.

.

SOLUTION F = ma = m

dv dv d v dx =m = mv dx dt dx dt

k

F=

Recoil of a Gun

x2 k x

2

LAW OF CONSERVATION OF LINEAR MOMENTUM

= mv

dv dx

k x m v2 k = c 2 mx

v dv =

v dv

2

k mx 2

dx

vb

dx

mb vb + mg vg

(i) v=0 k c= . m

c x v2 k k = 2 mx m 2k 1 v= 1 m x

mb vb + mg vg = 0 or

1/ 2

mb vg

vg = –

mb v b mg

mg

5.3

vg =

5.5

mb vb mg

IMPULSE A

=

v2

v2

2v 2 cos

=

2v 2 (1 cos )

2v cos

2

B

t

5.6

CONTACT FORCES

(1) Normal Reaction Impulse of a force is the product of the average force and the time for which the force acts and it is equal to the change in momentum of the body during that time. –1 or N s. I = Fav t = p

normal reaction (R). m R = mg

5.5

[Fig.

v

m

R = mg cos

SOLUTION v1 v2

= mv2 – mv1 = m(v2 – v1)

Fig. 5.2

= m[v2 + (– v1 m v v = v2 + (– v1

v2

v1

(2) Tension T

Fig. 5.1

v is [

v1

v2 = v v=

v12

v22

2v1v2 cos

5.7

FRICTION

5.4 Comprehensive Physics—JEE Advanced

or

–1

=

5.8

( )

BANKING OF ROUND TRACKS

limiting friction limiting static friction (fs s s

=

fs R

R

F

F= kinetic or sliding friction (fk ( k k

=

m R

fk R

mv 2 R v

F s.

k

F

Angle of Friction

= mg v v

N; N

= ( Rg)1/2

BANKING OF CURVES

f R

-

=

f R

Fig. 5.3 –1

( )

Angle of Repose

m Fig. 5.5

R N1

Car on a banked curved road

N2 N = N1 + N2 N cos

mg N sin

Fig. 5.4

f ) = mg sin R) = mg cos =

f = R

mv 2 – F cos R N cos = mg + F sin F= N N sin

=

5.5

F

5.10 =

v2

Rg

Rg

v2

v2 = Rg

1

MOTION IN A VERTICAL CIRCLE m

n n

v R

O A T1. Since

v R

mg centre O

R

. R v = (Rg

(

=

5.9

)1/2

v2 Rg

A CYCLIST NEGOTIATING A CURVED LEVEL ROAD

Fig. 5.7

F = T1 + mg = T1 =

or N

B

N = mg m

string is T2

mv 2 R

mv 2 – mg R

(i)

OB OB

mg

B

F = N = mg F = T2 =

mv 2 R

(ii)

At C

mg T3

C F = T3 – mg = T3 =

or

mv 2 + mg R

C

Fig. 5.6 2

mv /R

mg

T3

mv 2 R

F mv 2 > mg R R

A T 1. m

v=

A gR

mv 2 R

5.6 Comprehensive Physics—JEE Advanced

m v 2A = mg R

SOLUTION

v 2A = Rg

or

vC C vC2 = v 2A + 2

2 Rg

v2 = u2 + 2gh vC2

Fig. 5.8

= Rg + 4 Rg = 5 Rg

vC =

or

h = 2R

dx is m =

5 Rg

M dx L

L

m

2

x

o

vC2 R

T1 = m

g

= m (5g + g) = 6 mg

=

5.6

= m L

M L

2 L

1 ML 2

xdx o 2

5.8

k=

–1

l m r

SOLUTION 30 = 60 F = kx

=2

m(L + x)

2

mr = kx

x= =

2

m

0.02

.

SOLUTION

x

2

0.5

3.14 3.14

0.05

2

2

Fig. 5.9

0.1 N

5.7

OQ = r M

AB

–2

2

2 0.02

r= L + x F = kx = 2

Tension

g

.

r = L + x. =F

mL k

–1

=2

L

A i.e. B.

PQ = l O = T sin

mv 2 = T sin r

(i)

5.7

mg = T cos

(ii)

SOLUTION OP OC 1 = = OP 2

v2 rg v=

rg n

=

0.2 10

t

–2

g

mg 0.2 10 = =4N T= cos cos 60

.

OC = 60°. r = CP = OP sin

60 =

–1

5

= 5 sin 60°

3 2

5.9 m

SOLUTION Fig. 5.11

mg

2

m 2

r

N

r

g

C mr = N sin mg = N cos

N sin

2

2

2

r

g 30 60 r 4

2 2

v r g

–2

4

3.14

mg 0.5 10 = 10 N cos cos 60 R sin = N sin

N=

g =

m

.

2

0.5 9.8

2

0.3

= 0.3

2

T= R OC O

Fig. 5.10

N mR

=

5.10 m

(i) (ii)

5.11

2

=

10 0.5 5

2 ds

1

second = 3.14 s

SOLVING PROBLEMS IN MECHANICS BY FREE BODY DIAGRAM METHOD

5.8 Comprehensive Physics—JEE Advanced

R F – R = m 1a

(i)

R = m 2a

(ii)

free body diagram (1) Two masses tied to a string going over a frictionless m2 (m1 pulley m1 m2

F = (m1 + m2)a a=

F (m1

m2 )

m2 is F2 = m2 a = (3) Three masses in contact m1 m2

m2 F (m1 m2 ) m3 F

m1 a

a

m2

m3

Fig. 5.12

m1 is m1g – T. m1 m 1g – T = m 1a

(i)

T – m 2g = m 2a

m2 is (T – m2g). m2 is (ii) Fig. 5.14

a=

T=

m1 m1

m2 g m2

R m2

R

2m1m2 g m1 m2

m3 m2

m3 F – R = m 1a R – R = m 2a R = m 3a

(2) Two masses in contact m2 m1 F

m2 m1

m1

(i) (ii) (iii)

a. To a

m2 a=

(m1

F m2

m3 )

m2 is F2 = R F 2 = R = R + m 2a Fig. 5.13

F 2 = m 3a + m 2a

5.9

= (m2 + m3)a F2 =

F

m1 a

(m2 m3 ) F (m1 m2 m3 )

T m2

m1

m3 is F 3 = R = m 3a =

(m1

a

m3

m2 m2

m3 F m2 m3 )

T

m 3.

(4) Two masses connected with a string m2 m1 F

m2 a

a

m1 T

Fig. 5.17

Fig. 5.15

T = m 1a

(i)

F – T = m 2a

(ii)

a=

(i)

T – T = m 2a

(ii)

F – T = m 1a

(iii)

a=

F m1

T = m 3a

(m1

F m2

m3 )

m2

T = m 1a =

m1

m1 F m1 m2

T = (m2 + m3)a =

(m2 m3 ) F (m1 m2 m3 )

m2 T = m3 a Fig. 5.16

F

m1 a=

T=

m3 is T

m3 F m2 m3 )

(6) Two masses connected by a string and suspended m2 from a support m1 m2

F m1

(m1

m2 is T

m2

m2 F m1 m2

(5) Three masses connected by strings m3 m1 m2

F tension T

T

m1 m2 m1

T = T + m 1g

(i)

T = F + m 2g

(ii)

T = F + (m1 + m2)g

5.10 Comprehensive Physics—JEE Advanced

T = m 1a =

m1m2 g (m1 m2 )

m 2 < m 1. m1 f = R = m 1g

Fig. 5.20

m1 T – f = m 1a

Fig. 5.18

T – m 1g = m 1a

(7) Two blocks connected by a string passing over a

(8) Two blocks connected by a string passing over a

table

plane

m1

m2

T

(a) Mass m1 moving up along the incline with acceleration a

a. 1

2

(a) F.B.D. of

F.B.D. of

1

2

2

1

(b)

2

Fig. 5.19 Fig. 5.21

m1

m2

m1 T = m 1a m 2g – T = m 2a

a=

m2 g (m1 m2 )

(i) (ii)

m2 [see Fig.

= m 1a

(i)

m 2g – T = m 2a

(ii)

T – mg sin

a=

(m2 m1 sin ) g (m1 m2 )

5.11

T = m2 (g – a) m1 f = R = m1g cos m1 T – m1g sin T – m1g sin

– f = m 1a

– m1g cos

= m 1a

(b) Mass m1 moves down the incline with acceleration a T – m 2g m1g sin – T = m1a = m 2a a=

m1 sin m1

m2 g m2

T = m2 (g + a) (9) Two blocks connected by a string passing over a plane

m1 Fig. 5.23

1

m2

f

T

2

m1

m2 T – m1g sin m2g sin

2

1

m2

– T = m 2a

m1

f

m1 f = m 1a

m2 sin

2

m1

m1 sin m2

T = m1(a + g sin

1)

1

= m2 (g sin

R

(i)

2

m2

– a)

R + m 2g = R

F

F F Fig. 5.22

(10) One block placed on top of the another m2 m1 . Case 1: The maximum force that can be applied on the lower block so that the upper block does not slip [Fig. F

g=a

m2

F

m2

m 1g

a

g

F

F

m2

= R = m 1g

= m 1a

T a=

m1

m1 does

(ii)

– f = m 2a – R = m 2a

(iii)

– m 1g = m 2a = (m1 + m2) g

F

m1

F < F

m1

m2 m2

Case 2: The maximum force that can be applied on the upper block so that it does not slip on the lower block. [Fig. 5.24(a)] F F m1 m2 m1 m2 f = R = m 1g m1 on m2

5.12 Comprehensive Physics—JEE Advanced

–2

g

.

SOLUTION

Fig. 5.24

R

m2

a m1 F or

F

= R = m 1a

R = m 1g

– m 1g = m 1a m2 f = m 2a a=

F

=

(i) 1m 1g

= m 2a

m1 g m2 (m1

(ii)

m2 )m1 g m2

F < F F

Fig. 5.26

F 5.11

Fcos m

F = 10 N

= ma a=

F cos m

10 cos 60 1

5

2

Fcos – f = ma Fcos – mg = ma a= = Fig. 5.25

F cos

mg m

10 cos 60 –2

0.2 1 10 1

5.13

Fsin

F sin < mg

= mg – R. Since F sin

+ R = mg or

SOLUTION

5.12 m1

m2

AB CD Fig. 5.27. String AB CD Fig. 5.29

P

AB Q

CD

R cos R sin

CQ –2

g

Fig. 5.27

.

SOLUTION

= mg = ma a= g g

=g

CD is m = 0.5 Since string AB

AB

3

5.14 m

P = m1 + m + m2 P = 5.25

a

10 = 52.5 N

Q

QD

+ m2 = 0.5 Q = 3.4

10 = 34 N

Fig. 5.30

a. 5.13

SOLUTION

m a

Fig. 5.31

Fig. 5.28

T sin direction is

5.14 Comprehensive Physics—JEE Advanced

T sin

= ma

(i)

T cos

= mg

(ii)

a g a= g =

= 9.8

= 9.8

3 17

2

Fig. 5.33

5.15 m1

m2

T2 = 2T1

(iii)

-

m2

m1 .

–2

g

a1 = 2a2

m2g – 2T1 =

m2 a1 2

T1 a1 =

2m2 g 2 2 10 = m2 4m1 2 4 1.5

a2 =

a1 5 = 2 2

T1 = m1a1 = 1.5 T2 = 2T1 = 2

–2

–2

5 = 7.5 N 7.5 = 15 N

5.16 Fig. 5.32

m1 m1

SOLUTION a1 a2

m1 x1

m1

m2 m1 m2

x2 t

m2

x1 or x1 = 2x2 2

x2 =

d 2 x1 dt 2 m2 T2

= 2

d 2 x2 dt 2

(i) m1 m1

T1

m 2.

m1

m2

a1 = 2a2

m2 is F

m1 x1

m2

m2

Fig. 5.34

m 2. m1

SOLUTION T 1 = m 1a 1

m2 m 2g – T 2 = m 2a 2

(i) (ii)

m1 T m1

m2

m2 is f = R1 = m1g.

5.15

R = mg = 50 R 1 = m 1g

(i)

F – f – T= 0

T – f= 0

(ii)

T=f

(iii)

R 2 = R 1 + m 2g Using (iii) in (ii) F – f – f= 0 F = 2f = 2 m1g m2 F m1 F F = 2 m 1g =2

0.5

0.1

R1 – mg = ma R1 = m(g + a) = 50 (c) mg – R2 = ma R2 = m(g – a) = 50

(9.8 + 2.2) = 600 N (9.8 – 2.2) = 380 N

NOTE a

5.18 AB

10 = 1 N

L

A

M B

F1 = 4 N

F2 P

x

A.

Fig. 5.37

SOLUTION Since F1 F2 a

Fig. 5.35

T AP

5.17 m –1

a

–2

AP is m1 =

Mx L

PB is m2 =

M (L L

. AP

SOLUTION

P we con-

PB

x)

PB.

R -

Fig. 5.38

AP F1 – T = m1a

(i)

PB T – F2 = m2a

(ii)

F1 T m = 1 T F2 m2 Fig. 5.36

Mx L M L L

x x

L

x

5.16 Comprehensive Physics—JEE Advanced

F1 L

T= =

x L

F2 x

5.20 m

4 1 0.2

3 0.2 1 F

= 3.8 N

5.19 m

F

–2

a

–2

g M

Fig. 5.41

.

SOLUTION R

-

f –2

g

.

Fig. 5.39

SOLUTION A T A A – a)

Fig. 5.42

f = mg F= R f Fmin =

Fig. 5.40

mg – T = m(A – a) T – Mg = MA

R or mg

(i) (ii)

mg

=

F or F

mg

2 10 = 80 N 0.25

5.21 m M

mg – Mg = m (A – a) + MA

F

M

m g a Mg A= M m =

30 10 5 15 10 15 30 –2

T 250 N

A–a = M(g + A) = 15

–2

(10 + 6.7)

Fig. 5.43

5.17

F –2

g

m

M

v

.

SOLUTION Fig. 5.45

a SOLUTION P P

Q is f = mg Q.

P is aP = Q is aQ =

f m

mg m

f M

mg M P is

Q aQP = aQ – aP = Fig. 5.44

m

=

F – R = ma

(i)

f = mg

(ii)

R = Ma

(iii)

M

g

mg M

g m M

g 1

= 0.3 10 1

1 6

–2

Mg + f = R 5.23

a R=

MF M m

f

R

or

mg

or

F F

m = 500 g

a

MF M m mg M m M =

mg M m M

=

2 10 10 2 0.4 10

SOLUTION g

= 60 N

mg

P M

m

= mg

sin

cos

5.24

Q

m M

P –1

-

P Q

=g+a

0.58

5.22

v

–2

Q is P.

-

5.18 Comprehensive Physics—JEE Advanced

1 cos 45

=

sin 45

0.2

2

2 1 cos 45

5.25

Fig. 5.46

1

a

2

2

SOLUTION -

Fx Fy

SOLUTION A

Fx = (mg cos ) sin Fy = (mg cos ) cos

Mg

a is (A cos

Fig. 5.47

–a

m (A cos – a) = Ma F = mg cos sin A= N = Mg + mg cos2 F = N

=

m cos sin M

m cos 2

M m a m cos 2 1 1/ 2 1 cos 45 –2

I Multiple Choice Questions with Only One Choice Correct 1.

m

(c) g m

3 mg 2 (c)

2u cos

(d) u cos

5 mg 3

7 mg 6

(d)

9 mg 7

2.

P Q u

B

C

M u cos

u cos

Fig. 5.48

5.19

3.

P Q

M

7.

m

m F

-

P v P

Q

Q

P is g (c)

g 1

mg (c) F

mg M m M

(d)

g 1

mg

8.

m M

m = 10–2 x F= – k = 10–2

k 2 x2 2

v

x x

Fig. 5.49

4.

–1

–1

–1

–1

a 9.

= (c)

= 1

5.

a g

(d)

a g

= 1

a g

m

(c)

m -

(c) 0.3

(d)

=

2 n

10.

(d) A

=

1 n

B

1 2

A (c) A (d) A

B B B

11.

Fig. 5.50

6. a

2 a

12. N

m v

2 2 a nNmv

(c)

3 2 a

(d)

4 2 a

–1

n N mv n

5.20 Comprehensive Physics—JEE Advanced

(c) 13.

nN m v

(d)

nN v m

r

(c) 18.

( (c) 10 N 14. A

-

r v r2 v

(d)

r v2 r2 v2

m v

(d) 20 N

mv cos (

B

(c) 2 mv cos

mv sin ( ) (d) 2 mv sin

2

19.

2

x–t

A B.

B A. 15. h

2h g 1 (c) sin

Fig. 5.52

2h g ·

16.

2h g M

1 (d) · cos

(c) 8 s

2h g

(d) 16 s

20. (c) 0.8 Ns

(d) 1.6 Ns

21.

v

M s

F M g sin

Mg cos

(c) 2 Mg cos

(d) 2 Mg sin

M

(c) M 22.

v2 2s

g

2v 2 s

2 g

M

(d) M

v2 2s

g

2v 2 s

2 g

m u

Fig. 5.51

17.

v

5.21

u2 2 g sin (c)

u2 2 g cos

u2 4 g sin

(d)

u2 4 g cos

23. –2 –2

g

Fig. 5.55

27.

n

Fig. 5.53

(c) 60 N

(d) 600 N

24. (c)

k

= 1/(1 – n2

k

=

1 / 1 n2

(d)

k

= 1 – 1/n2

k

=

1 1 / n2

28. –2

–2

–2

–2

x k

25.

(x) = kx

x x0

–2

g

–2

(c)

.

k

k

. At

cot k

(d) k cot

29. = 0.5 x

x x x

(c) x = 26.

m1

(d) x =

g l2 1

m2

g

(c) –2

.

2 cot

in l

m2 m1

g

2 cot

30.

Fig. 5.54

2

x=

l

2

g (l2 – 1) (d)

1

g l

2

1

5.22 Comprehensive Physics—JEE Advanced

31. (c)

R h

32.

sin sin

1

(d)

2

R 10 R (d) 30

–2

g

)

(c) 40 N

m

(d) 60 N

37.

M m

33.

M

m

a. M

M Ma M m Ma (d) m m M

ma M m ma (c) M

t2 =1 t1

36.

10 = 0.95) R ) 5 R (c) 20

t2 t1

F –2

g

M

.

M F

m is a M Fig. 5.57

(c) 24 N 38.

Fig. 5.56

F

ma M

F

F M

(c)

(d)

(d) 96 N m2

m1

-

ma M

m1

am M

m2

m1 m

34.

m1 + m

m1 g

g 3 (c) g

2g 3

35. 1

t1 Fig. 5.58

2

t 2.

two t2 is

t1 t2 t1

sin sin

1/ 2 1 2

t2 t1

sin 2 sin

1

2 2

39. -

5.23

tion

a g

(c) t 1 is

to it.

1/ 2

(d) t 1

a g

1/ 2

45. –2

g –2

–2

–2

–2

) F

40. x 1. x2

(c)

x1/x2 is 2 (c)

3

2F 3 5F (d) 6

F

-

3F 5

46.

F

m a

(d) 2 F1 = F2 = F3 = F

41. a a g 2

g g

(c)

g

(d)

2

42.

3

m F1 -

Fig. 5.59

F2

2 1 a F1

1/ 2 3 F2 = F1

F2 is

F2 = 2F1

(c) F2 = 3F1

(c)

2 1 a

2 a

(d) a l1 l2

47.

(d) F2 = 4F1

43.

l1 + l2 1 (l1 + l2) 2 (c) 3l2 – 2l1

44. h

t

(d) 3l1 – 2l2 48.

-

a

A

h t

a t g

Fig B

B A Fig. 5.60

5.24 Comprehensive Physics—JEE Advanced

10 2 49.

2 10

53.

M strings A B

C R tension in string B is

W

-

(c) 0.75

(d) 0.85

54. 50

–1

–1

.

(c) 300

(d) 480

55.

M

Fig. 5.61 –2

100 g (c) 100 2 g newton

100

(d)

2

(c) 60°

g newton

)

(d) 90°

56.

50.

F = 600 – 2 F

(c) 0.9 Ns

i

105 t t

j –5

(d) 1.8 Ns

(2 i

3j)

10–5

(c) (3 j

2i )

105

(2 i (d) (2 i

3j) 3j)

105 10–5

57.

51.

–1

L A A 58.

(c)

52.

1

–2

F = 6 i – 8 j + 10 k –2

(c) 6 59.

(d) 8

g

–2

)

5.25

(d) cot

(c) sec

60.

=3

(d) cosec

=3 =3

64.

(c) 45°

(d) 60°

descending

descending 61.

m1

m2

1

2

t m1

1

1

t0 m2

2)

(m1

1

m2

2

2

is

Fig. 5.64

65.

m1 + m2)gt0 (c) 2(m1 + m2)gt0

(d)

m

1 (m1 + m2)gt0 2

62.

M

2 Mg

F

2 mg g

–2

)

(See Fig. 5.62) (c) 12 N

(c)

(M

m) 2

m2 g

(d)

(M

m) 2

M2 g

(d) 15 N

Fig. 5.62

63.

Fig. 5.65

66.

Fig. 5.63

m L

M

5.26 Comprehensive Physics—JEE Advanced

mL M mL (c) M m 67.

ML m mL (d) M m u

m

72.

m f

m

u 2 sin 2 g (c)

(c) g +

3u 2 sin 2 2g

2u 2 sin 2 g

(d)

f m

g f m

(d) g –

73.

3u 2 sin 2 g

68.

f m

–2

g a

-

(c) 500 N

(d) 600 N

74. m2

m1

v

av2 (c) 2 av2

av2 (d) av2

cos sin

cos sin

v1 x1

69. t seconds to

t

m1 m2

(c) 0.5 (c) 0.5

0.75 (d) 0.75

x2 x1/x2 is

m2 m1

m1 m2

-

v2

m2 m1

(d)

1

70. v

Fig. 5.66

v/n

-

75.

n = 1 = 1 (c)

= 1

(d)

= 1

71.

M tion a Ma M

1 n2 1 n

2

1 n

cot

(c)

-

m mg m

ma Mg M m

Ma M (d)

ma Mg M m

76.

1/ 2

-

2

1 n2

1/ 2

cot n –1

n g

–2

)

mg m

(c) 3

(d) 4

5.27

77.

m F

(c)

= sin

F = cot = cos

(d) F

78. mg 1 2

(c)

(d) N

mg 1

mg

mg

(d)

2

1

is N = F

2

1

79. T Fig. 5.68

82.

80.

g 4 3g (c) 4

g 2

–1

R

–2

g

R is

(d) g P

m Q

83. A

P k

–1

g

–2

.

A P

Q P

mg

Q is kA 2

84. –1

(c) kA

–2

g

–1 –1

(c)

–1

(0.35)

–1

(d)

(0.25) (0.45)

85. R

gR 2 (c)

gR (d) 2 gR

2gR

86.

Fig. 5.67

81.

-

–2

g

m F –1

f = mg

(c) 10 3

–1

2 –1

–1

5.28 Comprehensive Physics—JEE Advanced

87.

A

B

m

90.

m

3 -

m m g g g 2

g (c)

g g 2

(d)

g 2

g 2

Fig. 5.69

88.

x–y

91.

p (t) = A[ i cos (kt) – j sin (kt A k

(c) 45°

.

-

y = kx2 (y

(d) 90°

89.

m on

x

a

m y

a a gk

a’ P (c) F

F a 2 2m a x2 (c)

F x 2m a

2a gk

92.

x

(d)

a 4 gk

m L

F x 2 2m a x2 (d)

a 2 gk

-

F a2 x2 2m x

Fig. 5.71

(c) 27 Fig. 5.70

(d) 36

5.29 ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67. 73. 79. 85. 91.

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68. 74. 80. 86. 92.

(c) (c)

(c) (d)

(c) (c) (c)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63. 69. 75. 81. 87.

(c) (c) (c) (c) (c) (c)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64. 70. 76. 82. 88.

(d) (c)

(d) (d) (d) (d) (c)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65. 71. 77. 83. 89.

(d) (d) (c)

(c)

(c) (c) (d)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66. 72. 78. 84. 90.

(d) (c) (c) (d) (c) (c) (d)

(c) (c) (c) (c) (c) (c) (d) (c) (d) (d) (d)

(d)

SOLUTIONS 1.

T

a

Q is aQ =

3g 2

f = M

Q

a .

mg M P = aP + aQ =

T – mg = ma T–

(1)

mg m = 2 2

3g 2

a

(2) m (g – a) sin

= =

F = mg cos

OA = y Now F = =

dr dy = 0 + 2y dt dt

dy dt

cos sin 2

cos 2

6.

A tion a is (A cos – a

. P

Q is f = mg Q.

P is aP =

F mg cos sin = N 2 mg mg cos 2

For

u cos P

N

=

dr = u y = r cos dt M

3.

sin

N = 2 mg + mg cos2

dy r dr = dt y dt Now

m(g – a) cos

5.

t 2r

m M

4. g = (g – a

T= 7 mg 6 2. AB = AC = r OB = OC = x AOB r2 = x2 + y2

g 1

f = m

mg = g m

N N sin

= (2 m) a = m (A cos 3ma A= m cos

– a)

5.30 Comprehensive Physics—JEE Advanced

3a = cos 45

2aA sA = v2A

= 3 2 a

7. f = mg

f)

sA + sB

= F

sA

mB mA

70 50

7 . Since 5

sB

11.

12.

k x–2 dx 2m v2 k = +C 2 2 mx

v dv = –

(1) n

C v x we get C = – k/2m v2 k 1 = 2 2m x

sA sB

mA aA = mB aB

= N

dv d v dx dv 8. F = ma = m =m = mv dt dx dt dx k F=– 2 x2 k dv k – = mv v dv = – x–2 dx 2 dx 2m 2x

a B s B = v 2B

Nm

n

Nmv

m AB = L

AC = h

n -

nNmv . v=

1

k = 10–2

2

m = 10–2 v

13.

1/ 2

k 1 m x

1 x=

–1

.

9. g sin .

14.

-

g sin

g cos

– g cos g sin

mg cos R mg sin .

g sin = – (g sin cos = 2 sin

or

10.

mB

mA aA

– g cos )

is a = g sin

A

B

aB vA v = mB B t t 2 2 mA vA = mBvB or mA vA = m2B v2B

mA aA = mB aB or mA or

vA

(i)

vB

Fig. 5.72

t sA

A

sB

B v2 – u2 = 2 as

5.31

or

v2 – 0 = 2aL

or

v=

2aL

2 g sin L

v=

2 gh

2 0 2 0 s

–1

At t h = L sin

–1

v = u + at or v = 0 + at. v t= a

2 gh g sin

15. 16.

= mv – (– mv) = 2 mv –1 =2 –1 = 0.8 Ns

2h g

1 sin

Mg Mg sin

Mg sin . To -

21. v

s R = Mg a v u s v2 – u2 = 2as v2 = 2 as or a = v2/2s

nent Mg sin Mg sin

= 2 Mg sin

17.

A) = v

F1

r2 -

=

Av = r2v r 2v

M v2 2s F2 = Mg F1 + F2 = M

= r2 v2

v2 2s

22.

mg sin F = mg sin

r 2 v 2.

mg

a = – g sin – g sin = – 2g sin

r 2 v 2. 18. 19.

t

t = 2s is

x = 0 to x t = 0 to 2s. Between t = 2s

t

t x

Fig. 5.73

t

s t 0 – u2 = 2

20. Between t is v t

t x–t

(– 2g sin ) s=

t

g

v2 – u2 = 2as s or

u2 4 g sin

-

5.32 Comprehensive Physics—JEE Advanced

23.

t –2

a

a=

–2

s

mg

u = 0. Now t s = ut +

a s F = ma = 60

1.0 = 60 N

t=

u

1 2 at 2 a=a

–2

10 s. t =

24.

s

Since

=

f mg

n n

nor

60

s

1 2 at 2 1 =0+ 3 2 m2 = 6

10 = 120 N 26. m2

mac f 120 60

–2

25.

–2 a is F = ma = 10 3 = 30 N mg R

F

T = 60 N

nent F1 = m1g sin m1g. Now F1 = m1g sin = 5 10 sin 30° = 25 N. Since F1 T T F m1 f m1 m1

ac

f mac = f or ac = m

u = 0) s = ut +

mg =

s

f= 0.2

10 s is (

F = 30 N

f f f R mg f = mg = 0.2 = 20 N

Fig. 5.74

=

or

10

or

10

T – m1g sin 30° – f = 0 f = T – m1g sin 30° = 60 – 5

10

sin 30°

= 60 – 25 = 35 N F f

27.

-

F =F–f= 30 – 20 = 10 N a =

F m

10 10

–2

a1 = g sin 45° = a2 = (g sin 45° – g = (1– k) 2

g 2 k g cos 45°)

5.33

t22

= n2

t12

a1 a2

1

or

1

k

=1–

k

1 n2

31.

A

. h. Now OA = OC = R mg sin

28.

F = mg x is

cos mg sin = mg cos OAB =

R2

mg cos mg sin = OB OB = x AB AB

2 x2 .

Fig. 5.75

f (x) = mg sin – mg cos = mg (sin – cos ) = mg (sin – kx cos ) x = x0 sin – kx0 cos = 0

f (x

x0 =

n k Fig. 5.77

29. g sin – g cos 0.5x cos x x

= g (sin

–

cos ) = g (sin

–

R x

or

30.

AB = 1 AC = l BC =

l

2

x

=

a a cos g sin

g

x2

x R2

(

x2

=

3R

= 0.95 R 10 R – x = R – 0.95 R = 0.05 R x =

= AB/BC = 1/ l 2 1 . A

1

3=

2

h =

32. m is F

a cos or

=

m

g

F on

l2 1

M.

F = ma a = 33. M

Fig. 5.76

ma. m.

M

= g sin a=g

1 ) 3

M = F – f = F – ma F ma M = M

F M

on m2 = M is

ma M

m is f = ma m f = ma.

5.34 Comprehensive Physics—JEE Advanced

34.

a

m

a=

mg – T = ma

ss m

or T = mg – ma T=

2 mg 3

m Using (ii) in (i) we get

2 mg = mg – ma 3 a=g

F=

(M + m)g +

1

+

=(

h s1 = s2 =

h sin

h sin

2)

t22 t12

a2 = g sin

1

10

38 .

1

-

2

a1 s2 a2 s1

=

g sin = g sin sin 2

=

sin

m tension T

1 2

h sin

2

sin h

m 2g (m1 + m)g (m1 + m)g = m2g or m =

m

m1

m 6 – m1 = –4 0.4

k

F=

s

mg = 0.7

5

10 = 35 N.

1

2

f

2

= F – mk mg = 35 – 0.5

mg l

a=

(l – x) or

3 1 = 40 N

37.

40.

m (M + m)g M

2

=

F M

f 10 m 5

–2

v = 30° is a1 = g sin 1 1 v2 – u2 2 = 60° is a2 = g sin 2 u2 = 2a1x1 = 2a2x2 x1 a = 2 x2 a1

= 2ax

1

2

10 = 35 – 25

u

m

M

5

= 10 N

x

6 10 3

m2

s

1

l

T=

m1

(m1 + m

39.

l–x

1

(5 + 3)

= 96 N

v12 = 2a1 s1 v22 = 2a2 s2 v 1 = a 1t 1 v2 = a2t2 or a12 t11 2 2 a2 t2 = 2a2 s2

mg

(M + m)g

m

a1 = g sin

M=

2

(ii)

(M + m) g

2

= 2a1 s1

1g

=

1

= (0.5 + 0.7)

35.

36.

1mg

=

g sin g sin

2 1

sin 60 sin 30

3

41. a

a M + m)a F = (M + m)a +

2

(M + m)g m is

1mg

(i) -

g sin g sin Now

a cos g a cos

=

or a = g a = g

g 3

.

5.35

F

a=

M

T= M

m F 2 M m

=

F1 F1 = mg sin F2 F2 = mg sin

mg cos

–

+

m=

(i)

mg cos

(ii)

n

t

30

1/ 2 3

M 2

46.

F2 + F1 = 2 mg sin F2 – F1 = 2 mg cos F2 F1 = F2 F1 or F2 = 3 F1 43. m l L L–l

m a 2

M

Fig. 5.78

42.

m

T=

a= F2

5F 6

F m

F1

F

F = (F12 + F22)1/2 =

2F(

F1 = F2 = F)

F

F 3.

=2 =

F – F3 2 F–F

= ( 2 – 1) F 2 – 1) l = ml

F = ( 2 – 1)a. m

L– l) mlg

= m (L – l m(L – l) mlg = m (L – l) g L 1

l=

or

44.

h=

g+a

0.5 120 0.5 1 Fig. 5.79

1 2 gt 2 a

t

g = h=

1 gt2 2

45.

t = t 1 M

a g

F x is F = kx

1 1 2 gt2= gt or (g + a)t 2 = gt2 2 2 or

47.

1/ 2

m

k

l0 F = k(l – l0)

l F 2 = k(l1 – l0) k(l2 – l0)

(i) (ii)

5.36 Comprehensive Physics—JEE Advanced

3 l2 l0 2 l1 l0 Which gives l0 = 3l1 – 2l2. Using this value of l0 in 1 . either (i) or (ii) we get k = l2 l1 When a stretching force of 5 N is applied, let l3 be the length of the spring. Then 5 = k(l3 – l0) Substituting the values of l0 and k, and solving we get l3 = 3l2 – 2l1 Hence the correct choice is (c). 48. If the springs A and B are massless or their mass is negligible compared to the mass with which they are loaded, the tension is the same everywhere on the spring. Hence each spring balance will read 4 kg. Thus the correct choice is (c). 49. Let T be the tension in string C and T in string B. The y-component T cos balance with the weight Mg and the x-component T sin balances with tension T . Thus (see Fig. 5.80) T = T sin and Mg = T cos Dividing the two we get T = Mg tan = 100 g tan 45° = 100 g newton Hence the correct choice is (a).

= (600 t

105 t 2

t 3 10

3

s

0 5

= 600 3 10–3 – 10 (3 10–3)2 = 1.8 – 0.9 = 0.9 Ns Hence the correct choice is (c). 51. The radius of the circular motion of the bead is r = L. The linear acceleration of the bead is a = r = L. If m is the mass of the bead, then Force acting on the bead = ma = m L Reaction force acting on the bead is R = m L The bead starts slipping when frictional force between the bead and the rod becomes equal to centrifugal force acting on the bead, i.e. m v2 r L = mr 2 = mL 2 = 2 = ( t)2 R=

m

or or

( (

v=r ) = t)

= ( t)2 or t =

or

52. The magnitude of the force is F = F F = [(6 i – 8 j + 10 k ) (6 i – 8 j + 10 k )]1/2 = {(6)2 + (8)2 + (10)2]1/2 = (200)1/2 = 10 2 N F 10 2 N = = 10 2 kg. Hence the a 1 ms 2 correct choice is (b). 53. Refer to Fig. 5.81. Since the block moves with a constant velocity, no net force acts on it. Therefore, the horizontal component F cos of force F must balance with the frictional force, i.e. fr = F cos . Mass =

Fig. 5.80

50. Now F = 600 – 2 105 t. F will be zero at time t given by 600 – 2 105 t = 0 or t = 3 10–3 s

Fig. 5.81

t

F dt

Therefore, impulse =

Also

0 t

(600 – 2

= 0

105 t)dt or

fr = (mg – F sin ) = (f – F sin ) (f – F sin ) = F cos (200 – 100 sin 30°) = 100 cos 30°

Laws of Motion and Friction 5.37

or

200 100

1 2

= 100

0.866

= 86.6 86.6 = = 0.58, 150

or

which is choice (b). 54. Change of momentum of one bullet = m (v – u) = 0.03 {50 – (–30)} = 2.4 kg ms–1 Average force = rate of change of momentum of 200 bullets = 200 2.4 = 480 N, which is choice (d). 55. Let the body leave the surface at point B as shown in Fig. 5.82. When the body is between points A and B, we have M v2 Mg cos – N = r

Fig. 5.82

When the body leaves the surface at point B, the normal reaction N becomes zero. Thus M v2 Mg cos = r or

cos

=

v2 (5)2 = rg 5 10

1 or = 60° 2 Hence the correct choice is (c). 56. Mass of each piece (m) = 1 kg. Initial momentum = 0. Final momentum = p1 + p2 + p3. From the principle of conservation of momentum, we have p1 + p2 + p3 = 0 or p3 = – (p1 + p2) = – (mv1 + mv2) = – m (v1 + v2) =

(2 i

= –1 kg = (2 i Force

(2 i + 3 j) kgms 10

5

1

s

= (2i + 3 j) 105 newton Hence the correct choice is (b). 57. Magnitude of recoil momentum of the gun = forward momentum of the bullet = mbvb = (40 10–3 kg) 1200 ms–1 = 48 kg ms–1 n bullets per gun is F

=n =n

change in momentum time

48 kg ms 1 1s = 48 n newton Given F = 144 N. Thus 144 = 48 n which gives n = 3. Hence the correct choice is (d). 58. Force F = ( s – k) mg = (0.8 – 0.6) 4 10 = 8 N =

n

F 8N Acceleration = = 2ms–2, which is m 4 kg choice (a). 59. No net force acts on the block as it moves at a constant velocity. Therefore, downward force = upward force or mg sin = mg cos or = tan , which is choice (c). 60. When a cylinder rolls up or down an inclined plane, its angular acceleration is always directed down the plane. Hence the frictional force acts up the inclined plane when the cylinder rolls up or down the plane. Thus, the correct choice is (b). 61. Linear momentum of the system at time t = 0 is p1 (m1 v1 m2 v2 ) and at time t = 2t0 it is p2 (m1 v1 m2 v2 ) . Change in linear momentum in time 2t0 = p2 p1 . The rate of change of linear momentum is p2 p1 /2t0. From Newton’s second law of motion, the rate of change of momentum equals the force acting on the two particles, which is (m1 g + m2 g ). Hence ( p2 p1 ) = (m1 + m2) g 2t 0

3 j) ms–1

3 j) kg ms–1

p F= 3 t

=

(m1 v1 + m2 v2 ) – (m1 v1 + m2 v2 ) = (m1 + m2) g (2t0) Hence the correct choice is (c). or

5.38 Comprehensive Physics—JEE Advanced

62. The horizontal component of F parallel to the surface is F sin F is given by F sin = mg 3 10 or F sin 60° = 0.5 3 = 0.5 3 10 2 which gives F = 10 N. Hence the correct choice is (b). 63. As shown in Fig. 5.83, the insect will crawl without slipping if the value of is not greater than that given by the condition: force of friction f = mg sin . Now f = N, where N is the normal reaction. Thus N = mg sin or mg cos = mg sin or 1 = 3, which is choice (a). cot = F

or

64. Let T be the tension in the string. When the system is in equilibrium, then for the two equal masses m, we have T = mg (1) and for the mass 2 m, we have =

2 mg

(2) 1

or = 45°, Dividing (2) by (1), we get cos = 2 which is choice (c). 65. The force F on the pulley by the clamp is given by the resultant of two forces: tension T = Mg acting horizontally and a force (m + M)g acting vertically downwards. Thus F=

( Mg )2

{(m

mL M m Hence the correct choice is (c). 67. At the highest point of trajectory, the projectile has only a horizontal velocity which is u cos . After L =

or

horizontal velocity. If u is the horizontal velocity of the other fragment, the law of conservation of momentum gives (2m) u cos = m 0 + mu which gives u = 2u cos Now, the time taken to reach the highest point (as well as the time taken to fall down from this point) u sin is . Therefore, the horizontal distance travg elled by the other fragment is u sin u sin u cos + 2 u cos g g

Fig. 5.83

2T cos

mv – (M + m)V = 0 V m = or v M m Since the distance moved is proportional to speed, the displacement L of the plank is given by L V m = M m L v

M ) g}2

= [M2 + (m + M)2]1/2 g which is choice (d). 66. Before the boy starts walking on the plank, both the boy and the plank are at rest. Therefore, the total momentum of the boy–plank system is zero. If the boy walks with a speed v on the plank and as a result if the speed of the plank in the opposite direction is V, then the total momentum of the system is mv – (M + m)V. From the principle of conservation of momentum, we have

3u 2 sin 2 u 2 sin 2 u 2 sin 2 2g 2g g Hence the correct choice is (b). 68. The mass of water stream striking against the wall in 1 second = av . Hence, the change in its momentum per second is (av )v – (– av )v = 2a v2. The normal component of the rate of change of momentum and, therefore, force is 2a v2 cos . Hence the correct choice is (a). 69. The acceleration of the block sliding down the smooth inclined plane is a1 = g sin and down the rough inclined plane is a2 = g sin – g cos . Given t1 = t and t2 = 2t. If the length of the inclined plane is s, we have 1 1 s = a1t12 = a2t 22 2 2 =

a1t 21 = a2t 22

or or

t2 = (g sin

g sin

or

sin

= 4 (sin

3 3 tan = 4 4 Hence the correct choice is (d).

which gives

=

– g cos ) –

(2t)2

cos ) (

= 45°)

Laws of Motion and Friction 5.39

70. We use the relation v2 – u2 = 2as. Since u = 0, we have v2 = 2as. s ( v1= v) Now v 21 = 2a1s or v2 = 2g sin 2 v and v22 = 2 a2s or 2 = 2(g sin – g cos ) s n Dividing, we get or n2 (sin – cos ) = sin 1 which gives = 1 tan , which is choice (a). n2 71. The force of friction between the block and the belt is f = mg, where m is the mass of the object. This force produces an acceleration of the block which is given by force mg = g a= mass m The block will slide on the belt without slipping until its speed (v) becomes equal to the speed of the belt. Since u = 0, we have v2 = 2 as 52 v2 v2 or s = = 2.5 m 2a 2 g 2 0.5 10 Hence the correct choice is (b). 72. Normal reaction R = f. Therefore, force of friction = R = f. The net downward force F = mg F mg f = – f. Hence, the acceleration a = m m f g – . Hence the correct choice is (d). m 73. As the boy is climbing the pole at a constant speed (no acceleration), the force of friction must be just mg balanced by his weight, i.e. R = mg or R = = 40 10 = 500 N. Hence the correct choice is (c). 0.8 74. From the principle of conservation of momentum, we have m2 v m1v1 = m2v2 or 1 (i) v2 m1

75. The forces acting on the balloon are its weight acting downwards and upthrust F acting upwards. Thus F – Mg = Ma (i) When mass m is removed, we have F – (M – m) g = (M – m)a

(ii)

where a is the new acceleration. Eliminating F from (i) and (ii) and simplifying we get Ma mg a = M m which is choice (a). 76. Let m be the mass per unit length of the rope. Let x t. The weight of this part is F1 = mgx Now, if a small part dx dt, F2 = rate of change of momentum mdx v = mv2 = dt dx Now = v, where v is the velocity of that part dt of the rope at that instant. But v2 = 2gx. Hence F2 = mv2 = m (2gx) = 2mgx. Total force F = F1 + F2 = mgx + 2mgx = 3mgx = 3F1 Hence the correct choice is (c). 77. Refer to Fig. 5.84. Vertical component of F is F sin and the horizontal component is F cos .

Fig. 5.84

opposite force F on each block. Let a1 and a2 be the accelerations of blocks m1 and m2 respectively. Then a m1 F = m1a1 = m2a2 or 2 (ii) a1 m2 Also v21 = 2a1x1 and v22 = 2a2x2, which give x1 = x2

v12 a2 . v22 a1

m2 m1

2

[Use Eqs. (i) and (ii)] Hence the correct choice is (b).

m1 m2

=

m2 m1

Thus R + F sin

= mg R = mg – F sin

or Frictional force R =

(mg – F sin ). Also

(mg – F sin ) = F cos or F mum, i.e. if

F=

mg sin cos

(i)

5.40 Comprehensive Physics—JEE Advanced

d ( sin + cos ) = 0 d cos – sin = 0

or or

= tan , which is choice (a).

78. Now tan

= . Therefore, cos

1

= 1

sin

=

2

and

2

1

Using these in Eq. (i) above and simplifying, we get mg F= 2 1 Hence the correct choice is (d). 79. Force required to accelerate the body of mass m is F = ( s – k) mg = (0.75 – 0.5) mg = 0.25 mg F 0.25 mg Acceleration = = = 0.25 g, which is m m choice (a). 80. Let a be the acceleration at a time t of the blocks

v2 10 10 = = 12.5 m g 0.8 10 This is the minimum radius the curve must have for the car to negotiate it without sliding at a speed of 10 ms–1. Hence the correct choices is (b). 83. Speed of train (v) = 36 km h–1 = 10 ms–1 Radius of the curve (R) = 200 m Distance between rails (x) = 1.5 m Let the outer rails be raised by a height h with respect to the inner rails so that the angle of banking is (Fig. 5.85). or

Rmin =

Then

tan

=

h=

or

h v2 = Rg x xv 2 1.5 (10)2 = Rg 200 10

= 0.075 m = 7.5 cm Thus, the correct choice is (a).

acceleration is F = (m + m) a = 2 ma F =2ma

displacement, i.e. F = kA

(1)

Fig. 5.85

(2)

Equating (1) and (2), we get a

=

kA 2m force of friction = ma

kA kA 2m 2 Hence the correct choice is (b). 81. Since the block is held stationary, it is in translational as well as rotational equilibrium. Hence no net force and no net torque acts on the block. No net force will act on the block if f = mg and N = F. No net torque will act on the block, if torque by frictional force f about centre O = counter torque by normal reaction N about centre O. Hence choice (d) is false. 82. Speed of car (v) = 36 km h–1 = 10 ms–1 mum centripetal force that friction can provide is =m

f

=

mg =

mv 2 R

84. Now v = 54 km h–1 = 15 ms–1, R = 50 m. The required angle of banking is given by v2 15 15 = = 0.45 Rg 50 10 Thus, the correct choice is (d). 85. The horizontal velocity v must be such that the centripetal force equals the weigth of the body, i.e. mv 2 = mg or v = gR , which is choice (b). R 86. The motor cyclist can leave the ground only at the highest point on the bridge. At this point, the centripetal force is mv2/R. He will not leave the ground if the centripetal force equals the weight mg. Thus mv 2 = mg or v = gR = 10 10 = 10 ms–1. R Hence, the correct choice is (a). 87. When the system is in equilibrium, the spring force = 3 mg. When the string is cut, the net force on block A = 3 mg – 2 mg = mg. Hence the acceleration of this block at this instant is tan

=

Laws of Motion and Friction 5.41

force on block A mg g = = mass of block A 2m 2 When the string is cut, the block B falls freely with an acceleration equal to g. Hence the correct choice is (c). dp F = 88. Force dt d = [A{ i cos (kt) – j sin (kt)] dt a=

Fig. 5.87

The block will topple if the torque due to normal reaction N about O mg sin about O, i.e. N OA = mg sin OB

= Ak [– i sin (kt) – j cos (kt)] Now F p = Ak[– i sin (kt) – j cos kt] A[ i cos (kt) – j sin (kt)]

mg cos

= A2k[– sin (kt) cos (kt) + cos (kt) sin kt] =0(

F = T sin

+ T sin

T cos

15 cm 2

=

for sliding, the

91.

= 2 T sin

F 2sin = mf

T= Also

2 34°. 3 Since for toppling is less than correct choice is (b). tan

i i = j j = 1 and i j = 0)

Hence the angle between F and p is 90° which is choice (d). 89. Refer to Fig. 5.86. Let f be the force producing the acceleration of each mass. It follows from the

5 cm = mg sin

(1) (2)

Using (1) and (2), we get f=

F cos 2m sin

=

F 2m tan

=

Fx

2m a 2 Hence the correct choice is (b).

x2 Fig. 5.88

For the bead to stay at rest, (see Fig. 5.88) N cos = mg N sin

= ma a which give tan = . Now g dy d tan = slope of the curve = = (kx2) = 2 kx dx dx 2kx = Fig. 5.86

90. The block will just begin to slide if the downward force mg sin just overcomes the frictional force, i.e. if mg sin = N = mg cos tan = = 3 = 60° (see Fig. 5.87)

a g

x=

a 2 gk

92. Radius of the circular path is BC = r = L sin , where L = AB is the length of the string. The vertical component T cos of tension T balances with the weight mg and the horizontal component T sin provides the necessary centripetal force for circular motion. Hence

5.42 Comprehensive Physics—JEE Advanced

T sin = mr

2

T = mL

2

T

= mL

324 = 0.5

= m(L sin )

2

2

0.5

2

= 36 rad s–1

Fig. 5.89

II Multiple Choice Questions with One or More Choices Correct 1. Which of the following statements are true? No net force acts on (a) a drop of rain falling vertically with a constant speed

(c) can go up to a speed of 100 ms–1 in 10 s. (d) after acquiring a speed of 50 ms–1, can come to rest, with the engine shut off and brakes not applied, in a time of 10 s. 4.

(c) a car moving with a constant velocity on a rough road (d) a body moving in a circular path at constant speed. 2. In which of the following situations would a force of 9.8 N act on a stone of mass 1 kg? Neglect air (a) Just after it is dropped from the window of stationary train (b) Just after it is dropped from the window of a train running at a constant speed of 36 km h–1. (c) Just after it is dropped from the window of a train accelerating at 1 ms–2. which is accelerating at 1 ms–2. 3. a car and the ground is 0.5. The car starts from rest and moves along a perfectly horizontal road. If g = 10 ms–2, the car 5 ms–2 without slipping. (b) can attain a speed of 20 ms–1 in a minimum distance of 40 m.

speed of 15 ms–1 gushes out of a tube of cross sectional area 1 cm2, strikes against a kachha ver-

N on it. The impact of the water stream on the wall will ( to the water stream is only about 1500 N. (b) will break it. 105 pascal on the wall. 5. A block is released from the top of a smooth inclined plane. Another block of the same mass is allowed to fall freely from the top of the inclined plane. Both blocks reach the bottom of the plane (a) in equal time (b) with equal speed (c) with equal momentum (d) with equal kinetic energy 6. A body of mass 200 g is moving with a velocity of 5 ms–1 along the posotive x- direction. At time t = 0 when the body is at x = 0, a constant forse of 0.4 N

Laws of Motion and Friction 5.43

7.

8.

9.

10.

directed along the negative x-direction is applied to the body for 10s. (a) At time t = 2.5 s, the body will be at x =1.25 m (b) At time t = 2.5 s, the speed of the body will be zero. (c) At time t = 30 s, the body will return to x = 0. (d) At time t = 30 s, the speed of the body will be 15 ms–1 A train starts from rest with a constant acceleration a = 2 ms–2. After 5 seconds, a stone is dropped from the window of the train. If g = 10 ms–2. (a) the magnitude of the valocity of the stone 0.2 second after it is dropped is 2 ms–1. (b) the angle between the resultant velocity vector of the stone and the horizontal 0.2 second after it is dropped is = tan–1 (0.2). (c) the acceleration of the stone after it is dropped is a = 2 ms–2. (d) the acceleration of the stone after it is dropped is g = 10 ms–2. A man of mass m His weight when the lift is (a) stationary is mg (b) moving up with a uniform speed of 2 ms–1 is 5 mg. (c) moving up with a uniform acceleraiton a (< g) is m (g + a). (d) moving down with a uniform acceleration a (< g) is m (g – a). Two identical blocks, each of mass m, connected by a light string, are placed on a rough horizontal surface. When a force F is applied on a block in the horizontal direction, each block moves with an acceleration a. Assuming that the frictional forces on the two blocks are equal, (a) the tension in the string will be F. (b) the tension in the string will be F/2. (c) the frictional force on each block will be (F – ma). (d) the frictional force on each block will be F ma . 2 A body of weight W is suspended from a rigid support P by means of a massless string as shown in Fig. 5.90. A horizontal force F is applied at point O of the rope. The system is in equilibrium when the string makes an angle with the vertical. If the tension in the string is T, (a) F = T sin

(b) W = T sin

W2

(c) T =

F2

(d) F = W tan

Fig. 5.90

11. Two blocks of masses m1 and m2 are connected by a string of negligible mass which passes over a plane as shown in Fig. 5.91. m1 and the plane is . (a) If m1 = m2, the mass m1 up the inclined plane when the angle of inclination is , then = tan . (b) If m1 = m2, the mass m1 up the inclined plane when the angle of inclination is , then = sec – tan . (c) If m1 = 2 m2, the mass m1 down the plane if = 2 tan . (d) If m1 = 2 m2, the mass m1 1 down the plane if = tan – sec . 2

m1 m2 q

Fig. 5.91

12. Two blocks A and B are connected to each other by a string and a spring of force constant k, as shown in Fig. 5.92. The string passes over a frictionless pulley as shown. The block B slides over the horizontal top surface of a stationary block C and the block A slides along the vertical side of C of friction between the surfaces of the blocks is . If the mass of block A is m,

5.44 Comprehensive Physics—JEE Advanced

(a) (b) (c) (d)

the the the the

mass of block B is m. mass of block B is m/ . energy stored in the spring is m2g2/2k. energy stored in the spring is m2g2/k. B

T T k

C A

Fig. 5.92

13. Two blocks A and B of equal masses m each are connected to each other by a string passing over a frictionless pulley as shown in Fig. 5.93. The A and the surface below is 0.5. When the system is released, (a) the acceleration of the blocks is 3g/4. (b) the acceleration of the blocks is g/4. (c) the tension in the string is 3mg/4. (d) the tension in the string is mg/4.

(b) the tension in the rope at t = 6 s will be 10,000 N. (c) the tension in the rope at t = 11 s will be 8,000 N. (d) the height upto which the lift takes the passengers is 40 m. 15. Two blocks of masses m1 = 2 m and m2 = m are connected by a light string passing over a friction less pulley as shown in Fig. 5.95. When they are released, g (a) the acceleration of m1 is 3 vertically downwards. (b) the acceleration of m2 is 2g vertically upwards. 3 (c) the tension in the string is 4g Fig. 5.95 . 3 (d) the distance through which m2 rises in t secgt 2 onds after the blocks are released is . 6 16. Two blocks of masses m1 and m2 connected by a frictionless surface as shown in Fig. 5.96. A force F is applied to m2 in the horizontal directon as shown. F . (a) the acceleration of each block is (m1 m2 )

Fig. 5.93

14. A lift is going up. The variation of the speed of the lift with time is shown in Fig. 5.94. The total mass of the lift and passengers is 1000 kg. If g = 10 ms–2,

m1 F . (m1 m2 ) (c) the tension in the string is F.

(b) the tension in the string is

(d) the force on mass m1 is

m2 F . (m1 m2 )

Fig. 5.96

Fig. 5.94

(a) the tension in the rope of the lift at t = 1 s will be 12,000 N.

17. Ten coins, each of mass m, are placed on top of each other on a horizontal table. (a) The force on the 7th coin (counted from the bottom) due to all the coins above it is 3mg vertically downwards. (b) The force on the 7th coin by the 8th coin (both counted from the bottom) is 3mg vertically downwards.

Laws of Motion and Friction 5.45

(c) The force on the 7th coin by the 8th coin (both counted from the bottom) is 2mg vertically upwards. (d) The reaction force of the 6th coin from the bottom on the 7th coin from the bottom is 4mg vertically upwards. 18. Two blocks of masses m1 and m2 are placed in contact on a horizontal frictionless surface as shown in Fig. 5.97. A force F is applied to mass m1 as shown. (a) The acceleration of mass m2 is F/m2 m2 F m2 is . (m1 m2 ) m2 is F. F (d) The acceleration of mass m1 is . (m1 m2 )

Fig. 5.97

19. Two blocks of masses m1 and m2 (m2 < m1) are placed on an inclined plane of inclination and joined by a string as shown in Fig. 5.98.

l = 20% L l = 25% (d) If = 0.25, L 21. A block starts sliding from the top of an inclined plane of inclination between the block and the plane varies as = kx where x is the distance moved down the plane and k is a positive constant. (a) The block has a uniform acceleration along the plane. (b) The acceleration of the block increases for tan . x< k (c) If

= 0.25,

tan . k tan (d) The block starts decelerating for x > . k 22. A block of mass m is lying at x = 0 on a smooth horizontal surface. A variable force F = kx is applied to it as shown in Fig. 5.99 where k is a constant. Then (a) the block will move on the surface with a uniform acceleration. (b) the block will move on the surface with a variable acceleration. (c) the block will lose contact with the surface mg . after travelling a distance x0 = k sin (d) the block will always remain in contact with the surface. x=

Fig. 5.98

and the plane is and the blocks are released, (a) the acceleration of the blocks is g(sin – cos ). (b) the acceleration of the blocks is zero. (c) the tension in the string is zero. (d) the tension in the string is (m1 + m2)g (sin – cos ). 20. A uniform chain of length L is placed on a rough tween the chain and the table is length of the chain that can hang from the edge of the table is l. Then L (a) l = (1 ) L (b) l = (1 )

Fig. 5.99

23. A pendulum of length L and bob mass M is oscillating in a plane about a vertical line between angular limits – and + . For an angular displacement | | < , the tension in the string is T and the velocity of the bob is V. Then (a) T cos = Mg MV 2 (b) T – Mg cos = L (c) The magnitude of the tangential acceleration of the bob is |at| = g sin (d) T = Mg cos IIT, 1986

5.46 Comprehensive Physics—JEE Advanced

24. When a bicycle is in motion, the force of friction that it acts (a) in the backward direction on the front wheel and in the forward direction on the rear wheel. (b) in the forward direction on the front wheel and in the backward direction on the rear wheel.

(c) in the backward direction on both the front and the rear wheels. (d) in the forward direction on both the front and the rear wheels. IIT, 2007

ANSWERS AND SOLUTIONS 1. Choice (a) is true. Since the direction of motion as well as the speed of the drop of rain are constant, its velocity is constant and hence no net force acts on the drop. The downward gravitational force (i.e. weight) of the drop is completely cancelled by the upward buoyant force on it due to air and the viscous force due to its motion. Choice (b) is also true. The weight of the cork is balanced by the upward buoyant force due to water. Choice (c) is also true. The force due to the car’s acceleration is balanced by the backward force of friction between its tyres and the rough road. Choice (d) is false. Although the speed of the body is constant, its velocity is changing all the time because the direction of motion keeps changing. Hence, the motion of the body is being accelerated. A force (called centripetal force) must act on the body to produce the acceleration. Thus, choices (a), (b) and (c) are correct. 2. In choice (a), a force F = mg = 1 9.8 = 9.8 N acts on the stone vertically downwards. In choice (b), the velocity of the train is constant. Hence, there is no acceleration (and therefore, force) along the direction of motion. When the stone is dropped, the only force acting on it is F = mg = 9.8 N vertically downwards. In choice (c), before the stone is dropped, a force = ma = 1 kg 1 ms–2 = 1 N acts on it in the horizontal direction. But, after it is dropped, this force ceases to act because the stone is no longer located in the accelerating system (i.e. train). Hence, in this case also, the net force on the stone is 9.8 N vertically downwards. In choice (d), the weight of the stone is balanced by the normal accelerated in the forward direction along with the train. Hence, the force acting on the stone = ma = 1 kg 1 ms–2 = 1 N in the direction of motion of the train. Thus, the correct choices are (a), (b) and (c). 3. ping is given by ma = mg or a = g = 0.5 10 = 5 ms–2 tance s covered to acquire a speed v = 20 ms–1 is

given by v2 – u2 = 2as or (20)2 – 0 = 2 5 s or s = 40 m. Further, since the frictional retardation is 5 ms–2, the time needed to come to rest from a speed of 50 ms–1 is t = v/a = 50/5 = 10 s. Hence, choice (a), (b) and (d) are correct. 4. The rate of change of momentum of water stream = av2 – 0 = av2 . This is the force F water stream on the wall. F = av2 = (1 10–2) (15)2 1000 = 2250 N The pressure on the wall is F 2250 = 2.25 105 pascal a 1 10 2 Hence, the correct choices are (b), (c) and (d). 5. Since the acceleration along the inclined plane (g sin ) is less than g, the blocks take different times to reach the bottom. The speed of each block on reaching the bottom is v = 2g h , where h is the height of the inclined plane. Thus choice (b) is correct. The directions of the velocity are different, hence their momenta are not the same. Their 1 kinetic energy mv2 is the same. Hence the cor2 rect choices are (b) and (d). 6. Given u = + 5 ms–1 along positive x-direction F = – 0.4 N along negative x-direction P=

m = 200 g = 0.2 kg F 0.4 = = –2 ms–2. The The acceleration a = m 0.2 negative sign shows that the motion is retarded. The position of the body at time t is given by 1 x = x0 + ut + at2 2 At t = 0, the body is at x = 0. Therefore, x0 = 0. Hence 1 x = ut + at2 2 Since the force acts during the time interval from t = 0 to t = 10 s, the motion is accelerated only between t = 0 and t = 10 s. The position of the body t = 2.5 s is given by

Laws of Motion and Friction 5.47

1 (– 2) (2.5)2 = 1.25 m 2 The velocity of the body at t = 2.5 s is v = u + at = 5 + (– 2) (2.5) = 5 – 5 = 0 t = 0 to t = 10 s) the motion is accelerated. During this time a = – 2 ms–2. Putting u = 5 ms–1, a = – 2 ms–2 1 and t = 10 s in equation x = ut + at2. We have 2 1 x1 = 5 10 + (– 2) (10)2 = – 50 m (i) 2 The velocity of the body at t = 10 s is v = u + at = 5 + (– 2) 10 = – 15 ms–1 During the remaining 20 seconds, i.e. from t = 10 s to t = 30 s, the acceleration a = 0, because the force ceases to act after t = 10 s. The velocity of the body remains constant at – 15 ms–1 during the last 20 seconds. The distance covered by the body during the last 20 seconds is x2 = – 15 20 = – 300 m Position of the body at t = 30 s is x = x1 + x2 = – 50 – 300 = – 350 m The magnitude of the velocity (i.e. speed) of the body at t = 30 s is 15 ms–1. Hence the correct choices are (a), (b) and (c). 7. Given u = 0, a = 2 ms–2. Since the stone is located in the train, the acceleration of the stone is a = 2 ms–2. At time t = 5 s, the velocity of the stone is v = u + at = 0 + 2 5 = 10 ms–1. Before the stone is dropped, its motion is accelerated with the train. But, the moment it is dropped, its acceleration due to the motion of the train ceases. Therefore, after the stone is dropped, it has the following two motions: (a) a uniform motion with velocity 10 ms–1 parallel to the ground, i.e. vx = 10 ms–1 (the horizontal velocity) (b) an accelerated motion vertically downwards due to gravity. In time t = 0.2 s, the vertical velocity of the stone is vy = 0 + gt = 10 0.2 = 2 ms–2. The resultant velocity of stone at t = 0.2 s is x=5

v=

v 2x

2.5 +

v 2y =

(10) 2

( 2) 2 =

104

= 2 26 ms–1 The angle, which the resultant velocity vector, makes with the horizontal is given by vy 2 = = 0.2 tan = vx 10

After the stone is dropped, the horizontal velocity vx remains unchanged because the acceleration is zero along the horizontal direction. The only acceleration of the stone is the acceleration due to gravity. Hence the correct choices are (b) and (d). 8. mg) on the on him an upward reaction for (R) which it measures. When the lift is stationary, R = mg. When the lift is moving up or down with a uniform speed, it has no acceleration of its own. Hence, the reading of the machine will still be mg. When the machine is moving downwards with an acceleration a, a force F = ma acts downwards. But the reaction R = mg acts upwards. Hence, the effecting reading will be Reff = F – R = mg – ma = m(g – a) In case (c), R and mg both act in the same direction (upwards). Hence, the effective reading will be Reff = mg + ma = m(g + a) Hence the correct choices are (a), (c) and (d). 9. Refer to Sec. 5.11 and Fig. 5.15 on page 5.9 Equations of motion of m1 and m2 are m 1a = T – f (i) and m 2a = F – T – f (ii) Subtracting the two equations, we have (m1 – m2)a = 2T – F Since m1 = m2, we get, we get 0 = 2T – F F 2 Putting T = 10 N in Eq. (i) above, we have f = T – m 1a F – ma = 2 Hence the correct choices are (b) and (d). 10. The mass is in equilibrium at point O under the action of the concurrent forces F, T and W = mg. Therefore, as shown in Fig. 5.100, the horizontal component T sin of tension T must balance with force F and the vertical component T cos must balance with weight W = mg. or

T=

Thus

F = T sin

(i)

and

W = T cos

(ii)

Squaring Eqs. (i) and (ii) and adding we get T 2 = W 2 + F 2. Dividing (i) by (ii) we get F = W tan . Hence the correct choices are (a), (c) and (d).

5.48 Comprehensive Physics—JEE Advanced

and T – mg = ma (ii) Adding (i) and (ii) we get 1 1 0 .5 g g= g= a= 2 2 4 Using a = g/4 in Eq. (i) gives T = 3mg/4 Hence the correct choices are (b) and (c). 14. Between t = 0 and t = 2 s, the acceleration of the lift is a=

Fig. 5.100

11. The block m1 will just begin to move up the plane if the downward force m2g due to mass m2 trying to pull the mass m1 up the plane just equals the force (m1g sin + m1g cos ) trying to push the mass m1 down the plane, i.e. when m2g = m1g(sin + cos ) Now, it is given that m1 = m2 = m. Therefore, we have 1 = sin + cos = sec – tan The block m1 will just begin to move down the plane if the downward force (m1g sin – m1g cos ) on m1 just equals the upwards force m2g acting on m1 due to m2, i.e. if m2g = m1g(sin

–

cos )

m1 1 = m2 sin cos m1 = 2m2, then we have

or If

2 =

PE =

–

1 1 mg kx2 = k 2 2 k

1

= 2 ms–2

BE 4 ms 1 = 2 ms–2 EC 2s Tension = m (g – a) = 1000 (10 – 2) = 8,000 N The height to which the lift rises = area of OABC = 40 m. Hence all the four choices are correct. 15. The correct choices are (a), (c) and (d). The acceleration of each mass is the same and is given by a=

(m1 m2 ) g (m1 m2 )

2m1 m2 g (m1 m2 ) Since the masses start from rest, the distance moved by m1 in time t is 1 2 1 2 1 2 s = ut + at = 0 + at = at 2 2 2 16. The correct choices are (a), (b) and (d). The acceleration of each mass is given by The tension in the string is T =

cos

1 sec 2 Hence the correct choices are (b) and (d). 12. Since the blocks slide at the same uniform speed, no net force acts on them. If M is the mass of block B, then the tension in the string is T = Mg. Also T = mg. Equating the two, we get M = m m or M = F mg x= = . Therefore, k k potential energy stored in the spring is = tan

4 ms 2s

Since the lift is accelerating upward, the tension in the rope at t = 1 s (between t = 0 and t = 2 s) is T = m (g + a) = 1000 (10 + 2) = 12,000 N Between t = 2 s and t = 10 s, the speed of the lift is constant. Hence a = 0 and T = mg = 1000 10 = 10,000 N Between t = 10 s and t = 12 s, the lift is decelerating. Its deceleration is given by

a=

1 sin

AD OD

2

2

=

m g 2k

2

Hence the correct choices are (b) and (c). 13. If the acceleration of the blocks is a, then we have mg – T = ma (i)

a=

F (m1

m2 )

The tension in the string is T = m1 a. The force on mass m1 is m2 F (m1 m2 ) th 17. The 7 coin from the bottom has 3 coins above it. Hence, the force on the 7th coin = weight of 3 coins = 3 mg, vertically downwards. Since the 8th coin has 2 coins above, it supports the weight of two coins. Hence the force on the 7th F1 =

Laws of Motion and Friction 5.49

coin by the 8th coin = weight of 8th coin + weight of two coins above it = weight of three coins = 3 mg vertically downwards. From Newton’s third law, the reaction force th coin on the 7th coin is equal and th coin th th on the 6 th th coin on the 6 coin = weight of 7 coin + weight of 3 coins above it = weight of 4 coins = 4 mg vertically downwards. Hence, the reaction of the 6th coin on the 7th coin = 4 mg vertically upwards. Hence the correct choices are (a), (b) and (d). 18. The correct choices are (b) and (d). The acceleration of both the masses is the same and is given by net force F = a= total mass (m1 m2 ) Force on mass m2 = m2 a. 19. Figures 5.101 (a) and 5.101 (b) show the free body diagram of the two blocks.

Solving Eqs. (i) and (ii), we get T = 0 and a = g(sin

–

cos )

Hence the correct choices are (a) and (c). 20. Let M be the mass of the chain and L its length. If a length l hangs over the edge of the table, the Ml force pulling the chain down is g. The force L of friction between the rest of the chain of length M (L l) (L – l) and the table is g. L For equilibrium, the two forces must be equal, i.e. Ml M (L l) g= g L L or

l=

or

l=

(L – l) L 1

l 0.25 1 = = = or 20%. L 1 1 0.25 5 Hence the correct choices are (a) and (c). 21. The acceleration of the block is given by a = g(sin

–

cos

)

= g(sin – kx cos ) Acceleration a is not uniform; it varies with x. For sin > kx cos , a is positive, i.e. a is positive for x < tan /k. For x > tan /k, a is negative. For x = tan /k, a after which it begins to decrease as the block is decelerated. Hence the correct choices are (b), (c) and (d). 22. The horizontal and vertical components of F are F cos and F sin (see Fig. 5.102).

Fig. 5.101

T is the tension in the string and f1 and f2 are the frictional forces. It follows from the diagrams that N1 = m1g cos and f1 = m1g cos N2 = m2g cos and f2 = m2g cos If a is the acceleration of the blocks down the plane, the equations of motion are m1a = m1g sin – T – f1 = m1g sin – T – m1g cos (i) + T – f2 and m2a = m2g sin = m2g sin + T – m2g cos (ii)

Fig. 5.102

N + F sin

= mg N = mg – F sin

= mg – kx sin where N is the normal reaction.

5.50 Comprehensive Physics—JEE Advanced

The block will lose contact with surface at x = x0 for which N = 0. Putting N = 0 and x = x0, we mg have 0 = mg – kx0 sin x0 = k sin The acceleration a of block along the horizontal surface is given by ma = F cos

It follows from Fig. 5.103, that MV 2 = T – Mg cos L Tangential force ft = Mg sin Tangential acceleration at = g sin . So the correct choices are (b) and (c).

= kx cos

24.

k x cos a= m which depends upon x. Hence the correct choices are (b) and (c).

the force of friction acts in the opposite direction. But when a body itself applies a force in order to move, the force of friction acts in the direction of plied to the rear wheel and as a result the front wheel moves by itself. So, while pedalling, the choice (a) is correct. When pedalling is stopped, the choice (c) is correct as long as bicycle remains in motion.

23.

Fig. 5.103

III Multiple Choice Questions Based On Passage Question 1 to 3 are based on the following passage of reference; it holds only for inertial frames of reference. Passage I An inertial reference frame is a frame which moves with Reference Frames a constant velocity. 1. A reference frame attached to the earth reference. Rest and motion are relative; there is nothing like absolute rest or absolute motion. The position or state (b) cannot be an inertial frame because the earth of motion of a body may appear different from different is revolving round the sun (c) is an inertial frame because Newton’s laws everything else in a moving train are at rest in a reference of motion are applicable in this frame (d) cannot be an inertial frame because the earth frame situated in the train but they are in motion in a reference frame situated on the platform. Similarly, a stone dropped by a passenger from the window of a railway 2. Which of the following observers are inertial? (a) A child revolving in a merry-go-round carriage in uniform motion appears to him to fall vertically (b) A driver in a car moving with a constant downwards but to a person outside the carriage, it appears velocity to follow a parabolic path.

Laws of Motion and Friction 5.51

(c) A pilot in an aircraft which is taking off (d) A passenger in a train which is slowing down to a stop. 3. Choose the correct statements from the following. (a) An inertial frame is non-accelerating. (b) An inertial frame is non-rotating.

(c) A reference frame moving at a constant velocity with respect to an inertial frame is also an inertial frame. (d) Newton’s laws of motion hold for both inertial and non-inertia frames.

SOLUTION 1. The velocity of the earth changes with time (due to a change in its direction) as it revolves round the sun. Therefore, a frame attached to the earth is accelerated. Accelerated frames and rotating frames of reference are not inertial frames. Hence the correct choices are (b) and (d). Question 4 to 6 are based on the following passage Passage II Two bodies A and B of masses m and 2 m respectively are moving with equal linear momenta. They are subjected to the same retarding force. 4. If x1 and x2 are the respective distances moved by them before stopping, then x1/x2 is 1 (a) (b) 1 2 (d) 2 (c) 2

2. The observers in (a), (c) amd (d) are all accelerating. Hence, they are non-inertial. Only the driver in (b) is inertial since his motion is not accelerated. 3. The correct choices are (a), (b) and (c).

5. If t1 and t2 are the respective times taken by them to stop, then t1/t2 is (a) 1 (b) 2 1 1 (c) (d) 2 4 6. If a1 and a2 are their respective decelerations, then a1/a2 is (a) 1 (c)

(b) 2 2

(d)

1 2

SOLUTION (mu ) 2 p 2 = m m where F = ma is the retarding force p = mu is linear momentum. Thus

4. 2 ax = u2

2 max = mu2

x=

2 Fx =

p2 2 Fm

1 Since p and F are constants, x . Hence the m correct choice is (d). u mu p 5. 0 = u + at t=– =– =– . Hence a ma F t1 = t2. Thus the correct choice is (a). 6. a1 = F/m and a2 = F/2m. Hence the correct choice is (b).

Questions 7 to 9 are based on the following passage Passage III Two bodies of masses m and 2 m respectively are moving with equal kinetic energies. They are subjected to the same retarding force. 7. If x1 and x2 are the respective distances moved by them before stopping, then x1/x2 is (a) 2 (b) 2 1 (c) (d) 1 2

8. If t1 and t2 are the respective times taken by them to stop, then t1/t2 is 1 1 (a) (b) 2 2 (c) 2 (d) 2 9. If a1 and a2 are their respective decelerations, then a1/a2 is (a) 4 (b) 2 1 (c) (d) 1 2

SOLUTION 1 mu2. If K 2 is the kinetic energy, then Fx = K x = K/F.

7. 2ax = u2

2 max = mu2

Fx =

Since K and F are constants, the correct choice is (d).

5.52 Comprehensive Physics—JEE Advanced

8. 0 = u + at

t=–

u mu p =– =– . Now a ma F

p2 = m2u2 = (2m)

p = mu

1 2 mu 2

= 2mK.

Passage IV A block of masses m is initially at rest on a frictionless horizontal surface. A time-dependent force F = at – bt2 acts on the body, where a and b are positive constants. 10. t1 given by a 2a (a) (b) b b a a (c) (d) 2b 2b 11. F is given by

(c)

(d)

I

(a)

(c)

a3

(b)

3b 2 a3 9b 2

13.

d 2F

dt negative.

2

=

a3

a3

(b)

4mb 2 a3

(d)

12mb 2

a3 8mb 2 a3 16mb 2

12. t1

I

a –b 2b Hence the correct choice is (b).

a 2b

bt 2 )dt

0

=

2

=

a2 . 4b

a a 2 b a 3 a3 – = 2 2b 3 2b 12b 2 Hence the correct choice is (d). 13. Now impulse = change in momentum = mv – 0 = mv I a3 , which is choice (c). v = ma = m 12mb 2

(c) 2

Passage V A body of mass m is initially at rest. A periodic force F = a cos(bt + c) is applied to it, where a, b and c are constants.

at12 bt 3 – 1 2 3

=

Questions 14 to 17 are based on the following passage

14. The time period T of the force is 1 2 (b) (a) b b

(at

=

d (a – 2 bt) = – 2 b, which is dt

= at1 – bt21 = a

Fdt

= 0 t1

Hence the correct choice is (c). 11. F

6b 2

12b 2 attained by the block is

v (a)

a3

(d)

dF d Now = (at – bt2) = a – 2bt dt dt dF = 0 and t = t1, we get Putting dt a 0 = a – 2 bt1 t1 = 2b Also

imparted to the block

is given by

4a 2 b dF d 2F = 0 and dt dt 2

< 0.

1 Hence the correct choice is (b). m

(c)

SOLUTION 10.

F , i.e. a m

12.

a2 (b) 4b

2a 2 b

m . Hence the

9. a =

Questions 10 to 13 are based on the following passage

a2 (a) 2b

2mK . Thus t F correct choice is (a).

Hence t = –

a b

(d) 2

b a

15. a mb c (c) ma (a)

b mc b c (d) ma

(b)

Laws of Motion and Friction 5.53

16. The smallest value of t after t = 0 when the velocity of the body becomes zero is given by a (b) t1 = (a) t1 = a c c

(c) t1 =

b

(d) t1 =

b

17. The distance travelled by the body from time t = 0 to t = t1 is given by (a)

(c)

a mb

cos c

2

(b)

a2 cos c mb

(d)

2a mb 2

sin c

2a 2 sin c mb

SOLUTION 14. F will repeat itself at values of t given by cos(bt + c) = + 1, i.e. bt + c = 0, 2 , 4 , t=–

c 2 c 4 c , , , b b b

2 . Hence the The smallest time interval is T = b correct choice is (b). dp 15. From Newton’s second law of motion, F = dt dv =m dt

16. From Eq (i) it follows that v = 0 at values of t given by sin(bt + c) = 0 or (bt + c) = 0, , 2 , ... c c 2 c or t = – , , . Therefore, b b b –

b

c b

=

t1

t

vdt =

x= 0

a cos(bt + c)dt m

a 1 sin (bt + c)dt mb 0 =–

a a cos (bt + c)dt = sin(bt + c) (i) m0 mb

=–

bt + c) = 1, a v = . mb Hence the correct choice is (a).

=–

a mb 2 a mb 2 a mb

2

cos(bt1 + c) cos b

b

cos( + c) =

c a cos c mb 2

Hence the correct choice is (a).

Questions 18 to 20 are based on the following passage Passage VI

(c) a1 > a2 = a3

(d) a1 = a2 = a3

19. The tension in the string between masses m2 and m3 is

Three masses m1 = m, m2 = 2 m and m3 = 3 m are hung on a string passing over a frictionless pulley as shown in Fig. 5.104. The mass of the string is negligible. The system is then released. 18. If a1, a2 and a3 are the accelerations of masses m1, m2 and m3 respectively, then

(a) mg

Fig. 5.104

(b) 3 mg

5mg 3 20. The tension in the string between masses m1 and m2 is 2mg (a) 4 mg (b) 3 5mg (c) (d) 2 mg 3 (c) 4 mg

(a) a1 < a2 < a3 (b) a1 > a2 > a3

, which is choice (d).

dx dx = v dt. Therefore, the distance dt moved between t = 0 and t = t1 is

t

v=

b

17. Now v =

dv Thus m = a cos(bt + c) dt dv =

c

t1 =

(d)

SOLUTION 18. When the masses are released, mass m1 moves upward and masses m2 and m3 move downward with a common acceleration given by

(m2 m3 m1 ) g (2m 3m m) g 2g = = (m1 m2 m3 ) (m 2m 3m) 3 The correct choice is (d).

a=

5.54 Comprehensive Physics—JEE Advanced

19. The let T be the tension in the string between m1 and m2 and T be the tension in the string between m2 and m3 [see Fig. 5.105 (a)]. Figure 5.105 (b) shows the free-body diagram of mass m3.

m 3 g – T = m 3a T = m3(g – a) = 3 m

g

2g 3

= mg

Hence the correct choice is (a). 20. Figure 5.105(c) shows the free-body diagram of mass m1. T – m 1g = m 1a 2g 3 Hence the correct choice is (c). T = m1(g + a) = m

g

=

5mg 3

Fig. 5.105

Questions 21 to 23 are based on the following passage Passage VII Three blocks of masses m1 = m, m2 = 2m and m3 = 3m connected by two strings are placed on a horizontal frictionless surface as shown in Fig. 5.106. A horizontal force F is applied to mass m1 as shown.

F F (d) 5m 6m 22. The force on mass m2 is 5F (a) (b) 2 F 6 (c)

2F (d) F 3 23. The force on mass m3 is F (a) F (b) 2 F F (c) (d) 3 6 (c)

Fig. 5.106

21. The common acceleration of the blocks is F F (b) (a) m 3m

SOLUTION 21. If F2 and F3 are the forces on masses m2 and m3 respectively, then the free-body diagrams of m1, m2 and m3 are as shown in Fig. 5.107 where a is the common acceleration of the system.

Adding Eqs. (1), (2) and (3) we get a=

(m1

F m2

m3 )

=

F F = (m 2m 3m) 6m

Hence the correct choice is (d). 22. Adding Eqs. (2) and (3) we have F2 = (m2 + m3)a = (2m + 3m) Fig. 5.107

F – F 2 = m 1a

(1)

F2 – F 3 = m 2a

(2)

F 3 = m 3a

(3)

which is choice (a). 23. From Eq. (3), we have F F = F3 = 3ma = 3m 6m 2 Hence the correct choice is (b).

F 5F = , 6m 6

Laws of Motion and Friction 5.55

Questions 24 to 26 are based on the following passage Passage VIII Two blocks of masses m1 = m and m2 = 2 m are connected by a light string passing over a frictionless pulley. The mass m1 is placed on a smooth inclined plane of inclination = 30° and mass m2 hangs vertically as shown in Fig. 5.108.

(c)

g 2

(d) g

25. If the system is released the tension in the string is 3mg (a) mg (b) 2 (c) 2mg

(d)

2mg 3

26. If the inclined plane was rough, it was found that when the system was released, block m1 remained at rest. The frictional force between block m1 and the inclined plane is (a)

3mg 2

(b) 3 mg

(c)

4mg 3

(d)

Fig. 5.108

24. If the system is released, the blocks move with an acceleration equal to g g (a) (b) 4 3

2mg 2

SOLUTION 24. Since the inclined plane is smooth and m2 > m1, block m1 will up the plane and block m2 will move vertically with a common acceleration a. If T is the tension in the string, the free-body diagrams of masses m1 and m2 are as shown in Fig. 5.109

a=

(m2

m1 sin ) g (2m m sin 30 ) g g = = m1 m2 m 2m 2

Hence the correct choice is (c). 25. From Eqs. (1) and (2), we get g = mg 2 Hence the correct choice is (a). 26. Since the blocks remain at rest, the equations of motions of blocks m1 and m2 are (here f is the frictional force on m1) T = m2(g – a) = 2m

T – m1g sin and

f = m2g – m1g sin

The equations of motion of the blocks are and

–f=0 T = m 2g

These equations give

Fig. 5.109

T – m1g sin

g

= m 1a

(1)

m 2g – T = m 2a

(2)

Equations (1) and (2) give Questions 27 to 29 are based on the following passage Passage IX Two blocks of masses m1 = 3 m and m2 = 2 m are suspended

=2m

g – m g sin 30° mg 3mg = 2 mg – = , 2 2 which is choice (b). A and B. Wire A has negligible mass and wire B has a mass m3 = m, as shown in Fig. 5.110. The whole system of blocks, wires and the support have an upward acceleration a.

5.56 Comprehensive Physics—JEE Advanced

27. The tension 1 (a) m(g 2 3 m(g (c) 2 28. The tension

at the mid-point C of wire B is 3 + a) (b) m(g – a) 2 5 + a) (d) m(g + a) 2 at point O of wire B is

(a) 3m(g + a)

(b) 3m(g – a)

(c) 2m(g + a)

(d) 2m(g – a)

29. The tension at the mid-point D of wire A is

Fig. 5.110

(a) 2m(g + a)

(b) 4m(g – a)

(c) 6m(g + a)

(d) 8m(g – a)

SOLUTION 28. Let T1 be the tension in wire A. Since this wire has negligible mass, the tension is the same (= T1) at every point on this wire. Let T2 be the tension at point O of wire B. Then, we have for wire A.

27. Refer to Fig. 5.111. Let T be the tension at the midpoint C of wire B. Then T– m2

m3 g 2

= m2

m3 a 2

T = m2

m3 (g + a) 2

= 2m

m (g + a) 2

=

5 m(g + a), 2

T 1 – T 2 – m 1g = m 1a where T2 is given by T2 – (m2 + m3)g = (m2 + m3)a T2 = (m2 + m3) (g + a) = (2m + m) (g + a) = 3m(g + a) Hence the correct choice is (a).

Fig. 5.111

29. Putting T2 = 3m(g + a) in Eq. (1), we get T1 = 6 m(g + a).

which is choice (d).

Hence the correct choice is (c).

IV Matching 1. Three blocks of masses m1 = 3 m, m2 = 2 m and m3 = m are placed in contact on a horizontal frictionless surface as shown in Fig. 5.112. A horizontal force F is applied to m1 as shown. Match items in column I with those in column II Column I Column II (a) (b) (c) (d)

Force acting on m2 if F = 12 N Force acting on m2 if F = 6 N Force acting on m3 if F = 12 N Forcer acting on m3 if F = 6 N

(p) (q) (r) (s)

1 3 2 6

N N N N

Fig. 5.112

(1)

Laws of Motion and Friction 5.57

SOLUTION The contact forces acting on m2 and m3 respectively are (m2 m3 ) F F2 = (m1 m2 m3 ) F3 =

and

(m1

m3 F m2 m3 )

Hence the correct matching is as follows (a)

(s)

(b)

(q)

(c)

(r)

(d)

(p)

2. Two blocks of masses M = 5 kg and m = 3 kg are placed on a horizontal surface as shown in Fig. 5.113. 1 = 0.5 and that between the blocks M and the horizontal surface is 2 = 0.7. Taking g = 10 ms–2, match items in column I with those in column II Column I Column II (a) Frictional force between the blocks (p) 5 ms–2 (b) Acceleration of the upper block (q) 96 N F applied to M so that the two blocks move together without slipping. (r) 4 ms–2 (d) The common acceleration of the Fig. 5.113 blocks if F = 32 N. (s) 15 N

SOLUTION 1mg

(a) Force of frictional between blocks =

= 0.5

(b) Acceleration of the upper block of mass m is a =

3

10 = 15 N.

1mg/m

=

1g

= 0.5

10 = 5 ms–2.

(c) The force of friction between block M (with block m placed on top of it) and the horizontal surface = (M + m) 2g = (5 + 3) 0.7 10 = 56 N a = 5 ms–2. The force due to this acceleration = (M + m)a = (5 + 3) F

5 = 40 N

= 56 + 40 = 96 N

(d) If F = 32 N, the common acceleration of the blocks is a =

F (M

m)

=

32 = 4 ms–2 (5 3)

Hence the correct matching is as follows: (a)

(s)

(b)

(p)

(c)

(q)

(d)

(r)

V Assertion-Reason Type Questions In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each question has four choices out of which only one choice is correct

(a) Statement-1 is true, Statement-2 is true and ment-1.

5.58 Comprehensive Physics—JEE Advanced

(b) Statement-1 is true, Statement-2 is true but Statement-1. (c) Statement-1 is true; Statement-2 is false. (d) Statement-1 is false; Statement-2 is true. 1. Statement-1 A block is pulled along a horizontal frictionless surface by a thick rope. The tension in the rope will not always be the same at all points on it. Statement-2 The tension in the rope depends on the acceleration of the block-rope system and the mass of the rope. 2. Statement-1 A truck moving on a horizontal surface with a uniform speed u is carrying sand. If a mass m of the sand ‘leaks’ from the truck in a time t, the force needed to keep the truck moving at its uniform speed is u m/ t. Statement-2 Force = rate of change of momentum. 3. Statement-1 Two blocks of masses m and M are placed on a horizontal surface as shown in Fig. 5.114. The 1

and that between the block M and the horizontal surface is 2 applied to block M so that the two blocks move without slipping is F = ( 1 + 2) (M + m)g.

Fig. 5.114

Statement-2 acceleration. 4. Statement-1 A shell of mass m three fragments having masses in the ratio 2:2:1. mutually perpendicular directions with speed v.

The speed of the third (lighter) fragment will be 2 2 v. Statement-2 The momentum of a system of particles is conserved 5. Statement-1 F such that the block shown in Fig. 5.115 does not move is mg/cos , where block and the horizontal surface.

Fig. 5.115

Statement-2 normal reaction. 6. Statement-1 A ball of mass m is moving towards a batsman at a speed v it by an angle without changing its speed. The impulse imparted to the ball is zero. Statement-2 Impulse = change in momentum 7. Statement-1 A cloth covers a table. Some dishes are kept on it. The cloth can be pulled out without dislodging the dishes from the table. Statement-2 For every action there is an equal and opposite reaction. IIT, 2007 8. Statement-1 When a ball dropped from a certain height hits the momentum. Statement-2 reaction forces, being equal and opposite, cancel each other.

ANSWERS/SOLUTIONS 1. The correct choice is (a). 2. the leaking sand on the truck = rate of change of momentum = u m/ t. The sand falling vertically

the vertically upward direction. This perpendicular force can do no work on the truck. Since the truck moves with a uniform velocity, the force

Laws of Motion and Friction 5.59

3. The correct choice is (c). The force of friction between block m and block M = 1mg, where

From the principle of conservation of momentum, the momentum of the third (lighter) fragment of m must be 2 p but opposite in direction. mass 5 Thus, if V is the speed of the lighter fragment, we have mV 2mv = 2p = 2 5 5

1

blocks. Now, the force of friction between block M (with block m on top of it) and the horizontal surface = 2(M + m)g, where 2 of friction between block M and surface. The F applied to block M must be enough to overcome this force of friction and the force due to acceleration of the system. If the acceleration of the system is a then this force = (M + m)a. Thus (i) F = (M + m)a + m2(M + m)g Now, since the force on block m is 1mg, its acceleration a is mg force on mass m = 1 = 1g (ii) a= m mass m Using (ii) in (i) we get F = 1(M + m)g + 2(M + m)g = ( 1 + 2) (M + m)g

or V = 2 2v 5. The correct choice is (a). The component of F parallel to the horizontal surface is F cos . F F cos just overcomes the frictional force f = mg. Thus mg cos 6. The correct choice is (d). Refer to the solution of Q.27 of section I. F

= mg

F

=

7. motion also called the law of inertia. The dishes are not dislodged even when the cloth is suddenly pulled because the dishes have the inertia of rest. Statement-2 is Newton’s third law of motion, it

4. The correct choice is (a). The mass of two frag2m ments of equal masses = each. The mass 5 m . The momenta of of the lighter fragment = 5 2mv . The resultant of heavier fragments are p = 5 momenta p and p is p = (p2 + p2)1/2 =

cos

choice is (b). 8. The assertion is true but the reason is not correct because action and reaction forces do not act on the same body and hence do not cancel each other. Hence the correct choice is (c).

2p

VI Integer Answer Type 1. The magnitude of force f (in newton) acting on a body varies with time t (in millisecond) as shown in Fig. 5.116. Find the magnitude of the total impulse (in Ns) of the force on the body from t = 4 ms to t = 16 ms. IIT, 1994

2. A block of mass 0.2 kg is held against a wall by applying a horizontal force of 5 N on the block. is 0.5. Find the magnitude (in newton) of the frictional force acting on the block. Take g = 10 ms–2. IIT, 1995 3. Block A of mass m and block B of mass 2m are massless string and a frictionless pulley as shown

Fig. 5.116

block A and the wedge is 2/3 and that between block B and the wedge is 1/3. If the blocks are A (in ms–2). IIT, 1997

5.60 Comprehensive Physics—JEE Advanced

Fig. 5.117

4. A piece of uniform string hangs vertically so that its free end just touches the horizontal surface of a

table. The upper end of the string is now released. At any time during the falling of the string, the total force on the surface of the table is n times the weight of the part of the string lying on the surface. Find the value of n. IIT, 1989 5. A uniform rope of mass M and length L is pulled by a constant force of 10 N. Find the tension (in newton) in the rope at a point at a distance L/5 from the end where the force is applied. IIT, 1978

SOLUTIONS 1. Impulse from t = 4 ms to t = 16 ms = area under the F – t graph = area of EBCD = area of trapezium EBCF + area of CDF =

1 2

(200 + 800) N +

1 2

(2

800 N

10–3 s) (10

10–3 s)

= 1 + 4 = 5 Ns 2. Normal reaction R = 5 N. At equilibrium, the force of friction = weight of the block (see Fig. 5.118)

Fig. 5.119

The equations of motion of blocks A and B are T – mg sin 45° – A mg cos 45° = ma, where A = 2/3 and 2 g sin 45°– B 2 mg cos 45° – T = 2 ma, where B = 1/3. Adding these equations and solving we get a= –

Fig. 5.118

= mg = 0.2 10 = 2 N 3. Case (a): Let us assume that block A moves up the plane and block B moves down the plane. The free body diagrams of the blocks are as follows (See Fig. 5.119)

g 9 2

Case (b): If we assume that block A moves down 7g . and block B moves up, we would get a = – 9 2 Thus in both cases, the acceleration has a negative value which implies that the blocks will decelerate. This is not possible because the blocks start from rest. Hence when the blocks are released, they move with zero acceleration. Thus acceleration of block A = 0. 4. Let x be the length of the string lying on the surface of the table at an instant of time t. If an additional length dx of the string falls on the surface in time dt, the velocity v of this element when it strikes the surface is given by ( u = 0) v2 = u2 + 2gx = 0 + 2gx or

v2 = 2gx

(1)

Laws of Motion and Friction 5.61

The total force on the surface is F = rate of change of momentum of element of length dx + weight of a length x of the string lying on the table. If m is the mass per unit length of the string, then d dx (mdxv) + mxg = mv F= dt dt + mxg = mv2 + mxg (2) dx v dt Using (1) in (2) we get F = 2 mgx + mgx = 3 mgx But mx = M, the mass of the string lying on the table. Hence F = 3 Mg Thus n = 3. M . Let us 5. Mass per unit length of the rope is m = L P at a distance x from the end x = 0. Let T be the tension in the rope at point P. A

P

B

x=

x=L x

L–x

Fig. 5.120

x

For part AP the tension is towards the positive x-direction and for part BP the tension is towards the negative x-direction. If a is the acceleration produced in the rope by the constant force F, then for part AP, T = (mass of AP) a or T = mxa (1) For part BP, we have F – T = (mass of BP) a = m(L – x)a T From (1), we have a = mx Using this in (2), we get T ( L x)T F – T = m(L – x) mx x (L

or

F= T

At

x= L – T=

F L

x) x

L 5 4L 5

1

T

L or T x

(2)

Fx L

4L , 5 4F 5

4 10 5

8N

6

Work, Energy and Power

Chapter

REVIEW OF BASIC CONCEPTS 6.1 1.

WORK

force F in moving the body from a position r1 to a position r2 is given by r2

Work done by a Force

(a) Work done by a constant force When a constant force F acting on a body produces a displacement S, then the work done by the force is given by W = F S = FS cos where is the angle between the force vector F and the displacement vector S [see Fig. 6.1]. F and S are the magnitudes of F and S respectively.

F dr = area under the (F – r) graph

W= r1

6.1 makes an angle of 60° with the horizontal. Find the work done if a force of 150 N is applied to drag the box through a distance of 10 m. SOLUTION W = FS cos = 150

10

cos 60° = 750 J

6.2 A horizontal force F pulls a 20 kg box at a constant Fig. 6.1

(i) If is acute, cos is positive. Hence work is positive for acute . In this case the force increases the speed of the body. (ii) If = 90°, W = 0, i.e. if the force is perpendicular to displacement work done by the force is zero. (iii) If is obtuse, W is negative. In this case the force decreases the speed of the body. (iv) If = 0, i.e. force F is in the same direction as displacement S, then W = FS (v) If =180°, force F is opposite to S (example frictional force), W = – FS. Work done by frictional and viscous force is always negative. (b) Work done by a variable force Suppose a force F is not constant but depends on the position vector r of the body, then the work done by the

the work done by force F in moving the box through a distance of 2 m. SOLUTION Since the box is moved at a constant velocity, the applied force F just overcomes the frictional force f, i.e. F = f = mg Work done W = FS cos = 0.25

= mgS cos 0° 20

9.8

2 = 98 J

6.3 A block of mass m = 5 kg slides down from the top of an inclined plane of inclination = 30° with the

6.2 Comprehensive Physics—JEE Advanced

the block and the plane is 0.25. The length of the plane is 2 m. Find the work done by the (a) gravitational force, (b) frictional force and (c) normal reaction if the block slides to the the bottom of the plane. SOLUTION Refer to Fig. 6.2.

Fig. 6.2

Displacement S = AB = 2 m from A to B. (a) Angle between F and S is (90° – ) = 90° – 30° = 60° Work done by gravitational force is W1 = FS cos 60° = mgS cos 60° = 5 9.8

2

1 = 49 J 2

(b) Work done by frictional force is W2 = f S cos 180° = – f S = – RS = – mg (cos )S = – 0.25 5 9.8 cos 30° = – 21.2 J

2

(c) Since the normal reaction R is perpendicular to displacement S, work done by normal reaction is W3 = RS cos 90° = 0 6.4 A block of mass m = 2 kg is raised vertically upwards by means of a massless string through a distance of S = 4 m with a constant acceleration a = 2.2 ms–2. Find the work done by (a) tension and (b) gravity.

= 2 (2.2 + 9.8) = 24 N Work done by tension is ( T and S are in the same direction) W1 = TS cos 0° = 24 4 1 = 96 J (b) Since the gravitational force mg and displacement S are in opposite directions, work done by gravity is W2 = mgS cos 180° Fig. 6.3 = – 2 9.8 4 = – 78.4 J (c) Net work done W = W1 + W2 = 96 – 78.4 = 17.6 J 6.5 A block of mass m = 2 kg in suspended by a light string from the ceiling of a lift. The lift starts moving down with an acceleration a = 1.8 ms –2. Find the 5 seconds. SOLUTION Tension T = m(g – a) = 2 (9.8 – 1.8) = 16 N Distance moved in t = 5 s is 1 1 S = at 2 = 1.8 5 2 2 2 = 22.5 m Since the tension and displacement are in opposite directions, the work done by tension is W = TS cos180° = – TS = – 16 6.6 A constant force F = (2 i

T = m(a + g)

3 j) newton displaces

a body from position r1 = (4 i r2 = ( i

5 j) metre to

3 j) metre. Find the work done by the force.

SOLUTION Displacement S = r2 – r1 = (i

SOLUTION (a) From the free body diagram (Fig. 6.3) T – mg = ma

22.5 = – 360 J

3 j)

W = F S = (2 i

(4 i

5 j)

3 j) ( 3 i

3i

8j

8 j)

= – 6 + 24 = 18 J [ i i = j j 1 and i j = 0]

Work, Energy and Power 6.3

6.7 A body of mass m = 0.5kg travels in a straight line with a velocity v = 5x3/2 where v is in ms–1 and x is in metre. Find the work done in displacing the body from x = 0 to x = 2 m. SOLUTION Acceleration

dv d = dt dt 3 1/ 2 = 5 x 2 15 1/ 2 = x 2 75 2 = x 2

5 x3 / 2

a=

dx dt 5 x3 / 2

x 0

= 0.5

v

2

Fdx =

= 0.5

madx 0

75 2 75 2

2

x 2 dx 0

x3 3

2

0

0.5 75 8 0 2 3 = 50 J =

6.8 A block of mass 5 kg slides down from the top of an inclined plane of angle of inclination 30°. The the plane is 0.3. The length of the plane is 2 m. Find work done by (a) by gravity, (b) frictional force and (c) normal reaction. SOLUTION

= – 14.7 3 = – 25.5 J (c) Work done by normal reaction = RS cos 90° = 0 ( R S)

6.2

x 2

Work done W =

dx dt

Displacement S = 2 m down the inclined plane. (a) Work done by gravity = mg sin AB = (mg sin ) S = 5 9.8 sin 30° 2 = 49 J (b) Work done by friction = f S cos 180° =–fS = – RS = – mg cos S = – 0.3 5 9.8 cos 30° 2

ENERGY

So, energy is measured in the same units as work, namely, joule. Like work, energy is a scalar quantity. Energy can exist in various forms, such as heat energy, electrical energy, sound energy, light energy, chemical energy, nuclear energy, mechanical energy, etc. Mechanical energy is of two types, and Kinetic Energy: Energy due to Motion A moving object can do work on another object when it strikes it. In other words, an object in motion has the ability to do work

An initially motionless body can move and acquire a velocity only if a force acts on it. The work done by the force in causing the body to move measures the kinetic energy (written as KE) of the moving body, i.e. KE = W The kinetic energy of a body of mass m, moving with a velocity v is given by 1 KE = mv2 2 This relation holds even if the force is variable, i.e. if the force varies both in magnitude and direction. Work-Energy Principle Suppose a body of mass m moves with an initial velocity . A force F acts on it, as v. The work done by the force is given by Fdx

W=

v

=m

Fig. 6.4

=

ma dx

dv 1 dx = m(v2 – dt 2

1 1 mv2 – 2 2

2

2

)

6.4 Comprehensive Physics—JEE Advanced

Thus, This is the work-energy principle. Thus, when a force does work on a body, its kinetic energy increases; the increase in kinetic energy being equal to the amount of work done. The converse of this is also true. When the kinetic energy of a body is decreased by a retarding force, the decrease is equal to the work done by the body against the retarding force. Thus kinetic energy and work are equivalent quantities and are, therefore, measured in the same units, namely, joule. Potential Energy: Energy due to Position or An object can have energy not only by virtue of its motion, but also because of its position or

the reaction of the spring and is called the which is proportional to the displacement x and acts in a direction opposite to the displacement, i.e. F – x or F = – where is the force constant of the spring. The negative sign indicates that the force acts in a direction opposite to displacement. To stretch a spring by a displacement x, we must exert a force F on it, equal but opposite to the force F exerted by the spring on us. Therefore, the applied force is F =–F= Notice that F is a variable force as it depends on x. Therefore, the work done by the applied force in stretching the spring through a distance x is given by x

x

F dx =

W= An object held at a position above the surface of the earth has potential energy by virtue of its position. Consider a body of mass m. It is lifted vertically to a height above the earth by applying a force F vertically upward. The force F must be just enough to overcome the gravitational attraction, i.e. F = mg where g is the acceleration due to gravity at that place. For bodies not too far above the surface of the earth, the value of g is practically constant. Hence the work done by a constant force F in displacing a body by a height can be calculated by the product F = . Thus gravitational potential energy of a body of mass m at a height above the surface of the earth is . Gravitational PE = Consider a perfectly

a frictionless horizontal surface as shown in Fig. 6.5. We assume that the mass of the spring is negligible compared to the mass of the block.

Fig. 6.5

If we stretch the spring by a distance x, the spring will exert a force on us during stretching. This force is due to

0

( )dx 0

x

x dx =

= 0

x2 2

x

0

1 2

2

It is evident that the work done in compressing the 1 2 spring by an amount x is also given by W = . 2 Law of Conservation of Energy The total energy of an isolated system remains constant, the energy can only change from one form to another. 6.9 A block of mass 0.5 kg is taken from the bottom of an inclined plane to its top and then allowed to slide down to the bottom. The length of the inclined plane friction between the block and the plane is 0.2. Find (a) work done by the gravitational force over the round trip, (b) work done by the applied force over the upward journey, (c) work done by the frictional force over the round trip and (d) the kinetic energy of the block when it reaches the bottom of the plane. What conclusion will you draw from your answers to (b), (c) and (d)? Take g = 10 ms–2. SOLUTION Refer to Fig. 6.6. m = 0.5 kg, = 0.2, (= AC) = 2.5 m and (= AB) = 1.5 m

Work, Energy and Power 6.5

W4 = –

Fig. 6.6

sin

=

cos

=

AB 1.5 = = 0.6 AC 2.5 1 sin 2

= 0.8

(a) Work done by gravitational force to take the block from C to A is W1 = (– mg sin ) The negative sign indicates that the gravitational force mg sin is opposite to the displacement which is from C to A. Work done by the gravitational force to move the block from A to C is (because now the gravitational force is in the direction of the displacement) W2 = (+ mg sin Total work done by the gravitational force over the round trip is WG = W1 + W2 = 0 This shows that gravitational force is conservative. (b) Frictional force f = R = mg cos . The applied force to move the block from C to A is F = mg sin + f = mg sin + mg cos = mg (sin +

cos )

Work done by applied force to move the block from C to A is Wa = F AC = mg (sin + cos ) = 0.5 10 (0.6 + 0.2 0.8) 2.5 = 9.5 J (c) Friction always opposes motion. When the block is moved from C to A, frictional force f is opposite to direction. Hence work done by frictional force in the upward journey is W3 = – When the block slides from A to C, f acts upwards along the plane and is opposite to the displacement. Hence work done by frictional force in the downward journey is

Total work done by frictional force over the round trip is Wf = W3 + W4 = – 2 = – 2 mg cos = – 2 0.2 0.5 10 0.8 2.5 =– 4 J (d) When the block is at A, its initial velocity = 0. Let v be the velocity when it reaches C. Since f acts upwards, the net force on the block when it slides down is F = mg sin – f = mg (sin – cos ) F = g (sin – cos ) m = 10 (0.6 – 0.2 0.8) = 4.4 ms –2 = 2 , we have

Acceleration a =

From v2 –

2

v2 – 0 = 2 2

4.4

2.5

2 –2

v = 22 m s

1 mv2 2 1 = 0.5 2 = 5.5 J

Kinetic energy at C =

22

Conclusion Initial kinetic energy at A = 0. Therefore, change in K.E.= 5.5 – 0 = 5.5 J. Now total work done is W = WG + Wa + Wf = 0 + 9.5 – 4 = 5.5 J Thus, work done = change in kinetic energy. This is the work-energy principle. 6.10 A block of mass m = 500 g is placed at the top of an inclined plane of inclination = 60°. The length of the plane is 2 m. The block is released from rest. Find its speed when it reaches the bottom of the plane if (a) the inclined plane is smooth the plane is 0.4. Take g = 10 ms–2. SOLUTION Refer to Fig. 6.6 of Example 6.9 above. Height of the inclined plane is = AB = AC sin = 2 sin 60° = 3 m = 1.73 m

6.6 Comprehensive Physics—JEE Advanced

(a) As the block slides down the plane, it loses potential energy and gains kinetic energy. From the principle of conservation of energy, Gain in K.E = loss in P.E. 1 mv2 = or 2 = 2 10 1.73 5.9 ms–1 v= 2 (b) As the block slides down, loss of P.E. = gain in K.E. + work done against friction, i.e., 1 mg cos AC = m v2 2 1 1.73 10 = v 2 0.4 10 cos 60 2 2 v 5.2 ms–1 6.11 An elastic spring of negligible mass has a force constant 4 Nm–1 the wall and the other end touches a block of mass m = 250 g placed on a horizontal surface. The spring is compressed by an amount x = 5 cm as shown in and the horizontal surface is

= 0.2. If the system is

the spring.

the same plank before coming to rest if it were moving with a speed of 200 ms–1 ? SOLUTION Since the bullet and the plank are the same, the resistive force F exerted on the bullet is the same in the two cases. The kinetic energy is spent in doing work against friction. Hence 1 mv12 = Fx1 2 1 mv22 = Fx2 2 These equations give and

x2 = x1

v22 v12

200 100

= 2 cm

2

= 8 cm 6.13 A uniform chain of length L and mass M lies on a frictionless horizontal table with a very small part hanging from the edge of the table. The chain begins to fall under the weight of the hanging part. Obtain the expression for the velocity of the chain at the instant when the length of the hanging part becomes L/n where n > 1. SOLUTION M Mass per unit length of chain = . The mass of L L of the chain is length n

Fig. 6.7

SOLUTION Loss in P.E. of spring + work done against friction = gain in K.E. 1 2 1 mgx = mv 2 or 2 2 1 4 0.05 2 0.2 0.250 10 0.05 2 1 = 0.250 v 2 2 which gives v = 0.5 ms–1 6.12 A bullet moving with a speed of 100 ms–1 travels a distance of 2 cm in a plank of wood before coming to rest. How much distance will the same bullet travel in

m=

M L

L n

M n

As the chain slips down from the table, gravitational L (hanging part) depotential energy of the length n creases while the part of the chain left on the table does not lose any gravitational potential energy. The loss of P.E. of the hanging part gets converted into K.E. of the entire chain, i.e. Gain of K.E. of the complete chain = loss of P.E. of the hanging part. The mass m of the hanging part can be assumed to be concentrated at its centre of mass L below the edge of the which is at a height = 2n table. If v is the velocity of the slipping chain, then 1 M v2 = 2

=

M n

g

L 2n

MgL 2n 2

Work, Energy and Power 6.7

v2 =

gL n2 gL n

v=

6.3

CONSERVATIVE AND NON-CONSERVATIVE FORCES

(a) Conservative force A force is conservative if (i) the work done by it on a body in moving it from one position to another depends only on the initial

NOTE F is negative if r is opposite to F and positive if r is in the same direction as F.

6.14 The force between two point charges q1 and q2 1q2 separated by a distance r is F = where is a r2 constant. Find the potential energy of the system of charges. SOLUTION F=

followed by it between the two positions. or (ii) the net work done by the force on a body that moves through any closed path is zero. The above two conditions are equivalent. Examples of conservative forces are gravitational force, electrostatic force and spring force. (b) Non-conservative force A force is non-conservative if (i) The work done by it on a body in moving it from one position to another depends on the path followed by the body between the two positions. or (ii) The work done by the force on a body that moves through a closed path is non zero. Examples of non-conservative forces are frictional and viscous forces. Conservative Force and Potential Energy For a conservative force F that depends upon position r, there is a potential energy function U which also depends on r. When a conservative force does positive work, the potential of the system decreases, i.e. Work done = decrease in potential energy or or

Fdr = – dU dU F= dr

Hence

The change in potential energy when the body is displaced from r = a to r = b is b

Ub – Ua =

Fdr a

dU dr

dU = – Fdr. Integrating

r

r

Fdr =

U= 0

U=

1

r 2 dr

1 q2 0

q2

r

6.15 The potential energy U with position r as a b U= 2 r r where a and b are positive constants. Find the position r0 where the particle will be in stable equilibrium. SOLUTION If no force acts on a particle, it will be in stable equilibrium, i.e. F = 0 or dU =0 F= dr d a b =0 d r r2 r 2a

b

3

2

r = r0 =

=0

2a b

r r Stable equilibrium corresponds to minimum potendU d 2U 2a tial energy, i.e. = 0 and > 0. If r = r0 = , 2 dr b dr dU = 0, then U can be minimum or maximum. If dr d 2U d r2 Now

> 0 at r = r0, U will be minimum. dU = dr

2a

b

3

r2

r

6.8 Comprehensive Physics—JEE Advanced

d 2U d r2 At

r4

r0

3 b4 8 a3

=

4

– 2b

b4 4a

3

b 2a

3

b4

, which is posi8a tive.

NOTE For stable equilibrium U is minimum and for unstable equilibrium, U is maximum.

dU d 2U = 0 and >0 dr d r2

For unstable equilibrium;

6.4

m v2A – mg (i) r The body will revolve in the circle if the string does not sag, i.e. TA 0. From Eq. (i) it follows that m v2A – mg r v2A

3

2a corresponds to stable equilibHence r = r0 = b rium.

For stable equilibrium;

TA = tension in the string when the body is at A. TA =

b 2a

= 6a at r

and

2b

r3 2a r = r0 = , b

d 2U d r2

6a

=

dU d 2U = 0 and < 0. dr d r2

MOTION IN A VERTICAL CIRCLE

Consider a body of mass tied to a string of length r revolved in a vertical circle with centre O at the other end of the string as shown in Fig. 6.8. At all positions of the body, there are two forces acting on it: its own weight and the tension in the string. v

0 rg

Therefore, the minimum speed at the top that the body must have so that it can complete the circle is given by (vA)min =

rg

(ii)

Case (b): When the body is at bottom B of the circle When the body is at B, the net force on the body towards the centre is (TB – mg). Hence m vB2 r where vB = speed of the body at bottom B and TB = tension in the string when the body is at B. TB – mg =

m vB2 + mg (iii) r The minimum speed at B that the body must have so that it can complete the circle is obtained from the conservation of energy. As the body goes up from B to A, it K.E. decreases and P.E. increases. Loss in K.E. = gain in P.E. TB =

1 2 1 mvB – mv2A = mg 2 2 vB = (vB)min =

v2A

AB = mg

2r

4gr

( vA )2min

4 gr

(iv)

Using (ii) in (iv) we get (vB)min = v

Case (a): When the body is at top A of the circle When the body is at top A of the circle, the net force towards centre O is TA + mg. Hence [see Fig. 6.8] m vA2 r where vA = speed of the body at top A TA + mg =

(v)

The net force towards the centre acting on the body is obtained from (iii) by using (v). TB =

Fig. 6.8

5gr

m r

5gr + mg = 6 mg

NOTE 1. When a body moves in a vertical circle, the speed decreases as it goes up and increases as it goes down. Hence the body has a non-uniform circular motion. 2. The tension in the string is different positions of the body on the circle. The tension is minimum when the body is at the top of the circle and maximum when it is at the bottom of the circle.

Work, Energy and Power 6.9

6.16 A small block of mass m, starts from rest at A and slides on a frictionless track which ends in a circular loop of radius r. If = 6r when it reaches C as shown in Fig. 6.9. What is the force exerted on the block by the track when it is at C so that the block is able to complete the circle.

min

6.5

=

1 2 mvmin 2

min

= 2.5r.

POWER i.e.

Work Time The faster a given amount of work is done, the greater is the power of the agent that does the work. In the SI system, the unit of power is the watt (symbol W). i.e. Power =

1 W = 1 Js–1

Fig. 6.9

SOLUTION Let v be the speed of the block when it reaches C. From conservation of energy, gain in K.E. = loss in P.E., i.e. 1 OB = – mgr mv2 0 = 2 1 2 v = g 6r – gr = 5gr 2 v = 10gr When the block is at C, the track exerts a normal reaction N on the block. Since the block is moving in a circular path, the necessary centripetal force for circular motion is provided by the normal reaction (Fig. 6.10).

Fig. 6.10

m v2 m 10 gr = = 10 mg r r Thus the track exerts a force on the block equal to 10 times the weight of the block. To complete the circle, the minimum speed at D must be vmin = 5gr . Hence N=

Since the watt is a small unit for the measurement of power, larger units, namely kilowatt (kW) and megawatt (MW) are often used. 1 kW = 1000 W = 103 W 1 MW = 1,000,000 W = 106 W The power of an agent can also be expressed in terms of the force applied and the velocity of the object on which the force is applied. Now, power is given by W F S = = Fv t t S ( = rate of change of displacement = v) t Power is a scalar quantity as it is the ratio of two scalars W and t, or a scalar product of two vectors F and v. 6.17 An engine pulls a car of mass 1000 kg on a level road at a constant velocity of 5 ms–1 . If the frictional force is 500 N, what power does the engine generate? What extra power must the engine supply to maintain the same speed up an inclined plane having a gradient of 1 in 10? SOLUTION Since the car moves at a constant velocity, its acceleration is zero. Hence the engine has to do work only to overcome the frictional force f . Power = f v = 500 5 = 2500 W = 2.5 kW For an inclined plane having a gradient of 1 in 10, sin 1 . To maintain the same speed up the inclined plane, = 10 the engine has to do extra work against the force mg sin .

6.10 Comprehensive Physics—JEE Advanced

Therefore, v

Extra power = mg sin 9.8

= 100

1 3 v 3

1 10

5 = 4900 W

3

=

=

with water. If the tank is 40 m above the ground and Take g = 10 ms–2 . SOLUTION Volume of tank V = 2000 litre = 2000 10–3 m3 = 2 m3 Mass of water m = V = 1000 2 = 2 103 kg Work done to lift this mass to a height = 40 m is W= = 2 103 10 40 = 8 105 J W 8 105 4 Power needed = = = 103 W t 10 60 3 is the total power consumed, the useful power 0.4 . Hence 4 103 3 = 3.33 103 W = 3.33 kW

3000

5

3

2

3

SOLUTION Since F is constant, acceleration a =

= Fv = mav dv a= dt

dv d x dx dt

dv = dx mv 2

v dv =

m

v

a= vd v dx

At time t, the velocity of the car is v=

+ at = 0 + at = at =

Kinetic energy E at time t = =

F2 2 t . m

Since F is constant, E Power Thus

m

Ft 1 1 mv2 = m m 2 2

Ft = m

at time t = Fv = F

F2 t m

mv

displacement in time t is proportional to (b) t (a) t1/2 3/2 (d) t2 (c) t SOLUTION = Fv = mav = m

dx

2

t. So the correct choices are (b) and (c).

dx

0

Ft . m

t2 .

x

v2 d v =

F is constant. m

6.21 A body, initially at rest, moves in a straight line

SOLUTION

v

3x

6.20 A car of mass m starts from rest at time t = 0 and is driven on a straight horizontal road by the engine which exerts a constant force If friction is negligible, the car acquires kinetic energy E at time t and develops a power . Which of the following is/are correct ? (a) E t (b) E t2 (c) t (d) t2

=

6.19 A constant power is supplied to a car of mass m = 3000 kg. The velocity of the car increases from = 2 ms–1 to v = 5 ms–1 when the car travels a distance x = 117 m. Find the value of . Neglect friction.

Now

3

3 117 = 1000 W = 1 kW

6.18

0.4

v3

=

= 4.9 kW

If

m

vdv =

m

dt

dv dt

v

Work, Energy and Power 6.11

v

t

vd v =

Integrating 0

m0

dt

(

dx =

= constant) x

v2 = t 2 m

2 1/2 t dt m t

2 dx = t1/2 dt m 0 0

v=

2 1/2 t m

dx = dt

2 1/2 t m

x=

4 3/2 t 3m

So the correct choice is (c).

I Multiple Choice Questions with Only One Choice Correct 1. A particle of mass m is moving in a circular path of a constant radius r such that its centripetal acceleration ac is varying with time t as ac = 2 r t2 where is a constant. The power delivered to the particle by the force acting on it is 2 2 2 2 rt (b) rt (a) 2 4 2 5

(c)

r t 3

(d) zero

IIT, 1994 2. A stone is tied to a string of length L and whirled in a vertical circle with the other end of the string at the centre. At a certain instant of time, the stone is at the lowest position and has a speed . The magnitude of the change in its velocity as it reaches a position where the string is horizontal is (a)

2

2 L

(b)

2 gL

(c)

2

L

(d)

2(

2

L)

IIT, 1998 3. The power supplied to a body initially at rest varies with time t as = 2 where is a constant. The velocity of the body at an instant of time t will be proportional to (a) t (b) t3/2 (c) t2 (d) t3 4. A force F = (3 i 4 j) newton acts on a particle moving along a line 4y + = 3. The work done by the force is zero if the value of is (a) 1 (b) 2 (c) 3 (d) 4

5. Force F acting on a body moving in a straight line varies with the velocity v of the body as F = /v where is a constant. The work done by the force in time t is proportional to (a) t (b) t3/2 –1/2 (c) t (d) t–3/2 6. The force F acting on a body varies with its displacement x as F = –2/3. The power delivered by the force will be proportional to (b) x–1/2 (a) x–3/2 (c) x1/2 (d) x3/2 7. of 200 ms–1. After passing through the plank, its speed reduces to 180 ms–1. Another bullet, of the same mass and size but moving with a speed of 100 ms–1 speed of this bullet after passing through the plank? Assume that the resistance offered by the plank is the same for both the bullets? (a) 48 ms–1 (b) 49 ms–1 (c) 50 ms–1 (d) 51 ms–1 8. If the mass of either bullet in Q. 10 is 7 g and the thickness of the wooden plank is 1 m, what is the average resistance offered by the plank? (a) 36 N (b) 38 N (c) 40 N (d) 42 N 9. An engine pulls a car of mass 1500 kg on a level road at a constant speed of 5 ms–1. If the frictional force is 1500 N, what power does the engine generate?

6.12 Comprehensive Physics—JEE Advanced

(a) 5.0 kW (b) 7.5 kW (c) 10 kW (d) 12.5 kW 10. In Q. 9, what extra power must the engine develop to maintain the same speed up an inclined plane having a gradient of 1 in 10? Take g = 10 ms–2. (a) 2.5 kW (b) 5.0 kW (c) 7.5 kW (d) 10 kW 11. Two identical cylindrical vessels, with their bases at the same level, each contain a liquid of density . The height of the liquid in one vessel is 1 and that in the other is 2. The area of either base is A. What is the work done by gravity in equalizing the levels when the vessels are interconnected? (b) A g ( 1 + 2)2 (a) A g ( 1 – 2)2 2

2

(c) A g

1

2

2

(d) A g

1

2

2

17. ing a thickness of 3.5 cm. The total thickness penetrated by the bullet is (a) 8 cm (b) 10 cm (c) 12 cm (d) 14 cm 18. etrating a thickness x of the plank. What is the total thickness penetrated by the bullet? (a) 2x (b) 4x (c) 6x (d) 8x 19. A body of mass m, having momentum , is moving on a rough horizontal surface. If it is stopped in a distance x body and the surface is given by 2

12. 3

with water. If the tank is 60 m above the ground the tank? Take g = 10 ms–2. (a) 100 kW (b) 150 kW (c) 200 kW (d) 250 kW 13. The distance x moved by a body of mass 0.5 kg by a force varies with time t as x = 3t2 + 4t + 5 where x is expressed in metre and t in second. What (a) 25 J (b) 50 J (c) 75 J (d) 100 J 14. In a hydroelectric power station, the height of the dam is 10 m. How many kg of water must fall per second on the blades of a turbine in order to generate 1 MW of electrical power? Take g = 10 ms–2. (a) 103 kgs–1 (b) 104 kgs–1 (c) 105 kgs–1 (d) 106 kgs–1 15. A uniform steel rod of mass m and length is pivoted at one end. If it is inclined with the horizontal at an angle its potential energy will be 1 1 cos (b) sin (a) 2 2 (c) cos (d) sin 16. A bullet, incident normally on a wooden plank, loses one-tenth of its speed in passing through the plank. The least number of such planks required to stop the bullet is (a) 5 (b) 6 (c) 7 (d) 8

(a)

=

(c)

=

2

2 gm 2 x

(b)

=

(d)

=

2mgx

2 gm 2 x 20. A uniform chain of mass M and length L is held on 1 a horizontal frictionless table with th of its length n hanging over the edge of the table. The work done is pulling the chain up on the table is Mg Mg (b) (a) 2n n Mg Mg (c) (d) 2 n 2 n2 21. A body of mass m = 1 kg is dropped from a height 2mgx

base, as shown in Fig. 6.11. As a result the spring is compressed by an amount x = 10 cm. What is the force constant of the spring. Take g = 10 ms–2. (a) 600 Nm–1 (b) 800 Nm–1 –1 (c) 1000 Nm (d) 1200 Nm–1

Fig. 6.11

Work, Energy and Power 6.13

22. A force acts on a particle of mass 3 g in such a way that the position of the particle as a function of time is given by x = 3t – 4t2 + t3 where x is in metres and t is in seconds. The work (a) 570 mJ (b) 450 mJ (c) 490 mJ (d) 530 mJ 23. A sphere of mass m is tied to one end of a string of length and rotated through the other end along a horizontal circular path with speed v. The work done in one full horizontal circle is (a) zero (b) mg 2 (c)

mv 2

2

(d)

mv 2

24. The kinetic energy acquired by a mass m in travelling a certain distance d, starting from rest, under the action of a constant force is (a) directly proportional to m (b) independent of m 1 (c) directly proportional to m (d) directly proportional to m 25. A position dependent force F = 7 – 2x + 3x 2 newton acts on a body of mass 2 kg and displaces it from x = 0 to x = 5 m. The work done in joules is (a) 70 (b) 270 (c) 35 (d) 135 26. A particle is moved from a position r1 = (3 i + 2 j – 6 k ) metre to a position r2 = (14 i + 13 j + 9 k ) metre under the action of a force F = (4 i + j + 3 k ). What is the work done? (a) 10 J (b) 100 J (c) 0.01 J (d) 1 J

(a) zero

(b) 20 2 m/s

(c) 20 3 m/s

(d) 40 m/s

29. A force F = – K ( y i + x j ) where K is a positive constant, acts on a particle moving in the x-y plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force F on the particle is (b) 2 Ka2 (a) – 2 Ka2 2 (c) – Ka (d) Ka2 IIT, 1998 30. A wind-powered generator converts wind energy into electrical energy. Assume that the generator tercepted by its blades into electrical energy. For wind speed v, the electrical power output will be proportional to (a) v (b) v2 3 (c) v (d) v4 IIT, 2000 31. A body of mass 6 kg is acted upon by a force which t2 metre causes a displacement in it given by x = 4 where t is the time in second. The work done by the force is 2 seconds is (a) 12 J (b) 9 J (c) 6 J (d) 3 J 32. A ladder 2.5 m long and of weight 150 N has its centre of gravity 1 m from its bottom. A weight of 40 N is attached to the top end. The work required to raise the ladder from the horizontal position to the vertical position is (a) 190 J (b) 250 J (c) 285 J (d) 475 J 33. A body of mass 5 kg rests on a rough horizontal

27. energy increases by 28. A particle of mass 0.1 kg is subjected to a force which varies with distance as shown in Fig. 6.12. If it starts its journey from rest at x = 0, its velocity at x = 12 m is

Fig. 6.12

pulled through a distance of 10 m by a horizontal force of 25 N. The kinetic energy acquired by it is (take g = 10 ms–2) (a) 200 J (b) 150 J (c) 100 J (d) 50 J 34. A body is moving up an inclined plane of angle with an initial kinetic energy E friction between the plane and the body is . The work done against friction before the body comes to rest is: E cos (b) E cos (a) cos sin (c)

E cos cos sin

(d)

E cos cos sin

6.14 Comprehensive Physics—JEE Advanced

35. 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) x2 (b) ex (c) x (d) loge x 36. A body of mass m accelerates uniformly from rest to velocity v1 in time t1. The instantaneous power delivered to the body as a function of time t is mv1t (a) t1 (c)

(b)

mv12t 2

(d)

t12

37. A force F

5 i 3 j 2k

mv12t t12 mv1t 2 t1

newton is applied to

a particle which displaces it from its origin to the point r

(2 i

j) metre. The work done on the

particle (in joule) is (a) – 7 (b) + 7 (c) + 10 (d) + 13 38. Two masses M and m (with M > m) are connected by means of a pulley as shown in Fig. 6.13. The system is released. At the instant when mass M has fallen through a distance , the velocity of mass m will be (a) (b) (c)

(d)

2 M

2 m

M

2 M

m m

M

2

Fig. 6.14

40. An escalator is moving downwards with a uniform speed . A man of mass m is running upwards on it at a uniform speed v. If the height of the escalator is , the work done by the man in going up the escalator is (a) zero (b) mg mg v (c) (d) v v 41. The potential energy (in joule) of a body of mass 2 kg moving in the x – y plane is given by U = 6x + 8y where the position coordinates x and y are measured in metre. If the body is at rest at point (6 m, 4 m) at time t = 0, it will cross the y-axis at time t equal to (a) 1 s (b) 2 s (c) 3 s (d) 4 s 42. In Q. 41 above, the speed of the body when it crosses the y-axis is (a) zero (b) 5 ms–1 (c) 10 ms–1 (d) 20 ms–1 43. If W1, W2 and W3 represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 (as shown in Fig. 6.15) in the gravitational m between W1, W2 and W3.

M

Fig. 6.13

m m

39. A mass m, lying on a horizontal frictionless sur-face is connected to mass M as shown in Fig. 6.14. The system is now released. The velocity of mass m when mass M as descended a distance is (a)

(c)

2 Mg m 2 Mg M m

(b)

(d)

2 mg M 2

Fig. 6.15

(a) W1 > W3 > W2

(b) W1 = W2 = W3

(c) W1 < W3 < W2

(d) W1 < W2 < W3 IIT, 2003

Work, Energy and Power 6.15

44. A particle, which is constrained to move along the x-axis, is subjected to a force in the same direction which varies with the distance x of the particle from the origin as F(x) = – + ax3. Here and a are positive constants. For x 0, the functional form of the potential energy U(x) of the particle is (see Fig. 6.16)

Fig. 6.17

(a) 3 ms

–1

(b) 4 ms–1

(c) 5 ms–1 (d) 6 ms–1 48. A body of mass m is dropped from a height above the ground. The velocity v of the body when it has lost half its initial potential energy is given by (b) v =

(a) v =

2

(c) v =

Fig. 6.16

IIT, 2002 45. A raindrop of radius r falls from a certain height above the ground. The work done by the gravitational force is proportional to (a) r (b) r2 (c) r3 (d) r4 46. A smooth steel ball is moving to and fro about the lowest position O of a frictionless hemispherical bowl. The ball attains a maximum height of 20 cm on either side of O. If g = 10 ms–2, the speed of the ball when it passes through O will be (a)

(d) v = 2 2 49. A body of mass m is thrown vertically upwards with a velocity v. The height at which the kinetic energy of the body is half its initial value is given by v2 v2 (b) = (a) = g 2g (c)

=

v2 3g

(d)

=

v2 4g

50. A car of mass m moving at a speed v is stopped in a distance x by the friction between the tyres and the road. If the kinetic energy of the car is doubled, its stopping distance will be (a) 8x (b) 4x (c) 2x (d) x 51. A body is allowed to fall freely under gravity from

2 ms–1

(b) 2 ms–1 (c) 0.2 ms–1 (d) 0.02 ms–1 47. The bob of a pendulum is released from a horizontal position A as shown in Fig. 6.17. The length of of the bob is dissipated as heat due to the friction of air, what would be the speed of the bob when it reaches the lowermost point B? Take g = 10 ms–2.

impact with the ground, to what height will it rise after one impact? (a) 2.5 m (b) 5.0 m (c) 7.5 m (d) none of these 52. In Q. 51, to what height will the body rise after two such impacts with the ground? (a) 2.5 m (b) 5.0 m (c) 7.5 m (d) none of these 53. A body of mass m thrown vertically upwards attains a maximum height . At what height will its kinetic

6.16 Comprehensive Physics—JEE Advanced

displaced towards wall 1 by a small distance x (a)

(b)

6

5

a maximum distance y towards wall 2. Displacements x and y are measured with respect to the equix is librium position of the block B. The ratio y (a) 4 (b) 2 1 1 (c) (d) 2 4

(c)

(d) 4 3 54. A body, having kinetic energy , moving on a rough horizontal surface, is stopped in a distance x. The force of friction exerted on the body is (a)

(b)

x

(c)

x

(d)

x

55. force F = , where is a positive constant. If the potential energy U is zero at x = 0, the variation of potential energy with the coordinate x is represented by [see Fig. 6.18]

Fig. 6.19

IIT, 2008 57. A bob of mass m is suspended by a massless string of length L. The horizontal velocity v at position A B. The angle at which the speed of the bob is half of that at A

Fig. 6.18 v

IIT, 2004 56. A block (B) is attached to two unstretched springs S1 and S2 with spring constants and 4 , respectively (see Figure 6.19). The other ends are attached to identical supports M1 and M2 not attached to the walls. The springs and supports have negligible mass. There is no friction anywhere. The blook B is

Fig. 6.20

(a) (c)

=

(b)

4 3 4

2

(d)

4 3 4

IIT, 2008

ANSWERS

1. 7. 13. 19. 25. 31. 37.

(b) (b) (c) (a) (d) (d) (b)

2. 8. 14. 20. 26. 32. 38.

(d) (b) (b) (d) (b) (b) (c)

3. 9. 15. 21. 27. 33. 39.

(b) (b) (b) (c) (a) (b) (c)

2

4. 10. 16. 22. 28. 34. 40.

(c) (c) (b) (d) (d) (d) (d)

5. 11. 17. 23. 29. 35. 41.

(a) (c) (a) (a) (c) (a) (b)

6. 12. 18. 24. 30. 36. 42.

(b) (a) (b) (b) (c) (b) (c)

Work, Energy and Power 6.17

43. (b) 49. (d) 55. (a)

44. (d) 50. (c) 56. (b)

45. (c) 51. (c) 57. (d)

46. (b) 52. (d)

47. (d) 53. (c)

48. (a) 54. (a)

SOLUTIONS v2 = 2 r t2 v= r dv Tangential acceleration at = = dt Tangential force F = m at =

1. ac =

2

Work done is zero if the force is perpendicular to the displacement, i.e. if lines (1) and (2) are perpendicular each other. Thus the product of their slopes = – 1, i.e. 4 =–1 =3 3 4

r t2

Power = F v = 2. Refer to Fig. 6.21.

2 2

rt

=

5. Power = F v = v in time t is

v = . Therefore, work done

t

t =

W=

(

= constant)

0

Hence the correct choice is (a). 6. Given F

x–2/3. Therefore, acceleration x–2/3, i.e. dv v = K x–2/3 (K = constant) dx vdv = K

Fig. 6.21

v2

From conservation of energy, 1 1 2 = mv2 + mgL 2 2 v=

2

2 L

Change in velocity v = v = v ( ) . Thus v is the resultant of v and – . It follows from Fig. 6.21 (b) that v =

v2

(

=

2

2

)2 2

=

2(

2

L)

3

t =

t 2 dt =

2 v= t3 3m Hence the correct choice is (b). 4. The force F is parallel to the line 4 y= x + C 3 The particle moves along the line 3 y= – 4 4

3

(1)

(2)

or

2/3

dx

v

x1/6

= F v x–2/3 x1/6 x–1/2 7. Let m be the mass of each bullet. Since the resistance offered by the plank is the same for the two bullets, the amount of work done by the plank is the same for the two bullets. From work-energy principle, the decrease in the kinetic energy is the same for the two bullets. 1 1 2 mv 12 1– 2 2 1 1 = m(200)2 – m (180)2 (i) 2 2 If v2 is the speed of the second bullet after passing through the plank, then Decrease in KE of second bullet 1 1 mv 22 = m 22 – 2 2 1 1 = m(100)2 – mv 22 (ii) 2 2 Equations (i) and (ii) we have, 1 1 m(200)2 – m (180)2 2 2 1 1 = m(100)2 – mv 22 2 2 =

3. From work-energy principle, change in K.E. = work done. 1 mv2 = 2

x1/3

x

6.18 Comprehensive Physics—JEE Advanced

which gives v 22 = 2400 or v

49 ms

–1

Hence

Hence the correct choice is (b).

PE after connection =

8. Thickness of plank (S ) = 1.0 m Mass of bullet (m) = 10 g = 10–2 kg If F is the average resistive force exerted by the plank on the bullet, the work done by the plank on the bullet is W = FS Work done = decrease in KE of the bullet. Hence FS =

= Change in PE =

Putting S = 1 m and m = 10–2 kg, we get F = 38 N. Hence the correct choice is (b). 9. Since the car moves at a constant velocity, its acceleration is zero. Hence the engine has to do work only to overcome the frictional force (f). Since the distance moved in 1 second is v metres, the work done per second or the power of the engine is = f v = 1500 5 = 7500 W = 7.5 kW Hence the correct choice is (b). 10. When the car is being pulled along an inclined plane 1 , the engine has of gradient 1 in 10 i.e. sin = 10 to do extra work against the component Mg sin of the weight Mg of the car. The extra work per second or extra power the engine has to develop to maintain the same speed v is = Mg sin v 1 5 = 1500 10 10 = 7500 W = 7.5 kW Hence the correct choice is (c). 11. The work done by gravity equals the change in the potential energy of the system after the vessels are interconnected. We may regard the liquid in each vessel as equivalent to a point mass kept at their respective centres of gravity. Remembering that the mass of the liquid is given by ( ) and that the PE of a mass at a height in earth’s gravity is , we have Total PE at start = (A

1

)g

1

2

+ (A

2

)g

2

2

A g ( 21 + 22) 2 After the vessels are connected, the height of liquid in each vessel is ( 1 + 2)/2. =

A g ( 4

2

g

1

– 2(

2 1

2 +

1

A g {( 4

=–

1 1 m(200)2 – m(180)2 = 3800 m 2 2

1

A

+

1

A g ( 4

2)

1

2

2)

–

2

2

2

2)

+

2 2)}

2

2 1

=–A g

2

2 This must be equal to the work done ‘by’ gravity on the liquid. Thus the work done ‘by’ gravity is 2

A g

1

2

2 Hence the correct choice is (c). 12. Volume (V) = 30 m3, density of water ( ) = 1000 kg m–3. Therefore, mass of water to be lifted is m=

V = 1000

30 = 3

104 kg

Work done to lift this mass of water to a height = 60 m is W= = 3 104 10 60 = 1.8

107 J

work done by the pump is W 100 1.8 107 100 = W = 30 30 = 6

107 J

Time taken t = 10 min = 600 s. Therefore, power consumed is W 6 107 = = t 600 = 100,000 W = 100 kW Hence the correct choice is (a). dx d = (3t2 + 4t + 5) = 6t + 4. dt dt dv d Acceleration is a = = (6t + 4) = 6 ms–2. dt dt Therefore, applied force is F = ma = 0.5 6 = 3 N. Now t = 2s, the distance moved is

13. Velocity (v) =

x =3

(2)2 + 4

Work done W = Fx = 3 correct choice is (c).

2 + 5 = 25 m 25 = 75 J. Hence the

Work, Energy and Power 6.19

14. Let M kg of water fall per second. The power is = rate at which work is done = mass per second g = But = 1 MW = 106 W, = 10 m. Therefore M=

=

106 = 104 kg s–1, 10 10

15. The weight of the rod acts at the centre of gravity which is at a distance of /2 from the pivoted end. When the rod makes an angle with the horizontal, the vertical height of the centre of gravity is sin 2 i.e. the centre of gravity rises by an amount . Therefore =

1 2

=

sin

Hence the correct choice is (b). 16. Let v 9v plank. Its speed after it passes the plank = . If x 10 is the thickness of the plank, the deceleration a due to the resistance of the plank is given by 9v 10

2ax = v2 –

2

19 v 100

2

=

(i)

Suppose the bullet is stopped after passing through n such planks. Then the distance covered by the bullet is = nx. Thus, we have v2 – 0 = 2 n=

or

= 2anx

v2 v 2 100 100 = = = 5.26 2 2a x 19 19 v

– v2 = 2ax or

2

–

3 4

2

= 2a

3.5

2

cms–2. The bullet will come to 16 rest when its velocity v = 0. If x is the thickness

which gives a =

2a

2

. But a = 2

16

2

2

16

cms–2.

= 8 cm

Hence the correct choice is (a). 18. Since the wood offers a constant deceleration and hence a constant retardation force, the bullet after penetrating a further distance of 3x. Therefore, the total distance penetrated by the bullet before it comes to rest = x + 3x = 4x. Hence the correct choice is (b). 19. Force of friction = mg. Therefore, retardation a = mg/m = g. Also 2ax = v2 or 2am2x = m2v2. But = mv. Therefore, 2 am 2x = 2 But a = g. Therefore, 2

g m 2x =

2

2

or

=

2 g m2 x

. Hence the correct choice is (a).

M . L The mass of the hanging portion of the chain is mL m = . This mass can be assumed to be conn centrated at the centre of the hanging portion of the

20. The mass per unit length of the chain m =

L from the edge 2n of the table. Therefore, the work done in pulling the hanging portion of the chain on to the table top is mL L W = m gx = g n 2n chain which is a distance of x =

Thus the minimum number of planks required is 6. Hence the correct choice is (b). 17. Let cms–1 be the speed of the bullet. Since the mass of the bullet remains unchanged, its speeds 3 cms–1 after it penetrates a distance becomes v = 4 x = 3.5 cm. The retardation a due to the resistance of the wooden plank is given by 2

2

x =

or

Therefore x =

which is choice (b).

PE =

penetrated by the bullet, then 2 – v 2 = 2ax

=

mg L2 2n

2

=

M gL 2n2

Hence the correct choice is (d). 21. Since the platform is depressed by an amount x, the total work done on the spring is mg ( + x). This work is stored in the spring in the form of 1 2 . Equating the two, we potential energy 2 have 1 2m g 2 = mg ( + x) or = 2 x2

6.20 Comprehensive Physics—JEE Advanced

Given, = 0.4 m, x = 0.1 m, m = 1 kg and g = 10 ms–2. Substituting these values, we get = 1000 Nm–1. Hence the correct choice is (c). 22. The instantaneous velocity of the particle is dx d = (3t – 4t2 + t3) = 3 – 8t + 3t2 v= dt dt The instantaneous acceleration of the particle is dv d a= = (3 – 8t + 3t2) = – 8 + 6t dt dt

26. The displacement of the particle is r = r2 – r1 = (14 i + 13 j + 9 k ) – (3 i + 2 j – 6 k ) = (11 i + 11 j + 15 k ) Work done = F r = (11 i + 11 j + 15k ) (4 i + j + 3k ) = 4 11 + 11 + 3 15 = 100 J Hence the correct choice is (b).

4

4

Fd x =

W=

ma

0

0

dx dt dt

4

(– 8 + 6t) (3 – 8t + 3t 2) dt

=m 0 4

(– 24 + 82t – 72t 2 + 18t3) dt

=m 0

9 44 t 2 0 = m (– 96 + 656 – 1536 + 1152) =m

24t

41t 2

10–3 ( m = 3 10–3 kg) = 528 mJ ( 1 mJ = 10–3 J)

= 176 m = 176 10–3

= 528

24t 3

3

The closest choice is (d). 23. The centripetal force, being directed towards the centre is always perpendicular to the direction of displacement which is the direction of the velocity. Thus the dot product F S = 0, i.e. work done is zero, which is choice (a). 1 24. Kinetic energy K = mv 2. If a is the accelera2 tion, then v2 = 2ad. But a = force/mass = F/m. 2F d 1 2F d . Hence K = m Therefore, v2 = m 2 m = Fd, which is independent of m. Thus the correct choice is (b). 25. Work done is

i.e K increases by 0.44 K. The percentage increase 0.44 K in K is K 28. Work done = area under the (F – x) graph =

1 2

0

= 7x

7 2 x 3x d x x3

5 0

= 7 5 – (5)2 + (5)3 = 135 J Hence the correct choice is (d).

4+

1 2 4 = 80 J

Now, work done = increase in kinetic energy. If v is the velocity at x = 2 m, then increase in K.E. = 1 mv 2. Therefore 2 1 80 2 80 2 mv 2 = 80 or v2 = = 2 m 0.1 = 1600 or v = 40 ms–1, which is choice (d). 29. In going from (0, 0) to (a, 0), the x-coordinate varies from 0 to a while the y-coordinate remains zero. Work done by force F along this path is ( y = 0) a

F dx =

W1 = 0

0

x2

4 + 10

a 2

Fdx =

10

10

5

5

W=

2 1 = 27. Momentum = mv or 2 = m2v2 or 2 m 1 mv2. Thus the kinetic energy is 2 2 K = 2m If = 1.2 . Therefore the new kinetic energy will be 2 1.2 2 = 1.44 = 1.44 K K = 2m 2m

( K x j) dx i = 0 0

(

j i = 0)

In going from (a, 0) to (a, a), the x-coordinate remains constant at x = a while the y-coordinate changes from 0 to a.

Work, Energy and Power 6.21

Work done by force F along this path is ( a

F dy =

W2 =

x = a)

=

a

0

a

–K

yi

aj

dy j

1 m 2

2

=

(ii)

0

dy = – Ka2

= – Ka 0

(

i j = 0, j j = 1)

Since work is a scalar quantity, the total work done is W = W1 + W2 = 0 – Ka2 = – Ka2 Hence the correct choice is (c). 30. The power output of the generator is directly proportional to (i) the velocity v of the air molecules in wind and (ii) the average kinetic energy KE = 1 mv2 of the striking molecules (which is propor2 tional v2). Hence Power output v v2 v3 which is choice (c). 31. The velocity of the body at time t is given by dx d t2 t v= = = dt dt 4 2 At t = 0, v = = 0 and t = 2 s, v = 1 ms–1, Now, work done = increase in KE 1 1 1 2 mv2 – = mv2 – 0 = 2 2 2 1 1 = mv2 = 6 (1)2 2 2 = 3 J. Hence the correct choice is (d). 32. Work done = increase in potential energy in (i) raising the weight 150 N of the ladder through a height 1 m and (ii) raising a weight 40 N through 2.5 m = 150 N 1 m + 40 N 2.5 m = 250 Nm = 250 J Hence the correct choice is (b). 33. Friction force = mg = 0.2 5 10 = 10 N. Effective force F = applied force – frictional force = 25 – 10 = 15 N. Kinetic energy = work done by force F in pulling the body through a distance S (= 10 m) = 15 10 = 150 J, which is choice (b). 34. The retardation is given by [see Fig. 6.22] a = g ( cos + sin ) (i) Let be the initial velocity of the body. If it is stopped after moving a distance up the plane, then 2 =2 1 2 Kinetic energy = E = 2

Fig. 6.22

Now, work done is W = gain in PE = It is clear from fore, W=

= sin

sin . There(iii)

From (i), we have g= Also

a ( cos

= tan

or

sin

=

=

sin )

(iv)

sin cos

cos

(v)

Using (iv) and (v) in (iii), we have ma ( cos ) ( cos sin ) Using (ii) in (vi), we get W=

W=

(vi)

E cos , which is choice (d). ( cos sin )

35. Retardation (– a) is proportional to displacement (x), i.e. – a x or a – x. Hence the motion of the particle is simple harmonic. When the displace1 ment is x, the kinetic energy = m 2 (A2 – x2), 2 where m, and A are the mass, angular frequency and amplitude respectively. When displacement 1 x = 0, the kinetic energy = m 2 A2. Therefore, 2 the loss of kinetic energy for a displacement x is 1 1 1 m 2A2 – m 2 (A2 – x2 ) = m 2x2 2 2 2 which is proportional to x2 (since m and are constants of the motion. Hence the correct choice is (a). 36. Power delivered in time t1 is 1 = F v1 = m a v1 v1 Now, acceleration vector is a . Therefore t1 1

=

m v1 v1 t1

mv12 t1

(

v1 v1 = v12)

6.22 Comprehensive Physics—JEE Advanced

Power delivered per unit time = Power delivered at time t =

1

t=

Therefore, the x and y components of acceleration are F 6 = 3 ms–2 ax = x m 2

1

t1 mv12t

t12 t1 Hence the correct choice is (b). Notice that choices (a), (c) and (d) do not have the dimensions of power. 37.

W = F r = 5i

3j

2k

2i

and ay =

Fy

8 = 4 ms–2 2

m

Resultant acceleration a =

j

= = 10 i i

3j j

i j

k i

k j

0

= 10 – 3 = 7 Hence the correct choice is (b). 38. If mass m falls through a distance , mass m rises up through the same distance . Let v be the common velocity of the masses when this happens. Now, loss in PE = gain in KE, i.e. 1 (M + m) v2 – = 2 2 M m which gives v = , which is choice M m (c). 39. When M has descended a distance , loss of PE = . If v is the common velocity of the masses, 1 (M + m) v2. Hence gain in KE = 2

Work done = mg

v v

v

, which is choice (d)

41. Given U = 6x + 8y joule and mass m = 2 kg. Force along x-axis is dU d (6x + 8y) = 6 newton |Fx| = dx dx Force along y-axis is dU d (6x + 8y) = 8 newton |Fy| = dy dy

a 2y

2

4

3

2

= 5 ms–2

The x and y coordinates of the body at time t are 1 1 x = x0 – 3 t2 a x t2 = 6 – 2 2 =

3 2 t 2

6

metre

1 ay t 2 2

y = y0

and

2t 2

= 4

4

1 2

4 t2

metre

The body will cross the y-axis when x = 0, i.e. at 3 2 t = 0 or t = 2 s. Hence the time t given by 6 2 correct choice is (b). 42. vx = ax t = 3 ms–2 2 s = 6 ms–1 vy = ay t = 4 ms–2 2 s = 8 ms–1

1 (M + m) v2 = 2 2Mg , which is choice (c). or v= M m 40. Relative speed of man with respect to escalator = (v – ). Actual displacement of man per second = (v – ). Hence, the actual displacement of man in going up v . Therefore, the escalator of hight is (v )

ax2

v 2x

v=

v 2y

6

2

8

2

= 10 ms–1

Hence the correct choice is (c). 43. Gravitational force is conservative. The work done by a conservative force on a particle moving between two points does not depend on the path taken by the particle. Hence the correct choice is (b). 44. The potential energy of the particle is given by U=–

Fdx = –

(–

+ ax3) dx

x2 x4 x2 = (2 – ax2) (1) a 2 4 4 From Eq. (1) it follows that U = 0 at two values of x which are x = 0 and x = 2 / . Hence graphs (b) and (c) are not possible. Also U is maximum or dU minimum at a value of x given by = 0, i.e. dx 2 d a x4 0= = – ax3 4 dx 2

or

U=

= x( – ax2) or

x=

/ . At this value of x,

Work, Energy and Power 6.23

U is maximum if Now

d 2U dx

=

2

x=

At

d 2U dx

d 2U d x2

d ( dx

< 0,

– ax3) =

– 3ax2.

or

/ ,

=

– 3a

=

– 3 = – 2 , which is negative.

2

a

/ . Hence U is maximum at x = Hence graph (a) is also not possible. Also U is negative for x >

2 / . Therefore, the correct graph

is (d). 45. Mass of the drop m = volume density of water = 4 3 r , where is the density of water. Work done 3 by gravitational force is 4 3 r 3 Thus W r3. Hence the correct choice is (c). 46. At the highest point, the energy of the ball is entirely potential = and at the lowest point, the 1 mv2. Since friction energy is entirely kinetic = 2 is absent, the principle of conservation of energy requires 1 mv2 = 2 W=

1 49. Initial KE = mv2. Now, gain in PE = loss in KE. 2 Thus 1 = mv2 4

=

or v = 2 2 10 correct choice is (b).

0.2 = 2 ms–1. Hence the

=

Hence the correct choice is (d). 50. If a is the deceleration due to the force of friction f, then 2ax = v2 1 or mv 2 = max 2 or KE = f x ( f = ma) Thus if KE is doubled, x is also doubled. Hence the correct choice is (c). 51. Height = 10 m. PE at this height = . On reaching the ground, KE = . Since the body after one impact = 0.75 . If v1 is the initial upward velocity after the impact, we have 1 3 mv 21 = 0.75 = 2 4 v 21 = 1.5

or The height 1

90 = 0.9 100

v=

Hence the correct choice is (a).

2g

1

to which the body will rise is =

1.5 2g

= 0.75

1 mv 22 = 2 . Therefore,

1 mv2 = 0.9 2 = 1.8 10 2 = 36 or v2 = 1.8 which gives v = 6 ms–1. Hence the correct choice is (d). 48. Initial PE = . Now, gain in KE = loss in PE. Thus 1 1 mv2 = 2 2 or

=

v12

= 0.75 10 = 7.5 m ( = 10 m) Hence the correct choice is (c). 52. After the second impact, the initial KE of body = 3 3 2 = , i.e. 4 4

47. PE at A = at point B =

v2 4g

3 4

2

9 8 The height 2 to which the body will rise after the second impact is v 22 =

or

2

=

v22 9 9 = = 2g 8 2g 16

=

9 10 45 = m 16 8

Hence the correct choice is (d). 53. As the body rises, the initial kinetic energy is converted into potential energy. At the maximum height , the energy is entirely potential = , which is equal to the initial kinetic energy. Let its initial value. At this height, the potential energy

6.24 Comprehensive Physics—JEE Advanced

, PE = 0.25

Thus

= 0.25

or

1 2

=

. Hence the correct choice is (c). 4 54. Let f be the force of friction and m be the mass of the body. The retardation a = f/m. If v is the initial speed of the body, then 2ax = v2 1 or max = mv2 = 2 But ma = f. Therefore fx = or f = /x. Hence the correct choice is (a). 55. Potential energy function is x

U(x) =

x

Fd 0

xd 0

2

which givens

=2

2

y 1 = , which is choice (c) x 2

57. Refer to the Fig. 6.23. Here OA = OB = OC = L and OD = OC cos = L cos Therefore = OA – OD = L – L cos

v

1 2 x 2

v

The value of U(x) is always negative for both positive and negative values of x. Thus the variation of potential energy with x is an inverted parabola as shown in choice (a). 56. Potential energy stored in spring S1 when the block 1 2 B is moved through a distance x is U1 = 1x 2 1 2 = . When the block is released, it moves to the 2 left, compressing the spring S2 through a distance y. The potential energy stored in spring S2 when its 1 1 2 (4 ) y2 = 2 2. compression is y is U2 = 2y = 2 2 Since y is the maximum compression of spring S2, from conservation of energy, we have U1 = U2, i.e.

Fig. 6.23

or = L (1 – cos ). From conservation of energy, total energy at A = total energy at C, i.e. 1 1 v mv2 = m 2 2 2

2

+ mgL (1 – cos

)

8 gL (1 – cos ) (1) 3 The minimum velocity the bob must have at A so v2 =

as to reach B is v = 5gL . Putting this in Eq. (1), 7 3 we get cos = – . Therefore lies between 8 4 and

II Multiple Choice Questions with One or More Choices Correct 1. In which of the following is no work done by the force? (a) A man carrying a bucket of water, walking on a level road with a uniform velocity. (b) A drop of rain falling vertically with a constant velocity. (c) A man whirling a stone tied to a string in a circle with a constant speed (d) A man walking up on a staircase. 2. The work done by a force on a body does not depend upon (a) the mass of the body

(b) the displacement of the body (c) the initial velocity of the body (d) the angle between the force vector and the displacement vector 3. A simple pendulum of length L and having a bob of mass M is oscillating in a plane about a vertical line between angular limits – and + . At a time when the angular displacement is (< ), the tension in the string is T and the velocity of the bob is v. Which of the following relations will hold? (a) T cos = Mg (b) T – Mg cos = Mv2/L

Work, Energy and Power 6.25

(c) The magnitude of the tangential acceleration of the bob is |aT| = g sin (d) T = Mg cos 4. A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle. The motion of the particle takes place in a plane. It follows that (a) the velocity of the particle is constant (b) the acceleration of the particle is constant (c) the kinetic energy of the particle is constant (d) the particle moves in a circular path. IIT, 1987 5. Two inclined frictionless tracks of different inclinations meet at A from where two blocks and Q of different masses are allowed to slide down from rest at the same time, one on each track, as shown in Fig. 6.24.

Fig. 6.24

(a) Both blocks will reach the bottom at the same time (b) Block Q will reach the bottom earlier than block (c) Both blocks reach the bottom with the same speed (d) Block Q will reach the bottom with a higher speed than block 6. Choose the correct statements from the following: (a) When a conservative force does positive work on a body, its potential energy increases. (b) When a body does work against friction, its kinetic energy decreases. (c) The rate of change of total momentum of a many–particle system is proportional to the net external force acting on the system. (d) The rate of change of total momentum of many–particle system is proportional to the net internal force acting on the system. 7. Which of the following forces are conservative? (a) Coulomb force between charged particles at rest (b) Force of a compressed elastic spring (c) Gravitational force between two masses (d) Frictional force. 8. A body of m is moving in a straight line at a constant speed v. Its kinetic energy is and the magni-

tude of its momentum is . Which of the following relations is/are correct? (a)

=

2

(c) 2 = v

(b)

=

(d) v =

2 m 2

9. A block of mass m is taken from the bottom of an inclined plane to its top and then allowed to slide down to the bottom again. The length of the inclined plane is L block and the plane is . The inclination of the plane is . (a) The work done by the gravitational force over the round trip is zero. (b) The work done by the applied force over the upward journey is mgL (sin + cos ). (c) The work done by the frictional force over the round trip is zero. (d) The kinetic energy of the block when it reaches the bottom is mgL (sin – cos ). 10. A uniform rod has a mass m and a length . The potential energy of the rod when (a) it stands vertically is zero. (b) it stands vertically is /2 (c) it is inclined at an angle with vertical is 1 cos . 2 (d) it is inclined at an angle with the vertical is 1 sin . 2 11. A body is subjected to a constant force F in newton given by F = – i + 2 j + 3k where i , j and k are unit vectors along x, y and z axes respectively. The work done by this force in moving the body through a distance of (a) 4 m along the z-axis is 12 J. (b) 3 m along the y-axis is 6 J. (c) 4 m along the z-axis and then 3 m along the y-axis is 18 J. (d) 4 m along the z-axis and then 3 m along the y-axis is (12)2 (6) 2 J. 12. Figure 6.25 shows the force F (in newton) acting on a body as a function of x. The work done in moving the body (a) from x = 0 to x = 1 m is 2.5 J. (b) from x = 1 m to x = 3 m is 10 J. (c) from x = 0 to x = 4 m is 15 J. (d) from x = 0 to x = 4 m is 12.5 J.

6.26 Comprehensive Physics—JEE Advanced

Fig. 6.25

13. A block of mass 2 kg, initially at rest on a horizontal g = 10 ms–2 (a) the work done by the applied force in 4 s is 240 J. (b) the work done by the frictional force in 4 s is 96 J. (c) the work done by the net force in 4 s is 336 J. (d) the change in kinetic energy of the block in 4 s is 144 J. 14. In which of the following cases is no work done by the force? (a) A satellite revolving around the earth in a circular orbit. (b) The electron revolving around the proton in a hydrogen atom. (c) A charged particle moving in a circle in a (d) A man carrying a load on his head and walking on a level road at a uniform velocity. 15. A raindrop falls from a certain height above the ground. Due to the resistance of air, its acceleration gradually decreases until it becomes zero when the drop is at half its original height. If W1 and W2 are the amounts of work done by the of the journey, then (b) W1 < W2 (a) W1 > W2 (c) W1 = W2 0 (d) W1 = W2 = 0 16. A particle of mass m is moving in a horizontal circle of radius r, under a centripetal force F = /r2 where is a constant. (a) The kinetic energy of the particle is /2r. (b) The potential energy of the particle is – /2r. (c) The total energy of the particle is – /2r. (d) The total energy of the particle is zero. 17. The displacement x (in metres) of a particle of mass 100 g moving in a straight line under the action of

a constant force is related to time t (in seconds) as x =t–2 (a) The acceleration of the particle is 1 ms–2. (b) The acceleration of the particle is 2 ms–2. (c) The velocity of the particle at t = 3 s is 2 ms–1. (d) The work done by the force in 5 s is 1.8 J. 18. Two springs 1 and 2 have spring constants 1 and 2 ( 1 > 2). W1 and W2 are the amounts of work done to increase the lengths of springs 1 and 2 by the same amount and W3 and W4 are the amounts of work done when springs 1 and 2 are stretched with the same force. Then (b) W1 < W2 (a) W1 > W2 (c) W3 > W4 (d) W3 < W4 19. A box of mass m is dragged along a horizontal surface with a uniform speed with a force F directed at an angle with the horizontal as shown between the box and the surface is . (a) The normal reaction on the box is F cos . R= (b) The normal reaction on the box is R = (mg – F sin ) (c) The work done on the box in dragging it through a distance x is mgx . W= (sin cos ) (d) The work done on the box in dragging it through a distance x is mgx . W= (cos sin )

Fig. 6.26

20. A force F = (3 i + 4 j ) newton acts on a particle located at the origin O. (a) The work done in taking the particle from O along the x-axis to a point A (2 m, 0) is 6 J. (b) The work done in taking the particle from A parallel to the y-axis to a point B (3 m, 2 m) is 8 J. (c) The work done in taking the particle from O to A and then to B is 14 J.

Work, Energy and Power 6.27

(d) The work done in taking particle from O directly to B is 10 J. 21. The potential energy of a system varies with distance x as U = ax2 – bx where a and b are positive constants. Then (a) the potential energy is minimum at x = b/2a

(c) T1 = (d) T1 =

mg , T2 = 2

3mg 2

3mg mg , T1 = 2 2

b2 . 4a (b) the potential energy is maximum at x = b/2a and Umin = –

b2 . 4a (c) The force acting on the system decreases linearly with x. (d) The force acting on the system is proportional to x2. IIT, 2005 22. A car is of mass m moving along a circular track of radius r with a speed which increases linearly with time t as v = , where is a constant. Then (a) the instantaneous power delivered by the 3 3 t /r. centripetal force is (b) the power delivered by the centripetal force is zero. (c) the instantaneous power delivered by the 2 t. tangential force is (d) the power delivered by the tangential force is zero. 23. A body of weight mg is suspended from a rigid support by two light strings AB and AC as shown in Fig. 6.27. The tension in string AB is T1 and in string AC the tension is T2. Then and Umax =

(a) T1 =

3 T2

(b) T2 =

3 T1

Fig. 6.27

24. The potential energy function of a particle executing linear simple harmonic motion is given by 1 2 U (x) = 2 where x is the displacement of the particle from the equilibrium position x = 0 and is the force constant of the oscillator. Figure 6.28 shows the graph of U(x) against x for = 0.5 Nm–1. If the total energy of the particle is 1 J, it will turn back when it reaches the position (a) x = – 1 m (b) x = – 2 m (c) x = + 2 m (d) x = + 1 m

Fig. 6.28

ANSWERS AND SOLUTIONS 1. In choices (a) and (c), no work is done because the force of gravity in choice (a), and the centripetal force in choice (c) are perpendicular to the direction of the displacement. In choice (b), no net force acts on the raindrop since it is falling with a uniform velocity, hence no work is done. In choice (d), the man has to do work against gravity. Hence the correct choices are (a), (b) and (c). 2. The correct choices are (a) and (c). 3. Referring to Fig. 6.29, the net force along the string is F = T – Mg cos and this force provides the centripetal force Mv2/L necessary for circular motion of the bob. Hence choice (b) is correct.

Relation (a) i.e. T cos = Mg cannot hold for all values of because if = 0, then T = Mg and the net force along the string is F = T – Mg cos 0 = T – Mg = 0 Since F = 0, there would be no centripetal force and the bob would not oscillate. Hence choice (a) is incorrect. The tangential acceleration aT of the bob is caused by the tangential component Mg sin . Therefore, aT = Mg sin /M = g sin . Hence choice (c) is correct. The relation (d) cannot hold because if T = Mg cos , the net force along the string will be zero. Therefore, there will be no centripetal force.

6.28 Comprehensive Physics—JEE Advanced

Hence the correct choices are (b) and (c).

Fig. 6.30 Fig. 6.29

4. The particle moves in a circular path with a uniform speed, because the force has a constant magnitude and is perpendicular to the velocity of the particle. The velocity and acceleration of the particle are not constant because their directions are changing as the particle moves in a circular path. The magnitude of the velocity (i.e. speed) is 1 mv2 is constant. Therefore, the kinetic energy 2 constant. Hence choices (c) and (d) are correct and choices (a) and (b) are incorrect. 5. Refer to Fig. 6.30. The accelerations of blocks and Q are m g sin a1 = 1 m1

1

= g sin

1

m2 g sin 2 = g sin 2 m2 Since 2 > 1; a2 > a1. Now PE of block P at A = 1 m1 . Its KE on reaching the bottom = m1v 21. 2 Equating the two we get 1 m1v 21 = m1 2

and

or

a2 =

v1 =

6. The work done by a conservative force is equal to the negative of the potential energy. When the work done is positive, the potential energy decreases. Thus choice (a) is incorrect. Friction always opposes motion. Hence, when a body does work against friction, its kinetic energy decreases. Thus choice (b) is correct. The rate of change of total momentum of a many–particle system is proportional to the net force external to the system; the internal forces between particles cannot change the momentum of the system. Hence the correct choices are (b) and (c). 7. The correct choices are (a), (b) and (c). 2

8. Now

= mv or

2

= m2v2 or

2m

=

1 mv2 = 2

or

1 mv2. 2 Dividing the two we get 2 = v. Hence the correct relations are (a) and (c). 9. The gravitational force is conservative. Therefore, the work done by the gravitational force over the round trip is zero. Refer to Fig. 6.31. =

2

. Also

= mv and =

2

Similarly, for block Q, v2 = 2 . Since v1 = v2, both blocks will reach the bottom with the same = 0) and v2 = a2t2. But speed. Now, v1 = a1t1 ( v1 = v2. Therefore a 1t 1 = a 2t 2 t1 a = 2 t2 a1 Since a2 > a1; t1 > t2; i.e. block takes a longer time to reach the bottom. Hence the correct choices are (b) and (c).

Fig. 6.31

When the block is moved from C to A, the force of friction f acts along the plane in the downward direction which is in the direction of the component mg sin of the gravitational force. Hence the applied force F is F = mg sin + f

Work, Energy and Power 6.29

frictional force f f = = normal reaction R mg cos f = mg cos Therefore F = mg sin + mg cos = mg(sin + cos ) Work done by the applied force over the upward journey is Wa = F L = mg(sin + m cos ) L The frictional force is non-conservative. Hence the work done by the frictional force over the round trip is not zero. When the block is at a point A, it is at rest and its initial velocity = 0. It is allowed to slide down the plane. Let v be the velocity when it reaches the bottom C of the plane. Since the frictional force now acts upwards, the net force acting on the block when it slides down is Fn = mg sin – f = mg sin – mg cos Now

=

= mg(sin

–

cos )

fn m = g(sin – cos ). The velocity v is given by v2 – 2 = 2aL (where L = AC) or v2 = 2aL ( = 0) Kinetic energy, 1 KE = mv2 2 1 = m 2AL = maL 2 or KE = mg(sin – cos )L Acceleration of the block, a =

Hence the correct choices are (a), (b) and (d). 10. The potential energy in the vertical position = work done in raising it from horizontal position to vertical position. In doing so, the mid-point of the rod is raised through a height = /2. Since the entire mass of the rod can be assumed to be concentrated at the mid-point (centre of gravity), the work done = = /2. Refer to Fig. 6.32. AD = AB = . In the inclined position, let the centre of gravity C of the rod be at a height above the ground, so that AC = /2. In triangle ACE, Fig. 6.32 we have = AC sin(90° – ) =

2

cos

PE =

=

1 2

cos . The correct choices are

(b) and (c). 11. The correct choices are (a), (b) and (c). Displacement along the z-axis is S = 4 k metres. Therefore, work done is W1 = F S = (– i + 2 j + 3 k ) (4 k ) = – 4 i k + 8 j k + 12 k k Now i k = 0 and j k = 0 because j and j are perpendicular to k . But k k = 1. Therefore, W1 = 12 J. Displacement along the y-axis is S = 3 j metres. Therefore, the work done is W2 = F S = (– i + 2 j + 3 k ) (3 j ) = 6 j j = 6 J Since work is a scalar, the total work done is just the algebraic sum of W1 and W2, i.e. W = W1 + W2 = 12 + 6 = 18 J. 12. The correct choices are (a), (b) and (c). Work done = area under the F–x graph. Work done by the force in moving the body from x = 0 to x = 1 m is (see Fig. 6.33) W1 = area of triangle OAD 1 Fig. 6.33 = AD OD 2 1 = 5 N 1m = 2.5 J 2 Hence the correct choice is (a). Work done in moving the body from x = 1 m to x = 3 m is W2 = area of rectangle ABDE = AD DE = 5 N 2 m = 10 J, which is choice (c) Work done in moving the body from x = 0 to x = 4 m is W3 = area of (triangle OAD + rectangle ABDE + triangle BEC) = 2.5 + 10 + 2.5 = 15 J, which is choice (d). 13. The correct choices are (a), (b) and (d). Force of friction (f) = mg = 0.2 2 10 = 4N. Applied force (F) = 10 N. Since friction opposes motion, the net force acting on the body when it is moving is

6.30 Comprehensive Physics—JEE Advanced

dv d = (2t – 4) = 2 ms–2 dt dt Now F = ma = 0.1 kg 2 ms–2 = 0.2 N Now distance moved in t = 5 s = (5)2 – 4 5 + 4 = 9 m. Work done W = 0.2 N 9 m

F = F – f = 10 – 4 = 6 N 6N F Acceleration a = = = 3 ms–2 M 2 kg The distance travelled by block in 4s is 1 2 1 at = 0 + 3 (4)2 = 24 m S= + 2 2 Work done by applied force in 4 s is

= 1.8 Nm or J

W = applied force distance moved in 4 s = 10 N 24 m = 240 J Work done by the force of friction in 4 s is W = f S = 4 N 24 m = 96 J. Work done by the net force in 4 s is W = F = 6 N 24 m = 144 J Velocity acquired by the block in 4 s is v = + at = 0 + 3 4 = 12 ms–1 Kinetic energy of the block at t = 4 s is 1 1 mv2 = 2 (12)2 = 144 J KE = 2 2 Since the initial KE = 0, the change in KE = 144 J. 14. All the four choices are correct. In choices (a), (b) and (c), the centripetal force (being radial) is perpendicular to the velocity (and hence displacement) vector. In choice (d), the gravitational force (being vertically downwards) is perpendicular to the displacement. 15. The correct choice is (c). Work done is each half of the journey = where = H/2; H being the original height from which the drop fell. 16. The correct choices are (a) and (c). mv 2 = 2 or mv2 = . Therefore, F= r r r KE =

1 2 18. If the increase in length is x, W1 = 1x and 2 W1 1 2 W2 = = 1 . Since 1 > 2; 2x . Therefore, W2 2 2 W1 is greater than W2. Let a force F x1 and the x second by x2. Then F = 1x1 = 2 x2 or 1 = 2 . x2 1 1 1 2 2 Now W3 = x and W = x . Therefore, 1 1 4 2 2 2 2 2

PE =

=–

r

Fdr =

r

r

2

dr =

1r =– r r

dr Fig. 6.34

r2

Work done W = Fx Thus the correct choices are (a), (b) and (d). 20.

Total energy = KE + PE =

2

W3 x1 2 = 1 = 1 = 2 < 1. x2 W4 2 2 1 1 Thus the correct choices are (a) and (d). 19. The different forces acting on the block are shown in Fig. 6.34. It follows that R = F cos and R + F sin = mg Eliminating R we get mg F= cos sin

1 mv2 = 2 2r r

Now

a=

Acceleration

–

=–

2r r 2r 17. The correct choices are (b), (c) and (d). x = (t – 2)2 = t2 – 4t + 4 dx d 2 v= = (t – 4t + 4) = 2 t – 4 dt dt

WO

= (3 i + 4 j ) (2 i ) = 6 J

WO

A

WA

B

A

B

= (3 i + 4 j ) (2 j ) = 8 J = 6 + 8 = 14 J

WO

B

= (3 i + 4 j ) (2 i + 2 j ) = 6 + 8 = 14 J

Hence the correct choices are (a), (b) and (c) dU d 2U = 0 and > 0. Now 21. U is minimum if dx dx 2

Work, Energy and Power 6.31

dU d dU = (ax2 – bx) = 2ax – b. So = 0 at x dx dx dx d 2U d = b/2a. Also = (2ax – b) = 2a, which is 2 dx dx

Equations (1) and (2) give T1 = =

mg and T2 = 2

3 T1

3mg . Thus the correct choices are (b) and (c). 2

positive. Hence choice (a) is correct and choice (b) is wrong. For x = b/2a, b 2 b b2 –b =– . 4a 2a 2a Force acting on the system is dU F= – = b – 2ax dx Hence choice (c) is correct and choice (d) is wrong. Umin = a

Fig. 6.35

24. At any instant during its motion, the energy of the particle is partly kinetic and partly potential, but the total energy remains constant. If v is the velocity of the particle and x its displacement from the mean position at an instant of time, then at that instant the total energy is 1 1 2 E = K.E. + P.E. = mv2 + 2 2 where m is the mass of the particle. The particle will turn back when v = 0 momentarily. At

2 2 mv 2 t = . Since Fc is r r = Fc v = 0. Tangential force

22. Centripetal force Fc =

perpendicular to v, md v d Ft = = m ( ) = . Since Ft is parallel to dt dt v, = Ft v = = 2t. Hence the correct choices are (b) and (c). 23. and

T1 sin 60° = T2 sin 30°

(1)

T1 cos 60° + T2 cos 30° = mg

(2)

1 2E 2 1 2 x= = 2 0.5 = ± 2 m. Hence the correct choices are (b) and (c). that moment E =

III Multiple Choice Questions Based on Passage Questions 1 to 3 are based on the following passage Passage I Invariance Newton’s laws of motion are applicable in all inertial reference frames. Some physical quantities, when measured by observers in different reference frames, have exactly the same value. Such physical quantities are called ant. In Newtonian mechanics mass, time, and force are invariant quantities. On the other hand, some physical quantities, when measured by observers in different reference frames, do not have the same value. Such physical quantities are called not invariant. In Newtonian mechanics displacement, velocity and work (which is the dot product of force and displacement) are not invariant. Also kinetic 1 2 mv is not invariant. energy 2

Physicists believe that all laws of physics are invariant in all inertial reference frames. For example, the workenergy principle states that the change in the kinetic energy of a particle is equal to the work done on it by the force. Although, work and kinetic energy are not invariant in all reference frames, the work-energy principle remains invariant. Thus even though different observers measuring

energy principle holds in their respective frames. 1. Choose the invariant quantities from the following (a) mass (b) time (c) velocity (d) displacement 2. Which of the following quantities is/are not invariant (a) Work (b) Kinetic energy (c) Torque (d) Displacement

6.32 Comprehensive Physics—JEE Advanced

3. Choose the correct statements form the following. (a) Kinetic energy is not invariant. (b) Potential energy is not invariant.

(c) Laws of conservation of energy and momentum are invariant. (d) All laws of physics are invariant.

ANSWERS 3. The correct choices are (c) and (d).

1. The correct choices are (a) and (b). 2. All choices are correct. Questions 4 to 8 are based on the following passage Passage II Mechanical energy exists in two forms: kinetic energy and potential energy. Kinetic energy is the energy possessed by a body by virtue of motion. Potential energy is the energy These two forms of energy are inter-convertible. If no other form of energy is involved in a process, the sum of kinetic energy and potential energy always remains constant. 4. Two particles of masses m1 and m2 have equal linear momenta. The ratio of their kinetic energies is (a) 1

(b)

m2 m1 2

m1 m2

SOLUTION

(d)

2 1 1 mv2 = (mv)2 = , 2m 2 2m = mv is the linear momentum. Thus 2

K1 =

2m1

2

and K2 =

(b) 1 : 2

(c) 1 : 3 (d) 1 : 4 7. In Q.6, what is the original speed of the heavier particle? (a) 2(1 +

2 ) ms–1

(b) 2(1 –

2 ) ms–1

(d) (2 2 – 1) ms–1

8. A uniform rod of mass m and length is made to stand vertically on one end. The potential energy of the rod in this position is mg mg (b) (a) 4 3 (c)

mg 2

(d)

2

m1 m2

4. Kinetic energy K = where

(a) 1 : 1

(c) (2 2 + 1) ms–1

m2 m2 (c) (d) m1 m1 5. Two particles of masses m1 and m2 have equal kinetic energies. The ratio of their linear momenta is m1 (a) 1 (b) m2 (c)

6. A particle of mass m has half the kinetic energy of another particle of mass m/2. If the speed of the heavier particle is increased by 2 ms–1, its new kinetic energy equals the original kinetic energy of the lighter particle. The ratio of the original speeds of the lighter and heavier particles is

2m2

K1 m = 2 , which is choice (c). K2 m1 = 2mK . Hence the correct choice 5. K = 2/2m is (b). 6. Let v and v be the original speeds of the heavier and the lighter particles respectively. We then have 1 1 1 m mv2 = v2 2 2 2 2 1 v2 = v or v = 2v 4

Hence the correct choice is (b). 7. When the heavier particle is speeded up by 1 2.0 ms–1, its kinetic energy becomes m(v + 2)2. 2 Since this equals the original kinetic energy of the lighter particle, we have 1 1 m(v + 2)2 = (m/2)(4v2) 2 2 2 v + 4 + 4v = 2v2 or v2 – 4v – 4 = 0 v= =

4

16 16 2

4 2 8 = 2 2 2 2

The positive root is v = 2+ 2 2 = 2(1 + Hence the correct choice is (a).

2 ).

Work, Energy and Power 6.33

8. The potential energy in the vertical position = work done in raising it from horizontal position to vertical position. In doing so, the mid-point of the rod is raised through a height = /2. Since the entire Questions 9 to 11 are based on the following passage Passage III A light rod of length L having a body of mass M attached to its end hangs vertically. It is turned through 90° so that it is horizontal and then released. 9. The centripetal acceleration when the rod makes an angle with the vertical is (a) g cos

(b) 2g cos

(c) g sin

(d) 2g sin

mass of the rod can be assumed to be concentrated at the mid-point (centre of gravity), the work done = = /2. Hence the correct choice is (c).

10. The tension in the rod when it makes an angle with the vertical is (a) Mg cos (b) 2 Mg cos (c) 3 Mg cos (d) zero 11. The value of when the tension in the rod equals the weight of the body is given by 1 1 (b) = cos–1 (a) = cos–1 2 3 1 2

= sin–1

(c)

(d)

= sin–1

1 3

SOLUTION 9. The rod is released from the horizontal position OA. Let OB be the position of the rod when tension in the rod is T (Fig. 6.36).

v 2 2gL cos = L L = 2g cos ,

centripetal acceleration =

which is choice (b). 10. The centripetal force when the body is at B is M v2 Fc = L Thus, we have M v2 (2) T – Mg cos = L Using (1) in (2), we get Fig. 6.36

Let be the angle with the vertical at this position. The loss of PE when the body falls from A to B = Mg OC = MgL cos . If v is the velocity of the body at B, then 1 Mv2 = MgL cos 2

or v2 = 2gL cos

(1)

Questions 12 to 14 are based on the following passage Passage IV A small roller coaster starts at point A with a speed on a curved track as shown in Fig. 6.37. The friction between the roller coaster and the track is negligible and it always remains in contact with the track.

M 2gL cos = 2 Mg cos L or T = 3 Mg cos Thus the correct choice is (c). 11. T = Mg. Therefore, 1 Mg = 3 Mg cos or cos = , which is choice (b). 3 T – Mg cos

=

12. The speed of the roller coaster at point B on the track will be 1/ 2 2 2 (b) (a) ( 2 + )1/2 3 (c) (

2

)1/2

+2

2

(d)

2 13. The speed of the roller coaster at point C on the track will be 1/ 2

(a)

Fig. 6.37

1/ 2

3

(c)

2

3 2

2

(b)

1/ 2

2 3

1/ 2

4 3

(d) (

2

+2

)1/2

6.34 Comprehensive Physics—JEE Advanced

14. The speed of the roller coaster at point D on the track will be

SOLUTION 1 2 + . If vb 2 1 is the speed at point B, the total energy at B = mv2b 2 + mg(2 /3). From the principle of conservation of energy, we have 1 1 2mg 2 + = mv2b + 2 2 3

12. Total energy at A = KE + PE =

which gives vb =

2

1/ 2

2 3

(a) (

2

+

(c) (

2

+3

Questions 15 to 17 are based on the following passage Passage V The displacement x of a particle moving in one dimension, under the action of a constant force is related to time t by the equation t = x +3 where x is in metre and t is in second. 15. The displacement of the particle when its velocity is zero is

SOLUTION 15. Given t =

x + 3 or x = t – 3 or (1) x = (t – 3)2 Differentiating (1) with respect to t, we get dx = 2 (t – 3) dt or v = 2 (t – 3) (2) From (2) it follows that v = 0 at t = 3 s. Using t = 3 s in (1), we get x = 0. Thus, the displacement of the particle is zero when its velocity is zero. Thus the correct choice is (a). 16. From Eq. (2), we have dv d = [2(t – 3)] = 2 ms–2. a = dt dt Hence the correct choice is (d). Questions 18 to 20 are based on the following passage Passage VI The work done by a constant force acting on a body is given by W=F r where F is the force vector and r is displacement vector. The displacement vector r = r2 – r1 where r1 is the initial

)1/2

(b) (

2

+2

(d)

(

2

+4

)1/2 )1/2

13. Similarly, the speed at point C is given by 1 1 mg 2 + = mv2c + which gives 2 2 3 vc =

4

2

1/ 2

, which is choice (c). 3 14. At point D, the energy is entirely kinetic. If the speed of the roller coaster at point D is vd, then we have

,

which is choice (b).

)1/2

or

1 mv2d = 2 vd = (

(a) zero

+ 2

1 2

2

)1/2, which is choice (b).

+2

(b) 1 m

(c) 2 m

(d) 3 m

16. The acceleration of the particle (a) increases with time (b) decreases with time (c) increases with time up to t = 3 s and then decreases with time. (d) remains constant at 2 ms–2. 17. (a) 1 J

(b) 3 J

(c) 6 J

(d) zero

17. From Eq. (2), the initial velocity, i.e., velocity at t = 0 is v0 = 2(0 – 3) = – 6 ms–1 Final velocity, i.e., velocity at t = 6 s is v = 2(6 – 3) = 6 ms–1

=

1 1 1 mv2 – mv20 = m(v2 – v02 ) 2 2 2

=

1 m[(6)2 – (– 6)2] = 0, which is choice (d). 2

position vector and r2 If the force is variable, the work done in moving a body from a position r1 to a position r2 is given by r2

F dr

W = r1

where dr

Work, Energy and Power 6.35

20. A body of mass m is projected from the ground with a velocity at an angle above the horizontal. The work done by the gravitational force in time sin t= is g

18. A particle is moved from a position r1 = (3 i + 2 j – 4 k ) metre to a position r2 = (5 i + 6 j + 9 k ) metre under the action of a force F = ( i + 3 j + k ) newton. The work done is (a) zero

(b) 13 J

(c) 27 J

(d) 35 J

(1 + cos )

(c)

(b)

sin2

(b)

2

sin2

2

sin 2 (d) zero 2 21. If = 45°, the work by the gravitational force in 2 sin is Q.20 above in time t = g (c)

19. A body of mass m is projected from a tower of height at an angle above the horizontal. The work done by the gravitational force during the time it takes to hit the ground is (a)

2

(a) 2

2

(1 + sin )

(a)

(d) zero

2

(b)

4

2 (d) zero

2

(c)

SOLUTION 18. W = F r = F (r2 – r1)

Work done W = sin 20. t = g

= ( i + 3 j + k ) [(5 i + 6 j + 9 k ) – (3 i + 2 j – 4 k )]

, which is choice (c).

the body attains maximum height

= ( i + 3 j + k ) (2 i + 4 j + 13 k )

2

max

=

sin 2 2g

2 sin 2 = which is Work done = max 2 choice (c). 21. Net displacement = 0. Hence W = 0. Thus the correct choice is (d).

= 2 + 12 + 13 = 27 J, which is choice (c) 19. Net displacement from the time the body is projected to the time it hits the ground is = vertically downwards.

IV Matching 1.

Column I (a) (b) (c) (d)

Force Impulse Energy stored in a spring Force constant of a spring

ANSWER (a) (s) (c) (r)

.

Column II (p) (q) (r) (s)

Slope of force-extension graph Area under force-time graph Area under force-extension graph Slope of linear momentum-time graph

(b) (d)

(q) (p)

6.36 Comprehensive Physics—JEE Advanced

V Assertion-Reason Type Questions In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each questions has the following four choices out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true but Statement-2 is the correct explanation for Statement-1. (c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. 1. Statement-1 A simple pendulum of length is displayed from its mean position O to position A so that the string makes an angle 1 with the vertical and then released. If air resistances is neglected, the speed of the bob when the string makes an angle 2 with the vertical is v =

2 (cos

2

cos 1 ) .

Statement-2 The total momentum of a system is conserved if no external force acts on it. 2. Statement-1 A uniform rod of mass m and length is held at an angle with the vertical. The potential energy of 1 cos . the rod in this position is 2 Statement-2 The entire mass of the rod can be assumed to be concentrated at its centre of mass. 3. Statement-1 A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of 30° with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in the mechanical energy in the second situation is smaller than that in the first situation. Statement-2 the surface decreases with the increase in the angle of inclination. IIT, 2007

4. Statement-1 A man carrying a bucket of water and walking on a rough level road with a uniform velocity does no work while carrying the bucket. Statement-2 The work done on a body by a force F in giving it a displacement S W = F S = FS cos where is the angle between vectors F and S. 5. Statement-1 A crane lifts a car up to a certain height in 1 min. Another crane Q lifts the same car up to the same height in 2 min. Then crane consumes two times more fuel than crane Q. Statement-2 Crane supplies two times more power than crane Q. 6. Statement-1 Two inclined frictionless tracks of different inclinations 1 and 2 meet at A from where two blocks and Q of different masses m1 and m2 are allowed to slide down from rest, one on each track as shown in Fig. 6.38. Then blocks and Q will reach the bottom with the same speed.

Fig. 6.38

Statement-2 Blocks and Q have equal accelerations down their respective tracks. 7. Statement-1 In Q.6 above, block will take a longer time to reach the bottom than block Q. Statement-2 Block Q has a greater acceleration down the track than block . 8. Statement-1 Comets move around the sun in highly elliptical orbits. The work done by the gravitational force of the sun on a comet over a complete orbit is zero.

Work, Energy and Power 6.37

Statement-2 The gravitational force is conservative. 9. Statement-1 The total energy of a system is always conserved irrespective of whether external forces act on the system. Statement-2 If external forces act on a system, the total momentum and energy will increase. 10. Statement-1 The rate of change of the total linear momentum of a system consisting of many particles is proportional to the vector sum of all the internal forces due to inter-particle interactions.

Statement-2 The internal forces can change the kinetic energy of the system of particles but not the linear momentum of the system. 11. Statement-1 An elastic spring of force constant is stretched by a small length The work done in extending the spring by a further length x is 2 2. Statement-2 The work done in extending an elastic spring by a length x is proportional to x2.

SOLUTIONS 1. The correct choice is (b). It is clear from Fig. 6.39 = cos 2. Therefore, 1 = that = cos 1 and – cos 1 = (1 – cos 1) and 2 = (1 – cos 2). Let m be the mass of the bob and v be its speed when it reaches position B. Then, from the principle of conservation of energy, K.E. at B = loss of P.E. as the bob moves from A to B. Hence 1 mv2 = 1– 2 2 = mg[ (1 – cos 1) – (1 – cos 2)] = (cos 2 – cos 1) v=

2 (cos

2

cos 1 )

Fig. 6.39

2. The correct choice is (a). Let C be the centre of mass of the rod AB so that AC = /2. Let be the height of C above the ground. In triangle ACD, we have CD = AC sin (90° – ) (see Fig. 6.40). Or

=

cos . Since the en2 tire mass of the rod can be assumed to be concentrated at the centre of mass, therefore, potential energy = work done

Fig. 6.40

to raise the rod from horizontal position on the ground to the position show = 1 cos . 2 3. Statement-1 is true. The decrease in mechanical energy is smaller when the block made to go up on the inclined surface because some part of the kinetic energy is converted into gravitational potential friction does not depend on the angle of inclination of the plane. Hence the correct choice is (c). 4. The correct choice is (a). Since the velocity is uniform, the man exerts no net force on the bucket in the direction of motion. The only force he exerts on the bucket is against gravity (to overcome) the weight mg of the bucket) and this force is perpendicular to the displacement (i.e. = 90°). Hence W = FS cos 90° = 0. 5. The two cranes do the same amount of work = . Hence they consume the same amount of fuel. Crane does the same amount of work in half the time. Hence crane supplies two times more power than crane Q. Thus the correct choice is (d). 6. The acceleration of blocks and Q respectively are m g sin 1 a1 = 1 = g sin 1 m1 and

a2 =

m2 g sin m2

2

= g sin

2

Since 2 > 1; a2 > a1. The potential energy of block at A = m1 . When it reaches the bottom B, its 1 kinetic energy is m1v21 where v1is its speed when 2 it reaches B. Now P.E. at A = K.E. at B. Hence 1 v1 = 2 . m1 = m1v21 2

6.38 Comprehensive Physics—JEE Advanced

9. Statement-1 is false; the total energy of an isolated system is conserved. Statement-2 is true. Hence the correct choice is (d). 10. Statement-1 is false and Statement-2 is true. The rate of change of momentum is proportional to the net external force acting on the system. Hence the correct choice is (d). 11. The correct choice is (d). Potential energy stored in 1 2 the spring when it is extended by x is U1 = 2 Potential energy stored in the spring when it is further extended by x is 1 U2 = (x + x)2 = 2 2 2 Work done = gain in potential energy = U2 – U1 1 2 3 2 =2 2– = 2 2

1 m2v22 v2 = 2 = v 1. 2 Hence the correct choice is (c). 7. The correct choice is (a). If t1 and t2 are the times taken by and Q to reach the bottom, then Similarly

m2

=

v1 =

1

+ a 1t 1 = a 1t 1

(

1

= 0)

and

v2 =

2

+ a 2t 2 = a 2t 2

(

2

= 0)

Now

v1 = v2. Hence a1t1 = a2t2. Thus

t1 a = 2 t2 a1 Since a2 > a1; t1 > t2. 8. The correct choice is (a). For a conservation force, the work done in moving a body from one point to another does not depend on the nature of the path and the work done over a closed path is zero, irrespective of the nature of the path.

VI Integer Answer Type 1. A particle of mass 1g executes an oscillatory motion on the concave surface of a spherical dish of radius 2 m placed on a horizontal plane. If the motion of the particle begins from a point on the dish at a height of 2 cm from the horizontal plane and the in metre covered by the particle before it comes to rest. IIT, 1990 2. A light inextensible string that goes in Fig. 6.41 connects two blocks of masses 0.36 kg and 0.72 kg. Taking in g = 10 m/s2 joules) by the string on the block

system is released from rest. IIT, 2009 3. A block of mass 0.18 kg is attached to a spring of

block is at rest and the spring is unstretched. An impulse is given to the block as shown in Fig. 6.42. The block slides a distance of 0.06 m and comes to rest for Fig. 6.42

initial velocity of the block in m/s is V = N/10. Then N is

IIT, 2011

Fig. 6.41

SOLUTION

1. Refer to Fig. 6.43 in which is the angle which the normal to the surface at the location of the particle makes with the verti-

cal. The particle will keep on oscillating about O till its initial potential energy is completely used up in doing work against the frictional force f and the particle comes to rest. It follows from the f= Since Fig. 6.43

mg cos

<< R, 0. Therefore, cos f = mg

1. Hence

Work, Energy and Power 6.39

If is the distance travelled by the particle before it comes to rest, then the work done against friction is W= = Now, = or

=

=

3. Refer to Fig. 6.45.

2 cm = 200 cm = 2m 0.01

2. Refer to Fig. 6.44. T – m1 g = m1 a m2 g – T = m2 a which give a=

(m2 m1 ) g (m1 m2 )

Fig. 6.45

10 m s–2 3 T = m1(a + g) =

and

Fig. 6.44

10 10 = 4.8 N 3 Distance moved in t = 1 s is 1 1 10 = + at2 = 0 + (1)2 2 2 3 5 = m 3 5 Work done = T S = 4.8 =8J 3 = 0.36

1 mV 2 2 Work done against friction = mgx 1 2 Gain in potential energy = 2 From work-energy principle, 1 1 2 mV 2 = mgx + 2 2 Loss of kinetic energy =

(0.72 0.36) 10 = (0.72 0.36)

1 2

0.18

V2 = 0.1

0.18 +

V = 0.4 =

10 1 2

0.06 2

(0.06)2

4 ms–1. Hence N = 4. 10

7

Chapter

Conservation of Linear Momentum and Collisions

REVIEW OF BASIC CONCEPTS

7.1

LAW OF CONSERVATION OF LINEAR MOMENTUM

The law of conservation of linear momentum may be stated as ‘when no net external force acts on a system consisting of several particles, the total linear momentum of the system is conserved, the total linear momentum being the vector sum of the linear momentum of each particle in the system’. According to Newton’s second law of motion, we have for a system of particles, dp Fext = dt where p = p1 + p2 + p3 + = m1 v1 + m2 v2 + m3 v3 + In a system consisting of many particles, there can be two kinds of forces: (i) internal forces and (ii) external forces. Internal forces are the forces that the particles exert on each other during their interactions (e.g. collisions). From Newton’s third law, these forces always occur in pairs of action and reaction. Since these forces are equal and opposite, they bring about equal and opposite changes in the momentum of the particles. Thus, internal forces cannot bring about any net change in the momentum of a particle. The external forces, on the other hand, are the forces exerted from outside the system. If Fext = 0, we have

or

dp =0 dt p = constant

Thus, the vector sum p = p1 + p2 + p3 + of the linear momenta of the particles remains constant if the net external force is zero.

7.2

RECOIL OF A GUN

The gun and the bullet constitute a two-body system. at rest. Therefore, the total momentum of the gun-bullet forward and the gun recoils backwards. Let mb and mg be the masses of the bullet and the gun. If vb and vg are their mb vb + mg vg). From the law of conservation of momentum, the total momene the same, i.e. mb vb + mg vg = 0 m v or vg = – b b mg The negative sign indicates that the gun recoils in a direction opposite to that of the bullet. In terms of magnitudes, we have mv vg = b b mg

7.3

COLLISIONS

Elastic Collisions: If there is no change of kinetic energy during a collision it is called an elastic collision. The collision between subatomic particles is generally elastic. The collision between two steel or glass balls is nearly elastic. Inelastic Collisions: If there is a loss of kinetic energy during a collision, it is called an inelastic collision. Since there is always some loss of kinetic energy in any collision, collisions are generally inelastic. If the loss is negligibly small, the collision is very nearly elastic. Perfectly elastic collisions are not possible. If two bodies stick together, after colliding, the collision is perfectly inelastic, e.g. a bullet striking a block of wood and being embedded

7.2 Comprehensive Physics—JEE Advanced

in it. The loss of kinetic energy usually results in heat or sound energy. In may be remembered that the total momentum remains conserved in both elastic and inelastic collisions. Further, since the interacting forces become effectively zero after the collision, the potential energy remains the same both before and after the collision.

One-dimensional or Head-on Collision Consider two bodies of masses m1 and m2 moving with velocities u1 and u2 in the same straight line (with u1 > u2) colliding with each other. Let v1 and v2 be their respective velocities after the collision. If velocities u1, u2, v1 and v2 are all along the same straight line, the collision is known as one-dimensional or head-on collision (Fig. 7.1)

v

v

Fig. 7.1

From the law of conservation of momentum m 1 u 1 + m 2 u 2 = m 1v 1 + m 2 v 2

Coefficient of Restitution

If momentum along positive x-axis is taken to be positive, the momentum along the negative x-axis is taken to be negative. Two-dimensional or Oblique Collision If the velocities of the colliding bodies are not along the same straight line, the collision is known as two-dimensional or oblique collision (Fig. 7.2) v

Newton proved experimentally that, when two bodies collide, the ratio of the relative velocity after collision to the relative velocity before collision is constant for the two bodies. This constant is known as restitution and is denoted by letter e. velocity of separation after collision e =– velocity of approach before collision

(i) (ii) (iii)

(iv) v

v 2 v1 u 2 u1 For a perfectly elastice collision, e = 1. For a perfectly in inelastic collision, e = 0, because the two bodies stick together and hence v2 = v1. Perfectly elastic or perfectly inelastic collisions do not occur in nature. Hence, for any collision, e lies between 0 and 1. For a head-on collision (Fig. 7.1) velocity of separation e =– velocity of approach or

e =–

v2 v1 u2 u1 (v) For an oblique collision (Fig. 7.2) Velocity of approach = u1 cos 1 – u2 cos 2 Velocity of separation = v2 cos 2 – v1 cos 1 v cos 2 v1 cos 1 e=– 2 u1 cos 1 u2 cos 2 =

Fig. 7.2

In this case, we apply the law of conservation of momentum separately for x and y components of momenta. The components of momentum along the positive x-axis and positive y-axis are taken to be positive and components of momentum along negative x-axis and negative y-axis are taken to be negative. Momentum conservation of x-components gives m1u1 cos 1 + m2u2 cos 2 = m1v1 cos 1 + m2v2 cos 2 Momentum conservation of y-components gives – m1u1 sin 1 + m2u2 sin 2 = m1v1 sin 1 – m2v2 sin 2

Velocities after Head-on Elastic Collision Refer to Fig. 7.1 again. From the law of conservation of momentum, we have m 1u 1 + m 2u 2 = m 1v 1 + m 2v 2 (1)

Conservation of Linear Momentum and Collisions 7.3

v1 = – u1 v v2 e= 1 u2 u1 v1 – v2= e (u2 – u1)

and

m1 em2 u1 m1 m2 Using (3) in (2), we get

m2 1 e u2 m1 m2

(3)

m1 1 e u1 m1 m2

m2 em1 u2 m1 m2

(4)

v2 =

height from where it was dropped while the much more massive ground remains at rest. Finally, if the body at rest is much lighter than the colliding body, i.e. if m2 << m1, we have v1 u1 v2 2u1

Perfectly Elastic Collision For perfectly elastic collision, e = 1. Putting e = 1 in Eqs. (3) and (4) we get

and

v1 =

m1 m1

m2 u1 m2

v2 =

2m1 u1 m1 m2

2m2 u2 m1 m2

(5)

m2 m1

(6)

m1 u2 m2

Special Cases (i) If both bodies have the same mass, then m1 = m2 = m In this case, v1 = u2 and v2 = u1

i.e. the velocity of the massive body remains practically unchanged on collision with the lighter body at rest and the lighter body acquires nearly twice the initial velocity of the massive body. (iii) Kinetic energy delivered by incident body to a stationary body in perfectly elastic head-on collision. 1 Kinetic energy of m1 before collision is Ki = m1u12 2 1 and after collision is Kf = m1v12 . Therefore 2 Kf v12 = 2 Ki u1 Ki

or

This means that in a one-dimensional elastic collision between two bodies of equal mass, the bodies merely exchange their velocities after the collision. (ii) If one of the bodies, say m2, is initially at rest, then u2 = 0

0

Thus, if a very light body suffers an elastic collision with a very heavy body at rest, the velocity of the lighter body is reversed on collision, while the heavier body remains practically at rest. A common example of this type of collision is the dropping of a hard steel ball on a hard concrete

(2)

Eliminating v2 from (1) and (2), we get v1 =

v2

Kf

= 1

Ki u12 The fractional decrease in kinetic energy of m1 is v2 K = 1 12 Ki u1 If

u2 = 0,

In this case, v1 =

m1 m1

m2 u1 m2

2m1u1 m1 m2 If, in addition, m1 = m2 = m, these equations give v 1= 0 v 2 = u 1 and v2 =

Thus, if a body suffers a one-dimensional elastic collision with another body of the same mass at rest, . However, if the body at rest, namely B, is much more massive than the colliding body A, i.e. m2 >> m1, such that m1 is negligibly small, then

v12

K =1 Ki

v1 m = 1 m1 u1 m1 m1

m2 m2

m2 . Therefore, m2 2

=

4m1m2 m1

m2

2

NOTE

The fraction of kinetic energy lost by mass m1 is maximum . if m1 = m2 and minimum if m2 (iv) Change in kinetic energy of a system in a perfectly inelastic head-on collision. In a perfectly inelastic collision, the two bodies stick together after the collision. Hence v1 = v2 and e = 0. Putting e = 0 in Eqs. (3) and (4), we get v1 =

m1 u1 m1 m2

m2 u2 m1 m2

7.4 Comprehensive Physics—JEE Advanced

m2 m1 u2 u1 m1 m2 m1 m2 If mass m2 is stationary, u2 = 0. Then

e=

m1 u1 m1 m2

e=

and

v2 =

v1 =

(7) vcos

m1 u1 and v2 = m1 m2 Notice that v1 = v2 = v (say) Total K.E. of the system before collision is 1 Ki = m1u12 2 and after the collision is 1 Kf = m1 m2 v 2 2 Loss in K.E. of the system is 1 1 Ki – Kf = m1u12 m1 m2 v 2 2 2 From Eq. (7)

v = u1

j

v cos

j

u cos

=

v cos u cos

= eu cos

(10)

Since the impulsive force acts along the normal, the momentum along the normal is not conservved. Since the component of the impulsive force along the horizontal is zero, the momentum along the horizontal is conserved. Hence u sin = v sin (11) From Eqs. (10) and (11), we get v = (e2 cos2

(8)

m1 . Using this in m1 m2

Eq. (8) we get

+ sin2 )1/2 u

(12)

tan (13) e For a perfectly elastic collision, e =1 and Eqs. (12) and (13) give v =u and tan

=

and

=

i.e. for a perfectly elastic collision, the body re-

m1m2u12 Ki – Kf = 2 m1 m2 In general, if u2 0, we have Ki – Kf =

velocity of separation velocity of approach

m1m2 2 m1 m2

and at the same angle on the other side of the normal. 2

u1 u2

(9)

Oblique Impact on a Fixed Horizontal Plane Consider a body of mass m moving with a velocity u making an angle with the normal ON

Direct Impact on a Fixed Plane after impact, then, in this case (10) we get [Fig. 7.4] v = eu

=

= 0. Using this in Eq.

v

v making an angle with the normal. Since the horizontal place along the normal. The normal component of u is u cos along – y direction and the normal component of v is v cos along the +y directon. Now Fig. 7.4 v

7.1 A body is dropped from rest from a height h = 5.0 m.

is 0.8? Fig. 7.3

Conservation of Linear Momentum and Collisions 7.5

SOLUTION Refer to Fig. 7.5. = 2gh S e 2 gh . This is also the speed just before the second impact. Therefore,

v

speed just after second impact = e2 2 gh . This is the initial speed for the upward motion of the body

v

after the second impact, i.e. u = e2 2 gh . Therefore, height attained after two impacts is h2 =

u2 1 2 e 2 gh = 2g 2g

= (0.8)4

2

= e 4h

From conservation of momentum, mu = mv1 + mv2 u = v1 + v2 (i) Taking the scalar product of u with itself, we have u u = (v1 + v2) (v1 + v2)

5 = 2.05 m

NOTES (1) Height attained after n impacts is hn = e2nh (2) Speed of rebound after nth impact is vn = e n 2 gh (3) Total distance travelled before the body comes to rest = h

1 e

2

1 e2

.

7.2 A steel ball of mass m moving with velocity u1 undergoes a perfectly elastic head-on collision with another identical steel ball moving with velocity u2. Show that, after the collision, they merely exchange their velocities.

Refer to Fig. 7.1 again. From conservation of momentum, mu1 + mu2 = mv1 + mv2 u1 + u2 = v1 + v2

(i)

v2 – v1 = e (u1 – u2) For a perfectly elastic collision, e = 1. Hence v2 – v1 = u1 – u2 From (i) and (ii) we get

Fig. 7.5

(ii)

v1 = u2 and v2 = u1 7.3 A steel ball of mass m moving with a velocity u undergoes a perfectly elastic oblique collision with another indentical steel ball initially at rest. Show that, after the collision, they move at right angles to each other.

u2 = v12

2 v1 v 2

v22

(ii)

Since kinetic energy is also conserved in an elastic collision, we have 1 2 1 1 2 mu = mv12 mv2 2 2 2 u2 = v12 + v22

(iii)

Using (iii) in (ii), we get 2 v1 v2 = 0 v1 v2 = 0 v1 v2 cos

=0

cos

=0

= 90°

7.4 Two steel balls of the same mass m moving in opposite directions with the same speed u collide head-on. If the collision is perfectly elastic, predict the result of the collision.

m1 = m2 = m, u1 = u and u2 = – u. Let v1 and v2 be their velocities after collision. Total momentum before collision = m1u1 + m2u2 = m (u – u) = 0 Total momentum after collision = mv1 + mv2 = m (v1 + v2) From conservation of momentum, 0 = m(v1 + v2) v2 = – v1

7.6 Comprehensive Physics—JEE Advanced

Since e = 1, we have v2 – v1 = u1 – u2 = u – (– u) = 2u Putting v1 = – v2, we get v2 = u. Also v1 = – u. Thus, after the collision, the two balls move in opposite directions with equal speeds, each equal to u but their directions are reversed. 7.5 A ball of mass 2 kg moving with a velocity of 8 ms–1 collides head-on with another ball of mass of 4 kg moving with a velocity of 2 ms–1 moving in the same restitution is e = 0.5. (a) Find the velocities of the balls after the collision. (b) Calculate the loss of kinetic energy due to collision.

7.6 In Example 7.5, what is the loss of kinetic energy if the ball of mass 4 kg is moving towards the mass of mass 2 kg, their speeds being the same?

In this case u2 = – 2 ms–1. Equations (i) and (ii) become (iii) 4 = v1 + 2v2 (iv) and v2 – v1 = 5 Equations (iii) and (iv) give v1 = – 2 ms–1 and v2 = 3 ms–1 1 1 Ki = 2 (8)2 + 4 (–2)2 = 72 J 2 2 1 1 2 (–2)2 2 2 Loss of K.E = 72 – 22 = 50 J Kf =

Refer to Fig 7.1 again. (a) Given m1 = 2 kg, m2 = 4 kg, u1 = 8 ms–1, u2 = 2 ms–1 and e = 0.5 From conservation of momentum m 1u 1 + m 2u 2 = m 1v 1 + m 2v 2 2

8+4

2 = 2v1 + 4v2

12 = v1 + 2v2

(i)

Since e = 0.5, we have v2 – v1 = e (u1 – u2) = 0.5 (8 – 2) = 3 (ii) Eliminating v2 from (i) and (ii) we get v1 = 2 ms–1. Using this in (i) or (ii), we get v2 = 5 ms–1 (b) Kinetic energy before collision is Ki =

1 2 2 = 72 J

=

between blocks B and C.

1 2

(8)2 +

Fig. 7.6

4

(2)2

Kinetic energy after collision is Kf =

7.7 Two blocks B and C of masses 1 kg and 2 kg respectively are connected by a massless elastic spring of spring constant 150 Nm–1 and placed on a horizontal frictionless surface as shown in Fig. 7.6. A third block A of mass 1 kg moves with a velocity of 3 ms–1 along the line joining B and C and collides with B. If the collision is perfectly elastic and the natural length

1 1 m 1u 12 + m 2u 22 2 2

=

1 1 m 1v 12 + m 2v 22 2 2 1 2

2

(2)2 +

1 2

4

(3)2 = 22J

(5)2 = 54 J

Loss of K.E. = Ki – Kf = 72 – 54 = 18 J

Given mA = mB = 1 kg, mC = 2 kg and u = 3 ms–1. Block A will collide with block B. Since they have equal masses and the collision is perfectly elastic, A will come to rest and B will move to the right with a velocity u. Block B will compress the spring. Hence block C will accelarate and block B will retard until both B and C move with the same velocity. Let this common velocity be v. Since no external force acts, the momentum of B and C is conserved, i.e. mB u = (mB + mC) v 1

3 = (1 + 2) v

v = 1 ms–1

Conservation of Linear Momentum and Collisions 7.7

If x is the maximum compression, then from the principle of conservation of energy, 1 1 1 mAu2 = (mB + mC) v2 + kx2 2 2 2 1 1 1 (3)2 = (1 + 2) (1)2 2 2 1 + 150 x2 2 which gives x = 0.2 m = 20 cm Minimum separation between B and C = 80 cm – 20 cm = 60 cm 7.8 A block of m1= m is moving on a frictionless horizontal surface with velocity u1 = 2u towards another block of mass m2 = 3m moving on the same surface with velocity u2 = u in the same direction. A massless spring of force constant k is attached to m2 as shown in Fig. 7.7. When block m1 collides with the spring, show that the maximum compression of the spring is given u 3m by x = . 2 k

Fig. 7.7

SOLUTION When block m1 collides with spring, it begins to get compressed. As a result m2 gains speed. The compression of the spring is maximum at the instant when the relative velocity of m1 with respect to m2 is zero, i.e. when both m1 and m2 have equal velocities. Let v be the common velocity of the blocks. From conservation of momentum, m1u1 + m2 u2 = (m1 + m2)v

1 2

m

1 m2u22 2 (2u)2 +

– 1 2

1 m1 2 3m

5u 4

u 3m 2 k

7. Useful Formulae and Tips 1. A body of mass m is dropped from a height h. Due to the friction of air, it will hit the ground with a speed less than 2gh . If v is the speed with which it hits the ground, the work done by friction is 1 1 mv2 – mgh = m(v2–2gh) Wf = 2 2 If friction is absent, Wf = 0, then v = 2gh . 2. Two block A and B of masses m1 and m2 are released from the same height at the same time. Block A slides along an inclined plane of inclination and block B falls vertically downwards (Fig. 7.8) If the inclined plane is frictionless, gain in KE = loss in PE, i.e. 1 m1v21 = m1gh v1 = 2gh 2 If the air friction is neglected. 1 m2v22 = m2gh v2 = 2gh 2 Thus both block will hit the ground with the same speed independent of the mass. But the times taken to reach the ground will be different.

v

v

2h For block A, t1 =

g sin 2

2h g 3. If a block of mass m in contact with a spring compressed by a distance x is released, the block will leave the spring with a velocity v determined from For block B, t2 =

m2 v 2 = u2 –

x=

Fig. 7.8

2mu + 3mu = (m + 3m)v 5u v= 4 From the law of conservation of energy Loss in K.E. = gain in P.E of spring If x is the maximum compression, then 1 m1u12 2

3 mu2 = kx2 4

1 2 kx 2

1 (m + 3m) 2 2

=

1 kx2 2

1 1 kx2 = mv2 2 2 which gives v = constant.

k x, where k is the spring m

7.8 Comprehensive Physics—JEE Advanced

4. If a block of mass m moving with speed u comes in contact with a relaxed spring of spring constant k, its velocity v when the spring is compressed by an amount x is obtained from. 1 1 1 mu2 = mv2 + kx2 2 2 2 k x2 m

which gives v =

7. A chain has a length L and mass M. A part L/n is hanging at the edge of the table. The length of the chain lying on the table is (L – L/n). Then work done against gravity to pull the hanging part on the MgL table = 2n 2 8. If a body of mass m moving with velocity v is stopped in a distance x by a retarding force F, then 1 mv2 = Fx 2 (a) If two bodies of masses m1 and m2 moving with the same velocity are subjected to the same retarding force, the ratio of the stopping distance is x1 m1 = x2 m2 (b) If the two bodies are moving with equal kinetic energy and are stopped by the same retarding force, then x1 = x2

1/ 2

u

2

5. If two springs of spring constants k1 and k2 are stretched by the same force F, then F = k1x1 = k2 x2. Potential energy stored is 1 1 U1 = k 1x 21 = Fx1 2 2 1 1 and U2 = k 2x 22 = Fx2 2 2 which give

U1 x = 1 U2 x2

k2 k1

6. If the two springs are stretched by the same amount x, then F1 = k1x and F2 = k2x. U1 =

(c) If the two bodies are moving with equal liner momentum and are stopped by the same force, then. p2 = Fx 2m

1 1 k1x2 and U2 = k 2x 2. 2 2

U1 k = 1 U2 k2

F1 F2

and

m2 x1 = m1 x2

I Multiple Choice Questions with only One Choice Correct 1. A ball P moving with a velocity u suffers a onedimensional collision with another ball Q of the same mass but at rest. After the collision the velocity of Q is found to be three times that of P. The (a)

1 2

(b)

1 3

(c)

1 4

(d)

2 3

2.

= 45°.

the collision is 1 (a) 2

(b)

1 2 2

3 5 (d) 8 8 3. Two particles of the same mass m moving in different directions with the same speed v collide and stick together. After the collision, the speed of the composite particle is v/2. The angle between the velocities of the two particles before collision is (a) 60° (b) 90° (c) 120° (d) 150° 4. Two blocks of masses m1 = m and m2 = 3 m are connected by a spring of force constant k and placed on a horizontal frictionless surface as shown in Fig. 7.9. The spring is stretched by an amount x and released. The system executes simple harmonic motion. The relative velocity of the blocks when the spring is at its natural length is (c)

Conservation of Linear Momentum and Collisions 7.9

(a) x (c)

x 2

3k 2m

(b) 2 x

k m

k 3m

(d) 2 x

k 3m

Fig. 7.9

5. A block P moves with an initial velocity of 4 ms–1 towards a block Q of the same mass at rest at a distance of 2 m on a rough horizontal surface. The cois 0.2. An elastic one-dimensional collision occurs between the two blocks. If g = 10 ms–2, the separation between the blocks when both have come to rest is (a) 1 m (b) 2 m (c) 3 m (d) 4 m 6. A perfectly elastic oblique collision occurs between a ball A moving along the x-axis and a ball B at rest and of the same mass as ball A. After the collision, ball A moves at an angle of 30° with the x-direction and ball B at an angle with the x-axis. The value of is (a) 15° (b) 30° (c) 45° (d) 60° 7. A particle A moving with momentum p suffers a one-dimensional collision with a particle B of the same mass but at rest. During the collision B imparts an impulse I to A tion between A and B is 2I 2I (b) –1 (a) p p 2I I +1 (d) 2 1 p p 8. A man of mass m stands on one end of a wooden plank of length L and mass M kept initially at rest on a horizontal frictional surface. If the man walks from one end of the plank to the other end at a constant speed, the resulting displacement of the plank is mL ML (b) (a) M m (c)

(c)

mL ( M m)

(d)

mL ( M m)

9. A body P of mass m moving with kinetic energy k and momentum p undergoes a one-dimensional

elastic collision with a body of mass 2 m at rest. After the collision, the ratio of the kinetic energy of P to that of Q is 1 1 (b) (a) 8 4 1 8 (c) (d) 9 9 10. A shell of mass m initially at rest explodes into three fragments of masses in the ratio 2 : 2 : 1. mutually perpendicular directions with speed v. The speed of the third (lighter) fragment will be 2v

(a) v

(b)

(c) 2 2 v

(d) 3 2 v

11. A shell is fired with a speed of 100 ms –1 at an angle of 30° with the vertical (y-direction). At the highest point of its trajectory, the shell explodes into two fragments of masses in the ratio 1 : 2. The lighter fragment moves vertically upwards with an initial speed of 200 ms–1. The speed of the heavier fragment immediately after the explosion is (b) 150 ms–1 (a) 125 ms–1 (c) 175 ms–1 (d) 200 ms–1 12. A body of mass m moving with a certain speed suffers an inelastic collision with a body of mass M system to the initial kinetic energy is M m (b) m M (a) m M (c)

m

M

(d)

m

M

m M 13. A neutron moving at a speed v undergoes a head-on elastic collision with a nucleus of mass number A at rest. The ratio of the kinetic energies of the neutron after and before collision is (a) (c)

A 1 A 1

2

A

2

(b) (d)

A 1 A 1

2

A

2

A 1 A 1 14. A radioactive nucleus of mass number A, initially at rest, emits an -particle with a speed v. What will be the recoil speed of the daughter nucleus? 2v 2v (a) (b) A 4 A 4 4v 4v (c) (d) A 4 A 4

7.10 Comprehensive Physics—JEE Advanced

15. A ball is dropped from a height of 10 m. It is embedded 1 m in sand. In this process (a) only momentum is conserved (b) only kinetic energy is conserved (c) both momentum and kinetic energy are conserved (d) neither momentum nor kinetic energy is conserved. 16. n small balls, each of mass m, impinge elastically each second on a surface with velocity u. The force experienced by the surface will be (a) mnu (b) 2 mnu 1 mnu (c) 4 mnu (d) 2 17. A rubber ball is dropped from a height of 5 m on a planet where the acceleration due to gravity is not known. On bouncing it rises to 1.8 m. The ball loses its velocity on bouncing by a factor of 16 2 (b) (a) 25 5 3 9 (d) 5 2 18. An isolated particle of mass m is moving in a horizontal plane (x – y), along the x-axis, at a certain height above the ground. It suddenly explodes into two fragments of masses m/4 and 3m/4. An instant later, the smaller fragment is at y = + 15 cm. The larger fragment at this instant is at (a) y = – 5 cm (b) y = + 20 cm (c) y = + 5 cm (d) y = – 20 cm IIT, 1997 19. A body P strikes another body Q of mass that is p times that of body P and moving with a velocity 1 that is of the velocity of body P. If body P comes q (c)

p (a) p (c)

q q

p q p q 1

(c)

h(1 e2 ) 2(1 e2 )

(d)

h(1 e2 ) 2(1 e2 )

23. A bullet of mass m velocity v on a wooden block of mass M suspended from a support and gets embedded in it. The kinetic energy of the bullet + block system is (a)

1 mv2 2

(b)

1 (M + m)v2 2

(c)

Mmv 2 2 M m

(d)

m2 v2 2 M m

24. A body of mass m moving with a speed v suffers an inelastic collision and sticks with another body of mass M = 2m at rest. The speed of the composite body will be v (a) 3v (b) 3 (c)

2v 3

(d)

3v 2

of the system to the initial kinetic energy is 1 2 (a) (b) 3 3

p q p q 1

20. Two equal spheres A and B lie on a smoot horizontal circular groove at opposite ends of a diameter. Sphere A is projected along the groove and at the end of time T impinges on sphere B. If e is the occur after a time equal to (a) T (b) eT 2T (d) 2 eT (c) e

K 1 M KM (d) M M 4 22. A particle falls from a height h plate and rebounds. If e tution, the total distance travelled by the particle before it stops rebounding is h(1 e2 ) h(1 e2 ) (b) (a) (1 e2 ) 1 e2 (c)

25.

p q (b) q p 1 (d)

21. A nucleus of mass M amu emits an -particles with a energy K MeV. The total energy of disintegration (in MeV) is KM (a) K (b) M 4

3 2 26. A body of mass 5 kg is moving along the x-axis with a velocity 2ms–1. Another body of mass 10 kg is moving along the y-axis with a velocity 3 ms–1. They collide at the origin and stick together. The (c) 3

(a) IIT, 1997

(c)

(d)

3 ms–1 4 ms–1 3

(b)

3 1 ms–1

(d) none

Conservation of Linear Momentum and Collisions 7.11

27. A block of wood of mass M is suspended by means of a thread. A bullet of mass m into the block with a velocity v. As a result of the impact, the bullet is embedded in the block. The block will rise to vertical height given by mv 1 (a) 2g M m

2

mv 1 (b) 2g M m

2

mv 2 mv 2 1 1 (d) 2g M m 2g M m 28. A moving particle of mass m makes a head-on collision with a particle of mass 2m initially at rest. If the collision is perfectly elastic, the percentage loss of energy of the colliding particle is (a) 50% (b) 66.7% (c) 88.9% (d) 100% (c)

29. A body of mass m moving with a velocity v in the x-direction collides with a body of mass M moving with a velocity V in the y-direction. They stick together during collision. Then (a) the magnitude of the momentum of the composite body is

mv

2

MV

2

(b) the composite body moves in a direction MV making a angle = tan–1 with the mv x-axis. (c) the loss of kinetic energy as a result of col1 Mm (V2 + v2) lision is 2 M m (d) all the above choices are correct. 30. A body falls from a height h on a horizontal surface and rebounds. Then it falls again and again rebounds 1 , the 3 total distance covered by the body before it comes to rest is h 5h (b) (a) 4 4 (c) 2h (d) 3h 31. In Q. 30 above, the total time taken by the body to come to rest is (a) (c) 3

2h g 2h g

(b) 2

2h g

(d) 4

2h g

32. A body of mass m moving with a velocity v in the x-direction collides and sticks with another body of

mass M moving with a velocity V in the y-direction. The magnitude of the momentum of the composite body is (a) (mv MV )

(b) (m + M) (v + V)

(c) [(mv)2 + (MV)2]1/2

(d) (Mv + mV)

33. In Q. 32 above, the angle subtended by the velocity vector of the composite body with the x-axis is given by mv MV (b) = tan–1 MV (a) = tan–1 mv mV Mv (c) = tan–1 (d) = tan–1 Mv mV 34. A body P of mass m1 moving with a certain velocity collides head-on with a stationary body Q of mass m2. It the collision is elastic, the fraction of kinetic energy transferred from body P to body Q is (a) (c)

2(m1m2 ) (m1

m2 )

2

2m12 (m1

m2 )2

(b) (d)

4(m1m2 ) (m1

m2 )2

2m22 (m1

m2 )2

35. Two particles, each of mass m, moving along different directions with a velocity u making the same angle with the x-axis collide and stick together. The composite particle moves along the x-axis with a velocity u/2. The angle between their directions of motion before collision is (a) 60° (b) 90° (c) 120° (d) 150° 36. A bullet of mass m moving with a horizontal velocity u strikes a stationary wooden block of mass M suspended by a string of length L = 50 cm. The bullet emerges out of the block with speed u/4. If M = 6 m, the minimum value of u so that the block can complete the vertical circle is (take g = 10 ms–2) (b) 20 ms–1 (a) 10 ms–1 –1 (c) 30 ms (d) 40 ms–1 37. A compound pendulum consists of a uniform rod of length L of negligible mass. A body of mass m1 = m m2 = 2 m middle of the rod as shown in Fig. 7.10 The horizontal velocity v that must be given to mass m1 to rotate the pendnlum to the horizontal position OC is

Fig. 7.10

7.12 Comprehensive Physics—JEE Advanced

2 gL 3

(a) 2 (c)

(a) 4 (c) 2

(b) 2 gL

2gL

(d)

IIT, 2009 40. A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, travelling with a velocity V m/s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The velocity V of the bullet is (see Fig.7.13) (a) 250 m/s (b) 250 m/s (c) 400 m/s (d) 500 m/s

gL

38. A ball is thrown will a velocity v1 towards a vertical wall at an angle with the wall. It rebounds with a velocity v2 making an angle with the wall as between the ball and the wall is e, then v2 is given by (a) v2 = v1(cos + e sin ) (b) v2 = v1(sin + e cos ) (c) v2 = v1 sin 2

e2 cos 2

(d) v2 = v1 cos 2

e2 sin 2

(b) 3 (d) 1

Fig. 7.13

IIT, 2011 41. A ball is thrown from a point O with a velocity u = 20 ms–1 at an angle = 30° with the horizontal. It hits a vertical wall which is at a distance x from O as shown in Fig. 7.14. After rebounding from the wall, the ball returns to O without retracing its path. e = 0.5, If g = 10 ms–2 the value of x is (a) 9.6 m (b) 10.3 m (c) 11.5 m (d) 12.8 m

Fig. 7.11

39. Two small particles of equal masses start moving in opposite directions from a point A in a horizontal circular orbit. Their tangential velocities are v and 2v, respectively, as shown in Fig. 7.12. Between collisions, the particles move with constant speeds. After making how many elastic collisions, other than that at A, will these two particles again reach the point A? v

v

Fig. 7.12

Fig 7.14

ANSWERS 1. 7. 13. 19. 25. 31. 37.

(a) (b) (a) (d) (a) (b) (a)

2. 8. 14. 20. 26. 32. 38.

(c) (c) (c) (c) (c) (c) (d)

3. 9. 15. 21. 27. 33. 39.

(c) (a) (a) (b) (a) (a) (c)

4. 10. 16. 22. 28. 34. 40.

(d) (c) (b) (a) (c) (b) (d)

5. 11. 17. 23. 29. 35. 41.

(b) (a) (b) (d) (c) (c) (c)

6. 12. 18. 24. 30. 36.

(d) (a) (a) (b) (b) (d)

Conservation of Linear Momentum and Collisions 7.13

SOLUTIONS 1. Let v1 and v2 be the velocities of P and Q after the collision. v2 v1 = v2 v1 (1) u 0 u From conservation of momentum, we have mu + 0 = mv1 + mv2 (2) u = v1 + v2 From Eqs. (1) and (2), we get u u (1 – e) and v2 = (1 + e) 2 2 v2 1 e = v1 1 e 1 e 1 Given v2 = 3v1. Hence 3 = e= 1 e 2 2. Let u be the speed of the ball before the collision. After the collision, its speed will be v1 =

u

v = =

2

2 u2 2

2

eu

5 u 8

e

1 2

1 1 mu 2 mv 2 2 2 Fraction of K.E. lost = 1 mu 2 2 v2 5 3 =1– 2 =1– = 8 8 u 3. Let be the angle between the velocities of the two particles before collision. If p1 and p2 are the momenta of the particles before collision and p is the momentum of the composite particle, then the conservation of momentum gives p2 = p 21 + p 22 + 2 p1 p2 cos 2m

v 2

2

= (mv)2 + (mv)2 + 2 (mv) (mv) cos

1 or = 120° 2 4. If v is the relative velocity of the two blocks when the spring is at its natural length, then from the conservation of energy, we have 1 1 v2 = kx2 (1) 2 2 where is the reduced mass of the system and is given by cos

=–

m1 m2 m = (m1 m2 ) (m

3m 3m = 3 m) 4

Substituting in Eq. (1) we get 3 k mv2 v = 2x 4 3m 5. Frictional force f = mg. The retardation due to friction is f = g = 0.2 10 = 2 ms–2 a= m Since the blocks have the same mass and the collision is elastic, after the collision, block P will come to rest at the position previously occupied by block Q and Q will begin to move with the velocity at which P strikes Q which is given by kx2 =

v2 = 2as = 2

2

v=

2=8

8 ms–1

Moving with velocity of 8 ms–1, the block Q will come to rest after travelling a distance x given by v2 8 = =2m 2a 2 2 Hence the correct choice is (b). 6. In an oblique elastic collision between two body of the same mass, they move at right angles to each other after the collision. Hence the correct choice is (d). 7. Let p1 and p2 be the linear momenta of A and B after the collision. Now, impulse = change in momentum. For particle B : I = p1 For particle A : I = p – p2 p2 = p – I x=

2 u2 = 8

=

e=

v1

v2 u

=

=

m v1 p1

m v2 mu p2

p where m is the mass of each particle and u is the velocity A before collision and v1 and v2 are the velocities of A and B after the collision. Hence p p I I p I 2I e= 1 = = –1 p p p So the correct choice is (b). 8. Total initial momentum of the man-plank system is zero. If he walks with a speed v on the plank, as a result, the plank moves with a speed, say, v in the system = mv – (M + m) v . From conservation of momentum,

7.14 Comprehensive Physics—JEE Advanced

0 = mv – (M + m) v v = (M v

V = 2 2 v.

m m)

Since the distance moved in proportional to speed (since there is no acceleration), the displacement L of the plank is given by choice (c). 9. Total momentum before collision is p = mu + 0 1 = mu and kinetic energy is K = mu2. If v1 and 2 v2 are the velocities of P and Q after the collision, then, from momentum conservation, mu = mv1 + (2 m) v2 p = p1 + p2

(1)

From conservation of kinetic energy, K = K1 + K2 where K1 = Thus

Fig. 7.15

m p1 = v1, where v1 = 200 ms–1. If v2 is the ve3 locity of the heavier fragment, its momentum is 2 mv2 /3. Conservation of momentum along x and y-directions gives p = p2 cos

1 1 mv 12 and K2 = (2 m) v 22 2 2 1 1 mu2 = mv12 + mv 22 2 2

p2 p2 = 1 2m 2m

mv = (3)

From Eq. (1), p2 = p – p1. Using this in Eq. (3) and solving we get p 4p . Hence p2 = p1 = – 3 3

K2 =

and

p2 1 p2 mv 12 = 1 = 2m 18 m 2 p22

2

1 4p mv 22 = = 4m 2 9m

K1 1 = , which is choice (a). K2 8 10. Mass of each heavier fragment = 2 m/5 and of lighter fragment = m/5. Momentum of each heavier fragment is p = 2 mv perpendicular directions, their resultant momentum =

p2

p2 =

2 m v2 cos 3 3v = 2v2 cos

p22 4m

2p2 = 2p 21 + p 22

Now K1 =

11. Velocity of the shell at the highest point is v = 100 sin 30° = 50 ms–1 parallel to x-direction. Its momentum is p = mv (Fig. 7.15).

2 p . If V is the speed of the lighter

fragment, from the conservation of momentum, we have 2 mv mV = = 2 2 p 5 5

and

(1)

p1 = p2 sin 2 m v2 m v1 = sin 3 3 v1 = 2v2 sin

From (1) and (2), we get v2 = =

(2) 1 (v12 + 9v2)1/2. 2

1 [(200)2 + 9(50)2]1/2 2

= 125 ms–1 12. Initial momentum of the system = mv, since body of mass M is at rest. After the inelastic collision, the bodies stick together and the mass of the composite body is (m + M). If V is the speed of the composite body, its momentum will be (m + M)V. From the principle of conservation of momentum, we have mv = (m + M)V or V = Initial KE =

m v m M 1 1 mv 2. Final KE = (m + M)V2. 2 2

Conservation of Linear Momentum and Collisions 7.15

Therefore, 2

Final KE m+M V m = = 2 Initial KE m m M v Hence the correct choice is (a). 13. Mass of neutron (m1) = 1 unit. Mass of nucleus (m2) = A units. Refer to page 7.2. Here u1 = u and u2 = 0. Therefore the velocity of the neutron after the collision is v1 =

m1 m1

m2 u= m2

1 A u 1 A

KE of neutron after collision = =

1 A 1 A

1 2

KE of neutron before collision = =

1 2

u2 =

1

1 m1v 12 2 2

u2 1 m u2 2

1 2 u. 2

1 A 2 , which is choice (a). 1 A 14. The total number of nucleus (i.e. protons + neutrons) in a nucleus is called its mass number. An -particle is a helium nucleus having 2 protons and 2 neutrons. So the mass number of an -particle = 4. When a nucleus of mass number A emits an -particle, the mass number of the daughter nucleus reduces to (A – 4). If V is the recoil speed of the daughter nucleus, we have, from the law of conservation of momentum, (A – 4)V – 4v = 0 Their ratio is

4v A 4 Hence the correct choice is (c). V=

or

15. The collision is inelastic because the two bodies stick to each other after collision. In an inelastic collision, only the momentum is conserved; there being a loss in kinetic energy. Hence the correct choice is (a). 16. Since the collision (impact) is elastic, the ball rebounds with the same speed. Therefore, the change in momentum of each ball = 2 mu. The change in momentum per second due to n balls = 2 mnu. But the change in momentum per second is the force. Hence the correct choice is (b). 17. A ball dropped from a height h1 on reaching the planet’s surface will have a velocity given by v1 =

2 gh1

Let v2 be the velocity with which the ball bounces. It will attain a height h2 given by v22 = v2 = v1

2 gh2 h2 h1

18 . = 0.6 5

or

1–

v2 = 1 – 0.6 v1

or

v1

v2 v1

= 0.4 =

2 5

Hence the correct choice is (b). 18. Let m and M be the masses of the lighter and heaviers fragments respectively. Since the particle is moving along the x-axis, the y-component of momentum will be zero immediately after and before explosion, i.e. mvy + MVy = 0 where vy and Vy are the velocities of the lighter and heavier fragments respectively immediately after explosion. Thus m m/ 4 vy = vy Vy = M 3m / 4 = 1 vy 3 Since y = + 15 cm, the direction of vy is along the positive y-axis and that of Vy will be along the negative y-axis. An instant later (say, at time t), it is given that y = 15 cm = vy t 1 1 Y = Vy t = vy t = y 3 3 1 15 cm = – 5 cm = 3 19. Given mQ = p mP and vQ = vP /q. From the principle of conservation of momentum, we have (since body P comes to rest after collision) mP vP + mQ vQ = mQ v where v is the velocity of body Q after collision. Thus v mP vP + p mP P = p mPv. q v p q which gives = (i) vP pq e=

v vp

v vQ

vp

vp q

7.16 Comprehensive Physics—JEE Advanced

which gives

v e = (q – 1) vP q

(ii)

22. The total distance travelled is S = h + e2h + 2e4h + 2e6h + = h + 2h(e2 + e4 + e6 +

p q which is Equating (i) and (ii), we get e = p q 1 choice (d). 20. Refer to Fig. 7.16. If sphere A is projected with velocity v, the time taken by it to strike B is equal r = T or r = Tv. Now, to v titution is given by v vA e= B v

-

= h + 2h

= 1

e 1 e2

2e 2

=

h(1 e2 )

1 e2 1 e2 23. Initial momentum (p) = momentum of bullet + momentum of block = mv + 0 = mv. From the momentum of bullet + block system of mass (M + m) = Initial momentum p. Now

r A

)

KE =

B

2

p2 M m

m2 v2 2 M m

Hence the correct choice is (d). 24. The correct choice is (b). In an inelastic collision, Fig. 7.16

where vA and vB are the velocities of A and B after the collision. Thus, vB – vA = ev. The spheres travel with this relative velocity. It is clear that one will overtake the other after travelling a distance = 2 r. 2 r 2 r 2T v 2T Time taken = vB v A ev ev e (since r = Tv). Hence the correct choice is (c). 21. Let V be the velocity of the nucleus and v that of the -particle after disintegration, then from the principle of conservation of momentum, we have (since the mass of an -particle is 4 amu) 4v (i) (M – 4) V = 4v or V = M 4 Total KE = Now

1 2

4

Total KE =

1 1 (M – 4) V 2 + 2 2

4

v2 (ii)

v2 = K. Using (i) in (ii), we have 1 (M – 4) 2

16v 2 M

4

2

=

8v 2 + 2v2 M 4

=

4K MK +K= ( M 4 M 4

Hence the correct choice is (b).

+ 2v2

composite body, use the principle of conservation of linear momentum. 25. The correct choice is (a). Find kinetic energies before and after the collision. 26. Momentum of 5 kg mass (p1) = 5 2 = 10 kg ms–1 along the x-axis. Momentum of 10 kg mass (p2) = 10 3 kg ms–1 along the y-axis. These two momenta are perpendicular to each other. Therefore, the resultant initial momentum is p=

p12

p22

10

2

10 3

2

= 20 kg ms–1

If v ms–1 is the velocity of the combined mass, v = 15 v kg ms–1. Now, from the principle of conservation 4 of momentum, we have 15 v = 20 or v = ms–1, 3 which is choice (c). 27. Let V be the velocity of the block with the bullet embedded in it at the time of impact. Then from the principle of conservation of momentum, we have mv = (M + m) V mv (i) V= M m If the block, with the bullet embedded in it, rises to a vertical height h, then from the principle of conservation of energy, we have 1 (M + m) V 2 = (M + m) gh 2 or

2v2 = K)

or

V=

2gh

(ii)

Conservation of Linear Momentum and Collisions 7.17

Using (ii) in (i), we get mv 2gh = M m

31. Total time =

2h g

2e

2h g

=

2h g

2

2h e g

=

2h g

2

2h e g 1 e

=

2h g

h is given correctly by choice (a). 4mM

28. Percentage loss of energy =

=

4m

M

2m

m

100 =

2m m 2 Hence the correct choice is (c).

2

100

800 = 88.9% 9

29. Refer to Fig. 7.17. Here p = mv and P = MV. The resultant of p and P is pr =

p2

P2

mv

2

MV

1 3 1 1 3

2e 2

1

2

2h g

e2 2h 1 e g 1 e

2h , g

which is choice (b). 32. Refer to Fig. 7.18.

2

y

pr

P

P

q

x

p

Fig. 7.17

which is choice (a). The angle which the resultant momentum pr subtends with the x-axis is given by P MV tan = , which is choice (b). p mv Loss of KE =

1 mv 2 2

1 m 2 v 2 M 2V 2 M m 2

1 MV 2 2

1 Mm = (V2 + v2), which is choice (c). 2 M m 30. Total distance = h + 2 e2 h + 2 e4 h + ... = h + 2 e2 h (1 + e2 + ...) 2e2 h

=h +

h

1 e2 1

=h 1 which is choice (b).

1 3 1 3

1 e2 1 e2

2

2

5h , 4

Fig. 7.18

Let v be the velocity of the composite body at an angle with the x-axis (Fig. 7.11). Equating the x and y-axes, we have mv= (m + M) v cos = p cos (i) and

MV = (m + M) v sin

= p sin

(ii)

where p = (m + M) v is the momentum of the composite body. Find p by squaring and adding (i) and (ii). The correct choice is (c). 33. Divide (ii) by (i). The correct choice is (a). 34. Let u be the velocity of P before collision and v1 and v2 the velocities of P and Q after collision. From conservation of momentum, we have m1u + 0 = m1v1 + m2v2 which gives (1) m1(u – v1) = m2v2 From the conservation of kinetic energy we have 1 1 1 m1u12 = m1v12 + m2v22 2 2 2 2 2 m1(u – v1) = m2v22 m1(u – v1)(u + v1) = m2v22

(2)

7.18 Comprehensive Physics—JEE Advanced

Dividing Eq. (2) by Eq. (1) we get (3) u + v1 = v2 From Eqs. (1) and (3) we get m1 m2 u (4) v1 = m1 m2 1 Initial K.E. of P(K) = m1u2, 2 1 m 1v 12 P= 2 1 1 Decrease in K.E. of P( K) = m1u2 – m1v12. 2 2 Fractional decreases in K.E. of P is 1 1 m1u 2 m1v12 K 2 2 = 1 K m u2 2 1 =

u2

v12 u2 m1 m1

=1 – =

=1– m2 m2

2

u2 [use Eq. (4)]

= 2mv

u = mu ( 2

=2 m 2 cos which gives cos

2

2 which is choice (c)

2 =

3mu = 4M

5gL

4M 5 gL 3m 4 = 6 5 10 0.5 = 40 ms–1. 3

u=

2

37. Since mass m2 is at a distance L/2 from the axis of rotation, it speed will be v/2 (half that of mass m1). From the principle of conservation of energy we have 1 m 1v 2 + 2 1 mv2 + 2

1 v 2 m2 = m1gOB + m2gOA 2 2 1 v2 L 2m = mgL + 2mg 4 2 2

2 gL , which is choice (a). 3 38. The component of velocity parallel to the wall remains unchanged and the component of velocity perpendicular to the wall reduces by e times its value before collision. Thus we have v2 cos = v1 cos which gives v = 2

v = u/2)

=1 1 2

5gL

Hence the correct choice is (d).

2

4m1m2

+ mucos

V=

v12

(m1 m2 )2 Thus the correct choice is (b). 35. Refer to Fig. 7.19. From conservation of x-component of momentum, we have mucos

36. Let V be the speed of the block after the bullet emerges out of it. From conservation of momentum we have u mu = MV + m 4 3mu which gives V = 4M Now refer to Q. 27 of Section III of Chapter 4. The minimum speed the block must have to complete the vertical circle is

60°

= 120°,

and v2 sin

= ev1 sin

Squaring and adding we get v2 = v1 (cos2

e2 sin 2 )

Hence the correct choice is (d). 39. Refer to Fig. 7.20. First collision will occur when angle r 2 = v 2v which gives Fig. 7.19

r

=120°.

B, m2 will move back with a speed v and will make a second collision with m1

Conservation of Linear Momentum and Collisions 7.19

Horizontal velocity after rebounding from the wall is

at C. After the second collision at C, m1 will move back with a speed v and meet m2 at A. If the third collision at A is neglected, the particles will make two collisions before they reach A. Hence the correct choice is (c). v

u x = e ux = e u cos

Horizontal displacement from O to A or from A to O = x. Time taken to go from O to A is x (i) u cos Time taken to return from A to O is x t2 = (ii) eu cos Since the horizontal and vertical motions are independent of each other, the net vertical displacement Sy = 0 since the ball returns to O. If t is the total time taken by the ball to go from O to A and return to O, then from 1 2 gt we have S y = u yt – 2 1 2 gt 0 = (u sin ) t – 2 2u sin t= (iii) g Now t = t1 + t2. Using (i) and (ii) in (iii), we have t1 =

v

v

v

v

v

Fig. 7.20

40.

2h 2 5 = = 1s g 10 Horizontal range (R) = horizontal velocity t f) =

time

Horizontal velocities of the bullet and of the ball after the collision respectively are 100 = 100 ms–1 (v)bullet = 1 20 (v)ball = = 20 ms–1 1 From conservation of momentum,

2u sin g

= x=

(m)bullet

V = (m)bullet

0.01 V = 0.01

.

(v)bullet + (m)ball (v)ball 100 + 0.2

20

=

V = 500 ms–1 41. Refer to Fig 7.14 on page 7.12. Horizontal velocity before hitting the wall is ux = u cos

=

x u cos

x eu cos

u 2 sin(2 ) 1 1 e 20 2 sin 60 1 1 0.5 20 3

11.5 m, which is choice (c)

II Multiple Choice Questions with one or More Choices Correct 1. In an inelastic collision of two bodies, which of the following do not change after the collision? (a) total kinetic energy (b) total linear momentum (c) total energy (d) total angular momentum 2. Which of the following statements are true?

(a) In a elastic collision of two bodies, the momentum and energy of each body is conserved. (b) The total energy of a system is always conserved irrespective of whether external forces act on the system. (c) The work done by a force in nature on a body, over a closed loop, is not always zero.

7.20 Comprehensive Physics—JEE Advanced

(d) In an inelastic collision of two bodies, the kinetic energy of the system. 3. A molecule in a gas container hits the wall with speed v at an angle with the normal and rebounds with the same speed as shown in Fig. 7.21. Which of the following statements are true? (a) The momentum of the system is conserved in the collision. (b) The momentum of the molecule before collision with the wall is equal to the momentum of the molecule after collision. (c) The collision is elastic. (d) The collision is inelastic. Wall

v

v

Fig. 7.21

4. A particle A suffers an oblique elastic collision with a particle B that is at rest initially. If their masses are the same, then, after the collision (a) they will move in the opposite directions (b) A continues move in the original direction while B remains at rest (c) they will move in the mutually perpendicular directions (d) A comes to rest and B starts moving in the direction of the original motion of A 5. Choose the correct statements from the following: (a) The general form of Newton’s second law of motion is Fext = ma. (b) A body can have energy and yet no momentum. (c) A body having momentum must necessarily have kinetic energy. (d) The relative velocity of two bodies in a head– on collision remains unchanged in magnitude and direction 6. A ball of mass m moving horizontally at a speed v collides with the bob of a simple pendulum at rest. The mass of the bob is also m. (a) If the balls stick together, the height to which the two balls rise after the collision is

v2 . 8g

(b) If the balls stick together, the kinetic energy of the system immediately after the collision becomes half of that before collision. (c) If the collision is perfectly elastic, the bob of v2 . the pendulum will rise to a height of 2g (d) If the collision is perfectly elastic, the kinetic energy of the system immediately after the collision is equal to that before collision. 7. A body of mass 1 kg, initially at rest explodes into three fragments of masses in the ratio of 1 : 1 : 3. lar to each other with a speed of 30 ms–1, one along the + x direction and the other along the + y direction. Then (a) the speed of the heavier fragment will be 10 2 ms–1. (b) the speed of the heavier fragment will be 15 2 ms–1. (c) the direction of motion of the heavier fragment will be at angle of 135° with the + y direction (d) the direction of motion of the heavier fragment will be at an angle of 45° with the + x direction. 8. Two bodies A and B of masses m and 2m respecnected by a spring of spring constant k. A third body C of mass m moves with a velocity v0 along the line joining A and B and collides elastically with A as shown in Fig. 7.22. At a certain instant of time t0 after the collision, it is found that A and B have the same velocity v and at this instant, the compression of the spring is x0. Then (a) v =

v0 2

(b) v =

v0 3

(c) k =

2 mv02 3 x02

(d) k =

3 mv02 2 x02

v0

Fig. 7.22

9. A block of mass M attached to a light spring of force constant k rests on a horizontal frictionless surface as shown in Fig. 7.23. A bullet of mass m moving with a horizontal velocity v strikes the block and gets embedded in it. The velocity of the block with the bullet in it just after impact is V. If the impact compresses the spring by an amount x, then

Conservation of Linear Momentum and Collisions 7.21

(a) v = [k(M + m)]1/2 (b) v =

x m

1/ 2

2k M m

x

(c) V = [2k (M + m)]1/2 (d) V =

1/ 2

k M

x m

Fig. 7.25

x

m

Fig. 7.23

10. A body P of mass 1 kg moving with a velocity of 3 ms–1 along the + x direction collides head-on with a body Q of mass 2 kg at rest. The collision is elastic. After the collision (a) P moves along the +x direction with a velocity of 1 ms–1. (b) P moves along the –x direction with a velocity of 1 ms–1. (c) Q moves along the +x direction with a velocity of 2 ms–1. (d) Q moves along the –x direction with a velocity of 2 ms–1. 11. A block of mass M with a massless spring of force constant k is resting on a horizontal frictionless surface (Fig. 7.24). A block of mass m projected horizontally with a speed u collides and sticks to the spring at the point of maximum compression of the spring.

Fig. 7.24

If v is the velocity of the system after mass m sticks to the spring and n is the fraction of the initial kinetic energy of mass m that is stored in the spring, then (a)

v u

(c) n =

M (M

m) M

(M

m)

(b)

v u

(d) n =

The track is frictionless and the collision is elastic. If vA and vB are the velocities of A and B after the collision, then (a) vA = 0, vB = 1.4 ms–1 (b) vA = 1 ms–1, vB = 1 ms–1 (c) vA = – 1.4 ms–1, vB = 1.4 ms–1 (d) vA = – 1 ms–1, vB = 1 ms–1 13. Two blocks, each of mass m, moving in opposite directions with the same speed u, on a horizontal frictionless surface, collide with each other, stick together and come to rest. Then (a) work done by external force on the system is zero. (b) work done by the external force on the system is mu2. (c) work done by the internal force on the system is zero. (d) work done by the internal force on the system is – mu2. 14. A ball P of mass m1 moving with velocity u collides head-on with a stationary ball Q of mass m2. The collision is perfectly elastic. After the collision (a) if m1 = 2 m2, balls P and Q move in the same direction with speeds in the ratio of 1 : 4. (b) if m1 = 3 m2, balls P and Q move in the same direction with speeds in the ratio of 1 : 3. (c) if m2 = 2 m1, balls P and Q move in opposite directions with speeds in the ratio of 1 : 2. (d) if m2 = 3 m1, balls P and Q move in opposite directions with equal speeds. 15. A ball P of mass m1 moving with a velocity u collides obliquely with a stationary ball Q of mass m2. The collision is perfectly elastic. After the coloriginal direction of ball P as shown in Fig. 7.26.

m (M

m) m

(M

m)

12. A small ball A slides down the quadrant of a circle as shown in Fig. 7.25 and hits the ball B of equal mass which is initially at rest.

Fig. 7.26

7.22 Comprehensive Physics—JEE Advanced

(a) if m1 = m2, balls P and Q angles to each other with the same speed. (b) if m1 = m2, balls P and Q 60° with each other with the same speed. (c) if m1 = 2 m2, balls P and Q angles to each other with speeds in the ratio of 1 : 2. (d) if m1 = 2 m2, balls P and Q angle of 60° with each other with speeds in the ratio of 1 : 2. 16. Which of the following statements is/are incorrect in the case of an elastic collision between two bodies? (a) If two balls of the same mass moving with the same speed in opposite directions, collide head-on, then after the collision they move in opposite direction with the speed each ball had before collision. (b) If a body suffers a head-on collision with another body of the same mass but at rest, then dead and the second body moves with the e = 1 for a perfectly elastic collision. (d) If a body P collides head-on with a body Q of the same mass but at rest, then the percentage fraction of kinetic energy transferred from P to Q is 50%. 17. Which of the following statements is/are true in the case of an inelastic collision between two bodies? (a) The vector sum of the linear momenta of the two bodies before collision is equal to the vector sum of the linear momenta after the collision in the case of both one-dimensional and two-dimensional collisions. (b) The total energy of the system is conserved. (c) The two bodies stick together after inelastic collision. (d) If a body collides with another body of the same mass but at rest and two bodies stick together, the ratio of the total kinetic energy before and after collision is 2 : 1. 18. A body P of mass 1 kg moving a velocity of 15 ms–1 collides head-on with a stationary body Q 1/3, then (a) velocity of P after collision will be 5 ms–1. (b) velocity of Q after collision will be 10 ms–1. (c) the loss of kinetic energy of the system is 50 J. (d) the percentage fractional decrease in the kinetic energy of body P is 50%.

19. A body P of mass m1 moving with a velocity u1 = (a i + b j ) collides with a stationary body Q of m2. After the collision body P is found to move with velocity v1 = (c i + d j ) where a, b, c and d are constants. Then (a) Impulse received by P is m1[(a – c) i + (b – d) j ] (b) Impulse received by P is m1[(c – a) i + (d – b) j ] (c) Impulse imparted to Q is m1[(a – c) i + (b – d) j ] (d) Impulse imparted to Q is m2[(a – c) i + (b – d) j ] 20. A billiards ball C of mass m moving with velocity u collides two identical balls A and B in contact and at rest. After the collision, ball C is stopped dead and balls A and B move along directions shown in Fig. 7.27 with the same speed v. Then

Fig. 7.27

(a) v = (b) v =

u 3 u

2 1 (c) Loss of kinetic energy = mu2 3 1 (d) Loss of kinetic energy = mu2 6 21. A U-238 nucleus emits an alpla particle and changes into Th-234. In this process (a) the momentum of Th-234 is equal and opposite to that of the alpha particle. (b) the magnitude of the momentum of Th-234 is greater than that of the alpha particle. (c) the kinetic energy of the alpha particle is equal to that of Th-234. (d) the kinetic energy of the alpha particle is greater than that of Th-234.

Conservation of Linear Momentum and Collisions 7.23

22. collision. In this case (a) the momentum of the ball just after the collision is the same as that just before collision, (b) the mechanicial energy of the ball remains the same in the collision. (c) the total momentum of the ball and the earth is conserved. (d) the total energy of the ball and the earth is conserved. IIT, 1986 23. Two balls having linear momenta p1

p i and

is no external force acting on the balls. If p 1 and p 2 are option (s) is/are not allowed for any non-zero value of p, a1, a2, b1, b2, c1 and c2 (a) p 1 = a1 i

b1 j

(b) p 1 = c1 k ; p 1

c1 k ; p 2

a2 i

b2 j

c2 k ;

(c) p 1 = a1 i

b1 j

c1 k ; p 1

(d) p 1 = a1 i

b1 j ; p 2

a2 i

a2 i

b2 i

c1 k

b1 i IIT, 2008

p2 = – p i undergo a collision in free space. There SOLUTIONS

1. The correct choices are (b) and (c). 2. Choice (a) is false. In an elastic collision of two bodies, the speeds of the bodies change due to collision. Therefore, the momentum and energy of each body will change but the total momentum and total energy of the system of two bodies are conserved. Choice (b) is also false. The total energy of an isolated system is conserved. If external forces act on the system, the total momentum and energy will change. Choice (c) is true. For a non-conservative force such as friction, the work done over a closed loop is not zero. Choice (d) is also true. In an inelastic collision, the two bodies stick together after colliding. This results in heat or sound energy which is dissipated at the expense of kinetic energy. Hence choices (c) and (d) are correct. 3. The system consists of the molecule and the wall. Let us assume that initially the wall is stationary so that its momentum and kinetic energy are both zero before the collision. Therefore, the total momentum of the wall + molecule system before the collision is P = 0 + mv, where m is the mass of the molecule. After the collision the wall acquires a recoil velocity, say, V and a recoil momentum MV where M is the mass of the wall. After the collision, the recoil momentum of the wall + momentum of the outgoing molecule = momentum of the incoming molecule so that the total momentum of the system is conserved. Notice that the momenta of outgoing and incoming molecules are not the same, their directions are different. mentum produces a negligible velocity so that the kinetic energy of the wall is negligible after the collision. Since the speed v of the molecule is the same before and after collision, its kinetic energy

remains unchanged. Hence the total kinetic energy is also conserved. Therefore, the collision is elastic. Remember, in an inelastic collision, although the total momentum is conserved, the total kinetic energy is not conserved, it decreases. Hence the correct choices are (a) and (c). 4. From the principle of conservation of momentum, mentum. Momentum conservation is possible in cases (c) and (d). In case (c), the two masses should move in mutually perpendicular directions with velocity v/ 2 each inclined at 45° with the original direction of motion of particle A. In case (d), particle B must move with velocity v in the original direction of motion of A. Hence the correct choices are (c) and (d). 5. The general form of Newton’s second law is Fext =

dp d dv dm = (mv) = m +v dt dt dt dt

The form Fext = ma is valid only if

dm = 0, i.e. dt

if mass does not change with time. Hence choice (a) is incorrect. Choice (b) is correct because a body at rest may have potential energy and yet no momentum. Choice (c) is also correct. A body has momentum if it has mass and velocity and a body having a mass and velocity must have kinetic energy. Choice (d) is incorrect because the relative velocity remains unchanged in magnitude and gets reversed in direction; (v2 – v1) = – (u2 – u1). Hence the correct choices are (b) and (c).

7.24 Comprehensive Physics—JEE Advanced

6. In the inelastic collision, two bodies stick together. After the collision, the speed of the ball and the bob (sticking together) is v = v/2. The height to which they will rise is given by v = or

h =

2gh v2 2g

v2 8g

Mass of the ball and the bob sticking together is 1 1 2m m = 2 m. KE after collision = m v 2 = 2 2 v 2 1 1 = mv2. KE before collision = mv2. 2 4 2 Therefore, their ratio is 1: 2. In an elastic collision between two bodies of the same mass with one of them initially at rest, the moving body is brought to rest and the other moves in the same direction with the same speed. Thus the ball will come to rest and the bob of the pendulum acquires a speed v. At this speed, it will rise to height h given by h = v2/2 g. Thus all four choices are correct. 7. Let m1, m2 and m3 be the masses of the three fragments. As the total mass is 1 kg and m1 : m2 : m3 = 1 : 1 : 3, we have m1 = m2 = 0.2 kg and m3 = 0.6 kg. The linear momentum of m1 is p1 = m1 v1 = 0.2 30 = 6 kg ms–1 and let it be directed along the x-axis (Fig. 7.28).

Fig. 7.28

The linear momentum of m2 is p2 = m2 v2 = 0.2 30 = 6 kg ms–1 and let it be directed along the y-axis. The magnitude of the resultant momentum is p = ( p 12 + p 22)1/2 = [(6)2 + (6)2]1/2

or = 45° with the x or y axes. From the principle of conservation of linear momentum, the magnitude of the momentum of the third fragment is (here v3 is the speed of the heavier fragment) p3 = m3 v3 = 6 2 But m3 = 0.6 kg. Therefore, 6 2 10 2 = 14.1 ms–1 0.6 The direction of the velocity of the heavier fragment is inclined with x or y axes at an angle of 135° (see Fig. 7.28). The correct choices are (a) and (c). 8. Initially (i.e. before collision) bodies A and B are at rest and C is moving to the right (towards A) with a velocity v0. At a certain instant, say t = 0, C collides with A. Since the collision is elastic and A and C have equal masses, the entire momentum (mv0 ) and 1 kinetic energy m v02 of C are transferred to A 2 and hence C comes to rest. Thus at t = 0, A moves to the right with a velocity v0 and at this instant the spring is uncompressed and B is at rest. Hence the momentum of the system at t = 0 is (mv0). When A moves to the right, it compresses the spring and as a result body B begins to move to the right. It is given that at time t = t0, the compression of the spring is x0. Let v be the common velocity of A and B at this instant. From the principle of conservation of linear momentum, we have momentum of C before collision = momentum of A after collision + momentum of B after collision v (1) or mv0 = mv + (2m)v or v = 0 3 From the principle of conservation of energy, we have KE of C before collision =(KE of A + KE of B) after collision + PE in stored spring v 3=

1 1 1 1 mv20 = mv2 + (2m)v2 + kx 20 2 2 2 2 where k is the spring constant. Thus mv 20 = 3mv 2 + kx 02 Using (1) in (2), we get or

= 6 2 kg ms–1 The direction of the resultant momentum is given by p tan = 2 = 1 p1

mv 20 = 3m

v0 3

2

+ kx 02

2 m v02 2 m v02 = kx 02 or k = 3 3 x02 Thus the correct choices are (b) and (c). or

(2)

Conservation of Linear Momentum and Collisions 7.25

9. PE stored in the spring =

1 kx2. 2

or

mv = (M + m)V or V =

mv ( M + m)

(1)

After collision, KE of block + bullet in it = PE of the spring. Thus 1 1 (M + m)V 2 = kx2 2 2 k which gives V = x ( M m) Using Eq. (2) in (1), we have v= =

(M

m) V m

(M

m) m

1/ 2

k M

(2)

m

x m Thus the correct choices are (a) and (d). 10. From conservation of momentum, we have m 1u 1 + m 2u = m 1v 1 + m 2v 2 u1 + 0 = 1

v1 + 2

v2

1 (u1 – v1) (1) 2 From conservation of kinetic energy, we have v2 =

1 1 1 1 m1u12 + m2u22 = m1v12 + m2v22 2 2 2 2 2 2 2 u1 = v1 + 2v2 v12 = u21 – 2v22

(2)

Solving Eqs. (1) and (2) we get v1 = – 1 ms–1 and v2 = + 2 ms–1. Hence the correct choices are (b) and (c). 11. For conservation of momentum and conservation of total energy, we have mu = (M + m) v (1) 1 1 1 mu2 = (M + m)v2 + kx2 2 2 2 1 mu2, we get Dividing Eq. (2) bv 2 Also

1=

(M

m) v 2 mu 2

1 2 kx 2 1 mu 2 2

(3)

v m u ( M m) Using this in Eq. (3), we get

From Eq. (1), we have

1 2 kx M n= 2 = 1 ( M m) mu 2 2 Thus the correct choices are (b) and (c). 12. If uA is the velocity with which A strikes B, then 1 muA2 = mgh 2 uA = 2gh = 2 9.8 0.1 = 1.4 ms–1. Since the masses of the balls are equal and the collision is elastic, they exchange their velocities after collision. Hence the only correct choice is (a). 13. Total initial momentum before collision = mu + (– mu 0, as the blocks come to rest. Since the change in momentum is zero, no external force acts on the system. Hence no work is done by the external force on the system. From work-energy principle,

x

= [k (M + m)]1/2

1

1 2 kx ( M m) v 2 2 =1– 1 mu 2 mu 2 2

(2)

1 2 1 2 mu mu = – mu2. Hence 2 2 the correct choices are (a) and (d). 14. Let v1 and v2 be the velocities of balls P and Q respectively after the collision. Then we have (1) m 1u = m 1v 1 + m 2v 2 initial K.E = 0 –

and

1 1 1 m1u12 = m1v21 + m2v22 2 2 2 2 2 m 1u 1 = m 1v 1 + m 2v 22

(2)

(a) Putting m1 = 2m2 in Eqs. (1) and (2) and u 4u and v2 = . solving them, we get v1 = 3 3 Thus v1 = v2/4. u 3u (b) For m1 = 3m2, we get v1 = and v2 = . 2 2 Thus v1 = v2 /3. u 2u (c) For m2 = 2 m1, we get v1 = – and v2 = 2 3 giving v1 = – v2 /3 (d) For m2 = 3m1, we get v1 = –

u u and v2 = 2 2

giving v1 = – v2 Hence all the four choices are correct.

7.26 Comprehensive Physics—JEE Advanced

15. From conservation of x and y components of momentum we have m1u = m1v1 cos + m2v2 cos 0 = m1v1 sin

and

(1)

m 1v 1 = m 2v 2

(2)

Percentage fractional decrease in K.E. of P is 1 2 1 2 mu v K 2 1 = 2 1 2 K mu 2

If m1 = m2, then from Eqs. (1) and (2), we get v1 = v2 and u = 2v1 cos

= =

1 2

=

Hence choice (a) is correct and choice (b) is wrong.

v2 3 and cos = = 30°. 2 2 Hence choice (d) is correct and choice (c) is wrong. 16. The only incorrect statement is (d). Refer to the solution of Q.34 of Section I. v1 =

4m1m2 K 4m m = = = 1 or 100 % 2 K ( m m) 2 (m1 m2 ) 17. The only incorrect statement is (c). Statements (a), (b) and (d) are true. 18. Given m = 1 kg, u = 15 ms–1 and e = 1/3. Let v1 and v2 be the velocities of P and Q after the collision. mu + 0 = m v1 + m v2 u = v1 + v2 (1) v1

(2)

u

From Eqs. (1) and (2), we get v1 =

u 15 (1 – e) = 2 2

u 15 and v2 = (1 + e) = 2 2 Total K.E. before collision =

1

1 = 5 ms–1 3

1 = 10 ms–1 1 3 1 mu2 2

v12

100

u2 (15) 2

(5)2

19.

Similarly putting m1 = 2 m2 in Eqs. (1), (2) and (3), we get

v2

u2

100

100 89% (15) 2 Hence the correct choices are (a), (b) and (c).

= 45°.

e=

1 1 m v12 + m v22 2 2

1 1 (52+ 102) = 62.5 J 2 Loss in K.E. = 112.5 – 62.5 = 50 J.

(3)

Using these in Eq. (3), we get cos2

(15)2 = 112.5 J

1

=

From conservation of kinetic energy, we have 1 1 1 m1u2 = m1 v12 + m2 v22 2 2 2 m1 u2 = m1 v12 + m2 v22

1 2

Total K.E. after collision =

– m2v2 sin

which give m1u = (m1v1 + m2v2) cos and

=

v2 is the velocity of ball Q after the collision, then from the conservation of momentum, we have m 1u + 0 = m 1 v 1 + m 2 v 2 m2 v2 = m1(u – v1) = m1[(a i + b j ) – (c i + d j )] = m1[(a – c) i + (b – d) j ] Impulse received by P is IP = m1 v1 – m1 u = m1(v1 – u) = m1[(c – a) i + (d – b) j ] Impulse imparted to Q is IQ = m2 v2 – 0 = m1[(a – c) i + (b – d) j ] Hence the correct choices are (b) and (c). 20. Conservation of x-component of momentum gives mu = mv cos 30° + mv cos 30° = 2 mv v=

u 3

3 = 2

3 mv

. Hence choice (c) is wrong.

K.E. before collision =

1 mu2 2

Conservation of Linear Momentum and Collisions 7.27

K.E. after collision = =

1 1 mv2 + mv2 = mv2 2 2

=

mu 2 3

1 1 1 mu2 – mu2 = mu2. 2 3 6 Hence the correct choices are (a) and (d). 21. Initially U-238 is at rest and has zero momentum. When it emits an -particle, the sum of the momenta of -particle and Th-234 nucleus must be zero, i.e. v m m v + mTh vTh = 0 = – Th m vTh

Hence

K K TH

mTH m

2

=

mTH m

Since mTH > m ; K > KTh. Hence the correct choices are (a) and (d). 22. In a collision, the momentum of indivdual bodies is not conserved. In an inelastic collision, there is a loss of kinetic energy. Hence choices (a) and (b) are incorrect. The correct choices are (c) and (d). 23.

Loss of K.E. =

Now KTh =

m mTH

momentum is pi

p1

p2

pi

p i = 0. There-

c1 = 0 and in choice (b) and (c) for non-zero values of the

1 1 2 mTh vTh and K = m v2 2 2 m v 2 = mTH vTH

i and j . Hence the choices (a) and (d) are not allowed.

III Multiple Choice Questions Based on Passage Questions 1 to 3 are based on the following passage Passage I Collisions: In physics we come across many examples of collisions. The molecules of a gas collide with one another and with the walls of the container. The collision of a neutron with an atom is well known. In a nuclear reactor fast neutrons down. They are, therefore, made to collide with hydrogen atoms. The term collision does not necessarily mean that a particle or a body must actually strike another. In fact, two particles may not even touch each other and yet they the other. When two bodies collide, each body exerts an equal and opposite force on the other. The fundamental conservation laws of physics are used to determine the velocities of the bodies after the collision. Collision may be elastic or inelastic. Thus a as an event in which two or more bodies exert relatively strong forces on each other for a relatively short time. The forces that the bodies exert on each other are internal to the system. Almost all the knowledge about the sub-atomic particles such as electrons, protons, neutrons, muons, quarks, etc. is

obtained from the experiments involving collisions. There are certain collisions called nuclear reactions in which new particles are formed. For example, when a slow neutron collides with a uranium-235 nucleus, new nuclei baruim-141 and krypton-92 are formed. This collideuterium and trituim collide (or fuse) to form a helium nucleus with the emission of a neutron. 1. Which one of the following collisions is NOT elastic? (a) A hard steel ball dropped on a hard concrete (b) Two balls moving in the same direction collide and stick to each other. (c) Collisions between molecules of an ideal gas (d) Collisions of fast neutrons with hydrogen 2. Which one of the following statements is true about inelastic collisions? (a) The total kinetic energy of the particles after collision is equal to that before collision. (b) The total kinetic energy of the particles after collision is less than that before collision. (c) The total momentum of the particles after collision is less than that before collision.

7.28 Comprehensive Physics—JEE Advanced

(d) Kinetic energy and momentum are both conserved in the collision. 3. In elastic collisions (a) only energy is conserved

(b) only momentum is conserved (c) neither energy nor momentum is conserved (d) both energy and momentum are conserved.

ANSWERS

1. The correct choice is (b) 2. The correct choice is (b) Questions 4 and 5 are based on the following passage Passage II Two balls marked 1 and 2 of the same mass m and a third ball marked 3 of mass M are arranged over a smooth horizontal surface as shown in Fig. 7.29. Ball 1 moves with a velocity v1 towards balls 2 and 3. All collisions are assumed to be elastic.

3. The correct choice is (d) 4. If M < m, the number of collision between the balls will be (a) one (b) two (c) three (d) four 5. If M > m, the number of collisions between the balls will be (a) one (b) two (c) three (d) four

Fig. 7.29

SOLUTION 4. both have the same mass, after the collision ball 1 with come to rest and ball 2 will move with speed v1. The ball will collide with the stationary ball 3. After this second collision, let v2 and v3 be the speeds of balls 2 and 3 respectively. Since the collision are elastic, v2 and v3 are given by (see page 7.2) m M v2 = v1 (i) m M 2m v1 (ii) m M If M < m, it follows from (i) and (ii) that v2 < v3 and both have the same direction. Therefore, ball and

v3 =

Questions 6 to 8 are based on the following passage Passage-III A body of mass m1 = m moving with a velocity v1 = v in the x-direction collides with another body of the same mass m2 = m moving in the y-direction with the same speed v2 = v. They coalesce into one body during the collision. 6. The magnitude of the momentum of the composite body is (a) mv (b) 2 mv (c) 2mv

(d) 2 2 mv

2 cannot collide with ball 3 again. Hence there are only two collisions. Thus the correct choice is (b). 5. If M > m, we have from Eq. (i) M m v1 M m The negative sign indicates that, after the second collision, ball 2 will move in opposite direction lision. Therefore, ball 2 will make another collision with ball 1. Hence, in this case, there are three collisions in all between the balls. Thus the correct choice is (c). v2 = –

7. The angle which the direction of the momentum vector of the composite body makes with the x-axis is (a) 30° (b) 45° (c) less than 30° (d) greater than 45° 8. The fraction of initial kinetic energy transformed into heat during the collision is 1 1 (b) (a) 2 4 (c)

2 3

(d)

1 3

Conservation of Linear Momentum and Collisions 7.29

SOLUTION 6. Refer to Fig. 7.30. Let v be the velocity of the composite body and let be the angle which the velocity vector v makes with the x-axis.

7. Diving Eq. (2) by Eq. (1), we have tan 45°, which is choice is (b). 8. Initial kinetic energy is 1 1 m1v21 + m 2v 22 Ki = 2 2 1 1 mv2 + mv2 = 2 2 From Eq. (3), we have v = v/ energy is 1 Kf = (m1 + m2)(v )2 2 1 v2 = 2m = 2 2 Loss in K.E. = Ki – Kf =

Fig. 7.30

Conservation of x and y components of momentum gives m1v1 = (m1 + m2)v cos mv = 2mv cos (1) m2v2 = (m1 + m2)v sin

mv = 2mv sin

(2)

Squaring and adding Eqs. (1) and (2), we get 2(mv)2 = (2mv )2

2mv =

2 mv

(3)

Momentum of composite body is 2 mv . Hence the correct choice is (b). Questions 9 to 11 are based on the following passage Passage IV A ball P moving with a velocity u strikes an identical stationary ball Q such that after the collision, the direction of motion of balls P and Q make an angle of 30° with the original direction of motion of ball P, as shown in Fig. 7.31.

= mv2 –

=

mv2 2 . Final kinetic

1 mv2 2

1 mv2 2

1 mv2 2 Fraction of initial K.E. transformed into heat is 1 2 mv K 1 = 2 2 = . Ki 2 mv Hence the correct choice is (a). =

9. The speed v1 of ball P after the collision is u u (b) (a) 2 3 (c)

u

(d)

u

2 3 10. The speed v2 of ball Q after the collision is u 3 3 2u 2u (c) (d) 3 3 11. The ratio of the total kinetic energy of the balls after collision to that before collision is 1 1 (b) (a) 3 3 2 1 (c) (d) 3 2 (a)

Fig. 7.31

=1

u

(b)

SOLUTION From conservation of x and y components of momentum, mu = mv1 cos 30° + mv2 cos 30°

3 2 0 = mv1 sin 30° – mv2 sin 30°

u = (v1 + v2) and

(1)

7.30 Comprehensive Physics—JEE Advanced

v1 = v2 9. Eqs. (1) and (2) give v1 = 10. v2 = v1 =

(2) u 3

, which is choice (d).

u

and after collision 1 1 Kf = mv21 + mv22 2 2

. Hence the correct choice is (a). 3 11. Total kinetic energy before collision is 1 Ki = mu2 2

Kf Ki

Questions 12 to 14 are based on the following passage Passage V A body P of mass m moving with a velocity u along the + x-direction makes a head-on elastic collision a body Q of mass 2 m at rest. 12. After the collision, body P moves along the (a) positive x-direction with speed u/3. (b) negative x-direction with speed u/3. (c) positive x-direction with speed 2u/3. (d) negative x-direction with speed 2u/3.

SOLUTION Let V be the velocity of body Q after the collisions. From the principle of conservation of linear momentum, we have mu = mv + (2m)V or

u – v= 2 V

(1)

The conservation of kinetic energy gives

or

u2 – v2 = 2V2

or (u – v) (u + v) = 2 V 2

(2)

Using Eq (1) in Eq (2), we have 2V(u + v) = 2V2 or u + v = V or

2(u + v) = 2V

=

2 , which is choice (c). 3

Passage VI A body A of mass m1 moving with a velocity u makes a headon elastic collision with a body B of mass m2 initially at rest.

=

mu 2 3

14. What fraction of its kinetic energy does body P lose after the collision? (a)

8 9

(b)

7 8

(c)

6 7

(d)

5 6

1 mu2 2 1 Final kinetic energy, Kf = mv2 2 Loss in kinetic energy is 1 1 K = Ki – Kf = mu2 – mv2 2 2 Ki =

K Ki

1 2 1 2 mu mv 2 = 2 1 2 mu 2

(3)

Questions 15 to 17 are based on the following passage

+

13. After the collision, body Q moves along the (a) positive x-direction with speed u/3. (b) negative x-direction with speed u/3. (c) positive x-direction with speed 2u/3. (d) negative x-direction with speed 2u/3.

=

12. From Eqs. (1) and (3), we get v = – u/3. Hence the correct choice is (b). 13. Using v = – u/3 in Eq. (1), we get V = 2u/3. Hence the correct choice is (c). 14. Initial skinetic energy of the colliding mass is

1 u2 m 3 2

1 m 2

Fractional loss =

1 1 1 mu2 = mv2 + (2m)V2 2 2 2

u2 3

=

u2

v2 u2

=1– =

1 3

v u

=1–

2

2

(

v = – u/3)

8 , which is choice (c). 9

15. After the collision, body B will move with the greatest speed if (b) m1 << m2 (a) m1 >> m2 (c) m1 = m2 (d) m1 = 2 m2

Conservation of Linear Momentum and Collisions 7.31

16. After the collision, body B will move with the greatest momentum if (b) m1 << m2 (a) m1 >> m2 (c) m1 = m2 (d) m1 = 2 m1

17. After the collision, body B will move with the greatest kinetic energy if (b) m1 << m2 (a) m1 >> m2 (c) m1 = m2 (d) m2 = 2 m1

SOLUTION

Since the collision is elastic, both linear momentum and kinetic energy are conserved. Thus (1) m1 u1 = m1 v1 + m2 v2 1 1 1 m1 u21 = m1v21 + m2v22 2 2 2 Solving for v1 and v2, we get and

and

v1 =

m1 m1

m2 m2

v2 =

2m1 u1 m1 m2

m1

2

(4)

2

– 2 m1m2 + 2 m1m2

Questions 18 to 20 are based on the following passage Passage VII A body P of mass m moving along the positive x-direction with velocity u collides with a body Q of mass M initially at rest. After the collision body P moves along the positive y-direction and body Q moves along a direction making an angle below the x-axis. The collision is assumed to be elastic. 18. The angle is give by M m (b) = tan–1 (a) = tan–1 m M (c)

= tan–1

M M

m m

(d)

2

+ 2 m1m2

2m1m2u m1

m2

2

2 m1m2 2

(3)

m2

m2

Momentum p2 will be maximum if the denominator

u1

+

m1

p2 =

(2)

15. Equation (4) gives the recoil speed of body B which can rewritten as 2u1 v2 = m2 1 m1 Now v2 will be maximum if the denominator is the m minimum, i.e. if 2 << 1 or m2 << m1. Thus body m1 B will move with the greatest speed if its mass is very small compared to the mass of body A. Hence the correct choice is (a). 16. The momentum of body B after the collision is 2m1m2u p2 = m2 v2 = m1 m2 The denominator (m1 + m2) can rewritten as m1 + m2 =

=

= tan–1

M M

m m

is the minimum, i.e. if m1 m2 = 0 since a perfect square can never be negative. Thus for p2 to m2 or m1 = m2. Hence body be maximum m1 B will move with the greatest momentum if its mass is equal to the mass of body A. Hence the correct choice is (c). 17. The kinetic energy of body B after the collision is K2 =

2m1u 1 1 m2v22 = m2 m1 m2 2 2 =

2

2m12 m2u 2

m1 m2 2 The denominator (m1 + m2)2 can be rewritten as (m1 + m2)2 = (m1 + m2)2 – 4m1 m2 + 4m1 m2 = (m1 – m2)2 + 4 m1 m2 K2 =

2m12 m2u 2 m1

m2

2

4m1m2

Energy K2 will be maximum if the denominator is the minimum, i.e. if (m1 – m2)2 = 0 or m1 = m2. Hence body B will move with maximum energy if ita mass is equal to the mass of body A. Thus the correct choice is (c). 19. If M = 2m, the speed of body P after the collision will be u 2 u (c) 3

(a)

(b) (d)

u 2 u 3

20. If M = 2m, the kinetic energy gained by body Q due to collision is 2 mu2 3 1 (c) mu2 2

(a)

(b)

1 mu2 3

(d) mu2

7.32 Comprehensive Physics—JEE Advanced

SOLUTION 18. Using Eqs. (1) and (2) in Eq. (3), we get

Refer to Fig. 7.32.

M m 1/ 2 M m and cos = 2M 2M Hence the correct choice is (c). sin

1/ 2

=

19. If M = 2m, then sin

2m m 2 2m

=

1/ 2

=

1 2

= 30° Dividing Eq. (2) by Eq. (1), we get v = u tan

Fig. 7.32

Conservation of x and y components of momentum gives mu = MV cos (1) 0 = mv – MV sin mv = MV sin (2) From conservation of kinetic energy we have (3)

Questions 21 to 24 are based on the following passage Passage VIII A ball of mass m is dropped from a height h on a smooth e. 21. The total distance covered by the ball before it comes to rest is

(c) h

1 e (b) h 1 e

1 e2

2

1 e2

2h 1 e g 1 e

(b)

20. Similarly we get V = in K.E. of Q is

u 3

3

. Therefore, the increase

1 1 MV2 – 0 = MV2 2 2

1 u 2 1 2m = mu2. 2 3 3 Hence the correct choice is (b). 1/ 2

2h 1 e 2 2h 1 e 2 (c) (d) g 1 e g 1 e2 23. The total momentum imparted by the ball to the

(a) m 2hg

1 e2 1 e

2

(c) m 2hg

1/ 2

(d) h 1 e2 1 e2 22. The total time taken by the ball to come to rest is (a)

Hence the correct choice is (d).

u

=

1 1 1 mu2 = mv2 + MV2 2 2 2

1 e (a) h 1 e

= u tan (30°) =

2h 1 e 2 g 1 e2

(b) m 2hg

1 e 1 e

(d) me 2hg

24. The average force exerted by the ball on the wall is 1 e (a) mg (b) mg 1 e 1 e2 (c) mg (d) zero 1 e2

SOLUTION 21.

v = 2gh . It will rebounce with a speed v1 = ev and will rise to a height h1 given by v21 = 2g h1

v2 e2 v2 e2 2 gh which gives h1 = 1 = = = e2h. 2g 2g 2g 2

h1 = 2e h.

Between the second and the third impact, the distance covered = 2e4 h and so on. Therefore, the total distance covered by the ball before it stops is H = h + 2e2 h + 2e4 h + ... = h + 2e2 h(1 + e2 + ....) =h +

2e2 h

=h

1 e2

1 e2 1 e2 Hence the correct choice is (c).

Conservation of Linear Momentum and Collisions 7.33

22. Total time taken t = =

2h + 2e g

2h + 2e2 g

2h + g

2h 2h +2 e (1 + e + ....) g g

2h 2h e +2 = g g 1 e Hence the correct choice is (a). =

2h g

1 e 1 e

=m v

= mv

1 e 1 e

1 e 1 e Thus the correct choice is (b). = m 2gh

24. Average force =

momentum transferred time taken

p t

1 e 1 e = mg = 2h 1 e g 1 e Hence the correct choice is (a).

23. Total momentum transferred is p = m[v + 2(v1 + v2 + ...)]

m 2 gh

= m[v + 2(ev + e2v + ...)] = m[v + 2ev(1 + e + ...)] Questions 25 to27 are based on the following passage Passage IX

25. The speed of the block at point B immediately after it strikes the second incline is (a)

A small block of mass M moves on a frictionless surface of an inclined plane, as shown in Fig. 7.33. The angle of the incline suddenly change from 60° to 30° at point B. The block is initially at rest at A. Assume that collisions between the block and the incline are totally inelastic (g = 10 m/s2)

2ev 1 e

60 ms–1

(b)

45 ms–1

(c) 30 ms–1 (d) 15 ms–1 26. The speed of the block at point C, immediately before it leaves the second incline is (a)

120 ms–1

(b)

105 ms–1

(c)

90 ms–1

(d)

75 ms–1

27. If collision between the block and the incline is completely elastic, then the vertical (upward) component of the velocity of the block at point B, immediately after it strikes the second incline is (b) 15 ms–1 (a) 30 ms–1

Fig. 7.33

IIT, 2008

(c) 0

(d)

15 ms–1

SOLUTION 25. Let vB be the speed of the block just before it strikes the second incline. 1 mvB2 = mg AD = mg BD tan 60° 2 vB = 2 10

3 tan 60

1/ 2

=

60 ms–1

This velocity can be resolved into two components. vB cos 30° along the second incline and vB sin 30° perpendicular to it (see Fig.7.34) In an inelastic collision, the perpendicular component becomes zero after the collision. Hence the speed of the block at point B immediately after the collision is vB cos 30° =

60

3 2

45 ms–1.

Fig. 7.34

26. Let vC be the speed at C. From conservation of energy, gain in K.E. = loss in P.E., i.e. 1 2 1 mvC – mvB2 = mg 2 2

BE = mg

EC tan 30°

7.34 Comprehensive Physics—JEE Advanced

vC2 = vB2 + 2g = 45 + 2 vC =

3 3 tan 30° 1 10 3 3 = 105 3

105 ms–1

27. If the collision is perfectly elastic, the perpendicu1 lar component vB sin 30° = 60 = 15 ms–1 is 2 reversed after the collision as shown in the Fig.7.35. The resultant vertical component is 15 sin 60° –

Fig. 7.35

45 sin 30° = 0, which is choice (c).

IV Matching Match the Statements in Column I with the Processes in Column II Column I Column II (a) Collision of two light nuclei to form a heavier nucleus (p) Elastic collision (b) A speeding bullet getting embedded (q) Inelastic collision in a wooden plank (d) Collision in which there is no loss of kinetic energy

(s) Nuclear fusion

ANSWER (a) (c)

(s) (r)

(b) (d)

(q) (b)

V Reason-Assertion Type Questions In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each question has the following four choices out of which only one choice is correct. (a) Statement-1 is true; Statement-2 is true and Statement-2 is the correct explanation for Statement-1. (b) Statement-1 is true; Statement-2 is true but Statement-2 is not the correct explanation for Statement-1. (c) Statement-1 is true; Statement-2 is false. (d) Statement-1 is false; Statement-2 is true.

1. Statement-1 Two identical balls B and C lie on a horizontal smooth straight groove so that they are touching. A third identical ball A moves at a speed v along the groove and collides with B (see Fig. 7.36). If the collisions are perfectly elastic, then after the collision, balls A and B will come to rest and ball C moves with velocity v to the right.

Fig. 7.36

Conservation of Linear Momentum and Collisions 7.35

Statement-2 In an elastic collision, linear momentum and kinetic energy are both conserved. 2. Statement-1 Two bodies A and B of masses m and 2 m respecnected by a spring. A third body C of mass m moves with a velocity u and collides elastically with A as shown in Fig. 7.37. At a certain instant t0 after the collision, it is found that the velocities of A and B are the same = u/3.

8.

9.

v0

Fig. 7.37

Statement-2 In an elastic collision, the kinetic energy of the system is conserved. 3. Statement-1 In an inelastic collision between two bodies, the total energy does not change after the collision but the kinetic energy of the system decreases. Statement-2 The loss of kinetic energy appears as heat in the system. 4. Statement-1 In a collision between two bodies, each body exerts an equal and opposite force on the other at each instant of time during the collision. Statement-2 The total energy of the system is conserved. 5. Statement-1 The term ‘collision’ between two bodies does not necessarily mean that the two bodies actually strike against each other. Statement-2 In physics, a collision is said to take place if the one 6. Statement-1 In an inelastic collision, the two colliding bodies stick to each other after the collision and move with a common velocity. Statement-2 There is a lost of total kinetic energy in an inelastic collision. 7. Statement-1 In a collision between two bodies, the linear momentum each body remains constant.

10.

11.

Statement-2 If no external force acts, the total linear momentum of a system is conserved. Statement-1 In an elastic collision between two bodies, the energy of each body is conserved. Statement-2 The total energy of an isolated system is conserved. Statement-1 The total energy of a system is always conserved irrespective of whether external forces act on the system. Statement-2 The total energy of an isolated system is always conserved. Statement-1 A body P of mass M moving with speed u collides head-on and elastically with a body Q of m initially at rest. If m << M, body Q will have a maximum speed equal to 2 u after the collision. Statement-2 In an elastic collision, the momentum and kinetic energy are both conserved. Statement-1 A block of mass m starts moving on a rough horizontal surface with a velocity v. It stops due to friction between the block and the surface after moving through a certain distance. The surface is now tilted to an angle of 30° with the horizontal and the same block is made to go up on the surface with the same initial velocity v. The decrease in the mechanical energy in the second situation is smaller than that in the first situation. Statement-2 the surface decreases with the increase in the angle of inclination. IIT, 2007

12. Statement-1 In an elastic collision between two bodies, the relative speed of the bodies after collision is equal to the relative speed before the collision. Statement-2 In an elastic collision, the linear momentum of the system is conserved. IIT, 2007

7.36 Comprehensive Physics—JEE Advanced

SOLUTIONS 1. The correct choice is (a). Linear momentum will be conserved if A comes to rest and B and C move to the right with a velocity v/2 each or A, B and C all move to the right with velocity v/3 each. It is easy to see that in these two cases, the kinetic energy is not conserved. Hence the only result of the collision is the one given in Statement-1. 2. The correct choice is (b). Let C collide with A at t = 0. Since the collision in elastic and A and C have equal masses, C will come to rest and A will move to the right with velocity u and at this instant the spring is uncompressed and B is at rest. Hence at t = 0, the momentum of the system = mu. When A moves to the right, it compresses the spring and as a result B beings to move to the right. Let v be the common velocity of A and B at time t0. From the principle of conservation of linear momentum, we have Momentum of C before collision = momentum of A after collision + momentum of B after collision u or mu = mv + (2m) v v= . 3 3. The correct choice is (a). The total energy (which includes all forms of energy) is conserved in any process. 4. The correct choice is (b). Statement-1 follows from Newton’s third law of motion. 5. The correct choice is (a). 6. The correct choice is (d). The two colliding body need not get stuck after an inelastic collision. 7. The correct choice is (d). Since the velocities of the two bodies change due to collision, the linear momentum of each body will change but the total linear momentum of the system of two bodies is conserved. 8. The correct choice is (d). Due to change in velocity, the energy of each body changes on collision

but the total energy of the system of two bodies is conserved. 9. The correct choice is (d). If an external force acts on a system, it is accelerated which will increase the total energy. 10. The correct choice is (a). If v and V are the velocities of Q and P after the collision, then from conservation of momentum and kinetic energy, we have Mu = mv + MV M(u – V ) = mv (1) 1 1 1 Mu2 = mv2 + MV2 2 2 2 M(u – V) (u + V) = mv2 (2) From Eqs. (1), (2), we get 2u m 1 M If M >> m, then v is maximum equal to 2u (since m zero). M 11. Statement-1 is true. The decrease in mechanical energy is smaller when the block is made to go up on the inclined surface because some part of the kinetic energy is converted into gravitational potential energy. Statement-2 is false. The coefv=

2Mu M m

=

inclination of the plane. 12. For elastic collisions both the statements are true but statement-2 is not the correct explanation of statement-1. Linear momentum is conserved in elastic as well as inelastic collision. But in an elastic collision the total kinetic energy of the system is also conserved. Statement-1 is obtained if we use both the conservation of linear momentum as well as the conservation of kinetic energy in the case of an elastic collision.

VI Integer Answer Type 1. A shell of mass 4 kg, initially at rest explodes into three fragments. Two of the fragments, each of mass 1 kg are found to move with a speed of 2 ms –1 each in mutually perpendicular directions. Find the total energy (in joule) released in the explosion. IIT, 1987 2. A simple pendulum is suspended from a peg on a vertical wall. The pendulum is pulled away from

the wall to a horizontal position and released (see Fig. 7.38). The bob hits the wall, the restitution 2 / 5 . What is co the minimum number of impacts with wall after which the amplitude becomes less than 60°? IIT, 1988

Fig. 7.38

Conservation of Linear Momentum and Collisions 7.37

3. A cart is moving along the + x direction with a velocity of 4 ms–1. A person on the cart throws a stone with a velocity of 6 ms–1 relative to himself. In the reference frame of the cart, the stone is thrown in the y-z plane making an angle of 30° with vertical z-axis. At the highest point of its trajectory, the stone hits an object of equal mass hung vertically from the branch of a tree. A completely inelastic collision occurs in which the stone gets embedded in the object. Find the speed in ms–1 of the combined mass immediately after the collision with respect to an observer on the ground. IIT, 1997

4. Three objects A, B and C are kept in a straight line on a frictionless horizontal surface. These have masses m, 2m and m, respectively. The object A moves towards B with a speed 9 m/s and makes an elastic collision with it. Thereafter B makes completely inelastic collision with C. All motions occur in m/s) of the object C. [see Fig. 7.39]

Fig. 7.39

IIT, 2009

SOLUTIONS 1. Refer to Fig. 7.40. p1 = 1 kg

2 ms

–1

p2 = p1 = 2 kg ms p12

p=

= 2 kg ms

where e of the bob after the second rebound will be v2 = ev1 = e2v. Thus, the speed of the bob after n rebounds will be v n = env If the ball rises to a position C at a height h = BD after n rebounds, then from the principle of conservation of energy, we have 1 mvn2 = mgh 2

–1

–1

p22 = 2 2 kg ms–1

From the conservation of momentum, we have p3

p =0 p3 = – p =

2

v=

2 2 kg ms–1

2 2

2 ms–1

v=–

Total energy released 1 1 1 22 1 2 2 2 = 2+2+2=6J

=

2

1 2

2

en v

vn2 h= 2g 2

2

2

e2n v2 2g

2g

(2)

Using Eq. (1) in (2), we get 2 gle2 n (3) le2 n 2g For n is the angle the string subtends with vertical after n h = OB – OD = l – l cos n = l (1 – cos n) (4) h=

From Eqs. (3) and (4), we have le2n = l (1 – cos

n)

or e2n = 1 – cos

n)

Fig. 7.40

2. Refer to Fig. 7.41. When the bob is at the horizontal position A, its height above B = length of the string = l. Therefore, potential energy at A = mgl. When the bob is released, it hits the wall at B and the entire potential energy is converted into kinetic energy. If v is the velocity with which the bob hits the wall, then 1 mv2 = mgl or v = 2gl (i) 2 v1 = ev

Fig. 7.41

For n to be less than 60°, i.e. cos n is greater than 1 1 , (1 – cos n) must be less than . Thus 2 2 e2n <

1 2

7.38 Comprehensive Physics—JEE Advanced

Given, e =

4. Elastic collision of A with B. (Fig. 7.43)

2

. Therefore,

5

2n

2 5

<

1 or 2

4 5

n

<

1 2

1 or n log (0.8) < log (0.5) 2 0.3010 or or – (0.0969)n < – 0.03010 or n > 0.0969 n > 3.1 or (0.8)n <

The least integer greater than 3.1 is 4. Hence n = 4. 3. The x, y and z components of the initial velocity of the stone are (Fig. 7.42) ux = 4 ms–1 uy = (6 ms–1) sin 30° = 3 ms–1

Fig. 7.43

Momentum conservation gives mu1 + 2 mu2 = mv1 + 2 mv2 9m + 0 = mv1 + 2 mv2 9 = v1 + 2 v2

uz = (6 ms–1) cos 30° = 3 3 ms–1 When the stone reaches the highest point A of its parabolic path, the vertical i.e. z component of its velocity becomes zero but the x and y components remain unchanged, i.e. vx = ux = 4 ms–1 vy = uy = 3 ms–1 vz = 0 The speed of the stone at the highest point on its trajectory with respect to an observer on the ground is v = (v2x + v2y + v2z)1/2

v

v

(1)

Conservation of kinetic energy gives 1 m 2

(9)2 =

81 = v12

1 mv12 2

1 2m v22 2

2v22

(2)

Solving (1) and (2), we get we get v1 = – 3 ms–1 and v2 = 6 ms–1 Inelastic collision between B and C (Fig. 7.44) In a perfectly inelastic collision the two bodies stick together.

= (42 + 32 + 02)1/2 = 5 ms–1 v

v 6 ms–1

v 30°

Fig. 7.44

4

Momentum conservation gives 2 mv2 + 0 = (2m + m) v

ms–1

v= Fig. 7.42

2 2 v2 = 3 3

6 = 4 ms–1

8

Rigid Body Rotation

Chapter

REVIEW OF BASIC CONCEPTS 8.1

CENTRE OF MASS OF DISCRETE PARTICLES

that point where the entire mass of the system is imagined to be concentrated, for considerations of its translational motion. If r1, r2, r3, ... rn are the position vectors of masses m1, m2, m3, ...mn respectively, the centre of mass of the system is m r m2r2 m3r3 ......... mn rn rCM = 1 1 m1 m2 m3 ........ mn n

=

N 1 n

n

mn rn

N 1

= mn

N 1

8.1 Three particles of masses m1 = m, and m2 = 2m and m3 = 3m are placed at the corners of an equilateral triangle of side a as shown in Fig. 8.1. Locate the centre of mass of the system.

mn rn

Fig. 8.1

M

where M is the total mass of the system of particles. rCM is the weighted average of all the position vectors of the particles of the system, the contribution of each particle being proportional to its mass. For a system consisting of two particles, the centre of mass is m r m2r2 rcm = 1 1 m1 m2 If the masses are equal, i.e. m1 = m2, then 1 (r1 + r2) 2 which means that the centre of mass lies exactly in the middle of the line joining the two masses.

SOLUTION Take any one particle to be located at the origin O. The x and y coordinates of m1, m2, and m3 respectively a are x1 = 0 and y1 = 0, x2 = a and y2 = 0 and x3 = and 2 3a y3 = 2 The x and y coordinates of the centre of mass are xCM = m1 x1 m2 x2 m3 x3 m1 m2 m3

rCM =

=

m 0

2m m

2m

a

3m 3m

a 2

7a 12

8.2 Comprehensive Physics—JEE Advanced

and

yCM =

= =

m1 y1 m2 y2 m3 y3 m1 m2 m3 m 0

3a 2

2m 0 3m m

2m

3m

8.3 Locate the centre of mass of a uniform rod of mass M and length L.

3a 2

Fig. 8.3

8.2 Four particles of masses 1 kg, 1 kg, 2 kg, and 2 kg, are placed at the corners of a square of side 12 cm as shown in Fig. 8.2. Find the position vector of the centre of mass of the system.

SOLUTION We assume that the rod lies along the x-axis with one end at origin O. M Mass per unit length of the rod = L Mass of element of length dx is [Fig 8.3] M dm = dx L The x-coordinate of the centre of mass is L

x dm xCM = Fig. 8.2

1 0 1 12 2 12 xCM = 1 1 2 2

dm 2

0

= 6 cm

1 0 1 0 2 12 2 12 = 8 cm 1 1 2 2 rCM = 6 i 8 j cm

M x dx L 0

L 2

8.4 A non-uniform rod of length L is lying along the x-axis with one end at origin O as shown in Fig. 8.4. The linear mass density (i.e. mass per unit length) varies with x as = a + bx, where a and b are constants. Find the distance of the centre of mass from origin O.

CENTRE OF MASS OF A BODY HAVING CONTINUOUS DISTRIBUTION OF MASS

If a body has continuous distribution of mass, the position of its centre of mass is determined by dividing the body into a very large number of extremely small elements. If dm is the mass of the element and it is at a distance x and y from the origin of chosen coordinate system, then x and y coordinates of the centre of mass are given by xCM =

and

L

The centre of mass is at the centre of the rod.

The position vector of the centre of mass is

8.2

L

1 M

0

SOLUTION

yCM =

0

yCM =

Fig. 8.4

SOLUTION Mass of element is dm =

x dm dm y dm dm

dx = (a + bx) dx

L

x dm xCM =

(i)

0 L

dm 0

Rigid Body Rotation 8.3 L

L

x dm = 0

a

bx x dx

0

yCM =

2 = a x 2

L 0

L2 3a x dm = 6 0

x3 b 3

L 0

aL2 2

bL3 3

And

dm =

0

2bL 2

a

bx dx

aL

0

bL 2

L 2a 2

sin

bL (iii)

Using (ii) and (iii) in (i), we get L 3a 2bL xCM = 3 2a bL 8.5 Locate the centre of mass of a uniform semicircular ring of radius R and linear mass density . SOLUTION Let us take the centre of the ring at origin O. consider a small element of arc length dl of the ring. Let be the angle which the radius vector of the element makes with the x-axis as shown in Fig. 8.5.

d

0

dm

R d

R2 | cos |0 R

0

=–

L

R2

(ii)

L

L

ydm

R

cos

cos 0

2R

Thus, the centre of mass is at a distance of from origin O on the y-axis.

8.3

2R

FINDING CENTRE OF MASS OF A SYSTEM WHEN A PART OF ITS MASS IS REMOVED

Consider of system of mass M. If a mass m is removed, the remaining mass = M – m which may written as M + (–m). Then the x and y coordinates of the centre of mass of the remaining portion are given by xCM = Mx mx M m and yCM = My my M m where x and y are the coordinates of the centre of mass of the complete part and x and y are the coordinates of the centre of mass of the removed part. 8.6 Four particles, each of mass 1 kg, are placed at the corners of a square of side 12 cm. If mass m3 is system (see Fig. 8.6).

Fig. 8.5

Let d be the angle subtended by the element at the centre. Then dl = R d . Mass of the element is dm = dl = Rd The x and y components of radius vector R are x = R cos and y = R sin . Then

xCM =

x dm dm

R2

cos

d

0

R d 0

=

R

(sin

– sin 0) = 0

R2 | sin |0 R

Fig. 8.6

SOLUTION As shown in Example 2, the x and y coordinates of the original system are xCM = 6 cm and yCM = 6 cm If mass m3 is removed, the x and y coordinates of the centre of mass of the remaining system are x CM = 1 0 1 12 1 0 = 4 cm 1 1 1

8.4 Comprehensive Physics—JEE Advanced

y CM = 1 0 1 0 1 12 = 4 cm 1 1 1 Thus

r = 6i

and

The x and y coordinates of the remaining portion (shown shaded) are xCM =

6j

r = 4i

4j

=

Mx mx M m M

0 M

shift

r = r r

6i

6j

= 2i

4i

4j

M 4 M 4

8.7 From a uniform thin disc of radius R and mass M, a circular portion of radius r =R/2 is removed as shown in Fig. 8.7. Find the centre of mass of the remaining part of the disc.

R 6

M 0 4 yCM 0 M M 4 The negative sign of xCM indicates that the centre of mass of the remaining portion is located at a distance R/6 towards the left of the center O of the complete disc. = My my M m

2j

R 2

M

0

Alternative Method It is clear that centre of mass of the remaining portion of the disc will shift to the left of O. Let G be the centre of mass of the remaining portion of the disc (Fig. 8.8). Let OG = x. Equating the moments of Mg and mg about G, we have

Fig. 8.7

SOLUTION Let the centre of the disc be at origin O. The centre of mass of the complete disc will be at O (by symmetry) and its x and y coordinates are (0, 0). Mass per unit M . Therefore, mass of removed area of the disc = R2 portion is m=

M

=

M

R

2

2

r2 . R 2

2

M . 4

R The x and y coordinates of the centre of mass of the removed portion are x = r

R and y 2

0

Fig. 8.8

OG = mg

Mg

OG

M R x 4 2 R x= 6 Thus the centre of mass shifts by R/6 to the left of O. M

x=

8.8 From a thin uniform square lamina of side a a square of side a/2 is removed from its corner as shown in Fig. 8.9. Find the centre of mass of the remaining portion (shown shaded) of the lamina.

Rigid Body Rotation 8.5

The velocity of the centre of mass is given by drCM vCM = dt dr dr dr m1 1 m2 2 ... mn n dt dt dt = m1 m2 ... mn vCM

m1 v1 m2 v 2 ... mn v n m1 m2 ... mn

The acceleration of the centre of mass is given by d v CM dt dv m1 1 dt =

aCM =

Fig. 8.9

SOLUTION Let M be the mass of the complete lamina. Mass of M the removed portion is m . The centre of mass 4 of the complete lamina is taken to be at origin O (x = 0, y = 0). The coordinates of the centre of mass O of a a the removed portion are x = and y . 4 4 The x and y coordinates of the centre of mass of the remaining portion are xCM =

=

and

yCM =

=

8.4

Mx mx M m M

M 4 M M 4 0

a 4

a 12

My my M m M

M 4 M M 4 0

a 4

If r1, r2, ... rn are the position vectors of masses m1, m2, ... mn, respectively, th\e position vector of the centre of mass of the system of particles is given by mn rn mn

If

Fext

aCM =

F1 F2 m1 m2

Fn mn

mn

d vn dt

mn mn a n mn Fext M

0, then aCM = 0, i.e vCM = constant. Hence if no

net external force acts on a system, its centre of mass will remain at rest or will move with a constant velocity. 8.9 A boy of mass m = 50 kg stands at the end A plank AB of wood of mass M = 100 kg and length l = lake. The end B of the plank is at a distance of 30 m from the shore of the lake as shown in Fig. 8.10. The boy walks a distance of 6 m on the plank towards the shore. How far is the boy from the shore now? Neglect viscosity of water.

a 12

VELOCITY AND ACCELERATION OF CENTRE OF MASS OF A SYSTEM OF PARTICLES

rCM = m1r1 m2r2 m1 m2

m1 aCM = m1a1 m2a 2 m1 m2 or

d v2 dt m2

m2

Fig. 8.10

SOLUTION Initially the centre of mass of the system (plank + boy) is at rest. To walk, the boy exerts a force in the backward direction. The plank in turns exerts a reaction force on the boy in the forward direction.

8.6 Comprehensive Physics—JEE Advanced

These forces are internal to the system. Since no external force acts, the centre of mass of the system remains at rest even when the boy walks on the plank. Let the shore be at the origin O (x = 0). Initially let x be the distance of the centre of mass of the plank from O. Then the distance of the centre of mass of the system (plank + boy) from O will be xCM =

M

x

m 30 10 M m

100 x 50 40 100 x 2000 = = 100 50 150 Since the boy moves towards the shore and the centre of mass of the system has to remain at rest, the plank will move away from the shore. If x is the distance moved by the plank, the distance of the centre of mass from the shore when the boy walks 6 m on the plank is given by xCM = = Since

100( x

x ) 50(40 6 100 50

100 x 150 x 150

x)

1700

xCM = xCM,

100 x 2000 100 x 150 x 1700 = 150 150 300 x = =2m 150 Distance of the boy from the shore = 40 – 6 + 2 = 36 m. 8.10 A car of mass 1000 kg is moving with a velocity of 10 ms–1 towards another car of mass 1500 kg moving with a velocity of 15 ms–1 in the same direction. Find the velocity of the centre of mass of the two cars. SOLUTION vCM = m1v1 m2 v2 m1 m2 =

8.5

1000 10 1500 15 = 13 ms–1 1000 1500

MOMENTUM CONSERVATION AND CENTRE OF MASS MOTION

dP dt

where P = MvCM is the total linear momentum of the system of particles which is equal to the product of the total mass of the system and the velocity of the centre of mass. dP If Fext = 0, =0 P = constant. dt Thus, if no net external force acts on a system, the total linear momentum of the system remains constant; the total linear momentum being the vector sum of linear momentum of individual particles, i.e. + Pn P = P1 + P2 + 8.11 A boy of mass m = 40 kg is standing on a stationary long plank of mass M in a lake. He starts running with a velocity v = 6 ms–1 relative to the plank. Find the velocity of the boy relative to a stationary observer on the bank of the lake. SOLUTION Since the system (boy + plank) is initially at rest, its momentum is zero. Since no external force acts on the system, the momentum of the system will remain zero. Let us assume that the boy runs in the positive x-direction. Velocity of boy relative to the plank is vbp = v i Now

vbp= vb – vp

where

vb = velocity of boy relative to ground

and

vp = velocity of plank relative to ground

Hence

vb= vbp + vp = vi

vp i

= (v

vp ) i

Total momentum of plank + boy = 0 Mvp + mvb= 0 Mvp i + mvb i = 0 Mvp i + m (v + vp) i = 0

We have seen that Fext = MaCM

Fext =

M

d v CM dt

d M v CM dt

which gives vp =

mv M m

Rigid Body Rotation 8.7

Velocity of boy is vb = v

mv Mv i = i M m M m =

260 6 i 260 40

= 5.2 i ms

1

Hence the velocity of boy relative to a stationary observer is 5.2 ms–1 in the direction along which the boy is running.

8.6

TORQUE

If a force F acts on a particle P whose position vector with respect to the origin of an inertial reference frame is r, the torque acting on the particle with respect to the origin is =r

Fig. 8.12

= magnitude of either force perpendicular distance between the two antiparallel forces Work done by torque If a force F acts on a rigid body at perpendicular r from the axis of rotation, the work done by the force in rotating the body through an angle is given by = magnitude of torque

F

In terms of magnitudes, = r F sin

= F (r sin ) = Fr

where is the angle between vectors r and F. Torque is a vector quantity. Its magnitude is given by = rF sin ; its direction is normal to the plane containing vectors r and F and can be determined by the right–hand screw rule. Unit of Torque Torque has the same dimensions as those of work (both being force times distance) viz. ML2T–2. The two are, however, very different quantities. Work is a scalar, torque is a vector. To distinguish between the two we express work in joule and torque in newton–metre (N m).

=

W = Fr

angular displacement

dW d = = dt dt is the angular velocity.

Power = where

8.12 A force F = 2 i

3 j newton acts on a particle

whose position vector with respect to origin O is r = 4 i 5 j metre. Find the magnitude and direction of the torque. SOLUTION =r = 4i =4i

F 5j 2i

2i 12 i

= 0 12 k 10 k

3j j 10 j

i

15 j

j

0

= 22 k newton metre The magnitude of torque is 22 Nm and its direction is along the positive z-axis.

Fig. 8.11

Couple Two equal antiparallel forces having different lines of action constitute a couple. The moment of couple or torque = Fr (Fig. 8.12)

8.13 A rectangular plate OPQR of dimensions 2 m

3m

lies in the x-y plane as shown in Fig. 8.13. A force F = 3i

5 j newton is applied at point Q. Find the

8.8 Comprehensive Physics—JEE Advanced

torque of F (a) about origin O, (b) about point P and (c) about x-axis, y-axis and z-axis.

where r is the position vector of the particle and p its linear momentum. In terms of magnitudes, L = rp sin r p where is the angle between vectors r and p. The dimensions of angular momentum are (ML2 T–1) and its SI unit is kg m2s–1. The direction of L is perpendicular to the plane containing the vectors r and p and its sense is given by the right–hand rule.

Fig. 8.13

SOLUTION (a) r = OQ =

3i

2 j metre

Torque about O is O

=r

F

= 3i = 9i

2j

3i

5j

j

6j

15 i

i

= 0 15 k

6k

Fig. 8.14

i

10 j

j

0 9 k Nm

(b)Torque about P is P

= PQ = 2j

8.14 A body of mass m = 200 g is moving parallel to the x-axis with a velocity v = 30 cms–1 in the x-y plane as shown in Fig. 8.15. Calculate the magnitude of its angular momentum about origin O at any time t.

F F

= 2j

3i

2j

= – 6 k Nm

(c)Torque about x-axis is x=

Torque about y-axis is y

=

j = 9k j = 0

Torque about z-axis is z

8.7

= z O k = 9k k = 9 Nm

ANGULAR MOMENTUM

The angular momentum L of a particle P with respect to 8.14] L= r

Fig. 8.15

i = 9k i = 0

p

SOLUTION Magnitude of angular momentum = r p = r mv = OB m v = 0.15 0.2 0.3 = 9 10–3 kg m2s–1 8.15 A body of mass m = 200 g is projected with a velocity u = 5 ms–1at an angle = 30° with the horizontal. Calculate the magnitude of the angular momentum of the body about the point of projection when it is at the highest point of its trajectory. Take g = 10 ms–2.

Rigid Body Rotation 8.9

dp dt dt = d p = change in linear momentum

SOLUTION

=

In rotational motion, angular impulse J dt J= dL dt dt = dL = change in angular momentum =

Fig. 8.16

Refer to Fig. 8.16. At the highest point A, the body has only horizontal velocity v = u cos u 2 sin 2 2g Magnitude of angular momentum of the body about O when it is at point A is L = mv OB = mv hmax hmax =

u 2 sin 2 2g

= m u cos = =

=

mu3 sin 2 cos 2g 0.2

5

3

0.2 125

sin 2 30 cos 30 2 10 1 4

20 = 0.27 kg m2s–1

8.8

3 2

RELATION BETWEEN TORQUE AND ANGULAR MOMENTUM

In linear motion, the relation between force F and linear momentum p is dp F= dt In rotational motion, the relation between torque and angular momentum L is dL = dt which states that the torque acting on a particle is equal to the rate of change of angular momentum.

8.9

ANGULAR IMPULSE

In linear motion, impulse I I=

F dt

8.10

LAW OF CONSERVATION OF ANGULAR MOMENTUM

If no external torque acts, the total angular momentum of a body or a system of particles is conserved. We have seen that the rate of change of angular momentum of a particle is equal to the torque produced by the total force. If, in a certain situation, the torque itself vanishes, then it follows that the angular momentum of the particle will remain constant. This is the law of conservation of the angular momentum of a particle. One trivial situation is when the force vanishes. Then the torque vanishes too. The particle then moves freely in a straight both linear and angular momenta are conserved. A general situation is when the torque vanishes without the force itself vanishing. The torque will vanish if the component F (the angular component) of F vanishes but the radial component FII does not. The radial component FII is the component of F along the radius (or position) vector r. Hence, if the force acting on the particle is purely radial (i.e. if it is directed along or against its position vector) then the torque acting on the particle vanishes and its angular momentum is conserved and so is its areal velocity.

8.11

MOMENT OF INERTIA

The moment of inertia of a rigid body about a particular axis may be of all the particles constituting the body and the squares of their respective distances from the axis of rotation, i.e. + mnr2n I = m1r 21 + m2r 22 + m3r 23 + =

n N=1

mn rn2

Its value depends upon the particular axis about which the body rotates and the way the mass is distributed in the body with respect to the axis of rotation. In the case of a body which does not consist of separate, discrete particles but has a continuous and homogeneous distribution of matter in it, the summation is replaced by integration, so that

8.10 Comprehensive Physics—JEE Advanced

I=

r 2d m

where dm of the body at a distance r from the axis of rotation. Moment of inertia is a scalar quantity. Its SI unit is kg m2 and its dimensions are (ML2).

8.12

RADIUS OF GYRATION

The radius of gyration of a body about its axis of rotation which, if the entire mass of the body were concentrated, its moment of inertia about the given axis would be the same as with its actual distribution of mass. It is usually denoted by the letter K. If M is the mass of the body, its moment of inertia I in terms of its radius of gyration K can be written I = MK2

8.13

MOMENT OF INERTIA AND ROTATIONAL KINETIC ENERGY

Kinetic energy of a rotating body is related to moment of inertia as 1 I 2 KE = 2 where is the angular velocity (or frequency) of the body.

8.14

MOMENT OF INERTIA AND TORQUE

The magnitude of torque is given by = I where is the angular acceleration of the body.

8.18

MOMENT OF INERTIA AND ANGULAR MOMENTUM

The magnitude of angular momentum of a rotating body is given by L = I

8.16

PARALLEL AXES THEOREM

If M is the total mass of a body and h the distance between two parallel axes, then according to parallel axes theorem I = ICM + Mh2

8.17

PERPENDICULAR AXES THEOREM

The theorem of perpendicular axes for a body of plane lamina states that the moment of inertia of a plane lamina about an axis perpendicular to the plane of the lamina is equal to the sum of the moments of inertia of the lamina about any two mutually perpendicular axes in its own plane and intersecting each other at the point where the perpendicular axis passes through it. If Ix and Iy are the moments of inertia of a plane lamina about the perpendicular axes x and y respectively which lie in the plane of the lamina and intersect each other at O, then the moment of inertia I of the lamina about an axis passing through O and perpendicular to its plane is given by I = Ix + Iy

EXPRESSIONS FOR MOMENT OF INERTIA OF BODIES OF REGULAR SHAPES ABOUT PARTICULAR AXES OF ROTATION Shape of body

1.

8.15

Circular ring of mass M and radius R

Axis of Rotation

Expression for Moment of Inertia M R2

(i) through centre, perpendicular to plane of ring (ii) any diameter

(1/2) M R2

(iii) any tangent in the plane of ring

(3/2) M R2 2 M R2

(iv) any tangent perpendicular to plane of ring 2.

3. 4.

Circular disc of mass M and radius R

(i) through centre, perpendicular to plane of disc

(1/2) M R2

(ii) any diameter

(1/4) M R2

(iii) tangent in the plane of the disc

(5/4) M R2

(iv) tangent perpendicular to plane of disc

(3/2) M R2

Sphere of mass M and radius R

(i) any diameter

(2/5) M R2

(ii) any tangent plane

(7/5) M R2

Cylinder of mass M,

(i) own axis

(1/2) M R2

radius R and length L

(ii) through centre perpendicular to length

M

R2 4

L2 12

(iii) through end faces and

M

R2 4

L2 3

to length

Contd.

Rigid Body Rotation 8.11

Shape of body 5.

Axis of Rotation

One dimensional rod of mass M and length L

6.

Rectangular lamina of mass M, length L and breadth B

Expression for Moment of Inertia

(i) centre of rod and (ii) one end and

M L 2/12

to length

M L2/3

to length

(i) length of lamina and in its plane

M B2/3

(ii) breadth of lamina and in its plane

ML2/3

(iii) centre of lamina and parallel

M B 2 or M L2 12 12

to length or breadth in its plane

Rectangular block of mass M, length L, breadth B and height H

B2 12

M

L2 12

B2 3

M

L2 3

B2 12

to its plane

M

(v) centre of length and

to its plane

(vi) centre of breadth and 7.

L2

(iv) centre of lamina and

to its plane

(i) through centre or block and parallel to length or breadth or height of the block

M

M

H 2 L2 12

(ii) through end face and M

parallel to length or breadth or height of the block M

8.19

KINEMATICS OF ROTATIONAL MOTION WITH CONSTANT ANGULAR ACCELERATION

Consider a body rotating with an initial angular velocity 0. It is given a constant angular acceleration (by applying a constant torque) for a time t angular velocity and suffers an angular displacement in time t. The equations of rotational motion are = 0+ t

and

2

=

0t

=

2

1 2 t 2 2 0

8.16 The moment of inertia of a uniform circular disc of mass M and radius R about an axis passing through its 1 centre and perpendicular to its plane is MR2 . Find 2 the moment of inertia of the disc

B2

L2 3

H2 3

H2 12

H2 12

or

2 or M L

B2 12

B2 12

or

2 or M B 3

L2 12

(a) about any diameter (b) about an axis passing through a point on the edge of the disc and perpendicular to the disc (c) about a tangent in the plane of the disc. SOLUTION The plane of the disc is the x-y plane. (a) Using perpendicular axes theorem [Fig. 8.17(a)] Ix + Iy = Iz Now Iz = IC = 1 MR2 (given). From symmetry 2 I y = I x. Ix = Iy = 1 MR2 2Ix = 1 MR2 2 4 (b) using parallel axes theorem [Fig. 8.17(b)] IAB = Iz + M(CD)2 = IC + MR2 1 3 = MR2 MR2 = MR2 2 2

8.12 Comprehensive Physics—JEE Advanced

(c) using parallel axes theorem [Fig. 8.17(c)] IEF = Iy + MR2 =

1 MR 2 4

5 MR 2 4

MR 2 =

Fig. 8.17

8.17 A thin uniform disc of mass M and radius R has concentric hole of radius r. Find the moment of inertia of the disc about an axis passing through its centre and perpendicular to its plane. SOLUTION Mass per unit area of the disc is [Fig. 8.18] m=

R

r2

Mass of the disc if it was complete (i.e. without hole) is R2 M1 = m =

M R

2

r

r4 r

2

=

1 M R2 2

r2

8.18 Find the moment of inertia of a thin uniform rod of mass M and length L about an axis passing through its centre and making an angle with the rod.

Fig. 8.18

R2 =

2

M R4 2 R2

SOLUTION Divide the rod into a very large number of extremely small elements each of length dx. Consider one such element at a distance x from the centre O of the rod (Fig. 8.19).

M 2

=

M R R2

2

r2

Mass of the removed portion is M2 = m

Mr 2

r2 =

R2

r2

Fig. 8.19

Since the two portions are concentric, the moment of inertia of the given disc about the given axis is I= =

1 M1R 2 2

1 M2 r2 2

1 MR4 2 R2 r 2

Mr 4 R2

r2

M dx L Perpendicular distance of the element from the axis of rotation is r = OA = x sin Moment of inertia of the rod about the given axis is M I = dm r 2 = dx x sin 2 L Mass of element is dm =

Rigid Body Rotation 8.13

M sin 2 = L

x

M sin 2 = L

x3 3

= If

= 90°, I =

where

L 2

x dx x

L 2

L/2 L/2

2

ML sin 2 12 ML2 12

Fig. 8.22

8.19 The radius of gyration K of a hollow sphere of mass M and radius R about a certain axis is equal to R. Find the distance of that axis from the centre of the sphere.

M A 2M R2

Area of strip is dA =

x

2M R

2

2

2

x2

2

= (xdx)

dA =

2M R2 2M

xd x

xd x R2 Moment of inertia of the section about the given axis is R 2M 2 I = dmx = x3 dx 2 R 0

x2

Fig. 8.20

R

=

1 MR 2 2

8.21 Two particles of masses m1= 1 kg and m2 = 2 kg are connected by a rigid bar of length L = 1.2 m of negligible mass. The system rotates about an axis perpendicular to the rod and at a distance x from mass m1. Find the value of x for which the moment of inertia about the given axis is minimum. what is the minimum moment of inertia?

3

8.20 Figure 8.21 shows a section (a part) of circular disc of radius R. The mass of the section is M. Find the moment of inertia of the section of the disc about an axis passing through its centre O and perpendicular to its plane.

SOLUTION From Fig 8.23, it follows that

Fig. 8.21

SOLUTION Area of the section A =

dx

=

2 MR 2 Mx2 3 Given K = R. Hence

x=

=

Mass of strip is dm =

MK2 =

2 2 R 3

Mass per unit area =

2 xdx

SOLUTION Let x be the distance of the axis from the centre of the sphere [Fig. 8.20]. From parallel axes theorem IXY = IAB + Mx2

R2 =

is in radian [Fig 8.22].

2

R2

2

=

R2 2

Fig. 8.23

8.14 Comprehensive Physics—JEE Advanced

I = m1x2 + m2 (L – x)2 I will be minimum if Differentiating (i)

(i) 2

dI d I = 0 and dx d x2

0

dI = 2m1x + 2m2 (L – x) (–1) = 0 (ii) dx m2 L 2 1.2 = = 0.8 m x= m1 m2 1 2 Differentiating (ii)

= a + b is applied to it, where is the angular displacement and a and b are positive constants. Obtain the expression for the angular velocity of the disc as a function of . SOLUTION =I or

=

d a b = dt I

2

d I

= 2m1 + 2m2, which is positive for any value of x. dx2 Substituting x in (i), we have Imin = m1 =

2

m2 L m1 m2

m2 L m1 m2

m2 L

m1m2 L2 1 2 1.2 = m1 m2 1 2

2

d d

2

=

a

b I

= 0.96 kg m2

d = 0 2

2

300 = 5 Hz. Therefore 60 = 10 rad s–1 and t = 10 s

= = =

a I

b I

d 0

a 2 I 2

a I

=

d

=

0

+

t, we have 10 = 0 + 10

rad s–2

(a) Torque required is = I =

1 MR2 2

1 100 22 2 = 200 = 628 Nm =

(b) K.E. =

1 I 2

1 2 = 104

2

=

2

1 100 22 2 = 9.9 104 J

10

d

b d I

d 0

b I

1 a I

2

2 MR

2

2b a

2

2b

= 0, = 300 r.p.m. = =2

Using

=

Integrating

SOLUTION 0

d a b = dt I d

8.22 A giant wheel of radius 2.0 m and mass 100 kg is initially at rest. (a) What torque should be applied to it so that it acquires an angular frequency of 300 r.p.m. in 10 s? (b) Find the kinetic energy when it is rotating at 300 r.p.m.

Given

I

I

1 MR 2 2

8.24 A thin uniform rod AB of mass M and length L is hinged at one end A it stands vertically. It is allowed to fall freely in a vertical plane. (a) What is the angular acceleration of the rod when it is at an angle with the vertical? (b) With what linear speed will the end B hit the

2

8.23 A stationary horizontal uniform disc of mass M and radius R is free to rotate about an axis passing through its centre and perpendicular to its plane. A torque

SOLUTION (a) The entire mass of the rod acts at its centre of mass C. AC = L/2. The magnitude of the torque due to weight Mg is [Fig. 8.24] = Mg r = Mg

AD = Mg

L sin 2

Rigid Body Rotation 8.15

(b) Find the magnitude of the angular momentum of the particle about O at this time t. dL (c) Show that, in this example, = dt SOLUTION

Fig. 8.24

ML2 3 MgL sin 3

Moment of inertia of the rod about A is I = Angular acceleration

=

I

= =

ML2

2 3g sin L

(b) When the end B tance through which C falls is L/2. From the law of conservation of energy, Loss in P.E. = gain in K.E Mg

L 1 = I 2 2

2

3g L Linear speed of end B = L

=

1 ML2 2 3

2

(a) The torque is due to force of gravity F = mg. The magnitude of the torque of F about O is = rF sin = r mg sin x0 x = rmg 0 sin r r = mg x0 From right hand rule, the direction of the torque is into the page . (b) The magnitude of angular momentum about O is L = rp sin = rmv sin From v = u + at we have v = 0 + gt = gt. Therefore L = r m gt (c)

8.20

x0 = mg x0t r

dL d = mgx0 t = mgx0 = dt dt

ROLLING MOTION WITHOUT SLIPPING

(1) Total Kinetic Energy

=

=

3gL

8.25 A particle of mass m is released from rest at point P located at a distance x0 from origin O on the x-axis as shown in Fig. 8.25. It falls vertically along the negative y-axis.

The total kinetic energy of a body which is moving as well as rotating is equal to the sum of translational K.E. and rotational K.E., i.e. 1 1 I 2 K = m v2 2 2 where m = mass of the body, v = linear velocity of its centre of mass, I = moment of inertia of the body about an axis passing through its centre of mass and = angular velocity of rotation. (2) Instantaneous Velocity of a point on a Rolling Body Consider a wheel of radius R(= AC = BC) rolling without slipping on a horizontal rough surface (Fig. 8.26).

Fig. 8.25

(a) Find the magnitude and direction of the torque acting on the particle at time t when it reaches point Q whose position vector with respect to O is r.

Fig. 8.26

8.16 Comprehensive Physics—JEE Advanced

Every point on the wheel has instantaneous velocity. For a point P at a distance r from the centre of mass C, the instantaneous velocity is the vector sum of velocity v of the centre of mass and tangential velocity vt of point P relative to the centre of mass, i.e. vP = v + vt where vt = r . Vector vt is directed along the tangent to the circle of radius r about C. For point A, vt = R and v = r . Since these velocities are in the same direction, the instantaneous velocity of A is vA = v + vt = R + R = 2 R

For a hollow cylinder I = MR2 For a solid sphere I =

2 MR 2 5

aCM =

F 2M

aCM =

5F 7M

2 3F MR 2 aCM = 3 5M If the force is applied tangentially to the body as shown in Fig. 8.28, then For a hollow sphere I =

For point B in contact with the horizontal surface, vB = v + vt = R – R = 0 If a body rolls on a surface without slipping, the instantaneous velocity of the point of contact with the surface is zero. (3) A body Rolling without slipping on a Rough Horizontal surface Horizontal force F is applied at the centre of mass of a body (disc, ring, cylinder or sphere) of mass M and radius R (Fig. 8.27) on a rough horizontal surface. If f is the frictional force, and the body rolls without slipping aCM = R. F – f = MaCM (i) = fR = I

Fig. 8.28

aCM =

2F

I MR 2 I F 1 MR 2 and f= I 1 MR 2 (4) A body Rolling without slipping on a Rough Inclined Plane M 1

A body (ring, disc, cylinder or sphere) of mass M and radius R is rolling (without slipping) down a rough inclined plane of inclination (Fig. 8.29).

Fig. 8.27 I aCM I = (ii) 2 R R From (i) and (ii), acceleration of centre of mass is

f=

aCM =

F

Fig. 8.29

F

and

f=

For a disc I =

1 MR2 2

For a ring I = MR2

I MR2

M 1

2

MR I

1

aCM = aCM =

For a solid cylinder I =

2F 3M

F 2M

1 MR2 2

aCM =

2F 3M

For linear motion parallel to the plane Mg sin – f = Ma (i) where a = linear acceleration of the centre of mass. For rotational motion about the axis through the centre of mass =I I I =( a=R ) Rf = I f= R R2 (ii) I is the moment of inertia about the centre of mass.

Rigid Body Rotation 8.17

Using (i) and (ii), we get g sin a= I 1 MR 2 g sin For a ring I = MR2 a= 2 1 2 g sin For a disc I = MR2 a= 2 3 1 2 g sin For a solid cylinder I = MR2 a= 2 3 g sin For a hollow cylinder I = MR2 a= 2 2 3g sin For a hollow sphere I = MR 2 a= 3 5 2 5 g sin For a solid sphere I = MR 2 a= 5 7 Frictional force is f =

Ia R

2

=

Mg sin 1

MR 2 I

Condition for rolling without slipping To prevent slipping, f μN, where μ of static friction between the body and the plane and N = Mg cos is the normal reaction. Hence to avoid slipping, Mg sin 1

MR 2 I

SOLUTION The acceleration of the cylinder is a=

1

MR 2 I

8.26 A solid cylinder of mass M and radius R is released from rest from top A of an inclined plane of height h and inclination as shown in Fig. 8.30. The cylinder rolls without slipping. Find (a) the speed at which it reaches bottom B of the plane and (b) the time it takes to reach B.

Fig. 8.30

1 MR 2 2

h sin 2 2 (a) Using v – u = 2as, we have v2 – 0 = 2

2 g sin 3

h sin

4 gh 3 Speed v can also be found from the law of conservation of energy. As the cylinder moves from A to B, it loses P.E. and gains K.E. Loss in P.E. = gain in K.E. v=

or Now

Mgh =

1 1 = Mv2 + I 2 2

I=

1 MR2 and 2

Mgh =

1 1 Mv2 + 2 2

= tan

I

Distance travelled is s = AB =

μ Mg cos

μ

g sin 2 g sin = I 3 1 MR 2

=

2

v . Therefore R

1 MR2 2

3 M v2 4

4 gh 3 (b) From v = u + at, we have v=

4 gh 2 g sin =0 + t 3 3 t=

3h 1 g sin

8.27 A billiard ball has mass M = 250 g and radius R = 2.5 cm and is initially at rest. A rod held horizontal at a height h above centre C hits the ball. The ball begins to roll without slipping. Find the value of h [see Fig. 8.31]. Fig. 8.31

v R

2

8.18 Comprehensive Physics—JEE Advanced

SOLUTION The horizontal force F imparts a linear impulse I=

Fdt = change in linear momentum

I = Mv – 0 = Mv = MR

(i)

where v is the velocity of the centre of mass of the ball. Since it rolls without slipping, v = R , where is the angular velocity. The torque due to F imparts an angular impulse J = I h = change in angular momentum =I or

Ih =

v

I

2 MR 2 5

(ii)

Dividing (ii) by (i) 2R 2 = = 5

2.5 cm 5

8.28 A turntable of radius R = 10 m is rotating making 98 revolutions in 10 s with a boy of mass m = 60 kg standing at its centre. He starts running along a radius. Find the frequency of the turntable when the boy is 4 m from the centre. The moment of inertia of the turntable about its axis is 1000 kg m2. SOLUTION Initial moment of inertia of the system is M1 = M.I. of turntable + M.I. of boy at the centre = 1000 + 0 = 1000 kg m2 Initial frequency 1 = 9.8 rev/sec Final moment of the system is M2 = M.I. of turntable + M.I. of boy at a distance 4 m from the centre of turn table = 1000 + 60 (4)2 = 1960 kg m2 Since no external torque acts, the angular momentum of the system is conserved, i.e. I2 2 = I1 1 I2 2 = I1 1 =

Fig. 8.32

SOLUTION

= 1.0 cm

2

v

–0=I

2 M R2 5

2 M R2 5 h= MR

8.29 A uniform rod AB of length L = 1 m is sliding along two mutually perpendicular surfaces OP and OQ as shown in Fig. 8.32. When the rod subtends an angle = 30° with OQ, the end B has a velocity 3 ms–1. Find the velocity of end A at that time.

I1 1 1000 9.8 = = 5 rev/s I2 1960 = 5 Hz

OB = x, OA = y and x2 + y2 = L2 and x = L cos Differentiating x2 + y2 = L2 with respect to t we have dx dy + 2y =0 2x dt dt 2xvB + 2yvA = 0 vA = – | vA | =

x vB y

x vB = vB cot y =

3

=

3

cot 30° 1 3

= 1 ms–1

8.30 A rope is wound around a hollow cylinder of mass M = 3 kg and radius R = 40 cm. If the rope is pulled with a force F of the cylinder and (b) the linear acceleration of the rope. SOLUTION For a hollow cylinder I = MR2 (a) Torque on cylinder is = FR I = FR. Therefore FR FR F 30 = = = a= 2 I MR 3 0.4 MR = 25 rad s–2

Rigid Body Rotation 8.19

(b) Linear acceleration of rope is a = R = 0.4 25 = 10 ms–2

8.32

8.31 A uniform rod AB of mass M = 0.4 kg and length L = 1 m lies on a horizontal frictionless table with its end A pivoted to the table. A ball of mass m = 0.2 kg moving along the surface of the table with velocity u = 4 ms–1 perpendicular to the rod collides with the free end B velocity of the ball immediately after the collision and (b) the angular velocity of the rod after collision.

A uniform rod AB of mass M and length L is hinged at one end A. It is released from rest at a horizontal position. Find the angular acceleration of the rod and the linear acceleration of its centre of mass as it falls. SOLUTION Refer to Fig. 8.34.

SOLUTION Refer to Fig. 8.33. Fig. 8.34

v

or

Fig. 8.33

(a) Let v be the velocity of the ball just after collision. Since the collision is perfectly elasitc, e = 1, i.e. Velocity of approach = velocity of separation or

Torque about A = Mg

v

(i) L Since there is no external torque, the angular mementum about A is conserved, i.e. mu L = mvL + I =

2

ML 3

= MvL +

3 (u v)m ML From (i) and (ii), we get =

v= =

3m 3m

(ii)

M u M

3 0.2 3 0.2

0.4 0.4

(iii)

I

=

MgL 3g = 2I 2L

MgL 2 I

ML2 3

Linear acceleration of centre of mass is aCM =

u= L–v u

=

L 2

8.21

L 2

=

L 2

3g 3g = 2L 4

TRANSLATIONAL AND ROTATIONAL EQUILIBRIUM

A body is said to be in equilibrium if its state of motion does not change with time. Translational Equilibrium A body is in translational equilibrium if the total force acting on it is zero; the total force is equal to the vector sum (resultant) of the individual forces acting on the body, i.e. Ftotal = F1 + F2 + … + Fn If a body is in translational equilibrium, the linear momentum does not change with time, i.e. p = constant Rotational Equilibrium

4 = 0.8 ms–1

(b) Using (iii) in (i), we get =

6 mu (3 m M ) L

=

6 0.2 4 = 4.8 rad s–1 (3 0.2 0.4) 1

A body is in rotational equilibrium if the total torque acting on the body is zero; the total torque is equal to the vector sum of the individual torques acting on the body, i.e. total = 1 + 2 +……+ n If the body is in rotational equilibrium, the angular momentum about the axis of rotation does not change with time, i.e. L = constant

8.20 Comprehensive Physics—JEE Advanced

Consider a uniform rigid rod AB of negligible mass. Two parallel forces of equal magnitude F are applied perpendicular to the rod at ends A and B as shown in Fig. 8.35(a). Let C be the mid-point of rod.

For translational equilibrium Ftotal = 0 R– F1 – F2 = 0 F 1 a 1 – F 2a 2 = 0 For rotational equilibrium total = 0 or F 2a 2 = F 1a 1 i.e. anticlockwise torque = clockwise torque. This is the principle of moments. 8.33 A uniform metal bar AB of length 100 cm and mass M = 2kg is supported on two knife-edges placed 20 cm from each end. A mass of m = 3kg is suspended at a distance of 40 cm from end A. Find the normal reactions at the knife-edges. Take g = 10 ms–2. SOLUTION Refer to Fig. 8.37

Fig. 8.35

Taking the anticlockwise torque to be positive and clockwise torque to be negative, anticlockwise torque about C = Fa and clockwise torque about C = – Fa. Total torque about C = Fa – Fa = 0. Total Force = F + F = 2F. Hence the rod is in rotational equilibrium but not in translational equilibrium. If the forces act in opposite directions as shown in Fig. 8.34(b), total torque = Fa + Fa = 2Fa and total force = F – F = 0. Hence, in this case, the rod is in translational equilibrium but not in rotational equilibrium. A pair of equal and opposite forces having different lines of action is known as a couple. A couple produces a torque which produces rotation without translation. For example, opening the cap of a bottle or opening a tap. Torque is also called moment of force. Principle of Moments Consider a uniform rod AB of negligible mass pivoted at a point along its length as shown in Fig. 8.36. Forces F1 and F2 act at ends A and B are such that the rod is in translational as well as rotational equilibrium. Let R be the normal reaction of the pivot.

Fig. 8.37

Let R1 and R2 be the normal reaction at K1 and K2. The rod is in translational as well as rotational equilibrium. For translational equilibrium, Ftotal = 0, Hence mg + mg – R1 – R2 = 0 R1 + R2 = (m + m)g = (3 + 2) 10 = 50 N (i) For rotational equilibrium, total = 0, Hence Clockwise moment about G = anticlockwise moment about G R1

(K1G) = R2 (K2G) + mg R1

30 = R2

30 + 3

(PG) 10

10

R1 = R2 + 10 R1 – R2 = 10

(ii)

From Eqs. (i) and (ii) we get R1 = 30 N and R2 = 20 N

Fig. 8.36

8.34 A uniform rod AB of length 1.0 m and mass 5.0 kg leans on a frictionless vertical wall and a rough horiB touching the wall and end A at a distance of 40 cm from the wall as shown in Fig. 8.38(a) Find

Rigid Body Rotation 8.21

(a) Normal reaction of the wall and normal reaction

For translational equilibrium, Ftotal = 0. Hence R1 = f (horizontal direction)

(b) the frictional force at end A,

and

end A, and (d) the reaction force at A.

R2 = Mg (vertical direction)

For rotational equilibrium, total = 0. Hence clockwise moment about A = anticlockwise moment about A R1 BC = R2 0 + Mg AD + f 0 R1

0.92 = Mg

0.2 = 5

10

0.2 = 10

R1 = 10.9 N Also

end A is

SOLUTION

f 10.9 = = 0.22 R2 50 (d) The reaction force F at A is the resultant of f and R2 μ=

Refer to Fig. 8.38(b). AB = 1.0 m, AG = BG = 0.5 m, AC = 0.4m. G is the centre of mass of the rod. AB

10 = 50 N

[Note that the perpendicular distance between R2 and the axis of rotation at A is zero and perpendicular distance of f from the axis of rotation at A is also zero] (b) Frictional force f = R1 = 10.9 N

Fig. 8.38

BC =

R2 = Mg = 5

2

AC

2

=

1.0

2

0.4

2

= 0.92 m

F=

The forces acting on the rod are (i) normal reactions R1 and R2 at B and A (ii) weight Mg of rod acting at G and (iii) frictional force f A of the rod. (a) Since the frictional force f prevents the rod from sliding away from the wall, force f must be directed towards the wall.

=

f2

R22

10.9

2

50

2

= 51.2 N

NOTE At end B, reaction force = normal reaction R1 because the wall is frictionless.

I Multiple Choice Questions with Only One Choice Correct 1. A gun of mass M is initially at rest on a horizontal m with a velocity v at an angle with the horizontal. After (a) moves with a velocity mv/M opposite to the direction of motion of the bullet.

(b) moves with a velocity mv cos /M in the horizontal direction (c) moves with a velocity mv sin /M in the horizontal direction (d) remains at rest

8.22 Comprehensive Physics—JEE Advanced

2. A thin uniform circular disc has a radius R. A square portion of diagonal equal to R is cut out from it. The distance between the centre of mass of the remaining portion of the disc from the centre of the complete disc is (a) (c)

R 2

(b)

R

R (2

1)

radius R 1 3 MgR (b) MgR (a) 2 4 5 7 (c) MgR (d) MgR 8 8 4. A ring of radius r has its mass non-uniformly distributed over its circumference with centre at x is the distance of the centre of mass (a) x = r (b) x < r (c) x > r (d) 0 x r 5. Two particles of equal mass have velocities v1 = a i and v2 = a j

6.

j) where a and b

tre of mass of the two particles moves along a (a) straight line (b) circle (c) ellipse (d) parabola

-

an inclined plane of inclination 1 sliding and reaches the bottom with speed v1 and its time of descent is t1. The same sphere is then released from rest from the top of another inclined plane of inclination 2 without sliding and reaches the bottom with speed v2 and its time of descent is t2 2> 1 (a) v2 > v1 ; t2 < t1 (b) v2 = v1 ; t2 < t1 (c) v2 < v1 ; t2 > t1 (d) v2 = v1 ; t2 = t1 7. The moment of inertia of a uniform rod of mass M and length L about an axis passing through its centre and inclined to it at an angle = 60° is (a)

ML2 3

ML2 (c) 12

(c)

R

(d) (2 1) 2(2 1) 3. A carpet of mass M is rolled along its length in the form of a cylinder of radius R and kept on a rough

a1 = b ( i

8. A solid cylinder of mass M and radius R is rolling without slipping on a horizontal plane with a speed v to a maximum height given by v2 v2 (b) sin (a) 2g 2g

(b)

ML2 4

ML2 (d) 16

3v 2 4g

(d)

3v 2 sin 4g

9. A uniform rod AB of mass m and length L is suspended by two strings C and D of negligible mass as shown in Fig. 8.39. When string D tension in string C will be

mg 4 (c) 2 mg

(b) mg

(a)

(d) 4 mg

Fig. 8.39

10. A body of mass m is projected with a velocity u at an angle of 60° with the horizontal. The magnitude of the angular momentum about the point of projection when the body is at the highest point of its trajectory is

mu 3 2mu 3 (b) (a) 2g 5g 3mu 3 mu 3 (c) (d) 4g 16 g 11. A cubical block of side L and mass m rests on a rough horizontal surface. A horizontal force F is applied normal to one of its faces at a point that is directly above the centre of the face at a height 2L/3 above the base. The minimum force F required to topple the block before sliding is (a)

2mg 3

(b)

3mg 2

(c)

3mg 4

(d)

4mg 3

12. duration of the new day will be (a) 6 hours (b) 12 hours (c) 18 hours (d) 30 hours

Rigid Body Rotation 8.23

of friction between the coin and the record is the minimum angular frequency of the record for

13. A circular ring of mass M and radius R is rotating about its axis at an angular frequency . Two m the opposite ends of a diameter of the ring. The angular frequency of the ring becomes . The ratio / is 2M M (a) (b) M 2m M 2m

2m M (d) M 2m 14. A solid sphere rolls down from the top of an in(c)

the bottom is v . The ratio v /v is

(c)

g r

(d) 2

(a)

g r

(b)

(b)

g r

(c) 2 20.

(d) 2

g 2r g r

1 (ai + bj) 3 2 (c) (ai + bj) 3

g 2 (d) g 3 3 16. A block of mass M is released from the top of an (c)

2 g r 2 g r

m corners of a right angled triangle as shown in Fig. OA = a and OB = b the centre of mass is (here i and j are unit vectors along x and y axes respectively).

1 (ai – bj) 3 2 (d) (ai – bj) 3

(a)

acceleration of the disc down the inclined plane is g (a) g (b) 2

of the plane is v. A circular disc of the same mass M velocity on reaching the bottom is v . The ratio v /v will be 1 2 (b) (a) 3 3 2 2 (c) 1 (d) 3 17. Two circular loops A and B of radii R and 2R respectively are made of the same wire. Their moments of inertia about the axis passing through the centre and perpendicular to their plane are IA and IB respectively. The ratio IA/IB is 1 (a) 1 (b) 2 1 1 (c) (d) 4 8 18. A small coin is placed at a distance r from the centre of a gramophone record. The rotational speed of

2 g r

19.

the plane is v. When the same sphere slides down

3 (b) 1 (a) 5 3 7 (d) (c) 5 5 15. A circular disc is rolling down an inclined plane

(a)

(b)

Fig. 8.40

21. A sphere of mass M and Radius R is released from the top of an inclined plane of inclination . The and the sphere so that it rolls down the plane without sliding is given by 2 tan (a) = tan (b) = 3 (c) 22.

=

2 tan 5

(d)

=

2 tan 7 M and length

L moment of inertia of the composite structure about an axis passing through the centre of mass of the structure and perpendicular to its plane is

8.24 Comprehensive Physics—JEE Advanced

(a)

ML2 2

(b)

ML2 8

(d)

ML2 4

26. I1 I2 I3 and I4 are respectively the moments of inertia of a thin square plate ABCD of uniform thickness

ML2 12 23. M and length L are welded to form a square ABCD as shown in Fig. 8.41. What is the moment of inertia of the composite structure about a line which bisects rods AB and CD (c)

ML2 6 ML2 (c) 2

the plate (Fig. 8.43). The moment of inertia of the plate about an axis passing through thecentre O and perpendicular to the plane of the plate is (b) 2(I3 + I4) (a) 2(I1 + I2) (d) I1 + I2 + I3 + I4 (c) I1 + I3

ML2 3 2 ML2 (d) 3 (b)

(a)

Fig. 8.43 Fig. 8.41

24 . A thin uniform metallic triangular sheet of mass M has sides AB = BC = L. What is its moment of inertia about axis AC (Fig. 8.42) (a) M L2

27. The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height h from rest without sliding is

12 (b)

25.

M L2 6

28.

2

(c)

ML 3

(d)

2M L2 3

Fig. 8.42

M and radius R rolling and partly sliding. During this kind of motion of the sphere (a) total kinetic energy is conserved (b) the angular momentum of the sphere about the point of contact with the plane is conserved (c) only the rotational kinetic energy about the centre of mass is conserved (d) the angular momentum about the centre of mass is conserved.

(a)

10 gh 7

(b)

gh

(c)

6 gh 5

(d)

4 gh 3

energy to the total kinetic energy is given by (a) 7 : 10 (b) 2 : 5 (c) 10 : 7 (d) 2 : 7 29. A cart of mass M is tied at one end of a massless rope of length 10 m. The other end of the rope is in the hands of a man of mass M. The entire system is on a smooth horizontal surface. The man is at x = 0 and the cart at x a point (a) x = 0 (b) x = 5 m (c) x = 10 m (d) they will never meet. 30. A mass m is moving with a constant velocity along a line parallel to the x angular momentum with respect to the origin

Rigid Body Rotation 8.25

(a) is zero (c) goes on increasing

(b) remains constant (d) goes on decreasing.

31. Let I be the moment of inertia of a uniform square plate about an axis AB that passes through its centre and is parallel to two of its sides. CD is a line in the plane of the plate that passes through the centre of the plate and makes an angle with AB. The moment of inertia of the plate about the axis CD is then equal to (a) I (b) I sin2 (c) I cos2

(d) I cos2

2

32. A smooth sphere A is moving on a frictionless horizontal surface with angular speed and centre of mass velocity v on with an identical sphere B at rest. Neglect friction are (a) (c)

A

and A < A =

B B

respectively. Then (b) A = (d) B =

Fig. 8.44

1 MR2 2 3 (c) MR2 2

(b) MR2 (d) 2 MR2

34. A cubical block of side a is moving with a velocity v on a horizontal smooth plane as shown in O. The angular speed of the block after it hits the ridge at O is 3v 3v (b) (a) 4a 2a (c)

3v 2a

(d) zero

35. A thin wire of length L and uniform linear mass density is bent into a circular loop with centre at O as shown in Fig. 8.46. The moment of inertia of the loop about the axis XX is (a) (c)

L3 8

(b)

2

5 L3 16 2

(d)

L3 16

2

3 L3 8 2

B

33. A disc of mass M and radius R is rolling with angular speed on a horizontal plane as shown in Fig. 8.44. The magnitude of angular momentum of the disc about the origin O is

(a)

Fig. 8.45

Fig. 8.46

36. A tube of length L incompressible liquid of mass M and closed at both the ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity . The force exerted by the liquid at the other end is (a)

M

(c)

M

2

2

2

L

(b) M

L

(d)

4

M

2

2

L

L2

2

37. An equilateral triangle ABC formed from a uniform wire has two small identical beads initially located at A. The triangle is set rotating about the vertical axis AO. Then the beads are released from rest simultaneously and Fig. 8.47

one along AB and the other along AC as shown in Fig. 8.47. Neglecting

8.26 Comprehensive Physics—JEE Advanced

(a) angular velocity and total energy (kinetic and potential). (b) total angular momentum and total energy. (c) angular velocity and moment of inertia about the axis of rotation. (d) total angular momentum and moment of inertia about the axis of rotation. 38. A cubical block of side L rests on a rough horizontal . A horizontal force F is applied on the block as shown in Fig. minimum force required to topple the block is mg/4 (c) mg/2 (d) mg(1 — )

(a) M

(c) A M (d) AM2 43. One end of a thin uniform rod of length L and mass M1 is rivetted to the centre of a uniform circular disc of radius ‘r’ and mass M2 so that both are coplanar. The centre of mass of the combination from the centre of the disc is: (Assume that the point of attachment is at the origin)

L ( M1 M 2 ) 2 M1 2 ( M1 M 2 ) (c) LM1 (a)

39. The angular velocity of a body changes from 1 to 2 without applying a torque but by changing the moment of inertia about its axis of rotation. The ratio of the corresponding radii of gyration is (b) (a) 1 : 2 1: 2 (c) (d) : : 2 1 2 1 40. A thin uniform rod AB of mass m and length L is hinged at one end A stands vertically and is allowed to fall freely to the

(a)

mg L

(b)

1/ 2

3 g 1/ 2 g (d) L L 41. Moment of inertia of uniform horizontal solid cylinder of mass M about an axis passing through its edge and perpendicular to the axis of the cylinder when its length is 6 times its radius R is: (c)

(a)

39 MR 2 4

(b)

39 MR 4

49 MR 49 MR 2 (d) 4 4 42. A is the areal velocity of a planet of mass M angular momentum is (c)

(b) 4 (d) 8

45. A body of mass ‘m’ is tied to one end of a spring and whirled round in a horizontal plane with a constant angular velocity. The elongation in the spring is one elongation in the spring is 5 cm. The original length of the spring is: (a) 16 cm (c) 14 cm

(b) 15 cm (d) 13 cm

46. A particle performs uniform circular motion with an angular momentum L

the rod when its end B

mg 3L

LM1 2( M1 M 2 ) 2 LM1 (d) ( M1 M 2 ) (b)

44. Two circular loops A and B of radii rA and rB respectively are made from a uniform wire. The ratio of their moments of inertia about axes passing through their centres and perpendicular to their planes is IB rB is equal to IA rA (a) 2 (c) 6

Fig. 8.48

(b) 2MA

2

ular momentum becomes: (a) 4L (b) 2L L (d) L (c) 2 4 47. A uniform rod of length 1 metre is bent at its midpoint to make 90° angle. The distance of the centre of mass from the centre of the rod is (a) 36.1 cm

(b) 25.2 cm

(c) 17.7 cm

(d) zero

48. A mass is whirled in a circular path with constant angular velocity and its angular momentum is L the string is now halved keeping the angular veloc(a) L 4 (c) L

(b) L 2 (d) 2L

Rigid Body Rotation 8.27

-

49.

(a) (b) (c) (d)

Moment of inertia Angular momentum Angular velocity Rotational kinetic energy

50. A solid sphere is rotating about its diameter. Due to

two beads are at the centre of the rod and the system is rotating with angular velocity 0 about its axis perpendicular to the rod and passing through its mid point (see Fig. 8.50). There are no external the angular velocity of the system is M 0 M 0 (b) (a) M 3m M 6m (c)

M

6m M

0

(d)

0

speed of the sphere will

1 % 3 1 (b) decrease by nearly % 3 1 (c) increase by nearly % 2 2 (d) decrease by nearly % 3 51. The height of a solid cylinder is four times its radit = 0 on a belt which is moving in the horizontal direction with a velocity v = 2.45 t2 where v is in ms–1 and t (a) increase by nearly

t equal to (a) 1 s (b) 2 s (c) 3 s (d) 4 s 52. A circular portion of diameter R is cut out from a uniform circular disc of mass M and radius R as shown in Fig. 8.49. The moment of inertia of the remaining (shaded) portion of the disc about an axis passing through the centre O of the disc and perpendicular to its plane is (a)

15 MR2 32

(b)

7 MR2 16

(c)

13 MR 2 32

(d)

3 MR2 8

Fig. 8.49

53. A smooth uniform rod of length L and mass M has m

Fig. 8.50

54. Two particles A and B each other under a mutual force of attraction. At the instant when the speed of A is V and that of B is 2V the speed of the centre of mass of the system is (a) 0 (b) V (c) 1.5V

(d) 3V

55. One quarter sector is cut from a uniform circular disc of radius R. This sector has mass M made to rotate about a line perpendicular to its plane and passing through the center of the rotation is (Fig. 8.51) 1 MR2 (a) 2 1 (b) MR2 4 1 (c) MR 2 8 (d) 2 MR 2

Fig. 8.51

56. 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 center of mass is (a) 30 m/s (b) 20 m/s (c) 10 m/s (d) 5 m/s

8.28 Comprehensive Physics—JEE Advanced

57. throughout these motions). The directions of the frictional force acting on the cylinder are: (a) up the incline while ascending and down the incline while descending (b) up the incline while ascending as well as descending (c) down the incline while ascending and up the incline while descending (d) down the incline while ascending as well as descending. 58. The angular momentum of a particle moving in a circular orbit with a constant speed remains conserved about (a) any point on the circumference of the circle (b) any point inside the circle (c) any point outside the circle (d) the centre of the circle

A

(a)

B

(b)

B

(c)

A

A

(d)

B

B A

63. with speed vr

out rolling and reaches the bottom with speed vs. Then (a) vr > vs (c) vr = vs 64.

(b) vr < vs (d) vr = vs = 0

ABCD shown in a = AB = BC/2. The moment of inertia of the lamina is the minimum along the axis passing through (a) BC (b) AB (c) HF (d) EG

59. A particle moves in a circular orbit with uniform orbit is itself rotating at a constant angular speed. We may then say (a) the angular velocity as well as the angular acceleration of the particle are both constant (b) neither the angular velocity nor the angular acceleration of the particle are constant (c) the angular velocity of the particle varies but its angular acceleration is a constant (d) the angular velocity of the particle remains constant but its angular acceleration varies 60. A cylinder of mass m and radius r is rotating about its axis with a constant speed v (b) mv2 (a) 2 mv2 1 1 (c) mv2 (d) mv2 2 4 61. A circular disc of mass m and radius r is rolling on a smooth horizontal surface with a constant speed v

1 1 mv2 (b) mv2 4 2 3 (c) mv2 (d) mv2 4 62. Two solid spheres A and B R made of materials of densities A and B respectively. Their moments of inertia about a diameter are IA and IB respectively. The ratio IA /IB is (a)

Fig. 8.52

65. A uniform rod of length L is suspended from one end such that it is free to rotate about an axis passing through that end and perpendicular to the length. What minimum speed must be imparted to the lower end so that the rod completes one full (a)

2gL

(b) 2 gL

(c) 6gL (d) 2 2gL 66. A circular disc of radius R is free to oscillate about an axis passing through a point on its rim and perpendicular to its plane. The disc is turned through when it reaches the equilibrium position will be (a)

g 3R

(b)

2g 3R

2g 2g (d) 2 R R 67. A massless and inextensible cord is wound round the circumference of a circular ring of mass M and radius R. The ring is free to rotate about an axis passing through its centre and perpendicular to its (c)

Rigid Body Rotation 8.29

plane. A mass m is attached at the free end of the cord and is at rest. The angular speed of the ring when mass m has fallen through at height h is

2 gh R2

(a)

(b)

2mgh MR 2

2mgh 2mgh (d) 2 M m R M 2m R 2 68. The moment of inertia of a hollow sphere of mass M and internal and external radii R and 2R about an axis passing through its centre and perpendicular to its plane is 3 13 MR 2 (b) MR 2 (a) 2 32 31 62 (c) MR2 (d) MR2 35 35 69. angular frequency with his arms outstretched. He (c)

71. A particle is moving in the x – y plane with a constant velocity along a line parallel to the x away from the origin. The magnitude of its angular momentum about the origin. (a) is zero (b) remains constant (c) goes on increasing (d) goes on decreasing 72. A thin uniform disc has mass M and radius R. A circular hole of radius R/3 is made in the disc as shown in Fig. 8.54. The moment of inertia of the remaining portion of disc about an axis passing through O and perpendicular to the plane of the disc is 1 2 MR2 (b) MR2 (a) 9 9 1 4 (c) MR2 (d) MR2 3 9

with folded arms is 75% of that with outstretched (a) (b) (c) (d)

increase by 33.3% decrease by 33.3% increase by 25% decrease by 25%

70. A uniform disc of radius R is rolling (without slipping) on a horizontal surface with an angular speed as shown in Fig. 8.53. O is the centre of A and C are located on its rim and R from O point B is at a distance 2 the points A B and C lie on the vertical diameter vA vB and vC are the linear speeds of points A B and C respectively at (a) vA = vB = vC (c) vA

(b) vA > vB > vC 4 vC = vB (d) vA vC = 2 vB 3

Fig. 8.53

Fig. 8.54

73. A solid metallic sphere of radius R having moment of inertia equal to I about its diameter is melted and recast into a solid disc of radius r of a uniform thickness. The moment of inertia of the disc about an axis passing through its edge and perpendicular to its plane is also equal to I. The ratio r/R is (a)

2 15

(b)

2 10

(c)

2 5

(d)

1 2

74. A small object of uniform density rolls up a curved surface with an initial velocity v 3v 2 a maximum length h = 4g initial position. The object is (see Fig. 8.55) (a) ring (b) solid sphere (c) hollow sphere (d) disc

8.30 Comprehensive Physics—JEE Advanced

Fig. 8.55

75. The mass per unit length of a non-uniform rod OP of length L varies as m=k x L where k is a constant and x is the distance of any point on the rod from end O. The distance of the centre of mass of the rod from end O is (b) 2 L (a) L 3 3 2L L (c) (d) 3 2 76. A tube of length L compressible liquid of mass M and closed at both ends. The tube is then rotated in a horizontal plane about one of its ends with a uniform angular velocity . The force exerted by the liquid at the other end is. (b) M 2L (a) 2 M 2L 1 3 (c) M 2L (d) M 2L 2 2 77. A uniform thin rod of mass M and length L is hinged by a frictionless pivot at its end O 8.56. A bullet of mass m moving horizontally with a velocity v strikes the free end of the rod and gets embedded in it. The angular velocity of the system about O just after the collision is mv 2m v (b) (a) L ( M m) L ( M 2m) (c)

3mv L( M 3m)

(d)

Fig. 8.56

78. A gramophone record of mass M and radius R is rotating at an angular velocity . A coin of mass m is gently placed on the record at a distance r = R/2 from its centre. The new angular velocity of the system is 2 M 2 M (b) (a) ( 2 M m) ( M 2m) (c)

M M 10 cm and height 15 cm (d)

79. A bolck of base 10 cm

friction between them is 3 . The inclination of this inclined plane from the horizontal plane is gradually increased from 0°. Then (a) at the plane (b) the bock will remain at rest on the plane up to certain and then it will topple (c) at the plane and continue to do so at higher angles (d) at the plane and on further increasing topple at certain .

mv LM

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49. 55.

(d) (d) (a) (a) (b) (a) (b) (b) (b) (a)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56.

(d) (c) (c) (a) (c) (c) (c) (a) (b) (c)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57.

(d) (a) (c) (d) (a) (c) (c) (b) (a) (b)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58.

(d) (c) (b) (a) (d) (a) (d) (d) (c) (d)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59.

(a) (c) (d) (d) (b) (d) (d) (c) (b) (c)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60.

(b) (a) (c) (a) (b) (a) (b) (a) (a) (d)

Rigid Body Rotation 8.31

61. 67. 73. 79.

(c) (c) (a) (b)

62. (c) 68. (d) 74. (d)

63. (b) 69. (a) 75. (b)

64. (c) 70. (c) 76. (c)

65. (c) 71. (b) 77. (c)

66. (b) 72. (d) 78. (a)

SOLUTIONS 1. Since there is no external force acting on the gun-

4. Since the mass is not distributed uniformly over the

remains at rest. 2. Let = mass per unit area of the disc. Mass of the cut-out portion m1 = R2 /2 and mass of the remaining portion is (see Fig. 8.57)

lie anywhere between its centre and the circumference. Hence the correct choice is (d). m v m2 v 2 5. vCM = 1 1 m1 m2

m2 =

R2 2

R2

Let O1 be the centre of mass of the remaining portion. The centre of mass of the square is at O2. Taking moments of m1 g and m2 g about O m1 g x1 = m2 g x2 2

R 2

x1

R2

g = x2 =

x2

R 2(2

(

1)

1 (v1 + v2) 2

=

a i 2

aCM =

2

R 2

=

g

j

m1a1 m2 a2 a = 1 m1 m2 2

is conserved. Since both the inclined planes are of v2 = v1. The acceleration of the sphere rolling down the plane is

Since

2

>

R

M 4

M and its P.E. = mg (R/2) = 4

g

MgR Decrease in P.E. = MgR – 8

R 2

MgR 8

7 MgR 8

1;

g sin I 1 MR 2 a2 > a1. Hence

t1 sin = t2 sin

Fig. 8.57

2

a2 = 0)

line.

a=

M ( R / 2) 2

(

b i j 2 Since vectors vCM and aCM are parallel to each 6.

m=

m 1 = m 2)

=

x1 = R /2)

3. The entire mass M of the carpet can be assumed to be concentrated at its centre of mass which is originally at a height R P.E. = MgR. When the carpet is unrolled to a radius R R/2 above

(

2 1

Thus t1 > t2. Hence the correct choice is (b). M dx. 7. Mass of element of length dx is dm = L Perpendicular distance of the element from the axis of rotation = x sin . Therefore moment of inertia about the axis of rotation AB is (Fig. 8.58) I=

L/2

dm( x sin ) 2

L/2

=

M sin2 L

L/2 L/2

x 2 dx

8.32 Comprehensive Physics—JEE Advanced

=

ML2 sin2 12

ML2 ML2 sin2 (60°) = 12 16

=

L 2

Also a =

(3) a=

3mg 4 So the correct choice is (a). mg – T =

10.

Fig. 8.58

8. K.E. =

1 1 Mv 2 + I 2 2

=

1 1 Mv 2 + 2 2

1 MR 2 2

=

1 1 3 Mv 2 + Mv2 = Mv2 2 4 4

v = u cos 60° = u/2 along the horizontal direction. Hence the perpendicular distance of its linear momentum from the point of projection = maximum height attained. r1 = hmax =

2

v R

2

L . 2

= F

1

u 2

2L 3

Since the weight mg torque of the weight about A is = mg

2

L 2

The minimum force required to topple the block is given by 2L L = mg ( 1)min = 2 or Fmin 3 2 3mg 4

Fig. 8.60

Fig. 8.59

Moment of inertia of the rod about A is I = ML2/3. A is I

3u 2 8g

11. Torque of F about A is [see Fig. 8.60]

Fmin =

g 2L

=

3 mu 3 16 g

=

(1)

Torque of force mg about A is = Mg

3u 2 8g

=m

mg – T = ma

=

u 2 sin 2 60 2g

Magnitude of angular momentum = m r1 v

Loss in K.E. = gain in P.E. 3 or Mv 2 = Mgh 4 3v 2 h= 4g 9. When string D A. Let a be the linear acceleration of the centre of mass and T the tension in string C

=

3g . Then from Eq. (1) 4 mg T= 4

(2)

12. Let M be the mass and R the initial radius of the is the angular velocity of the rotation of T of the day is T=

2

Rigid Body Rotation 8.33

Let R be the radius of the earth after contraction and its angular velocity. From the conservation a=g

I =I 2 2 M R 2 and I M R 2 are the where I 5 5 moments of inertia of the earth before and after 2 2 MR 2 = MR 2 5 5

or

R2

=

2

R (

=4

R = R/2)

The duration T of the new day will be T = T =

2

I=

=

2 4

=

T 4

24 hours = 6 hours 4

16. The acceleration of the block sliding down the plane is a = g sin where l is the length reaching the bottom is given by v2 = 2 al = 2g sin l or v = 2gl sin The acceleration of the disc rolling down the plane is (as shown above) 2 a = g sin 3 bottom is given by v 2 = 2a l =

13. From the law of conservation of angular momen-

I =I Here I = MR2 and I = (M + 2m) R2. Therefore I M = I M 2m

Hence the correct choice is (a). 1 1 14. For rolling : Mgh = Mv2 + I 2 2 1 1 = Mv2 + 2 2 =

2 M R2 5

For sliding : Mgh =

R2

1 7 Mv 2 = Mv2 2 10 v 7 or = 5 v 15. The acceleration down the plane is given by g sin a= I 1 MR 2

IB = MBR 2B = 4 R = 16

v2

1 Mv 2. Therefore 2

v = 2

and

2

1 1 7 Mv2 + Mv2 = Mv2 2 5 10 2 I = MR 2 and 5

4 gl sin 3

gl sin v 2 = 3 v 3 Hence the correct choice is (b). 17. Let be the mass per unit length of the wire. The mass of loop A is MA = 2 R and mass of loop B is MB = 4 R Their moments of inertia respectively are IA = MA R A2 = 2 R R2 = 2 R3 or

=

1 MR2. Using this and 2

=

v R

(2R)2

R3

IA 1 = IB 8 Hence the correct choice is (d). 18. Let m the centripetal force mr 2 just exceeds the force of friction mg. The minimum is given by mr

2

= mg or

=

g r

Hence the correct choice is (c). 19. The minimum angular frequency is independent of g/r the mass. Hence the correct answer is still which is choice (a). 20. The (x y) co-ordinates of the masses at O A and B respectively are (refer to Fig. 8.40 on page 8.23) y1 x2 = a y2 = 0) and (x3 y3 = b) (x1 The (x y) co-ordinates of the centre of mass are xCM =

m1 x1 m2 x2 m3 x3 m1 m2 m3

8.34 Comprehensive Physics—JEE Advanced

= yCM = =

m

0 m a m m m m

0

=

a 3

m1 y1 m2 y2 m3 y3 m1 m2 m3 m

0 m 0 m m m m

b

=

b 3

The position vector of the centre of mass is xCM i + yCM j =

a b 1 i+ j = (ai + bj) 3 3 3

Fig. 8.61

-

21.

g sin I 1 MR 2 where K is the radius of gyration of the sphere

PS = PQ sin 60° = L sin 60° =

a=

the sphere about its diameter is 2 I= MR2 5 a=

g sin 2 1 5

=

5 g sin 7

(i)

f provides the necessary torque which is given by = force moment arm = fR But = I is the angular acceleration of I = fR a= R f =

I Ia 2 = 2 = Ma R R 5

= Mg cos . Thus Mg cos 5 g cos 2 Equating (i) and (ii) we have or

a=

2 MR 2 5 normal reaction 2 = Ma 5 I

(ii)

5 5 2 g sin = g cos or = tan 7 2 7 Hence the correct choice is (d). 22. Given PQ = QR = RP = L. The centre of mass is located at centroid C which cuts lines PS QT and UR in the ratio 2 : 1. Let h = CS = CT = UC PQS

h=

PS 1 = 3 3

3 L. 2

L 3 L= 2 3 2

its moment of inertia about an axis passing through its centre of mass C and perpendicular to its plane Ic = 3 (I + Mh2) where I is the moment of inertia of each rod about the axis passing through its centre and perpendicuI=

M L2 L Also Mh2 = M 12 2 3

2

=

M L2 12

2 2 M L2 M L2 =3 ML = ML 12 12 6 2 Hence the correct choice is (a). 23. Refer to Fig. 8.41 on page 8.24. Moment of inertia is a scalar quantity. So the moment of inertia of the structure is the sum of the moments of inertia of the

Ic = 3

I = I1 + I2 + I3 + I4 where I1= moment of inertia of rod 1 about an axis passing through its centre E and perpendicular to its M L2 12 I2= moment of inertia of rod 2 about an axis passing through its centre F and perpendicular to its plane plane =

M L2 12 I3 = moment of inertia of rod 3 about a parallel axis L 2 L M L2 at a distance from it = M = 2 4 2 =

Rigid Body Rotation 8.35

I4 = moment of inertia of rod 4 about a parallel axis L M L2 at a distance from it = . 4 2 I= = 24.

2 ML2 3

M L2 12

M L2 12

M L2 4

linear velocity v and an angular velocity kinetic energy at any time t = translational KE + rotational KE 1 1 or KE = mv2 + I 2 2 2

M L2 4

Here

1 2 moment of inertia of a square sheet ABCD about 1 Is its diagonal AC or It = 2 sheet = M + M = 2M L2 M L2 = Is = (2M) 12 6

the moment of inertia of triangular sheet ABC =

It =

Is M L2 = 2 12

I=

2 MR2 and 5

KE =

1 1 Mv2 + 2 2

=

v R

2 MR2 5

v2 R2

1 1 7 Mv2 + Mv2 = Mv2 2 5 10

7 Mv2 which gives v = 10 Hence the correct choice is (a). or

28.

=

Mgh =

10 g h 7

the total (translational + rotational) kinetic energy respectively are 1 1 1 I 2 and Kt = mv2 + I 2 2 2 2 The moment of inertia of a sphere of mass m and 2 radius r about its centre is I = mr2. Also = 5 2 v2 2 v/r. Therefore I 2 = mr2 = mv2. Thus 2 5 5 r 1 1 1 7 Kr = mv2 and Kt = mv2 + mv2 = mv2. 5 2 5 10 Hence 1 m v2 Kr 2 5 = 7 Kt 7 m v2 10 Hence the correct choice is (d). Kr =

Fig. 8.62

Hence the correct choice is (a). 25. The correct choice is (b) because the angular momentum of the sphere about the point of contact with plane surface also includes the angular momentum about the centre of mass. 26. I 1 = I 2. I3 = I4. From perpendicular axes theoplate about an axis passing through the centre and perpendicular to the plane of the plate is I = I1 + I2 = I3 + I4 = 2I1 = 2I3 ( I 1 = I 2 I 3 = I 4) or I 1 = I 3. Thus I = I 1 + I 2 = I 3 + I 4 = I 1 + I 3. Hence the correct choice is (c). 27. Let the mass of the sphere be M and R its radius. energy is entirely potential given by PE = Mgh

29.

man will meet at their centre of mass. Since their at x = 5 m. Hence the correct choice is (b).

30. Refer to Fig. 8.63. The angular momentum of the mass at point P(x y) about origin O v = m (x i

L = mr

y j)

(v i )

= myv ( k ) (

i

i = 0 and j

i = –k )

8.36 Comprehensive Physics—JEE Advanced

32. Since there is no friction between the sphere and the horizontal surface and also between the spheres momentum from sphere A to sphere B due to the collision. Since the collision is elastic and the A only transfers its linear velocity v to sphere B. Sphere A will continue to rotate with the same angular speed Fig. 8.63

Now m and v are constants. Also y remains constant as the mass moves parallel to the x-axis. Hence L remains constant. Thus the correct choice is (b). 31. Refer to Fig. 8.64. According to the perpendicular about the z-axis is Iz = Ix + Iy

is (c). 33. Refer to Fig. 8.65. Let OC = RC and let vc be the velocity of the centre of mass of the disc. The linear momentum of the centre of mass is pc = Mvc Lc is the angular momentum of the disc about C then the angular momentum about origin O is L0 = Lc + Rc pc

with Ix = Iy. The square plate lies in the x-y plane. Since the directions of the x and y the only restriction being that the angle between will not change if the axes are rotated through any angle in the plane of the plate. This can be proved as follows. Fig. 8.65

Magnitude of L0 = Ic =

+ Rc

1 MR2 2

Mvc sin

+ M Rcvc sin 1 M R2 2

Ic =

Fig. 8.64

Consider a particle of the plate of mass mn located at a point P(xn yn) in the x-y plane. Then IAB = mn y2n The moment of inertia about line CD will be ICD = mn y 2n The coordinates (xn yn) and (xn yn) are related as xn = xn cos – yn sin yn = yn sin – yn cos Now ICD = mn y 2n = mn(xn sin – yn cos )2 = (mn x2n) sin2 + (mn y 2n ) cos2 – 2 (mn xn yn) sin cos =

mn y2n

and mn xn yn = 0. Hence

IAB =

( =

1 MR2 2 Rc sin

R

= R and vc = R )

3 MR2 2

Hence the correct choice is (c). 34. When the block hits the ridge at point O start rotating about an axis passing through O and perpendicular to the plane of the paper (Fig. 8.66).

mn x2n

ICD = IAB (sin2 + cos2 ) + 0 = IAB = I Thus the correct choice is (a).

+ MR

Fig. 8.66

Rigid Body Rotation 8.37

an-gular momentum is conserved. Angular momentum of the block before it hits the ridge is a 1 Li = Mv AC = Mv = Mva (1) 2 2 Angular momentum of the block after it hits the ridge is (2) Lf = IO where IO is the moment of inertia of the block about an axis passing through O and perpendicular to the plane of the block and is the angular speed of rotation of the block. From the parallel axes IO = Ic + Mr

a2 . Hence 2 1 1 2 IO = Ma2 + Ma2 = Ma2 6 2 3 2 Ma2 Lf = 3 a 2

2

+

a 2

2

1 Ma2 6

=

(3)

Li = Lf

3v 2 2 Mva = Ma2 or = 4a 3 3 Hence the correct choice is (a). 35. Let m be the mass of the loop and r its radius. The moment of inertia of the loop about an axis passing through the centre O is (Fig. 8.67)

Fig. 8.67

IO =

1 mr2 2

ertia bout XX is I = IO + mr2 = Hence I=

3 2

-

1 3 mr2 + mr2 = mr2 2 2 m = L and radius r = L/2 . L

M v 2 M (r )2 ML = Mr 2 = 2 r r Hence the correct choice is (a).

Fc = 37.

2

gular momentum L is conserved. As the beads slide

2

C is Ic = and r2 =

36. The entire mass of the liquid can be regarded as being concentrated at the centre of mass of the tube L from the axis of which is at a distance of r = 2 revolution. The force exerted by the liquid at the other end of the tube is the centripetal force of a L mass M revolving in a circle of radius r = . Thus 2

L 2

2

Thus the correct choice is (d).

3 L3 8

2

change. From L = I

will also

correct choice is (b). 38. Torque due to F about A is 1 = FL Since the weight mg acts through the centre of mass of the block (which is at a distance of L/2 from the slide of the block) the torque due to weight mg about A is L 2 = mg 2 The minimum force required to topple the block is obtained when 1 is slightly greater than 2 limit L ( 1)min = 2 or Fmin L = mg 2 mg or Fmin = 2 Hence the correct choice is (c). 39. The magnitude of angular momentum of a rotating body is given by L = I I = conK1 and K2 are the corstant. Hence I1 1 = I2 2 I1 = MK12 and I2 = MK22. Hence MK21 1 = MK22 2 or

K1 = K2

2 1

40. Since the centre of gravity of the rod is at its cenL 1 1 = mg L. Gain in KE = I 2 where 2 2 2 I is the moment of inertia of the rod about an axis

mg

passing through its end and perpendicular to its 1 length which is given by I = mL2. Now gain in 3

8.38 Comprehensive Physics—JEE Advanced

1 2

1 mL2 3

which gives

2

=

mB r = B mA rA

1 = mgL 2

r Using (ii) in (i) and putting B rA have

3g L

41. Given: l = 6R moment of inertia about the given axis is given by I= M

R2 4

l2 3

=M

R2 4

(6 R ) 2 3

R2 =M 4

36 R 2 3

2 = 49 MR 4

area swept by radius vector time taken Assuming that the orbit of the planet is a circle of radius R

42. Areal velocity A =

A= T= A= or

=

Angular momentum

2

2

. Hence

R2 2

= (MR ) = 2 MA

or

rB rA

45. Let L cm be the original length of the spring and k be the spring constant. Then m(L + x1) 21 = kx1

L L

2 2

m(L + x2) x1 x2

1

2

=

2

Given x1 choice (b). 46. L = I

= kx2 x1 x2

(i)

x2 = 5 cm and L

1 I = 2 2

2

=

K 2

2

=2

K=

1.

1 I 2

2

Using

is the

OP = OQ = l/2

2A R2

A

(0, l/4)

Hence the correct choice is (b). 44. IA = mA rA2 and IB = mB rB2 . Hence IB mB rB = (i) IA mA rA Let k be the mass per unit length of the wire. Then the masses of loops A and B are mA = (2 rA)k and mB = (2 rB)k

C (l/8, l/8)

D O

M L/2 0 LM1 = 1 = M1 M 2 2 ( M1 M 2 )

2

3

P

Hence the correct choice is (b). 43. The mass of the rod can be considered to be concentrated at its centre (x = L/2) where x = 0 is the origin. Hence RCM

rB rA

is doubled and K value of L becomes one-fourth. Hence the correct choice is (d). 47. Rod POQ of length l = 100 cm is bent at its midpoint O so that POQ = 90° (see Fig. 8.68).

R2 R2 = 2 / 2 2A

L =I

8=

and

Hence the correct choice is (d).

R T

(ii)

E

(l/4, 0)

Q

Fig. 8.68

The mass of part PO of length l/2 can be taken to be concentrated at its mid-point A whose coordinates l/4) and of part OQ of length l/2 at its midpoint B whose coordinates are (l of mass of these two equal masses is at mid-point C between A and B. The coordinates of C are (l l/8). OC =

(OE ) 2

(CE )2 =

l 8

2

l 8

2

Rigid Body Rotation 8.39

=

l

=

Now v = 2.45 t2 dv d a= 2 . 45t 2 = 4.9t dt dt

100 cm

32 32 2 48. L = mr . For given m and L r2 r the angular momentum L becomes one-fourth. Hence the correct choice is (a). 49. momentum L = (I I is the moment of inertia and is the angular velocity I and hence 1 2 I will also change. Hence 2 the correct choice is (b). 4 4 50. V = r3 or log V = log + 3 log r. 3 3 Differ We have V r r 1 V 1 1 3 or 0 . 5% % V r r 3 V 3 6 2 I = constant or 2 2 5 mr = constant or r = constant (c) kinetic energy

or 2 log r + log

= log c 2 r r

=0

r 1 1 2 % % r 6 3 The negative sign indicates that decreases. Hence the correct choice is (b). 51. The cylinder will topple when the torque mgr equals h (see Fig. 8.69) the torque ma 2 or

2

t=

Ic = I0 + mr2 = =

or 54.

2 gr h

g 2

(

h = 4r)

(i)

1 MR 2 + 32

M 4

R 2

2

3 MR 2 32

Moment of inertia of the shaded portion about 1 3 13 MR 2 – MR 2 = MR2 O is Is = I – Ic = 2 32 32 which is choice (c). 53. From the principle of conservation of angular I0 and 0 are the I0 0 = I moment of inertia and angular velocity when the beads are at the centre of the rod and I and those when the beads are at the ends of the rod. I0 =

ML2 12

I=

ML2 12

ML2 12

or a =

g 9 .8 = = 1 s. 2 4 .9 9 .8

Hence the correct choice is (a). 52. Moment of inertia of complete disc about O is M 1 MR 2. Mass of the cut-out part is m = . I= 4 2 The moment of inertia of the cut-out portion about 1 R 2 = its own centre I0 = mr2 = 1 M 2 2 4 2 1 2 MR because r = R/2. From the parallel axes 32 tion about O is

and

Fig. 8.69

(ii)

mL2 4 0

= =

M

mL2 4

L2 (M + 6m) 12

6m 12

L2

M 0 M 6m

Hence the correct choice is (b).

centre of mass will remain at rest. Hence the correct choice is (a). 55. Area of complete disc = R 2. Area of one quarter 1 R 2. Mass of this sector = M. sector OAB = 4 Mass per unit area of the sector is

8.40 Comprehensive Physics—JEE Advanced

M

4M

2

2

(1)

energy =

Divide the sector into a large number of cylindrical shells. Consider an element of mass dm at a distance r from the centre O and having thickness dr (see Fig. 8.70). Then 2 rdr Fig. 8.70 mo dm = 4 The moment of inertia of the sector about the axis of rotation is

KE =

mo =

R

I=

r 2 dm

0

R /4

R

R

2 m0 4

m0 R 4 8

r 3 dr

0

I=

(2)

1 I 2

2

1 1 mv 2 + I 2 2

or vCM = 57.

MV M m

10 kg 14 ms 10 4 kg

v r (

I=

IA MA = = IB MB 63.

A B

Hence the correct choice is (c).

= 10 ms–1

-

2 g sin 3 -

Hence the correct choice is (c).

celeration is

its angular acceleration is always directed down the plane. Hence the frictional force acts up the inclined plate when the cylinder rolls up or down

where is the inclination of the plane. h on reaching the bottom are given by

-

64. IBC =

m AB 3

.

IAB =

1 2 mr and 2

m BC 3

I =

IHF =

m AB 12

v r

1 2

IEG =

m BC 12

1 mr2 2

1 I 2

v r

2

2

1 mv2 4 61. The kinetic energy of a rolling disc consists of two 1 parts: translational energy = mv2 and rotational 2

2ar h and vs =

Since as > ar choice (b).

-

60. The kinetic energy (which is rotational) is

=

as = g sin

vr =

stant. Hence the correct choice is (c).

1 mr 2 ) 2

1 1 3 mv2 + mv2 = mv2 2 4 4 Hence the correct choice is (c). 4 62. Mass of sphere A MA = R3 A 3 4 2 2 B MB = R3 B IA = MAR2 and IB = 3 5 5 2 MB R

ar =

58. The correct choice is (d). 59.

2

=

MV = (M + m) vCM 1

2

1 2 mr 2

1 1 = mv2 + 2 2

1 MR2 2

(a). 56. The velocity vCM of the centre of mass can be obtained by using the principle of conservation of

.

=

65.

where M

2as h vs > vr

2

ma 2 3

2

4 ma 2 3

2

ma 2 12

2

ma 2 3

HF is the miniMgL

Rigid Body Rotation 8.41

MgL =

1 I 2

1 ML2 2 3

2

2

6g 6g . Now v = L = L 6 gL L L Hence the correct choice is (c). 66. PE at = 60° is Mgh (1 – cos ) where h is the distance between the axis of rotation and the centre of mass of the disc. Thus h = R. Gain in KE when 1 I 2 the disc reaches the equilibrium position = 2 1 3 2 2 2 where I = IC.M. + Mx = MR MR MR2. 2 2 Here x is the distance between the centre of mass x = R. Now PE = KE gives 3 MgR (1 – cos 60°) = MR 2 2 4 2g which gives = 3R 67. Loss in PE = gain in rotational KE. Thus 1 1 mgh = (I + mR 2) 2 = (MR2 + mR2) 2 2 2 1 2 = R (M + m) 2 2 2mgh or = . Hence the correct choice is (c). M m R2 or

=

68. We obtain the given hollow sphere as if a solid sphere of radius R has been removed from a solid sphere of radius 2R. The mass of the given hollow sphere is (here is the density of the material of the sphere) M = M1 – M2 4 4 (2R)3 and M2 = R3 are the 3 3 masses of spheres of radii 2R and R respectively.

where M1 =

M = 28 R3 (i) 3 The moment of inertia of the given hollow sphere is I= =

2 2 M1 (2R)2 – M2 R2 5 5 2 5

4 3

2R

3

2 4 = (32 – 1) 5 3 choice (d).

2R

2

2 5

4 3

R3

R5 62 MR 2 35

Given I2 = or

3 I1. Hence I1 4 =

2

4 3

K1 = K2 = =

R2

1

3 I1 4

=

1 I1 2

1 I2 2

2

2 1

1 2

=

2

4 1 I1 3 2

3I1 4

100 =

4 3

1

2

4 K1 3

2 1

Percentage increase in KE = 4 K1 K1 = 3 K1

2

1.

K2

K1 K1

100

100 = 33.3% 3

Hence the correct choice is (a). 70. The disc is rolling about the point O. Thus the axis of rotation passes through the point A and is perpendicular to the plane of the disc. From the relation v = r where r is the distance of the point vB = (AB) vA vC = (AC) = 2R

and

v B 3R vc 2 choice is (c).

1 2R

Hence

=

=

3R 2

3 . Thus the correct 4

71. Refer to Fig. 8.71. The angular momentum of the particle at point P(x y) about origin O is given by L=mr v =m

xi

yj Fig. 8.71

= – myv k ( (ii)

I =

69. Let I1 and 1 be the moment of inertia and angular frequency when his arms are outstretched and I2 and 2 those when his arms are folded. Then I1 1 = I2 2

i

i = 0 and j

i =– k)

m and velocity v are constant. Also y remain constant as the particle moves parallel to the x-axis. Hence L remains constant. Thus the correct choice is (b).

8.42 Comprehensive Physics—JEE Advanced

M

72. Mass per unit area of the disc =

2

. Therefore,

R mass of the removed portion (hole of radius R/3) is

R 2 M = R 3 9 The moment of inertia of the complete disc about an axis passing through its centre O and perpendicular to its plane is 1 I= MR2 2 Using the parallel axes theorem, the moment of inertia of the removed portion of the disc about the axis passing through O and perpendicular to the plane of the disc is I = MI of mass m about O + m OO M

m=

2

R 1 m = 3 2 1 M = 2 9

2

2R 3

+m

R2 M + 9 9

2

mR 2 . Hence the object is a 2 disc, which is choice (d). 75. Consider a rod OP of length L lying along the x-axis with O as the origin (Fig. 8.72). Consider a small element AB of length dx at a distance x from O. On solving, we get I =

Fig. 8.72

k (xdx) L The distance of the centre of mass from O is given by L

xCM =

4R2 1 = MR2 9 18

Therefore, the moment of inertia of the remaining 1 MR2 – 2 1 4 MR2 = MR2. Hence the correct choice is (d). 18 9 73. Let M be the mass of the sphere. The mass of the disc will also be M. The moment of inertia of the sphere about its diameter is 2 Is = MR2 5 The moment of inertia of the disc about its edge and perpendicular to its plane is (using parallel axes theorem) Id = Icm + Mh2 = 1 Mr2 + Mr2 = 3 Mr2 2 2 Given Is = Id. Hence, we have portion of the disc about O = I – I =

AB (= dm) = mdx =

Mass of element

=

k x2 d x L0

( d m) x

L

( d m) x3 3

L

0 2 L

x 2

k xd x L0 L3 / 3 2

L /2

0

The correct choice is (b). 76. When the tube AB is rotated about its end A in a horizontal plane with a uniform angular velocity, all points of the tube rotate with same angular velocity. Consider a small element of the liquid of length dr at a distance r from the axis of rotation. The mass of this element is (see Fig. 8.73). M dr m= L

2 3 MR2 = Mr2 5 2 2 r = , which is choice (a). 15 R 74. From the principle of conservation of mechanical energy, we have 1 1 mv2 + I 2 = mgh 2 2 which gives

1 1 mv2 + I 2 2

v R

2

= mg

3v 2 4g

2L 3

Fig. 8.73

Rigid Body Rotation 8.43

Force exerted by the element is

Since no external torque acts on the system, the angular momentum is conserved, i.e. L = L or (I + mr2) = I

M 2 r dr L Total force exerted by the liquid at end B is L M M 2 2 L F= r dr r dr 0 0 L L dF = m r

=

M L

2

r2 2

L

0

2

=

1 M 2

2

or

L ,which is choice (c).

or

=

1 M R2 2

I m r2

I

=

m r2

2 m r2 M R2

1

77. Let be the angular velocity acquired by the system (rod + bullet) immediately after the collision. Since no external torque acts, the angular momentum of the system is conserved. Thus mvL = I (1)

1 M R2 2

Putting r = R 79. The block will just begin to slide if the downward force mg sin just overcomes the frictional force, i.e. if mg sin = μN = μ mg cos tan = μ = 3 = 60° [see Fig. 8.74] The block will topple if the torque due to normal reaction N about O just exceeds the torque due to mg sin about 0, i.e.

where I is the moment of inertia of the system about an axis passing through O and perpendicular to the rod. Thus I = M.I. of rod about O + M.I. of bullet stuck at its lower end about O 1 1 ML2 + mL2 = (M + 3m)L2 (2) = 3 3 Using Eq. (1) in Eq. (2), we have 1 m vL = (M + 3m) L2 3 3m v or = L ( M 3m ) Hence the correct choice is (c).

N mg cos

OA = mg sin 5 cm = mg sin

2 3 Since for toppling is less than correct choice is (b). tan

78. The initial angular momentum of the rotating record is L =I 1 where I = MR2. 2 Let be the angular velocity of the record when the coin of mass m is placed on it at a distance r from its centre. The angular momentum of the system becomes L = (I + mr2)

=

OB 15 cm 2 34° for sliding, the

Fig. 8.74

II Multiple Choice Questions with One or More Choice Correct 1. In the HCl molecule, the separation between the nuclei of hydrogen and chlorine atoms is 1.27 Å. If the mass of a chlorine atom is 35.5 times that of a hydrogen atom, the centre of mass of the HCl molecule is at a distance of

35.5 1.27 Å from the hydrogen atom 36.5 35.5 1.27 Å from the chlorine atom (b) 36.5 (a)

8.44 Comprehensive Physics—JEE Advanced

1.27 Å from the hydrogen atom 36.5 1.27 (d) Å from the chlorine atom 36.5 2. Choose the correct statements from the following: (a) The position of the centre of mass of a system of particles does not depend upon the internal forces between particles. (b) The centre of mass of a solid may lie outside the body of the solid. (c) A body tied to a string is whirled in a circle with a uniform speed. If the string is suddenly cut, the angular momentum of the body will not change from its initial value. (d) The angular momentum of a comet revolving around a massive star, remains constant over the entire orbit. 3. Which of the following statements are correct? (a) When a body rolls on a surface, the force of friction acts in the same direction as the direction of motion of the centre of mass of the body. (b) During rolling, the instantaneous speed of the point of contact is zero. (c) During rolling, the instantaneous acceleration of the point of contact is zero. (d) A wheel moving down a perfectly frictionless inclined plane will slip and not roll on the plane. 4. In which of the following is the angular momentum conserved? (a) The planet Neptune moves in an elliptical orbit round the sun with the sun at one of the foci of the ellipse (c)

elliptical orbit round the nucleus (c) An -particle, approaching a nucleus, is scattered by the force of electrostatic repulsion between the two (d) A boy whirls a stone, tied to a string, in a horizontal circle 5. Four tiny masses are connected by a rod of negligible mass as shown in Fig. 8.75. (a) The moment of inertia of the system about axis AB is 50 ma2 (b) The radius of gyration od the system about axis AB is 5 a. (c) The moment of inertia of the system about axis CD is 10 ma2 (d) The radius of gyration of the system about axis CD is a.

Fig. 8.75

6. The moment of inertia of a uniform circular disc of mass M and radius R about its centre and normal 1 to its plane is MR2. Then 2 (a) its radius of gyration about the centre is R. (b) the moment of inertia of the disc about its di1 ameter is MR2. 4 (c) the moment of inertia of the disc about an axis passing through a point on its edge and normal 3 to the disc is MR2. 2 (d) the moment of inertia of the disc about a tangent 5 in the plane of the disc is MR2. 4 7. A molecule consists of two atoms, each of mass m, separated by a distance a. The rotational kinetic energy of the molecule is K and its angular frequency is . I is the moment of inertia of the molecule about its centre of mass. Then 1 (b) I = ma2 (a) I = ma2 2 1 K 2 K (d) = a m a m 8. A circular ring of mass m and radius r rolls down an inclined plane of height h. When it reaches the bottom of the plane its angular velocity is and its rotational kinetic energy is K. Then 1 1 (b) = gh gh (a) = r 2r 1 (c) K = mgh (d) K = mgh 2 9. A rope is wound round a solid cylinder of mass M and radius R. If the rope is pulled with a force F, the cylinder acquires an anglular acceleration and the rope acquires a linear acceleration a. Then F 2F (a) = (b) = MR MR 2F F (c) a = (d) a = M M (c)

=

Rigid Body Rotation 8.45

10. A solid sphere rotating about its diameter at an angular frequency has rotational kinetic energy K. 1 When it is cooled so that its radius reduces to of n its original value, the new values of and K become and K respectively. Then (a)

=n

(b)

= n2

K K =n (d) = n2 K K 11. Three forces act on a wheel of radius 20 cm as shown in Fig. 8.76. (c)

(c) Frictional force = mg cos , where m is the mass of the cylinder and friction between the cylinder and the plane. (d) Frictional force helps rotational motion of the cylinder but opposes its translational motion. 13. A spherical ball is released from rest from point A on a hemispherical surface and it rises up to a point C as shown in Fig. 8.78. Part AB of the surface is rough and the ball rolls from A to B without slipping. Part BC of the surface is frictionless. KA, KB and KC are kinetic energies of the ball at points A, B and C respectively. Which of the following is/are correct? (b) KB > KA, hA < hC (a) KA = KC, hA = hC (c) KB > KC, hA > hC (d) KC > KA, hA > hC

Fig. 8.78

Fig. 8.76

(a) The torque produced by the force of 8N is 0.8 Nm clockwise. (b) The torque produced by the force of 4 N is 0.8 Nm anticlockwise. (c) The torque produced by the force of 9 N is zero. (d) The net torque produced by the forces is 1.8 Nm clockwise. 12. A solid cylinder rolls down a rough inclined plane without slipping as shown in Fig. 8.77. Choose the correct statement (s) from the following.

Fig. 8.77

(a) If is decreased, the force of friction will decrease. (b) Frictional force is dissipative.

14. The moment of inertia of a uniform disc about its diameter is I. Then the moment of inertia about an axis (a) passing through its centre and perpendicular to its plane is 2I. (b) tangential to its plane is 5I. (c) tangential and perpendicular to its plane is 6I. (d) all the above choices are correct. 15. A wire of mass M, length L and density is bent to form a circular ring of radius R. Then, the moment of inertia of the ring about its diameter is 1 MR2 (b) R4L /2 (a) 2 ML2

R 2 L3 8 2 8 16. A uniform solid sphere and a solid cylinder of the same mass and the same diameter are released from rest on the top of an inclined plane of inclination . If they roll down the plane without slipping, then (a) the acceleration of each down the plane is g sin . (b) the ratio of the accelerations of the sphere and the cylinder is 15:14. (c) the ratio of the times taken by the sphere and the cylinder to reach the bottom of the plane is 14 : 15 . (d) the sphere and the cylinder will reach the bottom with the same speed. (c)

(d)

8.46 Comprehensive Physics—JEE Advanced

17. A rope is wound around a hollow cylinder of mass 3 kg and radius 20 cm. The rope is pulled with a constant force of 30 N. If is the angular acceleration of the cylinder and a the linear acceleration of the rope, then (b) = 40 rad s–2 (a) = 50 rad s–2 –2 (c) a = 30 ms (d) a = 10 ms–2 18. A rope of negligible mass is wound around the circumference of a bicycle wheel (without tyre) of diameter 1 m. A mass of 2 kg is attached to the end of the rope and is allowed to fall from rest. The mass falls 2 m is 4 s. The axle of the wheel is horizontal and the wheel rotates in the vertical plane. Take g = 10 ms–2 and neglect the friction due to air. (a) the linear acceleration of the wheel is 0.25 m s–2. (b) the angular acceleration of the wheel is 0.5 rad s–2. (c) the magnitude of the torque acting on the wheel is 10 Nm. (d) the moment of inertia of the wheel about the horizontal axle is 20 kg m2. 19. A smooth sphere A is moving on a horizontal frictionless surface with angular speed and centre of mass velocity v. It collides head-on with an identical sphere B at rest. After the collision their angular speeds are A and B respectively. If the collision is elastic and the friction is neglected, then (b) B = 0 (a) A = (d) A = B (c) A < B IIT, 1999 20. The torque acting on a body about a given point is given by

= A

L where A is a constant vector

and L is the angular momentum of the body about that point. It follows that dL is perpendicular to L at all instants of (a) dt time. (b) the component of L in the direction of A does not change with time. (c) the magnitude of L does not change with time. (d) All the above choices are correct. IIT, 1998 21. A uniform bar of length 6a and mass 8 m lies on a horizontal frictionless table. Two point masses m and 2m moving in opposite directions but in the same horizontal plane with speeds 2v and v respectively strike the bar at distance a and 2a from one end and stick to the bar after the collision. Then after the collision

(a) the velocity of centre of mass is zero. (b) the angular speed of the bar with the masses v . stuck to it is 5a (c) the moment of inertia of the bar with masses stuck to it about the axis passing through the end of the bar and perpendicular to its plane is 30 ma2. 3 (d) the total energy of the bar mv2. 5 IIT, 1991 22. A disc of mass M and radius R is rolling with angular speed on a horizontal surface as shown in Fig. 8.79. The magnitude of angular momentum of the disc about the origin O is (here v is the linear velocity of the disc)

Fig. 8.79

(a)

3 MR2 2

(c) MRv

(b) MR2 (d)

3 MRv 2

IIT, 1999 23. Which of the following statements is/are correct about a particle moving in a circle with a constant speed? (a) The linear velocity and acceleration vectors are perpendicular to each other. (b) The linear velocity vector is always perpendicular to the angular velocity vector. (c) The force acting on the particle is radial. (d) The force does no work on the particle. 24. The position vector of a particle with respect toorigin O is r . If the torque acting on the particle is zero, then (a) the linear momentum of the particle remains constant. (b) the angular momentum of the particle about O remains constant. (c) the force applied to the particle is perpendicular to r . (d) the force applied to the particle is parallel to r.

Rigid Body Rotation 8.47

25. A block of mass m is connected to a spring of spring constant k r as shown in Fig. 8.80. The mass of the pulley is 2m. The block is pulled down by x0 from the equilibrium position and released. The spring has negligible mass. Then 1 2 (a) the total energy of the system is kx0 . 2 (b) the velocity of the block when it is at a distance x from the equilibrium position is 1/ 2 k . x02 x 2 2m (c) the velocity of the block is maximum when x = x 0. (d) the velocity of the block is zero when x = x0.

Fig. 8.80

26. A rod of mass M = 0.9 kg and length L = 1 m is suspended at O as shown in Fig. 8.81. A bullet of mass m = 100 g moving with velocity v = 80 ms–1 in the horizontal direction strikes the end P of the rod and gets embedded in it. If I is the moment of inertia of the system and is the angular velocity immediately after the collision, then (a) I = 0.4 kgm2 (b) I = 0.3 kgm2 (c) = 20 rad s–1 (d) = 26.7 rad s–1

Fig. 8.81

27. A wheel is initially at rest. A constant torque acts on it for a time t. As a result, the wheel acquires an angular acceleration and angular velocity . If the angular displacement produced is , then (a) t2 (b) = constant (c) t (d) power 28. A string of length L and of negligible mass hangs from a support O. The other end of the string carries

a mass m which is moved in a horizontal circle of radius R to form a conical pendulum. If the string makes an angle = 60° with the vertical, then (a) the speed of the body along the circle is v = Rg . (b) the tension in the string is 2mg. (c) the horizontal component of the angular mom3 Lg mL . entum of the body point O is 2 2 (d) the magnitude of the torque acting on the 3 mgL. 2 29. A solid cylindrical roller of mass M and radius R is rolled on a rough horizontal surface by applying a horizontal force F. If aCM is the linear acceleration of the centre of mass and f is the frictional force between the roller and the surface, then F 2F (b) aCM = (a) aCM = M 3M F (c) f = (d) f = zero 3 30. If the resultant of all the external forces acting on a system of particles is zero, then for an inertial frame, one can surely say that (a) linear momentum of the system does not change in time (b) kinetic energy of the system does not change in time (c) angular momentum of the system does not change in time (d) potential energy of the system does not change in time IIT, 2009 31. horizontal plane surface. In Fig. 8.82, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then, body about point O is

(a) VC

VA = 2 V B

(b) VC

VB = VB

VC VA

(c) VC

VA

= 2 VB

(d) VC

VA

= VB

VC

Fig. 8.82

IIT, 2009

8.48 Comprehensive Physics—JEE Advanced

32. A thin ring of mass 2 kg and radius 0.5 m is rolling without slipping on a horizontal plane with velocity 1 m/s. A small ball of mass 0.1 kg, moving with velocity 20 m/s in the opposite direction hits the ring at a height of 0.75 m and goes vertically up with velocity 10 m/s. Immediately after the collision [See Fig. 8.83] (a) the ring has pure rotation about its stationary CM. (b) the ring comes to a complete stop. (c) friction between the ring and the ground is to the left.

(d) there is no friction between the ring and the ground.

Fig. 8.83

IIT, 2011

ANSWERS AND SOLUTIONS

1. Since, most of the mass of an atom is concentrated in its nucleus and the size of a nucleus (which is of the order of 10–15 m) is very small compared to the separation between the hydrogen and chlorine atom (which is 1.27 10–10 m), the atoms can be treated as point masses. The centre of mass of HCl molecule will be on the line joining the two atoms. Let us say that the H atom is located at, say, x = 0 and the Cl atom at x = x. Let xCM be the position of the centre of mass between x = 0 and x = x. If m1 and m2 are the masses of H and Cl of centre of mass, we have m 0 m2 x m2 x xCM = 1 = m1 m2 m1 m2 Now x = separation between H and Cl atoms = 1.27 Å and m2 = 35.5 m1. Hence 35.5 m1 1.27 Å xCM = = 35.5 1.27Å m1 35.5 m1 36.5 This gives the distance of the centre of mass from the hydrogen atom. The distance of the centre of mass from the chlorine atom is 1.27 Å – 35.5 1.27 Å = 1.27Å 36.5 36.5 Hence the correct choices are (a) and (d). 2. The only incorrect statement is (c). Since no external torque acts on the body even after the string is cut, the angular momentum will remain unchanged. 3. Statement (a) is correct. The direction of motion of the centre of mass is the direction along which the body rolls. Since the force of friction is opposite to the direction of the velocity of the point of contact, the force of friction acts in the direction of motion of the centre of mass. Statement (b) is also correct. At each instant of time, the point of contact is momentarily at rest. Statement (c) is incorrect. Since the body is rotating while it is rolling, the direction of the velocity is changing with time. Hence the

acceleration of the point of contact is not zero. Statement (d) is correct. Rolling cannot take place in the absence of friction because it is the frictional force that provides the necessary torque which makes the body roll on a surface. Hence the correct choices are (a), (b) and (d). 4. The angular momentum is conserved in the four cases. The object, in each case, is moving under the action of a central (radial) force. The torque dL ; the due to a radial force is zero. Since = dt angular momentum L does not change with time. 5. All four choices are correct. The moment of inertia about AB is I = m1r 21 + m2r 22 + m3r23 + m4r24 = m 0 + 2m (a)2 + 3m (2a)2 + 4m (3a)2 = 0 + 2 ma2 + 12ma2 + 36ma2 = 50ma2 The radius of gyration K is given by I = MK2 = (m1 + m2 + m3 + m4)K2 = (m + 2m + 3m + 4m)K2 = 10mK2 I = 50ma 2 = 5a 10m 10m The moment of inertia about CD is I = m (2a)2 + 2 m (a)2 + 3m 0 + 4 m = 4 ma2 + 2ma2 + 0 + 4ma2 = 10ma2 The radius of gyration is or

K=

(a)2

I 10ma 2 = =a 10ma 10m 6. The correct choices are (b), (c) and (d). Let us consider two perpendicular diameters, one along the x-axis and the other along the y-axis. Then 1 MR2 Ix = Iy = 4 According to the perpendicular axes theorem, the moment of inertia of the disc about an axis passing through the centre is =

Rigid Body Rotation 8.49

1 1 MR2 + MR2 4 4 1 = MR2 2

Ic = Ix + Iy =

which is choice (b). Since the disc is uniform, its centre of mass coincides with its centre. Therefore, the moment of inertia of the disc about an axis passing through its centre of mass and normal to its plane is 1 MR 2 ICM = IC = 2 According to the theorem of parallel axes, the moment of inertia of the disc about an axis passing through a point on its edge and normal to its plane is given by 1 MR 2 + MR2 Ie = ICM + M h2 = 2 ( h = R) 3 2 = MR . 2 From the parallel axes theorem, the moment of in1 ertia of the disc about a tangent is MR2 + MR2 = 4 5 2 MR . 4 7. The correct choices are (b) and (d). Since the two atoms have the same mass, the centre of mass is at a distance of a/2 from each atom. Therefore, the moment of inertia of the molecule about its centre of mass is

Kinetic energy is k = = 8.

mgh =

=

2

a 2

I=m

1 I 2

2

2

+

ma 2

2

=

ma 2 2

, which gives 2k

2k = I

1 1 mv2 + I 2 2 1 2 mr 2

a 2

=m

=

1 2 mr 2

2

2 2 = 1 a

k m

1 1 mv2 + (mr2) 2 2 2 ( I = mr2) 2

(

= mr2 2 1 gh . Also = r 1 2 1 gh 1 I = (mr2) mgh 2 = 2 r 2 2 Hence the correct choices are (a) and (d). 1 9. I = MR2 and torque = FR = I . Hence 2 K=

FR FR 2F = = 1 2 I MR MR 2 2F Also a= R= M Thus the correct choices are (b) and (c). 10. The correct choices are (b) and (d). r 2 I 2 2 mr2, I = m = 2. I =I ,I= n n 5 5 I Hence I = 2 n or = n2 I 1 1 1 K = I 2, K = I 2 = (n2 )2 2 n 2 2 2 1 2 2 = I n = n2K. 2 11. The torques produced by forces 8 N, 4 N and 9 N respectively are 1 = 8 N 0.2 m sin 30° = 0.8 Nm (clockwise) 2 = 4 N 0.2 m sin 90° = 0.8 Nm (anticlockwise) 3 = 9 N 0.2 m sin 90° = 1.8 Nm (clockwise) Net torque = 0.8 – 0.8 + 1.8 = 1.8 Nm clockwise Hence the correct choices are (a), (b) and (d). 12. Refer to the Fig. 8.84. Here f is the frictional force. The linear acceleration of the centre of mass of the rolling cylinder is given by g sin (1) acm = I cm 1 mR 2 where R is the radius of the cylinder and Icm is the moment of inertia of the cylinder about the centre of mass which is given by 1 Icm = mR2 2 =

v=r ) Fig. 8.84

Using this in Eq. (1), we have g sin 2 g sin acm = mR 2 = 1 3 2mR 2 Now, for linear motion, we have mgsin – f = macm

(2)

(3)

8.50 Comprehensive Physics—JEE Advanced

Using Eq. (2) in Eq. (3), we get 2mg sin mgsin – f = 3 mg sin which gives f = (4) 3 It follows from Eq. (4) that if is decreased, f will also decrease. Hence choice (a) is correct. As the cylinder is rolling down, the point of application of the frictional force is at rest at any given instant. Hence no work is done by the frictional force, i.e. the frictional force is not dissipative. Therefore, statement (b) is wrong. Statement (c) is also wrong as f is given by Eq. (4). Statement (d) is correct because the frictional force provides torque = f R to help rotational motion but it will oppose translational motion. Hence the correct choice are (a) and (d). 13. The ball is at rest at point A. Hence its kinetic energy KA (rotational + translational) is zero, it has only gravitational potential energy mghA. As it is released at A, it begins to roll (due to friction in part AB) thus acquiring kinetic energy at the expense of gravitational potential energy. When it reaches point B, its kinetic energy consists of both rotational and translational energy. At point B its potential energy 1 is zero. At point B, the rotational kinetic energy is 2 I 2, where is the angular velocity at point B and I is the moment of inertia of the ball about its centre. Since part BC is frictionless, the torque on the ball is zero. Hence its angular momentum L = I remains constant in part BC. Hence, the angular velocity of the ball remains constant in part BC. The translational kinetic energy is converted into gravitational potential energy. At point C, the translational kinetic energy of the ball is zero; it has rotational kinetic 1 energy I 2 and gravitational potential energy 2 mg hC. Thus KA = 0 (1) 1 1 KB = mv2 + I 2 (2) 2 2 where v is the linear velocity of the centre of mass of the ball and 1 KC = I 2 (3) 2 It follows from (1), (2) and (3) that KB > KC > KA. Now total energy at A is EA = 0 + mghA = mghA and 1 I 2 + mg hC. From the total energy at C is EC = 2

law of conservation of energy EA = EC, i.e. mg hA 1 1 = I 2 + mg hC or mg (hA – hC) = I 2. Since the 2 2 right hand side of this equation is positive, hA > hC. Hence the correct choices are (c) and (d). 14. From perpendicular axes theorem, Ix + Iy = Ic. Hence Ic = I + I = 2 I. [see Fig. 8.85(a)] 1 we knows that Ic = MR2 MR2 = 2Ic = 4I. 2

Fig. 8.85

From parallel axes theorem [see Fig. 8.84 (b)], we have IAB = Ix + MR2 = I + 4 I = 5 I. Using parallel axes theorem [see Fig. 8.84 (c)], we have IPQ = Ic + MR2 = 2I + 4I = 6 I. Hence the correct choice is (d). M M = R2L (1) 15. = R2 L L L =2 R R= (2) 2 The moment of inertia of a ring about an axis passing through its centre and perpendicular to its plane = MR2. From the parallel axes theorem, the moment of inertia of the ring about its diameter = 1 MR2 2 four choices are correct. g sin 16. Acceleration a = I 1 MR 2

Rigid Body Rotation 8.51

If s is the distance along the inclined plane, the velocity when the body reaches the bottom of the plane is given by v2 – 0 = 2as v = 2as and the time taken to reach the bottom is 1/ 2 I 2s 1 MR 2 t = 2s = g sin a For cylinder: For sphere: Hence giving

1 MR2 2 2 Is = MR2 5 2 g sin ac = 3 as 15 = ac 14 Ic =

dL dL ; L = 0. Hence dt dt

dL = 0 L = constant. Hence choice dt dL (c) is correct. Since A , choice (b) is also cordt rect. Thus the correct choice is (d). 21. Since no external force is applied, the linear momentum is conserved. Hence L

and as =

5 g sin 7

vs as t a = and s = c vc ac tc as Thus the correct choices are (b) and (c). 17. I = MR2 and = FR. FR F 30 = = = 50 rad s–2 = 2 = MR I MR 3 0.2 a = R = 0.2 50 = 10 ms–2 The correct choices are (a) and (d). 1 1 2=0+ a (4)2 18. s = ut + at2 2 2 a = 0.25 ms–2. a 0.25 = = 0.5 rad s–2 = R 0.5 = mg moment arm = 2 10 0.5 = 10 Nm 10 I= = = 20 kg m2 0.5 Hence all the four choices are correct. 19. Since the collision is elastic and head-on and the spheres have the same mass, they will exchange their velocity after the collision, i.e. A comes to rest and B moves with velocity v. Since there is no friction, the torque on each sphere is zero. Hence their angular speeds remain unchanged on collision. Thus the correct choices are (a) and (b). (1) 20. Given = A L dL dL . Hence = A L . This = dt dt

dL is perpendicular to both L and A . dt Hence choice (a) is correct. Now L L = L2, where L is the magnitude of L . means that

dL dL dL L =2L + dt dt dt dL dL 2L =2L dt dt

L

Since L

Also

We know that

Differentiating, we have d d 2 (L L) = (L ) dt dt

dL =0 dt

(8 m + m + 2 m)vCM = 2m(– v) + m(2v) + 8m

0

where vCM is the velocity of the centre of mass. This gives vCM = 0. The moment of inertia of the system is 1 8m (6a)2 I = 2ma2 + m(2a)2 + 12 = 2ma2 + 4ma2 + 24ma2 = 30 ma2 Since no external torque is applied, the angular momentum of the system is conserved. If is the angular speed, then 2 mv a + m 2v 2a = I 2 v 6mva = 30ma2 2 = 5a The system has no translational K.E. Hence Total K.E. = K.E. of rotation =

1 I 2

2

1 v 2 = 3 mv2 30 ma2 2 5 5a Hence all the four choices are correct. 22. The angular momentum about O is =

LO = LCM + M R v Its magnitude is

R v and LCM = I

LO = I + MRv 1 MR 2 + MR = 2 =

3 MR2 2

R

(

v=R )

8.52 Comprehensive Physics—JEE Advanced

v 3 3 MR2 = MRv R 2 2 Hence the correct choices are (a) and (d). 23. All the four choices are correct. The linear velocity v , angular velocity and radius vector r are related as r v = =

Hence v is perpendicular to both and r . Hence choice (b) is correct. The acceleration (and hence force) is centripetal (towards the centre) and v is r . Hence choice (a) and (c) are correct. Since F is r , the work done is zero. Hence choice (d) is also correct. dL 24. = r F and = . Hence the correct choices dt are (b) and (d). 1 25. Total energy before the block is released = kx20. 2 After the block is released, the total energy when it is at a distance x from the equilibrium position = K.E. of block + rotational K.E. of pulley + P.E. 1 1 1 stored in the spring = mv2 + Ip 2 + kx2. 2 2 2 1 1 m pr 2 = (2m)r2 = mr2. 2 2 From the principle of conservation of energy, we have ( v = r ) v 2 1 2 1 2 1 1 kx0 = mv2 + (mr2) + kx r 2 2 2 2 where Ip =

1/ 2

k . x02 x 2 2m Thus v is maximum when x = 0. Hence the correct choices are (a), (b) and (d). which gives v =

1 1 26. I = mL + ML2 = 0.1 (1)2 + 0.9 (1)2 3 3 = 0.4 kgm2 From conservation of angular momentum, we have mvL = I mv L 0.1 80 1 = = = 20 rad s–1. I 0.4 The correct choices are (a) and (c). = 0 + t = t. Since is constant 27. = 0+ t

28. Refer to Fig. 8.86. OA = OB = OC = L R = l sin , OD = L cos

Fig. 8.86

T sin

= mg

(1)

mv 2 R Dividing (1) by (2), we get T cos

=

(2)

v = Rg tan = Rg tan 60 = 1.73Rg Hence choice (a) is incorrect. From (1) and (2) we get T 2 = (mg)2 + = (mg)2 +

m2

(v2)2

R2 m2

(Rg tan 60°)2 R2 = (mg)2 + 3(mg)2 = 4(mg)2 T = 2 mg Horizontal component of angular momentum about O = mv OD = mv L cos Rg tan

=m

L cos

= mL cos

Lg tan sin

= mL sin

Lg cos

2

and = I , it follows that is constant. Hence t. 1 2 t . Hence t2. Also power P = . Also = 2 Hence P . Thus all the four choices are correct.

= mL sin 60°

Lg cos 60

3 Lg mL , which is choice (c). 2 2 Torque about O = mgR = mgL sin =

3 mgL 2 Hence the correct choices are (b), (c) and (d). 29. Since the frictional force acts opposite to the direction of motion, the equation of translational motion is = mgL sin 60° =

F – f = MaCM

(1)

Rigid Body Rotation 8.53

For rotational motion, we have =I Since the torque is due to frictional force, = f R. a CM 1 MR 2 Hence fR = 2 R aCM 1 I MR 2 , 2 R f=

or

MaCM 2

VB

V A = Vi

Hence the correct choice are (b) and (c) 32. Let M be the mass of the ring and m that of the ball and let V and v be their velocity before collision. The initial momentum of the system (ring and ball) in the horizontal direction is

(2)

2F F and f = . 3M 3 So the correct choices are (b) and (c).

Equations (1) and (2) give aCM =

pi = MV

mv

=2

1 + 0.1

(–20)

=2 – 2 = 0 tum of the system pf = 0 in the horizontal direction. Hence Vcm = 0 for the ring, i.e. the ring has pure rotation about its centre of mass. So choice (a) is correct. The total initial angular momentum of the system about the point of collision is Li = mvr – I

30. The only correct choice is (a) 31. Refer to Fig. 8.87.

V R = mvr – MRV

= mvr – MR2 Fig. 8.87

= 0.1

20

0.75 – 2 2

VA = 0, Let V B = V i , then V C = 2V i

= 1.5 – 1 = 0.5 kg m s

0.5

1

–1

V C – V A = 2V i – 0 = 2V i

From the conservation of angular momentum, the

V C – V B = 2V i V i = V i

the friction between the ring and the ground is to the left. So the correct choices are (a) and (c).

VB

VC = Vi

III Multiple Choice Questions Based on Passage Questions 1 to 2 are based on the following passage Passage I Quantization In physics some measurable quantities are quantized. A physical quantity is said to be quantized if it can have only discrete (not continuous) values. Some examples of quantities which are quantized are mass, charge and energy. For sub-atomic particles (called fundamental particles) the angular momentum is quantized. Fundamental particles such as electrons and protons have a certain intrinsic angular momentum of their own. This angular momentum is called spin angular momentum. The spin

angular momentum of a fundamental particle is quantized and its value is given by h S=n 2 where h is the Planck’s constant = 6.63 10–34 Js and n is a number called the spin quantum number. The value of n for electrons, protons, positrons and antiprotons can 1 1 be + and – . Pions have n = 0. 2 2 1. Which of the following is not quantized ? (a) Mass (b) Energy (c) Linear momentum (d) Charge

8.54 Comprehensive Physics—JEE Advanced

2. Choose the correct statements from the following. (a) Electric charge on a charged body can only be an integral multiple of the smallest possible charge. (b) Energy can have only discrete values. (c) The spin angular momentum of an electron

can be +

h h or – . 4 4

(d) The spin angular momentum of a pion is +

h h or – . 2 2

ANSWERS 1. The correct choice is (c). 2. The correct choices are (a), (b) and (c). The spin angular momentum of a pion is zero. Questions 3 to 5 are based on the following passage Passage II A hollow sphere of mass M and radius R is initially at rest on a horizontal rough surface. It moves under the action of a constant horizontal force F as shown in Fig. 8.88.

Fig. 8.88

3. The frictional force between the sphere and the surface

(a) retards the motion of the sphere (b) makes the sphere move faster (c) has no effect on the motion of the sphere (d) is independent of the velocity of the sphere. 4. The linear acceleration of the sphere is 10 F 7F (b) a = (a) a = 7M 5M 6F F (c) a = (d) a = 5M M 5. The frictional force between the sphere and the surface is F F (b) (a) 2 3 F F (c) (d) 4 5

SOLUTIONS 3. If the horizontal force F is applied at the centre of mass of the sphere, then the frictional force opposes the translational motion of the sphere. If force F is applied above the centre of mass, the torque due to frictional force tends to rotate the sphere faster. Hence, in this case, frictional force f acts in the direction of motion, as shown in Fig. 8.89. Thus the correct choice is (b).

Fig. 8.89

4. Let a and be the linear and angular accelerations of the sphere respectively. For translational motion,

F + f = Ma (1) The magnitude of the net torque acting on the sphere = FR – f R. Hence, for rotational motion the equation is Ia ( a = R) FR – f R = I = R 2 For a hollow sphere, I = MR2. Hence 3 2 a 2 FR – f R = MR2 = MRa 3 R 3 2 F – f = Ma (2) 3 6F , which is choice Equations (1) and (2) give a = 5M (c) Ma F 5. From Eqs. (1) and (2) we get f = = . Hence 6 5 the correct choice is (d).

Rigid Body Rotation 8.55

Questions 6 to 11 are based on the following passage Passage III Four solid spheres each of mass m and radius r are located with their centres on four corners of a square ABCD of side a as shown in Fig. 8.90.

8. The moment of inertia of sphere B about side AD is 2 2 m mr (b) (5a2 + 2r2) (a) 5 5 m m (c) (2r2 + a2) (d) (3a2 + 5r2) 5 5 9. The moment of inertia of sphere D about side AD is 2 2 2 mr (b) m(r2 + a2) 5 3 m m (c) (3a2 + 4r2) (d) (5a2 + 2r2) 5 5 10. The moment of inertia of the system of four spheres about diagonal AB is m m (8r2 + 5a2) (b) (7r2 + 4a2) (a) 5 5 m m (5r2 + 8a2) (d) (3r2 + 5a2) (c) 5 5 11. The moment of inertia of the system of four spheres about side AD is 2m m (2r2 + 5a2) (b) (7r2 + 5a2) (a) 5 5 (a)

Fig. 8.90

6. The moment of inertia of sphere A about diagonal AB is 2 2 2 2 mr (b) mr (a) 3 5 a2 a2 (d) m r2 4 4 7. The moment of inertia of sphere C about diagonal AB is 2 2 2 (a) mr (b) (2r2 + 3a2) 5 5 (c) m

(c)

r2

m (5r2 + 3a2) 5

(d)

m (4r2 + 5a2) 10

(c)

2m (4r2 + 5a2) 5

(d)

m (3r2 + 5a2) 5

SOLUTION Refer to Fig. 8.91.

2 mr2 each. The distance 5 of this axis (shown by broken lines) from the diagonal AB = a/ 2 . From the parallel axes theorem, the moment of inertia of spheres C and D about diagonal AB is a 2 2 2 2 m a2 mr2 + m(CO)2 = mr2 + m = mr2 + 2 2 5 5 5 6. The correct choice is (b). 7. The correct choice is (d). 8. The correct choice is (b). 9. The correct choice is (a). 10. The moment of inertia of the system of four spheres about diagonal AB is IAB = MI of A about AB + MI of B about AB + MI of C about AB + MI of D about AB 2 2 2 1 m r2 m r2 m a2 = m r2 5 5 5 2 2 1 + mr 2 ma 2 5 2

their centre and parallel to AB =

Fig. 8.91

The moment of inertia of spheres A and B about their 2 mr2 each. Also the moment of common diameter AB = 5 inertia of spheres C and D about an axis passing through

8.56 Comprehensive Physics—JEE Advanced

IAD = MI of A about AD + MI of D about AD + MI of B about AD + MI of C about AD

2

8r 8 a2 = mr2 + ma2 = m 5 5 The correct choice is (a). 11. Moment of inertia of sphere A about side AD = moment of inertia of sphere D about side AD = 2 mr2. Using the parallel axes theorem, moment of 5 inertia of sphere C about AD = moment of inertia 2 of sphere B about AD = mr2 + ma2. Hence the 5 moment of inertia of the system of four spheres about side AD is Questions 12 to 15 are based on the following passage Passage IV A small sphere rolls down without slipping from the top of a track in a vertical plane. The track has a elevated section and a horizontal part. The horizontal part is 1.0 m above the ground and the top of the track is 2.4 m above the ground. (See Fig. 8.92)

Fig. 8.92

12. If g = 10 ms , the horizontal velocity when the sphere reaches point A is (b) 2 5 ms–1 (a) 5 ms–1 (c) 7 ms–1 (d) 2 7 ms–1 –2 13. If g = 10 ms , the time taken by the sphere to fall through h = 1.0 m is –2

=

2 m r2 5

2 m r2 5

2 m r 2 m a2 5 2 m r 2 m a2 5

=

8 mr2 + 2 ma2 = m 5

8r2 5

2 a2

The correct choice is (c). 1

s

(b)

2

s 5 5 (c) 0.1 s (d) 0.2 s –2 14. If g = 10 ms , the distance on the ground with respect to point B (which is vertically below the end A of the track) is (a) 1.0 m (b) 1.4 m (c) 2.0 m (d) 2.8 m 15. Choose the correct statement/statements from the following. (a) During its motion as a projectile after the sphere leaves the track at A, it will stop rotating. (b) During its motion as a projectile after point A, the sphere will continue to rotate about its centre of mass. (c) Due to rotation, the horizontal range of the sphere will be less than that found in Q.14 above. (d) The rotation of the sphere has no effect on the horizontal range found in Q.14. (a)

SOLUTION 12. The loss in potential energy when the sphere moves from the top of the track to point A = gain in total kinetic energy (translational and rotation), i.e. 1 1 Mv2 + I 2 2 2 2 v where I = MR2 and = . Thus 5 R 1 1 2 Mg(H – h) = Mv2 + MR2 2 2 5

= or

Mg(H – h) =

v = 10( H

2

h) g

1/ 2

7

10 (2.4 1.0) 10 1 / 2 =2 5 ms–1 7 The correct choice is (b). 13. Since the vertical component of velocity is zero, the =

v R

1 1 7 Mv2 + Mv2 = Mv2 2 5 10

Rigid Body Rotation 8.57

2h = g

1 2 1.0 = s, which is choice (a). 5 10 1 14. Horizontal range = vt = 2 5 = 2.0 m, which 5 is choice (c). 15. A t=

-

and (c).

Questions 16 to 18 are based on the following passage Passage V

in case (b) is

M and radius R s and

(a)

2

(b)

2 3

(c)

5 7

(d)

7 3

inclination 16.

18.

(a) to that in case (b) is

(c)

the bottom in case (a) to that in case (b) as

2 3 7 (d) 9

(a) 1

(b)

5 7

(a) 1 (c)

17.

(b) (d)

2

3 2 2 3

SOLUTION 16. g sin 5 I = g sin a1 = 1 2 7 MR In case (b), the acceleration is a2 = g sin Hence the correct choice is (c).

I

2 MR 2 5

18. From s = is (d).

Questions 19 to 22 are based on the following passage Passage VI m1 = 3 m and m2 = m are attached to M = 2 m and radius R) as shown in Fig. 8.93. The masses are then released. 19. the system is (a) 2 g 2g (b) 3 2g (c) 5 3g (d) 7

17. Using v2 = 2as v1 = v2 1 2 at 2

20. Tension T1 is mg (a) 5

7 mg 5 21. Tension T2 is 9 mg (a) 5 (c)

(c) mg 22. (a) 2 mgR (c) Fig. 8.93

5 , which is choice (c). 7

2 mgR 5

3 mg 5 9 mg (d) 5 (b)

7 mg 5 3 mg (d) 5 (b)

(b)

2 mgR 3

(d) 3 mgR

8.58 Comprehensive Physics—JEE Advanced

SOLUTION = (T1 – T2)R Also,

= I

where I =

a R 1 = MR2 2

(3)

1 MR2 and 2

is the angular

=

a 1 = MRa = m Ra R 2 ( M = 2m)

T1 – T2 = ma

(5)

19.

2g . Hence the correct choices is (c). 5 20. T1 = 3m(g – a) = 2g 9 mg = ; which is choice (d). 3m g 5 5 2g 21. T2 = m(g + a) = m g 5 7 mg = so choice (b) is correct. 5 get a =

Fig. 8.94

tensions T1 and T2

and

a is the acceleration

m 1g – T 1 = m 1a T 2 – m 2 g = m 2a

3mg – T1 = 3ma T2 – mg = ma

(1) (2)

22.

The resultant tension (T1 – T2

= mRa = mR

2 g 2mgR = , which 5 5

is choice (c). Questions 23 to 26 are based on the following passage Passage VII m

As R at a height H

P

(a)

2gR

(b)

3gR

(c)

5gR

(d)

7gR

24. Q is

R

(a) mg

(b) 2 mg

(c) 3 mg

(d) 5 mg

25. Q is (a)

. mg

(c)

26 mg

(b)

5 mg

(d) 3 3 mg

26.

H so that the A

Fig. 8.95

23.

Q on the track is

R

(b) 2.5 R

(c) 2.6 R

(d) 2.7 R

Rigid Body Rotation 8.59

SOLUTION 23.

Q = Gain in K.E. 1 1 mg(H – R) = mv2 + I 2 2 mg

R – R) = = v=

1 2 1 mv + 2 2

Fh =

2

25.

2 mR 2 5

v R

mv2 m = R R

2

1 2 1 2 7 mv + mv = mv2 2 5 10

Fh2

Fv2

F= = Thus the correct choice is (c). 26. A A

5gR . Hence the correct choice

m v 2A = mg R

Q

vA =

26 mg

Rg

H

7 7 mRg m v 2A = 10 10 H = 2 R + 0.7R = 2.7R, which is choice (d).

Q the centre O

mg(Hmin – 2R) =

Questions 27 to 30 are based on the following passage Passage VIII

(c) Mg sin 29. (a) MgR

M and radius R with the horizontal. 27.

2 g sin 3 2 g cos (d) 3

(a) g sin (c)

mg

Fv

is (c). 24.

(5 gR) = 5 mg, which is choice (d).

(c) 30.

(a)

28.

(c)

Mg sin (a) 3

2 MgR sin 3

MgR sin 3

(d)

M and the same radius R

(b)

g sin 3

(b) MgR sin

2 Mg sin (b) 3

3 2

(b) 2 (d) 1

2

SOLUTION Mg sin f

f a is the linear acceleration

– f = Ma

(1)

is the angular acceleration and I acting on it is =I

is (see Fig. 8.96) Now

I=

1 MR2, 2

=

a R

fR= f=

or 27. Fig. 8.96

and

= f R. Hence

1 M R2 2

a 1 = M Ra R 2

1 Ma 2

(2) a=

is choice (b).

2 g sin , which 3

8.60 Comprehensive Physics—JEE Advanced

28. f =

MgR sin , which is choice (d). 3 30. For a ring, I = M R2. The correct choice is (a).

1 Mg sin Ma = , which is choice (a). 2 3

29.

Questions 31 to 33 are based on the following passage Passage IX M and radius R is mounted on

=fR=

mg ( M m)

(c)

2 Mmg ( M 2m)

(c)

m is hung

(b)

Mmg ( M m)

(d)

Mmg ( M 2m)

33.

Fig. 8.97

to hn, where h is the height through which the mass n is (a) zero (b) 1 1 (d) 2 (c) 2

rest. 31. (b)

2mg ( M 2m)

32. The tension in the string is (a) mg

(a) g

(d)

mg M

SOLUTION a

32.

ma = mg – T where T is the tension in the string.

(1)

T= is choice (d).

33.

= TR = I I T= R =

1 MR 2 2

mgh = =

a R2

=

1 = Ma 2

1 2 1 mv + I 2 2 1 m R2 2 1

2

+

2

1 2

(2m + M) R2

1 MR 2 2

2

2

(2) Fig. 8.98

31. a=

mM g , which ( M 2m)

2 mg , which is choice (d). ( M 2m)

=

mgh ( M 2m) R 2

1/ 2

h1/2. Hence the correct choice is (c).

Thus

Questions 34 to 36 are based on the following passage

x2. Both the discs rotate in the clockwise direction.

Passage X

IIT, 2007

Two discs A and B I and 2I A

34. The ratio x1/x2 is (a) 2

x1. Disc B

(c)

1 2 1 (d) 2 (b)

2

Rigid Body Rotation 8.61

35. When disc B is brought in contact with disc A, they t. The

36. is (a)

2I 3t 9I (c) t

9I 2t 3I (d) 2t (b)

(a)

(c)

2

I

(b)

2 I

2

(d)

2

I 3

2

I 6

SOLUTION 34.

A

A is

1 1 kx12 = I (2 )2 2 2 For disc B, the energy is

(1)

2I 3

(2)

From (1) and (2) we get

x12 x22

=2

=

x1 = x2

= I (2 ) + (2I)

=

2.

35. When disc B is brought in contact with disc A, let I the

changein angular momentum 2I = Time 3t

36. E1 =

I

=

1 1 I(2 )2 + (2I) 2 2

2

1 1 I ( )2 = (3I) 2 2 3 E 1– E 2 8 1 = 3I 2 – I 2= I 3 3

= 3I

2

E2 =

I I = I + 2I = 3I

3I

3

= A is

1 2 1 kx 2 = (2I)2 2 2

I

= I (2 ) – I

3

=

2

8 I 3

2

2

Questions 37 to 39 are based on the following passage Passage XI M and radius R constant k

Fig. 8.99

L

IIT, 2008

L

V0

V0 i .

38.

μ. 37.

to x (a) –kx 2kx (c) 3

(b) – 2kx kx (d) 3

(a)

k M

(b)

2k M

(c)

2k 3M

(d)

k 3M

8.62 Comprehensive Physics—JEE Advanced

39.

V0 (c) μg (a) μg

M k

(b) μg

3M k

(d) μg

5M 2k

M 2k

SOLUTION 37. Let translational motion (see Fig. 8.100) Ma = f – 2kx

38. Acceleration a = (1)

Let

= 39. When V0

kx =– 3M

2

x. Hence the motion

k . 3M x

x

) and f

= μMg. Ma = μMg – 2kx and

Fig. 8.100

– fR =

1 MRa 2

Eliminating a

=I or

– μMgR =

1 MR 2 2

a ( R

a=

R)

kx

(2)

=

3 Mg 2

Eliminating f a =–

(3)

kx 3M F = Ma =

kx . 3

V

= μg

(5)

3

MV2

= kx2

3M k

IV Assertion-Reason Type Questions Mg h/3. Statement-2 correct. -

throughout its motion. 2. Statement-1 Two bodies A and B and m

m A is v

Statement-2 1. Statement-1 M and radius R rolls down h. The rotational kinetic

B is 2v, v/3.

Rigid Body Rotation 8.63

3. Statement-1 B -

is

3g / L .

Statement-2

circle. Statement-2

8. Statement-1 4. Statement-1 M and m (with M > m) V

v

9. Statement-1

MV ( M m)

=

anticlockwise along the circle. Statement-2

Statement-2 Statement-2 5. Statement-1 L

M and closed at both the ends. The tube is then rotated in a horizontal

other end is ML Statement-2

2

.

angular momentum. 10. Statement-1 mass M and the same radius R

Statement-2 11. Statement-1

6. Statement-1

-

L and mass m is bent into a r as shown in Fig. 8.101. X X is 3mL2/8 2.

Statement-2 12. Statement-1

Fig. 8.101

same time. Statement-2

Statement-2

-

XX 2

about YY + mr . 7. Statement-1 AB hinged at one end A

13. Statement-1 M and length L is -

8.64 Comprehensive Physics—JEE Advanced

Statement-2

14. Statement-1 suddenly .

Fig. 8.102

Statement-2

Statement-2 17. Statement-1

15. Statement-1 -

zero. Statement-2

Statement-2

18. Statement-1 -

16. Statement-1

tact is non zero. Statement-2

SOLUTIONS 1. The correct choice is (a).

4.

-

Potential energy = Translational kinetic energy + Rotational kinetic energy or or

1 1 Mv2 + I 2 2 2 1 1 1 M R2 Mgh = MR2 2 + 2 2 2 3 = MR2 2

MV = (M + m)vCM MV vCM = ( M m)

Mgh =

5. 2

r=

gh 3R 2 1 Now the rotational kinetic energy = I 2. 2 2 and I gh 1 1 MR 2 Rotational kinetic energy = 2 2 3R 2

or

2

=

= 2.

M gh 3

L 2 M

is the circle or radius r = Thus Fc =

M v2 M (r )2 = = Mr r r

2

=

ML 2

2

6. O is IO =

initially at rest) will remain at rest. 3.

L . 2

1 mr2 2 -

ertia about X X is

1 2 3 mr + mr2 = mr2 2 2 L = 2 r or r = L/2 I = IO + mr2 =

Now

Rigid Body Rotation 8.65

I=

3m 2

L 2

2

=

3mL2 8 2

a=

7. The correct choice is (c). Loss in P.E. = gain in through a distance L/2, the loss in P.E. =

1 I 2

Gain in K.E. =

2

=

MgL ML2 2 = 6 2 8. The correct choice is (d). 9.

1 2

2

ML 3

MgL . 2

s

>

5 g sin 7

a = g(sin =

14.

2 MR 2 5

cos )

and . I

I

>

–

g,

3g L

s

I

13.

2

L = I is constant, .

15. The correct choice is (a). 16. A

-

at A 17. -

R

c

=

radius will reach at the same time.

angular momentum will remain unchanged. 1 MR2 10. The correct choice is (a). For cylinder, Ic = 2 2 2 Is = MR . Also = I or = . 5 I , I. Hence Ic s = Is c Ic > Is ; both, as > ac

g sin I 1 MR 2

Hence no work is done.

c.

18.

11.

ity is changing with time. Hence the instantaneous

12. The correct choice is (a). The linear acceleration is

V Integer Answer Type 1.

2.

-

L and mass 300 g rests on F is L

F g = 10 ms–2 8.103. Find the distance 3. Fig. 8.103

IIT, 1980

8.66 Comprehensive Physics—JEE Advanced

v

Fig. 8.106

IIT, 1997

Fig. 8.104

7.

g = 10 ms–2

0.5 m with a stick as shown in Fig. 8.107. The stick B

2

A IIT, 1987 4.

.

ring is large enough that rolling always occurs and the

m

P. Take g = 10 ms–2.

(P

the circle is r, the tension in the string is T = Ar–n, where A n. IIT, 1993 5. A block X m = 0.5 kg is held by a string on = 30°. drum Y

M = 2 kg and radius R = 0.2 m as

Fig. 8.107

IIT, 2011

X –2

8.

g = 10ms newton) in the string during the motion.

5 cm and -

N

10 kg m2, then N is (see Fig. 8.108)

Fig. 8.105

6.

M and radius R u = 6 ms–1 on a rough sliding motion at t

t0 v

–1

)

t 0. Fig. 8.108

IIT, 2011

SOLUTION 1. O O1

O2 W1

Rigid Body Rotation 8.67

W2

3L

Fmin

L 2

= mg

Fmin =

2mg 2 0.3 10 = = 2N. 3 3

Fig. 8.110

3.

Fig. 8.109

p

(21)2 cm2

A = gain in total kinetic energy (translational and rotation), i.e. 1 1 Mg(H – h) = Mv2 + I 2 2 2 2 v Where I = MR2 and = . Thus 5 R

cm2

Mg(H – h) =

(28)2 cm2 [(28)2 – (21)2 cm2 area. Hence

=

or

W1 mg m = 1 = 1 = W2 m2 g m2 Taking moments about O W1 OO1 = W2 OO2 OO2 =

W1 W2

v=

= 9 7

mg

( 1)min =

10 H

h g

2

12

7 .

= 2 5 ms

–1

t=

1.0

10

12

7

L 2

2

2h = g

1 2 1.0 = s 5 10

Horizontal range = vt = 2 5 = 2.0 m. 4. Angular momentum is mvr = constant (say k). k . v= mr mv 2 mk 2 = 2 3 = m–1k2r–3 = Ar–3 r m r –1 2 where A = m k T = Ar–n, we get n = 3 5. X is as shown in Fig. X, then ma = T – mg sin (1) Also

A is = mg

v R

OO1

9 7 cm = 9 cm 7 F about A is (Fig. 8.110) 3L 1 = F

2

2 MR2 5

1 1 7 Mv2 + Mv2 = Mv2 2 5 10

10

=

=

2.

1 1 Mv2 + 2 2

T=

8.68 Comprehensive Physics—JEE Advanced

f = μmg a =R

P is

= RT = I 1 MR 2 I 2 T= = R R 1 T = Ma 2 Using (2) in (1)

F =

1 a MR 2 R

1 ma = Ma – mg sin 30° 2 mg a= M 2m M mg T= 2 M 2m =

2 0.5 10 =5N 2 2 2 0.5

(2)

Where

I = mR2 + mR2 = 2mR2

and

a =R

Also

f = μmg

F

(3)

R – fR = I

a R

R – μmgR = (2 mR2)

= 2maR

F – μmg = 2 ma 2 – μ 2 10 = 2 2 0.3 0.8 . = . Hence P μ= 2 10 10 8. Let M R its radius. diagonal AB I = I1 + I2 + I3 + I where I1, I2, I3 and I AB. I1 = I2 =

2 MR2 5

I3 = I =

2 MR2 + Mr2 5

Fig. 8.111

6.

=

1 I 2 2 1 v = MvR + (MR2) 2 R v u =v + 2 2u 2 6 –1 v= = 3 3

I =2

MuR = MvR +

7.

Fig. 8.112

2

a

2 MR2 + M 5

2

2 MR 2 5

2

2 MR 2 5

Ma 2 2

= 8 MR 2 + Ma2 5 8 = 5

0.5

5 10 2

2 2

+ 0.5 10–2)2

=9

10

kgm2. Hence N = 9.

9

Gravitation

Chapter

REVIEW OF BASIC CONCEPTS 9.1

NEWTON’S LAW OF GRAVITATION

Newton’s law of universal gravitation states as follows: ‘Any two particles of matter anywhere in the universe attract each other with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them, the direction of the force being along the line joining the particles, i.e. (Fig. 9.1)

to all other masses is obtained from the principle of superposition which states that ‘the gravitational force experienced by one mass is equal to the vector sum of the gravitational forces exerted on it by all other masses taken one at a time.’

Fig. 9.1

F

m1 m2 r2

where F is the magnitude of the force of attraction between two particles of masses m1 and m2 separated by a distance r. In the form of an equation the law is written as F=

Gm1 m2 r

2

where G is a constant called the universal gravitation constant. The value of this constant is to be determined experimentally and is found to be G = 6.67 10–11 N m2 kg–2.

9.2

GRAVITATIONAL FORCE DUE TO MULTIPLE MASSES

If a system consists of more than two masses, the gravitational force experienced by a given mass due

Fig. 9.2

The gravitational force on mass m due to masses m1, m 2, m3, ... mn is given by (Fig. 9.2) + Fn F = F1 + F2 + F3 + NOTE (1) Gravitational force is always attractive. (2) Gravitational force between two masses does not depend the medium between them. (3) Gravitational force acts along the straight line joining the centres of the two bodies.

9.1 Two bodies A and B of masses m 1 = 1 kg and m 2 = 16 kg respectively are placed 1.0 m apart. A third body C of mass m = 3 kg is placed on the line joining A and B. At what distance from A should C be placed so that it experiences no gravitational force?

9.2 Comprehensive Physics—JEE Advanced

SOLUTION Let x metre be the distance between A and C (Fig. 9.3)

The magnitude of the resultant force is Fr =

F12

F22

2 F1 F2 cos

=

F2

F2

2 F 2 cos 60

=

3F =

3

Gm 2 a2

Fig. 9.3

Force exerted by A on C is G m 1m F1 = directed towards A x2 Force exerted by B on C is G m2 m F2 = directed towards B (1 x) 2 C will experience no force if F1 = F2, i.e. G m1m x

2

=

G m2 m

(1 x) 2

m2 (1 x) 2 = m1 x2 16 = 4=

(1 x) 2 x2 1 x x

Gm 2 a2

= F (say)

F = mg =

GmM

R2 where M is the mass of the earth and R its radius (nearly constant for a body in the vicinity of the earth) GM g= R2 All bodies near the surface of the earth fall with the same acceleration which is directed towards the centre of the earth.

VARIATION OF G

1. Variation with altitude The acceleration due to gravity of a body at a height h above the surface of the earth is given by gh = g

2

R R

h

where g is the acceleration due to gravity on the surface of the earth. If h is very small compared to R, we can use binomial expansion and retain terms of order h/R. We then get 2h gh = g 1 R Fig. 9.4

SOLUTION The forces exerted on the body at A by bodies at B and C are shown in Fig. 9.5. F1 = F2 =

ACCELERATION DUE TO GRAVITY

Considering the earth as an isolated mass, a force is experienced by a body near it. This force is directed towards the centre of the earth and has a magnitude mg, where g is the acceleration due to gravity.

9.4

which gives x = 0.2 m. Note that if body C is placed to the left of body A or to the right of body B, it will 9.2 Three bodies, each of mass m, are placed at the vertices of an equilateral triangle of side a as shown in Fig. 9.4. Find the magnitude and direction of the force experienced by the body at vertex A.

9.3

, directed vertically downwards.

Fig. 9.5

Thus, the acceleration due to gravity decreases as the altitude (h) is increased. 2. Variation with depth The acceleration due to gravity at a depth d below the surface of the earth is given by d gd = g 1 R This equation shows that the acceleration due to gravity decreases with depth. At the centre of the earth where d = R, gd = 0. Thus the acceleration due to gravity is maximum at the surface of the earth, decreases with increase in depth and becomes zero at the centre of the earth.

Gravitation 9.3

3. Variation with Latitude Due to the rotation of earth about its axis, the value of g varies with latitude, i.e. from one place to another on the earth’s surface. At poles, the effect of rotation on g is negligible. At the equator, the effects of rotation on g is the maximum. In general, the value of acceleration due to gravity at a place decreases with the decrease in the latitude of the place. The acceleration due to gravity at a place on earth where the latitude is is given by g = g – 2 R cos2 ; = angular velocity of rotation of earth. At equator, =0 g e = g – R 2 (minimum) ° gp = g (maximum) At poles, = 90 g = g – 0.0337 cos 2 Thus the value of g varies slightly from place to place on earth. Variation of g with altitude and depth is shown in Fig. 9.6.

SOLUTION g = g 1 = 2 1 2 which gives

=

2

R R

h 2

R R

h

R R

h=

h 2 1 R.

9.5 A body weighs 63 N on the surface of the earth. How much will it weigh at a height equal to half the radius of the earth? SOLUTION W = mg = 63 N W = mg = mg

R

h

h R R R/2

= 63 Fig. 9.6

R 2

R

W =W

2

R

2

= 28 N

Variation of g (gravitational acceleration)

9.3 The acceleration due to gravity on the surface of the moon is 1.67 ms–2. The mass of the earth is about 80 times that of the moon. Estimate the ratio of the radius of the earth to that of the moon.

9.6 Assuming the earth to be a sphere of uniform mass of a body when it is taken to the end of a tunnel 32 km below the surface of the earth. Radius of earth = 6400 km. SOLUTION

SOLUTION ge =

G Me

Re = Rm

Me Mm

=

Re2

80

and gm = gm gm 1.67 9.8

G Mm Rm2

1/ 2

d 32 =g 1 R 6400 Decrease in weight = mg – mg g =g 1

= mg 1 1/ 2

3.7

9.4 Assuming that the earth is a sphere of radius R, at what altitude will the value of acceleration due to gravity be half its value at the surface of the earth?

Percentage decrease =

199 200

=

=

199 g 200 mg 200

mg / 200 100 = 0.5% mg

9.7 At what depth below the surface of the earth will the value of acceleration due to gravity become 90% of its value at the surface? R = 6.4 106 m.

9.4 Comprehensive Physics—JEE Advanced

SOLUTION g = 0.9 g.

g =g 1

d R d = 0.1 R = 6.4

intensity is given by

d R

I=

Gravitational Field due to some continuous Mass Distributions

0.9 = 1 –

9.5

105 m

M and radius R at a point at a distance r from the centre and on the axis of the ring is given by GM r I= 2 (R r 2 )3 / 2

GRAVITATIONAL FIELD INTENSITY

of mass M and radius R at a point P at a distance r > R from the centre of shell, GM I = 2 (outside the shell) r Inside the shell, I = 0 M and radius R is GM I = 2 for r > R r GM I= for r = R R2 GMr I= for r < R R3

experienced by a unit mass placed at that point. M. a point P at a distance r from M, we place a small mass m at P. The gravitational force exerted on m by M is (Fig. 9.7) GM m F= r2

Fig. 9.7

M at

P is given by

F GM = 2 m r I is a vector quantity. In vector form GM I= – r r2 where r is a unit vector directed from M to P, i.e radially away from M. The negative sign indicates that I directed radially inwards towards M. The SI unit of I is N kg–1. In three dimensions, if mass M is located at the origin, P (x, y, z) is given by I=

I= –

GM r2

Fig. 9.8

of a solid sphere.

9.6

r

where r = x i + y j z k represents the position of point P with respect to mass M at the origin. For a many body system, the principle of superposition gravitational forces, i.e. I = I1 + I2 + I3 +

dI

GRAVITATIONAL POTENTIAL ENERGY

Gravitational potential energy of a system of two masses M and m held a distance r respective locations along any path and without any acceleration. (see Fig. 9.9)

+ In

where I1, I2, ... In point due to bodies of masses M1, M2, ... Mn. For continuous mass distributions (i.e rigid bodies), Fig. 9.9

Gravitation 9.5

Work done to bring mass M Work done to bring mass m from r =

A is W1 = 0. to r = r is

=–

r

W2 =

F dr

=

GMm – ( R h)

GmM h R ( R h)

r

=

F dr cos

If h < < R ;

Since mass M will attract mass m, angle and dr is zero. Hence r

W2 =

GM m r2 r

= GMm

GMm r

GMm r Gravitational potential energy of the system is

Total work

GmM h R2

W = W1 + W2 = –

GMm r The zero of potential energy is assumed to be at r = . The negative sign indicates the potential energy is negative U= W = –

Expression for Increase in Gravitational Potential Energy If the body m is moved away from M, the potential energy of the system increases. [see Fig. 9.10]

= mgh GM R

9.7

r 2dr = –

U=

between F

dr

GMm R

2

Fig. 9.11

g

GRAVITATIONAL POTENTIAL

Gravitational potential at a point P in M Fig. 9.12

that point along any path and without any acceleration, i.e., (Fig. 9.12) V=

W GM m =– m r m

GM r Potential V is a scalar quantity. Hence the gravitational potential at a point P due to a number of masses m 1, m2, ... mn at distances r 1, r 2, ... rn respectively from P is given by V = V1 + V2 + + Vn V= –

=–G

m1 r1

m2 r2

mn rn

The SI unit of V is J kg–1. Relation between Gravitational Field Intensity (I) and Gravitational Potential (V)

Fig. 9.10

P.E. at B = –

GMm r1

a point are related as dV dr Gravitational Potential due to a Spherical Shell I=

GMm P.E. at C = – r2 Increase in P.E. = –

GM m r1

GM m – r2

= GMm

1 r1

1 r2

If the body of mass M is the earth, then the increase in gravitational P.E. when a body of mass m is taken from the surface of the earth to a height h above the surface is given by (see Fig. 9.11), R = radius of the earth. U = P.E. at Q – P.E at P

GM (for r r > R) where M is the mass and R is the radius of the shell. GM (ii) At a point on the surface of the shell, V = R GM (iii) At a point inside the shell, V = (for r < R) R Figure 9.13 shows the variation V with r for a spherical shell. (i) At a point outside the shell, V =

9.6 Comprehensive Physics—JEE Advanced

P will be zero if Gm1 Gm2 I1 = I2 r 12 r 22 or

Fig. 9.13

9=

Gravitational Potential due to a Solid sphere of mass M and radius R (i) For points outside the sphere (r > R), GM r (ii) For points inside the sphere (r < R), V=

V=

3G M R 2 2 R3

m2 = m1

r2 6

(ii) At the centre of the sphere (r = 0) 3G M V= 2R (iv) On the surface of the sphere (r = R) GM V= R 9.8 Two masses m1 = 100 kg and m2 = 8100 kg are held 1 m apart. (a) At what point on the line joining them is the tational potential at that point. (b) Find the gravitational potential energy of the system. Given G = 6.67 10–11 Nm2 kg–2.

r1 r2

2

81 =

r1

r1 = 0.1 m

1 r1

Gravitational potential at P is V = V1 + V2 = – G =

6.67 10

= – 6.67

=

Gm2 r 22

8100 0.9

11

100 8100 1

= – 5.4

10–5 J

9.9 Three equal masses, each equal to m, are kept at the vertices of an equilateral triangle of side a. Find the centroid of the triangle. SOLUTION Refer to Fig. 9.15.

directed towards m1

directed towards m2

100 0.1

11

m2 r2

10–7 J kg–1

6.67 10

P due to m2 is I2 =

m1 r1

(b) Gravitational potential energy of the system is G m1 m2 G.P.E. = r

Fig. 9.14

r 12

1 r1

r2 = r – r1 and r = 1 m)

P due to m1 is [Fig. 9.14]

Gm1

2

(

SOLUTION

I1 =

r1

Fig. 9.15

Gravitation 9.7

The centroid O divides the lines AD, BE and CF in the ratio 2:1. Also AO = BO = CO = r (say). Now AO 2 = AD and 3 a2 = 4

a2

AD =

3a 2

2 3

3a a a = , i.e. r = 2 3 3 O due to masses m at A, B Gm and C are IA = IB = IC = 2 = I. Their directions are r shown in Fig. 9.15. The angle between any two of them is = 120°. The resultant of IB and IC is AO =

I2

I =

I2

9.11 Two particles of masses m1 and m2 are initially at rest wards each other due to gravitational attraction. Find the ratio of their accelerations and speeds when the separation between them becomes r. SOLUTION Since no external force acts on the system, the acceleration of their centre of mass is zero, i.e.

2 I 2 cos 120

Gm r

=

3Gm = r

Gm r 3 3

Gm a

a

r

3

9.10 The gravitational potential at a height h = 1600 km above the surface of the earth is – 4.0 107 J kg–1. Assuming the earth to be a sphere of radius R = 6400

0=

m1a1 m1

m2 a2 m2

m2 m1

The negative sign indicates that they move in opposite directions. Let v1 and v2 be the speeds of two masses when they are at a distance r. Due to gravitational attraction, they gain speed as they approach each other. Hence their kinetic energy increases and gravitational potential energy (G.P.E) decreases. From the conservation of energy, Loss in G.P.E. = gain in K.E. G.P.E.)i – (G.P.E.)f = (K.E.)f – (K.E.)i 0

Gm1m2 r

=

1 m1 v 12 2

G m1 m2 1 = m1v12 r 2

at that height.

V= g =

GM r GM r

2

g GM / r 2 1 1 = = = V GM / r r R h g =

V R

h

1 m2 v 22 2 1 m2 v 22 2

0 (i)

Since no external force acts, the total momentum of the system is conserved, i.e., pi = pf or (ii) 0 = m 1v 1 – m 2v 2 From Eqs. (i) and (ii), we get

SOLUTION Let r = R + h. Then

and

m2 a2 m2

a1 = a2

I will cancel with IA at O is zero. The gravitational potential at O is V = V1 + V2 + V3 Gm r

m1 a1 m1

m 1a 1 = – m 2 a 2

= I directed vertically down.

=

aCM =

=

7

and

4.0 10 J kg 6.4 1.6

1

106 m

= 5.0 J kg–1 m–1 = 5.0 ms–2

v1 =

2G m22 r m1 m2

v2 =

2G m 12 r m1 m2

v1 m = 2 v2 m1

1/ 2

1/ 2

9.8 Comprehensive Physics—JEE Advanced

9.8

ESCAPE VELOCITY

The escape velocity is the minimum velocity with which a body must be projected in order that it may escape the earth’s gravitational pull. The magnitude of the escape velocity is given by 2MG R where M is the mass of the earth and R its radius. Substituting the known values of G, M and R, we get ve = 11.2 kms–1. The expression for the escape velocity can be written in terms of g as ve =

ve =

2gR

The escape velocity is independent of the mass of the body and the direction of projection.

9.9

v=

Orbital Velocity Let us assume that a satellite of mass m goes around the earth in a circular orbit of radius r with a uniform speed v. If the height of the satellite above the earth’s surface is h, then r = (R + h), where R is the mean mv2 necessary radius of the earth. The centripetal force r to keep the satellite in its circular orbit is provided by GmM between the earth and the the gravitational force r2 satellite. This means that mv2 mM =G 2 r r where M is the mass of the earth. Thus v=

GM r

T=

GM = gR2

If h << R, we have v =

g

R h If the satellite is a few hundred kilometres above the earth’s surface (say 100 to 300 km), we can replace

r3 GM

2

( R h )3

2

gR 2

g R . Hence R h

T= 2

gR

3

2

2

R g

Total Energy of a Satellite Total energyE = K.E. + P.E. =

1 m v2 2

Gm M r

=

Gm M 2r

Gm M r

E=

GmM = 2r

v=

GM r

Gm M 2 R h

NOTE P.E. 2 (2) E = – (K.E.) (3) The total energy is negative which implies that the sat(1) E =

The binding energy =

Gm M . This energy must be 2r

Angular Momentum The magnitude of angular momentum of a satellite is given by L = mvr = m

Substituting for GM we get v= R

2 r v

GM ( R h)

Now the acceleration due to gravity on earth’s surface is given by GM g= R2

9.8 6.4 106

gR

= 7.9 103 ms–1 8 km s–1 Periodic Time The periodic time T of a satellite is the time it takes to complete one revolution and it is given by (since r = R + h)

SATELLITES

A body moving in an orbit around a much larger and massive body is called a satellite. The moon is the natural satellite of the earth.

or

(R + h) by R. The error involved in this approximation is negligible since the radius of the earth (R) = 6.4 106 m = 6400 km. Thus, we may write

GM r r

L = m GMr Geostationary Satellites A Geostationary satellite is a particular type used in communication. A number of communication satellites are launched which remain in

Gravitation 9.9

They are called geostationary or synchronous satellites used in international communication. certain place on the earth, it must corotate with the earth so that its orbital period around the earth is exactly equal to the rotational period of the earth about its axis of rotation. at a height of about 36,000 km above the earth’s surface. Trajectory of a Satellite for Different Speeds Orbital speed of a satellite is v0 =

GM r

Escape velocity is 2GM ve = r Let v be the speed given to a satellite. (i) If v < v0, the satellite will follow an elliptical path with the centre of the earth as the farthest focus. The satellite will not complete the orbit and will fall to the earth. (ii) If v = v0, the satellite will follow a circular path with the centre of the earth as the centre of the circular orbit. (iii) If v0 < v < ve, the satellite will follow an elliptical orbit with the centre of the earth as the nearer focus. (iv) If v = ve, the satellite will escape the gravitational pull of the earth following a parabolic path. (v) If v > ve, the satellite will escape the gravitational pull of the earth following a hyperbolic path.

9.10

Fig. 9.16

According to Kepler’s second law, if the time interval between P1 and P2 equals the time interval between P3 and P4, then area A1 must be equal to area A2. Also the planet has the greater speed in its path from P1 to P2 than in its path from P3 to P4. 3. Law of periods The squares of the periods of the planets are proportional to the cubes of their mean distances from the sun. If T1 represents the period of a planet about the sun, and r1 its mean distance, then T12 r13 If T2 represents the period of a second planet about the sun, and r2 its mean distance, then for this planet T 22 r32 These two relations can be combined since the factor of proportionality is the same for both. Thus T12

KEPLER’S LAWS OF PLANETARY MOTION

Johannes Kepler formulated three laws which describe planetary motion. They are as follows: 1. Law of orbits Each planet revolves about the sun in an elliptical orbit with the sun at one of the focii of the ellipse. The orbit of a planet is shown in Fig. 9.16(a) in which the two focii F1 and F2, are far apart. For the planet earth, F1 and F2 are very close together. In fact, the orbit of the earth is practically circular. 2. Law of areas A line drawn from the sun to the planet (termed the radius) sweeps out equal areas in equal intervals of time. In Fig. 9.16(b) P1, P2, P3 and P4 represent positions of a planet at different times in its orbit and S, the position of the sun.

T22

=

r13 r23

9.12 The mass of Jupiter is 318 times that of the earth and its radius is 11.2 times that of the earth. Calculate the escape velocity from Jupiter’s surface. Given the escape velocity from earth’surface = 11.2 km s–1. SOLUTION For Jupiter :

vJ =

2M J G RJ

For Earth :

vJ =

2M E G RE

9.10 Comprehensive Physics—JEE Advanced

vJ = vE

MJ ME

At the highest point, v = 0. If h is the maximum height attained, the energy of the rocket at height h is

RE RJ

Ef =

1 5.33 = 318 11.2 vJ = 5.33 vE = 5.33 11.2 = 59.7 km s–1

From conservation of energy, Ei = Ef , i.e.,

9.13 Calculate the escape velocity of a body at a height 1600 km above the surface of the earth. Radius of earth = 6400 km. SOLUTION Work required to move a body of mass m from r = R + h to r = is W=

dr r

2

=

Gm M R h

If ve is the escape velocity, then

h R

ve v . Hence e = ve 2 2

= ve

h R

R

h

h R

h

h=

R 3

9.15

2 GM R h

Gm M (i) R h Total energy when it hits the surface of the earth is

2 gR 2 R h GM 2

R 6 10 m and

106 m, and h = 1.6 ve =

Given v =

2GM R

Ei =

g

Given R = 6.4 g = 9.8 ms–2

v=

GmM ( R h)

SOLUTION Total energy of the body at height h is

1 Gm M mve2 = 2 R h ve =

GmM = R

R , where 2 R is the radius of the earth. Show that it will hit the surface of the earth with a speed v = ve / 3 , where ve is the escape velocity from the surface of the earth.

R h

R h

1 2 mv 2

A body is dropped from a height h equal to

Fdr

= Gm M

GmM ( R h)

1 Gm M m v2 2 R From conservation of energy, Ei = Ef , i.e. Ef =

Gm M 1 = m v2 R h 2

6 2

2 9.8 (6.4 10 ) (6.4 1.6) 106

= 10 10 ms = 10 km s–1

v=

–1

9.14 A rocket is launched vertically from the surface of the earth with an initial velocity equal to half the escape velocity. Find the maximum height attained by it in terms of R where R is the radius of the earth. Ignore atmospheric resistance. SOLUTION On the surface of the earth, the total energy of the rocket is 1 2 GmM Ei = K.E. + P.E. = mv R 2

For h =

GmM R

2GM R

= ve

(ii)

h R

h

R R

h

v R ,v= e . 2 3

9.16 Show that the minimum energy required to launch a satellite of mass m from the surface of the earth in a circular orbit at an altitude h = R, where R is the 3mgR where M is the mass of radius of the earth is 4 the earth.

Gravitation 9.11

SOLUTION Total energy of the satellite orbiting the earth is Gm M E1 = 2r Gm M 2( R h)

=

GmM 4R

(

h = R)

Total energy when the satellite was at rest on the surface of the earth is E2 = K.E. + P.E. Gm M R

Gm M = R Minimum energy required is Emin = E1 – E2 = 0

Gm M 4R

=

NOTE Since Fg = Fc, the satellite is a freely falling body and is, therefore, weightless.

9.18 A body projected vertically upwards from the surface of the earth with a certain velocity rises to a height of 10 m. How high will it rise if it is projected with the same velocity vertically upwards from a planet whose density is one-third that of the earth and radius half that of earth? Ignore atmospheric resistance. SOLUTION Since the kinetic energy of the body is the same in both the cases, loss in K.E. = gain in P.E. will be equal, i.e., mgphp = mgehe

GmR R

3Gm M 3 = = mg R 4R 4 9.17 A satellite of mass m = 100 kg is in a circular orbit at a height h = R above the surface of the earth where R is the radius of the earth. Find (a) the acceleration due to gravity at any point on the path of the satellite, (b) the gravitational force on the satellite and (c) the centripetal force on the satellite. SOLUTION (a) g = g

2

R R

h

2

R

= 9.8

R

R

=

9.8 4

(b) Gravitational force on satellite is Fg = mg = 100 2.45 = 245 N m v2 Gm M = r r2 g= or

Fc =

R

= mg

2

R2

r2

GM R

2

h

=

ge R = e gp Rp

G R

e

2

=2

4 R3 3

=

4 GR 3

3=6

p

hp = 6he = 6

10 = 60 m

9.19 A satellite of mass 2000 kg is orbiting the earth at an altitude R/2, where R is the radius of the earth. What extra energy must be given to the satellite to increase its altitude to R? Given R = 6.4 106 m.

r1 = R

= mg = 245 N.

R 3R = and in the second 2 2

case, r2 = R + R = 2R. Required energy = E2 – E1 =

Gm M 2 r2

=

Gm M 4R

=

GmM 12 R

=

mgR 12

2

R R

g=

GM

R2

Gm M

Now

SOLUTION

= 2.45 ms–2

(c) Centripetal force Fc =

ge gp

hp = he

Gm M 2 r1 GmM 3R

g=

GM R2

9.12 Comprehensive Physics—JEE Advanced

6.4 106

2000 9.8

=

SOLUTION From Kepler’s law of periods, T2

12 10 J 8

= 1.04

Tp2

9.20 Two bodies of masses m1 and m2 are held at a distance r apart. Show that at the point where the gravitational is given by G m1 r

V=

m2

2 m1m2

SOLUTION P (Fig.

9.17). Then G m1

=

G m2

r1 = r2

m1 m2

=

m1

r 12

r1 r

r1

r1 = Also

Fig. 9.17

= (10)3 = 1000 31.6 = 31.6 years

9.22 A satellite is revolving in a circular orbit close to the surface of the earth with a speed v. What minimum additional speed must be imparted to it so that it escapes the gravitational pull of the earth? Radius of earth = 6.4 106 m. SOLUTION g R and ve = 2 1

= 0.414

2g R

gR 9.8 6.4 106

= 3.28 103 ms–1 = 3.28 km s–1

m2 r m1 m1

(i)

m2

r m1 m1

m2

r m2 m1

(ii)

m2

Gm1 r1

Gm2 r2

(iii)

Using (i) and (ii) in (iii) and simplifying, we get V=

r 3e

1000 = 1 year

ve – v =

Gravitational potential at P is V=

rp3

Additional speed required is

r2 = r – r1

=

Tp = Te

v=

r 22

= r

Te2

=

r3. Therefore

G m1 r

m2

2 m1m2

9.21 The distance of a planet from the sun is 10 times that of the earth. Find the period of revolution of the planet around the sun.

9.23 A body of mass m is placed at the centre of a spherical shell of radius R and mass M. Find the gravitational potential on the surface of the shell. SOLUTION Gravitational potential on the surface of the shell due to body of mass m is Gm Vb = R Gravitational potential on the surface of the shell due to shell itself is GM Vs = R G V = Vb + Vs = m M R 9.24 A tunnel is drilled from the surface of the earth to its centre. A body of mass m is dropped into the tunnel. Find the speed with which the body hits the bottom of the tunnel. The mass of earth is M and its radius is R.

Gravitation 9.13

Hence

SOLUTION Let v be the required speed. Gain in K.E. = loss in P.E. = P.E. at the surface – P.E. at the centre. The poten3 GmM tial energy at the centre of the sphere = . 2 R

1 mv2 = 2 v=

Gm M R GM = R

3 GmM 2 R gR

g=

GM R2

I Multiple Choice Questions with Only One Choice Correct 1. A body of mass m is placed at the centre of a spherical shell of radius R and mass M. The gravitational potential on the surface of the shell is G G (M + m) (b) – (M – m) (a) – R R (c) –

G mM R M m

(d) –

G mM R M m

2. The magnitude of angular momentum of the earth revolving round the sun is proportional to R n where R is the distance between the earth and the sun. The value of n is (assume the orbit to be circular). (a) 1 2

(b) 1

(c) 3 (d) 2 2 3. A tunnel is drilled from the surface of the earth to its centre. The radius of the earth is R and g is the acceleration due to gravity on its surface. A body of mass m is dropped into the tunnel. If M is the mass of the earth, the speed with which the body hits the bottom of the tunnel is (a) (c)

m M

gR

2 gR

(b) M m (d)

gR

gR

4. A satellite is orbiting at a height R above the surface of the earth where R is the radius of the earth. By what percentage must the energy of the satellite be increased so that it orbits at a height 2 R above the surface of the earth? (a) 25% (b) 33.3% (c) 50% (d) 66.7% 5. If E1 is the energy required to raise a satellite to a height h = R (radius of the earth) above the surface

of the earth and E2 is the energy required to put it into a circular orbit at that height, then the ratio E1/ E2 is 1 (a) 1 (b) 2 1 2 (c) (d) 3 3 6. If g is the acceleration due to gravity on the surface of the earth, the gain in potential energy of a satellite of mass m raised from the earth’s surface to a height equal to the radius R of the earth is (a) mgR/4 (b) mgR/2 (c) mgR (d) 2 mgR 7. The escape velocity of a body projected vertically upwards from the surface of the earth is v. If the body is projected at an angle of 30° with the horizontal, the escape velocity would be (a) v/2 (b) 3 v/2 (c) 2 v (d) v 8. The escape velocity of a body at a height h above the surface of the earth is (g = acceleration due to gravity on the surface of the earth and R = radius of the earth) is (b) 2 g R h (a) 2 gR 2 gR 2 gR 2 (d) 2 R h R h 9. A small planet is revolving around a very massive star in a circular orbit of radius r with a period of revolution T. If the gravitational force between the planet and the star were proportional to r –5/2, then T would be proportional to (b) r 5/3 (a) r 3/2 7/4 (c) r (d) r 3 10. If both the mass and the radius of the earth decrease by 1%, the acceleration due to gravity on the surface of the earth will (c)

9.14 Comprehensive Physics—JEE Advanced

(a) decrease by 1% (b) increase by 1% (c) increase by 2% (d) remain unchanged. 11. Two masses M1 and M2 are separated by a distance r. A particle of mass m is placed exactly mid-way between them. The minimum speed with which the particle should be projected so as to escape to G M1 M 2 r

(a) v = 2 (b) v =

2 G M1 M 2 r G M1 M 2 mr

(c) v = 2 (d) v =

2 G M1 M 2 mr

1/ 2

1/ 2

2 1/ 2

2 1/ 2

12. The areal velocity of a planet of mass m moving along an elliptical orbit around the sun is (a) proportional to m (b) proportional to m 1 (c) proportional to (d) independent of m m 13. A planet of mass m revolves around the sun in an elliptical orbit of semimajor axis a. If M is the mass of the sun, the speed of the planet when it is at a distance x from the sun is (a) (c)

GM GM

1 x

1 2a

(b)

2 x

1 a

(d)

GM

1 2x

1 a

GM a 2 x2

14. A satellite revolves around a planet with a speed v in a circular orbit of radius r. If R is the radius of the planet, the acceleration due to gravity on its surface is v2 r v2 R (b) g = 2 (a) g = 2 R r v2 v2 (c) g = (d) g = R r 15. Three spheres, each of mass M and radius R, are kept such that each touches the other two. The magnitude of the gravitational force on any one sphere due to the other two is 3 GM 2 GM 2 (b) (a) 2 R2 2R2 (c)

3 GM 2 2 R2

(d)

3 GM 2 4 R2

16. An extremely small and dense neutron star of mass M and radius R is rotating at an angular frequency . If an object is placed at its equator, it will remain stuck to it due to gravity if R R2 2 (a) M > (b) M > G G 3 2 2 R R 3 (c) M > (d) M > G G 17. What is the minimum energy required to launch a satellite of mass m from the surface of the earth of radius R in a circular orbit at an altitude of 2R? 5GmM 2GmM (b) (a) 6R 3R GmM GmM (c) (d) 2R 3R 18. Two stars, each of mass m and radius R are approaching each other for a head-on collision. They start approaching each other when their separation is r >> R. If their speeds at this separation are negligible, the speed with which they collide would be (a) v =

Gm

1 R

1 r

(b) v =

Gm

1 2R

(c) v =

Gm

1 R

(d) v =

Gm

1 2R

1 r 1 r 1 r

19. Consider a particle of mass m suspended vertically by a string at the equator. Let R and M denote the radius and the mass of the earth respectively. If is the angular velocity of earth’s rotation about its own axis, the tension in the string is equal to mM mM (b) G (a) G 2R2 R2 (c) G 20.

mM R

2

–m

2

R

(d) G

mM R

2

+m

2

R

m, are placed along a straight line at distances of r, 2r, 4r, 8r, etc. from a reference point O. The gravitational O will be (a) (c)

5 Gm 4r

2

3 Gm 2r

2

(b) (d)

4 Gm 3r 2 2 Gm r2

Gravitation 9.15

21. In Q. 20, the magnitude of the gravitational potential at point O will be Gm Gm (b) (a) 2r r 3 Gm 2 Gm (c) (d) 2r r 22. A satellite in force-free space sweeps stationary interplanetary dust at a rate dM/dt = v, where M is the mass and v is the velocity of the satellite and is a constant. The acceleration of the satellite is 2 v v2 (b) (a) M M 2

v (d) – v2 M 23. The time of revolution of a satellite is T. Its kinetic energy is proportional to 1 1 (b) 2 (a) T T 1 (c) 3 (d) T –2/3 T 24. A solid sphere of uniform density and radius 4 units is located with its centre at origin O of coordinates. Two spheres of equal radii 1 unit, with their centres at A (– 2, 0, 0) and B(2, 0, 0) respectively are taken out of the solid leaving behind spherical cavities as shown in Fig. 9.18. Choose the incorrect statement from the following. (c)

Fig. 9.18

(a) the gravitational force due to this object at the origin is zero. (b) the gravitational force at point B (2, 0, 0) is zero. (c) the gravitational potential is the same at all points of the circle y2 + z2 = 36. (d) the gravitational potential is the same at all points of the circle y2 + z2 = 4. IIT, 1993 25. Two bodies of masses m1 and m2 are initially at rest move toward each other under mutual gravitational

attraction. Their relative velocity of approach at a separation distance r between them is (a)

1/ 2

2G m1 m2 r

2G m1 m2 r 2

(b) (c)

r 2G m1 m2

(d)

2G m1 m2 r

1/ 2

1/ 2

1/ 2

26. If the distance between the earth and the sun were half its present value, the number of days in a year would have been (a) 64.5 (b) 129 (c) 182.5 (d) 730 IIT, 1996 27. around the earth has a total (kinetic + potential) energy E0. Its potential energy is (a) – E0 (b) 1.5 E0 (c) 2 E0 (d) E0 IIT, 1997 28. 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. Which of the following statements is correct? (a) The acceleration of S is always directed towards the centre of the earth. (b) The angular momentum of S about the centre of the earth changes in direction, but its magnitude remains constant. (c) The total mechanical energy of S remains constant. (d) The linear momentum of S remains constant in magnitude. IIT, 1998 29. The distance between the sun and the earth is r and the earth takes time T to make one complete revolution around the sun. Assuming the orbit of the earth around the sun to be circular, the mass of the sun will be proportional to (a) (c)

r2 T r3

(b) (d)

r2 T2 r3

T2 T3 30. A meteor of mass M breaks up into two parts. The mass of one part is m. For a given separation r the

9.16 Comprehensive Physics—JEE Advanced

mutual gravitational force between the two parts will be the maximum if M M (a) m = (b) m = 2 3 M M (c) m = (d) m = 2 2 2 31. A body of mass m is raised to a height h above the surface of the earth of mass M and radius R until its gravitational potential energy increases by 1 mgR. The value of h is 3 R R (b) (a) 3 2 mR mR (c) (d) M m M 32. Two balls A and B are thrown vertically upwards from the same location on the surface of the earth gR 2 gR and respectively, with velocities 2 3 3 where R is the radius of the earth and g is the acceleration due to gravity on the surface of the earth. The ratio of the maximum height attained by A to that attained by B is (a) 2 (b) 4 (c) 8 (d) 4 2 33. A uniform sphere of mass M and radius R exerts a force F on a small mass m situated at a distance of 2R from the centre O of the sphere. A spherical portion of diameter R is cut from the sphere as shown in Fig. 9.19. The force of attraction between the remaining part of the sphere and the mass m will be 7F 2F (a) (b) 9 3 4F F (c) (d) 9 3

Fig. 9.19

34. The centres of a ring of mass m and a sphere of mass M of equal radius R, are at a distance 8 R apart as shown in Fig. 9.20. The force of attraction between the ring and the sphere is

(a) (c)

2 2 GmM 27 R 2

(b)

GmM 9R

(d)

2

GmM 8R 2 2 GmM 9 9R2

Fig. 9.20

35. Two objects of masses m and 4m are at rest at under mutual gravitational attraction. Then, at a separation r, which of the following is true? (a) The total energy of the system is not zero. (b) The force between them is not zero. (c) The centre of mass of the system is at rest. (d) All the above are true. IIT, 1994 36. A satellite is launched 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 the approximately (a) 0.5% (c) 1.5%

(b) 1.0% (d) 3.0%

IIT, 1995 37. A simple pendulum has a time period T1 when on the earth’s surface, and T2 when taken to a height R above the earth’s surface, where R is the radius of the earth. The value of T2/T1 is (a) 1 (b) 2 (c) 4 (d) 2 IIT, 2001 38. An ideal spring with spring-constant k is hung from the ceiling and a block of mass M is attached to its lower end. The mass is released with the spring initially unstretched. Then the maximum extension in the spring is (a) 4 Mg/k (b) 2 Mg/k (c) Mg/k (d) Mg/2k IIT,2002 39. A geo-stationary satellite orbits around the earth in a circular orbit of radius 36000 km. Then, the time period of a spy satellite orbiting a few hundred kilometers above the earth’s surface (REarth = 6400 km) will approximately be

Gravitation 9.17

(a) (1/2) h (c) 2 h

(b) 1 h (d) 4 h

IIT, 2002 40. A satellite of mass m is moving in a circular orbit of radius R above the surface of a planet of mass M and radius R. The amount of work done to shift the satellite to a higher orbit of radius 2R is (here g is the acceleration due to gravity on planet’s surface) mgR (a) mgR (b) 6 mMgR mMgR (c) (d) M m 6 M m 41. The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the earth is (here R is the radius of the earth) n n mgR (b) mgR (a) n 1 n 1 mgR (c) nmgR (d) n 42. A body of mass m is dropped from a height nR above the surface of the earth (here R is the radius of the earth). The speed at which the body hits the surface of the earth is (a)

2 gR n 1

(b)

gR

(a) –

(b) –

10 3gR

(c) –

2 gR 10 4 gR

(d) – 10 10 46. The radius of the earth is R and g is the acceleration due to gravity on its surface. What should be the angular speed of the earth so that bodies lying on the equator may appear weightless? (a)

g R

(b)

2g R

g g (d) 2 R 2R 47. If W1, W2 and W3 represent the work done in moving a particle from A to B along three different paths 1, 2 and 3 (as shown in Fig. 9.21) in m correct relation between W1, W2 and W3. (c)

(a) W1 > W3 > W2

(b) W1 = W2 = W3

(c) W1 < W3 < W2

(d) W1 < W2 < W3 IIT, 2003

2 gR n 1

2 gRn 2 gRn (d) n 1 n 1 43. Two solid spheres of radii r and 2r, made of the same material, are kept in contact. The mutual gravitational force of attraction between them is proportional to 1 1 (b) 2 (a) 4 r r (c) r2 (d) r4 44. A comet is moving in a highly elliptical orbit round the sun. When it is closest to the sun, its distance from the sun is r and its speed is v. When it is farthest from the sun, its distance from the sun is R and its speed will be (c)

(a) v

r R

1/ 2

(b) v 3/ 2

r R

48. A binary star system consists of two stars of masses M1 and M2 revolving in circular orbits of radii R1 and R2 respectively. If their respective time periods are T1 and T2, then (a) T1 > T2 if R1 > R2 (b) T1 > T2 if M1 > M2 (c) T1 = T2 (d)

2

r r (d) v R R 45. The value of the acceleration due to gravity at the surface of the earth of radius R is g. It decreases by 10% at a height h above the surface of the earth. The gravitational potential at this height is (c) v

Fig. 9.21

T1 T2

R1 R2

3/ 2

IIT, 2006 49. A spherically symmetric gravitational system of R 0 for r where particles has a mass density = 0 for r R 0 is a constant. A test mass can undergo circular of particles. Its speed v as a function of distance

9.18 Comprehensive Physics—JEE Advanced

50. A satellite is moving with a constant speed ‘V ’ in a circular orbit about the earth. An object of mass ‘m’ is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is

r(0 < r < ) from the centre of the system is represented by (see Fig. 9.22) IIT, 2008

1 mV 2 2 3 (c) mV 2 2

(b) mV2

(a)

(d) 2mV2 IIT, 2011

Fig. 9.22

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49.

(a) (d) (c) (c) (a) (b) (d) (d) (c)

2. 8. 14. 20. 26. 32. 38. 44. 50.

(a) (c) (a) (b) (b) (c) (b) (b) (b)

3. 9. 15. 21. 27. 33. 39. 45.

(d) (c) (d) (d) (c) (a) (c) (c)

4. 10. 16. 22. 28. 34. 40. 46.

(b) (b) (c) (b) (a) (a) (b) (a)

5. 11. 17. 23. 29. 35. 41. 47.

(d) (a) (a) (d) (c) (d) (a) (b)

6. 12. 18. 24. 30. 36. 42. 48.

(b) (d) (b) (b) (a) (c) (d) (c)

SOLUTIONS 1. Gravitational potential on the surface of the shell due to the body of mass m is Gm Vb = – R Gravitational potential on the surface of the shell due to shell itself is GM Vs = – R G V = Vb + Vs = – (M + m), which is choice (a). R GmM m v2 GM 2. = v= R2 R R Angular momentum L = mvR = m = m(GMR)1/2. i.e. L R1/2. So the correct choice is (a).

GM R

R

3. Let v be the required speed. Gain in K.E. = loss in P.E. = P.E. at the surface – P.E. at the centre of the earth 1 GmM mv2 = – – 2 R

3 GmM 2 R

GM GM g = gR R2 R 4. The total energy of a satellite in orbital radius r is 1 GMm E = K.E. + P.E. = mv2 – 2 r v=

GMm GMm 2r r GMm = 2r =

v

GM r

Gravitation 9.19

r = R + R = 2R. Hence GMm E= 4R

k r 3/ 2 m 2 r T= =2 r v v=

In the second case, r = R + 2R = 3R. Hence GMm 6R

E =

=

GMm 12 R

E Percentage increase = |E|

GMm 4R

100

P.E. at a height h (= R) = GMm 2R

GMm R

GMm 2R GMm R

GM GMm 1 g = mgR 2R R2 2 7. The correct choice is (d) because the escape velocity is independent of the direction along which the body is projected. 8. Escape velocity at height h = 2g R h , where =

2

R

. Hence the correct choice is (c). R h 9. Since the gravitational force provides the necessary centripetal force, g =g

mv2 r

r

R2

G m M1 G m M2 r/2 r/2 2Gm = (M1 + M2) r If v is the required velocity of projection, the total initial energy is Total P.E. =

1 2Gm mv 2 – (M1 + M 2) 2 r Ef = 0. Putting Ei = Ef , we get

Ei =

v= 2

G M1

M2

k = constant

1/ 2

r

dA L = , where L is the magnidt 2m tude of angular momentum of the planet about the sun. L = mvr sin . Hence,

12. A real velocity

dA mvr sin v r sin = = dt 2m 2 So the correct choice is (d). 13. Total energy of the planet in an elliptical orbit of semimajor axis a is GmM 2a Total energy of the planet when it is at a distance x from the sun (here v = speed of the plane at that instant) is E2 = K.E. + P.E. E1 = –

1 GmM mv2 – 2 x From conservation of energy, E1 = E2, i.e. =

–5/2

mv2 = kr–5/2, r

GM g m 2 R = – g M R = – 1% – 2 (– 1%) = + 1%

GMm 3 GMm = R 4R

E1 2 = E2 3 6. P.E. on the surface of the earth =

Gain in P.E. =

10. g =

3/ 2

Hence g will increase by 1%. 11. Distance of m from M1 or M2 = r/2. Therefore,

GMm / 12 R = 100 = 33.3% GMm / 4 R 5. E1 = P.E. at h = R – P.E. at h = 0 = P.E. at r = 2R – P.E. at r = R GMm GMm GMm = 2R R 2R E2 =

kr 7/4 r , which is choice (c)

i.e. T

Increase in energy E = E – E GMm GMm = – 6R 4R

m

–

GmM 1 GmM = mv2 – 2a 2 x

9.20 Comprehensive Physics—JEE Advanced

which gives v =

2 x

GM

1 , which is choice (c). a

mv 2 GmM 14. = r r2 GM = v2 r g=

GM

=

R2

v2 r R2

So the correct choice is (a). 15. Force between any two spheres is F=

GM 2 2R

2

=

GM 2 4R2

This is the force exerted on any sphere (say A) by the other two spheres B and C (Fig. 9.23). Thus the resultant force on sphere A is Fr = =

F2 3F =

F2

2 F 2 cos 60 3 GM 2 4R2

, which is choice (d).

Fg > Fc or

GmM

2

> mR

mv 2 GmM = r r2

v2 =

or

off if

2

GM r

1 GmM mv2 = 2 2r

or

GmM 2r Thus the KE of a satellite in a circular orbit is numerically half its PE but opposite in sign. The total energy of the satellite in orbit is

i.e.

KE =

GmM GmM – 2r r

=–

Fig. 9.23

R3 G

R2 Hence the correct choice is (c). 17. Consider a satellite of mass m moving with a speed v at an altitude r (measured from the centre of the earth). Then 1 mv 2 Kinetic energy (KE) = 2 GmM Gravitational potential energy (PE) = – r where M is the mass of the earth. For a satellite in circular orbit, we have

E = KE + PE =

16. An object of mass m, placed at the equator of the star, will experience two forces: (i) an attractive force due to gravity towards the centre of the star and (ii) an outward centrifugal force due to the rotation of the star. The centrifugal force arises because the object is in a rotating (non-inertial) frame; this force is equal to the inward centripetal force but opposite in direction. Force on object due to gravity is GmM Fg = R2 Centrifugal force on the object is Fc = mR 2

or M >

GmM 2r

It is given that r = 2R + R = 3R, where R is the radius of the earth. GmM E= – 6R GmM Now PE on the surface of the earth = – R Minimum energy required (E min) GmM GmM =– – R 6R 5GmM 6R Hence the correct choice is (a). 18. The speeds of stars at separation r are negligible. Therefore, their energy is entirely potential at this separation (since KE = 0) Gm1 m2 E1 = (PE at r) = – r =

Gm 2 r As the stars approach each other under gravitational attraction, they begin to acquire speed and hence =–

Gravitation 9.21

kinetic energy at the expense of potential energy. When they eventually collide, the separation between their centres is r = R + R = 2R At

r = 2R, the total energy is E2 = PE at (r = 2R) + KE at (r = 2R) =–

2

Gm 2R

Gm 2 + mv2 2R From the principle of conservation of energy, E1 = E2, i.e. Gm 2 Gm 2 – =– + mv2 2R r or

E2 = –

1 1 which gives v = Gm 2R r Hence the correct choice is (b). 19. The acceleration due to gravity at a place on the surface of the earth is given by g = g – 2R cos where is the latitude of the place. At the equator, = 0. Therefore g = g – 2R Therefore, the tension in the string will be mg = mg – m 2R GmM = – m 2R R2 Hence the correct choice is (c). 20. O will be I = Gm = =

Gm r

2

1

1

r2

2r

1

2

4r

1

1

1

2

2

2

2

Gm 1 r

2

2

1

0

4

2

1

1

1

2

4

26

2

2

8r

2

4Gm Gm 1 = 2 = 1 r 1 3r 2 r 1 1 2 2 4 Hence the correct choice is (b). 21. The gravitational potential, in magnitude, at point O is 1 1 1 1 V = Gm r 2r 4r 8r =

Gm 2

1

8

1

Gm 1 1 r 2

=

Gm 1 r 20

22.

1 4

1 8

1

1

1

1

2

23

2

2

1

2Gm = 1 r 1 2 Hence the correct choice is (d). From Newton’s second law of motion, force is the rate of change of momentum, i.e. =

1 1 mv2 + mv2 2 2

+

=

Gm r

d dM (Mv) = v= dt dt

F=

Retardation =

F M

v2 M

v2

dM v dt or acceleration =

v2 . Hence the correct choice is (b). M 23. The orbital speed of a satellite at a height r from the GM centre of the earth is given by v = r where M is the mass of the earth. If m is the mass of the satellite, its kinetic energy is K=

1 1 GM mv2 = m 2 2 r

1 . Now, the time period r r3 of the satellite is given by T = 2 g R2

Thus, K is proportional to

where R is the radius of the earth. Thus T r3/2. But 1 K . Hence K T–2/3, which is choice (d). r 24. The distance of each cavity from the centre O is the same. Since the two cavities are symmetrical with respect to the centre O and the mass of the sphere can be regarded as being concentrated at the centre O, the gravitational force due to the sphere is zero at the centre. Hence choice (a) is correct. For the same reason, the gravitational potential is the same at all points of the circle y2 + z2 = 36 whose radius is 6 units and at all points of the circle y2 + z2 = 4. Hence choices (c) and (d) are also correct. But the gravitational force at point B cannot be zero. 25. tance from each other, their gravitational potential energy is zero. When they are at a distance r from each other the gravitational P.E. is

9.22 Comprehensive Physics—JEE Advanced

G m1 m2 r The minus sign indicates that there is a decrease in P.E. This gives rise to an increase in kinetic energy. If v1 and v2 are their respective velocities when they are a distance r apart, then, from the law of conservation of energy, we have PE = –

G m1 m2 1 m1v21 = r 2 or

2 G m2 r

v1 =

G m1 m2 1 m2v22 = r 2

and

2 G m1 r Therefore, their relative velocity of approach is or

v2 =

2 G m2 r

v1 + v2 =

2G (m2 r Hence the correct choice is (a). 26. According to Kepler’s law of periods, =

T1 = T2 = T2 =

R1 R2 2

3/ 2

3/ 2

T1

2 2 27. For a satellite, we have

R1 R1 / 2

2 G m1 r

2 2

M=

Also v =

4 2 r3 2 r . Using this, we get M = or T 2G T

r3

. T2 Hence the correct choice is (c). 30. Mass of the second part = M m. Gravitational force between the two parts is M

F=

3/ 2

= 129 days.

GmM 2r GmM Potential energy = – r Total energy E0 = KE + PE Gm M Gm M Gm M = 2r r 2r Kinetic energy =

PE 2

or PE = 2E0. Hence the correct choice is (c). 28. For elliptical orbit, the earth is at one focus of the ellipse. For spherical bodies, the gravitational force is central (or radial). Hence statement (a) is correct. The gravitational force exerts no torque on the satellite. Hence the angular momentum of S remains constant in magnitude as well as direction. Hence choice (b) is incorrect. For elliptical orbit, the distance of the satellite from the earth varies periodically. Hence potential energy, kinetic energy and

v2 r G

or

m1 )

2 2 365 days

linear momentum vary periodically. Hence choices (c) and (d) are also incorrect. 29. Let m and M be the masses of the earth and the sun respectively and v the speed of the earth in circular orbit. To keep the earth in circular orbit, the graviGmM must balance the centripetal tational force r2 2 mv , i.e. force r GmM m v2 = r2 r

G(M

m) m r

2

=

G r

2

(Mm

m 2)

dF d 2F F will be maximum if = 0 and is negadm d m2 tive. dF dF G = 2 (M 2m). Setting = 0, we Now, dm dm r M . Now get M 2m = 0 or m = 2 d 2F 2G = – 2 , which is negative. d m2 r Hence the correct choice is (a). GM m 31. PE on the surface of earth = – R PE at a height h above the surface of earth = GM m – ( R h) GM m GM m Increase in PE = – – ( R h) R = GMm =

by

1 R

1 R h

g Rmh ( R h)

PE will increase by

= GMm

h R ( R h) g

GM R2

1 mgR at a value of h given 3

g Rmh 1 = mgR ( R h) 3

Gravitation 9.23

1 R or h = , which is choice (b). R h 3 2 32. If h is the maximum height attained, then we have GM m 1 GM m mv2 – =– ( R h) 2 R h

or

v2 =

which gives

2g h R ( R h)

GM

g

R2

For ball A, we have

2g hA R 4g R = R hA ) ( 3

hA = 4R

For ball B, we have

2g hB R 2g R = ( R hB ) 3

hB =

R 2

hA = 8, which is choice (c). hB 33. The force of attraction between the complete sphere and mass m is GmM GmM (i) F= 2 = (2 R) 4R2 4 3 Mass of complete sphere is M = R . Mass 3 4 R 3 of the cut out portion is m0 = . Thus, 3 2 M . The distance between the centre of the m0 = 8 R 3R cut out portion and mass m = 2R = . 2 2 Hence the force of attraction between the cut out portion and mass m is f=

G m0 m (3R / 2)

2

=

G ( M/8) m 2

=

GmM 2

2 9

9 R /4 4R 2F Using (i), we get f = . Therefore, the force of 9 attraction between the remaining part of the sphere 2F 7F and mass m = F f = F = which is 9 9 choice (a). 34. Refer to Fig. 9.24. Let be the mass per unit length of the ring. L = 2 R is the length of the ring. Consider a small element of length dx of the ring located at C. Then G M dx . Therefore, force Force along BC is f = (3R) 2 G M dx 8R along BA is dF = f cos = = 2 9R 3R 8 GM dx 27

because dx = × L = m, the mass of the ring. Hence the correct choice is (a).

=

R2

Total force =

8 GM 27 R 2

dx =

8 GMm 27 R 2

Fig. 9.24

35.

the system of two masses and the force between

rest initially and there is no external force, the centre of mass cannot move. Hence the correct choice is (d). 36. According to Kepler’s law of period, T2 = kR3 where k is a constant. Taking logarithm of both sides, we have 2 log T = log k + 3 log R Differentiating, we get 2 or

T R =0+3 T R T 3 = T 2

3 1.01 R R = × 2 R R

R

100

= 1.5% Hence the correct choice is (c). 37. The acceleration due to gravity at a height h above the surface of the earth is given by g2 = g1

R

2

R h where g1 is the value at the surface of the earth. Now T2 = 2 T2 = T1

l and T1 = 2 g2

l g1

g1 R h R R = = =2 g2 R R

(

h = R)

Hence the correct choice is (d). 38. Let x be the extension in the spring when it is loaded with mass M. The change in gravitational potential energy = Mgx. This must be the energy stored in the 1 2 kx . Thus spring which is given by 2 2Mg 1 2 kx = Mg x or x = , which is choice (b). k 2

9.24 Comprehensive Physics—JEE Advanced

39. For a satellite of mass m moving with a velocity v in a circular orbit of radius r around the earth of mass M, we have mv 2 GmM = or v = r r2 Now

v= T2 = T1

GM r

2 r 2 r . Thus = T T r2 r1

GM or T r

r3/2.

3/ 2

(1)

Given r2 = 6400 km and r1 = 36000 km. For a geostationary satellite T1 = 24 h. Using these values 64 3 / 2 = 1.8 h. in (1), we have get T2 = 24 × Hence the closest choice is (c). 360

40. PE at a distance r from the centre of the planet GM m =– r GM m GM m Initial PE = – =– R R 2R GM m GM m =– R 2R 3R Now, work done = increase in PE Final PE = –

GM m 1 1 GM m 1 = = mgR 2 3 R 6R 6 Hence the correct choice is (b). =

g

GM R2

GM m n GM m – = mgR ( n 1 ) R 1 n R Hence the correct choice is (a). 42. From the principle of conservation of energy, we have GM m 1 GM m mv2 =– ( R n R) 2 R 41. Change in PE =

which gives v2 =

2 n R GM 2n R g = 2 (n 1) R (n 1) g

GM R2

Hence the correct choice is (d). 43. If is the density of the material of each sphere, then the mass of the sphere of radius r is 4 3 and the mass of the sphere of radius M1 = r 3 4 (2r)3 . 2r is M2 = 3

Distance between their centres is d = r + 2r = 3r. 4 4 G r3 ( 2 r )3 G M1 M 2 3 3 = Now F = d2 9r 2 which gives F r4, which is choice (d). 44. The angular momentum of the comet is constant over the entire orbit. Hence vr = VR or r V=v , which is choice (b). R 45. gh =

GM ( R h) 2

Also g =

GM

. Thus

R2

Given gh = R2 ( R h) 2

gh R2 = . g ( R h) 2

90 g 100 =

9 or (R + h) = 10

10 R 3

GM GMR 2 gR 2 =– 2 =– ( R h) ( R h) R ( R h) 3gR =– 10 Hence the correct choice is (c). 46. At the equator, the value of g is g =g–R 2 Potential = –

where is the angular speed of the earth. For bodies to appear weightless at the equator, g = 0, i.e. g – R 2= 0 which gives

=

g . Hence the correct choice is (a). R

47. Gravitational force is conservative. The work done by a conservative force on a particle moving between two points does not depend on the path taken by the particle. Hence the correct choice is (b). 48. In a binary star system, the two stars move under their mutual gravitational force. Therefore, their angular velocities and hence their time periods are equal. Thus the correct choice is (c). 49. If M is the total mass of the system of particles, the orbital speed of the test mass is v=

For r

R, v =

GM r G

4 3 r 3 r

0

which gives v

r,

Gravitation 9.25

50. Let M be the mass of the earth and r be the orbital radius of the satellite. The energy needed so that the object of mass m escapes from x = r to x = is

i.e. v increases linearly with r up to r = R. Hence choices (b) and (d) are wrong. For r > R, the whole mass of the system is 4 R3 0, which is constant. Hence for M = 3 r > R, v= i.e, v

1 r

dx

E = GMm r

x

2

GMm r

(i)

The orbital speed V is given by

GM r

mV 2 GMm = r r2 Using (ii) in (i), E = mV2.

. Hence the correct choice is (c)

mV2 =

GMm r

(ii)

II Multiple Choice Questions with one or More Choices Correct 1. A satellite is moving around the earth in a stable circular orbit. Choose the correct statements from the following. (a) It is moving at a constant speed. (b) Its angular momentum remains constant. (c) It is acted upon by a force directed away from the centre of the earth which counter balances the attraction by the earth. (d) It behaves as if it were as freely falling body. 2. The escape velocity from a planet depends upon (a) the mass of the body (b) the mass of the planet (c) the average radius of the planet (d) the average density of the planet 3. The orbital velocity of a body in a stable orbit around a planet depends upon (a) the average radius of the planet (b) the height of the body above the planet (c) the acceleration due to gravity (d) the mass of the orbiting body 4. Choose the correct statements from the following: (a) The equivalence of inertial and gravitational mass has provided a clue to the deeper understanding of gravitation . (b) At poles, the effect of rotation of earth on the value of g is the minimum. (c) Very massive rockets and extremely tiny particles, such as the molecules of a gas, require the same initial velocity to escape from the earth. (d) A geostationary satellite, if imparted the necessary velocity, can be put in orbit at any height above the earth.

5. Choose the correct statements from the following: (a) The magnitude of the gravitational force between two bodies of mass 1 kg each and separated by a distance of 1 m is 9.8 N. (b) Higher the value of the escape velocity for a planet, the higher is the abundance of lighter gases in its atmosphere. (c) The gravitational force of attraction between two bodies of ordinary mass is not noticeable because the value of the gravitation constant is extremely small. (d) Force of friction arises due to gravitational attraction. 6. Choose the wrong statements from the following. (a) It is possible to shield a body from the gravishielding material between them. (b) The escape velocity of a body is independent of the mass of the body and the angle of projection. (c) The acceleration due to gravity increases due to the rotation of the earth. (d) The gravitational force exerted by the earth on a body is greater than that exerted by the body on the earth. 7. A comet is revolving around the sun in a highly elliptical orbit. Which of the following will remain constant throughout its orbit? (a) Kinetic energy (b) Potential energy (c) Total energy (d) Angular momentum

9.26 Comprehensive Physics—JEE Advanced

8. A satellite is orbiting the earth. If its distance from the earth is increased, its (a) angular velocity would increase (d) linear velocity would increase (c) angular velocity would decrease (d) time period would increase. 9. For two satellites at distance R and 7R above the earth’s surface, the ratio of their (a) total energies is 4 and potential and kinetic energies is 2 (b) potential energies is 4 (c) kinetic energies is 4 (d) total energies is 4. 10. A satellite is orbiting the earth in a circular orbit of radius r. Its (a) kinetic energy varies as 1/r (b) angular momentum varies as 1/ r (c) linear momentum varies as 1/ r (d) frequency of revolution varies as 1/r3/2. 11. If both the mass and radius of the earth decrease by 1%, the value of (a) acceleration due to gravity would decrease by nearly 1% (b) acceleration due to gravity would increase by 1% (c) escape velocity from the earth’s surface would decrease by 1% (d) the gravitational potential energy of a body on earth’s surface remains unchanged. 12. An object is taken from a point P to another point Q (a) assuming the earth to be spherical, if both P and Q lie on earth’s surface the work done is zero. (b) If P is on earth’s surface and Q above it, the work done is minimum when it is taken along the straight line PQ. (c) The work done depends only on the positions of P and Q and is independent of the path along which the particle is taken. (d) there is no net work done if the object is taken from P to Q and then brought back to P, along any path. 13. A satellite is moving in a circular orbit around the earth with a speed equal to half the escape velocity from the earth. The radius of the earth is R and h is the height of the satellite above the surface of the earth. If the satellite is suddenly stopped in its orbit and allowed to fall freely, it will hit the surface of the earth with a speed v. Then R (b) h = R (a) h = 2 (c) v =

2gR

(d) v =

gR

14. Two stars of masses m and 2 m are co-rotating about their centre of mass. Their centres are at a distance r apart. If r is much larger than the size of the stars, then their (a) common period of revolution is proportional to r3/2. (b) orbital velocities are in the ratio 2 : 1. (c) kinetic energies are in the ratio 1 : 2 . (d) angular momenta are in the ratio 1 : 4. 15. A space-ship is orbiting close to the surface of the earth at a speed v. The radius of the earth is R and g is the acceleration due to gravity close to the surface of the earth. An additional speed of v0 is to be imparted to the space-ship so that it overcomes the gravitational pull of the earth. Then (b) v = 2Rg (a) v = Rg (c) v0 =

Rg

2 1

(d) v0 =

Rg

2 1

16. A satellite of mass m is moving in a circular orbit of radius r around a planet of mass M. (a) The magnitude of angular momentum with respect to the centre of the orbit is m GMr , where G is the gravitation constant. (b) The magnitude of the angular momentum is mR 2gr where g is the acceleration due to gravity on the surface of the planet. (c) The direction of angular momentum is parallel to the plane of the orbit. (d) The direction of angular momentum is perpendicular to the plane of the orbit. IIT, 1997 17. Two bodies of masses m1 = m and m2 = 4 m are placed at a distance r is zero at a point P at a distance x from mass m1. The gravitational potential at point P is V. Then (U is the gravitational P.E. of the system) (a) x =

r 3

(c) U = –

(b) x = 7G r

2r 3

(d) U = –

9G r

18. A small planet is revolving with speed v around a very massive star in a circular orbit of radius r with a period of revolution T. If the gravitational force between the planet and the star were inversely proportional to r, then (a) v is independent of r. (b) v decreases if r is increased. (c) T increases if r is increased. (d) T is independent of r. 19. The total energy of a satellite of mass m moving with speed v around the earth of mass M in a

Gravitation 9.27

20.

21.

22.

23.

24.

circular orbit of radius r is directly proportional to (a) m (b) M (c) v (d) r Assuming the earth to be a sphere of radius R and uniform density, if the acceleration due to gravity at a point at a distance r from the centre of the earth is g, then 1 for r < R (a) g r for r < R (b) g r 1 (c) g r2 for r > R (d) g for r > R r2 A satellite is revolving around a planet in a circular orbit. During its motion, the satellite begins to experience a resistive force (possibly due to cosmic dust). As a result (a) its kinetic energy will decrease (b) its angular speed will decrease (c) its time period of revolution will increase (d) its angular momentum will decrease. If a satellite revolving around the earth is moved from one stable orbit to another stable orbit of a higher orbital radius, then its (a) time period will increase (b) centripetal force will decrease. (c) gravitational potential energy will increase (d) angular momentum will remain unchanged A comet is moving around the sun in a highly elliptical orbit. Its (a) kinetic energy and gravitational potential energy both change over the orbit (b) total energy remains constant throughout the orbit (c) linear momentum changes in magnitude as well as direction over the orbit (d) angular momentum remains constant over the orbit. IIT, 1998 A body of mass m is released from rest from above at a distance r from the centre of the earth of mass M and radius R. If the air resistance is neglected, it will strike the surface of the earth with a velocity v. If g is the acceleration due to gravity on earth s surface, (a) v =

2Rg if r >> R

(b) v =

Rg if r = 2R

(c) v =

Rg if r = 4R 2

(d) v =

1 Rg if r = 8R 2

25. A satellite of mass m is launched from the surface of the earth in a circular orbit at a height h = R above the surface of the earth; R being the radius of the earth. If g is the acceleration due to gravity on the surface of the earth and the air resistance is neglected, (a) the increase in gravitational potential energy is mgR (b) the velocity with which it was projected from the earth is gR (c) the total energy spent in launching the satel3 mgR lite in the circular orbit is 4 (d) the acceleration due to gravity at the site of the satellite is g/4. 26. Two satellites of the same mass are put in the same orbit round the earth but they revolve in opposite directions. They undergo an inelastic collision and stick together. (a) The ratio of the potential energy of the system before and just after the collision is 2 : 1 (b) The ratio of the total energy of the system before and just after the collision is 1 : 2 (c) After the collision, the combined mass will continue moving in the same orbit (d) After the collision, the combined mass will fall freely under the gravity of the earth. 27. A solid sphere of uniform density and radius 4 units is located with its centre at origin O of coordinates. Two spheres of equal radii 1 unit, with their centres at A (– 2, 0, 0) and B(2, 0, 0) respectively are taken out of the solid leaving behind spherical cavities as shown in Fig. 9.25.

Fig. 9.25

(a) the gravitational force due to this object at the origin is zero. (b) the gravitational force at point B (2,0,0) is zero. (c) the gravitational potential is the same at all points of the circle y 2 + z2 = 36. (d) the gravitational potential is the same at all points of the circle y2 + z2 = 4. IIT, 1993

9.28 Comprehensive Physics—JEE Advanced

28.

tances r1 and r2 from the centre of a uniform sphere of radius R and mass M are F1 and F2 respectively. Then F (a) 1 F2 (b)

F1 F2

r1 if r1 < R and r2 < R r2 r22 r12

(c)

F1 F2

r12

(d)

F1 F2

r2 if r1 > R and r2 > R r1

r22

if r1 < R and r2 < R

IIT, 1994

if r1 > R and r2 > R

ANSWERS AND SOLUTIONS The correct choices are (a), (b) and (d). The correct choices are (b), (c) and (d). The correct choices are (a), (b) and (c). The only incorrect statement is (d). A geostationary satellite can be put in orbit only at a height of 35,870 km above the earth. 5. Statement (a) is incorrect; the magnitude of the force is 6.67 10–11 N. Statement (b) is correct. If the escape velocity for a planet is high, the thermal speeds of lighter gases are less than the escape velocity. Therefore, lighter gases are not able to escape. Statement (c) is also correct. Statement (d) is incorrect; force of friction arises due to electrical forces. Hence correct choices are (b) and (c). 6. Statement (a) is incorrect; there is no method by which a body can be shielded from the gravi-

1. 2. 3. 4.

correct. Statement (c) is wrong. As the earth rotates about its axis, a body on the surface of the earth also rotates with it. Since the body is in a rotating (non–inertial) frame, it experiences an outward centrifugal force against the inward force of gravity. As a result, the acceleration due to gravity decreases due to rotation. Statement (d) is incorrect, the forces are equal in magnitude but opposite in direction. Hence choices (a), (c) and (d) are wrong. 7. The correct choices are (c) and (d). The kinetic energy of the comet changes because its speed in the orbit keeps changing. The potential energy changes because the distance of the comet from the sun keeps changing for an elliptical orbit. GM 8. Linear velocity or orbital velocity is v = , r where r is the distance of the satellite from the centre of the earth. Therefore v decreases as r is increased. Also v = r , where is the angular velocity. Therefore =

v 1 GM = = r r r

GM r3

Thus decreases with increase in r. The time period of the satellite is given by r3 GM Thus T increases as r is increased. Hence the correct choices are (c) and (d). 9. Distances of the two satellites from the centre of the earth are r1 = 2R and r2 = 8R respectively. R = earth’s radius. Their potential energies are T =2

and

V1 = –

GmM r1

V2 = –

GmM r2

V1 r2 8R = = = 4. V2 r1 2R The kinetic energy of a satellite can be obtained from relation mv 2 GmM = r r2 1 GmM or K= mv2 = 2 2r GmM GmM Thus K1 = and K2 = 2r1 2r2 The ratio of their kinetic energies is K1 r2 8R = = = 4. K2 r1 2R Their total energies are GmM GmM GmM E1 = – =– 2r1 r1 2r1 Their ratio is

and

E2 = –

GmM r2

GmM GmM =– 2r2 2r2

E1 r2 8R = = = 4. E2 r1 2R Hence the correct choices are (b), (c) and (d).

Their ratio is

Gravitation 9.29

1 GmM 1 mv2 = or KE . Thus 2 2r r choice (a) is correct. Angular momentum = mvr = GM r = m GM r , which is proportional m r to r . Hence choice (b) is wrong. Linear momenGM , which is proportional to tum mv = m r 1 . Hence choice (c) is correct. The frequency r of revolution is

10. Kinetic energy

1 GM 1 1 = i.e. 3 3/ 2 2 r r T Hence the correct choices are (a), (c) and (d). 11. The acceleration due to gravity is GM g= R2 The new value of g would be G 0.99 M GM g = 1.01 1.01 g 2 0.99 R R2 =

Thus g would increase by about 1%. The new escape velocity would be 2 0.99 M 0.99 R

2M G = ve R Thus the escape velocity will remain unchanged. The potential energy of a body of mass m on earth’s surface would be ve =

GM 0.99 M 0.99 R

G

=

GmM R Thus the potential energy will also remain unchanged. Hence the correct choices are (b) and(d). 12. Work done is independent of the path chosen and –

=–

object. Also the work done on any closed path in a the surface of the earth is at the same potential, no work is done for points on the surface of the earth. Hence the correct choices are (a), (c) and (d). 13. If M is the mass of the earth, the escape velocity is 2GM ve = R For a satellite of mass m and orbital radius r (= its distance from the centre of the earth), the orbital speed v is given by GmM mv 2 GM = or v = 2 r r r

But

1 ve = 2

v=

1 2

2GM R

GM 2R

=

GM GM = r 2R or r = 2R. Height above earth = 2R – R = R. Potential energy of the satellite in its orbit is GmM GmM =– ( r = 2R) E1 = – r 2R The kinetic energy is zero because the satellite is stopped. Potential energy of the satellite on the surface of the earth is GmM E2 = – R Loss of PE = E1 – E2 GmM GmM – 2R R GmM = 2R This is converted into kinetic energy. If v is the speed with which the satellite hits the surface of the earth, then from the law of conservation of energy, we have 1 GmM mv2 = 2 2R GM GM or v2 = = gR g R R2 The correct choices are (b) and (d). 14. The distance x of the star of mass m from the centre of mass is given by m 2m = x (r x) =–

r . The orbital speed v1 of the star 3 of mass m1 = m is given by (here m2 = 2 m)

which gives x =

Gm1 m2 r2

=

m1v12 m1v12 = r /3 x

Gm2 2Gm = 3r 3r 2 x 2 r Time period (T) of m = = v1 3

which gives v1 =

=

(i) 3r 2GM 2 (r)3/2 3GM

9.30 Comprehensive Physics—JEE Advanced

or T r3/2. The orbital speed v2 of the star of mass m2 = 2m is given by Gm1m2 m2 v22 2m v22 Gm(2m) = or = (r x) 2r / 3 r2 r2 or

v2 =

v

2Gm = v1 3r

[see Eq. (i)]

1 1 1 m1 v12 and K2 = m2 v22 = (2m) v22 2 2 2 2 K1 v1 1 = = , since v1 = v2. K2 2v22 2 mv1r Angular momentum of m1 is L1 = m1v1 x = 3 Angular momentum of m2 is L2 = m2v2(r – x) Now K1 =

= L1 L2

2 mv 2 2 r 3

1 4

(since v1 = v2)

The correct choices are (a), (c) and (d). 15. The orbital velocity of the space-ship of mass M and orbital radius r is given by GM r where r = R + h; R being the radius of the earth and h the height of the space-ship above the surface of the earth. For a space-ship close to the earth s surface r = R. Therefore v=

GM GM = Rg g R2 R The space-ship will overcome the gravitational pull of the earth if it is given an additional v0 such that v + v0 = ve where ve is the escape velocity which is given by v=

2GM = 2Rg R The required addition velocity is ve =

v0 = ve – v = =

2Rg –

The magnitude of angular momentum is L = mrv sin where is the angle between vectors r and v. For a circular orbit, = 90°. Therefore L = mrv (1) The gravitational force of attraction on the satellite is GMm/r2 which provides the necessary centripetal force mv2/r, i.e. GMm r2

=

m v2 r

or

(mrv) = m GMr From Eqs. (1) and (2), we have

Rg

(2)

L = m GMr This gives the magnitude of angular momentum. The direction of angular momentum is perpendicular to the plane of the orbit. The correct choices are (a) and (d). 17. Consider a point P at a distance x from mass m1; its distance from mass m2 is (r – x). The net gravitaP will be zero if Gm1 x

2

=

Gm2

x)2

(r

or m2 m1 = (r x) x which gives

2 1 The correct choices are (a) and (d). 16. At a certain instant of time let r be the radius vector of the satellite from the centre of its circular orbit. If the velocity of the satellite is v as shown in Fig. 9.26, its angular momentum is given by L = r (mv) Rg

Fig. 9.26

r m1

x=

m1

(1)

m2

r m2

(2) m1 m2 The gravitational potential at point P is given by Gm1 Gm2 (3) U=– x r x Using (1) and (2) in (3), we have (r – x) =

U = – G m1

m1 r m1

m2

m2

m1 r m2

m2

Gravitation 9.31

G m1 m1 m2 m2 m1 m2 r G =– (4) m1 m2 2 m1m2 r Putting m1 = m and m2 = 4m in Eqs. (1) and (4) we =–

18. F =

k , where k is a constant. r

mv 2 k = r r

v=

k , which is independent of r. m

2 r m = 2 r T r. v k Hence the correct choices are (a) and (c). T=

GmM . Hence the correct choic19. Total energy E = – es are (a) and (b). 2r 20. The correct choices are (a) and (d). 21. Due to friction, the satellite will slow down. Hence its kinetic energy will decrease. Now v = R . Therefore, a decrease in v results in a decrease in . Since T = 2 / , time period T will increase. The angular momentum will remain unchanged because the torque due to a radial (central) force is zero. Hence the correct choices are (a), (b) and (c). GM . If r increases, v will 22. Orbital speed v = r r3 2 r =2 . Hence decrease. Time period T = GM v T will increase. Centripetal force = gravitational GmM force = . Hence if r increases, centripetal r2 GM force will decrease. Gravitational P.E. = – . r If r increase, P.E. becomes less negative, i.e. it increases. Since no external torque acts, the angular momentum is conserved. Hence all the four choices are correct. 23. The speed of the comet is greater when it is closer to the sun than when it is farther from it. Hence both v and r change for an elliptical orbit. Thus choice (a) is correct. The total energy is conserved. The linear momentum changes because the velocity changes in magnitude as well as direction. Since the force is central, the angular momentum is conserved. Hence all the four choices are correct. 24. Since the initial velocity of the body is zero, its total energy is Mm Ei = – G r

When the body reaches the earth s surface, its velocity is v and its distance from the centre of the earth is the earth s radius R. Therefore 1 Mm Ef = mv2 – G 2 R From energy conservation, Ei = Ef, i.e. Mm 1 Mm mv2 – G =–G R 2 r which gives v = R 2g

1 R

choices are (a) and (b). 25. (a) Increase in P.E. = – =

1 r

1/ 2

GmM ( R h)

GmM R

GMm 2R

(

h = R)

GM 1 mgR g R2 2 (b) If v is the velocity of projection, then from energy conservation, we have 1 GMm GMm mv2 – =– 2 R R h =

Using h = R, this equation gives v = gR . (c) Total energy of the satellite orbiting at h = R is GMm GMm GMm – =– E1 = K.E. + P.E. = 4R 4R 2R Total energy when the satellite is at rest on the surface of the earth is GMm E2 = – R Energy required is E = E1 – E2 =–

GMm – 4R

3 3 GMm GMm = = mgR 4 4 R R 2

g . 4 R h Hence the correct choices are (b), (c) and (d).

(d) Value of g at h (= R) = g

R

=

GMm . Therer fore, P.E. of two satellites before collision is GMm 2GMm (P.E.)i = 2 =– r r

26. (a) P.E. of a satellite in orbit = –

After the collision, they stick together. So the combined mass is 2m. Therefore, P.E. just after collision is

9.32 Comprehensive Physics—JEE Advanced

the combined mass will not move in an orbit; it will fall freely under the gravity of the earth. Hence the correct choices are (b) and (c). 27. The distance of each cavity from the centre O is the same. Since the two cavities are symmetrical with respect to the centre O and the mass of the sphere can be regarded as being concentrated at the centre O, the gravitational force due to the sphere is zero at the centre. Hence choice (a) is correct. For the same reason, the gravitational potential is the same at all points of the circle y 2 + z 2 = 36 whose radius is 6 units and at all points of the circle y2 + z2 = 4. Hence choices (c) and (d) are also correct. But the gravitational force at point B cannot be zero.

GM 2m 2GMm =– r r Hence the ratio of P.E. before and after collision is 1 : 1. (b) Total energy of a satellite in orbit is (P.E.)f = –

GMm GMm GMm + =– 2r 2r r Therefore, the total energy of the two satellites before collision is GMm GMm Ei = 2 =– 2r r Let V be the velocity of the combined mass. Final momentum = (2m)V. Since the satellites have opposite velocities, initial momentum = mv – mv = 0.

K.E. + P.E. =

28. (i) For r > R, F =

2 mV = 0 or V = 0. Hence the combined mass has no kinetic energy; it has only potential energy. Thus, total energy of the system just after the collision is Ef = –

GM r2

. Therefore,

F1 r22 = F2 r12 r22 r12 So choice (b) is correct and choice (d) is wrong. GMr . Therefore, (ii) For r < R, F = R3 F1 =

GM (2m) 2GMm =– r r Ei 1 = Ef 2

GM

F1 =

and F2 =

GMr1 3

, F2 =

GM

. Hence

GMr2 3

. Hence

F1 r = 1 F2 r2

R R So choice (a) is correct and choice (c) is wrong.

Since the combined mass comes to rest, the centripetal force disappears ( velocity V = 0). Therefore,

III Multiple Choice Questions Based on Passage Questions 1 to 3 are based on the following passage Passage I A satellite of mass m is revolving in a circular orbit of radius r around the earth of mass M. The speed of the satellite in its orbit is one-fourth the escape velocity from the surface of the earth. 1. The height of the satellite above the surface of the earth is (R = radius of earth) (a) 2R (b) 3R (c) 5R (d) 7R

2. The magnitude of angular momentum of the satellite is m (b) (a) m GMR GMR 2 m (c) (d) 2m GMR GMR 2 2 3. If the total energy of the satellite is E, its potential energy is (a) – E (b) E (c) 2E (d) – 2E

SOLUTION 1. v =

GM = r

GM ( R h)

(1)

v=

ve 1 2GM = 4 4 R

(2)

Gravitation 9.33

Equations (1) and (2) give h = 7R, which is choice (d). 2. L = mvR = m

GM 8R

R=

m 2 2

The correct choice is (c). 3. The correct choice is (c).

GMR (

h = 7 R)

Questions 4 to 8 are based on the following passage Passage II Considering the earth as an isolated mass, a force is experienced by a body at any distance from it. This force is directed towards the centre of the earth and has a magnitude mg, where m is the mass of the body and g is the acceleration due to gravity. The value of the acceleration due to gravity decreases with increase in the height above the surface of the earth and with increase in the depth below the surface of the earth. Even on the surface of the earth, the value of g varies from place to place and decreases with decrease in the latitude of the place. 4. Assuming the earth to be a sphere of uniform mass density, which of the graphs shown in Fig. 9.27 represents the variation g with distance r from the centre of the earth (R = radius of earth)

5. A body weighs 63 N on the surface of the earth. How much will it weigh at a height equal to half the radius of the earth? (a) 63 N (c) 28 N

(b) 32.5 N (d) none of these

6. Assuming the earth to be a sphere of uniform mass density, the weight of a body when it is taken to the end of a tunnel 32 km below the surface will (radius of earth = 6400 km) (a) decrease by 0.5% (b) decrease by 1% (c) increase by 0.5% (d) increase by 1% 7. If gp is the acceleration due to gravity at the poles and ge that at the equator, then (a) gp < ge

(b) gp > ge

(c) gp = qe

(d) ge = 0

8. If a tunnel is dug along a diameter of the earth and a body is dropped from one end of the tunnel, (a) it will fall and come to rest at the centre of the earth where its weight becomes zero. (b) it will emerge from the other end of the tunnel. (c) it will execute simple harmonic motion about the centre of the earth. (d) it will accelerate till it reaches the centre and decelerate after that eventually coming to rest at the other end of the tunnel.

Fig. 9.27

SOLUTION 4. At a height h above the surface of the earth, gh =

g0 R 2

( R h) 2

At a depth d below the surface of the earth gd = g0 1

d R

where g0 = acceleration due to gravity at the surface of the earth. Hence the correct choices is (d).

5. W = W0

2

R R h

2

R

= 63 R

R 2

= 28 N

199 g0 32 d = g0 1 = 200 6400 R Decrease in weight = mg0 – mg

6. g = g0 1

199 mg 0 = 200 200 Hence the correct choice is (a). = mg0 1

9.34 Comprehensive Physics—JEE Advanced

7. Due to the rotation of the earth about its axis, g p > g e. Questions 9 to 12 are based on the following passage Passage III The escape velocity on a planet or a satellite is the minimum velocity with which a body must be projected from that planet so that it escapes the gravitational pull of the planet goes into outer space. We obtain the expression for the escape velocity by equating the work required to with the initial kinetic energy given to the body. The escape velocity from a planet of mass M and radius R is given by ve =

2MG = R

2gR

where g is the acceleration due to gravity on the surface of the planet and G is the gravitation constant. 9. Choose the only incorrect statement from the following. The escape velocity from a planet (a) is independent of the mass of the body which is projected. (b) is independent of the direction in which the body is projected. (c) depends on the mass and radius of the planet. (d) will be less than the value given by the ex2MG if the planet has an pression ve = R atmosphere.

SOLUTION 9. The only incorrect statement is (d). Due to atmospheric friction, a part of the initial kinetic energy is lost as heat. Hence the actual value of the escape velocity is greater than the value obtained from the given expression. vJ = 10. vE

MJ ME

1 = 29 319 11 Hence the correct choice is (a). 11. The escape velocity at a height h is given by RE = RJ

ve =

2g R h

8. The correct choice is (c). 10. The mass of Jupiter is about 319 times that of the earth and its radius is about 11 times that of the earth. The ratio of the escape velocity on Jupiter to that on earth is (a) (c)

29 1

(b) 29 (d)

1

29 29 11. If R is the radius of the earth and g the acceleration due to gravity on its surface, the escape velocity of a body projected from a satellite orbiting the earth at a height h = R from the surface of the earth will be (a)

gR

(b)

2gR

(c)

3gR

(d) 2 gR

12. A body is dropped from a height equal to half the radius of the earth. If ve is the escape velocity on earth and air resistance is neglected, it will strike the surface of the earth with a speed ve v (b) e (a) 2 2 (c)

ve 3

(d)

ve 3

where g is the acceleration due to gravity at height h, R 2 g =g R h For h = R, we get ve = gR , which is choice (a). 12. The correct choice is (c). Use conservation of energy, i.e. Total energy at h (= R/2) = Total energy when the body strikes the earth –

GmM 1 GmM = mv2 – ( R h) 2 R

Gravitation 9.35

IV Assertion Reason Type Questions In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each question has the following four choices out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Statement-1. (c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. 1. Statement-1 A body is projected up with a velocity equal to half the escape velocity from the surface of the earth. If R is the radius of the earth and atmospheric resistance is neglected, it will attain a height h = R/3. Statement-2 The gravitational potential is – GM/R on the surface of the earth and it increases with height; M being the mass of the earth. 2. Statement-1 The total energy (kinetic + potential) of a satellite moving in a circular orbit around the earth is half its potential energy. Statement-2 The gravitational force obeys the inverse square law of distance. 3. Statement-1 Two bodies of masses m1 = m and m2 = 3m are iniallowed to move towards each other under mutual gravitational attraction. Their relative velocity of approach when they are at a separation r is v =

2Gm r

Statement-2 The gain in the kinetic energy of each body equals the loss in its gravitational potential energy. 4. Statement-1 An astronaut inside a massive space-ship orbiting gravitational force. Statement-2 The centripetal force necessary to keep the spaceship in orbit around the earth is provided by the gravitational force between the earth and the spaceship. 5. Statement-1 The escape velocity varies with altitude and latitude of the place from where it is projected. Statement-2 The escape velocity depends on the gravitational potential at the point of projection. 6. Statement-1 A comet orbits the sun in a highly elliptical orbit. Its potential energy and kinetic energy both change over the orbit but the total energy remains constant throughout the orbit. Statement-2 For a comet, only the angular momentum remains constant throughout the orbit. 7. Statement-1 The acceleration due to gravity decreases due to rotation of the earth. Statement-2 A body on the surface of the earth also rotates with it in a circular path. A body in a rotating (non-inertial) frame experiences an outward centrifugal force against the inward force of gravity.

SOLUTIONS 2Gm and R total energy at r (= R + h) = total initial energy, i.e. GmM 1 GmM – = mv2 – r 2 R 2. The correct choice is (a). The centripetal force needed for circular motion is provided by the 1. The correct choice is (b). Use ve =

gravitational force. Since the gravitational force obeys the inverse square law of distance, the orbital velocity of the satellite is given by v=

GM r

where M = mass of earth and r = orbital radius.

9.36 Comprehensive Physics—JEE Advanced

Therefore

1 m1v12 2

Gm1m2 or v1 r

2Gm2 r

Gm1m2 or v2 r

2Gm1 r

1 GmM Kinetic energy = mv2 = 2 2r where m = mass of the satellite. From the inverse

and

the satellite is given by

Therefore, their relative velocity of approach is

GmM r Total energy E = K.E. + P.E. GmM GmM = + 2r r

Potential energy = –

GmM P.E. =– = 2r 2 3. The correct choice is (d). Initially when the two masses are at an distance from each other, their gravitational potential energy is zero. When they are at a distance r from each other the gravitational P.E. is Gm1m2 P.E = – r The minus sign indicates that there is a decrease in P.E. This gives rise to an increase in kinetic energy. If v1 and v2 are their respective velocities when they are at a distance r apart, then, from the law of conservation of energy, we have

1 m1v22 2

v = v1 + v2 = =

2Gm2 + r 2G (m2 r

2Gm1 r m1 )

Putting m1 = m and m2 = 3m, we get v = 2

2Gm r

4. The correct choice is (b). Because the centripetal force equals the gravitational force exerted by the earth on the space-ship, the astronaut does not experience any gravitational force of the earth. The only force of gravity that an astronaut in an orbiting space-ship experiences is that which is due to the gravitational force exerted by the space-ship. Since but very small. 5. The correct choice is (a). The gravitational potential at a point varies with the altitude and latitude of the place. 6. The correct choice is (c). 7. The correct choice is (a).

10

Elasticity

Chapter

REVIEW OF BASIC CONCEPTS 10.1

ELASTICITY

The ability of a body to regain its original shape and size when the deforming force is withdrawn, is known as elasticity.

10.2

STRESS

When a deforming force is applied to a body, it reacts to the applied force by developing a reaction (or restoring) force which, from Newton’s third law, is equal in magnitude and opposite in direction to the applied force. The reaction force per unit area of the body which is called into play due to the action of the applied force is called stress. Stress is measured in units of force per unit area, i.e. Nm–2. Thus F Stress = A where F is the applied force and A is the area over which it acts.

10.3

STRAIN

When a deforming force is applied to a body, it may ratio of the change in size or shape to the original size or shape of the body. Strain is a number; it has no units or dimensions. The ratio of the change in length to the original length is called longitudinal strain. The ratio of the change in volume to the original volume is called volume strain. The strain resulting from a change in shape is called shearing strain.

10.4 HOOKES’ LAW Hookes’ law states that, within the elastic limit, the stress developed in a body is proportional to the strain produced

in it. Thus the ratio of stress to strain is a constant. This constant is called the modulus of elasticity. Thus stress Modulus of elasticity = strain Since strain has no unit, the unit of the modulus of elasticity is the same as that of stress, namely, Nm–2.

10.5

YOUNG’S MODULUS

Suppose that a rod of length L and a uniform crosssectional area A is subjected to a longitudinal pull. In other words, two equal and opposite forces are applied at its ends. F Stress = A The stress in the present case is called linear stress, tensile stress, or extensional stress. If the direction of the force is reversed so that L is negative, we speak of compressional strain and compressional stress. If the elastic limit is not exceeded, then from Hooke’s law Stress strain or Stress = Y strain stress F L (1) strain A L where Y, the constant of proportionality, is called the Young’s modulus of the material of the rod and may be ratio of the linear stress to linear strain, provided the elastic limit is not exceeded. Since strain has no unit, the unit of Y is Nm–2.

or

10.6

Y =

BULK MODULUS

Solids, liquids and gases can be deformed by subjecting them to a uniform normal pressure P in all directions. Stress and pressure have the same dimension (force per unit area), but pressure is not the same thing as

10.2 Comprehensive Physics—JEE Advanced

stress. Pressure is the force per unit area acting on the surface of a system, the force being everywhere perpendicular to the surface so that, for a uniform pressure, the force per unit area is the same. Pressure is a particular kind of stress which changes only the volume of the substance and not its shape. The substance may be a solid, liquid or gas. A small increase in pressure P applied to a substance decreases its volume from, say, V to V – V so that V is the small decrease in volume. The volume strain is given by V Volume strain = – V

Suppose the lower face PQRS F is applied parallel to the upper face MNUV. As a result of this, the lines joining the two faces turn through an angle . We say that the face MNRS is sheared through an angle (measured in radians). The angle is called the shear strain or the angle of shear and is a measure of the degree of deformation. If A is the area of the face MNUV, the ratio F/A is the shearing stress. It is found that for small deformation, the shearing stress is proportional to the shear strain, i.e. F F F , = ,n= A A A

pressure and the corresponding volume strain, i.e. P PV B= (2) V V V If P is positive, V will be negative and vice versa. The B ensures that B is always positive. The SI unit of B is Nm–2. The reciprocal of B is known as compressibility. The bulk modulus of a gas depends on the pressure. Under isothermal conditions (i.e. when the temperature is kept constant), the bulk modulus of a gas is equal to its pressure P. Under adiabatic conditions (i.e. when heat is not allowed to leave or enter the system), the bulk modulus is equal to P, where (= Cp/Cv) is the ratio of the molar heat capacities of the gas at constant pressure and constant volume. Thus Isothermal bulk modulus = P Adiabatic bulk modulus = P.

The quantity n is called the shear modulus or the modulus of rigidity. Referring to Fig. 10.1, if is small,

10.7

SHEAR MODULUS OR MODULUS OF RIGIDITY

Shear is a particular kind of stress which only solids can withstand. The solid is deformed by changing its shape without changing its size. The body does not move or rotate as a whole: there is a relative displacement of its contiguous layers. Consider a solid in the form of a rectangular cube as in Fig. 10.1.

tan

L L

F L A L This equation looks similar to Eq. (1) for Young’s modulus with the difference that F/A here is the tangential stress and not longitudinal stress. so that

10.8

=

POISSON’S RATIO

When a wire is stretched with a force, apart from an increase in its length, there is a slight decrease in its diameter, i.e. both shape and volume change under longitudinal stress. The ratio of the decrease D in diameter to the original diameter D is called lateral strain, i.e. strain at right angles to the deforming force. Thus change in diameter D Lateral strain = original diameter D change in length L Also Longitudinal strain = original length L The ratio of the two is called Poisson’s ratio and is denoted by . Hence, =

Lateral strain Longitudinal strain

D/ D L/ L

D L D L Since it is a ratio between two types of strain, is dimensionless. Theoretically, one can show that it must be less than 0.5. For most solids it lies between 1/4 and 1/3, and for rubber it is very nearly 0.5. =

10.9 Fig. 10.1

=

ENERGY STORED IN A STRAINED WIRE: STRAIN ENERGY

If a wire is stretched, the energy stored per unit volume is given by

Elasticity 10.3

U = =

1 stress 2 1 S 2

where

S = stress,

Since

Y = U =

10.10

10.1 A steel wire (original length = 2 m, diameter = 1mm) and a copper wire (original length = 1m, diameter = 2 mm) are loaded as shown in Fig. 10.2. Find the ratio of extension of steel wire to that of copper wire. Young’s modulus of steel = 2 1011 Nm–2 and that of copper = 1 1011 Nm–2.

strain

= strain

S 1 S 2

=

1 Y 2

2

=

1 S2 2 Y

THERMAL STRESSES

then due to expansion or contraction, tensile or compressive stress is set up in the rod. These stresses are called thermal stresses. If a rod of length L is free to expand or contract and its temperature is changed by T, the change in its length is given by L =

L

F =

FL A L FL 4F L L= = YA Y d2 Y=

For steel wire Fs = (6 + 4) 9.8 = 98 N For copper wire Fe = 6 9.8 = 58.8 N

FL AY

( L)s =

AY T

Thus, the thermal stress in the rod is F = A

SOLUTION

T

where Now from Eq. (1), we have L =

Fig. 10.2

4 Fs Ls Ys

d Ls

Y T

Lc

its volume cannot change, then a change in temperature results in a change in pressure. The thermal stress is then given by P = B T where B

10.11

is its

TORQUE

If a rod or wire of length l and radius r and the other end is twisted through an angle (in radian) by applying a torque, the restoring torque is given by =

r4 2l

Where is the shear modulus of the material of the rod or wire.

and ( L)c =

2 s

F = s Fc

Ls Lc

=

98 58.8

=

20 3

2 1

Yc Ys 1 2

4 Fc Lc

d c2

Yc dc ds 2

2

2

10.2 When the pressure on all sides of a metal cube is increased by 107 Pa, its volume decreases by 0.015% . Find the bulk modulus of the metal. SOLUTION P = 107 Pa,

0.015 100

V = V B=

P = V /V

= 6.67

107 0.015 /100

1010 Pa or Nm–2

10.4 Comprehensive Physics—JEE Advanced

10.3 A rubber cube is subjected to a pressure of 106 Pa on all sides. As a result, each side of the cube decreases by 1%. Calculate the bulk modulus of rubber. SOLUTION Volume of cube is V = L3. Therefore V 3 L = =3 V L

– 1% = – 3% = 106 = 3.33 3/100

P = V/V

B=

3 100 107 Nm–2

10.4 The base of a rubber cube of side 3.0 cm is clamped. A horizontal force F of 2.7 N is applied on the top face. Calculate (a) the angle of shear and (b) the horizontal displacement of the top face of the eraser. Shear modulus of rubber = 1.5 105 Nm–2. SOLUTION Refer to Fig. 10.1 and also to Fig. 10.3.

Fig. 10.3

Area of top face (A) = 3 cm 3 cm = 9 cm2 = 9 10–4 m2 F 2.7 Shearing Stress = = 3 103 Nm–2 A 9 10 4 (a) Angle of Shear

= =

(b) tan

x = L

10.5 In this example, Statement-I is followed by Statement-II. Choose (a) if statement-II is true and statement-II is true and statement-II is the correct explanation for statement-I. (b) if statement-I is true and statement-II is true but statement-II is not the correct explanation for statement-I. (c) if statement-I is true and statement-II is false. (d) if statement-I is false and statement-II is true. (i) Statement-I: A wire of length L extends by an amount L when a block of weight W is hung from it as shown in Fig. 10.4(a). If the same wire goes over a frictionless pulley and two blocks each of weight W are hung at its ends as shown in Fig 10.4(b), the wire will extend by 2 L, Fig. 10.4 if the elastic limit is not exceeded. Statement-II: For a given load, the extension is proportional to the length of the wire. (ii) Statement-I: Steel is more elastic than rubber. Statement-II: For a given deforming force, a steel wire is extended by an amount less than rubber band of the same radius and the same length. (iii) Statement-I: Figure 10.5 shows the stress-strain curves for two materials 1 and 2. It follows from the graphs that material 1 has higher Young’s modulus then material 2. Statement-II: Within elastic limit, the slope of the stress-strain graph is greater for material 1 than for material 2.

shearing stress shear modulus 3 103 5

1.5 10

= 2.0

10–2 rad

x = L tan = L ( is small) = 3 cm 2.0 10–2 = 6 10–2 cm = 0.6 mm

Fig. 10.5

Elasticity 10.5

(iv) Statement-I: It follows from Fig. 10.5 than material 1 is more brittle than material 2. Statement-II: The plastic range is more for material 1 than for material 2. (v) Statement-I: A hollow shaft is stronger than a solid shaft both made of the same material and the same external dimensions. Statement-II: To produce a given twist, a greater torque is needed for the hollow shaft than for the solid shaft. SOLUTIONS (i) Statement-II is true. Since the length of the wire and tension in wire in cases (a) and (b) shown in Fig. 10.4 are the same, the extension in the wire will be the same equal to L in both case. So Statement-I is false. The correct choice is (d) (ii) Ys > Yr, i.e.

FL FL < or ( L)s < ( L)r AYs AYr

So Statement-II is true. This means than steel develops less strain that rubber and hence steel, can withstand greater stress than rubber. So Statement-I is also true. The correct choice is (a). (iii) Young’s modulus = slope of the straight portion of the stress – strain graph. Hence Young’s modulus of material 1 is greater than that of material 2. So the correct choice is (a). (iv) The plastic range AB of material 2 is less than that of material 1. So the breaking strain of material 2 is less than that of material 1. Hence material 2 is more brittle than material 1. So, Statement-I is false and Statement-I is true. (v) The torque per unit angular twist for a solid shaft is r4 (1) 2L If r1 and r2 are the external and internal radii of the hollow shaft, the torque per unit angular twist is s

h

=

=

2L

(r12

r22 )(r12

i.e.

h

=

r2 2 (r1 2L

=

s

r22 )

r22

1

r2

(

r1 = r)

> s. So the correct choice is (a).

10.6 A heavy and thick rubber rope is hung frame support. It is stretched by its own weight. The strain in the rope is (a) the maximum near the support. (b) the maximum near the free end of the rope (c) the maximum at the centre of the rope (d) the same everywhere. SOLUTION If M is mass of the rope and L its length, then the tension at a point P at a distance y from the free end is (see Fig. 10.6) Tp =

Mg L

y

Thus the tension is maximum at A and zero at B. Now

Fig. 10.6

L T L AY Hence the strain is maximum near A and minimum near B. So the correct choice is (a). Strain =

10.7 A metal rod of length 1.0 m and cross-sectional area 1.0 mm2 supports. The tension in the rod is zero when the temperature is 40°C. Find the tension in the rod when the temperature falls to 20°C. Young’s modulus of metal = 2.0 1011 Nm–2 pansion = 1.0 10–5 K–1. SOLUTION

(r14 r24 ) 2L

=

h

r22 )

Since the volumes are the same, = r2L. Using this eq. (2), we get

(2) (r12

r22 ) L

Thermal stress = Y T T = Y T A T =A Y T = (1.0 10–5) (1.0 10–4) (2.0 1011) (40 – 20) = 40 N

10.6 Comprehensive Physics—JEE Advanced

10.8

Now

A bob of mass 1 kg is suspended by a rubber cord 2.0 m long and of cross-sectional area 1.0 cm2 . It is revolved in a horizontal circle of radius 1.0 m at a speed of 4 ms–1. If Young’s modulus of rubber is 5 108 Nm–2, calculate the extension of the cord. Take g = 10 ms–2.

L=

TL AY

= 7.5

18.87 (1.0 10

4

)

2.0 (5 108 )

10–4 m = 0.75 mm

10.9 A wire of length L is suspended from a support. Its length becomes L1 when its is loaded with a block of mass m1 and L2 when it is loaded with a block of mass m2. Then 1 L1 L2 (b) L ( L1 L2 ) (a) L 2

SOLUTION Refer to Fig. 10.7. Let T be the tension in the cord.

(c) L

L1m2 (m1

L2 m1 m2 )

(d) L

L1m2 (m2

L2 m1 m1 )

v

SOLUTION L=

mgL AY

L1 =

m1 gL AY

L1 – L =

m1 gL AY

L2 =

m2 gL AY

L2 – L =

m2 gL AY

Fig. 10.7

Figure 10.7(b) shows the free body diagram of the bob. From the diagram, it follows that mv 2 (centripetal force) T sin = r and T cos = mg Squaring and adding these equations, we get T=

=

mv 2 r

2

1.0 42 1.0

(mg ) 2

and

2

(i)

(ii)

Dividing (i) by (ii) L1 L2

(1.0 10) 2

m1 L = m2 L

L

L1m2 m2

L2 m1 m1

So the correct choice is (d).

= 18.87 N

I Multiple Choice Questions with Only One Choice Correct 1. 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 = 50 cm, diameter = 0.5 mm

(b) Length = 100 cm, diameter = 1 mm (c) Length = 200 cm, diameter = 2 mm (d) Length = 300 cm, diameter = 3 mm IIT, 1981

Elasticity 10.7

2. When a wire of length L is stretched with a tension F, it extends by l. If the elastic limit is not exceeded, the amount of energy stored in the wire is 1 Fl (a) F l (b) 2 (c)

Fl 2 L

(d)

Fl 2 2L

IIT, 1990 3. A wire, suspended from one end, is stretched by attaching a mass of 2 kg to the lower end. The wire stretches by 1 mm. The ratio of the energy gained by the wire to the gravitational energy lost by the mass in dropping a distance of 1 mm is (Take g = 10 ms–2) (a) 1 (b) 1/2 (c) 1/3 (d) 1/4 4. When the pressure on a metal cube is increased by 107 Pa, its volume decreases by 0.015%. The bulk modulus of the metal (in Nm–2) is (a) 1.5 1010 (b) 3.33 1010 10 (c) 6.67 10 (d) 7.5 1010 5. A rubber cube is subjected to a pressure of 106 Pa on all sides. As a result, each side of the cube decreases by 1%. The bulk modulus of rubber is (in Nm–2) (a) 1 108 (b) 3 107 7 (c) 3.33 10 (d) 6.67 107 6. Two wires A and B made of the same material have their lengths in the ratio 1 : 2 and their diameters in the ratio 2 : 1. If they are stretched with the same force, the ratio of the increase in the length of A to that of B is 1 (a) 1 (b) 2 1 1 (c) (d) 4 8 7. A copper wire of length 1.0 m and a steel wire of length 0.5 m having equal cross-sectional areas are joined end to end. The composite wire is stretched by a certain load which stretches the copper wire by 1 mm. If the Young’s modulii of copper and steel are 1.0 1011 Nm–2 and 2.0 1011 Nm–2, the total extension of the composite wire is (a) 1.25 mm

(b) 1.50 mm

(c) 1.75 mm

(d) 2.0 mm

8. Two springs of equal lengths and equal crosssectional areas are made of materials whose Young’s modulii are in the ratio of 3 : 2. They are suspended and loaded with the same mass. When stretched and released, they will oscillate with time periods in the ratio of

(a) 3 : 2 (c) 3 3 : 2 2

(b) 3 : 2 (d) 9 : 4

9. A wire of length L and cross-sectional area A is made of a material of Young’s modulus Y. The work one in stretching the wire by an amount x is given by YA x 2 L YA L2 (c) x (a)

YA x 2 2L YA L2 (d) 2x (b)

IIT, 1987 10. A solid sphere of radius R and made of a material of bulk modulus K is completely immersed in a liquid in a cylindrical container. A massless piston of area A M is placed on the piston to compress the liquid, the fractional change in the radius of the sphere, R/R is given by Mg Mg (a) (b) KA 2K A Mg Mg (c) (d) 3K A 4K A 11. A steel wire of cross-sectional area 3 10–6 m2 can withstand a maximum strain of 10–3. Young’s modulus of steel is 2 1011 Nm–2. The maximum mass the wire can hold is (Take g = 10 ms–2) (a) 40 kg (b) 60 kg (c) 80 kg (d) 100 kg 12. A rubber eraser 3 cm 1 cm 8 cm is clamped at one end with the 8 cm edge vertical. A horizontal force of 2.4 N is applied at the free end (the top face). If the shear modulus of rubber is 1.6 105 Nm–2, the horizontal displacement of the top face will be (a) 1 mm (b) 2 mm (c) 3 mm (d) 4 mm 13. The density of water at the surface of the ocean is . If the bulk modulus of water is B, what is the density of ocean water at a depth where the pressure is nP0, where P0 is the atmospheric pressure? (a)

B

B n 1 P0

(b)

B

B n 1 P0

B B (d) B nP0 B nP0 14. One end of a uniform rod of mass M and crosssectional area A is suspended from rigid support and an equal mass M is suspended from the other end. The stress at the mid-point of the rod will be (c)

10.8 Comprehensive Physics—JEE Advanced

2M g A Mg (c) A (a)

(b)

3M g 2A

(d) zero

15.

of linear expansion 1 and 2 and Young’s modulii Y1 and Y2 rods are heated to the same temperature. There is no bending of rods. If 1 : 2 = 2 : 3, the thermal stresses developed in the two rods will be equal provided Y1 : Y2 is equal to (a) 2 : 3 (b) 1 : 1 (c) 3 : 2 (d) 4 : 9 16. One end of a uniform wire of length L and of weight W is attached rigidly to a point in the ceiling and a weight W1 is suspended from its lower end. If S is the area of cross-section of the wire, the stress in the wire at a height 3L/4 from its lower end is W (a) W1/S (b) W1 S 4 3W (c) W1 S (d) (W1 + W)/S 4 17. A uniform wire (Young’s modulus 2 1011 Nm–2) is subjected to a longitudinal tensile stress of 5 107 Nm–2. If the overall volume change in the wire is 0.02%, the fractional decrease in the radius of the wire is (b) 1.0 10–4 (a) 1.5 10–4 –4 (c) 0.5 10 (d) 0.25 10–4 IIT, 1993 18. An elastic spring of unstretched length L and force constant k is stretched by a small length x. It is further stretched by another small length y. The work done in the second stretching is (a)

1 ky2 2

(b)

1 k(x + y)2 2

(d)

(c) 2b –

a 2

(a) 2 (c) 3

(d) 4a – 3b IIT, 1987

1011 N/m2 1012 N/m2

(b) 2 (d) 3

10–11 N/m2 10–12 N/m2 IIT, 2003

22. A light rod of length L is suspended from a support horizontally by means of two vertical wires A and B of equal length as shown in Fig. 10.9. The crosssectional area of A is half that of B and the Young’s modulus of A is twice that of B. A weight W is hung as shown. The value of x so that W produces equal stress in wires A and B is (a)

L 3

(b)

L 2

(c)

2L 3

(d)

3L 4

1 k (x2 + y2) 2

1 ky(2x + y) 2 IIT, 1994 19. The length of an elastic string is a metre when the longitudinal tension is 4 N and b metre when the tension is 5 N. The length of the string (in metre) when the longitudinal tension is 9 N is (a) a – b (b) 5b – 4a (c)

20. The Poisson’s ratio of a material is 0.4. If a force is applied to a wire of this material, there is a decrease of cross-sectional area by 2%. The percentage increase in its length is : (a) 3% (b) 2.5% (c) 1% (d) 0.5% 21. Fig. 10.8 shows a graph of the extension ( l) of a wire of length 1 m suspended from the top of a roof at one end and with a load W connected to the other end. If the crosssectional area of the wire is 10–6 m2, the Young’s modulus of the material of the Fig. 10.8 wire is

A

B

TA

TB L

x

C C W

Fig. 10.9

23. In Q. 22 above, the value of x at which W produces equal strain in wires A and B is (a)

L 4

(b)

L 2

Elasticity 10.9

2L 4L (d) 3 5 24. A wire breaks when subjected to a stress S. If is the density of the material of the wire and g, the acceleration due to gravity, then the length of the wire so that it breaks by its own weight is (c)

(a)

gS

(b)

(c)

gS

(d)

(c)

mg cos AY

(b)

g S S g

1 ML 2 1 (c) ML 8

(a)

1 1 p/ B

mg AY cos

(b) 1

2

2

1 ML 4 3 (d) ML 8

(b)

2

2

28. A rubber cord of mass M, length L and cross-sectional area A is hung vertically from a ceiling. The Young’s modulus of rubber is Y. If the change in the diameter of the cord due to its own weight is neglected, the increase in its length due to its own weight is

mg sin mg (c) (d) AY AY sin 26. The density of a metel at a normal pressure is . Its density when it is subjected to an excess pressure p is . If B is the bulk modulus of the metal, the ratio / is (a)

(d) 1 + B/p

27. A thin uniform metallic rod of mass M and length L is rotated with a angular velocity in a horizontal plane about a vertical axis passing through one of its ends. The tension in the middle of the rod is

25. A stone of mass m is attached to one end of a wire of cross-sectional area A and Young’s modulus Y. The stone is revolved in a horizontal circle at a speed such that the wire makes an angle with the vertical. The strain produced in the wire will be. (a)

1 1 p/ B

(a)

MgL AY

(b)

MgL 2 AY

(c)

2MgL AY

(d)

MgL 4 AY

29. The volume of a wire remains unchanged when the wire is subjected to a certain tension. The Poisson’s ratio of the material of the wire is (a) 0.25 (b) 0.3 (c) 0.4 (d) 0.5

p B

ANSWERS 1. 7. 13. 19. 25.

(a) (a) (a) (b) (b)

2. 8. 14. 20. 26.

(b) (a) (b) (b) (a)

3. 9. 15. 21. 27.

(b) (b) (c) (a) (d)

4. 10. 16. 22. 28.

(c) (c) (c) (c) (b)

5. 11. 17. 23. 29.

(b) (b) (d) (b) (d)

6. 12. 18. 24.

(d) (d) (d) (d)

SOLUTIONS FL 4 FL = AY d 2Y Since the wires are made of the same material, Y is the same for all wires. For a given tension F, L L d2 L The value of 2 is the largest for L = 50 cm and d = 0.5 mm. d 1 2. Energy stored per unit volume = (stress strain). 2 If A is the cross-sectional area of the wire, the total 1. Extension

L=

energy store in the wire is

1 (stress strain) 2 1 F l = AL 2 A L

U=

volume of wire

1 Fl 2 3. Energy gained by the wire is 1 1 mg l U1 = Fl = 2 2 1 = 2 10 (1 2 =

10–3) = 0.01 J

10.10 Comprehensive Physics—JEE Advanced

Gravitational P.E. lost by the mass is U2 = mgh = 2 10 (1 10–3) = 0.02 J Hence the correct choice is (b). 0.015 V = = – 1.5 10–4 4. P = 107 Pa, 100 V Bulk modulus B = –

= Mg is applied. Now, the time period of vertical oscillations is given by T= 2

P V /V

107

= 6.67 1010 Nm–2 1.5 10 4 5. V = L3. Therefore V 3 L = = 3 (– 1%) = – 3% V L =–

P 106 =– = 3.33 V /V 3%

B=– 6.

L=

107 Nm–2

FL 4 FL = AY d 2Y

L

A

L

B

=

LA LB

dB dA

2

1 2

=

1 2

2

=

1 8

ls =

F Ls AYs

ls = lc = 1 mm

0.5 m 1.0 m

1 2

Y

x L

2

A

L

Y Ax 2 2L Hence the correct choice is (b). 10. Pressure exerted by the piston on the liquid when a mass M is placed on the piston, P = Mg/A. This pressure is exerted by the liquid equally in all directions. Therefore, the surface of the sphere experiences a force P per unit area. The stress on the sphere is P = Mg/A. Now, the volume of the sphere is =

7. When a wire of length L, cross-sectional area A and Young’s modulus Y is stretched with a force F, the extension l in the wire is given by FL l = AY Since F and A are the same for the two wires, we have F Lc For copper wire lc = AYc For steel wire,

ML YA

T1 Y2 3 = = T2 Y1 2 Hence the correct choice is (a). 9. Work done per unit volume of the wire 1 = (stress strain). 2 stress Y = strain stress = Y strain Work done per unit volume 1 1 x 2 Y (strain)2 = Y = 2 2 L But, volume of the wire = A L Work done =

Since F and Y are the same for wires A and B,

M =2 k

4 R3 3 Due to stress, the change in the volume of the sphere is V =

4 R3 4 = . 3R2 R 3 3 = 4 R2 R V 3 R Volume strain = V R V =

Ls Yc Lc Ys 1.0 1011 Nm

2

2.0 1011 Nm

2

= 0.25 mm Total extension = 1 + 0.25 = 1.25 mm. Hence the correct choice is (a). F L 8. Young’s modulus Y = A l F YA Force constant k = l L where l is the extension in the spring of original length L and cross-sectional area A when a force F

stress M g /A = strain 3 R/R R Mg or = R 3K A Hence the correct choice is (c). 11. Maximum stress = Young’s modulus strain K=

=2

1011

10–3 = 2

maximum 108 Nm–2

Elasticity 10.11

Maximum force (F) = maximum stress 8

=2

area

–6

10 3 10 = 600 N F 600 Maximum mass = = = 60 kg, which is g 10 choice (b). 12. Figure 10.10 shows a rubber eraser ABCD clamped at end AB and a horizontal F applied at the free face DC resulting in a displacement DD = CC . The shear angle is given by DD tan = AD

V P B Volume of the same mass M of water at the given depth is V P V =V– V=V– B V =

P B

=V 1

=

V (B – B

P)

Density of water at that depth is =

=

M V V = = V V V B P B B B

P

=

B B

n 1 P0

Hence the correct choice is (a). 14. Since the rod is uniform, half its weight can be taken to act at its mid-point. Therefore, stress at mid-point is weight of suspended mass + weight of half the rod cross-secttional area Fig. 10.10

DD AD Area of top face = 3 cm 1 cm = 3 cm2 = 3 10–4 m2

Since

is small, tan =

Shearing stress =

=

F 2.4 = A 3 10

=8 Shear angle

= =

1 Mg 3M g 2 = = , which is choice (b). 2A A stress l 15. Young’s modulus Y = = , where = strain L Now, if the temperature of a rod is increased by , the increase in its length due to thermal expansion is l = L l = . Now stress is Strain = L =Y =Y For the two rods, the stress is and 2 = Y2 2 1 = Y1 1 But 1 = 2 (given). Hence Y1 1 = Y2 2 Y1 3 or = 2 = . Hence the correct choice is (c). Y2 2 1 16. Since the wire is uniform, i.e. its mass per unit length is constant over its entire length L, the total donward weight = the weight of the sus3L of the wire or pended mass + weight of length 4 3W F = W1 + . 4 3 W1 W force F 4 = = Stress = area S S Hence the correct choice is (c) Mg +

4

103 Nm–2

shearing stress shear modulus 8 103 1.6 105

= 0.05 rad

DD = 0.05 or DD = 0.05 AD AD = 0.05 8 cm = 0.4 cm = 4 mm Hence the correct choice is (d).

or

13. Pressure at the surface of the ocean = P0, the atmospheric pressure. Pressure at a depth = nP0 (given). Increase in pressure ( P) = nP0 – P0 = (n – 1)P0 Let V be the volume of a certain mass M of water at the surface, so that M = V. The decrease in volume under pressure P is

10.12 Comprehensive Physics—JEE Advanced

17.

L L

stress Y

5 107

= 2.5

2 1011

10 –4. Given

V 0.02 = 2 10–4 V 100 Now, V = r2L. Therefore, ( r 2 L)

V = V

r2 L

=

2 r

= 2 r r or

2 r V = r V

rL

r2L L

r2 L L L

L = 2 10–4 – 2.5 L = – 0.5 10–4

10–4

1 kx2 U1 = 2 Potential energy stored in the spring when it is further extended by y is 1 U2 = k(x + y)2 2 1 1 Work done = U2 – U1 = k(x + y)2 – kx2 2 2 1 ky(2x + y) 2 19. If L is the initial length, then the increase in length by a tension F is given by FL l= r 2Y 4L Hence a= L + l = L = L + 4c (1) r 2Y =

b= L +

r 2Y

= L + 5c

(2)

L

. Solving (1) and (2) for L and r 2Y c, we get L = 5a – 4b and c = b – a. For F = 9N, we have where c =

x= L +

9L

log A = log

4 Differentiating, we have

r = – 0.25 10–4 r Fractional decrease in r = 0.25 10–4 which is choice (d). 18. Potential energy stored in the spring when it is extended by x is

and

d /d l d /d or (i) l /l l where d is the diameter and l is the length of the wire. The area of cross-section is d2 A = r2 = 4 =

or

or

9L

20.

= L + 9c r 2Y = (5a – 4b) + 9(b – a) = 5b – 4a Hence the correct choice is (b).

A d =2 A d d 1 A = d 2 A

or

+ 2 log d

1 × 2% = 1% 2 A = 2%, given) A

( Using this in (i) we have 1% = 2.5% l 0.4 Hence the correct choice is (b). l

=

force area lus is given by

(

= 0.4)

W l . Strain = . Young’s moduA l

21. Stress =

W/A W l stress = (i) l /l l A strain l 4 10 4 = m/N. Using Now, slope of graph is W 80 this in (i), we get (given l = 1 m and A = 10–6 m2) Y=

80

1

= 2 1011 N/m2 4 10 10 6 22. Let TA and TB be the tensions in wires A and B respectively. If aA and aB are the respective crosssectional areas, then Y=

4

Stress in wire A =

TA aA

TB aB The stress in wires A and B will be equal if Stress in wire B =

TA T T a = B or A = A = 1 (given). aA aB TB aB 2 Since the system is in equilibruim, the moments of forces TA and TB about C will be equal (see Fig. 10.3), i.e. TA x = TB (L – x)

Elasticity 10.13

TA 1 L x = L x or = TB 2 x x 2L which gives x = . Hence the correct choice 3 is (c). T Stress 23. Strain = = , aY Young s modulus Strain in wires A and B will be equal if TA T = B a AYA aBYB or

TA Y = A TB YB

or

aA =2 aB

1 =1 2

which gives TA = TB. Equating moments about C, we have TA x = TB (L – x) L ( TA = TB). Hence Which gives x = L – x or x = 2 the correct choice is (b).

24. Let L be the required length of wire. If A is its area of cross-section and is the density of its material, the weight of the wire is W = mg = Al g

26. Let V be the volume of the metal at normal pressure. p , the decrease in the volume when From B = – V/V the metal is subjected to an excess pressure p is given by pV | V| = B p pV New volume of metal is V = V – =V 1 B B Now mass of the metal is m = V. Therefore, its new density is m V = = = p p V V 1 1 B B 1 or = , which is choice (a). 1 p/ B 27. Let m be the mass per unit length of the rod. Then M = mL. Consider a small element of length dx located at C at a distance x from A. (Fig. 10.12)

weight AL g = =L g area A S = L g L = S/ g

Stress = or

25. Refer to Fig. 10.11. For vertical equilibruim T cos

= mg mg or T= cos If L is the original length of the wire, the increase in length is TL l= AY Strain =

l T mg = = L AY AY cos

Fig. 10.12

The mass of element of length dx = mdx. The centripetal force at C is (1) dF = (mdx) x 2 The force (or tension) at point C is due to the outer portion CB of the rod, i.e. portion from x = x to x = L. Therefore tension at point C is x L

(mdx) x

F=

2

x x

= m

2

x L

xdx = x x

1 m 2

M Now, m = . Therefore, L 1M 2 2 (L – x2) F= 2 L Fig. 10.11

=

1 ML 2

2

1

x2 L2

2

(L2 – x2)

10.14 Comprehensive Physics—JEE Advanced

If r is the radius of the rod and its density, the mass of the rod M = r2L . Therefore, x2 1 r2 L2 2 1 2 (2) 2 L Thus the tension in the rod varies with x, it is zero at x = L, i.e. at point B and maximum at x = 0, i.e. at A where the rod is pivoted. L of the rod is Tension in the middle x 2 1 1 r2 L2 2 1 F = 4 2 3 = r2 L2 2 8 Thus the correct choice is (d).

To obtain the total elongation l of the cord, we integrate from y = 0 to y = L. Thus

28. Let M be the mass of the rubber cord, L its length, A its cross-sectional area (assumed constant). Let tion dl of an element AB of length dy at a distance y (Fig. 10.13). The force due to the weight of the cord is maximum at P(y = 0) and zero at Q(y = L). Therefore, the force Fig. 10.13 acting at the lower end B of the element is Mg (L – y) (1) F= L This force is responsible for the elongation of element AB. Now stress F / A Fdy Y = strain dl / dy Adl Fdy or dl = (2) AY Using Eq. (1) in Eq. (2), we get dl =

Mg (L – y)dy LAY

Mg LAY

L

Mg = Ly LAY

y2 2

L

Mg 2 L LAY

L2 2

l =

F =

dl =

=

(L – y) dy 0

0

MgL 2 AY

The correct choice is (b). 29. Poisson’s ratio =

lateralstrain = longitudinalstrain

r /r l /l

where r is the radius of the wire and l its length and r is the change in r and l the change in l when the wire is subjected to tension. Volume of wire before elongation is V 1 = r 2l Volume of wire after elongation is V2 = (r – r)2 (l + l) Given

V1 = V2. Thus r 2l = (r – r)2 (l + l) = [r2 – 2r( r) + ( r)2] (l + l) = r 2(l + l) – 2 r r (l + l) + ( r)2 (l + l)

Since r and l are very small, terms of order r l and ( r)2 and higher can be ignored. Then, we have r2l = r 2l + r2 l – 2 rl r or

r l=2l r =

which is choice (d).

or

l l

=2

r /r 1 = = 0.5, l /l 2

r r

Elasticity 10.15

II Multiple Choice Questions with One or More Choices Correct 1. Figure 10.14 shows the stress–strain graphs for materials A and B.

Fig. 10.14

From the graph it follows that (a) material A has a higher Young’s modulus (b) material B is more ductile (c) material A is more brittle (d) material A can withstand a greater stress. 2. Two wires A and B have the same cross-section and are made of the same material, but the length of wire A is twice that of B. Then, for a given load (a) the extension of A will be twice that of B (b) the extensions of A and B will be equal (c) the strain in A will be half that in B (d) the strains in A and B will be equal. 3. Two wires A and B have equal lengths and are made of the same material, but the diameter of wire A is twice that of wire B. Then, for a given load (a) the extension of B will be four times that of A (b) the extensions of A and B will be equal (c) the strain in B is four times that in A (d) the strains in A and B will be equal. 4. In Q. 3, above (a) Young’s modulus of A is twice that of B (b) Young’s modulus is the same for both A and B

(c) A can withstand a greater load before breaking than B (d) A and B will withstand the same load before breaking 5. Choose the correct statements from the following: (a) Steel is more elastic than rubber. (b) The stretching of a coil spring is determined by the Young’s modulus of the wire of the spring. (c) The frequency of a tuning fork is determined by the shear modulus of the material of the fork. (d) When a material is subjected to a tensile (stretching) stress, the restoring forces are caused by interatomic attraction. 6. A uniform wire of Young’s modulus Y is stretched by a force within the elastic limit. If S is the stress produced in the wire and is the strain in it, the potential energy stored per unit volume is given by 1 1 2 S (b) Y (a) 2 2 S2 1 (d) Y S 2Y 2 7. A wire AB of length 2l and cross-sectional area a A and B (Fig. 10.15). The wire is pulled at the centre into shape ACB such that d << l. The tension in the string is T and the strain produced in it is . If the Young’s modulus of the material of wire is Y. Then d d2 (b) = 2 (a) = 2l 2l (c)

(c) T = aY

d 2l

(d) T = aY

d2 2l 2

Fig. 10.15

ANSWERS AND SOLUTIONS 1. The slope of the linear portion of the curve gives the Young’s modulus of the material. The slope of the linear portion OP for material A is greater than

that of the linear portion OR for material B. Hence statement (a) is correct. The plastic region for material A (from P to Q) is greater than that (from

10.16 Comprehensive Physics—JEE Advanced

R to S ) for material B, which indicates that material A is more ductile. Hence statement (b) is incorrect. The breaking stress for material B (i.e. stress corresponding to point S) is less than that for ma-terial A (i.e. stress corresponding to point Q), which implies that material B can break move easily than material A. Thus material B more brittle. Hence choice (c) is also incorrect. Material A is stronger than material B because it can withstand a greater stress before it breaks. The breaking stress is the stress corresponding to point Q for material A and to point S for material B. Hence the correct choice are (a) and (d). 2. We know that FL Y= Al FL AY Since the two wires are made of the same material, the Young’s modulus Y is the same for both. Since load F and the cross-sectional A are the same, l L i.e. extension is proportional to the length of the wire. Hence choice (a) is correct. The strain in a wire is given by l F = L AY l=

Since F, A and Y are the same, the strain in wires A and B will be equal. Thus the correct choices are (a) and (d). 3. Area of cross-section A = d 2/4; where d is the diameter of the wire. Therefore 4F L l= d 2Y Since F, L and Y are the same for wires A and B, l 1/d 2, i.e. the extension is inversely proportional to the square of the diameter. Hence choice (a) is correct. The strain is l 4F = L d 2Y 2 Thus, strain 1/d . Hence the correct choices are (a) and (c). 4. Since wires A and B are made of the same material they have the same Young’s modulus. Now break-

ing load = breaking stress cross-sectional area, the wire having a greater area of cross-section can withstand a greater load. Hence the correct choices are (b) and (c). 5. Statement (a) is correct because the Young’s modulus of steel is greater than that of rubber. Statement (b) is incorrect. If a spring is stretched, the total length of the wire in the coil and the volume of the wire, both do not change. Only the shape of the coils of the wire undergoes a change. Hence it is the shear modulus that determines the stretching of a coil. Statement (c) is also incorrect. The bending moment of the prongs of a tuning fork is determined by the Young’s modulus of its material. Hence the restoring force on the prongs depends on Young’s modulus, which determines the frequency of the fork. Statement (d) is correct. When the material is not subjected to any stress, its atoms are in their normal (equilibrium) positions. When a tensile stress is applied, the separation R between the atoms becomes greater than the equilibrium separation R0. For R > R0, the interatomic forces are attractive. 6. The correct choices are (a), (b) and (c). 7. AC + CB = 2 AC = 2(l2 + d2)1/2. Increase in length is L = AC + CB – AB = 2(l2 + d2)1/2 – 2l T Stress = . a Strain =

L 2(l 2 = L

If d << l, (l2 + d2 )1/2 = l 1 Therefore 2(l2 + d2)1/2 – 2l = 2l 1

d2

d 2 )1 / 2 2l 1/ 2

=l 1

l2 d2 2l 2

2l

– 2l =

or

L d2 = 2 L 2l stress T /a Y= = 2 2 strain d / 2l

T=aY

2l 2

d2 . Thus l

Strain = Now

d2

d2

2l 2 Hence the correct choices are (b) and (d).

.

Elasticity 10.17

III Multiple Choice Questions Based on Passage 2. In Q.1 above, the metal shows plastic behaviour Questions 1 to 4 are based on the following passage beyond point Passage I (a) A (b) B Elasticity: The atoms in solids are held together by inter(c) C (d) D atomic forces. The average locations of the atoms in a 3. Figure 10.17 shows the strain-stress graphs for lattice do not change with time. Since the atoms are almost materials A and B. From the graph it follows that lacking in mobility, their kinetic energy is negligibly small. (a) material A has a higher Young’s modulus than It is this lack of mobility which makes a solid rigid. This B. rigidity is the cause of elasticity in solids. In some solids (b) material B has a higher Young’s modulus than such as steel, the atoms are bound together by larger interA. atomic forces than in solids such as aluminium. Thus, the (c) for a given strain, material B can withstand greater stress than A. exhibit elasticity. All material bodies get deformed when (d) for a given stress, the strain produced in subjected to a suitable force. The ability of a body to regain material B is more than that in material A. its original shape and size is called elasticity. The deforming force per unit area is called stress. The change in the dimension (length, shape or volume) divided by the original dimension is called strain. The three kinds of stresses are tensile stress, shearing stress and volumetric stress. The corresponding strains are called tensile strain, shearing strain and volume strain. According to Hookes’ law, within the elastic limit stress is proportional to strain. The ratio stress/strain is called the modulus of elasticity. 1. Figure 10.16 is the load-extension curve for a metallic wire. Over which region is Hookes’ law obeyed? Fig. 10.17 (a) OA (b) AB (c) BC (d) CD 4. Choose the correct statements from the following. (a) Steel is more elastic than rubber. (b) Fluids have Young’s modulus as well as shear modulus. (c) Solids have Young’s modulus, bulk modulus as well as shear modulus. (d) Bulk modulus of water is greater than that of copper. Fig. 10.16

ANSWERS 1. Hooke’s law is obeyed in the linear portion of the graph. Hence the correct choice is (a). 2. The correct choice is (b).

Stress , which is, therefore, given Strain by the reciprocal of the strain versus stress graph. The correct choices are (b) and (c). 4. The correct choices are (a) and (c). 3. Young’s modulus =

10.18 Comprehensive Physics—JEE Advanced

Questions 5 to 7 are based on the following passage Passage II Two rods P and Q of different metals having the same area A and the same length L are placed between two rigid expansion of P and Q are 1 and 2 respectively and their Young’s modulii are Y1 and Y2. The temperature of both rods is now raised by T degrees.

5. The force exerted by one rod on the other is (a) F =

(c) F = TA(Y1 + Y2)

6. The new length of the rod P is (a) L1 = L 1

1T

F AY1

(b) L1 = L 1

1T

F AY1

(c) L1 = L 1

1T

F AY1

F AY1 7. The new length of rod Q is F (a) L2 = L 1 2T AY2

Fig. 10.18

TA ( 1 2) 1 1 Y1 Y2 (b) F = TA Y1Y2( 1 +

(d) None of these

2)

(d) L1 = L 1

1T

(b) L2 = L 1

2T

F AY2

(c) L2 = L 1

2T

F AY2

(d) L2 = L 1

2T

F AY2

1 2

SOLUTION L 1 = L 1T L1 +

L2 = LT(

and 1

+

L 2 = L 2T

2)

(1)

Since the walls are rigid, this increase in length cannot occur. Hence a force F appears between them. The decrease in length of the rods due to this force is

L1 +

L1 =

FL Y1 A

L2 =

FL 1 A Y1

and

L2 =

FL Y2 A

1 Y2

(2)

Questions 8 to 11 are based on the following passage Passage III One end of a string of length L and cross-sectional area A of mass m. The bob is revolved in a horizontal circle of radius r with an angular velocity such that the string makes an angle with the vertical. 8. The angular velocity is equal to (a)

g sin r

(b)

g cos r

(c)

g tan r

(d)

g cot r

5. Since the total length cannot change, the increase in total length due to heating = decrease in total length due to the compressive force F. Equating (1) and 6. New length of rod P = original length + increase in length due to heating – decrease in length due to F. Hence. FL L 1 = L + L 1T – AY1 So the correct choice is (c). 7. The correct choice is (c). 9. The tension T in the string is mg mg (a) (b) cos sin mg (d) m(g2 + r2 tan 10. The increase L in length of the string is (c)

(a)

TL AY

(b)

MgL AY cos

MgL MgL (d) AY sin AY 11. The stress in the string is (c)

4 1/2

)

Elasticity 10.19

(a)

mg A

mg r 1 A L

(b)

(c)

mg r 1 A L

(d)

mg r A L

SOLUTION Refer to Fig. 10.19. The vertical component T cos of tension T balances the weight mg and the horizontal component T sin provides the necessary centripetal force. Thus T cos = mg (1) 2 (2) T sin = mr

g tan r Hence the correct choice is (c). or

9.

tan

=

r

T L stress . Also strain = A L Y L=

Fig. 10.19

-

es are (a) and (d).

10. Stress =

8. Dividing (1) and (2) 2

=

TL mg = AY cos

L AY

Hence the correct choices are (a) and (b) 11. The correct choice is (a).

g

Questions 12 to 14 are based on the following passage Passage IV A thin rod of negligible mass and cross-sectional area 4 10–6m2, suspended vertically from one end, has a length of 0.5 m at 100°C. The rod is cooled to 0°C. Young’s modulus = 1011 Nm–2 10–5 K–1 and g = 10 ms–2. 12. The decrease in the length of the rod on cooling is (b) 3 10–4m (a) 2 10–4m

(c) 4 10–4m (d) 5 10–4m 13. What mass must be attached at the lower end of the rod so that the rod is prevented from contracting on cooling? (a) 40 kg (b) 30 kg (c) 20 kg (d) 10 kg 14. The total energy stored in the rod is (a) 0.1 J (b) 0.2 J (c) 0.3 J (d) 0.4 J

SOLUTION 12. The decrease in the length of the rod on cooling is L =L t = 0.5 10–5 100 =5

10–4 m

So the correct choice is (d). 13. If the rod is to be prevented from contracting, the mass m attached at the lower end must increase its length by an amount L = 5 10–4 m. Now mgL Y = A L

Questions 15 to 17 are based on the following passage Passage V Two blocks of masses m and M = 2 m are connected by means of a metal wire of cross-sectional area A passing system is then released.

or

m=

YA L 1011 4 10 6 5 10 = gL 10 0.5

= 40 kg, which is choice (a). 14. Energy stored in the rod is U=

1 F 2

L=

1 40 10 2 = 0.1 J. Thus the correct choice is (a). =

1 mg L 2 5

10–4

15. The common acceleration of the blocks is g (a) g (b) 3 (c)

2g 3

(d)

3g 2

4

10.20 Comprehensive Physics—JEE Advanced

16. The stress produced in the wire is (a)

mg A

(b)

2mg 3A

(c)

3mg 4A

(d)

4mg 3A

17. If m = 1 kg, A = 8 10–9 m2, the breaking stress = 2 109 Nm–2 and g = 10 ms–2, the maximum value of M for which the wire will not break is (a) 4 kg (b) 6 kg (c) 8 kg (d) 10 kg Fig. 10.20

SOLUTION If a is the common acceleration of the blocks and T tension in the wire. Then the equations of motion of blocks are Mg – T = Ma and T – mg = ma

the the

T 4 mg = , which is choice (d). A 3A 17. Breaking stress is the maximum stress the wire can withstand. From Eqs. (1) and (2) 2 mMg T = ( M m) Stress =

(1) (2)

15. Adding Eq. (1) and Eq. (2), we get a =

( M m) g ( M m)

Breaking stress =

For M = 2 m, we get a = g/3, which is choice (b). 16. From Eqs. (1) and (2), T = m(g + a) = m(g + g/3) = 4 mg/3.

T = A

2 mg . m A1 M max

Using the given values, we get Mmax = 4 kg, which is choice (a).

IV Assertion-Reason Type Questions In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each question has the following four choices out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Statement-1. (c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. 1. Statement-1 The length of an elastic string of initial length L is a metre when the tension is 4 N and b metre when the

tension is 5 N. The length of the string (in metre) when the tension is 9 N is (a + b – L) Statement-2 The extension of an elastic string is proportional to the initial length of the string. 2. Statement-1 Steel is more elastic than rubber. Statement-2 The Young’s modulus of steel is greater than that of rubber. 3. Statement-1 The stretching of an elastic spring is determined by the shear modulus of the material of the spring. Statement-2 For a given stretching force, the amount of stretching depends on the force constant of the spring.

Elasticity 10.21

4. Statement-1 Figure 10.21 shows that stress-strain curves for two different types of rubber. Rubber A rather than rubber B should be used as a car tyre.

Fig. 10.21

Statement-2 Rubber A dissipates larger amount heat energy than rubber B. 5. Statement-1 Two wires A and B have the same cross-sectional area and are made of the same material but the length of wire A is twice that of B. For a given load, the extension of A will be twice that of B. Statement-2 For a given load the extension of a wire is proportional to its length. 6. Statement-1 Two wires A and B have the same cross-sectional area and are made of the same material but the length of wire A is the twice that B. For a given load, the strain in wire A is twice that in B. Statement-2 For a given load, the extension in a wire is proportional to its length.

7. Statement-1 Two wires A and B have equal lengths and are made of the same material but the diameter of wire A is twice that of B. For a given load, the extension of B will be four times that of A. Statement-2 For a given load, the extension of a wire is inversely proportional to its area of cross-section. 8. Statement-1 Two wires A and B are made of the same material. The length of wire A is twice that of B but the diameter of A is half that of B. For a given load, the strain produced in B will be twice that in A. Statement-2 For a given load, the extension produced in a wire is directly proportional to its length and inversely proportional to the area of cross-section. 9. Statement-1 When a material is subjected to a tensile (stretching) stress, the restoring forces are caused by interatomic attraction. Statement-2 Restoring force is called into play in an elastic material by an inherent property of the material and not due to interatomic attraction. 10. Statement-1 When a material is subjected to a compressional stress, the restoring forces are caused by interatomic repulsion. Statement-2 The atoms of a material never repel.

ANSWERS 1. The correct choice is (d). If L is the initial length, then the increase in length by a tension F is given by FL l= r 2Y Hence

a=L+l=L+

and

b=L+ L

5L r 2Y

4L r 2Y

= L + 4c (1)

= L + 5c

(2)

. Solving (1) and (2) for L and c, r 2Y we get L = 5a – 4b and c = b – a. For F = 9 N, we have where c =

x =L+

9L r 2Y

= L + 9c

= (5a – 4b) + 9(b – a) = 5b – 4a 2. The correct choice is (a). 3. The correct choice is (b). If a spring is stretched, the total length of the wire of the coil and the volume of the wire, both do not change. Only the shape (or change. Hence the stretching of a spring is not determined by Young’s modulus or bulk modulus. It is determined by the shear modulus. 4. The correct choice is (c). The area of the hysteresis loop for rubber A much smaller than that for rubber B. This implies that rubber A dissipates a smaller

10.22 Comprehensive Physics—JEE Advanced

Hence the correct choice is (a).

amount of heat energy than rubber B. Consequently, tyres made of rubber A will not get heated to a high temperature. This prevents wear and tear of tyres.

8. Strain

4F 2

d Y

. Thus strain

1 d2

.

Hence strain in B will be four times that in A. Thus the correct choice is (d). 9. The correct choice is (c). When the material is not subjected to any stress, its atoms are in their normal (equilibrium) positions. When a tensile stress is applied, the distance R between atoms becomes greater than their equilibrium separation R0. For R > R0, the interatomic force is attractive and this force provides the restoring force under which the material regains its original shape and size when the stress is removed. 10. The correct choice is (c). When the material is subjected to a compressional stress, R becomes less than R0 and in this case the interatomic force is repulsive which causes the restoring force.

FL . Since the two wires are made of the AY same material, the Young’s modulus Y is the same. Since F and A also the same, L L. Hence the correct choice is (a). 6. The correct choice is (d).

5.

L = L

L=

L F = L AY Since F, A and Y are the same for the two wires, the strains in them are equal. 7. Area of cross-section A = d 2/4, where d is the diameter of the wire. Thus 4F L 1 or L L = 2 d2 d Y Strain =

V Integer Answer Type 1. A light rod AB of length 2m is suspended from the ceiling horizontally by means of two vertical wires as shown in Fig. 10.22. One of the wires is made of steel of cross-section 0.1 cm2 and the other of brass of cross-section 0.2 cm2. The Young’s modulus of brass is 1.0 1011 Nm–2 and of steel is 2.0 1011 Nm–2. A weight W is hung at point C at a distance x from end A. It is found that the stress in the two n wires is the same when x = metre. Find the value 3 of n. IIT, 1980

Fig. 10.22

2.

the strains in the two wires are the same.

IIT, 1980 3. A metal wire of negligible mass, length 1 m and cross-sectional area 10–6 m2 is kept on a smooth ball of mass 2 kg is attached to the other end of the wire. When the wire and the ball are rotated with angular velocity of 20 rad s–1, it is found that the wire is elongated by 10–3 m. If the Young’s modulus n. of the metal is n 1011 Nm–2 IIT, 1992 4. In Q.10 above, if the angular velocity is gradually increased to 100 rad s–1, the wire breaks. If the x. breaking stress is x 1010 Nm–2 IIT, 1992 5. A body of mass 3.14 kg is suspended from one end of a wire of length 10.0 m. The radius of the wire is changing uniformly from 9.8 10–4 m at one end to 5.0 10–4 m at the other end. Find the change in the length of the wire in mm. Young’s modulus of the material of wire is 2 1011 Nm–2. IIT, 1994 6. Steel wire of length ‘L’ at 40°C is suspended from the ceiling and then a mass ‘m’ is hung from its free

Elasticity 10.23

end. The wire is cooled down from 40°C to 30°C to regain its original length ‘L linear thermal expansion of the steel is 10–5/°C, Young’s modulus of steel is 1011 N/m2 and radius

of the wire is 1 mm. Assume that L >> diameter of the wire. Then the value of ‘m’ in kg is nearly. IIT, 2011

SOLUTIONS 1. If Ts and Tb are the tensions in steel and brass wires, then the stress in them will be the same if Tb Ts = Ab As

Ts Tb

As Ab

0.1 cm 2 0.2 cm

2

1 2

Since the system is in equilibrium, the moments of forces Ts and Tb about C will be equal, i.e. (see Fig. 10.23) Ts x = Tb (2 – x) x = 2(2 – x)

x=

Thus the value of n = 4

4 metre 3 ( Tb = 2Ts)

Given m = 2 kg, R = 1 m, = 20 rad s–1, A = 10–6m2 and L = 10–3 m. Substituting these values in (1) and solving, we get Y = 8 1011 Nm–2 Thus n = 8. 4. If the wire breaks at breaking stress is

max

mR 2 max 2 1 (100)2 Fmax 2 1010 Nm 2 . = A 10 6 A Thus x = 2. 5. Figure 10.24 shows a wire PQ of length L having radius r1 at end P increasing uniformly to a radius r2 at end Q. Consider an element AB of the wire of length dy at a distance y from end P. Let r be the radius of the wire at y and (r + dr) at (y + dy). As the radius increases uniformly from P to Q. dr = constant, say C dy where

C=

r2

dy =

dr C

r1 L

. Thus

Fig. 10.23

stress T Y AY The strain in the two wires will be equal if

2. Strain =

Ts Tb = AsYs AbYb Ts A = s Tb Ab

Ys Yb

0.1 0.2

2.0 1011 1.0 1011

1

Equating moments about C, we have Ts x = Tb (2 – x) which gives x = 1 metre ( Ts = Tb) 3. Let R = L = length of the wire = radius of the circle. The centrifugal force F = mR 2 produces the stress as at result of which the wire elongates. Thus

Y=

FL A L

mR 2 L A L

mR 2 2 A L

(1)

= 100 rad s–1, then the

Fig. 10.24

10.24 Comprehensive Physics—JEE Advanced

dl produced in the element of length dy due to a force F = Mg. Stress in the element = F/ r2 and strain in it = dl/dy. Hence stress Y= strain

F / r2 dl / dy

r 2 dl

dy =

l=

dl

=– =

F CY

r2 r1

F 1 = CY r r1 r2

F (r2 r1 ) CYr1r2

dr r2 F 1 CY r1

r2

r1 L

. Hence

MgL Yr1r2 Substituting the given values of M, L, Y, r1 and r2 and g = 9.8 ms–2, we have l=

Fdy

dr . Therefore, C Fdr Fdr Y= or dl = 2 r Cdl CYr 2 Integrating from r = r1 to r = r2, we obtain the total extension l produced in the wire by a force F which is given by But

Now F = Mg and C =

1 r2

l=

3.14 9.8 10.0 3.14

2 1011

9.8 10

4

5.0 10

4

= 10–3 m = 1 mm. 6. Change in length L = L T mgL mgL Also Y = L= A L YA Equation (i) and (ii) we get ( A = r2) m= =

TY g (10 5 )

= 3.2 kg

(i) (ii)

r2 (10) 3 kg

(1011 ) 3.14 9.8

(1 10 3 ) 2

Hydrostatics (Fluid Pressure and Buoyancy) 11.1

11

Hydrostatics (Fluid Pressure and Buoyancy)

Chapter

REVIEW OF BASIC CONCEPTS 11.1

FLUID PRESSURE 105 Pa

P0 P= SI

F A

11.5

GAUGE PRESSURE

–2 –2

11.2

P=

g g

PASCAL’S LAW

pressure

11.6 11.3

HYDR0STATIC PRESSURE

DENSITY AND RELATIVE DENSITY

P0 g

–3 –3

P = P0 +

–3

g

11.1 density density

e

g

11.4

ATMOSPHERIC PRESSURE

–2

SOLUTION 103

–3

g

11.2

Comprehensive Physics—JEE Advanced

4

10 Pa

103)

10

SOLUTION (

2

2

=5

104) 104 10 1 2

cal side =

10–4

1g

p1 = P0 +

sure)

104

A

2

p1 = 4

–3

1

–3

P2 = P0 +

104 P

1g

(P0

2g

P2 – P1 = ( 2 – P 2 – P 1) A = ( 2 – 1)

2

= (1500 – 1000)

104) 104

1)

5

10–4

11.4

11.2

OP A P

A

r

L O

O is P0

P L

SOLUTION

SOLUTION F = (P0 + P=

g) A +

F = P0 + A

g+

at

a distance r =

O

A

11.3

Fig. 11.2

2

=

2

=A

r = L/2 to r = L F=A

2

L

r =A L/2

=

Pressure at P = Fig. 11.1

2

r2 2

F 3 = A

L

= L/2 2 2

L

3

A

2 2

L

Hydrostatics (Fluid Pressure and Buoyancy) 11.3

L

11.5

A

= L A

acceleration a P1 A – P2

= ( L A) a

P1 and P2

–3 –3

A=

–3

P1 – P2 = La SOLUTION

(

1

–

2)

–

1

(i)

g = La 2

La g

=

(ii)

P1 > P2 tan

1

=

2

L

=

a g

11.6 A

°

Fig. 11.3

Pressure at A = P0 +

0 0g

Pressure at B = P0 +

g

0

0

+

0

0

+(

+

g

(ii)

= –

0)

=

a g

tan 30° =

= 1300

(

g

a g

a

tan 30° =

=

11.8

0

11.7

SOLUTION tan

0)

0

(i)

PRESSURE DIFFERENCE IN AN ACCELERATED LIQUID

3

–2

=

ARCHIMEDES' PRINCIPLE

‘

a Law of Floatation

11.9

APPLICATIONS V g

Fig. 11.4

V

11.4

Comprehensive Physics—JEE Advanced

Case

Vi

< V0 outside liquid and Vi inside liquid V g =V g Vi = V

and

V0 = V

(

V = V0 + Vi )

= V0 >

Case Case T a liquid ( > Case

V

and V1 in liquid 1 and V2 in liquid 2 < 2) 1< 1

T= V

g–V g

= V(

– )g

T =V (

– )g

2

Case Fig. 11.5

a g = g + a and T = V ( a (< g) g = g – a and T = V(

<

Fig. 11.7

– ) (g + a)

V g = V1 – ) (g – a)

Also

a = g and T = 0

T =V ( – ) g a T =V ( – ) g

Fig. 11.6

+ V2

(i)

2g

V1 + V2 = V

V1 = and

1g

V2 = V1

V

2 2

1

(ii)

V2 =

V

1 2

1

1 2

11.7

g = (g + a)

a T =V ( –

g = (g – a) T =0 Fig. 11.8

(a) is at rest

–2 –2

Take g

–2

Hydrostatics (Fluid Pressure and Buoyancy) 11.5

SOLUTION

SOLUTION Let

–3 –3

10

–4

3

3

V

W=

W)

= Vg

U

Vg

Fig. 11.10

(5 –

)

10

–2

–2 2

(5

10 )

10–2 k 10–2 U = (5 10–2 )3

k

1000

g (

Fig. 11.9

T+U=W

W+

T = W – U = Vg – Vg

=

(i)

10–2 )3

Fig.11.11

=k

10–2

10 –2

k

k

–1

– Vg 11.9

= ( – ) V (g + a) = (7000 – 100)

)

U 10–4)

= (7000 – 1000) g =g+a T = W – U = Vg

g

=

–2

g

=U +

W = (5

= ( – ) Vg

1000 g = (5 10–2 )3

10–4)

R

(10 + 2)

M is 1

g =g–a T = ( – ) V (g – a) = (7000 – 1000)

10–4)

(10 – 2)

SOLUTION

11.8 –3

constant k Take g

–2

Fig. 11.12

11.6

Comprehensive Physics—JEE Advanced

P1 =

1

g

P2 =

2

g

2

A

103) F = ( p2 – p1) A = ( =

2

–

1)

Ag

103g = 2)

103 103

2

g g

2

Ag = V g

V

11.12

11.10

radius R

V SOLUTION

Fig. 11.13

SOLUTION 103

V R2 )

=(

= g [V + R

g+V g 2

V=

ss = density

–3

103

11.11 103

–3

V 2

103

–2

=

10 3 2

1000

g

= 10–3 V 2 1000

g

SOLUTION 10 10 = 103 (a) Let

3

= 10–3

3

1

11.10 103g

1)

1

103) 103 g

3

SURFACE TENSION AND SURFACE ENERGY

g

Hydrostatics (Fluid Pressure and Buoyancy) 11.7 –1

–2

PQ

ABCD

AB and

L

CD PBCQ L PQ is

Fig. 11.15

SOLUTION F =2 L

=

Fig. 11.14

F=

F 1 5 10 2 = 2L 2 03 10–2

2L=2 L

11.11

PQ

W =F

PQ =

(2L

)=

–1

EXCESS PRESSURE

r is

P Q is A p=

A

2 r

–2

r is

11.13

3

p=

4 r

r in

10–5 p=

SOLUTION

2

=1

10–4

A=2 =

W 3 10 = A 2 10

5 4

2 P+ r = P0 +

2

10 –2

–4

2

10–2

inside

P

g+

2 r

–1

11.12

11.14

2 r

WORK DONE IN BLOWING A SOAP BUBBLE

r1 to r2 =2

(4 r22 )

(4 r21

11.8

Comprehensive Physics—JEE Advanced

or

(r22 – r21)

W

r is

(since r2 = r and r1 = 0) W r2

SOLUTION

is R = 2 r

r1 = r to r2 = 2r is W [(2r)2 – r2

4 r2 – 4 R2 = 32 r2 = 4 (2r)2 = 16 r2 Q= A = 16 r2 4 32 3 = R3 = r 3 3

r2 R into n r

R2 4 r2

n

Q=

W = 4 (nr2 – R2)

T 32 r2 = r3 s T 3

16

4 3 4 3 r n= R 3 3 or

T=

n1/3 r = R

3 2r s

11.17

W = 4 R2 (n1/3 –1) 11.15

–1 –3

105

SOLUTION

–1

P = P0 +

SOLUTION r=

R 2 A

2

g

105 +

R 2 R2 4 r

4

2 + R

2

0 05

1 10

105 Pa

2

11.13

ANGLE OF CONTACT

R2 R2 – 4 R2 = 4 R2

A 10–2)2

4

3

10–5 11.16 r s

Fig. 11.16

1200

Hydrostatics (Fluid Pressure and Buoyancy) 11.9

–2

SOLUTION (a)

2 cos r g

=

10

=

2

0 07

0 5 10

os 0

3

103

10

–2

NOTE

11.14

CAPILLARITY

and density

r is =

Fig. 11.17

2 cos rg

=

=2

cos 60

11.19 =

2 rg

SOLUTION

NOTE

1

2

11.18

= =

1 –1

r

1 1

cos

and

1

2

2

1

1

cos cos

2

r

1

2 cos

2

1 2

26 10 0

cos 0 cos135

=

26 10

1 0 707

Multiple Choice Questions with Only One Choice Correct z

=

=

I

1.

2

11.10

Comprehensive Physics—JEE Advanced

(a) A and B

7. (c)

(d)

eration a (< g

2.

A B L (a)

2 3

(a) –3

3.

La g

Lg a a g

(c) L 1

a g

(d) L 1

a and

4.

A Fig. 11.19

(a) (c)

A

a (d)

A

8.

a

5. 9.

g

k

(a) g (1 – k (c) g 6.

1

g (1 + k)

1 k

1 k

(d) g 1

10. A a (< g 1

(a) 2 Ag ( (c) Fig. 11.18

11.

1 2

2

Ag (

– 2

1)

–

2

2

1)

Ag ( 2

(d) r

1 4

2

–

Ag (

1) 2

–

>

1

2 1)

2

Hydrostatics (Fluid Pressure and Buoyancy) 11.11

(a) 12.

(c)

r 2 = 2r

=

17.

=r (d)

density

= 4r

>

r

2

2/ 1

2 r

4 (c) r

+

2 r

–

+

4 (d) r

–

(d) A and B

18.

(c)

is

2 A

4 A

(d)

B 2 3

19.

A

and (a)

14.

( > g

(c)

1

1 ( 2

20.

and

(d) g

15.

1

(a)

T

(c)

+

2

1 2 1

g

g

A and B is

2

(c) (a)

1/ 2

(c)

2

2

1 2

1

2

1 2

1 ( 2

1

+

2

1

1

2

r

T

r

1/ 2

1/ 2

T

(d)

T

and density is tied to a

22.

r

> (a) (c)

(d)

23.

(a) (c)

(d) 2

2)

1 2

(d)

1 2

+

1/ 2

T

16.

1 2

(d)

21. (a)

1 4

A

and area A

A

1

(c)

13.

(a)

is

1

(a) (a)

1

2

11.12

Comprehensive Physics—JEE Advanced

30.

24. 2

E=

31.

25. 32.

–3

26.

–3

–3

rises

33.

is

(c)

27.

2

(d)

34. (a)

28. 0

e–

(a) (c)

k is a con

2

2k

0g

k

(d)

2

2

and density

(a)

k (d)

g

r

or equal to

0g

0g

2

g

35.

0=

2

2

g

(c) 2 =

29.

2

2

(a) 2

0g

(c)

2k

2

2 g

g

(d)

2 g

36.

10–2

–1

)

Hydrostatics (Fluid Pressure and Buoyancy) 11.13 –5

–5

–7

–7

44.

r1 and r2 (r2 > r1)

37.

(a)

r1 r2 r2 r1

(c)

1 (r1 + r2) 2

38.

r1 r2 r2 r1

2

(c)

2

–1

–2

1 2

1

(d)

2

radius r R

(c) 40.

(a) g

D/ is ( (

1)

1) 1/ 3

1

(d)

1

1 s r

1 R

1 s r W

1 R

ion) 3 1 s r

1/ 3

(d)

3

1 s r

s its 1 R 1 R

47.

R

V is

V 2 W (d) (4 ) W R

(a) 2 W (c) (21/3) W

(a)

1/3

1

1

R

(c)

(a) R (c) 4 R

R R

R

(d)

48.

42.

1)

46.

(a)

41.

(2

D

(c) 39.

2

45.

(d) (r2 – r1)

–2

(a)

r 5r

r

43.

(a) R (c) 4 R

R R

49. a r1 and r2

a (a)

(a) 2 (c) 4

(r1 + r2) (r1 + r2)

(r1 – r2) (d) 4 (r1 – r2)

4 (c) 2 50. W

2 (d) 4 V

11.14

Comprehensive Physics—JEE Advanced

V W (d) 16 W

(a) 2 W W 51.

Fig. 11.21 2 4

6

54. R

52.

R

A

V M

densities

L L

H 2

and 2

H 2

A 5

Fig. 11.22

(a) Mg Mg – V g (c) Mg + R2 (d) g (V + R2 )

L 4

liq 3 2 5 (c) 4

4 3

(a)

55.

and

(d) 3d

Fig. 11.23

(a) Fig. 11.20

53.

decreases and increases increases and decreases and increase and decrease

56. R T P0 ABCD

Hydrostatics (Fluid Pressure and Buoyancy) 11.15

(a) | 2P0 P0 (c) | P0 R2 (d) | P0 R2

+ R2 – 2 RT| 2 +R – 2 RT| 2 +R – 2 RT| 2 +R + 2 RT|

r

Fig. 11.25

Fig. 11.24

57.

r

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55.

(d) (a)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56.

(d) (a) (a) (d)

(c) (c) (a) (c) (a) (c) (c)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57.

(c) (c) (a) (c)

4. 10. 16. 22. 28. 34. 40. 46. 52.

(c) (c)

(a) (d)

(c)

5. 11. 17. 23. 29. 35. 41. 47. 53.

(d) (c) (c)

6. 12. 18. 24. 30. 36. 42. 48. 54.

(a) (c) (d) (c) (a) (c)

(c) (a) (c) (d) (a) (d) (c) (c)

SOLUTIONS 1.

V – and

=V

g

(1)

– z =V

g

(2)

v=

2.

U W=V

V g

U

W

=

V

( g

a=

=

v=

a=

2

=

2 g 1 =

–1

2g V

g

V) V V

1 0 75 0 75

g

g=

2g 0 – 2g = 2

H 3. Let

1

=

/

g

/

g 3 H v2 – u2 = 2aS

g 3

H V

V=

–3

11.16

Comprehensive Physics—JEE Advanced

8.

is a=

ss V

=

v=

g V V 2

3

g

=

1000 g 250 g = =3g 250

2 3g 1 =

= 200 or

9.

V

6g

g=

k

H

k=

H H=

=

)

/k

a 1 k

1

= or

/k –

1 1 k

=

g= (

/

v2 6g = 2g 2g

4. Let

Vg =

a =g 1

F=

1 k

10. is F = A 5.

1 ( 2

A

2

–

1

1 ( 2

g

6.

1 ( 2

W=

= 7. Let p1 and p2 A and B p 1 – p 2= A

W= g

(

F = (p1 – p2) A = F= F=

= (1)

gA

2

–

1 2

2

( 2 – 1) A g

1 4

Ag(

2

–

1)

1 ( 2

V=

2

–

1)

–

1)

A

1 ( 2 – 1) 2

2

11.

r

= AL L ALa

p1 = p2 = 0

(2) =

La g

p=

p1

2

p2

1)

Hydrostatics (Fluid Pressure and Buoyancy) 11.17

or

p=

/2 area

=

2

2 r

=

2

=

or

g =

g

2

p1 r2 = r

or

r2 or

=

r2

T =2

=2

g

=r

1/ 2

=

12. 13.

r

p=

r

=

r=

/2

g

T

2

T

g

16.

2 r

/ /

g=

=2 =

F = 14. Let

=

=

W=

17.

V

= V W=

–

T=

2 A

=

U=

Vg

Vg

Vg

F =U – W = ( –

V

) Vg

= V

U

uid is

U =V g

a=

F

Vg

=

V

F = W – U = ( – ) Vg

=

g

is a= =

F

v=

Vg

=

2

(i)

1

V 2

v=

g

2

(ii)

2

a 15.

or

2 1

F = ( – ) Vg =

18. Let V1 (

=

V)

a = = g

=

1

=

2

1 A and

1

11.18

Comprehensive Physics—JEE Advanced

1

1 4

V1 g = 1

V1

r2

g

0

0

3 1 = 4

or

= r2 g = r2 g =( r

B 2

V2 g = 1

=

2

2V2 3 3 4

g or

2

2 3

=

=

3 = 2

=

1

=

2

3

2

3

n

3

n

2

1

2

1

=

2

= P0

1 2

1 2

=

P1V1 = P2V2 or (P0 +

2V1 =

2 1

=

2

V2

P2 = P0 V2 = 2V1

n

2

1

g

P2

1

2

2

1 2

2

P0 0 75 = g

1

2

V1 V2

1V1

=

V1 = V2 = V

g

=

r

22.

+

is

=

2

r = 2r

r r2

=2 r

1 ( 2

=

2

=2

1

+

2

25.

r2

=

=

(2r)2

r

26.

2 =

2

=2

V g=

or

=

=

r2 is

or

V 2 13 6

g+ 0 2

27. and

=

R

2 V2

V1 and V2

=

40 3

g p = 2 /R

V1 V2

21.

g) V1

24.

1

20. =

/2

P0

1 1

/2 =

P1 P1 = P0 +

total ss total lu 1

2 )

23. Let V1

19. =

2

2

28.

2 cos r g

V 2

g –3

Hydrostatics (Fluid Pressure and Buoyancy) 11.19

=(

) g

r r2 = r 21 + r 22 or r =

as

=

e–

=

0

=

0g

e–

(1)

33.

P

0g

=

0g

0g

k

k

2

=

PE =

1

4

2

(

= r2

g

g

or

=

2

2

2

2

g

gr = 2

r=

2

=

=

or

= 2r)

is

k

g

=

2

2

0g

(e– – e–0) =

g

4

g

34.

0

2

=

g

W =2

is 2

2

e

29. Let

W=

r22 W) =

= r2

=

0

=–

(4 0

=

= 0 to P=

3 0) 2

r12

2

PE =

1000 2

35.

30.

=

2g

2

=

(

g )2 2g

2

(2)

2

2 g

liquid is P1 =

P2 =

r =

(2 )

2

g

2

=

r g 2

P1 = P2 31.

r or

32. 4 4 and P2 = P1 = r1 r2

air r1

36. F = 2 =2

r g 2

=

2

r=

10

g W = FS = 2 10 7

2

37. Let r

r2

P1 V1 + P2 V2 = PV or

4 r1

4 3 r1 3

4 r2

4 3 4 r2 = 3 r

4 3 r 3

(1)

4 r1

–

4 r2

=

4 r

(1)

11.20

Comprehensive Physics—JEE Advanced

r1 r2 r2 r1

r= 38.

2 r 2

39.

1 2

and g 10 2 2

W =4

R (n

or

=

3

r

1/3

1) = 4

R

= 103

4 T= 3

Q=

R

3

s T

(2)

40. W

r2 W

3V 4

42. Tension = 43. Force = 2 r1

(R3

r 3)

or

(R3

r 3)

or

3

(R

r2 2

R=4

r ) = R3 1/ 3

1 A

R =

47.

=

0

or v

g 2 g = g

4 3

or

r3 = (10)

( + 10) = 10

=g 1

R g R

2

2

=g 0

2

4 3 125 64

a = r2

49. v2 =g 2

g 1

P1 V1 = P2 V2

=g 1 v

2

48.

or ( + 10)

v

g

=1–

=g 1 or

= R3

0

0

=g –

R3

0

3

= g

4 3

g=

V2/3

W

r=2 r + 2 r2 = 2 (r1 + r2)

2

r

3

46. Let

2/3

and

=

2

r3

44.

v

(R

4 3

R = r

L=2 R L=2

41.

0

4 3 R3

T 4 V= 3

1 2

4 3

(1)

R = n1/3 r)

(

1

3

R 1 r

2

2

45. Let

2 = r g

g or

v=

r=

a/

a

5r 4

3

0

g

Hydrostatics (Fluid Pressure and Buoyancy) 11.21

2 cos r g

=

2

=

cos

FB

a g a is

a r is

50. W =4

r

=

2

4 3

V=

=

r3

L 4

1 10

2 3 20

3L 4

A 5

g+ =

g

1 4

W = FB 2/3

3V 4

W=4

A 5

or

1 1 ALDG = 5 4

or D =

5 4

V

is V

W =4

53.

2/3

V

4 2/3

3V 4

=4

2/3

51. Let r

and

=4W

R

V = 106

6

a F T F

)

4 r3 4 R3 = 3 3

R3 = 106 r 3 or R = 100 r

r is

S = 4 r2 Fig. 11.26

E1 = 4 r 2

54.

106

E2 = 4 R 2 E2 = E1 52.

R r

2

density A = 5

1 106

L

=

100 r r

2

1 106

=

1 D= ALD 5 W) =

1 ALDg 5

1

55.

= Mg +

g = Mg + R2

g

102

56.

in ABCD

F

1

ABCD 2

due

11.22

Comprehensive Physics—JEE Advanced

57.

ABCD

1

= P0

g

2

p

2 r AB

2

=T

2R = 2RT

F = P0

g

2

+R

2

2

= [2P0

– 2RT – 2RT

II

1.

–1

Multiple Choice Questions with One or More Choices Correct 6.

T–2

2. 7.

3.

4.

8. (a) rises

5.

A B

9. (a) rises

A B A A B

10.

1 r

Hydrostatics (Fluid Pressure and Buoyancy) 11.23

(a) PA = PB (c)

1

>

PC = PD (d)

2

15.

1

2

2

1

L M

11. Mg n is Mg 1

12. 16.

1 n

loaded is L1 = L – (d) L1 = L – /n

1 2

103 (a) 13.

–3 1

and

2

103

are

1

1

2

2

–3

17. P is F

W1

(a) P P and F

reads W2

14.

(a) W1 (c) W2 densities 1 and PC and PD

W1 W2 2

F

18.

a

PA PB A B C and D F U

Fig. 11.28 Fig. 11.27

A until

11.24

19.

Comprehensive Physics—JEE Advanced

(a) F = 2a (c) U = 2a

F=a (d) U = a and

2

aera A 1

liquid are

1 and

and

=

2

2

and

g 2sin << a

=

ga 2

<< a

=

ga

2

Fig. 11.29

(a)

=

1

=

1

1

1 –

1

1

+

Fig. 11.30 2

2

1 2

2

A and B A and

21.

1

B

F

1

F= 1

cylinder is a =

A

F

A

F

(d)

2 1

20.

(a) (c)

2

2

a

(a)

Fig. 11.31

g = 2 cos

ANSWERS AND SOLUTIONS

1. 2. 3. 4. 5.

B

F

A

B

F

Hydrostatics (Fluid Pressure and Buoyancy) 11.25

C

AC is due to V

B 6. 7. Let M and

(2)

V =

(1)

V=

V = V V >V

V =

M V
s

( =

V =V

Vs = s

9.

V = V + Vs

( 1

=

1 2

1

+

2

(2)

M

V =(

s

V
s

>

1

+

2)/

1/ 2/ s 1

V = A

s

B

= V g or V =

>

1

1

s

V
AB V

2 s

con

8.

s

10.

(1)

11. g 10

12. (a) Let Fig. 11.32

1

10 = 103 = 10–3

3 3

11.26

Comprehensive Physics—JEE Advanced 3

1)

PC = PD 103 g

1)

1>

2

1<

103

10–3 1)

g

g

103

2 2

2

W =V

2

103

1000 g

Mg – g

Y=

13.

F L A L

Hence MgL1 Y= A L L1

–3

ss density

72 7 2 103

3

= 10–3

3

V 2

Y=

V 2

1000

Mg 1

1 L n 1

A L

1

(2)

L 1= L –

= (

V = 10–3

3

) =

14. Pressure at A and B

17.

P0 g and PD = P0 +

2

2

g

18.

in air in

loss lo lo

=

1 1

L = (L – L1 – )

16.

g

1 g = 2

PC = P0 +

L = (L – L1) (1)

1 and F = Mg 1 n

g =

g

1 Mg = Mg 1 n n

and F =

V=

g=

M

L1

2

103

1

Mg n

= n=

g=

2

V

2

2)

=

M

g g

1

–

1

or

2

V =

15.

is

=

1 1

Hence P and F

r

=

iquid

=

=6

Hydrostatics (Fluid Pressure and Buoyancy) 11.27

is

= 2a g

F=

= 2a

tan sin

19.

(a

) (

=

2

=a

1

+

2)

A g=

1 1A

+

=

a

<< a sin

=

ga 2

=

2 2A

(2)

a

21. Let V

T A is UA = FVg B is UB = FVg A is WA = AVg B is WB = BVg A and B

1

equal to in 2

1 1 2

=

is F

2

M=(

1+

UA = T + WA and UB + T = WB

2) 2g

F Acceleration a = = M

1

=T + and FVg + T = FVg

2

2

=

1

F=

2

Hence 20. Let

(i) (ii)

AVg BVg

T + Vg

F>

A

A

W= =

Fig. 11.33

cos

or or

2 sin 2 sin

Fig. 11.34

sin

B

=W = g = 2 sin

(1)

2

F

=

A

+

T B

=

T + Vg

F

B

>

F

11.28

Comprehensive Physics—JEE Advanced

III Multiple Choice Questions Based on Passage 3.

Questions 1 to 4 are based on the following passage

P

Passage I

5

as

P P = P0 e

1.

P

/

0

0)

P0

and

0 0 0 0

0

LT

0 is

L T0

Fig. 11.35

0 0

T

2.

/

n

1 2 (d) 2

(c) 1

n is

4. (a) (c)

= =

0

e(2)

0

(d)

SOLUTION 3. P

1. 0

2.

4. P = P0

=–

P0

e

/

0

=–

0

0

k k P= = V ( = /V)

PV = k

=–

P0 = P0 e 2

P

k 0

Questions 5 to 10 are based on the following passage Passage II

1 = e 2 2= e e(2)

/

/ /

0

0

= 0

=

0

e

0

=2 0 = 0e–1/2

Hydrostatics (Fluid Pressure and Buoyancy) 11.29

9.

5.

10. 1

and

2

1 and

2 is

6.

7.

Fig. 11.36 –3 3

3

3

3

8.

(a)

=

1 ( 2

=2

R r

1

1

+

+

(c)

=

1 (2 3

(d)

=

1 ( 3

2 1

1

2)

+

2)

+ 2 2)

SOLUTION 5.

or

=

5

=

30 30 5

7. 6. 3

1

is

g = ( – 10)

or

g = ( – 5) = ( – 5)

264

g

– 15 or 1

=

3 3

8. Let let g

its density and

=

4 (R3 – r3) 3

11.30

Comprehensive Physics—JEE Advanced

and

= =

4 3 r 3 R3

1 V 3

r3

(i)

r3 V=

V g=

4 R3 3

1 V 3

or

=

3

Let (1 –

V or

+

g = V 0g or

0

4 ( R3 3

4 3 r 3

r3 )

(R3 – r3) + r3

or

= /

=

=

( (

) 1)

1=

( (

) –1 1)

(1 (

) 1)

r3 R3 r R

or

3

3

r3 r

3

0

=

=V =

0

4 R3 3

= R3

and

R3

or

+

/

or

=

1 3

0

or

= 1 3

3 + (1 – ) = (1 – )

1 3

(iii)

)

10.

1 0 3) 2 4 1) V and V g

9.

+ (1 – ) =

(ii)

) 1)

g

0

= (1 = (

g + (1 – ) V

V g=

24 03

1

4

(3 )3 g = (3 )2 =

its den

1 (2 3

2

2 1

+

1g

+ (3 )2

2)

Questions 11 to 13 are based on the following passage Passage III H 4r liquid is

r B H

1

h1

h2

+

4r B H 2r

Fig. 11.37

h

2g

Hydrostatics (Fluid Pressure and Buoyancy) 11.31

11.

4

(a)

1

2 3 5 (c) 3

2 3

(c)

3 4

(a)

(d)

3 5

13.

2

(d) 2

12.

2

=

2

=

2

=

2 2

SOLUTION

11.

=4 r =A = 4A

(4r)2 = 16

W

r2

2

3A =

4A

3

Liquid level

A – A = 3A h2

Fig. 11.38

1

+ ) 3A –

W= U =W g(

1

3

4 + )3A – 4 g 1A = 3 5 1 = 3

1

W

U

U = F1 – F2 = g(

4 5 U equals

12.

(2r)2

2

3

Fig. 11.39

4A

4A

2

13.

=

4 2

g

IV Assertion-Reason Type Questions I 1. Statement-1

Statement-2

11.32

Comprehensive Physics—JEE Advanced

5. Statement-1

2. Statement-2 n

Statement-2 6. Statement-1

n Statement-2

Statement-2 3. Statement-3 7. Statement-1

R is

2 /R

R is 4 /

Statement-2 Statement-2 8. Statement-1 4. Statement-1

us R

n

P0

4 R

and P0 Statement-2

n Statement-2

SOLUTION 1.

5.

2. g 6.

g 3.

4.

7. 8.

.

g n

= P0 +

2 + R

g

12

Hydrodynamics (Bernoulli’s Theorem and Viscosity)

Chapter

REVIEW OF BASIC CONCEPTS 12.1

vt

VISCOSITY

r

r2 g

2

vt =

9 vt

F= –

A

r2

dv dx dv dx

A

–2

Fig. 12.1

[ML

–1

12.2

–1

NOTE

STOKES’ LAW

F= 6

12.3

-

v

r

rv

12.4

POISEUILLI’S FORMULA r

TERMINAL VELOCITY

p Q= F

W–U

W

U

l

pr4 8 l

12.2 Comprehensive Physics—JEE Advanced

Capillaries Connected in Series l1

l2

r1

r2 p

R F= R = R1 + R2 =

8 l1

8 l2

r14

r24

2

v 15 = x

–1 3

v x

A

–4

12.2 3

Q = Q1 = Q 2 = p1

p R1

–3

–5

R2

p2 P = p1 + p 2

SOLUTION

Capillaries Connected in Parallel p

vt =

1 1 = R1 R

1 R2

8 l1

R1 =

r 14

R=

R2 =

R1 R 2 R1

R2

=

2 r2 g 9 2

12 1 9

8 l2

–2

r1 p R1

p R2

–1

r vt –5 –3

8

4 r3 3

R

=

4 R3 3

R = 2r 2

SOLUTION 2

A

–1

–1

2

12.4 v x

–2

-

SOLUTION r

–1

–1

–1

–1

-

–2

18 1

-

12.1

–4

98

5

12.3

r2 Q2 =

3 2

r 42 F=6 =6

Q1 =

3

–1 –3

= 15

12.3

h1

–2

g

h2

SOLUTION –3 –3

V

m W mg = V g U V g F=W–U a=

F F = = m V

–

V

Vg

Vg V Fig. 12.3

g

=

A B

1 2 s = ut + at 2 1 a t2 2

h

2h = a

t=

h1 – h2

g

a 2h

h1 – h2 a g

AB F = a p d m = adx

1/ 2

x1

x1

1/ 2

2 1

=

p1 = h 1 g p2 = h 2 g p = p1 – p2

Fe =

2

dm x

2

=

= 12.5

a

xd x x2

x2

1 2

2

a x12

x 22

= –3

2

h1 – h2 = x1 g

–2

x2

2g

x12

x 22

2

2

2

2 1

]

–1

12.6 –3

3

–1

–3

–2

Fig. 12.2

SOLUTION AB dx

x

SOLUTION l

r

–2

–2

12.4 Comprehensive Physics—JEE Advanced

Q=

=

p=

3

3 14 1 3

12 3 14 1 12

12.8

1 3

the total energy of an incompressible and non-viscous liquid in a streamline

6

3 –1

energy being the sum of pressure energy, potential energy and kinetic energy of the liquid.

8 lQ

1 mv 2 2

PV + m g h +

r4 2

p

12.5

BERNOULLI’S THEOREM

P+

1 2

gh +

v2

=

m V

STREAMLINE OR LAMINAR FLOW 12.9

VELOCITY OF EFFLUX H h

12.6

CRITICAL VELOCITY AND REYNOLD’S NUMBER Fig. 12.4

v vc =

V

k r

A PA +

r k

B

1 V2 + 2

gH = PB +

1 2 v + g H–h 2 AV = a v

PA = PB = P A a k

P +

k k

1 2

a 2v 2 A

2

+ gH = P +

1 2 v + g H–h 2 1/ 2

12.7 a1

EQUATION OF CONTINUITY OF FLOW

2g h

v =

a2 A2

1

a2 v1

v2 A>>a v =

a 1 v1 = a2 v 2

2g h

av

vice versa

t=

H h g

R = vt = 2 h H

h

12.5 NOTE

SOLUTION H 2

h = R

=H

h

v

h

=

k r –1

=

12.7

1 1

2

–1

12.10 SOLUTION

–2

g 3

31 4 1

Q

SOLUTION

3

v =

–2 2

Q = A

3 14 1 3 14 1

25

–1

3 –1

r2

A=

2 1

4

3 14 1

=

2gh =

2

12.11

4 2 2

h = PQ

–1

A

a

12.8 –1

SOLUTION a 1= r 21

r 22

a2 =

Fig. 12.5

a 1 v 1 = a 2 v2 v2 =

=

= 12.9

a1 v1 = a2 r1 r2 1

r 12 r 22

SOLUTION V

v1

2

v1

1 1 V 2 = PQ + 2 2 PP = PQ = P

PP + 2

–1

–1

gh +

P +

gh +

AV = a v v =

g

1 1 V2 = P + 2 2

V 2 = v 2 – 2g h –2

v

P AV = a v

AV a

Q

1 2 v2

v2

12.6 Comprehensive Physics—JEE Advanced 1/ 2

2 g ha 2

V=

A2

AV = aA

av

Q=

a2 1/ 2

2g h A

2

a

= av = a 2gh

F = Qv = Q 2gh

2

12.12

12.13

A

-

a

a = 2

h

Q v Q A >> a av

2

L vQ

agh

Q 2gh

O Fig. 12.6

SOLUTION

SOLUTION

v= v

A >> a

2gh

Q a Q2 a

F = av2 = av av

3

36

Q v = av2

=6 O=F

–5

3

L 2

av2

F

=

Q L a 3

2

F = av = a

=

2gh = 2a gh v=

5 2

6 1 2 1

14

4

–3

2gh

I Multiple Choice Questions with Only One Choice Correct 1.

T g

T 2 T

–2

2 2 2.

3.

T 2 T 4 -

12.7

v v v3

4. 10.

v

v2 v4 r v

v 2 v

r

v

v 4

v

5.

v 2 v

h

v

11.

V vt V

R h g h g

vt 2 vt

h g h g

6.

vt vt

12.

A

L L

a

L

h

-

ghA gha

W1

ghA gha

7. r P

W2

t1 < t2 ; W1 > W2

t1 > t2 ; W1 < W2

t1 = t2 ; W1 = W2

t1 > t2 ; W1 = W2

13.

Q

r

P

Q Q 4

t1

t2

14.

Q Q 8

8.

A l

B

r

l P

-

A

Q 15.

R

R

l

-

P

Q

2Q 3

Q 3

3Q 4

16.

V P d

l d

9.

P a

12.8 Comprehensive Physics—JEE Advanced

h1

V 4 V 16

V V 8

h2 h1 h2

17. –1 –4

2

23.

d

–2

g –4

2

–5

2

–5

2

–5

2

h

R d

18. L

2

y R

y

R h

4 R2 h

R2 4h

h2 R

24.

R L

A

L

2

A

19.

-

A

L 2

L

B

25.

H

m

x y

h R h

R= 2 h H 20.

mg x y

R h H

R=

R= h

h H

R= 2 h H

mg y x

h

y x

mg 1

h

mg 1

26.

R H h= 4 H h= 2

r

H h= 3

r p

h=H

l

21.

1/4

r

r r

r 27. g

22.

–2

h

2

H

B B

x y l

12.9

2 1

3 2

R= V

2 3

1 2

R=

28.

2h g

aV A

g 2h

R= V

2h g

R=

AV a

2h g

A a m t ga A

gA a

2ga A

gA 2a

Fig. 12.7

32.

29.

2

3mg 4A

2mg A

mg A

mg 2A

r

r2 r4

30.

r3 r5

A 33.

r

a

h

a A g

h 2

–1

–1

–1

r3 r5

r r4

–2

–1

34.

31.

h A

–2

g

a V

R

2

–1

4

–1

5 6

–1

5 8

–1

5

5

ANSWERS

1. 7. 13. 19. 25. 31.

2. 8. 14. 20. 26. 32.

3. 9 15. 21. 27. 33.

4. 10 16. 22. 28. 34.

5. 11. 17. 23. 29.

6. 12. 18. 24. 30.

12.10 Comprehensive Physics—JEE Advanced

SOLUTIONS 1. W1 = mgh

V

gh =

P1 – P2 =

ghV

1 2

v 22 – v 12

1 2 3 = 2

v

P = W2 =

P

V = h gV

W = W1 + W2 = 2 h gV P= t=

=

2.

W t

= 6.

W 2 h gV = P P 2

3

3

3

h1

– v 2]

v2

3 2

2gh = 3 hg

h2

3

v1 =

av2

av1 2 gh 2 m1 = av1

3

2

2 gh 1

v2 =

m2 = av2

h v =

p1 = m1 v1

2gh

p2 = m2 v2 v 2

va =

gh 2

1 2 gh 2

F m 1v 1 – m 2v 2

V V T = Av a

2V

=

av 12 –

=

a

av22

2 gh1 – a

2gh2

= 2 ag h1 – h2

A gh

A

=2 a h V V = 2

T

=

2V / 2 A gh / 2

h h = 2

V A gh

Pr 4 ; 8 l l

7. Q =

l r

T 2

Q

Pr4

P

Q 8. A

2

3.

4.

–1

=

ML 1

1

B

]

Pa = 2P

Pb = P l

Q = Qa + Qb

r r

r v

5. v

v =

2gh

2

Pr 4 = 8 l 1 1 = l 3l

2P / 3 r 4 8 2l 1 3l

P /3 r 4 8 l

-

12.11

Q =

mgh = mgL

3l 2

l =

Pr 4 3l / 2

8

Pr 4 8 l

2 3

13.

2Q 3

l x 2

lx

9.

l

V

l

l 2 x g d x = l2 g

W= o

W t

P=

W V

V t

= l2

=

e 1 2 v 2

=

1 = 2 P 10.

g

l2 2

l4 g 14 = 2

=

xd x o

98 2

av 14.

av

-

3

v3 a 1v 1 = a 2v 2

r 21v1 =

v2 r2 = 12 v1 r2

r2 r 2

2

r 22v2 15.

=4

m

v2 = 4v1 = 4v 11. gr 2

2 vt = 9 k=

2 9

V=

4 3

r = vt = k

= mg = kr2 6

rvt = mg –

vt W–U F = 6 rvt

W

r3

g

4 r3 3

g

= mg 1

r3

3 4

-

-

g

3V 4

2

4 r3 U = 3

r

4 r3 3 m

2/3

2/3

m

vt

rv t

V2/3

1 r

V Vt = k

3 4

2/3

V [

12.

v 2

2/3

16.

= 4vt 2/3

= 4] -

v 2 v

t1 > t2

V= V

d4 l d V 2

P r4 8 l l

Pd4 128 l

r= l

d 2

12.12 Comprehensive Physics—JEE Advanced

V V

V

17. –1

v1

–4

a1

v 1a 1 = v 2 a 2

2

v2 v2

h v22 = v21 + 2 gh –1

v2 a2 =

v1 a1 v2

4

–5

2

18.

Fig. 12.8

V1 v a = 1 1 V2 v2 a2 v1 =

v1 L v2

2 gy 8 gy

R

R

2

R = vt

v2 =

2gy

v1 = v2

2

1 2

R=

1 L2 V1 = 2 R2 V2

19.

t 2 H h 2 g

2g h

20. R

h H

h dR dh

h d2 R

1 L2 2 R2

V1 = V2 R =

v

2g 4 y

d h2 hH – h2

R

1/2

L

h

2

dR =2 dh

h

=

v

1 hH – h2 2 H 2h hH

v=

2g h

t 1 2 gt 2

1 2 H–h= gt 2 t=

2 H h g

H – 2h

1/ 2

dR dh H – 2h H h = 2 d 2R

S=H–h

S=

h2

–1/2

h

d h2 h

h d 2R dh

2

=–

2h H

h

1/ 2

1

1 H 2h 4h H h

12.13

h= H

P1r14 P2 r24 = 8 l1 8 l2

2

d R d h2

=–

2H

P1 l = 1 P2 l2

h H /2

R h=H

R

r2 r1

4

1 2

=

24 = 8

25.

21. 1 2

-

v2 = p = gh

r 4 F= 3

v 2g 2

h= v=r =r

2

h=

26.

2 2 2

r

g r

r x

y g=

y x

y g = mg 1

p R4 p r4 = 8 l 8 l R4 = r4 + 16r4

22.

m x x

R

2

g h

2

3

1/4

R

p 2r 8 l

4

r

27. h R=2 h H

=

h H 2

h1

h1

1

H 2

h2

2

h1

H h1 = 2 2 h1 = h2

23.

v t=

H 2

h2

H 2

2

=

v2 v1

A1 A2

= 2v2 A1v1 = A2v2 A1 1 2 A1 2

h

h2 A

2 gd

m = Ah

2h g

Fig. 12.9

vt

R =

2 gd

2h = g

v=

4dh

2gh av

2

R2 = 4 dh

d=

P1r14 8 l1

Q2 =

24. Q1 =

1v1

h2

28. H 2

=

= v H 2

H 2

e

R 4h

av av v = av2

P2 r24 8 l2 Q1 = Q2

t

12.14 Comprehensive Physics—JEE Advanced

e

=

AV = av AV v = a

av2 a v2 = Ah Ah

a 2 gh Ah 2ga = A

v

=

t

2 gh

t R = vt 1 h = gt2 2

29. h =h v = h= v =

2g

h 4

gh 2 R = v

m A gm 2A

30.

2h g

t=

2gh

R=

v V

AV a

2h g

2h g

32. a v = AV av V= A

vt =

r2g

2 9

P = f vt

r

=6

Fig. 12.10

P P +

1 2

1 v2 + gh 2 v2 = V 2 + 2g H – h

V 2 + gH = P +

rvt

r vt × vt

P= 6

=

f=6

r2g

2 9

8 g2

2

2

r5

r5

33. mg

v = 2

2g H a A

1 v 31.

h

2 98

2

2 2

= 25

rv = mg

6 rv

–1

v

v m=

4 3

r3

6

rv =

4 3

r3 g

12.15

v2 –

h

2 r2 g 9

v= gh

ag + hk

k

ag

h

-

v=

2gh

h

-

=

k

h

r4

k

h

h=

v2 h = 2g h

34.

ag + hk kag

ag = h

hk a

2 1

8

64

–1

5

h

II Multiple Choice Questions with Two or More Choices Correct 1. 6.

2. 3.

7.

8.

4. 9.

OP a

L O P

5.

-

O L P

p v

12.16 Comprehensive Physics—JEE Advanced

3 p= 4

3 p= 8

2 2

L

3

p2 – p1

2 2

L

4 4

v=

3 L 2

3 L 2

v=

14. h

10.

–1

–2

g

–1

R –1

5 2 11.

PA vA vA

A

vB

PB

PA vB

vA

A B A B PA = PB

-

B

PB vB

Fig. 12.12

15.

-

Fig. 12.11

r

12. v1

p1 v1 > v2 p1 > p2 13.

–1

–3

v 2=

ANSWERS AND SOLUTIONS 1. 2. 3.

pc

Fig. 12.13

ps –1 2

3

h

p2

v1 < v2 p1 < p2 v1

p2 – p1

,

v2

pc > ps h=

r2 2 2g

pc < ps h=

r2 g

2

12.17 2

dF = mr

-

2

=a

rdr

F l r

L–l

r=

L v

av

L

F= a

a

2

-

L

= -

1 a 2

l

2

[L2

L

L

L – l 2] = a

l=L

4.

r2 2

2

rdr = a

F=

2

l

l

L

l 2

2 2

3a

L

8

-

5. A

B A

a1

v1

Fig. 12.14

p1

a2 v2 B

p2

P

1 2 1 2 v1 = p2 + v2 2 2 1 p1 – p2 v22 v12 2

v=

v1 > v2 10. 11.

6. p r4 8 l

Q=

F 3 = a 8

1 2 3 v =p= 2 8

p1 +

a2 < a1 a1v1 = a2v2 p2 < p1

p=

2 2

L

v

2 2

L

3 L 2

vA < vB 1 1 2 PA + v2A = PB + vB 2 2

-

vB > vA pB < pA

7.

12. rv

F= 6 8.

vt =

r2g

2 9

=

13. p = p2 – p1 =

9. dr m = a dr

r

O =

1 2 p

1 2

v21 – v22 2

2

3

12.18 Comprehensive Physics—JEE Advanced 3

4

14.

D2 ; D 4

A=

vs = r r

a=

d2 ;d 4 AV = av

v= t=

AV a

2h g

D

ps +

2

V

d2

2

–1

ps

vc vc

2 1 25

1 2 1 2 v = r 2 2 ps h

pc – ps =

R = vt = 5 t

vx = v v = gt

–1

pc

2

–1

v2x +vy2 1/2

2

= 5 2

+

2

]

pc – ps =

–1

15. 1 p+ 2

1 2 1 2 v s = pc + vc 2 2 pc

gh =

v2

1 2 r 2

gh

2

h=

r2 2 2g

III Multiple Choice Questions Based on Passage Questions 1 to 4 are based on the following passage Passage I

F=6 1.

F=6

rv rv

–1

–1

–1

–2

–2

–1

–2

–2

2.

-

terminal velocity r

v F

-

12.19

3.

4. -

ANSWERS 1. 2. 3.

mg

vt = -

r

6

4.

Questions 5 to 7 are based on the following passage Passage II

f

-

f F F 6.

-

dv F=– A dx dv dx

A

–1

–2

–2

–1 –1

] –2

]

7.

–1 –3

5.

–1

–2 –4

f

–2 –4

F

–2

–4

–2

–4

–2

SOLUTION 5. -

2

ML

=

1

2

L

= [ML–1

–1

]

L

7. 6. =

F A d v dx

F = A

Questions 8 to 11 are based on the following passage Passage IV

dv = dx a

–3

2 =4 5

–4

–2

A m

A.

h

12.20 Comprehensive Physics—JEE Advanced

8.

10. agh

Ag 2a

Agh

2ag A

agh

Agh

a mg A

A mg a

gh 2

gh

2a mg A

2A mg a

3gh 4

3gh 2

11.

9.

SOLUTIONS 8.

10.

P=h g F

P

a

h ga =

v=

2gh

2a mg A

e

av = av

2a g A

av v = av2 av2 = a

11. h =h

2gh = 2 agh

9.

v

agh = 2a mg A

m = Ah

=

Questions 12 to 14 are based on the following passage Passage V

gh 2

h

2 gH

13.

A

R

H

a

12.

h 4

g H2 + h2

2gh 2g H

h.

2g

2gh

hH

2hH

2 hH

h2

A h a 2g

A 2h a g

A H a 2g

a 2H A g

14.

Fig. 12.15

H2

1/2 1/2

]

=

12.21

SOLUTION 12.

vx =

A dx/dt

2gh -

vy =

x

u2 = v2 + 2 as

2gH

2g x = – A

a v=

v 2x

v 2y =

2g H

dt = –

h

13. 2H g R = vx

dx dt Ad x

a 2 gx

=– t=

A

A a 2g

x–1/2 dx

t x=h t=

2H g

2gh

x=h

2 hH

t= –

14.

-

2g x

= –

a =

= avx = a 2g x Questions 15 to 17 are based on the following passage Passage VI

2A a 2g

h 4

h/4

h 1/ 2

h1/ 2

A h a 2g

1 gh 2 3 gh 2

a1 = 3a2

15.

h

16.

h. a2

x–1/2 dx

x1 / 2 =– a 2 g 1/ 2

t

a1

a 2g

h/4

A

x

vx =

A

2 gh 2 2 gh

17. 2gh

gh

gh 2

1 gh 2

2h g 6

2h g

4

h g

8

2h g

SOLUTIONS 15.

v1

a 1v 1 = a 2v 2 v1 =

v2

a2 v v2 = 2 a1 3 A

P + gh + Fig. 12.16

1 2

1 2 v2 2 v12 = v22 – 2gh

v12 = P +

B,

12.22 Comprehensive Physics—JEE Advanced

v1 =

1 gh 2

h=h t=–

3 16. v2 = 3v1 = gh 2 17. h

=–

dh dt dh 2 dt = – = – v1 g

v1 = –

g

h

1/ 2

dh

h

2

2 h

g

4

h g

18. 2 2 gH

2 gH

2 gH

gH

d

H/2 a

2

h–1/2 dh

Questions 18 to 20 are based on the following passage Passage VII A d

h

19.

A

x

h=H

H

H

2

H 2 20.

H 2 2 h x

Fig. 12.17

2H 3

3H 2

3H 4

H 2

SOLUTION 18.

v1

v2 =

h = H v2 Av1 = av2

v1 =

a v2 A

a << A, v2

P +

1 2 dv1 + dg 2

t=

a

H 2

+

d g

=P +

1 2

v1 v2 =

19.

3 H 2

1/ 2

2h g

x = v2 t =

H 2 d v22+

2h g

h = H

1/ 2

3 H 2

= [h H – 4h d gh

gH

2h g

2h g 1/2

x =

H 2

12.23

20.

h

Hh – 4h2 x

x h

–1/2

dx dh

x h

xm h=

d [3Hh – 4h2]1/2 dh 1 [3Hh – 4h2]–1/2 2 H – 8h

3 H 8 h

H – 8h h=

xm =

3 H 3H 8

12 H 8

1/ 2

=

3 H 4

3 H 8

IV Assertion-Reason Type Questions 4. Statement-1

-

Statement-2

not 5. Statement-1

1. Statement-1 Statement-2 Statement-2

6. Statement-1

2. Statement-1 Statement-2 Statement-2

7. Statement-1

3. Statement-1 Statement-2 Statement-2

12.24 Comprehensive Physics—JEE Advanced

8. Statement-1

9. Statement-1

Statement-2

10. Statement-1

Fig. 12.18

Statement-2

Statement-2

SOLUTION 1.

rv

F=6 6. 7.

-

8. 2. P 3.

P 9. 10.

-

Av

4. 5.

V Integer Answer Type 1. 2

g -

–2

12.25

SOLUTION 1.

vc p

1 2 v 2

pc – ps = pc

pc – ps =

vs = r

1 2 1 2 vs r 2 2 ps h

2

gh

r gh =

1 2 r 2

=2 n ps +

1 2 1 2 v s = pc + vc 2 2 ps vc

Fig. 12.19

pc

h=

4

2 2 2

r n 2g

2

h=

r2 2 2g

n 2

4 1 2 1

2

2

13

Simple Harmonic Motion

Chapter

REVIEW OF BASIC CONCEPTS 13.1

SIMPLE HARMONIC MOTION

Simple harmonic motion (SHM) is the simplest kind of oscillatory motion in which a body, displaced from its stable equilibrium position, oscillates to and fro about the position when released. If the displacement (x) from the equilibrium position is small, the restoring force (F) acting on the body is given by F =–kx where k is a positive constant, known as the force constant. In the SI system k is expressed in newton per metre (N m–1). The acceleration (a) of the body is given by F k a = =– x=– 2x m m k m Thus acceleration (a) = – constant displacement x. The resulting motion is called simple harmonic motion. Thus, a simple harmonic motion is a motion in which the acceleration (i) is proportional to the displacement from the equilibrium position and (ii) is directed towards the equilibrium position.

where m is the mass of the body and

13.2

=

CHARACTERISTICS OF SHM

The displacement x in SHM at time t is given by x = A sin (

t+ )

where the three constants A, and characterize the SHM, i.e., they distinguish one SHM from another. A SHM can also be described by a cosine function as x = A cos ( t + )

We will use the sine function. A cosine function is equally valid. (1) Amplitude The amplitude of SHM is the maximum (positive or negative) value of the displacement from the equilibrium position. Quantity A function) is the amplitude of SHM. (2) Time period The smallest time interval during which the motion repeats itself is called time period or simply period of SHM. The angular frequency is related to time period T as 2 T Frequency of SHM is the number of complete oscillations completed in 1 second. =

Thus,

=

1 = T 2

or

=

2 T

Since

= =

=2

k , we have m 1 T

1 2

Angular frequency the sine or cosine function.

k m t in

(3) Phase The quantity ( t + ) is called the phase of SHM at time t; it describes the state of motion at that instant. The quantity is the phase at time t = 0 and is called the phase constant or initial phase or epoch of the SHM. The phase constant is the timeindependent term in the cosine or sine function.

13.2 Comprehensive Physics—JEE Advanced

13.3

VELOCITY AND ACCELERATION IN SHM

The velocity V or the particle in SHM is given by V = But

sin ( cos (

t+ ) =

dx =A dt

1/ 2

x2

=

A2

V =A =

t+ )

x A

t+ ) = 1

Hence velocity

cos (

1/ 2

x2

1

(A2 – x2)1/2 d2 x dt

2

sin (

t+ )=–

2

x

Notice that, when the displacement is maximum, i.e., when x = |A|, the velocity is zero but the acceleration is maximum = | 2 A|. But when the displacement is zero (x = 0), the velocity is maximum = | A| and the acceleration is zero.

13.4

ENERGY IN SHM

1 mA2 2. 2 1 (3) E is constant = mA2 2 maximum =

1 mA2 2

2

cos 2

t

At any displacement x from the equilibrium position, 1 m 2 ( A2 x 2 ) K.E. = 2 x2 because cos2 t =1 – sin2 t =1– 2 . A At that instant t, the potential energy of the oscillator is given by P.E. = because k = m displacement x,

2

P.E. =

1 2 1 kx = m 2 2

and x = A sin 1 2 1 kx = m 2 2

Total energy of the oscillator is E = K.E. + P.E.

t 2 2

x

2

A2 sin 2 ( t

)

. In terms of

2

for all values of x.

Figure 13.1 shows the variation K.E., P.E. and E with t for the case = 0. 1 2

Kinetic 2

Potential

Total energy

2

Energy

1 mV 2 2

1

(1) At the mean position, x = 0, K.E. is maximum = mA2 2 2 and P.E. = 0 (2) At the extreme position, x = A, K.E. = 0 and P.E. is

At any instant of time t, the kinetic energy of the oscillator is given by K.E. =

2

NOTE

2

=–A

1 mA2 2

)

Thus, although the kinetic energy and potential energy of a simple harmonic oscillator both change with time t and displacement x, the total energy is independent of both x and t and hence remains constant. This is expected because friction has been neglected.

A2

The acceleration a of the particle in SHM is given by a =

1 ) sin 2 ( t mA2 2 cos 2 ( t 2 1 = mA2 2 2 E is also obtained as follows: E = K.E. + P.E. 1 1 = m 2(A2–x2) + m 2x2) 2 2 =

0 4

2

3 4

time

Fig. 13.1

The energy of an oscillator may decrease with time not only by friction (damping) but also due to radiation. The oscillating body imparts periodic motion to the particles of the medium in which it oscillates, thus producing waves. For example, a tuning fork or a string produces sound waves in the medium which results in a decrease in energy. 13.1 The displacement x of a body varies with time t as x = a sin (ct) + b cos (ct) where a, b and c are constants.

Simple Harmonic Motion

(a) Show that the motion of the body is simple harmonic. (b) Find the amplitude A, time period T and phase constant of the motion in terms a, b and c.

(a) Given x = a sin (ct) + b cos (ct) (i) Differentiating (i) w.r.t. time t, we get velocity dx = ac cos (ct) – bc sin (ct) dt The acceleration is given by dV d a= ac cos(ct ) bc sin(ct ) dt dt = – c2 [a sin (ct) + b cos (ct)] V=

a = –c2x [use Eq. (i)] Since a (–x), the motion is simple harmonic. (b) Since the motion is simple harmonic; x = A sin ( t + ) = A [sin ( t) cos + cos( t) sin ] x = (A cos ) sin( t) + (A sin ) cos( t) (ii) Comparing (i) and (ii), we get A cos = a A sin

(iii)

=b

(iv)

2 2 =c T= T c Squaring and adding (iii) and (iv), we get and

=c

a2

A=

b2

tan

b = a

= tan

–1

b a

13.2 Which of the following functions of time represent (a) simple harmonic, (b) periodic but not simple harmonic, and (c) nonperiodic motion? Find the period for each periodic motion. Here k is a positive real constant. (i) sin kt + cos kt, (ii) sin t + 2cos 2 t + 3 sin 3 t,

(vi) log kt.

(i) sin kt + cos kt represents simple harmonic motion since we can write it as

3

, (iv) cos kt + 2 sin2kt, (v) e–kt and

2 sin

kt

4 t in the 4 2 argument of the sine or cosine function = , T 2 2 or T = , where T is the period. Hence k = T k 2 . i.e. the period of the function is k (ii) Each term in (sin t + 2cos 2 t + 3sin 3 t) represents SHM. The period T of the term sin t is given by 2 = or T = 2s. The period of the term T 2 cos 2 t is 1 s, i.e. T/2 and the period of the term 3 sin 3 t is 2/3 s, i.e. T/3. The sum of the three terms, however, does not represent SHM; it rep2 cos kt

term completes one cycle, the second term completes two cycles and the third term completes three cycles. This shows that the sum represents a periodic motion with period T = 2s. (iii) cos 2kt

represents an SHM whose period

3 T is given by 2k = or

Dividing (iv) by (iii) we get

(iii) cos 2kt

SOLUTION

or

SOLUTION

13.3

T=

2 T k

(iv) cos kt + 2 sin2kt = cos kt + (1 – cos2kt) = 1 + cos kt – cos 2kt. The period of cos kt is 2 T T= and that of cos 2kt is = . These k k 2 2 two terms together have a period T = k as explained in (ii). The other term 1 is a constant independent of t and hence does not affect the period of the sum. Hence (coskt + 2sin2kt) represents a periodic motion with period 2 T= . k (v) e–kt decreases monotonically to zero as t . It is an exponential function with a negative exponent of e, where e 2.71828. It is non-periodic.

13.4 Comprehensive Physics—JEE Advanced

(vi) Function log (kt) increases monotonically with time. Therefore, it never repeats itself and is a non-periodic function.

SOLUTION The displacement of the particle is given by x = 2 cos 0.5 t

13.3 An oscillator has a time period T = 4s. If it is oscilenergy. SOLUTION The kinetic energy of a simple harmonic oscillator 2 varies with time as sin2( t), where = . T 1 1 Now sin2 ( t) = (1 – cos 2 t) = (1 – cos t), 2 2 2 2 T = 2 T = . where = 2 T T 2 T = 2 s. Hence K.E. varies periodically with period 2

or

T=

2

=

2 0.5

a=– 2x –1.0 = – 2 0.01 = 10 rad s–1 Velocity when displacement x = 0.01 m is 2

2 1/2

V = (A – x ) = 10 [(0.05)2 – (0.01)2]1/2 = 0.49 ms–1 Vmax = A = 0.05 10 = 0.5 ms–1 13.5 The displacement x (in centimetres) of an oscillating particle varies with time t (in seconds) according to the equation x = 2 cos

0.5 t

3 Find (a) the amplitude of oscillation (b) the time period of oscillation (c) the maximum velocity of the particle (d) the maximum acceleration of the particle.

-

= 4s

(c) Maximum velocity Vmax = | A | = 2 0.5 =

cm s–1

= 3.14 cm s–1 (d) Maximum acceleration amax = | – 2 A | = 2A = (0.5 )2

T The P.E. also varies periodically with period . 2

SOLUTION

cm

tion, we compare this equation with x = A cos ( t + ) (a) Amplitude A = 2 cm (b) Angular frequency = 0.5 rad s–1

NOTE

13.4 A body oscillates harmonically with amplitude 0.05 m. At a certain instant of time its displacement is +0.01 m and acceleration is –1.0 ms–2. Find (a) the velocity of the oscillator at this instant and (b) the maximum velocity.

3

2

2

cm s–2 2 = 4.935 cm s–2 =

13.6 A particle executes SHM of amplitude 25 cm and time period 3s. What is the minimum time required for the particle to move between two points located at 12.5 cm on either side of the mean position? SOLUTION Let A and B be the two extreme positions of the particle with O as the mean position. Displacements to the right of O are taken as positive while those to the left of O are taken as negative (Fig. 13.2)

Fig. 13.2

Let the displacement of the particle in SHM be given by x(t) = A sin ( t + ) (i) 2 2 = rad s–1 T 3 Let us suppose that at time t = 0, the particle is at extreme position B. Setting x = A at t = 0 in Eq. (i) we have where

A = 25 cm and

=

Simple Harmonic Motion

A = A sin giving = /2. Putting = /2 in Eq. (i), we get x(t) = A cos t (ii) Now let us say that the particle reaches point C at t = t1 and point D at t = t2. At C, the displacement x (t1) = + 12.5 cm and at D, it is x(t2) = – 12.5 cm (see Fig. 10.2). So from (ii) we have + 12.5 = 25 cos t1 and – 12.5 = 25 cos t2 or cos t1 = + 0.5 or t1 = /3 2 and cos t2 = – 0.5 or t2 = 3 2 Hence (t2 – t1) = – = 3 3 3 2 T t2 – t1 = = T 3 6 3 or (t2 – t1)min= = 0.5 s 6 4 . This Notice that cos t2 = – 0.5 even for t2 = 3 value of t2 does not correspond to the minimum time because this is the time at which the particle, moving to left, reaches A and then returns to D. 13.7 A particle is executing linear simple harmonic motion of amplitude A. At what displacement is the energy half kinetic and half potential?

SOLUTION The block will not be detatched from the platform, if the frequency of platform’s SHM is such that the maximum acceleration of the platform equals the acceleration due to gravity, i.e. A 2max = g

1 m 2

2

max

=

max

=

1 m 2

2

x2

2

3.13 2 3.14

0.5 Hz.

SOLUTION (a) As long as the string remains taut, the restoring force will be proportional to displacement from the mean position. The motion of the body is simple harmonic whose amplitude is A and angular frequency is k m

kA m (b) If the tension in the string is T, the downward acceleration is The maximum acceleration is

2

A=

T m For the string to remain taut T > 0. Therefore, the maximum downward acceleration cannot exceed g, i.e. amax = g a =g –

A2 – x2 = x2 x=

max

(a) the maximum downward acceleration of the oscillating body and (b) the maximum amplitude for which the string remains taut.

=

(A2 – x2) =

9.8 = 3.13 rad s–1 1.0

g A

13.9 A body of mass m is attached by a string to a suspended spring of spring constant k. Both the string and the spring have negligible mass. The body is pulled down a distance A and released. Assuming that the string

SOLUTION The energy will be half kinetic and half potential at a value of displacement x when K.E. = P.E., i.e.

13.5

A 2

13.8 A horizontal platform is executing SHM in the vertical direction with an amplitude of 1.0 m. A block of mass 5 kg is placed on the platform. What is the maximum frequency of platform’s SHM so that the block is not detatched from the platform?

2

Amax = g Amax =

g 2

=

mg k

2

k m

13.6 Comprehensive Physics—JEE Advanced

13.10 At time t = 0, the displacement of a simple harmonic oscillator from the mean position is x0 and the velocity is v0. Obtain the expressions for the amplitude and phase constant of the oscillator in terms of x0, v0 and where is the angular frequency of the oscillator. SOLUTION For a simple harmonic oscillator x = A sin ( t + )

(i)

v=

From (iii) and (iv), we get

and

A=

x02

tan =

x0 v0

v02

a=

1/ 2

k x m

F = m

(i)

Since a (–x), the motion of the block is simple harmonic. Comparing Eq. (i) with a = – 2x, we get 2

dx = A cos ( t + ) (ii) dt Putting t = 0 and x = x0 and v = v0 in (i) and (ii), we have (iii) x0 = A sin (iv) and v0 = A cos and

The block is pulled to the right by a small distance x from the equilibrium position and released. The restoring force on the block is F = – kx. The acceleration of the block is

=

k m

Time period T =

k m

= 2

=2

m k

(2) Vertical Oscillations of a Mass-spring System Consider a massless spring suspended from a support. [Fig. 13.4 (a)]. A block of mass m is attached at the lower end, as a result, the string extends by an amount d given by [Fig. 13.4 (b)]. F = kd mg = kd (i) This is the equilibrium state of the system.

2

= tan

1

x0 v0

NOTE If at t = 0, x = x0 and v0 = 0, then A = x0 and

. On 2 v the other hand, if at t = 0, x = 0 and v = v0 then A = 0 and = 0

13.5

=

EXPRESSIONS FOR TIME PERIOD OF MASS-SPRING SYSTEM

(1) Horizontal Oscillations of a Mass-Spring System Consider a block of mass m placed one horizontal frictionless surface and attached to a spring of negligible mass and spring constant k as shown in Fig. 13.3.

Fig. 13.4

When the body is pulled through a distance y from this position and released [Fig. 10.4 (c)], the restoring force is F = – ky and the acceleration of the block is a=

F k =– y m m

Hence the motion is simple harmonic whose time period is T= 2

Fig. 13.3

m k

(ii)

Equation (i) determines k. Time period given by Eq. (ii) is the same as for horizontal oscillation. It depends only on m and k and is independent of gravity.

Simple Harmonic Motion

(3) Parallel Combination of Springs Figure 13.5 shows the equilibrium state of a block connected to two springs which are joined in parallel. If the block is pulled down through a distance x, the extension produced in each spring will be x. The restoring forces in the springs are F1 = – k1 x and Fig. 13.5 F2 = – k2 x. Total restoring force F = F1 + F2 = – (k1 + k2)x = – (kp) x where kp = k1 + k2 is the effective force constant of the parallel combination. The time period is given by m =2 kp

T= 2

m (k1

k2 )

(4) Series Combination of Springs Figure 13.6 shows the equilibrium state of a block connected to two springs which are joined in series. The block is pulled down by a distance x. Let x1 and x2 be the extensions produced in the springs. The restoring force in each spring will be the same equal to F = – k 1x 1 = – k 2x 2 so that Total extension

or

x1 = –

Fig. 13.6

= –F

1 k1

= –F

k1 k2 k1k2

F= –

k1k2 k1 k2

1 k2

x = – ks x

k1k2 is the effective force constant of the k1 k2 series combination. The time period is

where ks =

T= 2

m =2 ks

m(k1 k2 ) k1k2

(5) A Block Connected between Two Springs Figure 13.7 shows the equilibrium state of a block connected between two springs

Fig. 13.7

If the block is displaced through a distance x, say to the right, the spring k1 is extended by x and spring k2 is compressed by x so that the restoring force exerted by each spring on the block is in the same direction (along the left). If F1 and F2 are the restoring forces, F1 = – k1x and F2 = – k2 x, the total restoring force is F = – (k1 + k2)x = – kx where k = (k1 + k2) is the effective force constant of the system. The time period is T= 2

m =2 k

m k1

k2

13.11 If a spring of force constant k is cut into two equal halves, what is the force constant of each half. SOLUTION

F F and x2 = – k1 k2

x = x1 + x2

13.7

If a force F produces an extension x in the spring, then F = kx (i) Since the extension produced by a force is proportional to the length of the spring, if a spring is cut into two equal halves, the same force F will produce an extension x = x/2 in half the spring. If k is the force constant of half the spring, x (ii) F= k x =k 2 From Eqs. (i) and (ii), we get k = 2k. 13.12 Two blocks, each of mass m, are connected by a spring of force constant k and placed on a horizontal frictionless surface as shown in Fig. 13.8. Equal force F is applied to each block as shown. Find time period of the system when the force is removed. SOLUTION

Fig. 13.8

When a force F is applied at each end of a spring, every coil of the spring is not elongated. The coil at point O in the middle of the spring is not elongated. This situation is the same as two springs each of length l/2 (where l is the length of the complete spring) joined to each other at point O. If k is the force constant of the complete spring, the force constant of each half = 2k. Hence

13.8 Comprehensive Physics—JEE Advanced

T= 2

m 2k

13.13 A spring has a natural length of 50 cm and a force constant of 2.0 103 Nm–1. A body of mass 10 kg is suspended from the spring. (i) What is the stretched length of the spring? (ii) If the body is pulled down further stretching the spring to a length of 58 cm, and then released, what is the frequency and amplitude of oscillation. Neglect the mass of the spring.

The radius of the circle along which the trolley moves is r = 40 cm = 0.4 m When the table is rotated, the tension in the spring is equal to the centripetal force, i.e. mv 2 = mr 2 r = 2.0 0.4 (10 )2 = 790 N

T=

Now, extension produced in the spring by this force = 40 – 35 = 5 cm = 0.05 m

SOLUTION Natural length of the spring = 50 cm = 0.50 m k = 2.0 103 N m–1, m = 10 kg (i) The extension produced in the spring is mg 10 9.8 y= = k 2.0 103 = 0.049 m = 4.9 cm Stretched length of the spring = 0.50 + 0.049 = 0.549 m = 54.9 cm (ii) The frequency of oscillation is given by =

1 2

k m

1 2.0 103 2 10 = 2.25 Hz

Force constant of spring =

force extension

790 N 0.05 m

= 1.58 104 Nm–1 1.6 104 Nm–1 13.15 A tray of mass M = 12 kg is supported on two identical springs as shown in Fig. 13.9. When the tray is depressed a little and released, it executes an SHM of period 1.5s. (a) Find the force constant of each spring. (b) When a block of mass m is placed on the tray, the period of the SHM becomes 3.0s. Find the value of m.

The length of the spring in the equilibrium position = 54.9 cm. If it is pulled to a length of 58 cm, the maximum displacement from the equilibrium position = 58 – 54.9 = 3.1 cm, which is the amplitude of oscillation. 13.14 A small trolley of mass 2.0 kg resting on a horizontal frictionless turntable is connected by a light spring to the centre of the table. The relaxed length of the spring is 35 cm. When the turntable is rotated at a speed of 300 rev/min, the length of the spring becomes 40 cm. Find the force constant of the spring.

Fig. 13.9

SOLUTION (a) The time period of oscillation of M is given by M 2k

T =2

SOLUTION Mass of trolley (m) = 2.0 kg –1

300 = = 60

Frequency of rotation ( ) = 300 rev. min 5 rev. s–1 Angular frequency ( ) = 2 = 2 5 = 10 rad s–1

where k is the force constant of each spring. Therefore k=

2

2

M

T2

=

2 (3.142) 2 12

(1.5) 2 = 105.3 N m–1

Simple Harmonic Motion

(b) The new period T1 is now given by M m 2k

T1 = 2 so that m + M = or

m + 12 =

2

2

(3)2 105.3

2 (3.142) 2 m = 36 kg

or

13.6

T12 k

= 48

EXPRESSIONS FOR TIME PERIOD OF SOME OTHER SYSTEMS

(1) A Ball Oscillating in a Concave Mirror A small spherical steel ball is placed a little away from the centre of a concave mirror whose radius of curvature is R. When the ball is released, it begins to oscillate about the centre. Place a small steel ball at A, a little away from the Fig. 13.10 centre O of a concave mirror of radius of curvature R (= OC = AC) as shown in Fig. 13.10. Let ACO = . If m is the mass of the ball, its weight mg acts vertically downwards at A. This force is resolved into two rectangular components: mg cos (which is balanced by the reaction of the mirror) and mg sin (which provides the restoring force F). Thus F = – mg sin = – mg (since =

mg x R

(

is small, R being very large)

x = R ; x being the arc OA)

= – Kx where force constant K = mg/R. Thus the motion is harmonic and the angular frequency is given by = T= 2

k = m

g R

R g

(2) Oscillation of a Liquid in a U-tube The column of the liquid is displaced through y by gently blowing into the tube (Fig. 13.11). The columns exhibit

13.9

vertical oscillations. Let L, A and be respectively, the length of the liquid column, area of cross-section of the tube and density of the liquid. We shall neglect viscous effects. Since the right-hand side column is higher by 2y, with respect to the column on the left-hand side, the mass of this column of liquid is m = 2A y. The restoring force (which is a gravitational force) is given by F = – mg = – 2A gy = – Ky where the force constant K = 2A g. The angular frequency of the harmonic oscillation is K M where M = AL is the total mass of the liquid in oscillation. Thus =

=

Fig. 13.11

2A g AL

2g L

Oscillations of a liquid column

The time period of oscillation is T =2

L 2g

It is interesting to note that the period of oscillation does not depend on the density of the liquid or the area of crosssection of the tube. (3) Oscillation of Floating Vertical Cylindrical Body A cylindrical piece of cork of height h and density c a liquid of density l. The cork is depressed slightly and released. Let A be the cross-sectional area of the cork and M its mass. Fig. 13.12 (a) shows the static equilibrium, the weight of the cork being balanced by the Fig. 13.12 weight of the liquid it displaces. If the cork is depressed through a distance x, as shown in Fig. 13.12 (b), the buoyant force on it increases by l Agx, because l Ax is the mass of the liquid displaced by dipping, g being the acceleration due to gravity. If viscous effects are neglected, the restoring force on the cork is given by

13.10 Comprehensive Physics—JEE Advanced

F= –

l

Agx = – Kx

where K = l Ag. Since F – x, the motion of the cork is simple harmonic. The time period of the motion is M K where M is the mass of the cork = Ah c. Hence

If I is the moment of inertia of the system about the axis passing through O and perpendicular to the plane of the rod, the angular acceleration is =

T= 2

Ah c l Ag

T= 2

h g

2

Time period T = 2

T is independent of the mass of the bob.

restoring torque = – mgL sin = – mgL

=

mgL I

Thus

T= 2

where

I=

Thus

T= 2

If M << m, T = 2 pendulum.

mgL I

=

I mgL

ML2 1 + mL2 = (M + 3m)L2. 3 3 (M

3m) L 3mg

L , the same as that of a simple g

(6) Horizontal Oscillations of a Cylinder-Spring System A solid cylinder of mass M and radius R is connected to a spring of force constant k as shown in Fig. 13.15. Case (a): Cylinder slips without rolling. The total energy of the system Fig. 13.15 is translational K.E. + P.E. If x is the instantaneous displacement and v the velocity of the cylinder, the total energy is E=

NOTE

(5) A Compound Pendulum A compound pendulum is a rigid body capable of oscillating about an axis. The pendulum consists of a rod of mass M and length L which is pivoted at O and carries a bob of mass m at the other end as shown in Fig. 13.14. The rod is displaced and released.

2

l

L g

I

since – , the motion is simple harmonic whose angular frequency is (compare with = – 2 ) given by

c

(4) A Simple Pendulum A simple pendulum consists of a massless inextensible O and having a small bob at the other end (Fig. 13.13). When the bob is displaced from equilibrium position A to a position B and released, the component mg cos of its weight balances with tension T and it returns under a Fig. 13.13 restoring force F = – mg sin x If is small, sin . Also = L mg x F= L g F x. Acceleration a = = L m

mgL I

1 1 Mv2 + kx2 2 2

(i)

If friction is neglected, E = constant. Hence

dE = 0. dt

dE = 0, we Differentiating (i) w.r.t. time t and setting dt get dv dx 0 = Mv + kx dt dt 0 = Mva + kxv ( Fig. 13.14

(for small )

a= T= 2

k x M M k

a=

dv dx ,v= ) dt dt

Simple Harmonic Motion

Case (b): Cylinder rolls without slipping In this case, the total energy of the system is 1 1 1 Mv2 + I 2 + kx2 E= 2 2 2 =

Setting

E= Setting

I 1 1 Mv2 1 + kx2 2 2 2 MR

dE = 0, we get dt dv 0 = Mv dt

I

1

0 = Mva 1

MR I

kx M (m ) 2 2

= –

x

M 2

m

=

dx dt

v

k k m

1\ 2

T= 2

k

1 This the general formula. For a solid cylinder, I = MR2 2 and then we get T= 2

dv =– dt

M 2 If the pulley has a mass M << m, then

x

I R2

M T= 2

Acceleration a =

M dv dx v + kx 2 dt dt

m

T= 2

I R2

M

dE = 0, we have dt

dx dt

+ kxv

MR 2

k

a= –

2

+ kx

1 1 M v2 + kx2 2 2

1 m 2

0= v R

13.11

3M 2k

(7) Vertical Oscillations Mass-spring System Connected by a Pulley The block is pulled out through a small distance x and released. If M is the mass of the pulley and R its radius, the total energy of the system is [Fig. 13.16]

k m

(8) A Block Attached to Three Springs A block of mass m is connected to three identical springs as shown in Fig. 13.17. The block is pushed towards spring 1 through a small distance so that spring 1 is compressed by x and springs 2 and 3 are extended by x2 = x3 = x cos 45° = x / 2 . When the block is released, the restoring force acting on it is F =

F1

F2

F3

Fig. 13.17

The magnitude of F along the vertical direction is F = – (F1 + F2 cos 45° + F3 cos 45°)

Fig. 13.16

E= where

1 1 mv2 + I 2 2

1 I= MR2 and 2

2

+

1 2 kx 2

v = . Thus R

=

kx kx2

=

kx

= – 2 kx

kx 2

1 2 kx 2

kx3

1 2

13.12 Comprehensive Physics—JEE Advanced

Acceleration

a=

F = m

2

x. Hence

IMPORTANT TIPS AND ADDITIONAL FORMULAE

(i) Mass m suspended from a spring of force constant k

m T= 2 k (ii) If a spring of spring constant k is cut into n equal parts, the spring constant of each part becomes n k. If each part is loaded with a mass m, the time period of each will be m T= 2 nk (iii) If two springs of spring constants k1 and k2 are connected in parallel, the spring constant of the combination is k = k1 + k2. If the combination is loaded with mass m, then 1

m

T= 2

k1

2

k2

(iv) If two springs of spring constants k1 and k2 are connected in series, the spring constant of the combination is given by

1 1 = k1 k in which case T=

1 k2

m k1 k2 k1k2

1

2

where

=

T= 2

LC

(x) Simple pendulum immersed in a liquid

T= 2

l g

where g = g 1

= density of liquid, = density of bob. (xi) Simple pendulum in a lift moving up with acceleration a

T= 2

l g

a

If the lift is moving down with acceleration a (< g)

T= 2

l g

a

If the lift is falling freely a = g and T = , frequency = 0. (xii) Simple pendulum in a trolley moving with acceleration a in horizontal direction.

T= 2

l g

where g =

(xiv) For a seconds pendulum T = 2s. Hence l = 99 cm 1 m. (xv) An extended body pivoted at point P at a distance h from its centre of mass (Fig. 13.18) and oscillating in the vertical plane.

I mgh

(vi) Simple pendulum of length l

l g (vii) Liquid column of length L in a U-tube T= 2

L 2g

a2

l g sin

m1m2 is the reduced mass. m1 m2

T= 2

g2

(xiii) Simple pendulum in a trolley moving down an inclined plane of inclination

T= 2 k

M Ag

(ix) LC circuit consisting of inductance L and capacitance C

T= 2

(v) If two masses m1 and m2 and connected to the ends of a spring of force constant k, the time period of the oscillations of the system is

T= 2

(viii) A pole of mass M and area of cross-section A in a liquid of density

T= 2

m 2k

T= 2

13.7

2k x =– m

Fig. 13.18

Simple Harmonic Motion 13.13

where I = moment of inertia of the body about the pivot. (a) If the body is a rod of mass m and length l (Fig. 13.19) pivoted at its end, then l ml 2 h= and I = . 3 2 Hence

2l 3g

T= 2

Fig. 13.19

1

Fig. 13.21 (c)

Tc = 2

m 4k

Fig. 13.21 (d)

Td = 2

m k1 k2 4k1k2

(xvii) Horizontal oscillation (without slipping) of a disc, ring, cylinder or sphere of mass M and radius R rolling on a horizontal surface (Fig. 13.22).

2r g

For a disc pivoted at a point on its periphery,

4m k

l

h = r and I = mr2 + mr2 = 2mr2. T =2

Tb = 2

l

(b) If the body is a ring of mass m and radius r pivoted at a point P on its periphery (Fig. 13.20), then

Hence

Fig. 13.21 (b)

T= 2

Fig. 13.20

I R2

M k

3

h = r and I = mr2 + mr2 = mr2. 2 2 Hence

T= 2

13.8

3r 2g

13.9

FORCED OSCILLATIONS AND RESONANCE

external periodic force are called forced oscillations. The external force maintains the oscillations of a damped oscillator. The amplitude of these oscillations remains constant. If the frequency of the externally applied force is equal to the natural frequency of the oscillator, resonance is said to occur. If damping is small, the amplitude of resonant oscillations will become very large. At resonance, the oscillator absorbs maximum energy supplied by the external force.

Fig. 13.21

Ta = 2

DAMPED OSCILLATIONS

force are called damped oscillations. Due to damping the amplitude (and hence energy) of the oscillator keeps on decreasing with time and eventually the oscillator comes to rest. Damping also decreases the frequency of the oscillator.

(xvi) Some cases of oscillation of spring-mass-pulley system. The mass of pulley is negligible.

Fig. 13.21 (a)

Fig. 13.22

m k

I Multiple Choice Questions with Only One Choice Correct 1. A uniform rod AB of length L is pivoted at one end A and hangs vertically. The time period of small

oscillations about an axis passing through A and perpendicular to the rod is

13.14 Comprehensive Physics—JEE Advanced

(a) 2

L g

(b) 2

(c) 2

L 3g

(d)

6. A body is executing simple harmonic motion of time period T. Its kinetic energy varies periodically with time period equal to T (b) T (a) 2

L 2g L 2g

2. The displacement of a particle executing simple harmonic motion is given by x = a sin t + a cos t The total energy of the particle is 1 m a2 2 (b) m a2 2 (a) 2 (c)

1 m a2 4

2

(d) 2 m a 2

2

3. In Q.2 above, the phase constant of the simple harmonic motion is (a) zero (b) 30° (c) 45° (d) 90° 4. The displacement of a particle executing simple harmonic motion varies with time t as x = a sin

t + a sin

t

3 The amplitude of oscillation is (a) a

(b)

2a

(d) 2a (c) 3 a 5. A block of mass m is attached to a spring of force constant k by means of a string going over a frictionless pulley as shown in Fig. 13.23. The block is held in position so that the spring is unstretched. The block is then released and it begins to oscillate with a small amplitude. The maximum velocity of the block during oscillation is mg (a) k (c) g

(b)

k m

(d) g

mg k m k

(c)

(d) 2 T

2 T

7. A particle is executing a linear simple harmonic motion. At time t = 0, it is at one extreme position cm in the next second moving in the same direction. The amplitude of the motion is (a) 8 cm (b) 10 cm (c) 12 cm (d) 14 cm 8. A body of mass 200 g executing SHM has a velocity of 3 cm s–1 when its displacement is 4 cm and a velocity of 4 cm s–1 when its displacement is 3 cm. The total energy of the oscillator is (a) 2.5 10–4 J (b) 2.5 10–2 J (c) 2.5 J (d) 250 J 9. The percentage change in the time period of a simple pendulum if its length is increased by 2% is (a) 4% (b) 2% (c) 1% (d) 2 % 10. Two springs of equal lengths and equal crosssectional areas are made of materials whose Young’s modulii are in the ratio of 3 : 2. They are suspended and loaded with the same mass. When stretched and released, they will oscillate with time periods in the ratio of (a)

3 :

2

(c) 3 3 : 2 2

(b) 3 : 2 (d) 9 : 4

11. Two bodies A and B of equal masses are suspended from two separate springs of force constants k1 and k2 respectively. If the two bodies oscillate such that their maximum velocities are equal, the ratio of the amplitudes of oscillation of A and B will be (a) k1 / k2

(b)

k1 / k2

(c) k2 / k1

(d)

k2 / k1 IIT, 1988

12. A spring of force constant k is cut into three equal pieces. If these three pieces are connected in parallel, the force constant of the combination will be Fig. 13.23

(a) k/3

(b) k/9

(c) 3k

(d) 9k

Simple Harmonic Motion

13. The displacement x (in centimetres) of an oscillating particle varies with time t (in seconds) as x = 2 cos

0.5 t

3

The magnitude of the maximum acceleration of the particle in cms–2 is (a)

(b)

2 2

(c)

4 2

(d)

2 4 14. A particle is executing linear simple harmonic motion of amplitude A. What fraction of the total energy is kinetic when the displacement is half the amplitude? 1 1 (b) (a) 4 2 2 1 3 (c) (d) 2 4 15. In Q.14, at what displacement is the energy of the oscillator half potential and half kinetic? A A (a) (b) 4 2 A A (c) (d) 2 3 16. A horizontal platform is executing simple harmonic motion in the vertical direction of frequency . A block of mass m is placed on the platform. What is the maximum amplitude of the platform so that the block is not detached from it? (a) (c)

g 4

(b)

2 2

g 2

(d)

2 2

mg 4

mg 2

2 2

17. A spring of negligible mass having a force constant k extends by an amount y when a mass m is hung from it. The mass is pulled down a little and released. The system begins to execute simple harmonic motion of amplitude A and angular frequency . The total energy of the mass–spring system will be (a)

1 mA2 2

2

(b)

1 2 mA 2

2

+

1 k y2 2

1 2 1 1 ky (d) mA2 2 – ky2 2 2 2 18. A small trolley of mass 2 kg resting on a horizontal frictionless turntable is connected by a light spring to the centre of the table. The relaxed length of the (c)

spring is 35 cm. When the turntable is rotated with an angular frequency of 10 rad s–1, the length of the spring becomes 40 cm. What is the force constant of the spring? 103 Nm–1

(a) 1.2

(b) 1.6

103 Nm–1

(c) 2.0 103 Nm–1 (d) 2.4 103 Nm–1 19. A simple pendulum of length l and bob mass m is displaced from its equilibrium position O to a position P so that the height of P above O is h. It is then released. What is the tension in the string when the bob passes through the equilibrium position O? Neglect friction. (a) mg

(b)

(c) mg 1

h l

2 mg h l

(d) mg 1

2h l

20. When a mass m is hung from the lower end of a spring of negligible mass, an extension x is produced in the spring. The mass is set into vertical oscillations. The time period of oscillation is (a) T = 2

x mg

(b) T = 2

gx m

(c) T = 2

x g

(d) T = 2

x 2g

21. A small spherical steel ball is placed a little away from the centre of a large concave mirror of radius of curvature R = 2.5 m. The ball is then released. What is the time period of the motion? Neglect friction and take g = 10 ms–2. (a)

2 2

13.15

sec 2 (c) sec (d) 2 sec 22. The potential energy of a particle executing simple harmonic motion at a distance x from the equilibrium position is proportional to 4

(a) x (c) x2

sec

(b)

(b) x (d) x3

23. The displacement y of a particle executing simple harmonic motion is given by t sin (1000 t) y = 4 cos2 2 This expression may be considered to be a result of the superposition of how many simple harmonic motions? (a) two (b) three IIT, 1992

13.16 Comprehensive Physics—JEE Advanced

24. The kinetic energy of a particle executing S.H.M. is 16 J, when it is at its mean position. If the amplitude of oscillation is 25 cm and the mass of the particle is 5.12 kg, the time period of oscillation is (a)

5 (c) 20

sec.

(b) 2 sec.

sec.

(d) 5

sec.

25. The time taken by a particle executing simple harmonic motion of time period T to move from the mean position to half the maximum displacement is T 2 T (c) 8

T 4 T (d) 12 (b)

(a)

IIT, 1992 26. One end of a long metallic wire of length L is tied to the ceiling. The other end is tied to a massless spring of spring constant K. A mass m hangs freely from the free end of the spring. The area of crosssection and Young’s modulus of the wire are A and Y respectively. If the mass is slightly pulled down and released, it will oscillate with a time period given by (a) 2

m K

(b) 2

YA + KL m YAK

1/ 2

m YA KL

(c) 2

mL YA

(d) 2

IIT, 1993

27. A particle is executing simple harmonic motion along the x–axis with amplitude 4 cm and time period 1.2 s. The minimum time taken by the particle to move from x = + 2 cm to x = + 4 cm and back again is (a) 0.6 s (b) 0.4 s (c) 0.3 s (d) 0.2 s 28. If a spring extends by x on loading, then the energy stored in the spring is (T is the tension in the spring and k its force constant) (a) (c)

T2 2x 2k T

2

(b)

T2 2k

2 (d) 2 T k

29. A rigid cubical block A of mass M and side L is same dimensions and of modulus of rigidity such that the lower face of A completely covers the upper face of B. The lower face of B is rigidly held on a horizontal surface. A small F is applied perpendicular to one of the side faces of A. After the force is withdrawn, block A executes small oscillations, the time period of which is (a) 2

M L

(b) 2

M L

(c) 2

ML

(d) 2

M L

IIT, 1992 30. A particle free to move along the x-axis has potential energy given by x + U(x) = k[1 – exp (– x2)] for – where k is a constant of appropriate dimensions. Then (a) at points away from the origin, the particle is in unstable equilibrium x, there is a force directed away from the origin (c) if its total mechanical energy is k/2, it has its minimum kinetic energy at the origin (d) for small displacements from x = 0, the motion is simple harmonic. 31. The period of oscillation of a simple pendulum of length L suspended from the roof of a vehicle which moves without friction down an inclined plane of inclination , is given by (a) 2

L g cos

(b) 2

L g sin

(c) 2

L g

(d) 2

L g tan

IIT, 2000 32. A spring of force constant k is cut into two pieces such that one piece is double the length of the other. Then the longer piece will have a force constant of 2 3 k (b) k (a) 3 2 (c) 3 k (d) 6 k IIT, 1999 33. A body of mass 1 kg is executing simple harmonic motion. Its displacement x (in cm) at time t (in second) is given by x = 6 sin 100 t

4

Simple Harmonic Motion

The maximum kinetic energy of the body is (a) 6 J (b) 18 J (c) 24 J (d) 36 J 34. Two particles P and Q start from the origin and execute simple harmonic motion along x-axis with the same amplitude and time periods 3 s and 6 s respectively. The ratio of the velocities of P and Q when they meet is (a) 1 : 2 (b) 2 : 1 (c) 2 : 3 (d) 3 : 2 35. A body is executing simple harmonic motion. At a displacement x, its potential energy is E1 and at a displacement y, its potential energy is E2. The potential energy E at a displacement (x + y) is (a) E1 + E2

(b)

E12

(c) E1 + E2 + 2 E1 E2

(d)

E1 E2

40. A horizontal platform with an object placed on it is executing SHM in the vertical direction. The amplitude of oscillation is 2.5 cm. What must be the least period of these oscillations so that the object is not detached from the platform? Take g = 10 m s–2. (a) 0.1 (c)

sec

(b) 0.5

sec

(d) 2

sec sec

41. The ends of a rod of length l and mass m are attached to two identical springs as shown in Fig. 13.24. The rod is free to rotate about its centre O. The rod is depressed slightly at end A and released. The time period of the resulting oscillation is

E22

36. A body executes simple harmonic motion under the 4 action of a force F1 with a time period s. If the 5 force is changed to F2 it executes S.H.M. with time 3 period s. If both the forces F1 and F2 act simul5 taneously in the same direction on the body, its time period in seconds is: 12 24 (a) (b) 25 25 35 15 (c) (d) 24 12 37. If the displacement (x) and velocity (v) of a particle executing simple harmonic motion are related through the expression 4v2 = 25 – x2, then its time period is (a) (b) 2 (c) 4 (d) 6 38. The bob of a simple pendulum executes simple harmonic motion in water with a period t, while the period of oscillation of the bob in air is t0. If the density of the material of the bob is (4/3) × 1000 kg m–3, and the viscosity of water is neglected, the relationship between t and t0 is t (a) t = t0 (b) t = 0 2 (c) t = 2t0 (d) t = 4t0 39. A body at the end of a spring executes S.H.M. with a period t1, while the corresponding period for another spring is t2. If the period of oscillation with the two springs in series is T, then (b) T 2 = t12 + t22 (a) T = t1 + t2 1 1 1 1 1 1 (c) (d) 2 T t12 t22 T t1 t2

13.17

(a) 2 (c)

m 2k 2m 3k

(b) 2 (d)

2m k 3m 2k

Fig. 13.24

IIT, 2009 42. A uniform metre scale of length 1 m is balanced on shown in Fig. 13.25. One end of the scale is slightly depressed and released. The time period (in seconds) of the resulting simple harmonic motion is (Take g = 10 ms–2) (a) (c)

(b) (d)

3

2 4

Fig. 13.25

43.

A simple pendulum of length l is suspended from the ceiling of a train which is moving in the horizontal direction with a constant acceleration

13.18 Comprehensive Physics—JEE Advanced

a. The time period of the pendulum is given by l where geff is given by T=2 g eff (a) g

(b) (g + a)

48. A simple pendulum is oscillating without damping. When the displacement of the bob is less than maximum, its acceleration vector is correctly shown in (see Fig. 13.27).

(c) (g – a) (d) g 2 a 2 44. A simple pendulum of length l is suspended from the ceiling of a trolley which is moving, without friction, down an inclined plane of inclination . The time period of the pendulum is given by l , where geff is given by T=2 g eff (a) g

(b) g sin

(c) g cos

(d) g tan

45. One end of a massless spring of relaxed length 50 cm and spring constant k frictionless inclined plane of inclination = 30º as shown in Fig. 13.26. When a mass m = 1.5 kg is attached at the other end, the spring extends by 2.5 cm. The mass is displaced slightly and released. The time period (in seconds) of the resulting oscillation will be (a)

7 2 (b) 7 (c)

5

2 (d) Fig. 13.26 5 46. Two particles are executing simple harmonic motions of the same amplitude and the same frequency along the same straight line and about the same mean position. If the maximum separation between them is 2 times the amplitude, the phase difference between them is (a) (c)

(b) 3

(d)

Fig. 13.27

IIT, 2002 49. For a particle executing simple harmonic motion, the displacement x is given by x = A cos t. Identify the graph which represents the variation of potential energy (PE) as a function of time t and displacement x. (see Fig. 13.28). (a) I, III

(b) II, IV

(c) II, III

(d) I, IV IIT, 2003

2 4

47. A particle executes simple harmonic motion between x = – A and x = + A. The time taken for it to go from 0 to A/2 is T1 and to go from A/2 to A is T2. Then (a) T1 < T2

(b) T1 > T2

(c) T1 = T2

(d) T1 = 2T2 IIT, 2001

Fig. 13.28

50. The time period of a particle in simple harmonic motion is 8 seconds. At t = 0 it is at the mean position. The ratio of the distances travelled by it in

Simple Harmonic Motion

(a) (c)

1 2

(b) 1

(d)

1 2 1

2 1 3 51. A particle is executing linear simple harmonic motion about the origin x = 0. Which of the graphs shown in Fig. 13.29 represents the variation of the potential energy function U (x) versus x?

54. A block is kept on a horizontal table. The table is executing simple harmonic motion of time period T static friction between the block and the table is . The maximum amplitude of the table for which the block does not slip on the surface of the table is (a)

IIT, 2004 52. A simple pendulum has a time period of 3.0 s. If the point of suspension of the pendulum starts moving vertically upward with a velocity v = Kt where K = 4.4 ms–2, the new time period will be (Take g = 10 ms–2) 9 (a) s (b) 5 s 4 3 (c) 2.5 s (d) 4.4 s IIT, 2005 53. A simple pendulum is moving simple harmonically with a period of 6 s between two extreme positions B and C as shown in Fig. 13.30. If the angular distance between B and C is 10 cm, how long will the pendulum take Fig. 13.30 to move from position C to position D exactly midway between O and C. (a) 1 s (b) 2 s (c) 3 s (d) 4 s

gT 2

(b)

gT 2

gT 2 2

2

(d) gT2 4 2 55. A particle executes a linear simple harmonic motion of amplitude 25 cm and time period 3 s. What is the minimum time required for the particle to move between two points located at 12.5 cm on either side of equilibrium position? (a) 0.5 s (b) 1.0 s (c) 1.5 s (d) 2.0 s 56. One end of a light spring of force constant k to a block A of mass M placed on a horizontal fric(c)

Fig. 13.29

13.19

to a wall (Fig. 13.31). A smaller block B of mass m is placed on block A. The system is displaced by a small amount and released. What is the maximum amplitude of the resulting simple harmonic motion of the system so that the upper block does not slip tion between the two blocks is .

Fig. 13.31

(a) Amax = (c) Amax =

Mg k (M

(b) Amax = m) g k

mg k

(d) None of these

57. A block A of mass m is placed on a frictionless horizontal surface. Another block B of the same Fig. 13.32 mass is kept on A and connected to the wall with the help of a spring of force constant k, as shown in Fig. 13.32. The A and B is . The blocks move together executing simple harmonic motion of amplitude a. The maximum value of frictional force between A and B is

13.20 Comprehensive Physics—JEE Advanced

(a) ka (c) zero

(b) ka/2 (d) mg

(c)

IIT, 2005 58. A simple pendulum attached to the ceiling of a stationary lift has a time period T. The distance y covered by the lift moving upwards varies with time t as y = t2 where y is in metre and t in second. If g = 10 ms–2, the time period of the pendulum will be (a)

4 T 5

(b)

5 T 6

(c)

5 T 4

(d)

6 T 5

k1 A k1 k2

(d)

k2 A k1 k2 IIT, 2009

M

60. A wooden block performs SHM on a frictionless surface with frequency, v0. The block carries a charge +Q on its surface. If now a uniform electric E is switched-on as shown in Fig. 13.34, then the SHM of the block will be (a) of the same frequency and with shifted mean position. (b) of the same frequency and with the same mean position (c) of changed frequency and with shifted mean position. (d) of changed frequency and with the same mean position. IIT, 2011

IIT, 2007 59. The mass M shown in Fig. 13.33 oscillates in simple harmonic motion with amplitude A. The amplitude of the point P is

Fig. 13.33

(a)

k1 A k2

(b)

k2 A k1

Fig. 13.34

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49. 55.

(c) (a) (c) (d) (d) (a) (c) (d) (a) (a)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56.

(b) (a) (d) (c) (b) (b) (c) (b) (c) (c)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57.

(c) (c) (c) (c) (b) (b) (b) (a) (d) (b)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58.

(c) (a) (a) (c) (b) (b) (a) (b) (c) (b)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59.

(d) (d) (b) (b) (d) (c) (c) (a) (a) (d)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60.

(a) (d) (b) (a) (d) (a) (c) (b) (c) (a)

SOLUTIONS I MgL where M is mass of the rod and I is its moment of inertia about the axis passing through its end and perpendicular to its length.

1. T = 2

I=

1 ML2 3

L , which is choice (c). 3g 2. Given x = a sin t + a cos t (1) The displacement equation of a simple harmonic motion is x = A sin ( t + ) T=2

where A = amplitude and

= phase constant.

Simple Harmonic Motion

x = A sin t cos + A cos

t sin

(2)

Comparing (1) with (2) we get A cos = a and A sin = a

(3) (4)

Squaring and adding Eqs. (3) and (4) we get A2 = a2 + a2

A2 = 2a2

1 m 2 A2 = m 2 a2 2 So the correct choice is (b). 3. Dividing Eq. (4) by Eq. (3) we get tan = 1 = 45°, which is choice (c). Energy =

4. Given x = a sin = a sin

t + a sin

t + a sin

t

t cos

3

7 = A [1 – (2 cos2 – 1)] = 2 A (1 – cos2 ) =2 A 1

t sin

3a 3a x= sin t + cos t 2 2 For a simple harmonic motion x = A sin ( t + ) x = A sin t cos + A cos t sin Comparing (1) and (2) we get 3a A cos = 2 and A sin

=

3a Hence A = 2 2

3 (1)

(2)

3a 2

2

= 3a2

vmax = A =

mg k

6. Kinetic energy = =

1 m A2 4

2

m k = g k m 1 m A2 2

2

cos2 ( t + )

[1 + cos (2 t + 2 )]

Time period of kinetic energy =

So the correct choice is (a).

2 2

T 2

2 T

4 A

4 A

2

=8–

8 A

which gives A = 8 cm. (A 2 – x 2) (1) 9 = 2 (A2 – 16) and 16 = 2 (A2 – 9) (2) Solving Eqs. (1) and (2), we get A = 5 cm and = 1 rad s–1. 1 E= m A2 2 2 1 = (200 10–3) (5 10–2)2 (1)2 2 = 2.5 10–4 J

8. V =

9. T = 2

A = 3 a, which is choice (c). 5. When the block is released, the spring extends by an amount x given by mg kx = mg x = , which is also the amplitude of k mg the oscillation, i.e. A = k

2

2 A

1

7 = 2A 1 1

3a 2 2

7. Since the particle is at extreme position at time t = 0, x = A cos t At t = 0, x0 = A At t = 1 s, x1 = A cos At t = 2 s, x2 = A cos 2 2 If A is in cm, 2 = A – A cos cos = 1 – A and 2 + 5 = A – A cos 2

3 + a cos

13.21

or

2

L g

T2 = 4

2

L g

2 T L = T L T 1 L 1 = = T 2 L 2

2% = 1%

F L A l F YA Force constant k = = l L where l is the extension in the spring of original length L and cross-sectional area A when a force F = Mg is applied. Now, the time period of vertical oscillations is given by ML M =2 T =2 YA k

10. Young’s modulus Y =

T1 = T2

Y2 = Y1

3 2

Hence the correct choice is (a).

13.22 Comprehensive Physics—JEE Advanced

11. The velocity of an oscillating body is maximum when it is at the equilibrium position where the potential energy is zero and the energy is entirely kinetic. At the extreme positions, the kinetic energy is zero and the energy is entirely potential. Therefore, the kinetic energy at equilibrium position = potential energy at extreme positions = total energy. Since the maximum velocities (i.e. velocities at equilibrium position) are equal for the two equal masses, their kinetic energies are also equal = their potential energies at extreme positions where the displacement is maximum = amplitude. If x1 and x2 are amplitudes of bodies A and B, we have x k2 1 1 k1 x 21 = k2 x 22 or 1 x2 k1 2 2 Hence the correct choice is (d). 12. If a force F is applied to a spring of force contant k and the spring extends by an amount x, then F = kx The extension x produced in a spring is proportional to its length. Thus, if the spring is cut into three equal pieces, the same force F will produce an extension x/3 in a piece. If k is the force constant of the piece, we have F = k x/3 k = k or k = 3k. Thus, the force Therefore 3 constant of each piece is 3k. When springs are connected in parallel, the force constant of the combination is equal to the sum of the individual force constants of the springs so connected. Therefore, the force constant of the combination = 3k + 3k + 3k = 9k. Hence the correct choice is (d). x = 2 cos 0.5 t

13. Given

3

dx Velocity V = dt =–2

0.5

0.5

sin

0.5 t

0.5

cos

0.5 t

Maximum acceleration, amax = – 2 Magnitude is

Hence the correct choice is (a). Notice that Amax is independent of the mass of the block. 17. Let L be the relaxed length of the spring and y the extension produced in it due to force mg so that ky = mg (i) The displacement of the mass during oscillation is given by x = A sin ( t + ) (ii) At the instant when the displacement is x 2

3

dV Acceleration a = dt =–2

1 m 2(A2 – x2) 2 1 Potential energy (PE) = m 2x 2 2 1 Total energy (E) = m 2A 2 2 When x = A/ 2, A2 1 m 2 A2 KE = 4 2 3 2 2 = m A 8 1 KE 3 E= m 2A 2 2 E 4 15. The energy will be half kinetic and half potential at a value of x when KE = PE, i.e. 1 1 m 2(A2 – x2) = m 2x 2 2 2 A or A2 – x2 = x2 or x = 2 16. The block will not be detached from the platform, if the amplitude of the platform’s SHM is such that the maximum acceleration equals the acceleration due to gravity, i.e. g g 2 Amax = g or Amax = 2 4 2v2

14. Kinetic energy (KE) =

0.5

0.5

amax =

Hence the correct choice is (c).

=– 2

2

=

2

2 2

2

3 cms–2 cms–2

dx KE of mass = 1 mV2 = 1 m dt 2 2 = 1 mA2 2 cos2 ( t + ) 2 1 PE of spring = k(y + x)2 2 1 = k (y2 + 2yx + x2) 2 1 = k y2 + kyx + 1 kx 2 2 2 Using (i) and (ii) and

=

k , we have m

(iii)

Simple Harmonic Motion

PE of spring = 1 ky2 + mgx + 1 m 2 2 sin2 ( t + )

2

A2 (iv)

Taking gravitational PE at the mean position to be zero, Gravitational PE at x = – mg x (v) Adding (iii), (iv) and (v), we get Total energy of mass–spring system = 1 mA2 2

2

cos2 ( t + ) + 1 ky2 + 2

mgx + 1 mA2 2 sin2 ( t + ) – mgx 2 2 2 1 mA + 1 ky2 = 2 2 18. The radius of the circle along which the trolley moves is r = 40 cm = 0.4 m When the table is rotated, the tension in the spring is equal to the centripetal force, i.e. mv2 = mr 2 r = 2 0.4 (10)2 = 80 N The extension in the spring is x = 40 – 35 = 5 cm = 0.05 m F Force constant k = x 0 = = 1.6 103 N m–1 0.05 Hence the correct choice is (b). F =

13.23

At position O, the tension F in the string is given by mV 2 F – mg = centripetal force = l mV 2 or F = mg + l 2 mg h ( V 2 = 2gh) = mg + l 2h or F = mg 1 l Hence the correct choices is (d). 20. If k is the force constant, we have mg = k x m x = or k g T =2

m k

2

x g

Hence the correct choice is (c). 21. Let the steel ball be placed at A, a little away from the centre O of a concave mirror of radius of curvature R (= OC = AC), as shown in Fig. 13.36. Let ACO = . If m is the mass of the ball, its weight mg acts vertically downwards at A. This force is resolved into two rectangular components: mg cos (which is balanced by the normal reaction N of the mirror) and m g sin (which provides the restoring force F). Thus F = – m g sin =–mg ( is small; x/R being very small)

19. P.E. at point P = mgh. If friction is neglected, the potential energy is completely converted into kinetic energy when the bob reaches the equilibrium position O (see Fig. 13.35). If V is the velocity of the bob at O, then 1 mV 2 = mgh 2 or V 2 = 2gh

Fig. 13.36

or

x R ( x = R ; x being the arc OA) F = – kx

where

k =

=–mg

Fig. 13.35

mg is the force constant. R

13.24 Comprehensive Physics—JEE Advanced

m R =2 k g 2.5 =2 = seconds. 10 Hence the correct choice is (c). 22. The potential energy of a particle of mass m executing simple harmonic motion of angular frequency at a distance x from the equilibrium position is given by 1 1 m 2 x2 = k x2 P.E. = 2 2 where k = m 2 is constant. Hence the correct choice is (c). t t 23. We can write, 4 cos2 = 2 2 cos2 =2 2 2 (1 + cos t). Therefore, y = 2 (1 + cos t) sin (1000 t) = 2 sin (1000 t) + 2 cos t sin (1000 t) = 2 sin (1000 t) + sin (1001 t) + sin (999 t) Hence

T =2

Thus y is a superposition of three simple harmonic motions of angular frequencies 999, 1000 and 1001 rad s–1. Hence the correct choice is (b). But a superposition of two or more simple harmonic motions of different frequency does not produce a simple harmonic motion. The statement of the question is incorrect. 24. At the mean position, the velocity of the particle is v = A . Therefore 1 1 2 2 mA 2 2 = mA2 K.E. = 2 2 T = or

T=

2

2

m A2 2m K.E.

=

0.25

= 0.2

second

26. If the wire extends by an amount x when a force F is applied to it, then FL F/A Y = = Ax x /L YA YA x = k x; k = L L Thus the force constant of the wire is k. If K is the force constant of the spring, then the force constant of the series combination of the wire and the spring is given by 1 1 1 = k k K YA K kK L or k = = k K YA K L Y AK = (i) Y A KL The time period of the combination is or

F =

T =2

2 5.12 16

1/ 2

Hence the correct choice is (a). 25. Let the displacement of the particle be given by 2 t x = A sin t = A sin T i.e. when x = 0, t0 = 0. When x = A/2, the value of t is given by 2 t1 A = A sin T 2 2 t1 1 or sin = T 2

m k

(ii)

27. Let the displacement of the particle be given by x = A sin

T2 A

2 t1 T = or t1 = T 12 6 T T t1 – t0 = –0= . 12 12 Hence the correct choice is (d). or

2 t T

where A = 4 cm and T = 1.2 s. If t1 is the time taken by the particle to move from x = 0 to x = 2 cm, then 2 t1 2 = 4 sin T which gives t1 = T/12. If t2 is the time taken to move from x = 0 to x = 4 cm, then 4 = 4 sin

2 t2 T

T . Therefore, time taken to 4 T move from x = 2 cm to x = 4 cm is t2 – t1 = – 4 T T 1.2 s = 0.2 s. Therefore, time taken 12 6 6 by the particle to move from x = 2 cm to x = 4 cm and back = 0.4 s. Hence the correct choice is (b).

which gives t2 =

Simple Harmonic Motion

28. T = kx. Therefore x = T/k. Now energy stored in 1 1 T 2 T2 k x2 = k , which is the spring = 2k 2 2 k choice (b).

The effective acceleration due to gravity acting on the bob is given by g2eff = a 2x + (g – ay)2 = a 2x + g2 + a2y – 2gay

29. Let the force F produce a deformation x. When this force is withdrawn, the force that tries to restore block A to its equilibrium position is proportional to x and is given by f =– L x d2 x

Acceleration

dt

2

=

f M

L x M

2

= g2 sin2

cos2

+ g2 + g2 sin4 – 2g2 sin2

x

L . The angular frequency of this M 2 simple harmonic frequency is . Now T = .

where

13.25

=

Fig. 13.38

= g2 sin2

Therefore T =2

M L

30. Figure 13.37 shows the plot of U(x) versus x. At x = 0, potential energy U(0) = k[1 – exp (0)] = k(1 – 1) = 0 and it has a maximum value = k at x = ± since U (±

) = k[1 – exp (– ±

) 2]

= k (1 – 0) = k

Fig. 13.37

Since the total mechanical energy has a constant value = (k/2), the kinetic energy will be maximum at x = 0 and minimum at x = ± . At x = 0 dU dx

x 0

= [2 kx exp (– x2)]at

x=0

=0

Hence the particle is in stable equilibrium at x = 0 (origin) and would oscillate about x = 0 (for small displacements) simple harmonically. Hence (d) is the only correct choice. 31. The acceleration of the vehicle down the plane is g sin . The reaction force acting on the pendulum bob gives it an acceleration a = g sin up the plane. This acceleration has two rectangular components ax = a cos = g sin cos and y = a sin = g sin2 as shown in Fig. 13.38.

or

(cos2

= g2 (1 – sin2 geff = g cos

Now T = 2

L geff

+ sin2

) + g2 – 2g2 sin2

) = g2 cos2 L g cos

2

Hence the correct choice is (a). 32. The force constant of a spring is inversely proportional to its length. If a spring of length L is cut into two pieces of lengths x and (L – x), such that 2L , x = 2 (L – x) or x = 3 then the force constant of the spring of length x is related to the force constant k of the complete spring of length L as k1 L = k x

3 3 or k1 = k 2 2

L 2 L /3

which is choice (b). 33. Velocity of the body at time t is dx d 6 sin 100t v= dt dt

4

cm s–1 4 = 6 ms–1

= 600 cos 100t vmax = 600 cm s–1

1 1 mv2max = 1 (6)2 = 18 J 2 2 Hence the correct choice is (b). 34. Since the particles start from x = 0 and have the same amplitude but different time periods, they will meet again at x = 0 where their velocities are maximum equal to a 1 and a 2, i.e Maximum K.E. =

13.26 Comprehensive Physics—JEE Advanced

v1 T2 T2 2 = 1 v2 T 2 T1 2 1 Hence the correct choice is (b). 1 35. E1 = m 2 x 2 2 1 m 2

E1 = x

or

E2 =

1 m 2

E=

2

1 m 2

(1)

2 2

y

1 m 2

E2 = y

or

6 =2 3

2

2

(2)

(x + y)2

1 m 2 2 From (1), (2) and (3), it follows that E = (x + y)

or

E = or

E1

(3)

E2

E = E1 + E2 + 2 E1 E2

which is choice (c). 36. Let the body be displaced by a distance x. If the restoring force is F1, then the angular frequency of the resulting simple harmonic motion is given by K Kx F1 = (i) m mx mx where m is the mass of the body. For force F2, we have 2 1

2 2

=

F2 mx

(ii)

F1

(iii)

F2 mx From (i), (ii) and (iii) we get 2

2

or or

or

2

2 T 1 T

2

1 T2

= =

2 1

=

2 T1

=

=

T=

a= – 2x (ii) Comparing (i) and (ii), we have 1 = 2 2 1 = or T = 4 , which is choice (c). or T 2 38. Since the density of water is 3/4 of the density of the bob, the effective acceleration due to gravity when the bob is in water decreases (due to buoy3g g . Now ancy) from g to g = g 4 4

and

2

1

1

T12

T22

25 16

25 9

1 4 5

l g

t= 2

l g

t g g = = 2 or t = 2t0 t0 g g /4 Hence the correct choice is (c). 39. Let m be the mass of the body and k1 and k2 the force constants of the two springs. Then t1 = 2

2

2 T2

t0 = 2

Dividing, we get

2 2

+

144 25 25 which is choice (a).

or

37. Given 4v2 = 25 – x2. Differentiating with respect to time t, we have dv dx = 0 – 2x 8v dt dt dv dx ,v or 8va = – 2xv a dt dt where a is the acceleration. Thus 1 a= x (i) 4 For a simple harmonic motion

1 2

25 25 16 9 12 s, 25

3 5

2

25 25 144

m k1

(1)

m (2) k2 If the two springs are connected in series, the force constant of the combination is given by 1 1 1 = k1 k2 k and

t2 = 2

or

k=

k1k2 (k1 k2 )

T=2

m =2 k

m (k1 k2 ) k1k2

(3)

Simple Harmonic Motion

Squaring and adding (1) and (2), we get t12 + t22 = 4

2

=4

2

1 k1

m m

= T2

1 k2

(k1 k2 ) k1k2

or T =

13.27

2m , which is choice (c). 3k

[Use Eq. (3)]

40. The object will not detach from the platform, if the angular frequency is such that, during the downward motion, the maximum acceleration equals the acceleration due to gravity, i.e., 2 max A = g or

max

g A 2

=

Fig. 13.39

A g max Now A = 2.5 cm = 2.5 10–2 m and g = 10 ms–2. Substituting these values we get, T = 0.1 sec. Hence the correct choice is (a). 41. Let the rod be depressed by a small amount x (Fig. 13.39). Both the springs are compressed by x. When the rod is released, the restoring torque is given by l l + (kx) × = (kx)l = (kx) × 2 2 x 2x Now sin = = . Since is small, sin l/2 l , where is expressed in radian. Thus = 2x/l or x = l/2. Hence l k l2 ×l= =k 2 2 If I is the moment of inertia of the rod about O, then or

Tmin =

=2

d2 I

d2

or Since

d t2 dt

2

=–

kl 2 2

=–

kl 2 2I

d2

42. Refer to Fig. 13.40. The magnitude of the restoring torque = force × perpendicular distance = mg × AB = mg × R sin Since is small, sin . here is expressed in radian. The equation of motion of the scale is I

d2 dt 2 d2

or

dt =

Now I =

2

= – mgR =

mgR 2 or = I T mL2 . Hence 12 T=

mgR I mgR or T = 2 I

I . mgR

L

3gR Using the values L = 1 m, g = 10 ms–2 and R = 0.3 m, we get T = /3 second. Hence the correct choice is (c).

(– ), the motion is simple harmonic d t2 whose angular frequency is given by kl 2 2I

= Now 2 = T

=

2 ml2 and I = . Therefore, we have 12 T kl 2 2

12 ml

2

=

6k m

Fig. 13.40

43. There are two perpendicular acceleration vectors; g acting vertically downwards and a acting horizontally. The resultant acceleration is given by

13.28 Comprehensive Physics—JEE Advanced

g2

geff =

a2

Hence the correct choice is (d). 44. The correct choice is (b) since the component of the acceleration due to gravity along the plane is g sin . 45. The force which increases the length of the spring by x = 2.5 cm is F = mg sin . Therefore, the spring constant is F mg sin k= x x m k

Now time period T = 2

m mg sin / x

2

–2

–2

Putting x = 2.5 cm = 2.5 × 10 m, g = 9.8 ms and = 30º, we get T = /7 second, which is choice (a). 46. Let the motions be given by x1 = A sin t and x2 = A sin ( t + ). Separation between the particles at time t is given by (1) x = x2 – x1 = A sin ( t + ) – A sin t dx The separation will be maximum when = 0, i.e. dt when A cos ( t + ) – A cos t = 0 or cos ( t + ) = cos t or t+ =± t [ cos (– ) = cos ] The plus sign is not possible because then = 0 which implies that the separation x is always zero. Therefore, t + = – t or t = – /2. Using this value in (1), we have

Given xmax =

– A sin

2

= 2 A sin

2

PE =

1 m 2

=

1 m 2

sin

2

=

1

2

(A cos

t)2

2

A2 cos2

t

1 m 2 A 2. 2 Hence the correct graph is I. Thus, the correct choice is (a).

At t = 0, PE is maximum equal to

50. Since the particle is at the mean position at t = 0, the phase of the motion is zero. Therefore, the displacement of the particle is given by x = A sin ( t) = A sin = A sin

t 4

(

2 2 Hence the correct choice is (b).

4

or =

2

2

(i)

x1 = A sin

4

=

A 2 t

= 2 s in (i)]

2 =

T = 8s)

t=

x = A sin

or

2 t T

1s in (i)]

2 A. Hence 2 A = 2 A sin

or

1 m 2 x2 2 At x = 0, PE = 0. Hence the correct graph is III. As a function of t, the PE is given by PE =

x g sin

=2

xmax=A sin

48. In the simple harmonic motion of a pendulum, the restoring force vector (and hence the acceleration vector) is tangential to the path of the bob and is directed towards the mean position. Hence the correct choice is (b). 49. Given x = A cos t. As a function of x, the PE is given by

.

47. In simple harmonic motion, the speed of the particle is the maximum at the mean position x = 0, decreases as it moves towards the extreme position becoming zero at the extreme position x = A. Hence the particle will take shorter time to move A A from x = 0 to x = than to move from x = to 2 2 x = A. Thus, the correct choice is (a).

2

=A

Distance travelled in the IInd second is A =A 1 x2 = x – x1 = A – 2 A 2

1 2

1 = 1 2 1 A 1 2 Hence the correct choice is (c). 51. In simple harmonic motion, the force acting on the particle (restoring force) is given by F = – kx x1 = x2

Simple Harmonic Motion

where k is a positive constant. Now

2 2 rad s–1 T 6 3 Let us suppose that at t = 0, the pendulum is at C, i.e. at t = 0, x = A, so that A = A sin ( 0+ ) where A = 5 cm and

dU dx

Therefore, – kx = or

OD = 2.5 cm Let the displacement of the pendulum be given by x = A sin ( t + )

dU dx

F=

dU = kdx x

U (x) =

kdx = 0

1 kx 2 + c 2

where c is a constant of integration. In simple harmonic motion, the potential energy of the oscillator is zero at the mean position, i.e. U (0) = 0. Hence c = 0. Therefore U (x) =

which is the equation of a parabola. Since U (x) is positive for all values (positive and negative) of x, the correct graph is (d). 52. Original time period is l g

(1)

When the pendulum is moving upwards, the effective value of g is geff = g + a where a is the acceleration of the pendulum which is given by a=

dv d = (Kt) = K = 4.4 ms–2 dt dt

geff = g + a = 10 + 4.4 = 14.4 ms–2. Therefore, the new time period is l

T2 = 2

or

=

A = A sin

or sin

= 1 or

=

2 Thus the motion of the pendulum is given by putting

2 x = A sin

1 kx 2 2

T1 = 2

13.29

t

2 = A cos t = 5 cos t The value of t for which x = 2.5 cm is given by 2.5 = 5 cos t 1 or t = or cos t = 2 3 2 t T or = , or t = = 1 s. T 3 6 54. Refer to Fig. 13.41. Let the mass of the block be m and let, at a certain instant of time, the direction of acceleration a of the table (executing simple harmonic motion) be along the positive x-direction. As a result, the block will experience a force ma directed along the negative x-axis. Consequently, the force of friction mg will act along the positive x-axis. The weight mg of the block will be balanced by the normal reaction R. The block will not slip on the surface of the table, if the acceleration a of the motion of the table is such that mg ma or g a

(2)

g eff

From (1) and (2), we get T2 = T1 or

g geff

T T2 = 1 1.2

10 14.4

1 1.2

3 = 2.5 s 1.2

Hence the correct choice is (c). 1 53. Amplitude = OC = OB = BC 2 10 = = 5 cm 2

Fig. 13.41

Therefore, for no slipping, the table can have a maximum acceleration amax = g. We know that, for a simple harmonic motion, amax = 2A, where is the angular frequency and A the amplitude of the motion of the table. Therefore, the maximum amplitude is given by

13.30 Comprehensive Physics—JEE Advanced 2

or

Amax = g Amax =

g 2

=

gT 4

2

2

,

which is choice (c). 55. Let A and B be the two extreme positions of the particle with O as the equilibrium position. Displacements to the right of O are take as positive, while to the left of O are taken as negative (Fig. 13.42)

The upper block of mass m will not slip over the lower block of mass M if the maximum force on the upper block fmax does not exceed the frictional force mg between the two blocks. Now fmax = mamax = m 2Amax (2) where amax is the maximum acceleration and Amax is the maximum amplitude. Using (1) in (2), we get fmax =

mk Amax M m

For no slipping, fmax = mg mk Amax = mg or Amax = M m which is choice (c). or

Fig. 13.42

Let the displacement of the particle in SHM be given by x(t) = A sin ( t + ) (i) 2 2 = rad s–1 where A = 25 cm and = T 3 Let us suppose that at time t = 0, the particle is at extreme position B. Setting x = A at t = 0 in Eq. (i) we have A = A sin giving = /2 Putting = /2 in Eq. (i), we get x(t) = A cos t (ii) where A = 25 cm. Now let us say that the particle reaches point C at t = t1 and point D at t = t2. At C, the displacement x(t1) = + 12.5 cm and at D, it is x(t2) = – 12.5 cm (see Fig. 13.12). So from (ii) we have + 12.5 = 25 cos t1 and – 12.5 = 25 cos t2 or cos t1 = + 0.5 or t1 = /3 2 and cos t2 = – 0.5 or t2 = 3 2 Hence (t2 – t1) = 3 3 3 T 2 t2 – t1 = = 3 6 T 3 or (t2 – t1)min = = 0.5 s 6 4 Notice that cos t2 = – 0.5 even for t2 = 3 which gives t2 = 2 T/3 = 4 s which does not correspond to minimum value of (t2 – t1). Thus the correct choice is (a). 56. The angular frequency of the system is =

1/ 2

k M

m

(1)

M

m g k

,

57. The blocks will move together as long as the frictional force of block B = mass of block B maximum acceleration of its S.H.M., i.e. f = m 2a k

where

=

Thus

f=m

k 2m

( m m) k 2m

a

= ka/2, which is choice (b). 2

58. Given y = t . The velocity of the lift varies with t as v =

dy = 2t dt dv = 2 ms–2, directed updt

Acceleration a = wards, Hence

l

T =2 and

g l g

T=2 T = T

a

g g

a

=

10 = (10 2)

5 6

The correct choice is (b) 59. If a force F is applied to M, say to the right, let A be the distance moved by M. If the system is released, it executes simple harmonic motion of amplitude A. If A1 and A2 are the extensions in springs k1 and k2 then A = (A1 + A2) and F = k1 A1 = k2 A2

Simple Harmonic Motion 13.31

A1 =

The amplitude of point P = amplitude of oscillations of spring k1 which is

F F and A2 = k1 k2

1 A = A1 + A2 = F k1

1 k2

A1 =

60. The force exerted on charge +Q E is F = QE in the direction of E . Since F is constant, a constant force is added to the applied force. Hence only the mean position will change and the frequency of oscillation will remain the same.

F (k1 k2 ) = k1 k2 F=

k2 A F = (k1 k2 ) k1

k1 k2 A (k1 k2 )

II Multiple Choice Questions with One or More Choices Correct 1. Which of the following expressions represent simple harmonic motion? (a) x = a sin ( t + ) (b) x = a cos ( t + ) (c) x = a sin t + b cos t (d) x = a sin t cos t 2. Choose the correct statements from the following in which k is a real, positive constant. (a) Function f (t) = sin kt + cos kt is simple harmonic having a period 2 / k. (b) Function f (t) = sin t + 2 cos 2 t + 3 sin 3 t is periodic but not simple harmonic having a period of 2 s. (c) Function f (t) = cos kt + 2 sin2 kt is simple harmonic having a period 2 / k. (d) Function f (t) = e–kt is not periodic. 3. A simple pendulum of length l and bob mass m is displaced from its equilibrium position O to a position P so that the height of P above O is h. It is then released. What is the tension in the string when the bob passes through the equilibrium position O? Neglect friction. V is the velocity of the bob at O. V2 2 mg h (b) (a) m g l l h 2h (d) mg 1 l l 4. A trolley of mass m is connected to two identical springs, each of force constant k, as shown in (c) mg 1

Fig. 13.43. The trolley is displaced from its equilibrium position by a distance x and released. The trolley executes simple harmonic motion of period T. After some time it comes to rest due to friction. The total energy dissipated as heat is (assume the damping force to be weak) 1 kx2 (b) kx2 (a) 2 (c)

2

2

mx 2

(d)

T2

mx 2 T2

Fig. 13.43

5. In order to execute simple harmonic motion, a system must have (a) inertia (b) moment of inertia (c) elasticity (d) buoyancy 6. The time period of a system executing simple harmonic motion depends upon (a) mass of the system (b) force constant of the system (c) amplitude of the oscillator (d) phase constant of the oscillator. 7. The amplitude of a particle executing simple harmonic motion depends upon

13.32 Comprehensive Physics—JEE Advanced

(a) initial displacement (b) initial velocity (c) initial acceleration (d) initial phase. 8. The displacement (x) of a particle as a function of time (t) is given by x = a sin (bt + c) where a, b and c are constants of motion. Choose the correct statements from the following: (a) The motion repeats itself in a time interval 2 /b (b) The energy of the particle remains constant (c) The velocity of the particle is zero at x=±a (d) The acceleration of the particle is zero at x=±a 9. Figure 13.44 (a) shows a spring of force constant k m at the other end placed on a horizontal frictionless surface. The spring is stretched by a force F. Figure 13.44 (b) shows the same spring with both ends free and a mass m stretched by the same force F. The mass in case (a) and the masses in case (b) are then released. Which of the following statements are true? (a) While oscillating, the maximum extension of the spring is more in case (a) than in case (b). (b) The maximum extension of the spring is the same in both cases. (c) The time period of oscillation is the same in both cases. (d) The time period of oscillation in case (a) is 2 times that in case (b).

10. A simple pendulum is oscillating between extreme positions P and Q about the mean position O. Which of the following statements are correct about the motion of the pendulum? (a) At point O, the acceleration of the bob is different from zero (b) The acceleration of the bob is constant throughout the oscillation (c) The tension in the string is constant throughout the oscillation (d) The tension is the maximum at O and the minimum at A or B. 11. Two springs A and B have force constants k1 and k2 respectively. The ratio of the work done on A to that done on B in increasing their lengths by the same amount is and the ratio of the work done on A to that done on B when they are stretched with the same force is . Then k k (b) = 2 (a) = 1 k2 k1 (c)

=

k1 k2

=

k2 k1

12. A body of mass 50 g executing linear simple harmonic motion has a velocity of 3 cms–1 when its displacement is 4 cm and a velocity of 4 cms–1 when its displacement is 3 cm. (a) The amplitude of oscillation is 5 cm. (b) The angular frequency of oscillation is 1 rad s–1. (c) The maximum kinetic energy of the oscillator is 6.25 10–5 J. (d) The maximum potential energy of the oscillator is 6.25 10–5 J. 13. The displacement of an oscillating particle is given by x = a sin(ct) + b cos(ct) where a, b and c are constants. If A is the amplitude of the oscillations and T their time period, then (a) A = a + b

(b) A =

a2

b2

2 2 a (d) T = c bc 14. All the springs shown in Fig. 13.45 (a), (b) and (c) are identical, each having a force constant k. If Ta, Tb and Tc are the time periods of oscillations of the three systems respectively, then (c) T =

Fig. 13.44

(d)

Simple Harmonic Motion 1/ 2

L

(d) 2 g

13.33

v2 R

17. A solid sphere of density and radius R ing in a liquid of density with half its volume submerged. When the sphere pressed down slightly and released, it executes simple harmonic motion of time period T. If viscous effect is ignored, then (a)

=2

(c) T = 2

Tb

(b) Tb = 2Tc

2

y1 = 10 sin

Tc 2 15. Two masses m1 and m2 are suspended together by a massless spring of spring constant k (see Fig. 13.46). When the masses are in equilibrium, m1 is removed without disturbing the system. If is the angular frequency of oscillation and x2 the amplitude of oscillation of mass m2, then k (a) = m2 (c) Ta =

(b)

=

2 Tc

(d) Ta = 2Tb =

k (m2

m1 )

m1 g k m g (d) x2 = 2 k (c) x2 =

L g

g2

and

4

y2 = 5 (sin 3 t +

3R 2g

(12t + 1) 3 cos 3 t)

The ratio of their amplitudes is m and the ratio of their time periods is n. Then (a) m =

2 3

(b) m = 1

(c) n =

3 2

(d) n = 1 IIT, 1986

19. A solid cylinder of mass m and radius R is attached to a massless spring of force constant k as shown in Fig. 13.47. The cylinder is pushed to the right a little and released. It executes simple harmonic motion. The cylinder rolls on the horizontal surface without slipping.

Fig. 13.47

The time period of oscillation is T. When the instantaneous displacement of the cylinder from the mean position is x, the total energy of the system is E. Then (a) T = 2

1/ 2

L

(c) 2

(d) T = 2

Fig. 13.46

IIT, 1981 16. A simple pendulum of length L and mass m is suspended from the ceiling of the compartment of a train that is travelling at a constant speed v around a circular track of radius R. The tension in the string is T. The time period of the pendulum is mL (a) 2 T (b) 2

2R 3g

=2

IIT, 2004 18. Two simple harmonic motions are represented by the following equations

Fig. 13.45

(a) Ta =

(b)

v4 R2

1/ 2

(c) E =

2k 3m

3 2 1 2 mv + kx 4 2

(b) T = 2 (d) E =

m k

1 2 1 2 mv + kx 2 2

13.34 Comprehensive Physics—JEE Advanced

20. A man is standing on a weighing machine placed on a horizontal platform that is executing vertical simple harmonic motion of angular frequency . The maximum and minimum readings of the machine are m1 kg and m2 kg respectively. If the true mass of the man is m and A is the amplitude of the motion then m1m2 1 (b) m = (a) m = (m1 + m2) (m1 m2 ) 2 (c) A =

(m1 (m1

m2 ) g m2 ) 2

(d) A =

(m1 (m1

m2 ) g m2 ) 2

21. The displacement x of a particle varies with time t as x = A sin2 t + B cos2 t + C sin t cos t For what values of A, B and C is the motion simple harmonic? (a) All values of A, B and C with C 0. (b) A = B, C = 2B (c) A = – B, C = 2B (d) A = B, C = 0 IIT, 2007

ANSWERS AND SOLUTIONS 1. Simple harmonic motion is represented by a sine function or a cosine function or a linear combination of them. Hence the correct choices are (a), (b) and (c). The choice (d) which is a product of the two functions does not represent a simple harmonic motion. 2. Statement (a) is correct. The function f (t) = sin kt + cos kt can be written as f (t) = or

2 cos

kt

4

2 sin kt

4

both of which are simple

t in the argument of the sine or cosine function = 2 /T where T is the period. Hence k = 2 / T or T = 2 /k. Statement (b) is also correct. Each term represents simple harmonic motion. The period T of term sin t is = 2 / T or T = 2s. The period of term 2 cos 2 t is 1s, i.e. T/2 and the period of term 3 sin 3 t is 2/3 s, i.e. T/3. The sum of two or more simple harmonic motions of different periods is not simple harmonic. The sum, however, is periodic. second term completes two cycles and the third term completes three cycles. Thus the sum has a period of 2s. Statement (c) is incorrect. We can write f(t) = cos kt + 2 sin2kt as f (t) = cos kt + (1 – cos 2kt)

3. P.E. at point P = mgh. If friction is neglected, the potential is completely converted into kinetic energy when the bob reaches the equilibrium position O (see Fig 13.48). If V is the velocity of the bob at O, then 1 mV 2 = mgh 2 or V 2 = 2gh At position O, the tension F in the string is given by F – mg = centripetal force =

mV 2 l

mV 2 Fig. 13.48 l 2 mg h = mg + ( V 2 = 2gh) l 2h or F = mg 1 l Hence the correct choices are (a) and (d). 4. In general, the motion of a damped oscillator is not simple harmonic. If the damping forces are weak, the motion is very nearly simple harmonic and all formulae of SHM apply. The amplitude A = x. The time period T is or

F = mg +

T=2

= 1 + cos k t – cos 2kt The period of cos kt is T = 2 /k and of cos 2kt is /k which is T/2. As explained above, the period of the two terms together is T = 2 /k. The term 1 is a constant independent of time. Statement (d) is correct. Function e–kt decreases monotonically to zero at t ; it never becomes negative. Hence it is non-periodic.

m 2k

(i)

If the trolley eventually comes to rest, the entire energy of oscillation is dissipated as heat due to friction. Hence, the total energy dissipated as heat is E=

1 mA 2 2

2

=

1 mx2 2

2 T

2

=

2

2

mx 2

T2

(ii)

Simple Harmonic Motion

which is choice (c). Using (i) in (ii) we get E = kx 5. 6. 7. 8.

2

Hence choice (b) is also correct. The correct choices are (a) and (c). The correct choices are (a) and (b). The correct choices are (a) and (b). The motion of the particle is simple harmonic. The displacement at time t is x = a sin (bt + c) Displacement at time x at t

2 b

t

2 b

is

2 c b = a sin (bt + c + 2 )

= a sin

b t

= a sin (bt + c) = x at time t Hence statement (a) is correct. Statement (b) is also correct since the same displacement is recovered after a time interval of 2 / b. Statement (c) is correct because the velocity is zero when the displacement = ± amplitude a, i.e. at the extreme ends of the motion. Statement (d) is incorrect, the acceleration is maximum (in magnitude) at x = ± a. 9. The maximum extension x produced in the spring in Fig. 13.25 a is given by F = kx or x = F/k The time period of oscillation is mass m =2 force constant k In case (a) one end A wall. When a force F is applied to the free end B in the direction shown in Fig. 13.44a the spring is stretched exerting a force on the wall which in turn exerts an equal and opposite reaction force on the spring, as a result of which every coil of the spring is elongated producing a total extension x. In case (b) shown in Fig. 13.44b, both ends of the spring are free. Therefore, the reaction force is absent, as a result of which every coil of the spring is not elongated when force F is applied at each end in opposite directions. The coil at point O in the middle of the spring is not elongated. This situation can be visualized as two springs each of length l/2 (where l is the length of the complete spring) are joined to each other at point O. Since extension is proporT=2

13.35

tional to the length of the spring, the force F applied at end B produces an extension x/2 in the part OB of the spring and the force F applied at A produces an extension x /2 in the part OA. The total extension x x = x. in the spring is 2 2 Thus, the maximum extension produced in the spring in cases (a) and (b) is the same. Now, the force constant of half the spring is twice that of the complete spring. In case (b) the force constant = 2 k. Hence the time period of oscillation will be m T =2 2k T = 2 T Hence the correct choices are (b) and (d). 10. Statement (a) is correct. At any position between O and P or between O and Q, there are two accelerations–a tangential acceleration g sin and a centripetal acceleration v2/l (because the pendulum moves along the arc of a circle of radius l) where l is the length of the pendulum and v its speed at that position. When the bob is at the mean position O, the angle = 0, therefore sin = 0; hence the tangential acceleration is zero. But at O, speed v is maximum and the centripetal acceleration v2/l is directed radially towards the point of support. When the bob is at the end points P and Q, the speed v is zero, hence the centripetal acceleration is zero at the end points, but the tangential acceleration is maximum and is directed along the tangent to the curve at P and Q. The tension in the string is not constant throughout the oscillation. At any position between O and end points P or Q, the tension in the string is given by T = mg cos . At the end points P or Q, the value of is the greatest, hence the tension is the least. At the mean position O, = 0 and cos = 1 which is the greatest; hence tension is the greatest at the mean position. 11. F1 = k1x, F2 = k2x. 1 1 k1x2 and W2 = k2x2 Work done W1 = 2 2 W k = 1 = 1 W2 k2 When the springs are stretched by the same force F, the extensions in springs A and B are x1 and x2 respectively which are given by x1 k = 2 (i) F = k1 x1 = k2 x2 or x2 k1 Work done

W1 =

1 1 k1 x12 and W2 = k2 x22 2 2

13.36 Comprehensive Physics—JEE Advanced

W1 k = 1 W2 k2 Using (i) and (ii) we get

x12 x22

(ii)

k1 k22 W1 k2 = 2 k2 k1 W2 k1 Hence the correct choices are (a) and (d). 12. In SHM, the velocity V at a displacement x is given by =

(A2 – x2)1/2

V= or V2 = Now

2

V = 3 cm s–1 when 9=

Also

(A2 – x2) 2

(A2 – 16)

V = 4 cm s–1 16 =

2

x = 4 cm. Therefore,

when

(i) x = 3 cm. Therefore,

2

(A – 9)

Subtracting (1) and (2) we have m1g = kx2

Simultaneous solution of Eqs. (i) and (ii) gives Amplitude A = 5 cm Putting A = 5 cm in Eq. (i) above we get = 1 rad s–1. Maximum kinetic energy = maximum potential 1 mA2 2 energy = total energy which is E = 2 1 = (50 10–3) (5 10–2)2 (1)2 2 = 6.25 10–5 J Hence all the four choices are correct. 13. The correct choices are (b) and (c). 14. The correct choices are (a), (b) and (c). In case (b), the equivalent force constant of the series combination is ks = k/2 and in case (b), the equivalent force constant of the parallel combination is kp = 2k. 15. Let x1 be the extension produced in the spring when it is loaded with mass m2 alone and x2 be the further extension when mass m1 is added to mass m2 so that x = x1 + x2 is the total extension produced by m1 + m2 (see Fig. 13.49). Thus we have, For equilibrium state of m2 m2 g = kx1 (1) For equilibrium state of (m1 + m2) (m1 + m2)g = k(x) = k(x1 + x2)

Fig. 13.49

(ii)

(2)

When the mass m1 is removed, the mass m2 will move upwards under the unbalanced force = m1 g. Hence Restoring force (F) on m2 = – m1g

(3)

Hence, Restoring force on m2 = – kx2 F k x2 m2 m2 – displacement

accelerating of m2 = i.e.,

acceleration

k m2 Frequency of oscillation is

Angular frequency is

=

k 1 m2 2 2 It is clear that A is the equilibrium position of m2 and B its maximum displacement position. Hence AB = x2 is the amplitude of oscillation of m2 which from Eq. (3) is given by mg Amplitude = x2 = 1 k Thus the correct choices are (a) and (d). 16. Since the train is moving in a circle, it is in an accelerated (non-inertial) frame of reference. mv2/R is to be introduced as shown in Fig. 13.50. This force is horizontal. Consequently, the equilibrium position of the pendulum will not be vertical; it will be inclined at an angle with the vertical. If T is the tension in the string in this position, it follows from n=

T cos

= mg and T sin

=

m v2 R

Simple Harmonic Motion

F= –

13.37

R 2 gx

Fig. 13.50

Squaring and adding we get 2

2

T = (mg) +

m v2 R v4

2

2

1/ 2

Fig. 13.51

= mge R2 where the effective value of acceleration due to gravity is 1/ 2 v4 ge = g 2 R2 or

T= m

Time period = 2

g

L ge

1/ 2

=2

mL T

g

2

v4 R2

R2 g x 4 R3 3 Using Eq. (2) we get a=

1/ 2

The correct choices are (a) and (c). 17. Mass of sphere is 4 m= R3 = V (1) 3 where V is the volume of the sphere. The volume of sphere under water = volume of water displaced V . If is the density of water, the upthrust is = 2 1 U = (V g) 2 sphere, i.e. U = mg or 1 (V g) = V g 2 or =2 (2) If the sphere is pressed down through a small distance x (see Fig. 13.51), the volume of water displaced due to this pressing = volume of a disc of radius R (since the half the sphere is submerged) and thickness x which is R 2 x. Hence, upthrust due to this pressing is R 2 x g, which provides the restoring force. Hence the restoring force acting on the sphere when it is released is given by

3g 4

F m

a=

1/ 2

L

=2

Therefore, the acceleration of the sphere is [use Eq. (1)]

3g x =– 2R

which gives T = 2

2

x R

(3)

x

2R . Hence the correct choices 3g

are (a) and (c). 18. The two simple harmonic motions are represented by equations 12 t y1 = 10 sin 4 4 = 10 sin and

3 t

(1)

4

y2 = 5 sin 3 t + 5 3 cos 3 t

(2)

Refer to the solution of Q.13 of this section. The correct choices are (b) and (d). The amplitude of 2 each = 10 units and the time period of each = 3 second. 19. Total energy = translational K.E. + rotational K.E. + P.E. stored in spring 1 1 1 = mv2 + I 2 + kx2 2 2 2 = E=

1 2 1 1 mv + mR 2 2 2 2 3 1 mv2 + kx2 4 2

v R

2

+

1 2 kx 2 (1)

13.38 Comprehensive Physics—JEE Advanced

Now, the total energy of the system must remain dE constant, i.e. = 0. dt Differentiating Eq. (1) with respect to time t and dE setting = 0, we have dt

or

C sin2 t 2 Choice (a): Equation (1) can be written as +

x = x0 + a cos2 t + b sin2 t

dv dx dE 3 1 = 0 = m 2v + k 2x dt dt dt 4 2 Now acceleration a = Therefore, or

3 ma kx = 0 2

v

Since v 0, we have 3 ma + kx = 0 2 2k x=– or a=– 3m where

=

2k . Hence T = 2 3m

(2)

x0 =

x = x0 + A0sin(2 + ) 2

(3)

2 1/2

where A0 = (a + b ) and tan = a/b. Equation (3) represents a simple harmonic motion of angular frequency 2 , amplitude = x0 + A0 and phase constant . Choice (b): For A = B and C = 2B, Eq. (1) becomes x = B + B sin2 t = B(1 + sin2 t) 2

x

This equation represents a simple harmonic motion of amplitude 2 B and angular frequency 2 .

(2)

Choice (c): For A = – B and C = 2B, Eq. (1) becomes x = B cos2 t + B sin2 t

2k m

The correct choices are (a) and (c). 20. mg + m 2A = m1g and mg – m 2 A = m2g. Form

which represents a simple harmonic motion of amplitude B, angular frequency 2 and phase constant /4.

(a) and (d). 21. The displacement equation can be rewritten as x=

(1)

1 1 C (A + B), a = (B – A) and b = . 2 2 2 Equation (2) can be recast as

where

dv dx and velocity v = . dt dt

3 mva + k vx = 0 2

1 1 (A + B) + (B – A)cos2 t 2 2

x=

Choice (d): For A = B and C = 0, Eq. (1) reduces to x=A which does not represent simple harmonic motion. Hence the correct choices are (a), (b) and (c).

C A B (1 – cos2 t) + (1 + cos2 t) + sin2 t 2 2 2

III Multiple Choice Questions Based on Passage Questions 1 to 3 are based on the following passage Passage I Two light springs of force-constants k1 = 1.8 Nm–1 and k2 = 3.2 Nm–1 and a block of mass m = 200 g are arranged on a horizontal frictionless table as shown in Fig. 13.52. One end is free. The distance CD between the free ends of the springs is 60 cm and the block moves with a velocity v = 120 cm s–1 between the springs. IIT, 1985

Fig. 13.52

1. When the block moves towards the spring k2, the time taken by it to move from D upto the maximum compression of spring k2 is (in seconds)

Simple Harmonic Motion

(a)

(b)

13.39

(c)

2

(d) 3 4 2. When the block moves towards spring k1, the time taken by it to move from C upto the maximum compression of k1 is (in seconds) 2 (a) (b) 3 (c)

(d) 3 4 3. The period of oscillation of the block between the springs is (in second) (a) 1

5 6

(b) 1

7 6

(c) 1

5 12

(d) 1

7 12

SOLUTION 1. Time taken by block to move from D upto the maximum compression of spring k2 = half the time period of oscillation of the block if it were attached to the free end of k2, i.e. t2 = =

1 T2 = 2 2 1 2

2

m k2

2

0.2 = second 3.2 4

Passage II A uniform cylinder of length L and mass M having crosssectional area A is suspended, with its length vertical, half submerged in a liquid of density at equilibrium position. When the cylinder is given a small downward push and released, it starts oscillating vertically with a small amplitude. IIT, 1990 4. The extension x0 of the spring when it is in equilibrium is Mg g (a) (b) (M – LA ) k k 1 LA 2

(d)

g k

M

1 LA 2

5. If the cylinder is given a small downward displacement x from the equilibrium position and released, the restoring force F acting on it is

2

1 m = 2 k1

2

0.2 1.8

second 3 3. Time period of oscillation of block is T = time taken by the block to move freely from C to D and from D to C = t1 + t2 2 60 cm 1

+

120 cms which is choice (d).

Questions 4 to 6 are based on the following passage

g M k

T1 1 = 2 2

=

=

The correct choice is (d).

(c)

2. Similarly t1 =

3

4

(a) – Mgx (c) = – k

= 1

7 , 12

(b) – (k + A g)x 1 A g x 2

(d) – k

1 A g x 2

6. The time period T of the vertical oscillations of the cylinder is (a) 2

M k

(b) 2 k

M 1 A g 2

(c) 2

M (k A g )

(d) 2

M k A g

1/ 2

1/ 2

1/ 2

13.40 Comprehensive Physics—JEE Advanced

SOLUTION 4. The upthrust on the cylinder with half its length submerged in the liquid is given by

is

U = weight of the liquid displaced by a length L/2 of the cylinder L L g= (A g) 2 2 Let x0 be the extension of the spring when it is in equilibrium. Then L kx0 = Mg – (A g) (1) 2 The correct choice is (c).

F=–

or

x A g

(a) (c)

F = – (k + A g)x

(a) (b)

(M

m)

M M

m A m A

M

(c)

m

m

1/ 2

m

1/ 2

m M

M

(b)

M

m

(d)

M M

(b)

Am A ( M m)

(d)

A ( M m) AM

m m A m A

9. The ratio A /A is

/ is

1/ 2

M M

m

SOLUTION 7. The frequency of the system before the object is placed on the block is given by 1 k 1/ 2 2 M After the object of mass m is placed on the block, the new frequency of the system becomes =

1 = 2

1/ 2

k M

(2)

MA ( M m) A 1 / 2 (d) mA mA 8. If v and v are the velocities before and after the object is placed on the block, then the ratio v /v is

The spring-block system is executing simple harmonic motion of amplitude A and frequency . When the block is passing through the equilibrium position, an object of a mass m is gently placed on the block. As a result, the frequency of the system becomes and the amplitude becomes A .

m

L 2

Mg

(c)

One end of a light spring of force constant k a block of mass M placed on a horizontal frictionless

M

x)

Thus the correct choice is (b) 6. The acceleration of the cylinder is F k A g a= =– x (3) M M Comparing Eq. (3) with a = – 2x, where = 2 /T

Passage III

(a)

k ( x0

F = – (k x + A gx)

Questions 7 to 9 are based on the following passage

1/ 2

A g and the force in the spring is

Using Eq. (1) in Eq. (2), we have

5. Let x be the small downward displacement given to the cylinder so that the submerged length of L the cylinder is now x and the extension 2 of the spring is now (x0 + x). The upthrust now

M

x

k (x0 + x). Hence, the restoring force on the cylinder is

=A

7. The ratio

L 2

m

= 8.

1/ 2

M M

m

,

which is choice (a) From conservation of momentum, we have Mv = (M + m)v or

v M = . v ( M m)

So the correct choice is (a).

1/ 2

1/ 2

Simple Harmonic Motion

9. From the principle of conservation of energy we know that the kinetic energy of the block when it is passing through the equilibrium position = potential energy of the spring when the displacement is equal to the amplitude. Thus we have

=

1 1 Mv2 = k A2 2 2 and

(M

A v = A v

=

1 1 (M + m)v 2 = kA 2 2 2

13.41

1/ 2

m) M

M M

M m

M

1/ 2

M 1/ 2

M

m

m

,

which is choice (c).

Questions 10 to 12 are based on the following passage Passage IV Three simple harmonic motions in the same direction having the same amplitude a and the same period are superposed. Each motion differs from the preceding motion by a phase = 45°. IIT, 1999 10. The amplitude of the resulting simple harmonic motion is (b) (1 + 3 )a (a) (1 + 2 )a

(c) 3a

(d)

3a

2 11. The phase of the resultant motion relative to the (a) 30° (b) 45° (c) 60° (d) 90° 12. If the energy associated with each motion is E, the total energy of the resultant motion is (a) 3E (b) (1 + 2 )E (c) (3 +

2 )E

(d) (3 + 2 2 )E

SOLUTION 10.

x1 = a cos

a sin ( N / 2) and sin ( / 2) a sin (3 / 2) For N = 3, R = sin ( / 2)

t

x2 = a cos ( t +

where

)

x3 = a cos ( t + 2 ) where = 45°. From the principle of superposition, the resultant motion is given by

R=

a (sin cos / 2 cos sin / 2 sin / 2 = a (sin 45° cot 22.5° + cos 45°)

)+

It can be shown that the resultant displacement x for N collinear simple harmonic motions of the same amplitude a and differing in phase by is given by x = R cos ( t + ) Questions 13 to 15 are based on the following passage Passage V Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful

( N 1) 2

=

x = x1 + x2 + x3 x = a cos t + a cos ( t + a cos ( t + 2 )

=

= a ( 2 + 1) 11.

=

1 (3 – 1) 2

12. Total energy =

45° = 45° R2 a2

energy of any one motion

= ( 2 + 1)2E = (3 + 2 2 )E in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one-dimension. For such systems, phase space is a plane in which position is plotted along

13.42 Comprehensive Physics—JEE Advanced

horizontal axis and momentum is plotted along vertical axis. The phase space diagram is x(t) vs. p(t) curve in this plane. The arrow on the curve indicates the time moving with constant velocity is a straight line as shown in the Fig. 13.53. We use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative. IIT, 2011 13. The phase space diagram for a ball thrown vertically up from ground is Fig. 13.54

15. Consider the spring-mass system, with the mass submerged in water, as shown in Fig. 13.55. The phase space diagram for one cycle of this system is

Fig. 13.55

Fig. 13.53

14. The phase space diagram for simple harmonic motion is a circle centered at the origin. In Fig. 13.54, the two circles represent the same oscillator but for different initial condition, and E1 and E2 are the total mechanical energies respectively. Then (a) E1 =

2 E2

(b) E1 = 2E2 (c) E1 = 4E2 (d) E1 = 16E2

Fig. 13.56

SOLUTION 13. According to the given sign convention, position (x) remains positive and momentum is positive

when the body is moving upwards and becomes zero when it reaches the highest point after which

Simple Harmonic Motion 13.43

the momentum (p) becomes negative. Hence the correct graph is (d). 14. Energy of simple harmonic oscillator is

E1 = E2

A1 A2

2

= 4. So E1 = 4E2

15. Due to upthrust, the spring will be compressed.

1 2 kA 2 where k is the force constant and A the amplitude of the oscillator. Since the oscillator is the same, the value of k is the same. Hence 1 1 E1 = kA12 and E2 = kA22 2 2 E=

E1 = E2

2a a

will be smaller than the initial position. Hence choices (c) and (d) are not possible. Due to buoyancy, the block will move upwards. Hence, according to the given sign convention, position (x) is positive initially. When the system is released, x will decrease and momentum (p) will increase becoming maximum when the system reaches the mean position (x = 0) after which the momentum will decrease to zero when the oscillator reaches the extreme position, after which the momentum becomes negative. Hence the correct graph is (b).

2

Now A1 = maximum value of displacement of oscillator having energy E1 = 2a and A2 = a. Therefore

IV Matrix Match Type 1. Column I describes some situations in which a small object moves. Column II describes some characteristics of these motions. Match the situations in Column I with the characteristics in Column II. (a)

(b)

(c)

(d)

Column I The object moves on the x- axis under a conservative force in such a way that its “speed” and “position” satisfy v = c1 c2 x 2 where c1 and c2 are positive constants. The object moves on the x-axis in such a way that its velocity and its displacement from the origin satisfy v = – kx, where k is a positive constant. The object is attached to one end of a massless spring of a given spring constant. The other end of the spring is attached to the ceiling of an elevator. Initially everything is at rest. The elevator starts going upwards with a constant acceleration a. The motion of the object is observed from the elevator during the period it maintains this acceleration. The object is projected from the earth’s surface vertically upwards with a speed 2 GM e / Re , where Me is the mass of the earth and Re is the radius of the earth. Neglect forces from objects other than the earth.

(p)

Column II The object executes a simple harmonic motion

(q)

The object does not change its direction

(r)

The kinetic energy of the object keeps on decreasing.

(s)

The object can change its direction only once.

IIT, 2007

13.44 Comprehensive Physics—JEE Advanced

2. Column I gives a list of possible set of parameters measured in some experiments. The variations of the parameters in the form of graphs are shown in Column II. Match the set of parameters given in Column I with the graphs given in Column II. Column I

Column II

(a)

Potential energy of a simple pendulum (y axis) as a function of displacement (x axis)

(p)

(b)

Displacement (y axis) as a function of time (x axis) for a one dimensional motion at zero or constant acceleration when the body is moving along the positive x-direction

(q)

(c)

Range of a projectile (y axis) as a function of its velocity (x

(r)

(d)

The square of the time peroid (y-axis) of a simple pendulum as a function of its length (x axis).

(s)

IIT, 2008

ANSWERS 1. (a) The velocity v of a body executing simple harmonic motion is related to displacement x as v=

A2

x2

where ands A are positive constants. The given equation v = c1 c2 x 2 is similar to the above equation. Hence the body executes simple harmonic, motion, which is choice (p). (b) It follows from equation v = – kx that v = 0 at x = 0, i.e. object comes to rest at x = 0. Thus the object starting from a negative x value comes to rest at x = 0; its velocity (and hence kinetic energy) decrease with time. Hence correct choices are (q) and (r). (c) For an observer in the reference frame of the elevator, a constant force acts on the object. This is a pseudo force. Hence the motion of the object remains simple harmonic, which is choice (p). 2GM e / Re , the direction of motion (d) Since the speed of the object is 2 times the escape velocity ve of the object does not change (since the forces exerted by objects other than the earth are neglected) and the speed of the object keeps decreasing. Hence the correct choices are (q) and (r). (a)

(p)

(b)

(q), (r)

(c)

(p)

(d) (q), (r) 1 kx2 where x is the displacement and k 2. (a) The potential energy of a simple pendulum is given by U = 2 is the force constant. U is maximum where x = + A, where A is the amplitude and minimum at x = 0. Hence the correct graph is (p).

Simple Harmonic Motion 13.45

(b) For one-dimensional motion with constant velocity (a = 0), displacement S = vt (i.e. s t) and with 1 2 2 constant acceleration S = ut + at , i.e. S t . Hence the correct choices are (q) and (s). 2 (c) For a projectile, range R u2, where u is the speed of projection. Hence the correct graph is (s). 4 2l (d) For a simple pendulum, T2 = . Hence the correct choice is (q). g (a) (p) (b) (q), (s) (c) (s) (d) (q)

V Assertion-Reason Type Questions In the following questions, Statement-1(Assertion) is followed by Statement-2 (Reason). Each question has the following four choices out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Statement-1 (c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. 1. Statement-1 If a spring of force constant k is cut into two equal halves, the force constant of each half is 2k. Statement-2 When an elastic spring is extended by an amount x, the work done is

1

2

k x 2.

2. Statement-1 A particle executes simple harmonic motion between x = – A and x = + A. The time taken for it to go from x = 0 to x = A/2 will be less than the time taken for it to go from x = A/2 to x = A. Statement-2 In simple harmonic motion, the speed of the particle is the maximum at the mean position x = 0 and decreases as it moves towards the extreme position becoming zero at x = A. 3. Statement-1 A body is executing simple harmonic motion. At a displacement x, its potential energy is E1 and at a displacement y, its potential energy is E2. The potential energy at a displacement (x + y) is E = E12 E22 .

Statement-2 For a body executing simple harmonic motion, the potential energy is proportional to the square of its displacement from the mean position. 4. Statement-1 The time period of a simple harmonic oscillator depends upon its amplitude and force constant. Statement-2 The frequency of a simple harmonic oscillator is determined by elasticity and inertia. 5. Statement-1 The amplitude and phase constant of a particle in SHM are determined from its initial displacement and initial velocity. Statement-2 The amplitude and phase constant of SHM depend on the magnitude of the restoring force. 6. Statement-1 The displacement of a particle is given by x = a sin (bt + c) This equation suggests that the time period of motion of 2 /b. Statement-2 The same value of x is obtained at t = t and at t = t + 2 /b. 7. Statement-1 The displacement of a simple harmonic oscillator is given by x = A sin( t + ) This equation suggests that the energy of the oscillator remains constant.

13.46 Comprehensive Physics—JEE Advanced

Statement-2 The same value of x is obtained after an interval equal to the time period of the oscillator. 8. Statement-1 The time period of a simple pendulum is independent of the mass of the bob. Statement-2 The restoring force does not depend on the mass of the bob. 9. Statement-1 For small amplitudes, the motion of a simple pendulum is simple harmonic of time period T = 2 l / g . For larger amplitudes, the time period

Statement-2 For larger amplitude, the speed of the bob is greater when it passes through the mean position. 10. Statement-1 For an oscillating simple pendulum, the tension in the string is constant at T = mg for all positions of the bob. Statement-2 The tension in the string will not remain constant at T = mg because the speed of the bob is different at different positions.

l /g .

is greater than 2

SOLUTION 1. The correct choice is (b). The force required to extend the spring by an amount x is given by F = kx (1) If the spring cut into two equal halves, the same force F will produce half the extension because the extension is directly proportional to the length of the spring. Hence x (2) F =k x =k 2 From (1) and (2), we get k = 2k. 2. The correct choice is (a). 3. The correct choice is (d). E1 =

1 m 2

where

2

x2

k= E2 =

1 and E = m 2

E1 = x

(1)

m 2 2 1 m 2 2

2 2

y

(x + y)

2

From (1), (2) and (3), we get E =

m 2 = xk 2

E1

E2

E = E1 + E2 + 2 E1 E2

E2 = yk

(2)

E = (x + y)k (3)

4. The correct choice is (d). 5. The correct choice is (c). 6. The correct choice is (a). The value of x at t = t + 2 /b is x = a sin b t

2 b

c

= a sin[bt + 2 + c] = a sin(bt + c) = x [

sin( + 2 ) = sin ]

7. The correct choice is (a). 8. The correct choice is (c). The restoring force when the string makes an angle with the vertical is given by F = – mg sin , which depends upon m. 9. The correct choice is (b). Restoring force is F = – mg sin or F = – mge where ge = g sin . For small oscillations, is small so that sin (here is in radian) and the effective value of g is ge . For larger oscillations g sin is less than g because sin < . Hence T is greater than 2 l / g . 10. The correct choice is (d). The tension in the string is given by T = mg cos

+

mv2 l

where l = length of the pendulum, v its speed when the string makes an angle with the vertical.

Simple Harmonic Motion 13.47

VI Integer Answer Type 1. A mass M attached to a spring oscillates with a period of 1s. If the mass is increased by 3 kg, the period increases by 1s. Find the value of M (in kg) assuming that Hookes’ law is obeyed. IIT, 1979 2. An object of mass 0.2 kg executes simple harmonic motion along the x-axis with a frequency of 25/ Hz. At the position x = 0.04 m, the object has kinetic energy of 0.5 J and potential energy of 0.4 J. Find the amplitude of oscillations in cm. IIT, 1994

SOLUTION M M 3 and 2 = 2 k k Dividing these equations we get

1. 1 = 2

3. A block is kept on a horizontal table. The table is undergoing simple harmonic motion of frequency friction between the block and the table is 0.72. Find the maximum amplitude in cm of the table for which the block does not slip on the surface of the table. Take g = 10 ms–2. IIT, 1996 4. A particle executes simple harmonic motion between x = – A and x = + A. It takes time t1 to go from 0 to A/2 and t2 to go from A/2 to A. Find the ratio T2/T1. IIT, 2001 the table, if the acceleration a of the motion of the table is such that [Fig. 13.57] mg ma or g a

M 3 M = 1 kg M 25 2. Angular frequency = 2 = 2 = 50 rad s–1 1 Kinetic energy K = m 2 (A2 – x2) 2 4=

1 m Potential energy U = 2 Dividing we get

2

x

2

K A2 x 2 = U x2 0.5 A2 x 2 = 0.4 x2 3x 3 0.04 m A= = 2 2 = 0.06 m = 6 cm 3. Let the mass of the block be m and let, at a certain instant of time, the direction of acceleration a of the table (executing simple harmonic motion) be along the positive x-direction. As a result, the block will experience a force ma directed along the negative x-axis. Consequently, the force of friction mg will act along the positive x-axis. The weight mg of the block will be balanced by the normal reaction R. The block will not slip on the surface of

Fig. 13.57

Therefore, for no slipping, the table can have a maximum acceleration amax = g. Where amax = 2A. Therefore, the maximum amplitude is given by 2 Amax = mg or

= 4. From

Also

g

Amax =

2

=

g 4

0.72 10 4

2

(3)2

(

2 2

2

From (1) and (2), we get t1 t2 =2 t1

)

= 0.02 m = 2 cm

x = A sin t, we have A = A sin t1 t1 = 2 6 A = A sin (t1 + t2) (t1 + t2) =

=2

(1)

(2) t1 t2

=

1 which gives 3

14

Chapter

REVIEW OF BASIC CONCEPTS 14.1

WAVE MOTION

Wave motion involves the transport of energy without any transport of matter. In case of mechanical waves, the disturbance is the physical displacement of particles of a medium. In case of electromagnetic waves, the distur-

14.2

TYPES OF WAVES

There are two types of wave motions: (1) transverse and (2) longitudinal. (1) Transverse Waves In transverse waves the particles of the medium vibrate at right angles to the direction in which the wave propagates. Waves on strings, surface water waves and electromagnetic waves are transverse waves. (2) Longitudinal Waves In longitudinal waves the particles of the medium vibrate along the direction of wave propagation. Sound waves are longitudinal.

14.3

CHARACTERISTICS OF A HARMONIC WAVE

(1) Amplitude The amplitude of a wave is the maximum displacement of the particles of the medium from their mean position. (2) Period The time period of a wave is the period of harmonic oscillations of particles of the medium. The frequency of a wave is the reciprocal of the time period. (3) Wave Velocity Wave velocity is the distance travelled by the wave in one second. (4) Wavelength tance (measured along the direction of propagation

of wave) between two nearest particles which are in the same phase of vibration.

14.4

DISPLACEMENT EQUATION FOR A TRAVELLING WAVE

When a plane wave travels in a medium along the positive x-direction, the displacement y of a particle located at x at time t is given by y = A sin ( t – kx) where A = amplitude of the wave, (= 2 ) is the is the frequency angular frequency (in rad s–1), 2 (in Hz) and k = is the angular wave number and is the wavelength of the wave. For a wave travelling along the negative x-direction, y = A sin ( t + kx) The wave velocity is given by v=

14.5

=

k

PHASE AND PHASE DIFFERENCE

The argument of the sine (or cosine) function which represents a wave is called the phase of the wave. For a wave travelling along the positive x-direction, the phase at a space point x at time t is given by =

t – kx

It is clear that the phase changes with time t as well as space point x. Phase Change with Time The phase of a given particle (i.e. x time. As time changes from t to (t + t), the phase of a

14.2 Comprehensive Physics—JEE Advanced

particle oscillation changes from to ( + ) where is given by = { (t + t) – kx} – { t – kx} 2 t T where T is the time period of particle oscillation. If t = T, = 2 . Phase Change with Position At a given instant of time t, the phase of particles of the medium varies with position x of the particles. The phase difference at an instant t between two particles separated by x and (x + x) is given by =

t=

= { t – k(x + = –k

x=

x)} – ( t – kx) 2

x

The minus sign indicates that, for a wave travelling along the positive x-direction, the particles located at higher values of x lag behind in phase. If x = , | |=2 . Hence two particles whose phases differ by 2 . 14.1 When a plane wave travels in a medium, the displacements of particles are given by y = 0.01 sin [2 (2t – 0.01x)] where x and y are in metre and t in second. Find (a) the amplitude, wavelength, wave velocity and frequency of the wave, (b) the phase difference between two positions of the same particle in a time interval of 0.25s and (c) the phase difference at a given instant of time between two particles 50 m apart. SOLUTION (a) The given equation is y = 0.01 sin (4 t – 0.01 2 x) Comparing this equation with y = A sin ( t – kx), we get A = 0.01 m =4

2

k = 0.01

2

=4 2

= 2 Hz =

= 100 m v= (b)

=2

t=2

=2

2 100

100 = 200 ms 2

(0.25) =

–1

= 180°

(c)

2 50 = – = – 180° 100 The negative sign indicates that the particle at x = 50 m lags behind the particle at x = 0 by a phase angle of 180°. =

2

x=

14.2 A travelling wave on a string is given by y = 7.5 sin 0.005 x 12t

4

where x and y are in cm and t in second. (a) Find the displacement and velocity of the particle at a point x = 1 cm at t = 1s. Is this equal to the wave velocity? Given sin (12.7°) = 0.22. (b) Locate the points on the string which have the same transverse displacement and velocity as the x = 1 cm point has at t = 2s, 5s and 11s. SOLUTION Given

y = 7.5 sin 0.005 x 12t

(i)

4

Putting x = 1 cm and t = 1s in Eq. (i) we have y = 7.5 sin 0.005 1 12 1

3.142 4

= 7.5 sin (12.79 rad) = 7.5 sin (732.7°) ( rad = 180°) = 7.5 sin (4 + 12.7°) = 7.5 sin (12.7°) = 7.5 0.22 = 1.65 cm Particle velocity is obtained by differentiating (i) w.r.t. time t. dy = 7.5 12 cos 0.005 x 12t V= 4 dt Putting x = 1 cm and t = 1s, V = 7.5 12 cos 12.7° = 87.8 cm s–1 Comparing Eq. (i) with y = A sin (kx +

t+

0)

–1

we get k = 0.005 rad cm and = 12 rad s–1 giving 12 v= = = 2400 cm s–1 k 0.005 Particle velocity V is not equal to wave velocity v. ), all points on the string which are located at x = , 2 , ...

Waves and Doppler’s Effect 14.3

= n (n = 0, 1, 2, ...) away from x = 1 cm have the same displacement and velocity as those at x = 1 cm for all values of t. 2 = 0.005 = 12.67 cm. From k = 0.005 Hence the positions of the required points are x = 1 cm + (12.67 cm) n = (1 + 12.67 n) cm where n is an integer. 14.3 A tuning fork vibrating at 200 Hz produces a sound wave which travels in air at a speed 340 ms–1. Find the distance travelled by sound during the time the fork makes 60 vibrations. SOLUTION = 200 Hz, v = 340 ms–1 =

v

=

340 = 1.7 m 200

) is the distance travelled by the wave in 1 complete vibration (= time period) of the fork. Hence x = 60 1.7 = 102 m

14.6

EXPRESSIONS FOR WAVE VELOCITY IN DIFFERENT MEDIA

1. Velocity of sound in an elastic medium is given by v= where and

E

E = modulus of elasticity of the medium = density of the medium.

(a) For gases: E=

P

where g = Cp/Cv of the gas at constant pressure and that at constant volume and P is the pressure of the gas. Thus P

v=

(b) For solids: E = Y; the Young’s modulus of the solid. Thus v=

Y

(c) For liquids: E = B, the bulk modulus of the liquid. Thus B

v=

2. The velocity of sound in a gas is independent of the pressure but is directly proportional to the square root of the absolute temperature. vt T t 273 1 / 2 = = 273 v0 T0 where t is the temperature in °C. 3. The velocity of sound increases with increase in humidity. Sound travels faster in moist air than in dry air at the same temperature. 4. Velocity of a transverse wave on a stretched string is given by T m where T = tension in the string and m = mass per unit length of the string. For a string of diameter d and density , we have v=

d2 4

m =

Thus, v =

2 d

T

14.4 Uniform rope AB of mass M and length L hangs vertically with end A is created at the free end A. The speed of the pulse is (a) the same at every point of the rope (b) maximum near end A (c) maximum near end B (d) zero at the mid-point of the rope. SOLUTION ent at different points on the rope. Hence the speed of the pulse is different at different points. M . Mass per unit length is m = L Tension at point P (Fig. 14.1) is Mgx = mgx L Speed of the pulse is T=

v=

T m

mgx m

x

gx

Fig. 14.1

i.e. v x . Hence the correct choice is (b). The speed of the pulse is

14.4 Comprehensive Physics—JEE Advanced

14.5 A long uniform steel wire has a diameter of 2.0 mm. What should be the tension in the wire, so that the speed of the transverse wave on it equals the speed of sound at STP (= 320 m s–1)? The density of steel is 7800 kg m–3. SOLUTION The volume of a wire of length L and diameter d is V = r2L = d2L/4 The mass of the wire is d 2L M = volume density = 4 Mass per unit length (m) = Now

v=

M = L

Hence

T m

d 2 v2 T = mv2 = 4 Substituting the values of d, and v and solving we get T = 25.1 N 14.6 Transverse waves are generated in two uniform steel wires A and B of diameters 10–3 m and 0.5 10–3 m respectively, by attaching their free end to a vibrating source of frequency 500 Hz. Find the ratio of the wavelengths if they are stretched with the same tension.

now

vA =

T mA

and

vB =

T mB

vA = vB but vA = source.

A

and vB =

1P

v1 =

10

and

1

v1 = v2

3

3

= 0.5

2P

v2 =

2

1 2 2 1

Since density of a gas is proportional to its molecular weight, 2 1

= 44.01 = 21.83 2.016

v1 1.4 = v2 1.3

21.83

1/2

= 4.85

Velocity of sound in hydrogen is 4.85 times that in carbon dioxide. 14.8 At what temperature will sound travel in hydrogen with the same speed as in oxygen at 927°C? SOLUTION v1 = v2

1 2 2 1

Hydrogen and oxygen are both diatomic gases. Hence, 1 = 2. Also molecular mass of oxygen 2 = molecular mass of hydrogen 1 =

being the frequency of the

0.5 10

dB dA

SOLUTION

mB d = B mA dA B,

vA vB

14.7 Compare the velocities of sound in hydrogen (H2) and carbon dioxide (CO2). The ratio ( of H2 and CO2 is respectively 1.4 and 1.3.

SOLUTION The density of a wire of mass M, length L and diameter d is given by 4M 4m = = 2 d L d2

=

B

2

d 4

A

For oxygen

32 = 16 2

v1 = v2

16 = 4

vt = v0

t

vt = v2

927 273 273

273 273

1/2

1/2

=

1200 273

1/2

(i)

Waves and Doppler’s Effect 14.5

For hydrogen

1/2

273 t 273

vt = v1

(ii)

Dividing (i) and (ii) vt vt

v1 = v2

vt vt

4=

1200 273 t

1/2

1200 273 t

1/2

4= 16 =

1200 273 t

1/2

t = – 198°C 14.9 In a laboratory experiment (room temperature being 15°C) the wavelength of a note of sound of frequency 500 Hz is found to be 0.68 m. If the density of air at STP is 1.29 kg m–3 heats of air. SOLUTION Speed of sound at 15°C is v1 = = 500 0.68 = 340 ms–1 273 Speed of sound at 0°C is v0 = 340 273 15 = 331 ms–1 –3 At STP, 105 N m–2 0 = 1.29 kgm and P0 = 1.01 Now

v0 =

P0 0

;

=

Cp Cv

v2 (331) 2 1.29 = 0 0 = = 1.39 P0 1.01 105

14.7

The following three cases of superposition are of practical importance.

14.8

Given v t = vt. Hence 1200 273 t

to the algebraic sum of the individual displacements. This is called the principle of superposition. y = y1 + y2 + ... + yn

SUPERPOSITION OF WAVES (THE SUPERPOSITION PRINCIPLE)

When a wave reaches a particle of a medium, it imparts a displacement to that particle. If two or more waves arrive at a particle, the resultant displacement of the particle is equal to the vector sum of individual displacements. In the particular case when the waves travelling in the same straight line superpose, the resultant displacement is equal

INTERFERENCE

The superposition of two waves of the same frequency travelling in the same direction in a medium is called interference. Consider two waves of equal amplitude a, equal angular frequency and and equal angular wave number k but having a phase difference travelling along the positive x-direction. The displacements y1 and y2 of a particle located at x at time t are y1 = a sin ( t – kx) and y2 = a sin ( t – kx + ) According to the superposition principle, the resultant displacement is given by y = y1 + y2 = a [sin ( t – kx) + sin ( t – kx + )] Using sin get

+ sin

= 2 sin

y = 2a cos y = A sin

2 t

cos

2 sin kx

t

kx

2

, we

2

2

where A = 2a cos

is the resultant amplitude. The 2 frequency and wavelength of the resultant wave remain the same as those of individual waves. (a) Constructive Interference: If A is maximum (positive or maximum), the interference is constructive, the Amax = 2a. This happens if cos

= 1

2 2

= 0, , 2 , ... = 2n ;

(n = 0, 1, 2, ...)

(b) Destructive Interference: If A = 0, the interference is destructive. This happens if cos

2

=0 3 5 , , 2 2 2 2 = (2n + 1) ; =

,

n = 0, 1, 2, ...

14.6 Comprehensive Physics—JEE Advanced

14.10 Two waves each of amplitude 2 cm and wavelength 5 cm and frequency 10 Hz have a constant phase difference of 60°. Travelling in the same direction, they superpose at a particle of the medium. What is the resultant amplitude of the oscillations of the particle? tant wave.

Resultant amplitude is A = 2a cos =2

cos

Nodes There are certain points in the medium which are permanently at rest. These points are called nodes. The position of nodes is given by A =0

SOLUTION 2 (2 cm)

Equation (i) represents a standing wave. It does not represent a travelling wave since it does not involve the combination ( t kx) in the argument of the sine or cosine function. Equation (ii) tells us each particle has a simple harmonic motion and Eq. (iii) tells us that the amplitude of motion is different for different particles (i.e. for different values of x). Such simple harmonic motions of the particles of a medium are called normal modes.

60 2

2a sin (kx) = 0

= 2 3 cm

sin kx = 0

The frequency and wavelength of the resultant wave are 10 Hz and 5 cm respectively, the same as those of the individual interfering waves.

14.9

kx = 0, , 2 , ... 2

STANDING (OR STATIONARY) WAVES

Standing (or stationary) waves are produced when two waves of the same frequency travelling in opposite directions in a medium superpose. In actual practice, we do not send two independent waves in a medium in length or a rod or a column of gas or liquid. The wave rise to standing waves.

, , ... and so on 2 Distance between two consecutive nodes = x = 0,

yr = – a sin ( t – kx)

There are certain points in the medium which have maximum (positive or negative) amplitude. These points are called antinodes. Their position is given by A= 1

= a [sin ( t + kx) – sin ( t – kx)] 2

cos

sin kx = 1 3 2 2 2 3 x= , 2 2 kx =

2

, we get

y = 2a sin (kx) cos ( t) y = A cos t

(i) (ii)

A = 2a sin(kx)

(iii)

,

5 , 2 5 , , 2 ,

3 5 , , 4 4 4 Distance between two consecutive antinodes = x=

From the superposition principle, the resultant displacement is y = yi + yr

where

.

2a sin (kx) = 1

yi = a sin ( t + kx)

Using sin – sin = 2 sin

2

Antinodes

undergoes a reversal of amplitude (which implies a phase change of ). Consider a wave travelling in the negative x-direction towards a boundary at x = 0, where it is

and

x = 0, , 2 , ...

,

2

.

NOTE 1. Exactly mid-way between two nodes is an antinode and vice versa. 2. The distance between a node and the next antinode . 4 3. There is no transfer of energy along the medium. =

Waves and Doppler’s Effect 14.7

14.11 Standing waves are produced by the superposition of two waves y1 = 5 sin (3 t – 2 x) and y2 = 5 sin (3 t + 2 x) where y and x are in cm and t in second. Find the amplitude of the particle at x = 2 cm.

The frequency of the fundamental mode is 1

=

v

=

v 2L

T m (b) Second Harmonic. [Fig. 14.2 (b)] where

v=

For the second harmonic,

SOLUTION y = y1 + y2 = 5 [sin (3 t – 2 x) + sin (3 t + 2 x)] Using sin ( + ) + sin ( – ) = 2 sin cos , we get y = 10 cos (2 x) sin (3 t) y = A sin (3 t), where A = 10 cos (2 x) Amplitude A at x = 2 cm is 10 cos (2 2) = 10 cos 4 = 10 cm Normal Modes of a String Fixed at Both Ends Consider a uniform string of length L stretched with a tension T x = 0 and x = L. The string can vibrate in a number of modes. Figure 14.2

2

=

v

=

2

2

v =2 L

1

=L

= L.

=L

=

(c) Third Harmonic. [Fig.14.2(c)] For the third harmonic,

2

2

2

2L 3

3v =3 1 2L In general, for a string vibrating in the nth harmonic, the frequency of vibration is 3

=

v

n

=

nv n T = 2L 2L m

=

14.12 ends is given by

2 x cos (120 t) (i) 3 where y and x are in metre and t in second. The length of the string is 1.5 m and its mass is 3 10–2 kg. (a) What is the amplitude at point x = 0.5 m? (b) What is the velocity of the particle at x = 0.75 m at t = 0.25 s? (c) Write down the equation of the component waves whose superposition gives the vibration given in Eq. (i) above. What is the wavelength, frequency and speed of each wave? (d) Determine the tension in the string. (e) Do all points on the string vibrate with the same (i) frequency, (ii) phase and (iii) amplitude? y = 0.06 sin

SOLUTION Fig. 14.2

(a) Fundamental Mode (or First Harmonic) In this mode, the string vibrates in one segment. [Fig. 14.2 (a)] 2

=L

= 2L

(a) Displacement is maximum when cos (120 t) = 1. 2 x Hence the amplitude = 0.06 sin . 3 2 At x = 0.5 m, the amplitude = 0.06 sin 0.5 3 = 0.06 sin

3

= 0.052 m

14.8 Comprehensive Physics—JEE Advanced

(b) The velocity at point x at time t is obtained by differentiating Eq. (i) with respect to t. V=

dy dt

= – (0.06

120 ) sin

2 x sin(120 t) 3

Now at t = 0.25s, sin(120 t) = sin (30 ) = 0. Hence at t = 0.25s, the velocity is zero for all values of x including x = 0.75 m. (c) The stationary wave in the question may be written as y = 2A sin cos where A = 0.03 m,

2 x and 3

=

= 120 t.

Now 2A sin cos = A sin( + ) + A sin( – ) = 0.03 sin

2 x 3

120 t

2 x + 0.03 sin 3 y = y1 + y2

i.e.

y2 = 0.03 sin

2 x 3

120 t

120 t

= – 0.03 sin 120 t

2 x 3

Let be the wavelength, the frequency and v the speed of each wave. Then 2 x in the argument 2 of the sine function = 3 or = 3 m. Also

Young’s modulus Y=

which gives

Hence,

T Yl 0.9 1011 0.05 = = A L 100 = 4.5

= 60

3 = 180 m s–1

3.0 10 2 1.5 = 2.0 10–2 kg m–1

(d) Mass per unit length (m) =

107

Now mass per unit length m= = =

mass length volume density length area

= area =A

length density length density

Density = 9 g cm–3 = 9000 kg m–3. The lowest frequency is the frequency of the fundamental mode.

= 60 Hz. v=

stress T /A TL = = strain l/L Al

where T = tension, L = original length = 100 cm, l = extension = 0.05 cm and A = area of cross-section of the wire.

=2 t in the argument of the sine function = 120

Hence

14.13 A wire of density 9 g cm–3 is stretched between two clamps 100 cm apart while being subjected to an extension of 0.05 cm.. What is the lowest frequency of transverse vibrations of the wire, assuming that the Young’s modulus of the material of the wire = 0.9 1011 N m–2. SOLUTION

Hence the two component waves are 2 x y1 = 0.03 sin 120 t 3 and

T We know that v = , where T is tension in the m string. T = mv2 = 2.0 10–2 (180)2 = 648 N (e) In a stationary wave on a string, all points on the string vibrate with the same frequency and the same phase, but the amplitude is different at different points (i.e. different values of x). The amplitude is zero at nodes and maximum at antinodes.

1

= =

1 T 1 = 2L m 2L 1

2 1 = 35.3 Hz

T A

4.5 107 9000

Waves and Doppler’s Effect 14.9

14.14 A wire having a linear density of 0.05 g cm –1 is stretched between two rigid supports with a tension of 4.5 102 N. It is observed that the wire resonates at a frequency of 420 Hz. The next higher frequency at which the wire resonates is 490 Hz. Determine the length of the wire. SOLUTION Let 420 Hz be the nth harmonic, then 490 Hz is the (n + 1)th harmonic. Therefore

and

420 =

n T 2L m

(i)

490 =

(n 1) T 2L m

(ii)

Dividing (i) and (ii) we get n = 6. Putting n = 6 in (i) we get 6 4.5 102 420 = 2 L 0.05 10 1

(a) Fundamental Mode (or First Harmonic) [Fig. 14.3 (a)] =L

4 1

=

v

L=4 =

v 4L

P where v = is the speed of sound in the gas. (b) Third Harmonic (or First Overtone) [Fig. 14.3 (b)] 4L =L = 2 4 3 v 3v = =3 1 3 = 4L (c) Fifth Harmonic (or Second Overtone) [Fig. 14.3 (c)] 2

L = 2.14 m

2

=L

4 5

Normal Modes in Air Columns in a Pipe A gas column in a pipe can oscillate in a number of modes Case 1: Closed Pipe Consider a pipe of length L open at one end and closed at the other. The closed end is a node and the open end is

=

v

= =

4L 5

5v =5 4L

1

In general, for nth harmonic n

=

nv n = 4L 4L

P

where n = 1, 3, 5, ... etc. NOTE

closed pipe.

In a closed pipe only odd harmonics are present; all the even harmonics are absent.

Case 2: Open Pipe

Fig. 14.3

14.10 Comprehensive Physics—JEE Advanced

14.15 A pipe of length 20 cm is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 430 Hz source? Will the same source be in resonance with the pipe if both ends are open? Speed of sound = 340 ms–1. Fig. 14.4

(a) Fundamental Mode (or First Harmonic) [Fig. 14.4 (a)] 4

=L

4

= 2L

v 2L (b) Second Harmonic (or First Overtone) [Fig. 14.4 (b)] 1

4

2

=

v

=

=L

4 2

=

=

v =2 2L

1

(c) Third Harmonic (or Second Overtone) [Fig. 14.4 (c)] 2L =L = 4 2 2 4 3 v 3v = =3 1 3 = 2L For nth harmonic n

= n 1;

Let N be the frequency of the source and m that of the mth harmonic of the closed pipe, where m = 1, 3, 5, .... Resonance will occur if mv N= m= 4L 430 =

m 340 4 0.2

which gives m = 1.01

=L

v

SOLUTION

n = 1, 2, 3, ...

NOTE In an open pipe, all harmonics (even as well as odd) are present.

End Correction We have taken the open end of a pipe to be an antinode. This is not strictly true. In fact, the particles of air just at the open end are not perfectly free because of the restriction imposed by the pipe. The true antinode is slightly away from the open end as shown in Fig. 14.5.

Fig. 14.5

The distance e is called the end correction. The effective length of the pipe is (L + e).

1. The source of frequency

fundamental mode) of the closed pipe. For an open pipe, the condition of resonance with the same source will be nv , where n = 1, 2, 3, ... N= n= 2L N 2L 430 2 0.2 = = 0.5 n= v 340 which is not an integer. Hence the source will not be in resonance with any harmonic of the open pipe. 14.16 A half-metre long tube open at one end, with a movable piston shows resonance with a tuning fork of frequency 512 Hz when the tube length is 16.0 cm and 49.0 cm. Calculate the speed of sound at the temperature of the experiment and determine the endcorrection. SOLUTION The tuning fork is in resonance at two lengths of the pipe, viz. 16 cm and 49 cm. Since the second reso(i.e. the fundamental mode) when L1 = 16 cm and with the third harmonic when L2 = 49 cm. Thus we have v (i) N= 1= 4( L1 e) Also

N=

3

=

3v 4( L2 e)

(ii)

Waves and Doppler’s Effect 14.11

where N is the frequency of tuning fork and e is the end-correction. Dividing (ii) by (i) we get 3(16.0 + e) = 49.0 + e or e = 0.5 cm Then the speed of sound as obtained from (i) is v = 4N(L1 + e) = 4 512 (16.0 + 0.5) = 33792 cm s–1 = 337.92 m s–1 338 m s–1

14.10

BEATS

The periodic rise and fall of intensity of the wave resulting from the superposition of two waves of different frequencies is called the phenomenon of beats. Consider two waves of angular frequencies 1 and 2. For simplicity, we assume that they have equal amplitude a and that the observation point is at x = 0. Then y1 = a sin 1t; 1 = 2 1 and y2 = a sin 2t; = 2 2 2 Using the superposition principle, y = y1 + y2 = a (sin

1t

= 2a cos

1

y = A sin( where

av

+ sin

t); 1

A = 2a cos

2

2

2t)

av 2

2

1

t sin =

1 ( 2

2

2 1

+

t 2)

t

Now, intensity is proportional to A2. Therefore, the resultant intensity is maximum when 1

cos

1 1

2

2 2

2

t= 1

2

t = 0, , 2 , ...

2

t = 0, , 2 , ...

2

t = 0,

1 (

2)

1

,

2 (

1

2)

,

The time interval between two consecutive maxima is 1 tb = 1

=

1 = tb

1

–

2

Similarly, we can show that the frequency of minima =

Thus Beat frequency = difference between the frequencies of interfering waves. 14.17 Two tuning forks A and B produce 10 beats per second when sounded together. On loading fork A with a little wax it is observed that 5 beats per second are produced. If the frequency of fork B the frequency of fork A (a) before loading and (b) after loading. SOLUTION There are two possibilities (i) A < B or (ii) A > B. 480 – A = 10 Case (i) A < B ; B – A = b A = 470 Hz. On loading with a little wax, the frequency of a fork decreases slightly, i.e. A becomes slightly less than 470 Hz. Hence the number of beats per second must increase. But b decreases to 5. Hence A cannot be less than B. Case (ii) A > B. In this case A = B + b = 480 + 10 = 490 Hz. On loading A, A decreases. Hence b = A – B will decrease. Since b is observed to decrease to 5, A must be greater than B. (a) Hence before loading, A = 490 Hz (b) After loading A = 490 – 5 = 485 Hz 14.18 A metal wire of diameter 1.5 mm is held on two knife edges separated by a distance of 50 cm. The tension in the wire is 100 N. The wire vibrating with its fundamental frequency and a vibrating tuning fork together produce 5 beats per second. The tension in the wire is then reduced to 81 N. When the two are excited, beats are heard at the same rate. Calculate (a) the frequency of the fork, and (b) the density of the material of the wire. SOLUTION

2

Therefore frequency of maxima is b

Hence the frequency of beats is b = 1 – 2

b.

(a) Let N be the frequency of the tuning fork. Then, the frequency of the wire, when the tension is 100 N will be (N + 5) and when the tension is 81 N, it is (N – 5); since in each case 5 beats are heard per second. Hence

14.12 Comprehensive Physics—JEE Advanced

N+5=

and N – 5 =

1 T1 1 100 10 = = 2L m 2 0.5 m m (i) 1 T2 1 81 = = 2L m 2 0.5 m

9 m (ii)

Subtracting (ii) from (i) we have 1

10 =

m

or

m = 0.01 kg m–1

Using this values of m in (i) or (ii) gives N = 95 Hz. (b) Now m = r2 = Putting d = 1.5 we get = 5.7

14.11

d2 4 10–3 m and m = 0.01 kg m–1, 103 kg m–3

DOPPLER EFFECT IN SOUND

The apparent change in frequency of sound heard by an observer due to a relative motion between the observer and the source of sound is called the Doppler effect. The expressions for the apparent frequencies are as follows: 1. Source approaching a stationary observer

where

v

, v us = real frequency v = velocity of sound us = velocity of source 1

=

2. Source receding from a stationary observer v v + us 3. Observer approaching a stationary source of sound v + u0 , 3 = v where u0 = velocity of observer 2

=

4. Observer receding from a stationary source of sound v u0 4 = v 5. Both approaching each other 5

=

v v

u0 us

6. Both receding from each other v u0 v us 7. Source approaching a receding observer 6

=

v u0 v us 8. Observer approaching a receding source 7

=

v u0 v us When the source of sound goes pass stationary observer, the apparent change in the frequency of sound is given by 2 vus = v 2 us2 8

=

2us v When the observer goes past a stationary source of sound, the apparent change in frequency of sound is given by If us << v, then v2 – u2s

=

v2, then

=

2 u0 v

NOTE (1) If the source of sound moves, the apparent change in frequency is due to change in wavelength; the speed of sound remaining the same. (2) If the observer moves, the apparent change in frequency is due to change in the speed of sound relative to observer; the wavelength of sound remaining the same.

Effect of the Motion of the Medium The velocity of material or mechanical waves is affected by the motion of the medium. If the medium is moving with a velocity um in the direction of propagation of sound, the effective velocity of sound is increased from v to (v + um). In this case, v is replaced by (v + um) in the above expressions. On the other hand, if the medium is moving with a velocity um in a direction opposite to the direction of wave propagation, v is replaced by (v – um).

14.12

DOPPLER EFFECT IN LIGHT

If a star emitting light of frequency goes away from the earth with a speed v, the apparent frequency of the light reaching the earth is given by =

c

c v where c is the speed of light. Since we have

= c/ and

= c/ ,

Waves and Doppler’s Effect 14.13

c

or

=

c

=1+

c c

v

v or c

=

v c

v c The apparent change in wavelength is called the Doppler shift. a moving object decreases, the object is moving towards the observer and vice versa. The wavelength of light or

=

called the red shift which indicates that the galaxy is receding from us. The red shift indicates that the universe is expanding. 14.19 A train standing at the outer signal of a railway station is blowing a whistle of frequency 500 Hz. The speed of sound is 340 m s–1. (i) Find the frequency of the sound of the whistle heard by a man standing on the platform when the train (a) approaches the platform with a speed of 20 ms–1 and (b) recedes from the platform with a speed of 20 ms–1. (ii) What is the speed of sound in each case? (iii) What is the wavelength of sound heard by the man in each case? SOLUTION –1

= 500 Hz, us = 20 m s , v = 340 m s

–1

(i) (a) Train approaching 1

v 500 340 = = = 531 Hz v us 340 20

(b) Train receding 1

v 500 340 = = = 472 Hz v us 340 20

(ii) Since the observer is at rest, the speed of sound relative to him is v = 340 m s–1 in cases (a) and (b). (iii) Now, wavelength speed of sound relative to observer = frequency of sound heardd by him v 340 In case (a) = = 0.64 m 1= 531 1 v 340 = = 0.72 m In case (b) 1= 472 1

14.20 A train standing at a platform is blowing a whistle of frequency 500 Hz in still air. The speed of sound in still air is 340 m s–1. (i) What is the frequency of the sound of the whistle heard by a man on a trolley which is moving (a) towards the engine with a speed of 20 m s–1 and (b) away from the engine with a speed of 20 m s–1? (ii) What is the speed of sound heard by the man in each case? (iii) What is the wavelength of sound heard by the man in each case? SOLUTION = 500 Hz, u0 = 20 m s–1, v = 340 m s–1 (i) (a) Observer approaching 2

=

(v

u0 )

=

v

500

(340 340

20)

500

(340 340

20)

= 529 Hz (b) Observer receding 2

(v

=

u0 )

=

v = 471 Hz (ii) Since the observer is moving, the speed of sound heard by him in case (a) is (v + u0) = 340 + 20 = 360 m s–1 and case (b) it is (v – u0) = 340 – 20 = 320 m s–1, (iii) Wavelength = In case (a) In case (b) Notice that

2

2

=

2

=

=

speed of sound relative observer frequency of sound heard byy him v

u0

=

360 = 0.68 m 529

u0

=

320 = 0.68 m 471

2

v

2

2.

14.21 An ambulance blowing a siren of frequency 700 Hz speed of 2 ms–1. If the speed of sound is 350 m s–1, calculate the number of beats heard in one second by (a) the driver of the ambulance, (b) a person standing between the ambulance and the wall, and (c) a person standing behind the ambulance.

14.14 Comprehensive Physics—JEE Advanced

(c) A person standing at B will hear two sounds, one from the siren receding from him and the other from the mirror image of the siren approaching him with the same speed.

SOLUTION (a) The driver will hear two sounds, one coming the wall or coming from the acoustic image of the car. = frequency of direct sound = 700 Hz. ing from the mirror image (shown dotted in Fig. 14.6).

1

v

=

v

us

= 700

350 350 2

= 696 Hz 2

v

=

v

us

= 700

350 350 2

= 704 Hz Beat frequency = 2 – 1 = 704 – 696 = 8 Hz

14.13

INTENSITY, QUALITY AND PITCH OF SOUND

1. Intensity The intensity of a travelling wave is per unit area of a surface held perpendicular to the direction of propagation. Intensity I of a wave in a medium of density is given by

Fig. 14.6

I = 2 2 v A2 2 where is the frequency of particle oscillation, A their amplitude and v is the velocity of the wave in that medium. The SI unit of intensity is watt per square metre (W m–2).

The observer (driver) is approaching this image-source which is also approaching him with the same speed. Hence, the frequency of sound heard by him is given by (u0 = us) =

v v

u0 = 700 us

350 350

2 2

an arbitrary scale. The zero of the scale is taken at the sound wave intensity

= 708 Hz Number of beats heard per second = – = 708 – 700 = 8 Beat frequency = 8 Hz (b) A person standing at A will hear two sounds, one from the siren approaching him and the other from the mirror image of the siren also approaching him with the same speed. Hence he will hear no beats.

I0 = 1 10–12 W m–2 which corresponds to the faintest audible sound. The unit of intensity level is called decibel (dB). Intensity level in decibels of a sound of intensity I = 10 log

I I0

I Multiple Choice Questions with Only One Choice Correct 1. When a plane harmonic wave of wavelength travels in a medium, the particle speed will always be less then the wave speed if the amplitude of the wave is

(a) less then

2

(c) greater then

(b) less then (d) greater then

Waves and Doppler’s Effect 14.15

2. An open pipe and a closed pipe have the same length L. The ratio of the frequencies of their n th overtones is 2n n (a) (b) n 1 n 1 (c)

2 n 1 n 2

(d)

graphs shown in Fig. 14.7 correctly represents the variation of the apparent frequency of sound as heard by the observer with time t?

2 n 1 2n 1

3. also found that a vibrating source resonates with the n th harmonic of the closed pipe and with the m th harmonic of the open pipe. The ratio n/m is 2 3 (b) (a) 3 2 3 5 (d) 5 3 4. A standing wave is produced due to a superposition (c)

a boundary. It is observed that the amplitude at antinode is 9 times that at node. The percentage of is

(a) 36% (b) 64% (c) 72% (d) 81% 5. A source emitting a sound of wavelength

is in

Another source emitting a sound of wavelength is in resonance with the second overtone of the same pipe. The ratio / is 5 4 (b) (a) 3 3 3 5 (c) (d) 2 2 6. The displacements of a particle located at x at time t due to two waves are given by y1 = a sin ( t – kx) And y2 = a sin ( t – k x + ) If the amplitude of the resultant wave formed by the superposition of these two waves is a, the phase constant is equal to (a) zero

(b)

2

2 3 (c) (d) 3 4 7. Starting from rest, an observer moves with a constant acceleration a towards a stationary source emitting a sound of frequency 0. Which of the

Fig. 14.7

8. A train is moving with a constant speed u along a circular track. A siren in its engine is emitting a sound of frequency . If u = v/10, where v is the speed of sound, the apparent frequency of sound as heard by a passenger at the rear end of the train is 9 11 (b) (a) 10 10 9 (c) (d) 11 9. A progressive wave in a medium is represented by the equation y = 0.1 sin 10 t

5 x 11

where y and x are in cm and t in seconds. The maximum speed of a particle of the medium due to the waves is (a) 1cm s–1 (b) 10 cm s–1 (c) cm s–1 (d) 10 cm s–1 10. A tuning fork of frequency 340 Hz is sounded above a cylindrical tube 1 m high. Water is slowly poured into the tube. If the speed of sound is 340 ms–1, at what levels of water in the tube will the (a) 25 cm, 75 cm (b) 20 cm, 80 cm (c) 15 cm, 85 cm (d) 17 cm, 83 cm 11. A sonometer wire, 65 cm long, is in resonance with a tuning fork of frequency N. If the length of the wire is decreased by 1 cm and it is vibrated with the same tuning fork, 8 beats are heard per second. What is the value of N?

14.16 Comprehensive Physics—JEE Advanced

(a) 256 Hz (b) 384 Hz (c) 480 Hz (d) 512 Hz 12. Two organ pipes, each closed at one end, give 5 beats per second when emitting their fundamental notes. If their lengths are in the ratio of 50 : 51, their fundamental frequencies (in Hz) are (a) 250, 255 (b) 255, 260 (c) 260, 265 (d) 265, 270 13. Two sources A and B are sounding notes of frequency 680 Hz. A listener moves from A to B with a constant velocity u. If the speed of sound is 340 ms–1, what must be the value of u so that he hears 10 beats per second? (a) 2.0 ms–1 (b) 2.5 ms–1 (c) 3.0 ms–1 (d) 3.5 ms–1 14. Standing waves are produced by the superposition of two waves y1 = 0.05 sin (3 t – 2x) and y2 = 0.05 sin (3 t + 2x) where x and y are expressed in metres and t is in seconds. What is the amplitude of a particle at x = 0.5 m. Given cos (57.3°) = 0.54. (a) 2.7 cm (b) 5.4 cm (c) 8.1 cm (d) 10.8 cm 15.

both ends is given by 2 x y = 0.06 sin cos (120 t) 3 where x and y are in metres and t is in seconds. The length of the string is 1.5 m and its mass is 3.0 10–2 kg. What is the tension in the string? (a) 648 N (b) 724 N (c) 832 N (d) 980 N 16 . A pipe of length 20 cm is closed at one end. Which harmonic mode of the pipe is resonantly excited by a 425 Hz source? The speed of sound = 340 ms–1. (a) First harmonic (b) Second harmonic (c) Third harmonic (d) Fourth harmonic 17. A pipe of length 20 cm is open at both ends. Which harmonic mode of the pipe is resonantly excited by a 1700 Hz source? The speed of sound = 340 ms–1. (a) First harmonic (b) Second harmonic (c) Third harmonic (d) Fourth harmonic 18. Two sitar strings A and B are slightly out of tune and produce beats of frequency 6 Hz. When the tension in string A is slightly decreased, the beat frequency is found to be reduced to 3 Hz. If the original frequency of A is 324 Hz, what is the frequency of B?

(a) 318 Hz (b) 321 Hz (c) 327 Hz (d) 330 Hz 19. A metal wire of diameter 1 mm is held on two knife edges separated by a distance of 50 cm. The tension in the wire is 100 N. The wire vibrating with its fundamental frequency and a vibrating tuning fork together produce 5 beats per second. The tension in the wire is then reduced to 81 N. When the two are excited, beats are heard at the same rate. What is the frequency of the fork? (a) 90 Hz (b) 95 Hz (c) 100 Hz (d) 105 Hz 20. gation and Ranging) system which operates at an ultrasonic frequency of 42 kHz. An enemy submarine is moving towards the SONAR with a speed of 200 ms–1. If the speed of sound in seawater is 1400 ms–1, what is the frequency of sound received tion from the enemy submarine? (a) 36 kHz (b) 42 kHz (c) 48 kHz

(d) 56 kHz

21. A machine gun is mounted on a tank moving at a speed of 20 ms–1 towards a target with the gun pointing in the direction of motion of the tank. The muzzle speed of the bullet equals the speed of sound = 340 ms–1 target is 500 m away from the tank, then (a) the sound arrives at the target later than the bullet (b) the sound arrives at the target earlier than the bullet (c) both sound and bullet arrive at the target at the same time (d) the bullet will never arrive at the target. 22. Out of the four choices given in Q. 21 above, choose the correct choice, if the gun points in a direction opposite to the direction of motion of the tank. 23. Three sound waves of equal amplitudes have frequencies ( – 1), ( ) and ( + 1). They superpose to give beats. The number of beats produced per second will be (a) (c) 2

(b)

2 (d) 1

24. The speed of sound in hydrogen at STP is v. The speed of sound in a mixture containing 3 parts of hydrogen and 2 parts of oxygen at STP will be

Waves and Doppler’s Effect 14.17

(a)

v 2

(b)

v 5 v

(c)

7 v

(d)

(c)

2 v

(d) 2 v

7 25. The speed of sound in hydrogen at STP is v. What is the speed of sound in helium at STP? v v (b) (a) 2 2 26. Nine tuning forks are arranged in order of increasing frequency. Each tuning fork produces 4 beats per second when sounded with either of its neighbours. If the frequency of the 9th tuning fork is

(a) 32 Hz (c) 48 Hz

(b) 40 Hz (d) 56 Hz

27. A sonometer wire of length 120 cm is divided into three segments of lengths in the ratio of 1 : 2 : 3. What is the ratio of their fundamental frequencies? (a) 3 : 2 : 1 (b) 4 : 2 : 1 (c) 5 : 3 : 2 (d) 6 : 3 : 2 28. A tuning fork produces 4 beats per second when sounded with a sonometer wire of vibrating length 48 cm. It produces 4 beats per second also when the vibrating length is 50 cm. What is the frequency of the tuning fork? (a) 196 Hz (c) 375 Hz

(b) 284 Hz (d) 460 Hz

29. Two identical strings of a stringed musical instrument are in unison when stretched with the same tension. When the tension in one string is increased by 1%, the musician hears 4 beats per second. What was the frequency of the note when the strings were in unison? (a) 796 Hz (b) 800 Hz (c) 804 Hz (d) 808 Hz 30.

frequency 300 Hz at 27°C. If the temperature of the

beats heard per second will be (a) 1 (b) 2 (c) 3 (d) 4 31. The wavelength of light of a particular wavelength received from a galaxy is measured on earth and is found to be 5% more that its wavelength. It follows that the galaxy is

(a) approaching the earth with a speed 3 107 ms–1 (b) going away from the earth with a speed 3 107 ms–1 (c) approaching the earth with a speed 1.5 107 ms–1 (d) going away from the earth with a speed 1.5 107 ms–1 32. Radiowaves of frequency 600 MHz are sent by a radar towards an enemy aircraft. The frequency measured at the radar station is found to increase by 6 kHz. It follows that the aircraft is (a) approaching the radar station with a speed 1.5 kms–1 (b) going away from the radar station with a speed 1.5 kms–1 (c) approaching the radar station with a speed 3 kms–1 (d) going away from the radar station with a speed 3 kms–1. 33. An observer moves towards a stationary source of sound with a velocity one-tenth the velocity of sound. The apparent increase in frequency is (a) zero (b) 5% (c) 10% (d) 0.1% 34. A wave represented by the equation y = a cos (kx – t) is superposed with another wave to form a stationary wave such that the point x = 0 is a node. The equation of the other wave is (a) y = a sin (kx + t) (b) y = – a cos (kx – t) (c) y = – a cos (kx + t) (d) y = – a sin (kx – t) IIT, 1983 35. Two wires of the same material and radii r and 2r are welded together end to end. The combination is used as a sonometer wire and kept under tension T. The welded point is mid-way between the two bridges. When stationary waves are set up in the composite wire, the joint is a node. Then the ratio of the number of loops formed in the thinner to thicker wire is (a) 2 : 3 (b) 1 : 2 (c) 2 : 1 (d) 5 : 4 36. A uniform rope of length 12 m and mass 6 kg hangs vertically from a rigid support. A block of mass 2 kg is attached to the free end of the rope. A transverse pulse of wavelength 0.06 m is produced in the lower end of the rope. What is the wavelength of the pulse when it reaches the top of the rope?

14.18 Comprehensive Physics—JEE Advanced

(a) 0.06 m (c) 0.12 m

(b) 0.03 m (d) 0.09 m

IIT, 1984 37. A tube closed at one end containing air, produces, when excited, the fundamental note of frequency 512 Hz. If the tube is open at both ends, the fundamental frequency that can be excited is (in Hz) (a) 1024 (b) 512 (c) 256 (c) 128 IIT, 1986 38. Two sound waves of equal intensity I produce beats. The maximum intensity of sound produced in beats will be (a) I (b) 4I (c) 2I (d) I/2 39. Two travelling waves y1 = A sin [k(x + ct)] and y2 = A sin [k(x – ct)] are superposed on a string. The distance between adjacent nodes is ct ct (b) (a) 2 (c)

(d)

2k

k

IIT, 1992 40. Three waves of equal frequency having amplitudes 10 m, 4 m and 7 m arrive at a given point with a successive phase difference of /2. The amplitude of the resulting wave in m is given by (a) 7 (b) 6 (c) 5 (d) 4 41. A transverse wave is represented by y = y0 sin

2

vt

x

For what value of is the maximum particle velocity equal to twice the wave velocity? 2 y0 (b) = (a) = 2 y0 3 y0 (c) = (d) = y0 2 42. A train blowing its whistle moves with a constant velocity u away from the observer on the ground. The ratio of the actual frequency of the whistle to that measured by the observer is found to be 1.2. If the train is at rest and the observer moves away from it at the same velocity, the ratio would be given by (a) 0.51 (b) 1.25 (c) 1.52 (d) 2.05 IIT, 1993

43. An organ pipe P1, closed at one end vibrating in P2, open at both ends vibrating in its third harmonic, are in resonance with a given tuning fork. The ratio of the lengths of P1 and P2 is 8 1 (a) (b) 3 6 1 1 (c) (d) 2 3 IIT, 1988 44. The extension in a string, obeying Hookes’ law, is x. The speed of the wave in the stretched string is v. If the extension in the string is increased to 1.5 x, the speed of the wave in the string will be (a) 1.22 v (b) 0.61 v (c) 1.50 v (d) 0.75 v IIT, 1996 45. A travelling wave in a stretched string is described by the equation y = A sin (kx – t) The maximum particle velocity is (a) A

(b)

k x (d) t

d (c) dk

IIT, 1997 46. A transverse sinusoidal wave of amplitude a, wavelength and frequency f is travelling on a stretched string. The maximum speed at any point on the string is v/10, where v is the speed of propagation of the wave. If a = 10–3 m and v = 10 ms–1, then and f are given by 103 (a) =2 10–2 m, f = Hz 2 (b) = 10–2 m, f = 103 Hz (c) = 10–3 m, f = 104 Hz (d)

=

10 2

2

m,

103 Hz

f=2

IIT, 1998 47. The ratio of the speed of sound in nitrogen gas to that in helium gas at 300 K is (a)

2 7

(b)

1 7

(c)

3 5

(d)

6 5

IIT, 1999 48. Two vibrating strings of the same material but lengths L and 2L have radii 2r and r, respectively. They are stretched under the same tension. Both the

Waves and Doppler’s Effect 14.19

strings vibrate in their fundamental modes, the one of length L with frequency 1 and the other with frequency 2. The ratio 1/ 2 is given by (a) 2

(b) 4

(c) 8

(d) 1

pulses will be (a) zero (b) purely kinetic (c) purely potential (d) partly kinetic and partly potential IIT, 2001

IIT, 2000 49. Two sounds of wavelengths 5 m and 6 m, travelling in a medium produce 10 beats per second. The speed of sound in the medium is (b) 320 ms–1 (a) 300 ms–1 –1 (c) 350 ms (d) 1200 ms–1 50. The frequencies of tuning forks A and B are respectively 3% more and 2% less than the frequency of fork C. When A and B are simultaneously excited 5 beats per second are produced. The frequency (in Hz) of fork A is (a) 98 (b) 100 (c) 103 (d) 105 51. A metallic wire with tension T and at temperature 30ºC vibrates with its fundamental frequency of 1 kHz. The same wire with the same tension but at 10ºC temperature vibrates with a fundamental expansion of the wire is (a) 2 × 10–4 ºC –1 –4

(b) 1.5 × 10–4 ºC–1

–1

–4

–1

(c) 1 × 10 ºC (d) 0.5 × 10 ºC 52. A knife-edge divides a sonometer wire into two parts. The fundamental frequencies of the two parts are 1 and 2. The fundamental frequency of the sonometer wire when the knife-edge is removed will be 1 (b) ( 1 + 2) (a) 1 + 2 2 (c)

1 2

(d)

1 2

1

2

53. A sonometer wire is stretched by a hanging metal bob. Its fundamental frequency is 1. When the bob is completely immersed in water, the frequency becomes 2. The relative density of the metal is (a)

2 1

2 1

2 2

1

(c) 1

(b)

2 1

2 2

1

55. A siren placed at a railway platform is emitting sound of frequency 5 kHz. A passenger sitting in a moving train A records a frequency of 5.5 kHz while the train approaches the siren. During his return journey in a different train B he records a frequency of 6.0 kHz while approaching the same siren. The ratio of the velocity of train B to that of train A is (a) 242/252 (b) 2 (c) 5/6 (d) 11/6 IIT, 2002 56. A sonometer wire resonates with a given tuning between the two bridges when a mass of 9 kg is suspended from the wire. When this mass is replaced by a mass M, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of M is (a) 25 kg (c) 12.5 kg

2

54. Two pulses in a stretched string whose centers are initially 8 cm apart are moving towards each other as shown in Fig. 14.8. The speed of each pulse is 2 cm/s. After 2 seconds, the total energy of the

(b) 5 kg (d) (1/25) kg

IIT, 2002 57. A police van, moving at 22 ms–1, chases a motorcyclist. The policeman sounds his horn at 176 Hz, while both of them move towards a stationary si-

If the motorcyclist does not observe any beats, his speed must be (take the speed of sound = 330 ms–1) (a) 33 ms–1 (b) 22 ms–1 (c) zero

2

(d) 2

2 2

Fig. 14.8

(d) 11 ms–1

IIT, 2003 58. In the experiment for the determination of the speed of sound in air using the resonance column, it is observed that 0.1 m of air column resonates with a tuning fork in the fundamental mode. When the

14.20 Comprehensive Physics—JEE Advanced

length of the air column is changed to 0.35 m, the What is the end correction? (a) 0.0125 m (b) 0.025 m (c) 0.05 m (d) 0.075 m IIT, 2003 59. Transverse waves are generated in two uniform steel wires A and B of diameters 10–3 m and 0.5 10–3 m respectively, by attaching their free end to a vibrating source of frequency 500 Hz. The ratio of the wavelengths if they are stretched with the same tension is 1 2

(b)

(c) 2

(d)

(a)

1 2 2

60. A wire is stretched between two rigid supports with a certain tension. It is observed that the wire resonates in the nth harmonic at a frequency of 420 Hz. The next higher frequency at which the wire resonates is 490 Hz. The value of n is (a) 2 (b) 4 (c) 6 (d) 8 61. The fundamental frequency of a sonometer wire increases by 6 Hz if its tension is increased by 44%, keeping the length constant. The frequency of the wire is (a) 24 Hz (b) 30 Hz (c) 36 Hz (d) 42 Hz 62. Which of the following functons represents a travelling wave? Here a, b and c are constants. (a) (b) (c) (d)

y = a cos (bx) sin (ct) y = a sin (bx) cos (ct) y = a sin (bx + ct) – a sin (bx – ct) y = a sin (bx + ct)

63. A sonometer wire is vibrating with a frequency of 30 Hz in the fundamental mode. If the length of the wire is increased by 20%, the change in the frequency of the fundamental mode is (a) 5 Hz (b) 10 Hz (c) 15 Hz (d) 20 Hz 64. An organ pipe P1, closed at one end and containing a gas of density 1 Another organ pipe P 2, open at both ends and containing a gas of density 2 is vibrating in its third harmonic. Both the pipes are in resonance with a given tuning fork. If the compressibility of gases is equal in both pipes, the ratio of the lengths of P1 and P2 is

(a) 1 3 (c)

(b) 3

1 6

1

(d)

2

1 6

2 1

IIT, 2004

65. A source of sound of frequency 600 Hz is placed inside water. The speed of sound in water is 1500 ms–1 and in air it is 300 ms–1. The frequency and wavelength of sound recorded by an observer who is standing in air respectively are (a) 600 Hz, 0.5 m (c) 3000 Hz, 0.4 m

(b) 600 Hz, 2.5 m (d) 120 Hz, 2 m IIT, 2004 66. In the resonance tube experiment for determining the speed of sound in air using a tuning fork of at 17.7 cm of air column and the second at 53.1 cm. The maximum possible error in the speed of sound in air is (b) 96 cm s–1 (a) 192 cm s–1 –1 (c) 64 cm s (d) 48 cm s–1 IIT, 2005 67. A rod AB of length L is hung from two identical wires 1 and 2. A block of mass m is hung at point O of the rod as shown in Fig. 14.9. The value of x so that a tuning fork excites the fundamental mode in wire 1 and the second harmonic in wire 2 is (a)

L 5

(b)

L 4

(c)

L 3

(d)

2L 3

Wire 1

Wire 2 L A

x

B

O

m

Fig. 14.9

68. Two tuning forks, each of frequency , move relative to a stationary observer. One fork moves away from the observer with a speed u while the other fork moves towards him at the same speed. The speed of sound is v. If u << v, the observer hears beats of frequency

Waves and Doppler’s Effect 14.21

69.

(a) 344 (c) 117.3

u v

(a) zero

(b)

2 u (c) v

u (d) 2v

both ends are represented by x cos (96 t) 15 where x and y are in cm and t in second. The particle velocity at x = 7.5 cm and t = 0.25 s is y = 4 sin

(a) zero

(b) 10 cm s–1

(c) 100 cm s–1

(d) (4

(b) 336 (d) 109.3

IIT, 2008 73. A transverse sinusoidal wave moves along a string in the postive x-direction at a speed of 10 cm s–1. The wavelength of the wave is 0.5 m and its amplitude is 10 cm. At a particular time t, the snap-shot of the wave is shown in Fig. 14.10. The velocity of point P when its displacement is 5 cm is

96) cm s–1 IIT, 1985

70. A uniform rope of mass M hangs vertically from a rigid support. A block of mass m is attached to the free end of the rope. A transverse pulse of wavelength is produced at the lower end of the rope. The wavelength of the pulse when it reaches the top of the rope is (a)

m M m

(b)

(c)

m M

(d)

m M m

1/ 2

IIT, 1984 71. A band playing music at a frequency f is moving towards a wall at a speed u. A motorist is following the band with the same speed u. If v is the speed of sound, the beat frequency heard by the motorist is 2f u (a) f (b) (v +u) (c)

f (u v) (v u)

(d)

2 fu (v u)

72. A vibrating string of certain length L under a tension T resonates with a mode corresponding to the length 75 cm inside a tube closed at one end. The string also generates 4 beats per second when excited along with a tuning fork of frequency n. Now when the tension of the string is slighty increased the number of beats reduces to 2 per second. Assuming the velocity of sound in air to be 340 m/s, the frequency n of the tuning fork in Hz is

Fig. 14.10

(a)

3 j ms 50

(c)

3 i ms 50

1

(b) –

3 j ms 50

1

(d) –

3 i ms 50

1

1

IIT, 2009 74. A point mass is subjected to two simultaneous sinusoidal displacements in x-direction, x1(t) = A sin t 2 . Adding a third sinuand x2(t) = A sin t 3 soidal displacement x3(t) = B sin( t + ) brings the mass to a complete rest. The values of B and are (a)

2 A,

3 4

(b) A,

(c)

3 A,

5 6

(d) A,

4 3 3

IIT, 2011 75. A police car with a siren of frequency 8 kHz is moving with uniform velocity 36 km/hr towards a speed of sound in air is 320 m/s. The frequency of the siren heard by the car driver is (a) 8.50 kHz (c) 7.75 kHz

(b) 8.25 kHz (d) 7.50 kHz IIT, 2011

14.22 Comprehensive Physics—JEE Advanced ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67. 73.

(a) (d) (b) (b) (a) (d) (a) (b) (a) (b) (b) (a) (a)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68. 74.

(d) (d) (b) (d) (a) (a) (b) (a) (c) (a) (d) (c) (b)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63. 69. 75.

(b) (c) (a) (a) (d) (c) (d) (a) (d) (b) (a) (a) (a)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64. 70.

(b) (a) (a) (b) (a) (c) (c) (a) (d) (b) (d) (b)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65. 71.

(a) (d) (b) (d) (b) (b) (d) (c) (a) (a) (a) (d)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66. 72.

(c) (a) (a) (d) (b) (b) (b) (d) (b) (c) (a) (a)

SOLUTION 1. For a plane harmonic wave, the particle displacement is given by y = A sin ( t – k x) dy = A cos ( t – kx) Particle speed V = dt Vmax = A = 2 A Wave speed v = Particle speed will always less than wave speed if 2 A < or

A<

2 2. In an open pipe, all harmonics (even as well as odd) are present, but in a closed pipe, only odd harmonics are present. Thus, in an open pipe, the second is the second overtone and so on. In a closed pipe, harmonic is the second overtone and so on. If v is the speed of sound, the frequency of the n th overtone in an open pipe is v 0 = (n + 1) 2L In a closed pipe, the frequency of the nth overtone is v c = (2n + 1) 4L 0 c

=

2 n +1 2n 1

3. Let Lc be the length of the closed pipe and Lo of the open pipe. It is given that 3

v 4Lc

=2

v 2Lo

Lc 3 = Lo 4 Further, the frequency of the n th harmonic of the closed pipe = frequency of the mth harmonic of the open pipe, i.e, n 1, 3, 5, nv mv = ; m 1, 2, 3, 4 Lc 2 Lo

etc. etc.

2 Lc n 3 3 = =2 = Lo m 4 2 4. Let Ai and Ar be the amplitudes of the incident and Amplitude at antinode = Ai + Ar Amplitude at node = Ai – Ar Given Ai

Ar

Ai

Ar

Ar

=9

Ai

= 0.8

Intensity is proportional to the square of the amplitude. Hence Ir = Ii

Ar Ai

2

= (0.8)2 = 0.64 = 64%

Waves and Doppler’s Effect 14.23

5. Let L be the length of the pipe. L=

3 2

For second overtone; L =

=

given u0 = us = u. Thus, the source is receding from the observer with a speed u cos and the observer is approaching the source with the same speed u cos

2L 3

5 2

=

2L 5

Apparent frequency

5 , which is choice (a) 3 6. The resultant amplitude is given by

1 2 = 2 3 So the correct chose is (c). 7. Speed of observer at time t is u0 = 0 + at = at =

0

v + u0 v

=

0

v + at v

a 0 = 0+ t v This is the equation of a straight line with a positive a 0 and a positive intercept c = 0 slope m = v Hence the correct choice is (d).

5 x 11

dy = 0.1 10 cos 10 t dt Vmax = 0.1 10 = cms–1.

(1)

=–

Apparent frequency

=

9. The speed of particle is given by

Given A = a1 = a2 = a. Substituting these in Eq (1), We have a2 = a2 + a2 + 2a2 cos cos

v + u cos v + u cos

Hence the correct chose is (d)

=

A 2 = a 21 + a 22 + 2a1 a2 cos

=

V=

Hence the correct choice is (c). 10. Frequency of sound emitted by a closed pipe of v . For length L in the fundamental mode is = 4L resonance = where is the frequency of 340 or 4 L = 1 m or the tuning fork. Thus 340 = 4L 4 L=1 L = 25 cm. The next resonance occurs at 3 m or L = 75 cm. Hence the correct choice is (a). 1 2 65 1 Therefore

T 1 . Given N = m 2 65

=

11. N

T . m

N 2 65 65 = = Also N – N = 8 N 2 65 1 64 N = N + 8.

or

8 65 = which gives N = 512 Hz. N 64 Hence the correct choice is (d). N

8. Refer to Fig. 14.11.

12.

1

=

1 4 L1

1

=

2

,

2

=

1 4 L2

p

. Therefore

L2 50 = . It is given that L1 51

Therefore Hence

p

2

2

=

1

+ 5.

50 = which gives 1 = 250 Hz. 5 51 1 = 250 + 5 = 255 Hz which is choice (a). 1

13. The listener moves away from A and approaches B. Hence the apparent frequencies are Fig. 14.11

The components of u o and u s along the direction of the velocity of sound are uo cos and us cos

and

1

=

1

u v

2

=

1

u v

14.24 Comprehensive Physics—JEE Advanced

for any value of p = 1, 3, 5, .... Now for a closed pipe, we know that pv p = 4L Therefore, for resonance, pv N= 4L 4NL 4 425 0.2 or p= = =1 v 340 Hence the correct choice is (a).

u/v. It is given that 2 – 1 = 2 – = 10, v = 340 ms–1 and = 680 Hz. 2 1 Substituting these values we get 2 680 u 10 = or u = 2.5 ms–1. 340 Hence the correct choice is (b). 14. The resultant displacement is given by y = y1 + y2 = 0.05 {sin (3 t – 2 x) + sin (3 t + 2x)} Using the trigonometric relation sin ( + ) + sin ( – ) = 2 sin we have y = 0.1 cos 2 x sin 3 t or y = R sin 3 t

cos

where R, the amplitude of standing waves, is given by R = 0.1 cos 2 x with x = 0.5 m cos 2x = cos (2 0.5 rad) = cos (1 rad) = cos 57.3° = 0.54 Amplitude R at x = 0.5 is 0.1 0.54 = 0.054 m = 5.4 cm 15. Let be the wavelength, the frequency and v the speed of each wave. Then 2 x in the argument of the sine 2 function = 3 or

= 3 m.

Also

=2 = c t in the argument of the sine function = 120 which gives = 60 Hz. Hence v= = 60 3 = 180 m s–1 3.0 10 2 1.5 = 2.0 10–2 kg m–1

Mass per unit length (m) =

17. In an open pipe, the condition of resonance is pv ; p = 1, 2, 3, .... 2L 2NL 2 1700 0.2 or p= = =2 v 340 Hence, the correct choice is (b). N=

=

= 324 Hz, b = 6 Hz. The frequency of string B is B = A B = 324 ± 6 = 330 or 318 Hz Now, the frequency of a string is proportional to the square root of tension. Hence, if the tension in A is slightly decreased, its frequency will be slightly reduced, i.e. it will become less than 324 Hz. If the frequency of string B is 330 Hz, the beat frequency would increase to a value greater than 6 Hz if the tension in A is reduced. But the beat frequency is found to decrease to 3 Hz. Hence, the frequency of B cannot be 330 Hz; it is, therefore 318 Hz. When the tension in A is reduced, its frequency becomes 324 – 3 = 321 Hz which will produce beats of frequency 3 Hz with string B of frequency 318 Hz. Hence the correct choice is (a). 19. Let N be the frequency of the tuning fork. Then, the frequency of the wire, when the tension is 100 N will be (N + 5) and when the tension is 81 N, it is (N – 5); since in each case 5 beats are heard per second. Hence

18.

A

N+5=

T , where T is tension in m

We know that v = the string. T = mv2 = 2.0 10–2 = 648 N

p

1 2L

T1 1 100 = m 2 0.5 m

(180)2

16. Let N be the frequency of the source and p that of the pth harmonic of a closed pipe. The source will resonantly excite that harmonic mode of the pipe for which N= p

= and

N–5=

1 2L =

10 m

(i)

T2 1 81 = m 2 0.5 m 9 m

(ii)

Waves and Doppler’s Effect 14.25

Subtracting (ii) from (i) we have 1 10 = m or m = 0.01 kg m–1 Using this value of m in (i) or (ii) gives N = 95 Hz. Hence the correct choice is (b). 20. The frequency of ultrasonic (sound) waves sent out from the SONAR undergoes a change in two received by the enemy submarine which is approaching the SONAR with a speed u0 = 200 ms–1 is given by v + u0 = v =

3

42 10

1400 1400 103 Hz

= 48

200

(ii) The wave of frequency enemy submarine, which acts as a virtual image source of frequency , approaching the SONAR with a speed us = 200 ms–1. Hence the frequency of sound received back at the SONAR will be (observer stationary, source approaching) =

v v

us

1400 48 103 1400 200 = 56 103 Hz = 56 kHz Hence the correct choice is (d). 21. Since only the source of sound, i.e. the gun is in motion, the speed of sound remains unchanged at 340 ms–1. Therefore, the time taken by the sound of =

500 25 ts = = s 340 17 If the gun points in the direction of motion of the tank, the effective speed of the bullet = 340 + 20 = 360 ms–1. Therefore, the time taken by the bullet to reach the target is 500 25 = s 360 18 Since ts > tb, the correct choice is (a). 22. In this case, the effective speed of the bullet = 340 – 20 = 320 ms–1. Therefore, 500 25 tb = = s 320 16 Thus ts < tb. Hence the correct choice is (b). tb =

23. When the three waves superpose at a point, then from the superposition principle, the resultant particle displacement at that point is given by y = y1 + y2 + y3 = a sin {2 ( – 1)t} + a sin (2 t) + a sin {2 ( + 1) t} Now sin {2 ( 2 cos 2 t sin 2

– 1) t} + sin {2 t

(

+ 1) t} =

Therefore, y = y1 + y2 + y3 y = a (1 + 2 cos 2 t) sin 2 or

y = A sin 2

t

t

where A = a (1 + 2 cos 2 t) is the resultant amplitude. Now, the resultant intensity A2. Now A2 will be maximum when cos 2 t = + 1 or 2 t = 0, 2 , 4 , ... etc. or t = 0, 1s, 2s,... etc. Time period of beats = time interval between two consecutive maxima = 1 s. Hence the beat frequency is 1 Hz, i.e. one beat is heard per second which is choice (d). 24. Let the density of hydrogen be = 2 , then the density of oxygen will be 32 . The density of the mixture will be 3 2 2 + 32 = 14 = 5 5 Since the pressure is the same, we have 2 1 v = 14 v 7 =

v

. Hence the correct choice is (d). 7 25. The density of helium at STP = 2 times the density of hydrogen at STP. Since v 1/ , the speed of sound in helium will be v/ 2 . Hence the correct choice is (a). 26. . The frequency of the second will be ( + 4) and of the third will be ( + 8) and so on. Now + 8 = + (3 – 1) 4. Therefore, the frequency of the 9th tuning fork = + (9 – 1) 4 = + 32. It is given that + 32 = 2 . Hence = 32 Hz which is choice (a). 27. The frequency of the fundamental mode is given by =

1 2L

T m

14.26 Comprehensive Physics—JEE Advanced

1 1 1 1 . Hence 1 : 2 : 3 = : : . L L1 L2 L3 Now L1 = 20 cm, L2 = 40 cm and L3 = 60 cm. Therefore, 1 1 1 : : =6:3:2 1 : 2 : 3 = 20 40 60 Hence, the correct choice is (d). 28. Let 1 be the frequency of the wire when its vibrating length is L1 = 48 cm and 2 when L2 = 50 cm. 1 Since ; 1 > 2. If is the frequency of the L tuning fork, then + 4 and 2 = – 4 1 = 4 1 = 4 2 1

But

=

2

L2 50 = L1 48

25 24

25 4 = 24 4 which gives = 196 Hz. Hence the correct choice is (a). 29. Let T1 be the tension in each string when they are in unison. Let T2 be the tension in each string when T ; 2> 1 they are not in unison; then since such that T 2 = 2 T1 1 Now T2 = 1.01 T1. Therefore Thus

2

=

1.01 = 1

1

1 100

1/ 2

Also, or

1 100

2

1/ 2

1 201 = . Thus 200 200 201 2 = 200 1 – 1= 4 2 = 1 + 4. Therefore, we have 4

1 1

which gives choice is (b).

1

1+

201 = 200 = 800 Hz. Hence the correct

30. Let v1 be the speed of sound at 27°C and v2 at 31°C. Then v2 v1

273 31 273 27 1

1/ 2

1 2

304 1 / 2 1 300 4 151 300 150

4 300

Thus the correct choice is (b). 31. If a source emitting light of wavelength goes away from the earth, the apparent wavelength of the light reaching the earth is given by v =1+ c where v is the speed of the source of light and c the speed of light. The increase in wavelength = – is given by v = c 5 Here = 5% = and c = 3 108 ms–1. 100 Therefore, 5 = 1.5 107 ms–1 100 Hence the correct choice is (d). 32. If the aircraft is approaching the radar station with a speed u, the apparent frequency of radiowaves rev=3

108

is given by

2u c Apparent increase in frequency is 2u = – = c Given, = 600 MHz = 600 106 Hz and = 6 kHz = 6 103 Hz. Thus c 3 108 6 103 u= = 2 2 600 106 =

Expanding binomially, we have 1

Now, frequency speed of sound. Hence v 151 2 = 2 = v1 150 1 151 300 151 = = 302 Hz. 2 = 1 150 150 Hence beat frequency = 302 – 300 = 2

1/ 2

= 1.5

1

103 ms–1 = 1.5 kms–1

Hence, the correct choice is (a). u0 1 1 . Given u0/v = . Therefore 33. = v 10 11 = 10 11 or – = – = . 10 10 The percentage increase in is 100 = 10% which is choice (c).

Waves and Doppler’s Effect 14.27

34. To form a stationary wave, waves y and y must travel in opposite directions. Wave y = a cos (kx – t) travels along the positive x-direction. Waves y = – a cos (kx – t) and y = – a sin (kx – t) in choices (b) and (d) travel along positive x-direction. Hence choices (b) and (d) are not possible. Choice (a) is also incorrect because at x = 0 y = a sin t and y = a cos (– t) = a cos t Therefore, the resultant displacement at x = 0 which is y + y = a sin t + a cos t is not zero, i.e. these waves do not produce a node at x = 0. Choice (c) is correct because at x = 0, y + y = 0. 35. The frequency of a string of length L, mass m per unit length, stretched with a tension T and vibrating in p segments, is given by p T 2L m If the radius of the wire is r and mass = length

r2 L L

its density, then =

r2 .

p T 2Lr Since , T, L and are the same for both wires, p = constant or p r. Hence the number of loops r formed in the thicker wire will be two times that in the thinner wire. Hence the correct choice is (b). =

36.

the rope is different at different points on the rope. sion = weight of the rope + the weight attached to the free end of the rope = 6 kg + 2 kg = 8 kg wt. Tension at the free end of the rope = 2 kg wt. Since T , if the tension becomes 4 times, the frev 1 quency is doubled. Since = ; . Hence

the wavelength is halved. Thus the correct choice is (b). v 37. For a closed tube = . For an open tube = 4 L v . Hence = 2 = 2 512 = 1024 Hz. Thus the 2L correct choice is (a). 38. When two waves of amplitudes a1 and a2 superpose to produce beats, the resultant amplitude of the maxima of intensity is A = a1 + a2 2

=

. k 40. The amplitude of the three waves are a1 = 10 m, a2 = 4 m and a3 = 7 wave be zero. Then the wave of the second wave = /2 and of third wave =

2

. Therefore,

2

third wave is . Hence their resultant is

=

m=

i.e. a1 = a2 = a. Thus A = 2a. Therefore, A2 = 4a2 or Imax = 4I. Hence the correct choice is (b). 39. Distance between adjacent nodes (or antinodes) = /2. Also 2 x in the argument of the sine function = k 2 . or = k Hence, the distance between adjacent nodes

Now, intensity (amplitude) . Since the two waves have the same intensity, their amplitudes are equal,

a = a1 – a3 = 10 – 7 = 3 m The phase difference between this resultant (of /2. Therefore, the resultant of a and a3 is A = (a2 + a 22)1/2 = (32 + 42)1/2 = 5 m. Hence the correct choice is (c). 41. Wave velocity = v. Particle velocity is dy V= dt = y0 Vmax = y0

2 v

cos

2

vt

x

2 v 2 v

Now, Vmax = 2 v, if y0

= 2v, which gives

= y0. Hence the correct choice is (d). 42. If the train is going away from the observer, the apparent frequency is 1

=

v

v u

1

It is observed that = 1.2 apparent frequency is 2

or

= 2

v u v

=

1 1

u v

(i)

u v

1. In the second case, the

=

1

u v (ii)

14.28 Comprehensive Physics—JEE Advanced

u = 1 + u or 1.2 = 1 + , v v 1 u = 0.2. Using this in (ii), we get or u = 0.2 v or v 5 = 1.25. Hence the correct choice is (b). 4 2 v 43. For pipe P1 : 1 = 4 L1 Now, from (i) we have

For pipe P2 :

3

3v 2L2

=

It is given that 1 = 3. Therefore, L1/L2 = 1/6 which is choice (b). 44. The speed of the wave in the string is given by T m According to Hookes’law, tension (T) extension (x). x . Therefore Hence v

47. For a gas, adiabatic elasticity E = P where = Cp/Cv and P is the pressure. The speed of sound in the gas is given by RT M where M is the molecular mass and R is the gas constant. Thus vN M He 4 1 = vHe MN 28 7 48. Frequency of the fundamental mode is given by

46. Standing waves are formed on the string. Particle displacements are given by 2 x

y = a sin

cos (2

1 T ; 2L m T = tension and m = mass per unit length

= (– 2

fa) sin

2 x

or Now

fa =

f=

v 20 a

sin (2

ft)

=

v . Therefore, 10

10 3 m

20 v f

1

10 ms

1

3

10 /2 Hz

Hence the correct choice is (a).

103 Hz 2 =2

T / m1

1/ 4L

T / m2

Now

m1 =

M1 = L

and

m2 =

M2 2L

1

Hence

r2

= 2

49. Beat frequency

0

=

1

b

v=

(

2

2r 2 L L r2

4 r2

2

2L 2L

m2 m1 =4 =

r2 r2

=1

30 = 10 per second. Now 3 v v – 2=

=

1

2

5 6 10 (6 5) 1) = 300 ms–1

1 2 b 2

50. Let the frequency of fork C be n. Then nA = n + 0.03 n = 1.03n and nB = n – 0.02 n = 0.98 n. The beat frequency is nb = nA – nB or

v 10 10 ms

1/ 2L

Hence the correct choice is (a).

(V)max = 2 fa

2

=

2

or

dy V= dt

Given(V)max =

1

f t)

Particle velocity

P

=

v=

v 15 . x = 15 . = 1.22 v x Hence the correct choice is (a). dy d 45. Particle velocity V = [A sin (kx – t)] dt dt = – A cos (kx – t) Hence Vmax = A

E

v=

10–2 m

5 = 1.03 n – 0.98 n = 0.05 n

which gives n = 100 Hz. Hence nA = 1.03 100 = 103 Hz which is choice (c). 51. Given t1 = 10ºC, n1= 1.001 kHz, t2 = 30ºC, n2 = 1 kHz. Let l1 and l2 be the vibrating lengths of the wire at 10ºC and 30ºC respectively. Since tension T is kept constant. (i) n 1l 1 = n 2 l 2 l2 n = 1 l1 n2

1.001 1.001 1

Waves and Doppler’s Effect 14.29

If l2 = l1 (1 +

55. For train A: uA = 0.1 v

t)

l2 = 1 + a t = 1 + a (30–10) l1 = 1+ 20a From (i) and (ii) we have 1 + 20 = 1.001 or

(ii)

–4

which gives = 0.5 × 10 per ºC, which is choice (d). 52. Let L1 and L2 be the lengths of the two parts of sonometer wire. Given or Thus

1

1

=

L1 = L1 =

1 2 L1

T and m

L2 =

2

k

=

1 2 L2

T m

1 T = constant, say k. 2 m

and L2 =

1

2

k 2

The fundamental frequency of the complete sonometer wire is T 1 k = = 2( L1 L2 ) m ( L1 L2 ) or or

1

=

L2 1 = k 1

L1 k

1 2

= 1

1 2

, which is choice (d). 2

53. Let W1 be the weight of the bob in air and W2 when it is immersed in water. Given 1

=

W1 = W2

1 W1 and 2L m

2

=

1 W2 2L m

2 1 2 2

Relative density = =

weight in air loss of weight in water

W1 = W1 W2

Hence the correct choice is (a).

2 1

2 1

2 2

54. After 2 seconds, both the pulses will be at the same location on the string and will superpose on each other. Since their amplitudes are equal and opposite, they cancel each other and the string becomes straight. Hence the string has no potential energy, i.e. the total energy is purely kinetic. Thus the correct choice is (b).

For train B:

A

=

B

=

uB = v/5 = 0.2 v

v

uA v

or 5.5 = 5

v

uB v

or 6.0 = 5

v

uA v

v

uB v

uB 0.2 v = = 2, which is choice (b). uA 0.1v 56. Let L be the length of the wire between the bridges and let m be the mass per unit length of the wire. 5 Five antinodes on a length L implies that L = 1 2 2L or 1 = . Thus in this case we have 5 1

5 2L

v1

=

1

T1 , where T1 = M1 g m

Three antinodes on a length L implies that 3 2L . In this case, we have L= 2 or 2 = 2 3 2

=

3 T2 , where T2 = M2 g 2L m 2 = 2, M1 = 9 kg and M2 = M. Therefore,

v2

Given 1 we have

5 9g 3 Mg 2L m 2L m which gives M = 25 kg. Hence the correct choice is (a). 57. Given v = 330 ms–1 and up = 22 ms–1. The apparent frequency of the police man’s horn of frequency 176 Hz as heard by the motorcyclist is given by 330 um 176 = (330 – um) (i) 330 22 308 The apparent frequency of the stationary siren of frequency 165 Hz as heard by the motorcyclist is given by 330 um (ii) 2 = 165 330 Since the motorcyclist does not observe any beats, 1 = 2. Equating (i) and (ii) and solving for um we get um = 22 ms–1. Hence the correct choice is (b). 58. Let be the frequency of the tuning fork and e the end correction. Given L1 = 0.1 m and L2 = 0.35 m 1

= 176

= =

=

1

3

=

1 4 ( L1

P e)

3 4 ( L2 e)

P

(i) (ii)

14.30 Comprehensive Physics—JEE Advanced

Equating (i) and (ii), we get 1 3 = L1 e L2 e 1 3 = 01 . e 0.35 e which gives e = 0.025 m. Hence the correct choice is (b). 59. The density of a wire of mass M, length L and diameter d is given by 4M 4m = = 2 d L d2 or

Now

vA = vA = vB

T mA

and vB =

T mB

490 =

(n 1) 2L

1 T 1 1.44T 1.2 = = 2L m 2L m 2L Comparing this with Eq. (1), we have =

= 1.2 –

= 6 Hz. Hence

+ 6 = 1.2

1 2L

T 1 = m 1.2 2 L

T = m 1.2

30 = 25 Hz. Thus, 1.2 the frequency decreases by 5 Hz. Hence the correct choice is (a). nv ; n = 1, 3, 5, etc 64. For a closed pipe, n = 4L nv For an open pipe, n = ; n = 1, 2, 3, etc 2L Now = 30 Hz. Therefore,

where v =

P

=

.

(n = 1), we have

Dividing, we have 490 (n 1) = which gives n = 6. 420 n 61. The fundamental frequency of the vibration of a wire of length L, mass m per unit length and under tension T is given by 1 T (1) = 2L m If the tension is increased by 44%, the new tension is 44 T = T + 0.44 T = 1.44 T T =T+ 100 Since L is kept constant, the new fundamental frequency is

Given

1 T 2L m Since T and m are the same, the new frequency will be =

T m

=

=

mB d = B mA dA

but vA = n A and vB = n B, n being the frequency of the source. v d 0.5 10 3 A Hence = A = B = = 0.5, vB dA 10 3 B which is choice (a). 60. The frequency of the nth hormonic is 420 Hz. The frequency of the (n +1)th harmonic is 490 Hz. Therefore, n T 420 = 2L m and

6 = 30 Hz, which is choice (b). 0.2 62. A travelling wave is characterized by wave functions of the type y = f (vt + x) or y = f (vt – x) where f stands for sine or cosine function. Hence the correct choice is (d). Choices (a), (b) and (c) represent a stationary or standing wave. 63. If the length is increased by 20%, the new length is 20 L L =L+ = 1.2 L 100 The original frequency is which gives

T m

1 P (1) 4 L1 1 For the open pipe vibrating in the third harmonic (n = 3), we have 1

3 P 2 L2 2 . Equating (1) and (2), we get 3 3

Given

1

=

=

L1 1 = L2 6

2

(2)

.

1

Hence the correct choice is (d). 65. The frequency of sound is a characteristic of its source. Hence frequency of sound is the same in air as is water. Therefore, the observer in air will receive a sound of frequency 600 Hz. If a and w are the wavelengths in air and water resptively, then v vw n= a a

w

Waves and Doppler’s Effect 14.31

300 = 0.5 m 600 Hence the correct choice is (a). The wavelength is water is vw 1500 = = 2.5 m w= 600 66. Refer to Fig. 14.12. 3 L1 = and L2 = 4 4 v N= 1 = (1) 4 ( L1 e) which gives

a=

va

3v (2) 4 ( L2 e) where e is the end correction. Eliminating e from Eqs (1) and (2), we get (3) v = 2N (L2 – L1) and

N=

3

=

Lengths L1 and L2 are measured with a metre scale whose least count is 0.1 cm. Thus L1 = (17.7 0.1) cm and L2 = (53.1 ± 0.1) cm. The maximum error in (L2 – L1) is ± 0.2 (for maximum error, the errors in individual measurements add up). Thus L2 – L1 = (53.1 – 17.7) = 35.4 cm. Hence L2 – L1 = (35.4 ± 0.2) cm. Using this in Eq. (3), we have v = 2 cms–1

480

(35.4 ± 0.2) = (33984 ± 192)

Hence maximum error = 192 cm s–1, which is choice (a).

For rotational equilibrium of the rod about O, we have T1 AO = T2 BO or T1 x = T2 (L – x), T T ( L x) . But 1 = 4. Hence which gives 1 = T2 T2 x ( L x) 4= x L which gives x = . Hence the correct choice 5 is (a). 68. When the source moves towards from the stationary observer, its apparent frequency is v

(1) v u When the source moves away from the stationary observer, its apparent frequency is =

v

(2) v u When both the forks are moving relative to stationary observer, the number of beats heard by him per second = – Since us << v follows. u 1 u 1 1 = v v =

=

1

1

u v

1

u v

where terms of order u2/v2 have been neglected in the binomial expansion. Thus, –

=

1

u v

1

u v

= 2

u v

Thus the correct choice is (c) 69. Particle velocity V= Fig. 14.12

67. Let T1 and T2 be the tensions in wires 1 and 2 respectively. Let m be the mass per unit length of each wire and let l be the length of each wire. Given 1 = 2l which gives

T1 =4 T2

T1 2 T2 = 2m 2l m

dy d = dt dt

4 sin

x cos (96 t ) 15

x sin (96 t) 15 At x = 7.5 cm and t = 0.25 s, the particle velocity is =–4

96 sin

V= – 4

96 sin

= – (4

7.5 15

sin (96

0.25)

96) sin (0.5 ) sin (24 )

=0 The correct choice is (a).

[

sin (24 ) = 0]

14.32 Comprehensive Physics—JEE Advanced

70. The tension at the lower end of the rope is T = mg. T , = mass The speed of pulse at this point is v = per unit length of the rope. If the pulse, then v

=

1

=

is the frequency of T

(1)

The tension at the upper end of the rope is T = (m + M) g. Let be the wavelength of the pulse when it reaches the upper end of the rope. Then, since frequency of the pulse remains the same, 1

=

T

than 340. If n = 336 Hz, this will result in a greater number of beats per second. But the number of beats decreases to 2 per second. Hence n is not 336 Hz; it is 344 Hz. So the correct choice is (a). 73. In a transverse wave, the particle displacement and particle velocity are perpendicular to the direction of propagation of the wave. Hence choices (c) and (d) are wrong. The particle displacements is given by

(2)

sin

(

V=

v um =f v ub

Beat frequency (= f2 – f1) = f =

V=

Using sin

v u –f v u

2A sin

(2)

x)

3 ms–1 along the y-axis. 50

+ sin

t

4 3

– A sin

t

2 3

= 2sin cos t

+ B sin( t + ) =0 cos

2

4 3

+ B sin( t + ) + 0

B sin( t + ) = A sin

t

4 3

4 3

4 . 3 u = 36 km h–1 = 10 ms–1, v = 320 ms–1, = 8 kHz

which gives B = A and 75.

,

2

4 + B sin ( t + ) = 0 3

cos

3v 3 340 = = 340 Hz. 4L 4 0.75

frequency will slightly increase and become greater

(vt

we get

2fu v u

T . Hence if tension T is slightly increased, the

2

cos

A sin t + A sin

v u v u

Beat frequency fb = 4. Therefore n = f fb = 340 4 or n = 336 Hz or 344 Hz. The frequency of string

6

74. Since the mass is brought to rest, the total displacement is zero, i.e. x1(t) + x2(t) + x3(t) = 0

Thus the correct choice is (d). 72. The frequency of the third harmonic of a closed pipe is f=

(vt – x) =

Given a = 0.1 m, v = 0.1 ms–1, a = 0.1 m and 2 (vt – x) = . Putting these = 0.5 m. Also 6 values in Eq. (2), we get

source of sound are both approaching each other) f2 = f =

1 2

(vt x) =

dy 2 =a v dt

um = ub = u)

the mirror image which is approaching the motorist with a speed vb. Hence, the apparent frequency of

(1)

x)

Particle velocity is

from the wall. The apparent frequency of the direct sound is given by (the observer is approaching a receding source of sound) =f

2 2

T (m M ) g M m = = = T mg m Hence the correct choice is (b). 71. Let um = speed of motorist and ub = speed of the band. The motorist will hear two sounds–one com-

f1 = f

(vt

Putting y = 5 cm and a = 10 cm in Eq. (1), we get

From Eqs. (1) and (2), we have

v um v ub

2

y = a sin

=

1 2

Waves and Doppler’s Effect 14.33

imagined to be coming from the mirror image. The driver is approaching the image-source which is also approaching him with the same speed. Hence the frequency of sound heard by the driver is

=

v v

= 8 kHz

u u 320 10 320 10

= 8.5 kHz

II Multiple Choice Questions with One or More Choices Correct 1.

medium, which of the following will change? (a) Amplitude (b) Frequency (c) Wavelength (d) Phase

2. When a wave is refracted into another medium, which of the following will change? (a) Amplitude (b) Velocity (c) Frequency (d) Phase 3. Choose the correct statements from the following. (a) Any function of the form y (x, t) = f(vt + x) represents a travelling wave. (b) The velocity, wavelength and frequency of a wave do not undergo any change when it is (c) When an ultrasonic wave travels from air into water, it bends towards the normal to the air–water interface. (d) The velocity of sound is generally greater in solids than in gases at STP. 4. Which of the following statements are correct? (a) The decrease in the speed of sound at high altitudes is due to a fall is pressure. (b) The standing wave on a string under (c) The phenomenon of beats is not observed in the case of visible light waves. (d) The apparent frequency is 1 when a source of sound approaches a stationary observer with a speed u and is 2 when the observer approaches the same stationary source with the same speed. Then 2 < 1, if u < v, where v is the speed of sound. 5. Which of the following functions represent a stationary wave? Here a, b and c are constants. (a) y = a cos (bx) sin (c t) (b) y = a sin (b x) cos (ct)

(c) y = a sin (b x + ct) (d) y = a sin (b x + ct) + a sin (bx – c t) IIT, 1987 6. When a wave travels in a medium, the particle displacements are given by y = a sin 2 (bt – cx) where a, b and c are constants. (a) The wavelength of wave is c. b (b) The velocity of the wave is . c (c) The maximum particle velocity is twice the 1 wave velocity if c = a (d) The frequency of the wave is b. 7. Two persons A and B, each carrying a source of sound of frequency 90 Hz are standing a few metres B with a speed u = v/10, where v is the speed of sound. Then (a) A will hear 9 beats per second (b) A will hear 6 beats per second (c) B will hear 12 beats per second (d) B will hear 10 beats per second. 8. A wire of length L having linear density of 1.0 10–3 kg/m is stretched between two rigid supports with a tension of 40 N. It is observed that the wire, vibrating in p segments resonates at a frequency of 420 Hz. The next higher frequency at which the wire resonates is 490 Hz. The values of p and L are (a) p = 6 (b) p = 7 (c) L = (60/49) m (d) L = (10/7) m 9. Lo pipe (of length Lc) with a beat frequency of 10 Hz. The fundamental frequency of the closed pipe is 110 Hz. If the speed of sound is 330 ms–1, then

14.34 Comprehensive Physics—JEE Advanced

(a) (b) (c) (d)

Lc = 0.75 m Lo = (33/34) m or (33/35) m Lo = (33/32) m or 1 m Lo= (33/32) m or (33/34) m

IIT, 1997 10. In a resonance tube experiment, a tuning fork of frequency 480 Hz resonates in the fundamental mode with an air column of length 16 cm in a tube closed at one end. If the speed of sound in air is 336 ms–1, the diameter of the tube is (a) 5.6 cm (b) 5.0 cm (c) either 5.0 cm or 5.6 cm (d) neither 5.0 cm nor 5.6 cm. IIT, 2003 11. all points on the string vibrate with (a) the same frequency and the same phase but different amplitude. (b) the same frequency and the same amplitude but different phase. (c) the same frequency but different phase and different amplitude. (d) the same frequency, the same phase and the same amplitude. 12. Sound waves travel from location A at absolute temperature T1 to location B at absolute temperature T2 in time t. The air temperature increases linearly form T1 to T2. The speed of sound varies with absolute temperature T as v = k T where k is a positive constant. The distance between A and B is L. T T dT (a) The temperature gradient = 2 1. L dx dT 1 (b) The temperature gradient = (T2 – T1). dx 2L kt (c) L = T2 T1 2 kt T2 T1 2 13. A whistle emitting a sound of frequency 440 Hz is tied to a string of length 1.5 m and rotated with an angular velocity of 20 rad s–1 in the horizontal plane. An observer is stationed at a large distance from the whistle. If the speed of sound is 330 ms–1, (a) he will hear two sounds of frequencies 403 Hz and 440 Hz. (b) he will hear two sounds of frequencies 484 Hz and 440 Hz. (d) L =

(c) he will hear two sounds of frequencies 403 Hz and 480 Hz. (d) he will hear a range of frequencies between 403 Hz and 480 Hz. 14. A wave is represented by the equation y = A sin (10 x + 15 t +

/3)

where x and y are in metre and t in second. The expression represents a wave (a) travelling in the positive x-direction (b) travelling in the negative x-direction (c) of wavelength 0.2 m (d) of velocity 1.5 ms–1. IIT, 1990 15. Two identical wires as a stretched with tensions T1 and T2 with T1 > T2. They produce 6 beats per second when vibrated. If the tension in one of them is changed slightly, it is observed that the beat frequency remains unchanged. Which of the following is/are possible? (b) T1 was decreased (a) T1 was increased (d) T2 was decreased (c) T2 was increased IIT, 1991 16. A plane progressive wave of frequency 25 Hz, amplitude 2.5 10–5 m and initial phase zero propagates in a non-absorbing medium along the negative x-direction with a velocity of 300 ms–1. Then (a) Wavelength of the wave is 12 m (b) The phase difference between the oscillations at two points 6 m apart is (c) The corresponding amplitude difference is 2.5 10–5 m (d) The corresponding amplitude difference is zero. IIT, 1997 17. The (x, y) co-ordinates of the corners of a square plate are (0, 0), (L, 0), (L, L) and (0, L). The edges of the plate are clamped and transverse standing waves are set up in it. If u(x, y) denotes the displacement of the plate at the point (x, y) at a certain instant of time, the possible expression(s) for u(x, y) is/are (a = positive constant) x y cos (a) u(x, y) = a cos 2L 2L x y (b) u(x, y) = a sin sin L L x 2 y (c) u(x, y) = a sin sin L L 2 x y (d) u(x, y) = a cos sin L 2L IIT, 1998

Waves and Doppler’s Effect 14.35

18. A transverse sinusoidal wave of amplitude a, wavelength and frequency f is travelling on a stretched string. The maximum speed of any point on the string is v/10, where v is the speed of propagation of the wave. If a = 10–3 m and v = 10 ms–1, then (b) = 10–3 m (a) =2 10–2 m 103 Hz (d) f = 104 Hz (c) f = 2 IIT, 1998 19. As a wave propagates in a non-absorbing medium, (a) the wave intensity remains constant for a plane wave (b) the wave intensity decreases as the inverse of the distance from the source for a spherical wave (c) the wave intensity decreases as the inverse square of the distance from the source for a spherical wave (d) the total intensity of a spherical wave over a spherical surface centred at the source remains constant at all times. IIT, 1999 20. A moving pulse is represented by the expression 0.8 y(x, t) = [(4 x 5t ) 2 5] where x and y are in metre and t in second. Then (a) the pulse is moving in the + x direction (b) in 2s it will travel a distance of 2.5 m (c) its maximum displacement is 0.16 m (d) it is a symmetric pulse. IIT, 1999

21. In a wave motion y = a sin (k x – t), y can represent (c) displacement

(d) pressure IIT, 1999

22. Standing waves can be produced (a) on a string clamped at both the ends (b) on a string clamped at one end and free at the other wall (d) when two incident waves with a phase difference of are moving in the same direction. IIT, 1999 23. A student performed the experiment to measure the speed of sound in air using resonance air-column method. Two resonances in the air-column were obtained by lowering the water level. The resonance and that with the longer air-column is the second resonance. Then, resonance was more than that at the second resonance (b) the prongs of the tuning fork were kept in a horizontal plane above the resonance tube (c) the amplitude of vibration of the ends of the prongs is typically around 1 cm nance was somewhat shorter than 1/4th of the wavelength of the sound in air IIT, 2009

ANSWERS AND SOLUTIONS 1. When a wave travelling in a medium falls on the into the second medium. Therefore, the intensity refracted waves will be less than that of the incident wave. The velocity of a wave depends upon the medium in wave will be the same as that of the incident wave. But the velocity of the refracted wave will be different from that of the incident wave. The frequency same as that of the incident wave. From v = have.

, we

=

v

same as that of the incident wave. But the wavelength of the refracted wave will be different from that of the incident wave. Furthermore, when a wave travelling in a rarer medium, it undergoes a phase change of

or 180°.

a rarer medium, it does not undergo any phase change. The refracted wave, in both cases, does not undergo any phase change. Thus the correct choice are (a) and (d). 2. The correct choices are (a) and (b).

14.36 Comprehensive Physics—JEE Advanced

3. Statement (a) is correct. Let us write y (x, t) = f (vt + x) = f (z) Differentiating with respect to t, we have

u v Expression (i) may be written as

y f z f = =v t z t z Differentiating again w.r.t time t we have

u 1 v Expanding binomially and retaining terms upto order u2/v2, we have

2

y 2

= v2

2

and

f 2

t z Similarly differentiating twice with respect to x we have 2 2 y f = 2 2 x x Hence, 2

y 2

= v2

2

y 2

t x which is the standard equation (in differential form) of a travelling wave. Statement (b) is also correct. Because the wave is remains unchanged. The wavelength cannot change Statement (c) is incorrect. The ultrasonic wave bends away from the normal because the speed of the wave (being a sound wave) is greater in water than in air. Statement (d) is correct. The reason is that solids have a much higher modulus of elasticity than gases at STP. 4. Statement (a) is incorrect. A change in pressure has no effect on the speed of sound. The decrease in the speed of sound at high altitudes is due to a fall in temperature. Statement (b) is correct. Standing waves are produced due to superposi-

dent wave. Consequently, the resultant amplitude at nodes is not exactly zero. Thus the nodes are not Statement (c) is also correct. To observe beats the difference between the two interfering frequencies must be less than about 10–16 Hz. Since visible light waves have very high frequencies, beats are not observed due to persistence of vision. Statement (d) is correct. We know that 1

=

1

u v

2

(i)

=

1

1 =

1

=

1

1

u v

(ii)

u2

(iii)

v2

1 > 2. 5. A stationary wave is characterized by a function of the type y = f (t) g(x). Hence choices (a) and (b) represent a stationary wave. Choice (d) is a superposition of two oppositely travelling waves of the same amplitude and the same frequency which gives rise to a stationary wave. Hence choice (d) also represents a stationary wave. 6. Comparing y = a sin 2 (bt – c x) with 2 vt x y = a sin

we have 2 v 2 2 b= and 2 c = which give v = b and Particle velocity is

=

1 . Thus v = b/c. c

dy d = [a sin 2 (bt – cx)] dt dt = 2 ab cos 2 (bt – c x) Vmax = 2 ab. Now Vmax = 2v, if 2b 2 ab = c 1 which gives c = . a V=

b/c =b 1/ c Thus the correct choices are (b), (c) and (d). 7. Person A hears the sound of his own source whose frequency is . He also hears the sound of the source carried by person B, towards whom he is moving with a speed u. The apparent frequency of this sound is given by Frequency

=

v

=

= Beat frequency

u v

1 b

=

or –

= 9 Hz

–

=

u v

u 90v = v 10v ( u = v /10).

=

Waves and Doppler’s Effect 14.37

Person B hears the sound of his own source of frequency in . He also hears the sound of the source carried by person A, who is approaching with a speed u. The apparent frequency of this sound is given by v =

or

–

=

u v

1

v

=

Beat frequency

=

b

–

P 2L

T m

490 =

(2)

p T 600 = 2L m L

2v v = (3) 2 L0 L0 In a closed pipe, only odd harmonics are present. Hence, the frequencies of the overtones are 3, 5, 7, ... etc. times the fundamental frequency. Hence the is

=2

0

=

(4)

330 Lc = 0.75 m. 4Lc = ±10. There are the following two 2

= 10. Thus

3v = 10 4 Lc

33 m 32 Hence the correct choices are (a) and (d). 10. Including end correction, we have for the fundamental mode, L =

10 which gives L = m. Hence the correct choices are (a) and (d). 7 9. The fundamental frequencies of the open and closed pipes respectively are v (1) 0= 2 L0 v and (2) c= 4Lc where v is the speed of sound. In an open pipe, all harmonics are present. Hence the frequencies of the overtones are 2, 3, 4, ..... etc. times the fundamental frequency. Hence the fre-

1

3v 4 Lc

33 m 34 Case (b) : 1 – 2 = – 10. In this case, we get

490 p 1 = giving p = 6. 220 p Substituting this value of p in Eq. (1) we get 420 =

=

Lo =

(1)

p 1 T 2L m Dividing (1) and (2) we have

and

c

Putting v = 330 ms–1 and Lc = 0.75 m, we get

u = v u

90 v /10 = 10 Hz v v /10 Hence the correct choices are (a) and (d). 420 =

=3

110 =

v Lo

u v u

=

8.

Now Given 1 – 2 possibilities. Case a : 1 –

v v±u 1 =

v u

2

= L + 0.3 D or = 4 (L + 0.3 D) 4 where D is the diameter of the tube. Now v = v = or

4 (L + 0.3 D) =

v

(i)

Given L = 16 cm = 0.16 m, v = 336 ms–1 and = 480 Hz. Using these values in (i) and solving we get D = 5 10–2 m = 5 cm. The correct choice is (b). 11. The correct choice is (a). dx , dx = vdt = k T dt. As the temdt perature increases linearly, the rate of change of temperature with distance is given by

12. Since v =

T T dT = 2 1 L dx where L is the distance between locations A and B. Thus dx =

LdT T2 T1

But dx = k T dt. Therefore k T dt =

LdT T2 T1

14.38 Comprehensive Physics—JEE Advanced

or

dt =

If T1 is kept constant and T2 is increased such that the new value 2 becomes greater than 1 by 6, again 6 beats will be heard per second. Hence choice (c) is correct. If T2 is kept constant and T1 is decreased such that the new value of 1 becomes less than 2 by 6, again 6 beats will be heard per second. Hence choice (b) is also correct. v 300 16. = = = 12 m 25 2 path difference Phase difference =

LdT k (T2

T1 ) T

Integrating from T = T1 to T = T2, we have t=

t=

=

T2

L k T2

T

T1

T1 1/ 2

L

T T1 ) 1 / 2

k (T2

–1/2

2L k (T2 T1 )

T2

T1

T2

T1

2L

=

k T2 T1 Thus the correct choices are (a) and (c). 13. Linear velocity of source (whistle) is us = r = 1.5 20 = 30 ms–1. The observer will hear a range of frequencies lying between a minimum value vmin and a maximum value max, which are given by min

v

=v

v us

and vmax = v

v v us

; source receding ; source approaching

Substituting v = 440 Hz, v = 330 ms–1 and us = 30 ms–1 and solving we get vmin = 403.3 Hz and vmax = 484 Hz. The correct choice is (d). 14. The standard equation of a wave travelling in the negative x-direction is y = A sin ( t + kx + 0) 2 v 2 where = ,k= and 0 is the phase at x = 0 and t = 0. Comparing the given equation with this equation, we have 2 = 10 = 0.2 m k = 10 and

2 v

=15

= 15

15 15 0.2 = = 1.5 ms–1 2 2 Thus the correct choices are (b), (c) and (d). v=

15.

=

1 T . Since the wires are identical, 2L T . Since T1 > T2;

1

>

2. Also

2 6= 12 In a non-absorbing medium, the amplitude of the wave remains constant as the wave propagates. Thus the correct choices are (a), (b) and (d). 17. The expression for u(x, y) must satisfy the following boundary conditions: (i) u = 0 at x = 0 and at y = 0 (ii) u = 0 at x = L and at y = L. The choices (b) and (c) satisfy these conditions. 18. y = a sin ( t – kx) Speed at a point on the string is dy = a cos( t – kx) V= dt v Vmax = a = 2 af. Given Vmax = . 10 Hence 10 v v = 2 af f = 10 3 10 20 a 20 =

1

–

2

= 6.

= =

v 10 = f 103 2

=2

103 Hz 2 10–2 m

So the correct choices are (a) and (c). 19. The amplitude (and hence intensity) of a plane wave remains constant as the wave propagates in a non-absorbing medium. For a spherical wave, the same energy crosses a spherical surface of area 4 r2 where r is the distance from the source. The unit area per second. Hence, for a spherical wave, the intensity decreases as 1/r2. But the total intensity spread over the spherical surface is the same at all times. Hence the correct choices are (a), (c) and (d). 20. The displacement y(x, t) is maximum when the denominator is minimum, i.e.

Waves and Doppler’s Effect 14.39

[(4x + 5t)2 + 5] is minimum. Its minimum value is 5 when (4x + 5t)2 = 0 or 4x + 5t = 0 x = t

or

ymax =

5 4

22. Standing waves are produced due to a superposi-

5 ms–1 4

v=

0.8 = 0.16 m 5

Distance travelled by pulse in t = 2 s is

23.

5 2 = – 2.5 m 4 The negative sign shows that the pulse is travelling in negative x-direction. Since y is not a symmetric function of x, the form of the pulses changes as it travels. Hence the correct choices are (b) and (c). 21. In an electromagnetic wave, y represents electric pendicular to each other as well as perpendicular to the direction of propagation of the wave. In a mechanical wave, y represents displacement. In a sound wave, y represents pressure. Thus all the four choices are correct. vt =

clamped end of the string. In case (c) the incident ing in the same direction cannot produce standing waves; they give rise to interference. Thus the correct choices are (a), (b) and (c). tuning fork. At second resonance, the frequency of the third harmonic equals the frequency of the tuning fork. As the amplitude of oscillation of the fundamental mode is the highest, the intensity of Hence choice (a) is correct. Choice (b) is wrong, the prongs are kept in a vertical plane. Choice (c) is also incorrect as the amplitude of vibration of the ends of the prongs is typically around 1 mm. Choice (d) is correct. Due to end-correction (= e), (L1 + e) = choices are (a) and (d).

III Multiple Choice Questions Based On Passage Questions 1 to 3 are based on the following passage Passage I When two sound waves travel in the same direction in a medium, the displacements of a particle located at x at time t is given by y1 = 0.05 cos (0.50 x – 100 t) and

y2 = 0.05 cos (0.46 x – 92 t)

where y1, y2 and x are in metre and t in second IIT, 2006

1. What is the speed of sound in the medium? (b) 100 ms–1 (a) 332 ms–1 –1 (c) 92 ms (d) 200 ms–1 2. How many times per second does an observer hear the sound of maximum intensity? (a) 4 (b) 8 (c) 12 (d) 16 3. At x = 0, how many times between t = 0 and t = 1 s does the resultant displacement become zero? (a) 46 (b) 50 (c) 92 (d) 100

SOLUTION 1. The two displacements can be written as y1 = A cos (k1x –

1t)

and y2 = A cos (k2x –

2t)

where A = 0.05 m, k1 = 1=2

1 = 100

2

(1) (2) –1

= 0.50 m , 1

rad s1, k2 =

2

= 0.46 ms–1 and 2

is

2

=2

2

= 92

v1 = =

rad s–1. The speed of either wave

1 1

100 0.50

=2

1

2

1

= 200 ms–1

=

1

k1

14.40 Comprehensive Physics—JEE Advanced

92 k2 0.46 Hence the correct choice is (d). or

v2 =

2 2

2. Beat frequency = 1=

2

=

1

2

1

– =

=

2.

= A cos (k1x –

= 200 ms–1.

1t)

+ A cos (k2x –

2t)

For x = 0, we have y = A cos

Now

= 2 A cos

100 = 50 Hz 2

92 2 = = 46 Hz and 2= 2 2 Beat frequency = 50 – 46 = 4 Hz. Hence the correct choice is (a). 3. The resultant displacement is given by y = y1 + y2

Questions 4 to 6 are based on the following passage Passage II A string 25 cm long and having a mass of 2.5 g is under of tension. A pipe closed at one end is 40 cm long. When the pipe in its fundamental frequency, 8 beats per second heard. It is observed that decreasing the tension in the string decreases the beat frequency. The speed of sound in air is 320 ms–1. IIT, 1982

1t

+ A cos 1 ( 2

1

2t 2 )t

1 ( 1 2 )t 2 y = 0.10 cos (96 t) cos (4 t) cos

Between t = 0 and t = 1 s, cos (96 t) becomes zero 96 times and cos (4 t) becomes zero 4 times. Hence the resultant displacement y at x = 0 becomes zero 100 times between t = 0 and t = 1 s. The correct choice (d).

4. The frequency of the fundamental mode of the closed pipe is (a) 100 Hz (b) 200 Hz (c) 300 Hz (d) 400 Hz 5. overtone is (a) 92 Hz (b) 108 Hz (c) 192 Hz (d) 208 Hz 6. The tension in the string is very nearly equal to (a) 25 N (b) 27 N (c) 28 N (d) 30 N

SOLUTION 4. Frequency of fundamental mode of the closed pipe is v 320 np = = = 200 Hz 4L 4 0.40 The correct choice is (b). 5. Since the beat frequency is 8, the frequency of the ns = np ± 8 = 200 ± 8 = 192 Hz or 208 Hz 1 T ns = l m

(1)

Questions 7 to 11 are based on the following passage Passage III ends are represented by the equation y = 4 sin

x 15

cos (96 t)

where x and y are in cm and t in second.

IIT, 1985

It is given that the beat frequency decreases if the tension in the string is decreased. As the frequency decreases with decrease of tension, it is obvious that ns > np. Hence ns = 208 Hz and not 192 Hz. Thus the correct choice is (d). 6. Substituting the values of l, m and ns in Eq. (1), we get T = 27.04 N. Hence the correct choice is (b).

7. The frequency of vibrations of the string is (a) 48 Hz (b) 50 Hz (c) 96 Hz (d) 100 Hz 8. The maximum displacement of a point at x = 10 cm is (a) 2 cm (b) 4 cm (c) 2 3 cm

(d) 4 3 cm

Waves and Doppler’s Effect 14.41

9. How many nodes are formed on the string? (a) 2 (b) 3 (c) 4 (d) 5 10. In which harmonic mode is the string vibrating? (a) Fundamental (b) third

11. The velocity of the particle at x = 7.5 cm at t = 0.25 s is (a) zero (b) 320 ms–1 –1 (c) 60 ms (d) 96 ms–1

SOLUTION 7. 2 = 96 = 48 Hz. Thus the correct choice is (a) 8. Displacement is maximum when cos (96 t) = 1. 10 2 = 4 sin At x = 10 cm, ymax = 4 sin 15 3

10. The correct choice is (c) because 5 nodes are formed on the string. 11. The velocity of the string at a point x at time t is obtained by differentiating y = 4 sin

= 2 3 cm

The correct choice is (c). 9. At nodes the displacement is always zero. Hence nodes are located at values of x given by x x = 0 or =p sin 15 15 where p = 0, 1, 2, 3, ... etc. Thus x = 15 p = 0, 15, 30, 45 and 60 cm. Thus the correct choice is (d). Questions 12 to 15 are based on the following passage Passage IV The displacement of the medium in a sound wave is given by y1 = A cos (ax + bt) where A, a and b are positive constants. The wave is x = 0. The intensity of IIT, 1991 12. The wavelength and frequency of the incident wave respectively are (a)

2 b , a 2

2 b (c) , a 2

(b)

a 2 , 2 b

x 15

cos (96 t)

which respect to t. Velocity

dy = – (4 dt

96 )

sin

x sin (96 15

t)

At x = 7.5 cm and t = 0.25 s, the velocity is zero because at t = 0.25 s, sin (96 t) = sin (24 ) = 0. Hence the correct choice is (a).

13. (a) y2 = 0.8 A cos (– ax + bt) (b) y2 = – A cos (– ax + bt) (c) y2 = – 0.64 A cos (– ax + bt) (d) y2 = – 0.8 A cos (– ax + bt) 14. In the standing wave formed due to the superposimum values of the particle speed in the medium is (a) Ab (b) 1.64 Ab (c) 1.8 Ab (d) 2 Ab 15. The minimum value of the particle speed in the medium is

1 (d) a, b

(a) zero

(b) 0.2 Ab

(c) 0.64 Ab

(d) 0.8 Ab

SOLUTION 12. The incident wave is given by y1 = A cos (ax + bt) (1) The standard wave equation is y = A cos (kx + t) (2) where k is the wave number and , the angular frequency. Comparing (1) and (2) we get k = a and =

13.

2 2 b. Hence wavelength = = and frequency k a b = = . 2 2 The correct choice is (a). times that of the incident wave, the amplitude Ar of

14.42 Comprehensive Physics—JEE Advanced

0.64 = 0.8 times that

Differentiating Eq. (3) with respect to t, the particle

of the incident wave, i.e. Ar = 0.8 A

V2 =

suffers a reversal of amplitude (which implies a phase change of radian), i.e. Ar = – 0.8 A. Since the incident wave is travelling in the negative x dix wave can be obtained from Eq. (1) by replacing A by Ar = – 0.8 A and x by – x wave is given by (3) y2 = – 0.8 A cos (– ax + bt) Thus the correct choice is (d). 14. Differentiating Eq. (1) will respect to time t, we get the expression for the particle speed in the medium due to the incident wave, which is d y1 d {A cos (a x + bt)} V1 = dt dt = – Ab (a x + bt) (4) Maximum particle speed due to the incident wave is (V1)max = Ab Questions 16 to 19 are based on the following passage Passage V The air column in a pipe closed at one end is made to vibrate in its second overtone by a tuning fork of frequency 400 Hz. The speed of sound in air is 320 ms–1. The end correction may be neglected. Let P0 denote the mean pressure at any point in the pipe and P0 the maximum amplitude of pressure variation. IIT, 1998 16. The length L of the air column is (a) 20 cm (b) 60 cm (c) 1.0 m (d) 1.4 m 17. The amplitude of pressure variation at the middle of the air column is

d y2 dt d [– 0.8 A cos (bt – ax)] dt

= 0.8 Ab sin (bt – ax) (V 2)max = 0.8 Ab From the superposition principle, the maximum particle speed in the medium due to both the incisum of the individual maximum particle speeds, i.e. Vmax = (V1)max + (V2)max = Ab + 0.8 Ab = 1.8 Ab So the correct choice is (c). 15. Since A and b are positive constants, the minimum particle speed is Vmin = (V1)max – (V2)max = Ab – 0.8 Ab = 0.2 Ab The correct choice is (b). (b) 2 P0 P0 (c) 2 P0 (d) 2 18. The maximum and minimum pressures at the open end of the pipe respectively are (a) P0 + P0, P0 – P0 (b) P0 + P0, P0 (c) P0, P0 – P0 (d) P0, P0 19. The maximum and minimum pressures at the closed end of the pipe respectively are (a) P0 + P0, P0 – P0 (b) P0 + P0, P0 (c) P0, P0 – P0 (d) P0, P0 (a)

P0

SOLUTION 16. Figure 14.13 shows the longitudinal displacement y as a function of x for x lying between x = 0 and x = L, where L is the length of the pipe.

Fig. 14.13

The fundamental frequency of a closed pipe is given by v = 4L In a closed pipe, only odd harmonics are present, i.e. the frequency v1 the fundamental frequency, that of the second overtone v2 is 5 times the fundamental frequency and so on. Thus

Waves and Doppler’s Effect 14.43

5v 4L 5v 5 320 L= = = 1.0 m 4 2 4 400 The correct choice is (c). 17. Since the distance between two consecutive 2

nodes is

2

antinode is

=5 =

and that between a node and the next 4

, it follows from Fig. 14.7 that

5 2 2 4 4 We know that the pressure variation (excess pressure) is maximum at a node and minimum (equal to zero) at an antinode. Therefore, the pressure variation at a distance x from a node is given by 2 x P = P0 cos (1) L=

L 5 The centre C of the tube is at x = or x = 2 8 5 since L . The second node N2 is at x = 4 Questions 20 to 22 are based on the following passage Passage VI A wire of mass 9.8

10

–3

kg per metre passes over a

plane which makes an angle of 30° with the horizontal. Two masses M1 and M2 are tied at the two ends of the wire. Mass M1 rests on the inclined plane and mass M2 hangs freely vertically downwards. The whole system is in equilibrium. Now a transverse wave propagates along the wire with a speed of 100 ms–1.

. Therefore, the distance of C from N2 is x = 2 5 . Using this value of x in Eq. (1) we 8 2 8 have P at C =

P0 cos

=

P0 cos

2 8 4

=

P0

2 Thus the correct choice is (d). 18. At an antinode, the pressure variation is zero, i.e. P0 = 0. Hence, at an antinode Pmax = Pmin = P0 So the correct choice is (d). 19. At a node, the pressure variation is maximum equal to P0. Hence, at a node Pmax = P0 + P0 and Pmin = P0 – P0 Thus the correct choice is (a).

20. The tension T in the wire is (a) 0.98 N (b) (c) 98 N (d) 21. The value of mass M1 is (a) 2 kg (b) (c) 10 kg (d) 22. The value of mass M2 is (a) 5 kg (b) (c) 15 kg (d)

9.8 N 980 N 5 kg 20 kg 10 kg 20 kg

SOLUTION 20. Refer to Fig. 14.14. Let T be the tension in the string when the system is in equilibrium. position, the component M1g cos

Fig. 14.14

of weight M1 g

balances with the normal reaction N and the other component M1 g sin will balance with tension T in the string. Also weight M2 g of mass M2 will balance with tension T. Thus (1) M1 g sin = T and M2 g = T (2) Now, the speed of a transverse wave in a wire of mass m per unit length and stretched with a tension T is given by T v= m or T = v2 m (3)

14.44 Comprehensive Physics—JEE Advanced

Given m = 9.8 10–3 kg m–1 and v = 100 ms–1. Using these values in Eq. (3), we have T = (100)2 9.8 10–3 = 98 N 21. From Eq. (1), we get T 98 M1 = = g sin 9.8 sin 30

Thus the correct choice is (d). 22. From Eq. (2), we have M2 =

= 20 kg

Questions 23 to 25 are based on the following passage Passage VII A source of sound of frequency 90 Hz is moving towards a wall with a speed u = v/10, where v is the speed of sound in air. IIT, 1981 23. The beat frequency of the sound heard by an observer between the wall and the source is (a) 20 Hz (b) 10 Hz (c) 5 Hz (d) zero

T 98 = = 10 kg, which is choice (b). g 9.8

24. The beat frequency of the sound heard by an observer behind the source is 200 (a) Hz (b) 20 Hz 9 200 Hz (d) zero 11 25. The beat frequency of the sound heard by the observer moving with the source is (a) 11 Hz (b) 9.9 Hz (c) 10 Hz (d) 20 Hz (c)

SOLUTION 23. The observer hears two sounds–one coming directly from the approaching source and the other considered as coming from the mirror image of the source). The apparent frequency of the approaching source is v = 100 Hz v us v v/10 When the observer is between the wall and the =

v

v v v /10

= 90 Beat frequency =

–

= 100 – =

= 90

from the wall is also . Therefore, frequency of beats = – = 0. The observer will not hear any beats. So the correct choice is (d). 24. When the observer is behind the source, i.e. when the source is between the wall and the observer, the apparent frequency of the sound coming directly from the receding source is =

v v us

=

900 Hz 11

900 11

200 Hz. 11

Hence the correct choice is (c). 25. If the observer is moving with the source, the frequency of the direct sound is v = 100 Hz. The ap=

= 90

v us v us v + v /10 v v /10

= 110 Hz

Beat frequency = – = 110 – 90 = 20 Hz So the correct choice is (d).

Waves and Doppler’s Effect 14.45

IV Matrix Match Type 43. Column I shows four systems, each of the same length L, for producing standing waves. The lowest possible natural frequency of a system is called its fundamental frequency, whose wavelength is denoted as f. Match each system with statements given in Column II describing the nature and wavelength of the standing waves. Column I Column II (a) Pipe closed at one end (p) Longitudinal waves

(b)

Pipe open at both ends

(q)

Transverse waves

(c)

Stretched wire clamped at both ends

(r)

f

=L

(d)

Stretched wire clampeed at both ends and at mid-point

(s)

f

= 2L

(t)

f

= 4L

IIT, 2011

SOLUTION (a) The closed end of a pipe is a node and the open end is an antinode. The distance between a node and the next antinode is /4. Hence f = 4L. In pipes, the standing waves are due to superposition of oppositely travelling sound (longitudinal) waves. (a) (p, t) (b) Here L = (b) /2,

f

2 (p, s)

f

(c)

f

= 2L (

distance between consecutive antinodes = /2)

= 2L. Also standing waves on a string are due to superposition of transverse waves. (q, s)

L = f (d) Here the mid point is a node. Hence 2 2 (d) (q, r)

f

=L

ANSWERS (a)

(p, t)

(b)

(p, s)

(c)

(q, s)

(d)

(q, r)

14.46 Comprehensive Physics—JEE Advanced

V Assertion-Reason Type Questions In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each question has the following four choices out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation of Statement-1. (c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. 1. Statement-1 Only longitudinal mechanical waves can propagate in gases. Statement-2 Gases have only bulk modulus. 2. Statement-1 Two sound waves of equal intensity I produced beats. The maximum intensity of sound produced in beats is 4I. Statement-2 If two waves of amplitudes a1 and a2 superpose, the maximum amplitude of the resultant wave = a1 + a2. 3. Statement-1 A medium must possess elasticity in order to support wave motion. Statement-2 Restoring force does not exist in a medium which does not have elasticity. 4. Statement-1 Solids can support both longitudinal and transverse mechanical waves but only longitudinal mechanical waves can propagate in gases. Statement-2 Gases do not have shear modulus. 5. Statement-1 In standing sound waves, a displacement node is a pressure antinode and vice versa. Statement-2 In a standing wave, the restoring force is the maximum at a node and minimum at an antinode.

6. Statement-1 Our ears cannot distinguish two notes, one produced by a violin and other by a sitar, if they have exactly the same intensity and the same frequency. Statement-2 When a musical instrument is played, it produces a fundamental note which is accompanied by a number of overtones called harmonics. 7. Statement-1 Doppler’s effect does not occur in case of a supersonic source. Statement-2 A supersonic source produceds a shock wave. 8. Statement-1 If a source of sound moves always from a stationary observer, the apparent frequency of sound as heard by the observer is greater than the actual frequency. Statement-2 The cause of the apparent change in frequency is the change in the wavelength brought about by the motion of the source. 9. Statement-1 If an observer moves towards a stationary source of sound, the frequency of the sound as heard by him is greater than the actual frequency. Statement-2 The apparent increase in frequency is due to the fact that the observer intercepts more waves per second when the moves towards the source. 10. Statement-1 If a source of sound is in motion and the observer is stationary, the speed of sound relative to him remains unchanged. Statement-2 The apparent change in frequency is due to the change in the wavelength brought about by the motion of the source. 11. Statement-1 If the observer is in motion and the source of sound is stationary, the speed of sound relative to him is changed.

Waves and Doppler’s Effect 14.47

tionary observer with a certain velocity and (ii) observer approaching a stationary source of sound with the same velocity. Statement-2 The cause of the apparent change in the frequency is different in the two cases.

Statement-2 The wavelength of sound received by the observer does not change due to his motion. 12. Statement-1 The apparent frequency is not the same in the following two cases– (i) source approaching a sta-

SOLUTION 1. The correct choice is (a). Gases cannot withstand a shearing stress or longitudinal stress. Hence they do not have shear modulus and Young’s modulus; they have only bulk modulus. 2. The correct choice is (a). When two waves of amplitudes a1 and a2 superpose to produce beats, the resultant amplitude of the maximum of intensity is A = a1 + a2 Now, intensity (amplitude)2. Since the two waves have the same intensity, their amplitudes are equal, i.e. a1 = a2 = a. Thus A = 2a. Therefore, A2 = 4a2 or Imax = 4I. 3. The correct choice is (a). 4. The correct choice is (a). Gases cannot withstand a shearing stress. Hence gases do not have any shear modulus; they have only bulk modulus. Solids have Young’s modulus, bulk modulus and shear modulus. Therefore, solids can support both transverse and longitudinal waves. 5. The correct choice is (c). 6. The correct choice is (d). When a musical instrument is played, it produced a fundamental note

7.

8. 9. 10. 11. 12.

which is accompanied by a number of overtones called harmonics. The number of harmonics is not the same for all instruments. It is the number of harmonics which distinguishes the note produced by a sitar and that produced by a violin. The correct choice is (a). If the source of sound is moving at a speed greater than the speed of sound, then in a given time the source advances more than the wave. The resultant wave motion is a conical wave called a shock wave which produces a sudden and violent sound. The correct choice is (d). The correct choice is (a). The correct choice is (a). The correct choice is (a). The correct choice is (a). In case (i) the speed of sound relative to the observer remains unchanged; the change in frequency is due to a change in wavelength brought about by the motion of the source. In case (ii) the wavelength of sound remains unchanged; the change in frequency is due to a change in the speed of sound relative to the observer.

VI Integer Answer Type 1. An ambulance sounding a horn of frequency 256 Hz is moving towards a vertical wall with a velocity of 5 ms–1. If the speed of sound is 330 ms–1, how many beats per second will be heard by an observer standing a few metres behind the ambulance? IIT, 1981 2. A steel wire of length 1 m, mass 0.1 kg and uniform cross-sectional area 10–7 m2 ends. The temperature of the wire is decreased by 80/3°C. If transverse waves are set up in the wire,

3.

vibration in Hz. The Young’s modulus of steel = 2 1011 Nm–2 of steel = 1.2 10–5 per °C. IIT, 1981 at both the ends. The tension in the string is 0.5 N. The string is set into vibrations using an external vibrator of frequency 100 Hz. Find the separation (in cm) between the successive nodes on the string. IIT, 2009

14.48 Comprehensive Physics—JEE Advanced

SOLUTIONS 1. The observer will hear a sound of the source moving away from him and another sound after of these sounds are v 256 330 = = 252 Hz 1 = v u (330 5) v 256 330 = = 260 Hz 2 v u (330 5) No. of beats per second (beat frequency) = 260 – 252 = 8 2. Contraction L = L TL ; T = tension Young’s modulus Y = A L T=

1011

= 6.4 N

10–7

1 T ; 2L

1.2

10–5

80 3

= mass per unit length of wire

1 6.4 = 4 Hz 2 1 0.1

3. Mass per unit length of the string is 1.0 10

3

= 5 10–3 kg m–1 20 10 2 Speed of waves in the string is m=

T = m

v= Now

YA L = YA L

=2

= =

=

and

Frequency of fundamental mode is

v=

0.5 3

= 10 m s –1

5 10 v 10 = = = 0.1 m 100 = 10 cm

Separation between successive nodes =

2

= 5 cm

Thermal Expansion 15.1

15

Thermal Expansion

Chapter

REVIEW OF BASIC CONCEPTS 15.1

THERMAL EXPANSION

15.2 (i)

mass or = volume Since mass remains constant, Thus = 0 0 Density =

If the temperature of a body is increased, its length, surface area and volume all increase. (i) If the body is in the form of a rod, the increase in its length when the temperature is increased by T is proportional to ( ) the original length and (b) increase in temperature T, i.e. T = T Thus

=

(ii)

=

where =

0

(1 +

Strain Now

=

(iv)

0

(1 +

: i.e.

:

0 0

=

=

Young’s modulus (Y) = Thermal stress =

T)

and is (°C)–1 or K–1. , =1 : 2 : 3 =2 = 3 .

The SI unit of ,

1

(1

0

T)

If a rod is held between two rigid supports and its temperature is increased or decreased, the rigid supports prevent the rod from expanding or contracting. As a result, a stress (called thermal stress) is developed in the rod. The change in length of the rod is = T

T)

(iii)

=

=

= constant.

Thus density of a substance decreases with increase in temperature.

(ii) Similarly

0 0

0

=

where material of the body. Thus T) = 0 (1 +

APPLICATIONS

(iii)

T stress strain =Y

T

A linear metallic scale expands when heated and contracts when cooled. A reading of 1 unit of a heated scale is equivalent to a an actual length of 1 unit (1 + T) where T is the

15.2 Comprehensive Physics—JEE Advanced

rise in temperature. If the reading of the heated scale is units, the actual length = (1 + T) units. If the temperature is decreased by T, the actual reading = (1 – T) units. (iv) Consider two metal rods of lengths 1 and 2. Let 1 and 2 be their lengths when their temperature is increased by T, then and

1

=

1(1

+

1

T))

2

=

2(1

+

2

T)

where 1 and 2 linear expansion. The difference of their lengths is 2

–

1

=

2(1

+

T) –

2

1(1

+

1

15.1 At 20°C, a brass rod has a length 50.0 cm. It is joined to a steel rod of the same length and the same diameter at the same temperature. Find the change in the length of the composite rod when it is heated to 220°C. For brass = 2 10–5 K–1 and for steel = 1 10–5 K–1. SOLUTION For brass: (

T)

Their difference in lengths will remain constant if 2

i.e. if

–

1

=

2

–

1 1

=

2 2

1

)b =

b

(

(v)

= (2

10 )

=2

–3

10

) =

The time period of a simple pendulum is given by

=(

(1

=

= 1+

1 2

T(

T

stress =

1 2

T

T

1011) (2

(5

10–5)

10–6) [30 – (– 20)]

= 500 N 15.3 (60

60 24)

1 86400 s T 2 If the room temperature falls by T, then 1 86400 s T Time gained in one day = 2 =

) = 3 mm

T

= (1

=

Time lost in one day =

(220 – 20)

F

F=

T is small)

1 T T 2 T 1 = T T 2 This gives the time lost per second. T

10–5)

15.2 A metal wire of cross-sectional area 5 10–6 m2 is held taut at 30°C between two rigid supports with negligible tension in it. Find the tension developed in the wire if it is cooled to –20°C. Given of metal = 2 10–5 K–1 and Y = 1 1011 N m–2.

Also

T)1/2

= (1 +

0.5

m

)b + (

SOLUTION Thermal stress = Y

)

(220 – 20)

= 1 mm

T= 2

T = T

0.5

T

= (1

T = 2

–5

= 2 mm

For steel:

If the room temperature rises by T, the length of the pendulum increases. Hence the time period increases which implies that a metallic pendulum clock slows down. If is the length of heated clock, then its time period becomes

T

b

5

10–4 K–1. If its temperature is increased by 30°C,

SOLUTION = (1 +

T) =

(1 0

0

T) –

0

=–

0

T

Thermal Expansion 15.3

0

or 0

=–

T 10–4

=–5

30 = – 1.5

10–2.

The negative sign indicates that the density decreases with increase in temperature. Percentage change in density = (1.5 10–2) 100 = 1.5% 15.4 A block of mass 248 g and volume 205 cm3 liquid contained in a vessel. The density of the liquid at 0°C is 1.248 g cm–3. It is found that the block just sinks in the liquid if the temperature is raised to 50°C. expansion of the block is negligible compared to that of the liquid. SOLUTION The block will just sink in the liquid if its density becomes equal to the density of the liquid at 50°C. Density of block = 0

248 g 205 cm

= (1 +

= 1.210 g cm–3

3

T)

1.248 = 1.210 (1 + = 6.28

10

–4

50) (°C)–1 or K–1

The quantity T Hence the value of must be rounded off to two sig-

15.5 same material. At 25°C, the diameter of the shaft is 8.00 cm and the diameter of the hole is 7.99 cm. The shaft is cooled using ‘dry ice’. At what temperature of the shaft does the wheel slip into the shaft. Give for steel = 1 10–5 K–1. SOLUTION Decrease in temperature =

0.01cm 1 10

T2 – T1 = –125

SOLUTION T = 40 – 15 = 25°C Actual distance = measured distance (1 + T) –5 25) = 3152 (1 + 1.2 10 = 3152.9 cm

3153 cm

15.7 A metal pendulum clock gives correct time at a temperature of 20°C. How much time does it lose in a day if the room temperature is 35°C. Given of metal = 1.25 10–5 K–1. SOLUTION 1 T 2 1 (24 60 60)s Time lost in a day = T 2 1 = 1.2 10 5 15 2 (24 60 60) = 8.1 s

Time lost per second =

15.8

10–4 K–1

= 6.3

T=

15.6 A steel tape gives correct readings at a temperature of 15°C. On a day when the temperature is 40°C, this tape measures the distance between two points as 3152 cm. What is the actual distance between the two points? Given for steel = 1.2 10–5 K–1.

5

8.00 cm

= – 125°C

T2 = –125 + T1 = –125 + 25 = –100°C

bottle and water are heated to 80°C if (a) the expansion of bottle is neglected (b) the expansion of bottle is not neglected of glass of water = 6.0 10–4(°C)–1, –5 –1 = 0.6 10 (°C) . SOLUTION Volume of bottle = volume of water = 1 litre = 1000 cm3

(a) If the expansion of bottle is neglected, the = 1000 (1 + T) – 1000 = 1000 (1 + 6.0 10–4 60) – 1000 = 1036 – 1000 = 36 cm3

15.4 Comprehensive Physics—JEE Advanced

(b)

= 3 = 3 0.6 10–5 = 1.8 10–5(°C)–1 If the expansion of bottle is not neglected, the = 1000

(1 +

T) – 1000 (1 +

= 1000 = 1000

( (6.0

= 34.9 cm

T)

–

)

T –4

10

10–5)

– 1.8

60

3

I Multiple Choice Questions with Only One Choice Correct 1. When a solid metallic sphere is heated, the largest percentage increase occurs in its (a) diameter (b) surface area (c) volume (d) density 2. A steel scale measures the length of a copper rod as cm when both are at 20°C, the calibration linear expansion for steel and copper are and respectively, what would be the scale reading (in cm) when both are at 21°C? 1 (b) (a) 1 (c)

(d)

3. Two uniform brass rods and B of lengths and 2 and radii 2 and respectively are heated to the same temperature. The ratio of the increase in the length of to that of B is (a) 1 : 1 (b) 1 : 2 (c) 1 : 4 (d) 2 : 1 4. of thermal expansion 1 and 2 and Young’s modulii Y1 and Y2 walls. The rods are heated to the same temperature. If there is no bending of the rods, the thermal stresses developed in them are equal provided Y (a) 1 = Y2 (c)

Y1 = Y2

1 2 1 2

Y (b) 1 = Y2 (d)

Y1 = Y2

2 1 2 1

IIT, 1989 5.

of volume expansion of the metal and mercury are 1 and 2 respectively. If their temperature is increased by T, the fraction of the volume of metal submerged in mercury changes by a factor

(a) (c)

1 1 1 1

T T

(b)

T T

(d)

2 1 2 1

1 1

2 1

T T

2 1

IIT, 1991 6. A thin copper wire of length increases its length by 1% when heated from temperature T1 to T2. What is the percentage change in area when a thin copper plate having dimensions 2 is heated from T1 to T2? (a) 1% (b) 2% (c) 3% (d) 4% 7. direction ( -axis) is 2.0 10–6 K–1 and that in the other two perpendicular ( -and z-axes) directions is 1.6 10–6 K–1 expansion of the crystal? (b) 1.8 10–6 K–1 (a) 1.6 10–6 K–1 (c) 2.0 10–6 K–1 (d) 5.2 10–6 K–1 8. A metal ball immersed in alcohol weighs W1 at 0°C and W2 expansion of metal is less than that of alcohol. If the density of the metal is large compared to that of alcohol, then (a) W1 > W2 (b) W1 = W2 (c) W1 < W2

(d) W2 =

W1 2

IIT, 1980 9. A rod of length 20 cm made of a metal expands by 0.075 cm when its temperature is raised from 0°C to 100°C. Another rod of a different metal B having the same length expands by 0.045 cm for the same change in temperature. A third rod of the same length is composed of two parts, one of metal and the other of metal B. This rod expands by 0.060 cm for the same change in temperature. The portion made of metal has length

Thermal Expansion 15.5

(a) 20 cm (b) 10 cm (c) 15 cm (d) 18 cm 10. A metallic circular disc having a circular hole at its centre rotates about an axis passing through its centre and perpendicular to its plane. When the disc is heated, its (a) moment of inertia increases but angular speed decreases (b) moment of inertia decreases but angular speed increases (c) moment of inertia and angular speed both increase (d) moment of inertia and angular speed both decrease. 11. A uniform metal rod of length and mass M is rotating with angular speed about an axis passing through one of the ends and perpendicular to the rod. If the temperature increases by ºC, then the change in its angular speed is proportional to 2

(d)

16.

(c)

1

2

1

2

2

1 2

(d)

1

1 2 2

1

+ 1

2)

(b)

1 ( 2

(d) (

2

1

+

2)

2)

1

17. Three rods of the same length are arranged to form an equilateral triangle. Two rods are made of the 1

and the third rod which forms the base of the trian2. The altitude of the triangle will remain the same at all temperatures if the 1/ 2 is nearly 1 (a) 1 (b) 2 1 (c) (d) 4 4

12. A steel metre scale is to be ruled so that the millimetre intervals are accurate to about 5 × 10–5 m at a certain temperature. The maximum temperature

(c) (d) 2 14. Two spheres made of the same material have the same diameter. One sphere is hollow and the other is solid. If they are heated through the same range of temperature, (a) the hollow sphere will expand more than the solid sphere (b) the solid sphere will expand more than the hollow sphere (c) both spheres will expand equally (d) the hollow sphere will not expand at all. 15. Two rods of lengths 1 and 2 are welded together to make a composite rod of length ( 1 + 2). If the

1

-

neous rod changes linearly from 1 to 2 from one end to the other end of the rod. The effective coef(a) (

1

of linear expansion of steel = 10 × 10–6 K–1) (a) 2 ºC (b) 5 ºC (c) 7 ºC (d) 10 ºC 13. When the temperature of a rod increases from to + , the moment of inertia of the rod increases from I to I + I I sion of the rod is , the ratio is I 2 (b) (a)

1

(c)

(b)

(a) (c)

and 2 ear expansion of the composite rod will be 1 (a) ( 1 + 2) (b) 1 2 2

18.

19.

80% of its volume immersed in a liquid at 0ºC. When the temperature of the liquid is raised to 62.5ºC, the cal expansion of liquid is (b) 2 × 10 3 K 1 (a) 1 × 10 3 K 1 (c) 3 × 10 3 K 1 (d) 4 × 10 3 K 1 fraction 1 of its volume is submerged, while at the temperature 60°C, a fraction 2 is seen to be subiron is Fe and that of mercury is 1/ 2 can be expressed as (a) (c)

1 60 1 60 1 60 1 60

Fe

(b)

Hg Fe Hg

(d)

Hg,

then the ratio

1 60 1 60

Hg

1 60

Hg

1 60

Fe

Fe

IIT, 2001 20. Two rods, one made of aluminium and the other made of steel, having initial lengths 1 and 2 respectively are connected together to form a single rod of length ( 1 + 2 expansion for aluminium and steel are 1 and 2

15.6 Comprehensive Physics—JEE Advanced

respectively. If the length of each rod increases by the same amount when their temperature is raised by °C, then the ratio 1/( 1 + 2) (b) 2/ 1 (a) 1/ 2 (c) 2/( 1 + 2) (d) 1/( 1 + 2) IIT, 2003 21.

22.

18 10–5/°C. A thermometer has a bulb of volume 10–6 m3 and the cross-sectional area of the stem is 0.002 cm2 mercury when the temperature is 0°C. When the temperature rises to 100°C, the length of the mercury column in the stem will be (a) 9 cm (b) 18 cm (c) 9 mm (d) 18 mm when determined using two different vessels B are 1 and 2 linear expansion of vessel is linear expansion of vessel B is 1

(a) 1

(b)

1

2 2

2

3

(d)

2

(c) increase by 2 (d) decrease by 2 26. A vertical glass tube, closed at the bottom, contains a mercury column of length 0 at 0°C. If is the the of the mercury column when the temperature rises to °C is (assuming that not more than 100°C)

(b)

(a)

=

0[1

+ ( – 3 )]

(b)

=

0[1

+(

(c)

=

0[1

+ ( + 2 )]

is 20 10–6 per °C. The barometer reads 75 cm at 40°C. The atmospheric pressure at 40°C is (a) 75 cm of Hg (b) 74.94 cm of Hg (c) 75.06 cm of Hg (d) none of these 28.

. When the temperature is raised by T, the depth upto which the cube is submerged in the liquid remains unchanged. If the expansion of the beaker is ignored, the relation between and is (a)

(c)

1

(d)

1

24. A steel rod and a copper rod have lengths and respectively at a certain temperature. It is found that the difference between their lengths remains constants at all temperatures. If and are their / (a) (c)

is given by 1

(b)

1

(c)

(b)

=

3 =3

(d)

=

2 =2

IIT, 2004 29. A copper wire of length and cross-sectional area is held at the ends by two rigid supports. At temperature T the wire is just taut with negligible tension. If the temperature reduces to (T – T), the speed of transverse waves in the wire is (here Y is Young’s modulus, density and linear expansion of copper) Y T T (b) (a)

(d)

25. A uniform metallic circular disc, mounted on frictionless bearings, is rotating at an angular frequency about an axis passing through its centre and

+ 3 )]

(d) = 0[1 + ( – 2 ) ] 27. The brass scale of a barometer gives correct reading

1

and mercury respectively, the volume of mercury (a)

(b) increase by

and

2 + 3 23. contains some mercury. It is found that at different temperatures, the volume of and

(c)

1

2

expansion of the metal is . If the temperature of the disc is increased by , the angular frequency of rotation of the disc will (a) remain unchanged

(c)

T

(d)

T IIT, 1979

Thermal Expansion 15.7

30. A metal ball immersed in alcohol weighs W1 at 0°C and W2 expansion of metal is less than that of alcohol. If the density of the metal is large compared to that of alcohol, then

(a) W1 > W2

(b) W1 = W2

(c) W1 < W2

(d) W2 =

W1 2 IIT, 1980

ANSWERS

1. 7. 13. 19. 25.

(c) (d) (d) (a) (d)

2. 8. 14. 20. 26.

(c) (b) (c) (c) (d)

3. 9. 15. 21. 27.

(b) (b) (c) (a) (c)

4. 10. 16. 22. 28.

(d) (a) (b) (d) (b)

5. 11. 17. 23. 29.

(a) (b) (c) (a) (a)

Y1

1

6. 12. 18. 24. 30.

(b) (b) (d) (d) (b)

SOLUTIONS 1. On heating, the diameter, surface area and volume of the sphere will all increase. Since the mass remains unchanged, the density decreases. The percentage increase in the volume is the largest beexpansion. Hence the correct choice is (c). 2. The length of 1 cm division of the steel scale at 21°C is (1 cm) [1 + (21 – 20)] = (1 + ) cm Length of copper rod at 21°C will be ( cm) [1 + (21 – 20)] = (1 + ) cm 1 1

Scale reading at 21°C =

=

=

=2

.

(

=2)

1 2

and

stress = Y

Y1 = Y2

2 1

1

1 100

2

=2

=

2

1

1

2 100

=

102 100

Thus the area increases by 2%, which is choice (b).

Hence the correct choice is (b). 4. If a rod of length expansion is heated to a temperature , the increase in the length is given by = Strain =

or

1 2 100 Now 2 2 = area of the plate at temperature T2 and 2 2 = area of the plate at temperature T1. Therefore, 1 2 = 1 100 2

3. The increase in length due to heating is independent of the radius of the rod. The increase in the length of rod is = =

2

Hence the correct choice is (d). 5. The correct choice is (a). It follows from the fact that the volumes of metal and mercury increase to 0 (1 + 1 T) and 0 (1 + 2 T) respectively; 0 being the initial volume. 6. Length of wire at temperature T2 is

Hence the correct choice is (a).

and of rod B is

= Y2

= strain = Y

Since the value of is the same for the two rods, the stresses in them will be equal if

7. =

+

+ –6

z

=

+2 (

z)

= –6

= 2.0 10 + 2 1.6 10 = 5.2 10–6 K–1 Hence the correct choice is (d). 8. Let 0 be the volume of the metal at 0°C and its volume at °C. At temperature the upthrust is U = where is the density of alcohol at temperature . Now

15.8 Comprehensive Physics—JEE Advanced

= 0 (1 + ) where alcohol and 0 is the volume of alcohol displaced at temperature = 0°C. Now the density of alcohol at temperature is 0

=

where Hence

0

1 is the density of alcohol at U =

0

(1 +

)

=

0

0

= U0

1

= 0°C.

0

where is the length of metal and ( – ) that of metal B in rod C. Now ( )1 = 0.075 cm, ( )2 = 0.045 cm and ( )3 = 0.060 cm. Notice that 1 [( )1 + ( )2]. This is possible only if ( )3 = 2 20 cm = = 10 cm. Hence the correct choice 2 2 is (b). 10. Due to thermal expansion, the diameter of the disc as well as that of the hole will increase. Therefore, the moment of inertia will increase resulting in a decrease in the angular speed. Hence the correct choice is (a). 11. At ºC, the length of the rod becomes = (1 + ), where From the law of conservation of angular momentum, we have I =I

or

1 3

2

=

1 3 2

1

(1 )2 Now, for a given value of , (1 + constant, say . = or

=

–1

–

= ( – 1)

)

. Hence the corect choice is (b).

12. The maximum temperature variation T allowed is given by = T, which gives 5

5 10

= 5ºC 1 10 10 6 Hence the correct choice is (b). 13. The moment of inertia of a rod of mass M and length is given by 2 (1) I= 1 if the axis of rotation is through the 12 1 centre and = if the axis of rotation is through 3 the end of the rod. Partially differentiating (1), we have I= 2

where

=

Now

=

. Therefore,

I= 2

× 2

=2(

)×

= 2I

I =2 , which is choice (d). I 14. The correct choice is (c). 15. Total length of the composite rod at 0°C is 0 = 1 + 2. When the composite rod is heated to °C, its length at °C will be = 1 (1 + 1 ) + 2 (1 + 2 ) = ( 1 + 2) + 1 1 + 2 2 , or or

=

0

+(

1 1

+

2 2)

Effective coefficient of expansion of the composite rod is 0

=

1 1

2

1

2

0

2

,

which is choice (c). 16. Consider a small element of length at a distance from one end of the rod. Let be the length

2

=

i.e. (

–

T=

where U0 is the upthrust at 0°C. Since the upthrust is independent of temperature, W1 = W2. Hence the correct choice is (b). 9. Here ( )1 = 1 ( )2 = 2 ( )3 = +( – ) 2 1

or

or

)–2 is

of linear expansion by unit length of the rod is at the ( 2 – 1)/ . Therefore, the value of element located at is =

1

+

2

1

Increase in the length of the element = , where T is the rise in temperature. Therefore, the increase in the length of rod is

Thermal Expansion 15.9

=

18. Weight of cylinder = weight of the liquid displaced, i.e. (0.8 ) b = 0 5 b b or (1) 0 = 0 .8 4 Here 0 is the density of the liquid at 0°C. For the block to sink in the liquid at °C, the density of the liquid must change from 0 to at °C. Now

T 0

=

T

2

1

1

0

=

T

=

T

where

eff

2

2

1 ( 2

=

2

1

2

1 1

=

2

1

2

2

+

2).

0 (2) 1 For the block to sink = b. Using (1) and (2), we have 5 b /4 b = 1

=

1

T= 1

0

T

eff

Hence the correct

or

choice is (b). 17. Let be the length of each rod at 0°C and altitude at 0°C. Then (Fig. 15.1),

0

be the

1+

=

5 4

Given = 62.5°C. Using this value, we get = 4 10–3 (°C)–1 or K–1, which is choice (d). 19. Let 0 be the total volume of the block of iron at 0°C and its total volume at 60°C. Then =

(1 + 60

0

Fe)

(1)

Let v0 be the volume of the block submerged in mercury at 0°C and v the volume submerged at 60°C. Then (2) v = v0 (1 + 60 Hg) Fig. 15.1

0

2

=

Dividing (1) by (2), we have

2 1/ 2

4 When the rods are heated to a temperature , the altitude becomes 2

=

1

=

2

2

1

Since 1 and can write (1 + 1 + 2 2 . Thus

2

1

4

2

2

1/ 2

1

1 2

2 1

4

1 2

2 1/ 2

4

2

2

(2)

2

1

1 2

1

=

1

or

4

1/ 2

1 2

2

1 , which is choice (c). 4

and

v

=

(3)

Hg

2.

Using these in (3), we

2

or

1 2

+ 1=

1 2

1 2

=

Fe

20. The length of each rod increases by the same amount if 1 1 = 2 2 or

1/ 2

Equating (1) and (2), we get 2

v0

Given

0

are much less than unity, we )2 1 + 2 1 and (1 + 2 )2

2

1 60 1 60

/v / 0 v0

(1)

1

1

2

=

1

1

1

2 2

1

or

+1

2

2

= 2

1

, which is choice (c). 2

15.10 Comprehensive Physics—JEE Advanced

21. Increase in volume of mercury when the temperature increases by 100°C is v= T = (18 10–5) (10–6) 100 10–9 m3 v or =

= 18 Now v =

9

(18 10

=

m3 ) 4

= 0.09 m = 9 cm

2

i.e. For vessel ; given)

v

=

:

+3 =

For vessel B:

1

=

2

+3 +3

v 1

=

+3

=

which gives

2

+3 2

=

1

(

v

+3

(

= 3 v) 1

23. Let

3

2

=

is the same.

+ , which is choice (d).

J=I

(3)

Since no external torque acts, J remains constant. Partially differentiating (3), we have +

I

I=0

, which is choice (a).

= = The difference between their lengths will remain constant at all temperatures if = , i.e. if = =

Hence the correct choice is (d). 25. The moment of inertia of the disc about the given axis of rotation is 1 MR2 (1) I= 2 where M is the mass of the disc and R its radius.

I

=–

(4)

I Using (1) and (2) in (4) we get = –2

The negative sign indicates that the angular frequency decreases due to increase in temperature. Hence the correct choice is (d). 26. If 0 is the volume of mercury at 0°C and 0 the cross-sectional area of the tube at 0°C, then the length of the mercury column at 0°C is 0

24. When the rods are heated by °C, the increase in length of steel rod and copper rod are

or

(2)

Now, the angular momentum of the disc is given by

= or

M = constant)

. Therefore,

or

2 1

R=R

=

2

Since the liquid is the same, Hence 1

But

(

I = MR2 -

22.

+

or

1 M 2R R 2 I = MR R I=

(0.002 10 m ) Hence the correct choice is (a).

=

If the disc is heated, it expands. Hence R increases. The resulting increase in I is obtained by partially differentiating (1).

=

0

(1)

0

The cross-sectional area at °C is given by =

0

(1 +

)

where of glass. Since = 2 , we have = 0 (1 + 2 ) If is the volume of mercury at °C, the length of the mercury column at °C is 0 1 = = 0 1 2 Using (1) we have = 0 (1 + ) (1 + 2 )–1 Since is not too high (~ 100°C) and and are of the order of 10–5, and will be very small compared to unity. Hence we can expand (1 + 2 )–1 binomially and retain terms upto order . Thus (1 + 2 or

)–1 = (1 – 2 ) = 0 (1 + ) (1 – 2 =

0

[1 + ( –2 ) ]

)

Thermal Expansion 15.11 2 where we have neglected term of which is negligibly small. Thus the correct choice is (d). 27. Due to rise in temperature, the brass scale expands. It will give lower readings because the graduations on the scale will be farther apart. If H is the barometric height at 0°C, the error in the reading of the scale at 40°C is H= H = 75 20 10–6 40 = 0.06 cm Atmospheric pressure at 27°C = H + H = 75 + 0.06 = 75.06 cm of Hg. Thus the correct choice is (c). 28. Let the initial temperature be T and let M be the mass of the cube. Let 0, 0 and 0 respectively be the base area of the cube, the density of the material of the cube and the depth upto which it is submerged in the liquid, the upthrust = 0 0 0 . From

or

= M=

0 0 0

Using (2) and (3) we have 0 = 0 0 (1 T) which gives 2 = . Thus the correct choice is (b). 29. = T. Since the wire is free to contract, the tension F must increase to produce the same change in length. Now / Y= /

0

Also

=

(1

0

or

where Now

. From

= M=

or

=

0 0

(

=

0;

=

F

;

= mass per unit length = / = T

=

=

Y

is the density of alcohol at temperature . =

0

(1 +

)

0

U =

0

(1 +

)

=

0

0

= U0

1

0

where U0 is the upthrust at 0°C. Since the upthrust is independent of temperature, W1 = W2.

given)

II Multiple Choice Questions with One or More Choices Correct 1. Choose the wrong statements from the following. A metallic circular disc having a circular hole at its centre rotates about an axis passing through its centre and perpendicular to its plane. When the disc is heated.

T

1 where 0 is the density of alcohol at = 0°C. Hence

From (1) and (4), we get 0 0 0

F

=

(4)

=

v=

where alcohol and 0 is the volume of alcohol displaced at temperature = 0°C. Now the density of alcohol at temperature is

(3)

The upthrust at temperature (T + DT) =

T

30. Let 0 be the volume of the metal at 0°C and its volume at °C. At temperature the upthrust is U =

(2)

T)

=

Speed of transverse waves in the wire is

When the temperature is raised to (T + T), let , and be the base area, density and depth at this (area) expansion is = 2 . Hence T) = 0(1 + T) = 0 (1 + 2

T)

F=

(1)

0 0 0

(1 + 2

(a) (b) (c) (d)

its its its its

speed will decrease diameter will decrease moment of inertia will increase speed will increase

15.12 Comprehensive Physics—JEE Advanced

2. Choose the wrong statements from the following. Two spheres made of the same material have the same diameter. One sphere is hollow and the other is solid. If they are heated through the same range of temperature, (a) the hollow sphere will expand more than the solid sphere (b) the solid sphere will expand more than the hollow sphere (c) both spheres will expand equally (d) the hollow sphere will not expand at all. 3. Two rods and B of different metals have lengths and B at a certain temperature. It is observed that rod is 5 cm longer than rod B at all temperatures. If = 1.0 10–5 per °C and B = 1.5 10–5 per °C, then (a) = 10 cm (b) = 15 cm (c) B = 10 cm (d) B = 5 cm 4. A uniform metallic circular disc of mass M and radius R, mounted on frictionless bearings, is rotating an angular frequency about an axis passing through its centre and perpendicular to its plane. The temperature of the disc is then increased by . If . (a) the moment of inertia increases by MR2 (b) the moment of inertia remains unchanged.

(c) the angular frequency increases by 2 (d) the angular frequency decreases by 2

. .

5. A clock with a metallic pendulum gains 6 seconds each day when the temperature is 20°C and loses 12 seconds when the temperature is 40°C. Then (a) the clock will keep correct time at tempera80 °C ture 3 (b) the clock will keep correct time at tempera100 °C. ture 3 is 1.2

10–5 per °C.

is 2.1 10–5 per °C. 6. A bimetallic strip is formed out of two identical strips one of copper and the other of brass. The are C and B. When the temperature of the strip is increased by T, it bends to form an arc of radius of curvature R. Then R is (a) proportional to T (b) inversely proportional to T (c) proportional to ( C – B) (d) inversely proportional to ( C – B). IIT, 1999

ANSWERS AND SOLUTIONS 1. Due to thermal expansion, the diameter of the disc as well as that of the hole will increase. Therefore, the moment of inertia will increase resulting in an increase in the angular speed. Hence the correct choices are (a) and (c). 2. Statements (a), (b) and (d) are wrong. 3. When the rods are heated by °C, the increase in the length of the rods is = and ( – =

= B B B) will remain the same for all , if B, i.e. = B B B

=

=

B

= 1.5

B. Also

B

(

–

1.5 10 1.0 10 B)

5

= 1.5

= 5 cm. Thus

1.5 B – B = 5 cm B = 10 cm, and The correct choices are (b) and (c).

= 15 cm.

4. The moment of inertia of the disc about the given axis of rotation is 1 MR2 (1) I = 2 If the disc is heated, it expands. Hence R increases. The resulting increase in I is obtained by partially differentiating (1). 1 M 2R R ( M = constant) I= 2 or I = MR R But

R=R I = MR2

. Therefore, , which is choice (a)

(2)

Now, the angular momentum of the disc is given by =I (3) Since no external torque acts, remains constant. Partially differentiating (3), we have I

+

I= 0

Thermal Expansion 15.13

or

I

=–

At 20°C, the gain in time is

(4)

1 ( – 20) 2 At 40°C, the loss in time is

I Using (1) and (2) in (4) we get = –2 The negative sign indicates that the angular frequency decreases due to increase in temperature. Thus the correct choices are (a) and (d).

6 =

1 (40 – ) 2 Dividing (4) by (3), we have 12 =

5. Time taken for one oscillation of the pendulum is

or

2

T =4

which gives =

(1)

we have

Partially differentiating, we get 2

2T T = 4

(3)

86400

(4)

12 40 = 6 20

T=2 2

86400

1 2 which gives

80 °C. Using this value in Eq. (3), 3 80 3

6=

(2)

Dividing (2) by (1), we get T 1 = = = ( = ) T 2 2 2 where is the change in temperature. Now, one day = 24 hours = 86400 s. Therefore, gain or loss of time in one day is 1 86400 seconds T= 2 Let be the temperature at which the clock keeps correct time.

= 2.1

20

86400

10–5 per °C.

Thus the correct choices are (a) and (d). 6. C = 0 (1 + C T) and B = 0 (1 + B T) C – B = 0 ( C – B) T. Initially the bimetallic strip is straight, i.e. R is R decreases. The amount of bending is proportional to ( C – B) and T. Greater the bending, the smaller is the value of R. Hence the correct choices are (b) and (d).

III Multiple Choice Questions Based On Passage Question 1 to 3 are based on the following passage Passage I 20,000 J of heat energy is supplied to a metal block of mass 500 g at atmospheric pressure. The initial tempera400 J kg–1 °C–1 of volume expansion of metal = 8 spheric pressure = 105 Pa.

10–5 °C–1 and atmoIIT, 2005

1.

(a) 120°C (b) 130°C (c) 140°C (d) 150°C 2. Work done by the block on the surroundings is (a) 0.05 J (c) 1.0 J

(b) 0.1 J (d) 10 J

3. The change in internal energy is (a) zero (b) equal to 20,000 J (c) slightly greater than zero (d) slightly less than 20,000 J.

SOLUTION 1.

Q= T=

Final temperature = 100 + 30 = 130°C, which is choice (b).

T. Therefore, Q

=

20, 000 J (0.5 kg ) (400 J kg

1

C 1)

= 100°C

2. Density of metal ( ) = 8000 kg m–3. Volume of the block is

15.14 Comprehensive Physics—JEE Advanced

=

=

0.5 kg 8000 kg m

3

=

Increase in volume = = (8

1 10 16

10–5 )

1 16 =

3

=

10–3 m3

1 2

10–6 m3

T

1 10 2

= (105)

Work done W = P

6

= 0.05 J, which is choice (a). 3. Change in internal energy U = Q – W = 20,000 – 0.05 = 19999.95 J Thus the correct choice is (d).

100

(a) 2 : 1 (c) 3 : 1

Questions 4 to 6 are based on the following passage Passage II A rod of metal X of length 50.0 cm elongates by 0.10 cm when it is heated from 0°C to 100°C. Another rod of metal Y of length 80.0 cm elongates by 0.08 cm for the same rise in temperature. A third rod of length 50 cm, made by welding pieces of rods X and Y placed end to end, elongates by 0.03 cm when its temperature is raised from 0°C to 50°C. 4. X and of metal Y are in the ratio of

(b) 3 : 2 (d) 4 : 3

5. The length of the rod of metal X in the composite piece is (a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm 6. The length of the rod of metal Y in the composite piece is (a) 10 cm (b) 20 cm (c) 30 cm (d) 40 cm

SOLUTION = (1 + Therefore,

0.10 50.0 100 = 2.0 10– 6 per °C. For rod Y, we get = 1.0 10–6 per °C. So the correct choice is (a). 5. In the composite rod, + = 50.0 cm. When this rod is heated by 50°C, let the new lengths be and . Given + = 50.0 + 0.03 = 50.03 cm. Here

4. For rod X,

=

=

50) and

50.03 =

+

= 50 + (

=

+(

(1 + +

+

50). ) 50

) 50

Substituting the values of and and noting that + = 50 cm, we get = 10 cm, which is choice (a). 6. = 50 – = 50 – 10 = 40 cm, which is choice (d).

IV Integer Answer Type 1. A composite rod is made by joining a copper rod, end to end, with a second rod of a different material but of the same cross-section. At 25°C, the composite rod is 1 m in length of which the length of the copper rod is 30 cm. At 125°C the

length of the composite rod increases by 1.91 mm. = 1.7 10–5 per °C and that of the second rod is = 10–5 per °C. Find the value of n. IIT, 1979

SOLUTION 1. Length of the second rod at 25°C = 70 cm. Length of copper rod at 125°C = 30 (1 + 1.7 10–5 100) = 30.051 cm Increase in the length of copper rod = 0.051 cm

Increase in the length of second rod = 70 100 = 7000 cm Total increase in length = 0.051 cm + 7000 cm = 0.191 cm (given) which gives = 2 10–5 per °C. Thus the value of = 2.

16

Measurement of Heat

Chapter

REVIEW OF BASIC CONCEPTS 16.1

HEAT ENERGY

Heat is a form of energy. It is, therefore, measured in energy units. The SI unit of heat is joule (J). Another unit commonly used is the calorie. A calorie is the amount of heat required to raise the temperature of 1 g of water through 1°C. Experiments have shown that 4.18 J of mechanical work produce one calorie of heat. Thus 1 calorie = 4.18 joules or The ratio

1 cal = 4.18 J J=

work done = 4.18 J per cal heat produced

Specific Heat Capacity If m = 1 unit and T = 1 unit, then s = Q heat of a substance is the amount of heat required to raise the temperature of a unit mass of the substance through a unit degree. A commonly used unit of s is cal g–1 °C–1. In the SI system s is expressed in J kg–1 K–1. The two units are related as 1 cal g–1 °C–1 = 4.18 Jg–1 °C–1 ( 1 cal = 4.18 J) 1cal g–1 °C–1 = 4180 J kg–1 °C–1

or

Since the size of a degree on the celsius scale is equal to that on the kelvin scale, a temperature difference of, say, 1 °C is equal to a temperature difference of 1 K. Thus 1 cal g–1 °C–1 = 4180 J kg–1 K–1

is called the mechanical equivalent of heat. capacity of water = 1 cal g–1 °C–1 or 4180 J kg–1 K–1.

16.2

CALORIMETRY

If two substances having different temperatures are

energy will continue till the temperatures are equalized. The common temperature at thermal equilibrium is called the equilibrium temperature. If no heat energy is allowed to escape to the surroundings, the amount of heat energy gained by the initially colder body is equal to the amount of heat energy lost by the initially hotter body, i.e. Heat gained by one body = heat lost by the other body. This is the basic principle of calorimetry. The heat energy Q needed to raise the temperature through T of a mass m capacity s is given by Q = ms T

Molar Specific Heat C of a substance is the amount of heat energy required to raise the temperature of 1 mole of the substance through 1 K. It is expressed in J mol–1 K–1. s C) are related as C s = m where m is the number of kilograms per mole in the substance. ) is v

p

16.2 Comprehensive Physics—JEE Advanced

SOLUTION

Cp – Cv = R where R is the universal gas constant and its value is R = 8.315 J mol–1 K–1

16.3

LATENT HEAT

The heat energy supplied to a substance to change from solid to liquid state or from liquid to gaseous state is not registered by a thermometer as the heat energy is used up in bringing about a change of state. Hence it is called latent (or hidden) heat. A substance has two latent heats. of a substance is the heat energy required to convert a unit mass of a substance from the solid to the liquid state, without change of temperature. The latent heat of fusion of ice is 3.36 105 J kg–1 or 80 cal g–1. of a substance is the heat energy required to convert a unit mass of the substance from the liquid to the gaseous state, without change of temperature. The latent heat of vaporisation of steam is 2.26 106 J kg–1 or 540 cal g–1. 16.1 A 12 kW drilling machine is used to drill a hole in a metal block of mass 10 kg. Assuming that 25% power is lost in the machine, calculate the rise in tempera-

The amount of heat energy in copper block at 500°C is Q = ms T = (3.35 390 500) J Now L = 335 J g–1 = 335 103 J kg–1. The maximum mass of ice that can melt is Q 3.35 390 500 m= = L 335 103 = 1.95 kg 16.3 16 g of oxygen is heated at constant volume from 25°C to 35°C. Find the amount of heat energy required. Given Cv = 20 J mol–1 K–1. SOLUTION Mass of 1 mole of oxygen = 32 g Number of moles in 16 g of oxygen is 16 1 = n= 32 2 Heat energy required is Q = n Cv T =

of metal = 0.4 J g–1 K–1. SOLUTION

20

(35 – 25) = 100 J

16.4

3 Useful power available = 75% of 12 kW = 4 = 9 kW = 9000 W

12 kW

Heat energy consumed in 2 minutes (= 120 s) is Q = (9000 120) J Now

1 2

s = 0.4 J g–1 K–1 = 0.4

103 J kg–1 K–1

Q Rise in temperature is T = ms =

9000 120 10

3

(0.4 10 )

= 270°C

16.2 A copper block of mass 3.35 kg is heated to 500°C and then placed on a large block of ice. What is the maximum mass of ice that can melt? –1 K–1 and latent heat of fusion of water = 335 J g–1.

290 J of heat energy is required to raise the temperaheat of nitrogen at constant pressure. SOLUTION 7 1 Number of moles in 7 g of oxygen is n = = 28 4 Q = n Cp T Cp =

Q n T

=

290 1 4

= 29 J mol–1 K–1

40

16.5 200 g of water at 25°C is added to 75 g of ice at 0°C of the mixture?

Measurement of Heat 16.3 T2

SOLUTION

Q=

Heat energy of 200 g of water at 25°C is Q1 = ms T = 200 1 25 = 5000 cal Heat energy required to melt 75 g of ice at 0°C is Q2 = mL = 75 80 = 6000 cal

If 1 kg of ice at 0°C is mixed with 1 kg of steam at 100°C, the equilibrium temperature is 100°C, 0.665 kg of steam will be left and 1.335 kg of water will be formed. Water of mass m at °C is mixed with ice of mass mi at 0°C. L f mi (a) If m = temperature = 0°C. L f mi (b) If m < temperature = 0°C and mass of ice melted is mi=

. Amount of ice left = mi – m i.

Lf

(c) If m >

L f mi

temperature =

L mi m

m

> 0°C

16.6 very low temperatures (close to absolute zero), the s of a solid varies with absolute temperature T as s= 3 where is a constant whose value depends upon the material of the solid. Find the heat energy required to raise the temperature of 200 g of the solid from 1 K to 4 K. SOLUTION The amount of heat energy required to raise the temperature of mass m of the solid by dT kelvin is given by dQ = msdT Heat energy required to raise the temperature of the solid from T1 = 1 K to T2 = 4 K is

T 3 dT

T1

=

4

T1

T24

0.2 4 = 12.75 =

Since Q1 < Q2, the whole of ice will not melt. Hence, NOTES

T2

msdT = T14 ( 4) 4

(1)4

joule

16.7 Three liquids 1, 2 and 3 of masses m1 = m, m2 = 2m and m3 = 3m are at temperatures of 10°C, 18°C and 30°C. When liquids 1 and 2 are mixed, the equilibrium temperature is 16°C. When liquids 2 and 3 are mixed, the equilibrium temperature is 22°C. Find the equilibrium temperature when liquids 1 and 3 are mixed. Assume that there is no loss of heat to the surroundings. SOLUTION Let s1, s2 and s3 uids 1, 2 and 3 respectively. When liquids 1 and 2 are mixed, heat gained by 1 = heat lost by 2 m1s1(16 – 10) = m2s2(18 – 16) ms1(6) = 2ms2(2) 2 s 3 2 When liquids 2 and 3 are mixed heat gained by 2 = heat lost by (3) 2ms2(22 – 18) = 3ms3(30 – 22) s s3 = 2 3 s1 =

-

(1)

(2)

Let T be the equilibrium temperature. When liquids 1 and 3 are mixed. Heat gained by 1 = heat lost by 3 (3) ms1(T – 10) = 3ms3(30 – T) Using (1) and (2) in (3), we have 2 3

s2(T – 10) = 3

s2 (30 – T) 3

T = 22°C

16.4 Comprehensive Physics—JEE Advanced

I Multiple Choice Questions with Only One Choice Correct 1. 300 g of water at 25°C is added to 100 g of ice at (a) –

5 °C 3

(c) – 5 °C

(b) –

5 °C 2

(d) 0 °C

IIT, 1989 2. 100 g of ice at 0°C is mixed with 100 g of water (a) 0°C (b) 20°C (c) 40°C (d) 60°C 3. Steam at 100°C is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at 15°C till the temperature of the calorimeter rises to 80°C. The mass of steam condensed in kilogram is (a) 0.13 (b) 0.065 (c) 0.260 (d) 0.135 IIT, 1986 4. minute, from 27°C to 77°C. If the heat of combustion of LPG is 4.0 104 J g–1, how much fuel in grams is consumed per minute? (a) 15.25 (b) 15.5 (c) 15.75 (d) 16 5. A copper block of mass 2 kg is heated to a temperature of 500°C and then placed in a large block of ice at 0°C. What is the maximum amount of ice that –1

°C–1 and latent heat of fusion of water is 3.5 105 J kg–1. IIT, 2005 4 6 (a) kg (b) kg 3 5 8 10 kg (d) kg (c) 7 9 6. How much heat energy is joules must be supplied to 14 grams of nitrogen at room temperature to raise its temperature by 40°C at constant pressure. Molar mass of nitrogen = 28 and R J K–1 mol–1 is the gas constant. (a) 50 R (b) 60 R (c) 70 R (d) 80 R

7. The temperature of a liquid does not increase during boiling. The heat energy supplied during this process, (a) increases the kinetic energy of the molecules of the liquid (b) increases the potential energy of the molecules (c) increases both the kinetic and potential energy of the molecules (d) is merely wasted since no increase occurs in the total energy of the molecules. 8. A block of ice at – 10°C is slowly heated and converted to steam at 100°C. Which of the following curves represents the phenomenon qualitatively?

Fig. 16.1

IIT, 2000 9. 2 kg of ice at –20°C is mixed with 5 kg of water at 20°C in an insulating vessel having a negligible

and ice are 1 kcal/kg/°C and 0.5 kcal/kg/°C re-spectively and the latent heat of fusion of ice is 80 kcal/kg. (a) 7 kg (b) 6 kg (c) 4 kg (d) 2 kg IIT, 2003 10. A metal sphere of radius r S is rotated about an axis passing through its centre at a speed of n rotations per second. It is suddenly stopped and 50% of its energy is used in increasing

Measurement of Heat 16.5

its temperature. Then the rise in temperature of the sphere is: (a)

2

2

2 2 2

n r 5S

(b)

n2

10 r 2 S

7 5 ( rn)2 r2 n2 S (d) 8 14 S 11. If there are no heat losses, the heat released by the condensation of x grams of steam at 100°C into water at 100°C converts grams of ice at 0°C into water at 100°C. The ratio /x is (a) 1 (b) 2 (c) 3 (d) 4 12. 5 g of water at 30°C and 5 g of ice at –20°C are

15. A 1 kW electric kettle contains 2 litre water at 27°C. The kettle is operated for 10 minutes. If heat is lost to the surroundings at a constant rate of 160 J/sec, the temperature attained by water in 10 minutes (a) 57°C (c) 77°C

(c)

ice = 0.5 cal g–1 (°C)–1 and latent heat of fusion of ice = 80 cal g–1. (a) –5°C (b) 0°C (c) + 5°C (d) + 10°C 13. A metal block of mass 10 kg is dragged on a horizontal rough road with a constant speed of 5 ms–1. the road is 0.5, the rate at which heat is generated in (J s–1) is (a) 100 (b) 245 (c) 9.8 (d) 10 14. Liquid oxygen at 50 K is heated to 300 K at constant pressure of 1 atm. The rate of heating is constant. Which of the following graphs represents the veriation of temperature with time? (see Fig. 16.2). IIT, 2004

(b) 67°C (d) 87°C

IIT, 2005 16. In an industrial process 10 kg of water per hour is to be heated from 20°C to 80°C. To do this, steam at 150°C is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at 90°C. How many kg of steam are required per hour? heat of steam = 1 kilo cal kg–1 °C–1. Latent heat of steam = 540 kilo cal kg–1. (a) 1 kg (b) 2 kg (c) 3 kg (d) 4 kg 17. With what velocity should a lead bullet at an initial temperature of 30°C strike a target so that it just melts? Assume that 84% of the heat produced is cal kg–1 (°C)–1, latent heat of lead = 6 kilo cal kg–1 and melting point of lead = 330°C. (1 kilo calorie = 4.2 103 joule)

18.

(a) 100 5 ms–1

(b) 100 2 ms–1

(c) 100 10 ms–1

(d) 100 15 ms–1

ids at extremely low temperature (close to absos of a solid varies with absolute temperature T as s = cT 3 where c is a constant depending on the material of the solid. The heat energy required to raise the temperature of 0.1 kg of the solid from 0 K to 4 K is. (a) 4.2 c joule (b) 6.4 c joule (c) 8.4 c joule

(d) 12.6 c joule

19. 400 g of ice at 253 K is mixed with 0.05 kg of steam at 100°C. Latent heat of vaporisation of steam = 540 cal/g. Latent heat of fusion of ice = 80 cal/g. temperature of the mixture is (a) 273 K (b) 300 K (c) 330 K (d) 373 K IIT, 2007 Fig. 16.2

16.6 Comprehensive Physics—JEE Advanced

20. Two litres of water (density = 1 g/ml) in an openlid insulated kettle is heated by an electric heater of power 1 kW. The heat is lost from the lid at the rate of 160 J/s. The time taken for heating water

–1

to 75°C is (a) 340 s (c) 620 s

K–1) from 20°C

(b) 550 s (d) 760 s IIT, 2005

ANSWERS

1. 7. 13. 19.

(d) (b) (b) (a)

2. 8. 14. 20.

(a) (a) (c) (b)

3. (a) 9. (b) 15. (d)

4. (c) 10. (a) 16. (a)

5. (c) 11. (c) 17. (d)

6. (c) 12. (b) 18. (b)

SOLUTION 1. Let the temperature of the mixture be C. Heat lost by water in calories = 300 1 (25 – ) = 7500 – 300 Heat in calories required to melt 100 g of ice = 100 80 = 8000 Now Heat lost = heat gained 5 °C or 7500 – 300 = 8000 or = – 3 Since is negative, the water at 25°C cools to 0°C and melts a part of ice at 0°C. Heat lost = 300 1 (25 – 0) = 7500 cal. Hence only a part of the ice melts and resulting temperature is 0°C. Hence the correct choice is (d). 2. The amount of heat required to convert 100 g of ice at 0°C into water at 0°C = 100 80 = 8000 calories. This is precisely the amount of heat lost by 100 g of water at 80°C to bring its temperature down to 0°C. Therefore, the temperature of the mixture remains 0°C. Hence the correct choice is (a). 3. Let the mass of steam condensed be m kg. The latent of vaporisation of water = 556 k cal/kg. Therefore, heat lost by steam = 556 m + m (100 – 80) = 576 m kcal. Heat gained by calorimeter and water = (1.1 + 0.02) (80 – 15) = 72.8 kcal. Now, heat lost = heat gained, i.e. 576 m = 72.8 or m = 0.13 kg 4. min–1 = 3000 cm3 min–1 Density of water = 1 g cm–3.

Heat energy supplied to the geyser per minute = 3.0 4200 50 = 63

104 J min–1

Now, heat of combustion of LPG = 4.0 104 J g–1 = 4.0 107 J kg–1 Amount of fuel consumed per minute =

63 104 J min

1

4.0 107 J kg

1

= 15.75 10–3 kg min–1 = 15.75 g min–1 5. Heat energy in copper block = 2 400 500 = 4 105 J. The amount of ice that melts will be maximum if the entire heat energy of the copper block is used up in melting ice. Now, 3.5 105 J of heat energy is needed to melt 1 kg of ice into water. Therefore, the amount of ice melted by 4 105 J of heat energy is 4 105 J 8 = kg 5 1 7 3.5 10 J kg Hence the correct choice is (c). 6. Number of moles in 14 grams of nitrogen (n) = 1 14/28 = . Since nitrogen is diatomic Cp = 7R/2. 2 Therefore, amount of heat energy supplied 1 7R = n Cp = 40 = 70R joules 2 2 Hence the correct choice (c) 7. The correct choice is (b).

= 3000 g min–1 = 3.0 kg min–1

8.

Rise in temperature = 77 – 27 = 50°C –1

°C

–1

ice increases from – 10°C to 0°C. At 0°C, ice starts melting at a constant temperature. When the whole

Measurement of Heat 16.7

of ice has melted into water, the temperature of water will increase from 0°C to 100°C. At 100°C, again the temperature becomes constant due to the conversion of liquid water into water vapour (steam) at 100°C. Hence the correct graph is (a). 9. Let m kg be the mass of ice melted into water. Heat lost by 5 kg of water = 5 kg 1 kcal/kg/°C 20°C = 100 kcal. Heat gained = m kg 80 kcal/kg + 2 kg 0.5 kcal/kg/°C 20°C = 80 m kcal + 20 kcal. Now, heat gained = heat lost. Therefore, 80 m + 20 = 100 or m + 1 kg = 6 kg. Hence the correct choice is (b). 10. Moment of inertia of the sphere I =

2 Mr2. Given 5

= n rotations per second = 2 n rad s–1. The kinetic energy is 1 2 Mr2 (2 n)2 2 5 4 = M 2r 2n 2 5 Since half of KE is converted into heat energy, we have 1 2 dQ = KE = M 2 r2 n2 2 5 KE =

1 I 2

2

=

Heat required for step (ii) = 5 80 = 400 cal When water at 30°C is allowed to cool to 0°C, the heat given out = 5 1 (30 – 0) = 150 cal Thus it is clear that all the ice cannot melt and the system will remain at 0°C. Since only (150 – 50) = 100 cal are available for melting ice, the mass of ice melted = 100/80 = 1.25 g. 6.25 g of water at 0°C and (5 – 1.25) = 3.75 g of ice Hence the correct choice is (b). 13. Heat energy generated per second = work done per second against friction = force of friction distance moved in 1s = = 0.5 10 9.8 5 = 245 J s–1 Hence the correct choice is (b). 14. Liquid oxygen will undergo a change of phase when heated from 50 K to 300 K. During the phase change the temperature will not change. When all the liquid oxygen has changed to gaseous state, the temperature will increase because rate of heating is constant (see Fig.16.3). Hence the correct graph is (c).

Now dQ = MSdT which gives 2 M 2 r 2 n2 2 2 r 2 n2 dQ 5 dT = = = 5S MS MS Hence the correct choice is (a). 11. The latent heat of vaporisation of water is very nearly 540 calories per gram. Therefore heat released in the condensation of x gram of steam = 540 x calories. The latent heat of fusion of ice is very nearly 80 calories. Therefore, heat required to convert gram of ice at 0°C to water at 100°C = 80 + 100 = 180 calories. Thus 180 = 540 x

15. Power (P) = 1 kW = 1000 W, mass of water (m) = s) = 4200 J/kg °C. Time = 10 min = 600 s. Initial temperature T1 = 27°C. Heat energy supplied by kettle in 600 s = = 1000 600 = 600,000 J. Heat energy lost in 600 s = 160 600 = 96,000 J. Therefore, heat energy used up in heating the water is H = 600,000 – 96,000 = 504,000 J

=3 x Hence the correct choice is (c).

or

Now H = ms T = ms (T2 – T1) = 2 27), where T2

12. temperature of ice from – 20°C to 0°C and (ii) melt the ice at 0°C into water at 0°C. Heat required for step (i) = 5 50 cal

Fig. 16.3

0.5

{0 – (–20)} =

2

4200

4200

(T2 –

(T2 – 27) = 504,000

which gives T2 = 87°C, which is choice (d). 16. Let the mass of steam required per hour be m kg. Heat gained by water in boiler per hour is

16.8 Comprehensive Physics—JEE Advanced

=10 kg

18.

1 kilo cal kg–1 °C–1

(80 – 20)°C = 600 kilo cal (1) Heat lost by steam per hour is = heat needed to cool m kg of steam from 150°C to 100°C + heat needed to convert m kg of steam at 100°C into water at 100°C + heat needed to cool m kg of water from 100°C to 90°C = m 1 (150 – 100) + m 540 + m 1 (100 – 90)

constant. The amount of heat energy required to raise the temperature of the solid of mass m through dT kelvin is dQ = msdT Heat energy needed to raise the temperature of the solid from 0 K to 4 K is 4

0

= 50 m + 540 m + 10 m = 600 m kilo cal

17. Let the mass of the bullet be m kg and its velocity be v ms–1. Before striking the target, the kinetic en1 ergy of the bullet is mv2 joule which is converted 2 into heat energy when the bullet strikes the target. Thus heat energy produced is 1 mv2 joule Q = 2

8.4 103

kilo cal

Heat energy absorbed by the bullet is Q = 84% of Q = 0.84 Q =

0.84

mv

2

8 4 103

kilo cal

In order that heat energy Q melts the bullet, it bullet from 30°C upto its melting point (330°C) and then to supply the latent for melting. Hence 0.84

m v2

8. 4 103

=m

0.03

(330 – 30)

= 9 m + 6 m = 15 m

0

T 3dT = mc

= mc 0

44 4

= mc

T4 4 c

= 0.1

4

0

(4)3

= 6.4 c joule. So the correct choice is (b). 19. Heat required to melt the whole of ice is Q1 = ms T + mL = 400

0.5

20 + 400 (

80 253 K = –20°C)

= 4000 + 32000 = 36000 cal The maximum heat released by steam when the whole of it (= 50 g) is converted into water at 0°C is Q2 = ms T + mL = 50 1 100 + 540 50 = 5000 + 27000 = 32000 cal Since Q2 is less than Q1, the whole of ice will not

1 m v 2 joule 2 = 4.2 103 joule/kilo cal m v2

cT 3dT

4

(2)

Heat lost = heat gained. Equating (1) and (2) we have 600 m = 600 or m = 1 kg, which is choice (a).

=

4

msdT = m

Q =

m

6

will be 0°C or 273 K. 20. Mass of 2 litres of water = 2 kg. Heat energy needed to raise the temperature of 2 kg of water from 20°C to 75°C is Q = 2 (4.2 103) 55 = 4.62 105 J If is the time taken, heat energy supplied by the heater in time is Q1 = (power time) = 1000 joule Heat energy lost in time is Q2 = 160 joule

which gives v = 100 15 ms–1

Heat energy available for heating water is Q = Q1 – Q2 = 840 J

Thus the correct choice is (d).

Equating Q = Q , we get

550 s.

Measurement of Heat 16.9

II Multiple Choice Questions with One or More Choices Correct 1. A source of heat supplies heat at a constant rate to a solid cube. The variation of the temperature of the cube with heat supplied is shown in Fig. 16.4. (a) Portion BC of the graph represents conversion of solid into liquid. (b) Portion BC of the graph represents conversion of solid into vapour. (c) Portion DE of the graph represents conversion of vapour into liquid. (d) Portion DE of the graph represents conversion of liquid into vapour

(c) thermal capacity of liquid (d) thermal capacity of vapour. 4. In Fig. 16.4, it is observed that DE = 3 BC. This means that (a) the thermal capacity of the vapour is 3 times that of the liquid of the liquid (c) the latent heat of vaporisation of the liquid is 3 times the latent heat of fusion of the solid. (d) the latent heat of fusion of the solid is 3 times the latent heat of vaporisation of the liquid. 5. 800 J kg–1 K–1 is initially at a temperature 20°C. It –1

Fig. 16.4

2. In Q.1 above, the slope of portion CD of the graph shown in Fig. 16.4 gives (a) latent heat of fusion (b) latent heat of vaporisation (c) thermal capacity of liquid (d) thermal capacity of vapour 3. In Q.1 above, the slope of the portion EF of graph shown in Fig. 16.4 gives (a)

and on returning to the starting point strikes a lump of ice at 0°C and gets embedded in it. Assume that all the energy of the bullet is used up in melting. Neglect the friction of air. Latent heat of fusion of ice = 3.36 105 Jkg–1. (a) Energy of bullet used in melting is 1000 J. (b) The mass of ice method = 5 g (c) The mass of ice melted is slightly greater than 5 g. (d) The mass of ice melted is less than 5 g. 6. ity s falls from a height of 10 cm and bounces to a height of 4 m. If all dissipated energy is absorbed by the ball as heat, its temperature rises by 0.075 K. Take = 10 ms–2 . (a) Q = 60 J (b) Q = 100 J (c) s = 800 J kg–1 K–1 (d) s = 1333 J kg–1 K–1

ANSWERS AND SOLUTIONS 1. The correct choices are (a) and (d). The heat supplied is the latent heat. 2. The slope of portion CD of the graph gives the amount of heat supplied by unit rise in temperature of of the liquid. Hence the correct choice is (c). 3. The correct choice is (d). 4. The portion BC of the graph represents the conversion of solid into liquid, temperature remaining the

same. The portion DE represents the conversion of liquid into vapour at the same temperature. The heat supplied in the two cases is latent (hidden). Hence the correct choice is (c). 5. If the friction offered by air is neglected, the speed of the bullet on returning to the starting point will be equal to its initial speed v = 200 ms–1. The kinetic energy of the bullet is

16.10 Comprehensive Physics—JEE Advanced

K.E. =

x = 5.3 10–3 kg = 5.3 g. Hence the correct choices are (a) and (c).

1 mv2 2

6. Q =

1 = (50 2

–3

2

10 )

(200) = 1000 J

Also

Heat lost by bullet for its temperature to fall from 20°C to 0°C = (50 10–3) 800 20 = 800 J. If x kg is the mass of ice melted, then x

(3.36

(h – h ) = 1

s=

10

(10 – 4) = 60 J.

Q = ms T. Hence

Q 60 = m T 1 0.075

= 800 J kg–1 K–1. Thus the correct choices are (a) and (c).

105) = 1000 + 800

III Multiple Choice Questions Based on Passage Questions 1 to 3 are based on the following passage Passage I The basic principle of calorimetry is heat gained by one body = heat lost by the other body. This follows from the principle of conservation of energy according to which the total heat energy of the two substances must remain constant. Hence heat lost by one body must be gained by the other, provided no part of heat energy is allowed to escape. An aluminium container of mass 100 g contains 200 g of ice at – 20°C. Heat is added to the system at ice = 2100 J kg–1 K–1 aluminium = 840 J kg–1 K–1 and latent heat of fusion of ice = 3.36 105 J kg–1.

1. The time taken to raise the temperature of the container and ice from – 20°C to 0°C is (a) 12 s (b) 24 s (c) 36 s

(d) 48 s

2. The time taken to melt ice at 0°C into water at 0°C is (a) 40 s (b) 80 s (c) 120 s

(d) 160 s

3. The temperature of the system after 4 minutes is (a) 15.45°C (b) 20.45°C (c) 25.45°C

(d) 30.45°C

SOLUTION 1. Heat energy needed to raise the temperature of the container and ice from – 20°C to 0°C is Q1 = (100

10–3)

+ (200

840 –3

10 )

20

2100

20

= 1680 + 8400 = 10080 J 10080 Time needed is 1 = = 24 s 420 So the correct choice is (b). 2. Heat energy required to melt ice at 0°C into water at 0°C is Q2 = (200 10–3) (3.36 105) = 6.72

4

10 J

Time needed is

= 2

6.72 104 = 160 s, which is 420

choice (d). 3. Total time 1 + 2 = 24 + 160 = 184 s is less than 4 minutes (= 240 s). Hence heat energy supplied during (240 – 184) = 5 s will be used up in raising the temperature of the system (container + 200 g plied is 56 s = 420 56 = 23520 J. If temperature, then 23520 = (100 10–3) 840 + (200

10–3)

4200

which gives = 25.45°C. So the correct choice is (c).

Measurement of Heat 16.11

4. The rate at which heat is produced is (a) 35.8 Js–1 (b) 36.8 Js–1 –1 (c) 37.8 Js (d) 38.8 Js–1 5. The power of the drill is (a) 40 W (b) 42 W (c) 48 W (d) 56 W 6. The torque required to drive the drill is (a) 1.2 Nm (b) 2.2 Nm (c) 3.2 Nm (d) 4.2 Nm

Questions 4 to 6 are based on the following passage Passage II A steel drill making 180 revolutions per minute is used in during a hole in a block of steel. The mass of the steel block is 180 g. 90% of the entire mechanical energy is used up in producing heat and the rate of rise of temperature of of steel = 420 J kg–1 K–1.

SOLUTION 4.

Q

= mass

sp. heat

second = (180 10–3) 420 So the correct choice is (c). 5. Power of drill =

180 60 = 3 r.p.s. Angular frequency of rotation ( ) = 2 3 = 6 rad s–1 Now power = torque angular frequency or P = P 42 = = = 2.2 Nm 6 3.14

6. Number of revolutions per second =

rise in temperature per 0.5 = 37.8 Js–1.

37.8 = 42 W, which is choice 0.9

(b).

Thus the correct choice is (b).

IV Integer Answer Type 1. 2 kg of ice at – 20°C is mixed with 5 kg of water at 20°C in an insulating vessel having negligible heat

and ice are kcal/kg/°C and 0.5 kcal/kg/°C and latent heat of fusion of ice is 80 kcal/kg. IIT, 2003

SOLUTION 1. mi si [0 – (– 20)] + m L = m s (20 – 0) 2 0.5 20 + m 80 = 5 1 20

m = 1 kg. Mass of water in container = 5 + 1 = 6 kg

17

Chapter

Thermodynamics (Isothermal and Adiabatic Processes)

REVIEW OF BASIC CONCEPTS 17.1

EQUATION OF STATE

In the case of ideal gases, the equation of state is PV = RT, for one mole and PV = nRT, for n moles where P, V and T are respectively the pressure, volume and temperature of the gas and R is the universal gas constant.

17.2

MOLAR SPECIFIC HEAT

C of a substance is the amount of heat energy required to raise the temperature of 1 mole of the substance through 1 K. It is expressed in J mol–1 K–1. s C) are related as C s= m where m is the number of kilograms per mole in the substance. v) is

p

17.3

The Zeroth law of thermodynamics which states that

Thus if

17.4

when R is the universal gas constant and its value is R = 8.315 J mol–1 K–1

T = TC and TC = T , then T = T .

INTERNAL ENERGY

amount of energy. This energy is called the and is usually denoted by the symbol U. The internal energy of solid, liquid or gas consists of two rotational and vibrational) of the molecules, and of the molecules. If the intermolecular forces are extremely weak or absent, then the change in internal energy is given by U = mcv T where m is the mass of the gas, cv

T

heat Cv Mcv where M is the molecular mass), we have, for n moles of an ideal gas m Cv T U = nCv T = M

17.5 Cp – Cv = R

ZEROTH LAW OF THERMODYNAMICS

FIRST LAW OF THERMODYNAMICS

When heat energy is supplied to a system, a part of this energy is used up in raising the temperature of the system the rest is used up in doing external work against the

17.2 Comprehensive Physics—JEE Advanced

surroundings. Thus, if Q is the heat energy supplied to a gas and if W is the work done by it, then from the law of conservation of energy, the increase U in the internal Q – W) or Q= U+ W

law of thermodynamics which may be stated in words as ‘

’

where

=

Cp Cv

.

For an adiabatic process, Q thermodynamics, Q = U + W, it follows that U + W = 0 or W = – U W is positive. hence U is negative, i.e. the temperature of the gas falls. W is negative. Hence U is positive, i.e. the temperature of the gas increases. 3. Isochoric Process

Sign convention for

Q,

W and

U

1.

Q is positive if heat is supplied to the system and negative if heat is taken out of the system. 2. W is positive if work is done by the system and negative if work is done on the system. 3. U is positive if the temperature of the system increases and negative if the temperature of the system decreases.

17.6

THERMODYNAMIC PROCESSES

1. Isothermal Process

between P and V is PV = constant

. From PV = nRT, it follows n = constant), the relation

an isochoric process. The relation between P and T is P1 P2 P = constant P T T1 T2 T The work done in a process is W = P V. For an isochoric process, V = 0. Hence W law of thermodynamics it follows that Q= U Q is positive. Hence U is positive, i.e. the temperautre of the gas rises. Hence pressure also increases. Q is negative. Hence U is negative, i.e. the temperature of the gas falls. Henc pressure also falls. 4. Isoboric Process

P 1V 1 = P 2V 2 T = constant)

If T is kept constant T = 0. Hence U = 0, i.e. for an isothermal process, the internal energy of the gas remains that Q= W i.e. all the heat energy supplied to the gas is used up in doing work. gas. Hence both W and Q are positive. on the gas. Hence both W and Q are negative. 2. Adiabatic Process . For an adiabatic process, the realtion between P and V is PV = constant P 1V 1 = P 2V 2

V

. The relation between V and T is V1 V2 V = constant T T1 T2 T

from V T, that the temperature rises. Hence W and U thermodynamics, that Q is also positive. Hence Q, U and W are all positive in isobaric expansion. Q, U and W are all negative.

17.7

WORK DONE IN A PROCESS

from volume V1 to volume V2 is given by W =

V2 V1

1. Work Done in an Isothermal Process When an

17.3

at constant temperature) work is done by it. For an isothermal process, the equation of state for n moles of ideal gas is PV = nRT where T is the constant absolute temperature and R is the universal gas constant. The value of R for all gases is R = 8.31 J K–1 mol–1 P=

The work done W = nRT

V2 V1

or

V

n RT V = nRT ln

W = 2.303 nRT log

V2 V1

e)

4. Work Done in an Isochoric Process In an isochoric process, volume V is constant, i.e. = 0. Hence, work done W = 0.

17.8

INDICATOR DIAGRAM

P – V diagram) is a graph in which P) is plotted on the V) on the -axis. Figure 17.1 shows indicator diagrams for expansion, compression and for a closed cyclic process. The initial P1, V1) is represented by point P2, V2) by point . The intermediate states are represented by points between and on the curve .

V2 V1

where V1 is the initial volume and V2 P1 and P2 P 2V 2)

P 1V 1 =

P1 P2 2. Work Done in Adiabatic Process When a gas undergoes an adiabatic change, the pressurevolume changes obey the relation PV = constant = C where = Cp/Cv the gas at constant pressure to that at constant volume. W = 2.303 nRT log

C

, where C is a constant V The work done is given by 1 W= P 2V 2 – P 1V 1) 1 P=

= =

1 1 nR 1

P 1V 1 – P 2V 2) T 1 – T 2)

since P1V1 = nRT1 and P2V2 = nRT2, T1 and T2 being the absolute temperatures before and after the adiabatic change. 3. Work Done in an Isobaric Process For an isobaric process, pressure P is constant. Therefore, the work done is given by W =P

V2 V1

= P V 2 – V 1)

Fig. 17.1

The work done in a process is given by W = area enclosed by P – V curve and volume axis. Thus, the work done is given by the area of the shaded portion in Fig. 17.1.

17.9 given by

EFFICIENCY OF AN IDEAL HEAT ENGINE ) of an ideal reversible heat engine is =1 –

where

T2 T1

T1 = absolute temperature of the source which supplies heat and T2 = absolute temperature of the sink which takes in the part of heat not converted into useful work,

17.4 Comprehensive Physics—JEE Advanced

T1 is always greater than T2. If T1 = T2 = 0 implying that an engine working under isothermal conditions can produce no useful work. Complete conversion of heat into = 1) is possible only if T2 = 0, i.e. the sink is at absolute zero, which is unattainable.

17.10

Cv, Cp AND GAS

= Cp /Cv FOR AN IDEAL

1. For a monoatomic gas, 3R 5R , Cp = and Cv = 2 2 2. For a diatomic gas, 5R 7R , Cp = and Cv = 2 2 3. For a triatomic or polyatomic gas

5 = 1.67 3

=

7 = 1.4 5

= 1.4) at 127°C are expanded adiabatically to twice the original volume. done in the process. Given R = 8.3 J K–1 mol–1 and 0.4 = 0.76. SOLUTION T1 V1

and

W=

=

=

– 1)

V1 V2 nR 1

1

0.4

= 400

0.76 = 304 K

T 1 – T 2)

2 8.3 1.4 1

17.3

CP = Cv Cp = Cv Cv =

= T2 V2

T2 = T1

Cp – Cv = R

or

– 1)

4 = 1.33 3 RELATION BETWEEN Cp, Cv AND Cv = 3R, Cp = 4R and

17.11

=

17.2

takes up 746 J of heat from the high temperature res-

R 1

and Cp =

R 1

17.1 5 moles of an ideal gas are compressed to half the initial volume at a constant temperature of 27.0°C. Calculate the work done in the process. Given R = 8.3 J K–1 mol–1. Write your result up to appropriate

SOLUTION T1 = 100°C = 373 K, T2 = 0°C = 273 K, Q1 = 746 J T W =1– 2 T1 Q1

W = Q1

T1

T2 T1

= 746 = 200 J Q2 = Q1 – W = 746 – 200 = 546 J

SOLUTION Given T = 27.0°C = 300 K. Since T is constant, the process is isothermal W = nRT loge

V2 V1

=5

8.3

300

=5

8.3

300

373 273 373

=

W 200 = = 0.268 = 26.8% Q1 746

NOTE

loge

1 2

= – 8628 J Since the value of R s. .), the value of W must be rounded off to 2 s. . as W = – 8.6 103 J The negative sign indicates that work is done on the gas.

=1–

T2 273 =1– = 0.268 T1 373

17.4 Figure 17.2 shows the P-V diagram of a cyclic process . Calculate the work done in process to to C C to

17.5

17.6

8

= 1.4) are heated at constant volume. If 280 J of heat energy is supplied

6 (Nm–2)

4

gas and the work done.

2

2

4

6

8

SOLUTION Since V = 0, W modynamics, =

10 (litre)

= 280 J

Fig. 17.2

SOLUTION W + area of rectangle 1 2 1 = 2

=

SOLUTION

+ 4

6

10

17.7 The pressure P of an ideal gas varies with volume V as P = where is a constant. The volume of n moles of the gas is increased from V to mV. Find the work done and the change in internal energy.

–3

+4

6

10

–3

mV

mV

W=

1 litre = 10–3 m3)

V = V

V

10–3 J

= 36

W C = –24 10–3 J Negative sign shows that the work is done on the gas. volume is constant) WC W = 36 10–3 – 24 10–3 = 12 10–3 J which is equal to area enclosed by the closed loop . 17.5 = 1.4) are heated at constant pressure. If 280 J of heat energy is sup-

=

nCp n C

Q=

U+W

T=

U+W

T=

U+W

U=

U+W W

U=

1

V2 2

mV

= V

2

2

m2 – 1)

Cp C U=nC =

2

2

m2

T)

1 1

17.8 Two moles of a monoatomic gas undergo a cyclic process as shown in Fig. 17.3. 2 x 105

SOLUTION U=nC Given

(Nm–2)

C T=n Cp Q = n Cp U= W=

Cp T =

nCp T

T = 280 J. Hence

2.80 = 200 J 1.4 Q–

1 x 105

U = 280 – 200 = 80 J

0

1

2

3

4

5 (

3)

Fig. 17.3

Process is isobaric, process C is adiabatic and process C is isothermal. Find

17.6 Comprehensive Physics—JEE Advanced

Hence the sope of P-V graph for an adiabatic process is greater than for an isothermal process. Figure 17.4 shows two P-V curves, one for adiabatic expansion and the other for isothermal expansion.

C . SOLUTION P = P = 2 105 Nm–2 V = 2 m3, V = 5 m3 and C P V = PC V C For isothermal process C P V = PC VC P =P ) 1 1

V V

VC =

Substituting the values of V , V and , we get 3 VC PV PC = = 0.2 105 N m–2 VC W=W =P

+W

+ WC 1 V –V)+ 1 C

P V – P CV C)

+ P V loge Substituting the values, we get W = 6 105 N m–2

V VC

17.9 Show that the slope of P-V curve for an adiabatic process is greater than that for an isothermal process. SOLUTION For an isothermal process, PV = constant. Differentiating w.r.t. volume V we have P + =0 = V iso For an adiabatic process, PV = constant, Cp . where = Cv Differentiating w.r.t. volume V, we have +V =0 PV – 1 adia

Since

> 1,

adia

P V

=

iso

.

Fig. 17.4

17.10 Two moles of a diatomic ideal gas, initially at pressure 5.0 104 Pa and temperature 300 K are expanded isothermally until the volume of the gas is doubled and then adiabatically expanded until the volume is again doubled. Find end of the complete proces

process. P-V graph for the complete process. Given R = 8.3 JK–1 mol–1. SOLUTION Let P1, V1, T1 be the initial pressure, volume and temperature of the gas For isothermal process P1V1 = P2V2 where V2 = 2V1 P2 = P1

V1 V2

= 5.0

104

= 2.5

104 Pa

1 2

In an isothermal process, the temperature remains constant. Hence T2 = T1 = 300 K. Therefore change U)1 = 0. Work done is W1 = nRT1ln

V2 V1

=2

8.3

300

17.7

= 3.45 103 J For adiabatic process P 2V 2 = P 3V 3 P3 = P2

5 8.3 2 5R for diatomic gas) Cv = 2 = – 3.03 103 J =2

V2 V3

= 2.5

1 2

104

1.4

P3 103 Pa, T3 = 227 K W = W1 + W2 = 3.45 103 + 3.03 103 = 6.48 103 J U)2 = 0 – 3.03 103 = U U)1 – 3.03 103 J. The negative sign shows that there is a decrease in internal energy. P-V graph for the complete process is shown in Fig. 17.5.

3

10 Pa T 2V 2

– 1)

– 1)

= T 3V 3

1

V2 V3

T3 = T2

1 2

= 300

0.4

= 227 K

Work done is

Isothermal expansions

W2 = =

nR 1

T2

T3

2 8.3 1 4 1)

Adiabatic expansions

= 3.03 103 J Change in internal energy is )2 = nCv T = n

5R 2

1

T

2

3

Fig. 17.5

I Multiple Choice Questions with Only One Choice Correct 1. If 2 moles of an ideal monoatomic gas at temperature T are mixed with 3 moles of another monoatomic gas at temperature 2T, the temperature of the mixture will be 8T 6T 5 5 4T 3T 3 2 2. If heat energy Q is supplied to an ideal diatomic gas, the increase in internal energy is U and the work done by the gas is W. The ratio Q : U : W is 3. Figure 17.6 shows a cyclic process. When a given mass of a gas is expanded from state to state , it absorbs 30J of heat energy. When the gas is adiabatically compressed from state to state ,

the work done on the gas is 50 J. The change in internal energy of the gas in the process is

Fig. 17.6

4. Figure 17.7 shows a cyclic process for an ideal diatomic gas. The ratio of the heat energy absorbed in the process to the work done on the gas in the process C is

17.8 Comprehensive Physics—JEE Advanced

10. For a thermodynamic process, the P-V graph for a monoatomic gas is a straight line passing through the origin and having a positive slope. The molar heat capacity of the gas in this process is R R

11.

7 2 ln 2

5 ln 2 4

5 ln 2 2

5. The internal energy of 3 moles of hydrogen at temperature T is equal to the internal energy of n moles of helium at temperature T/2. The value of n ideal gases) 3 2 6. The temperature of n moles of an ideal gas is increased from T to 3T in a process in which the temperature changes with volume as T = 2 where is a constant. The work done by the gas in this process is n RT n RT 3 n RT n RT 2 7. In a certain process, pressure P, volume V and temperature T of a gas are related as PV = n where and n are constants. The work done by the gas when the pressure is kept constant, is proportional to T)n T )1/n T n –1) T n + 1) 8. If the pressure of an ideal gas in a closed container is increased by 2%, the temperature of the gas increases by 5°C. The initial temperature of the gas is 9.

pressure P in the balloon is related to the volume V as PV2/3 = , where is a constant. If T is the temperature of the mixture, volume V is proportional to T T2 3 T4 T

R

600 K has a work output of 800 J per cycle. How much heat energy is supplied to the engine from the source in each cycle?

Fig. 17.7

7 4 ln 2

3 R 2

12. should the temperature of the source be increased

13. The pressures and volumes corresponding to P = 3 104 Pa, –3 3 4 V = 2 10 m , P = 8 10 Pa, V = 5 10–3 m3. In process , 600 J of heat and in process , 200 J of heat is added to the system. The change in the internal energy in process would be

Fig. 17.8

14. One mole of an ideal gas requires 207 J heat to raise its temperature by 10 K when heated at constant pressure. If the same gas is heated at constant volume to raise the temperature by the same R, the gas constant = 8.3 JK–1 mol–1)

17.9

15. during the cycle is PV PV

PV

Fig. 17.10

21. In rising from the bottom of a lake to the top, the temperature of an air bubble remains unchanged, but its diameter is doubled. If h is the barometric density ) at the surface of the lake, the depth of the

Fig. 17.9

16. The equation of state corresponding to 8 g of O2 is 2 to be an ideal gas) RT PV = 8 RT PV = 4 RT PV = RT PV = 2 17. The equation of state of a gas is P

aT V

2

Vc

h h h h 22. Entropy of a thermodynamic system does not change when the system is used for cold reservoir

RT + )

where a, , c and R are constants. The isotherms can be represented by m n – P= where and depend only on temperature and m = – c, n m = c, n = 1 m = – c, n m = c, n = – 1

-

chorically

23. Heat energy absorbed by a system in going through a cyclic process shown in Fig.17.11 is 7

4

2

–3

18. When an ideal monoatomic gas is heated at constant pressure, the fraction of heat energy supplied which increases the internal energy of the gas is 2 3 5 5

19.

3 7

3 4

heats is 4150 J kg–1 K–1 and the ratio of the two gas at constant volume in units of J kg–1 K–1?

Fig. 17.11

24. will be value of

20.

to by three different paths 1, 2 and 3 as shown in Fig. 17.10. If W1, W2 and W3 respectively be the work done by the gas along the three paths, then W1 < W2 < W3 W1 > W2 > W3 W1 = W2 = W3

W1 < W2 W1 < W3

for the mixture?

= 5/3) is mixed = 7/5). What

25. If the ratio Cp/Cv = , the change in internal energy of the mass of a gas, when the volume changes from V to 2V at constant pressure P is R PV 1

17.10 Comprehensive Physics—JEE Advanced

PV PV 1 1 26. Two cylinders and equal amounts of an ideal diatomic gas at 300 K. The piston of is free to move, while that of is gas in each cylinder. If the rise in temperature of the gas in is 30 K, then the rise in temperature of the gas in is

Fig. 17.12

T 1,

30.

piston. The gas is allowed to expand adiabatically to a temperature T2 by releasing the pison suddenly. If L1 and L2 are the lengths of the gas column before and after expansion respectively, then T1/T2 is given by

27. Two identical containers and tionless pistons contain the same ideal gas at the same temperature and the same volume V. The mass of the gas in is m and that in is m . The gas in each cylinder is now allowed to expand V. The changes in pressure in and are found to be P and 1.5 P respectively. Then m

m =3m

m =2m

m =4m

m

28. Two monoatomic ideal gases 1 and 2 of molecular masses M1 and M2 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 M1 M2 M1 M2

L1 L2

2/3

L2 L1

L1 L2 L2 L1

2/3

IIT, 2000 31. Starting with the same initial conditions, an ideal gas expands from volume V1 to V2 in three different ways. The work done by the gas is W1 if the process is purely isothermal, W2 if purely isobaric and W3 if purely adiabatic. Then

M2 M1

W2 > W1 > W3

W2 > W3 > W1

W1 > W2 > W 3

W1 > W3 > W2 IIT, 2000

M2 M1

32.

IIT, 2000 29. T and volume V. Its volume is increased by V due to an increase in temperature T, pressure remaining constant. The quantity = V V T) varies with IIT, 2000

expansion. If its temperature falls by 2K, its internal energy will

7 initially at S.T.P. 5 are compressed adiabatically so that its temperature becomes 400ºC. The increase in the internal energy

33. 5 moles of Hydrogen

R = 8.30 J mol–1 K–1)

34. During an adiabatic process, the pressure of a gas is proportional to the cube of its absolute temperature. The value of Cp/Cv for that gas is:

17.11

35.

3 5 5 3

IIT, 2002

4 3 3 2 2

32) at a temperature T. The pressure of the gas is P T has a pressure of P P 8 P P Fig. 17.15

36. In a given process on an ideal gas, dW = 0 and dQ < 0. Then for the gas

ill increase 37. P-V plots for two gases during adiabatic processes are shown in Fig. 17.13. Plots 1 and 2 should correspond respectively to

IIT, 2001

40. Figure 17.16 shows the P-V mass of an ideal gas undergoing cyclic process. represents isothermal process and represents adiabatic process. Which of the graphs shown in Fig. 17.17 represents the P-T diagram of the cyclic procee?

Fig. 17.16

IIT, 2003

2 2

and He

2

and N2

Fig. 17.13

IIT, 2001 38.

C , as shown in Fig. 17.14. If the net heat supplied to the gas in the cycle is 5 J, the work done by the gas in the process C is

IIT, 2002

Fig. 17.17

Fig. 17.14

39. Which of the graphs shown in Fig. 17.15 correctly represents the variation of V/dp)/V with p for an ideal gas at constant temperature?

41. The pressure and density of a diatomic gas p1, 1) to p p2, 2). If 2 = 32, then 2 is p1 1 1 128

17.12 Comprehensive Physics—JEE Advanced

42.

43.

= 1.4) is adiabatically compressed so that its temperature rises from 27°C to 35°C. The change in the internal energy of the R = 8.3 J/mole/K)

Va V = Vc V

Va T = 2 T1 V

Va V = V Vc

Va T = 1 T2 V

V, pressure P and temperature T. The mass of each molecule of the gas is m is the P P T

44.

45.

mP

a monoatomic ideal gas which expands at constant pressure. What fraction of the heat energy is converted into work? 2 3 2 5 5 7

Fig. 17.18

49.

in the V – T diagram. Which of the diagrams shown in Fig. 17.20 shows the same process on a P – V diagram.

gas at 0°C and also contains an insulated piston of negligible weight and negligible thickness at the middle point. The gas on one side is heated to 100°C. If the piston moves through 5 cm, the length of the hollow cylinder is Fig. 17.19

46.

volume V1 to volume V2 and then compressed adiabatically to original volume V1. The initial pressure is P1 P3. If the net work done is W, then P 3 < P 1, W < 0 P 3 > P 1, W P 3 > P 1, W

47.

P 3 = P 1, W = 0 IIT, 2004

resistance 50 for 10 minutes by connecting it to a d.c. source of 10 V. The change in the internal energy is Fig. 17.20

48. Two different adiabatic paths for the same gas intersect two isothermals at T1 and T2 as shown in the P–V

50.

= 1.4) expands from 5 10–3 m3 to 25 10 m at a constant pressure of 1 105 Pa. The heat energy supplied to the gas in this process is –3

3

17.13

internal energy of the gas is

51. Three moles of an ideal gas are taken through a cyclic process as shown on T–V diagram in Fig. 17.21. The gas loses 2510 J of heat in the complete cycle. If T = 100 K and T = 200 K, The work done by the gas during the process is

. The change in the

IIT, 2005 54. The P–V diagram for n moles of an ideal gas The maximum temperature of the gas during the process is 3P0V0 P0V0 nR 4nR 2 P0V0 nR

3P0V0 2nR

Fig. 17.21

52. Liquid oxygen at 50 K is heated at 300 K at constant pressure of 1 atmosphere. The rate of heating is constant. Which of the graphs shown in Fig. 17.22 represent the variation of temperature T )? IIT, 2004

Fig. 17.23

55.

lid insulated kettle is heated by an electric heater of power 1 kW. The heat is lost from the lid at the rate of 160 J/s. The time taken for heating water –1 –1 K ) from 20°C to 75°C is

IIT, 2005 2

PT = constant.

56.

Fig. 17.22

53. One mole of a monoatomic ideal gas is contained in a insulated and rigid container. It is heated by

1 T

2 T

3 T

4 T IIT, 2008

ANSWERS

1. 7. 13. 19. 25.

2. 8. 14. 20. 26.

3 9. 15. 21. 27.

4. 10. 16. 22. 28.

5. 11. 17. 23. 29.

6. 12. 18. 24. 30.

17.14 Comprehensive Physics—JEE Advanced

31. 37. 43. 49. 55.

32. 38. 44. 50. 56.

33. 39. 45. 51.

34. 40. 46. 52.

35. 41. 47. 53.

36. 42. 48. 54.

SOLUTIONS 1. Let T0 be the temperature of the mixture. Since the total internal energy remains unchanged, we have U of mixture = U1 + U2 n 1 + n 2) C v T 0 = n 1 C v T 1 + n 2 C v T 2 n 1 + n 2) T 0 = n 1 T 1 + n 2 T 2 T0 = 2T + 3 T ) = 8T 8T Which gives T0 = 5 7R 5R 2. For a diatomic gas Cp = and Cv = 2 2 7 Q = n Cp T = nR T 2 5 U = n Cv T = nR T 2 7 5 nR T – nR T 2 2 = nR T Q: U: W=7:5:2 3. In the process , work is done on the gas. = –50 J. Since this process is W) Q) thermodynamics, the change in internal energy in this process is Q) W) U) W= Q– U=

Since the process is cyclic, there is no net change in internal energy. Hence U) = –50 J U) 4. In the process , V is proportional to T. Hence pressure P remains constant. Therefore, heat C p = 7 R/2 for a diatomic gas) 7R = n Cp T = n T 0 – T 0) Q) 2 7 n RT0 = 2 Process C is isothermal in which the gas is compressed. Hence work done on the gas in this process is V0 W ) C = – n R T0) ln 2V0

1 2

= – 2 n RT0 ln = 2 nRT0 Q

=

W

7 4 ln 2

5. The internal energy of n moles of an ideal gas at temperature T is given by U=

n RT 2 where = number of degrees of freedom. For hydrogen, = 5. Therefore 5 15 3 RT = RT U1 = 2 2 For helium, = 3. Therefore 3 3 n R T/2) = n RT U2 = 2 4 Given U1 = U2, i.e. 15 3 RT = nRT 2 4 which gives n = 10. 2 . Therefore =2 or 6. Given T = =

2 PV = n RT

nR T V

P=

Work done W =

3T

V T

= =

3T T 3T

nRT V nR 2

T

=

nR 2

2 T 2

3T T

= n RT n

7. Given V = P

. Since P = constant,

2

= T)

17.15

=

n – 1)

T

T2 =

P V=

Work done W =

n 1

T2 – T2

T

n = +c where c = constant of integration. Hence the

8. Since the volume of the gas is constant, P1 T = 1 P2 T2

13. Process is isochoric, i.e. the volume remains constant. Thus V = 0. Hence work done P V = 0. Process is isobaric, i.e. the pressure remains constant and external work has to be done. The 10–3 work done = P V – V ) = 8 104 – 2 10–3) = 240 J. Therefore, change in internal energy is

Now P2 = P1 + 0.02 P1 = 1.02 P1 and T2 = T1 + 5. P1 T1 = T1 5 1.02 P1

=

Heat energy required to raise the temperature n moles of a gas by T at constant volume is Qv C Qv = n Cv T, = v Qp Cp

n RT P= . Using V

V 2/3 =

T V –1/3 =

nR

or

= constant

Hence V T 3 10. Since the P – V graph is a straight line with a positive slope, P V or PV –1 = constant For a process in which PV n = constant, the molar heat capacity is given by R R C= 1 1 n Putting n = – 1 and = we have C=

11.

= 800 – 240 = 560 J

Qp = n Cp

Equation of state is PV = nRT

or

–

14. Heat energy required to raise the temperature of n moles of a gas by T at constant pressure is

T1 = 250 K

9. PV 2/3 =

n RT V

300 = 750 K 0.4

R 5 1 3

5 3 R 3R = 1 1 2

R = 2R 2

W T 300 =1– 2 =1– = 0.5. Therefore Q = 2W Q T1 600 =2

12. T2 = 300 K. Now = 1 – T2/T1. When 0.4, the value of T1 is given by T2 = 1 – 0.4 = 0.6 T1 T 300 = 500 K. When or T1 = 2 = 0.6 0.6 = 0.6, the value of T2 should be

= 40% =

= 60%

Qv = =

Cv Cp 3 5

Qp =

3R / 2 5R/2

207 = 124.2

Qp 124 J

15. Work done = area enclosed by the indicator diagram 1 = 2 1 = P – P) V – V) 2 = 3 PV 16.

1 . Now 4 the equation of state for n moles of an ideal gas is 1 RT RT = PV = n R T = 4 4 n) =

17. Expanding the equation of state we have PV c + aT2 V c–1 = RT + or or where given that

–c P = – aT 2 V –1 + RT V–c + –c –1 P= – = RT + and = aT2. We are

P=

m

–

n

17.16 Comprehensive Physics—JEE Advanced

Comparing the powers of V = – c and n

m

18. Now Qp = n Cp T and Qv = n Cv T Qv gives the heat energy which increases the internal energy of the gas. Thus the required fraction is Qv C 1 3 1 = v = = = Qp Cp 5 5/3 For monoatomic gas

=

5 3

19. Given Cp – Cv = 4150 and Cp/Cv = 1.4 or Cp = 1.4 Cv. Therefore, 1.4 Cv – Cv = 4150 or 20.

Cv =

4150 = 10375 J kg–1 K–1 0.4

m

=7h

23. Heat energy absorbed = work done = area of the loop r2 =

2

/4 =

4

U = n Cv T We know that Cp – Cv = R Cp

or

Cv

2

= 102

24. For a monoatomic gas, Cv = 3R/2 and for a diatomic gas, Cv = 5R/2. Since one mole of each gas is mixed together, the Cv of the mixture will 1 3R 5R Cv = = 2R 2 2 2 Now Cp – Cv = R. Therefore, for the mixture, Cp = R + Cv = R + 2R = 3R heats of the mixture is

Cp

R Cv

=1+ =1+

Cv

=

R , which gives Cv

R 1

PV = n RT volume changes by V, the change in temperature T is given by P V = nR T or

T=

P V PV = nR nR V = 2V – V = V

U=n

R

m

22. When work is converted into heat at a constant temperature, the entropy of the system remains con-

=

Cv

3R = 3 = 1.5 2R 2

=

Cv =

3 . Since the diameter of the 21. Volume bubble is doubled in rising from the bottom to the top of the lake, its volume becomes 8 times. Now PV = constant. Therefore, the pressure at the bottom of lake = 8 times that at the top. Let H be the depth of the lake. H h – h) m

H = 7h

Cp

25. Let T be the increase in temperature when the volume of the gas is changed by V at constant pressure. The change in internal energy of n moles of a gas is given by

Therefore,

P V P-V) curve, which is the largest for curve 3 and the smallest for

or

=

1

PV = nR

PV 1

26. Heat is given to the gas in cylinder at constant pressure while the same amount of heat is given to the gas in cylinder at constant volume. Heat given to gas in is Q = n Cp T Heat given to gas in Since nCp or

is Q = nCv T

Q = Q , we have T = nCv T Cp 7 T = 30 K = 42 K T = Cv 5 for a diatomic gas, Cp/Cv = 7/5)

27. The equation of state for an ideal gas of mass m and molecular mass M is m RT PV = M For an isothermal process, T = constant. DifferenT, we get

17.17

P V+V P=0

1 V 1 1 = or = V T T T Thus, the value of decreases as T is increased. or

V P= – P V mRT P= MV

or

m RT

P =–

MV

Hence

m P = m P

Given

P = 1.5

or

– 1)

30. For adiabatic process, T1 V 1

mRT P= – MV

V = 2V – V = V) and

P =–

P . Therefore,

T1 V2 = T2 V1 For a monoatomic gas, Hence

m RT MV

T1 = T2

m 1 = m 15

3m =2m .

28. The speed of sound in a gas of bulk modulus density is given by

and

L2 L1

5 3

= T2 V 2

– 1)

. Thus

1

V2/V1 = L2/L1. 1

=

L2 L1

2/3

31. Since the slope of the P–V graph for adiabatic expansion is times that for isothermal expansion, curves and represent isothermal and adiabatic expansions of V 2. the gas from initial volume V1

v= V P V Now, for a perfect gas, PV = nRT. Differentiating at constant T, we get V P P V + V P = 0 or = –P V is given by

Hence

=–

P

v=

If m is the mass of the gas and M its molecular mass, then m mRT RT or PM = = RT PV = M V or

P

=

RT M

v=

RT M

Hence v1 =

RT M1

v1 = v2

M2 M1

or

give

or v2 =

Fig. 17.24

between volumes V1 beand V2 is greater than the area under curve tween V1 and V2, it follows that W1 > W3. P–V graph for isobaric V2 ume V1 and pressure P1 sure remaining unchanged at P1 between volumes V1 and V2 is greater than the area under curves and . Hence W2 is greater than W1 and W3

RT M

and v2 =

RT which M2

29. For an ideal gas, PV = nRT. Since pressure P is kept constant, P V = nR T nRT V nR nRV V or = = = P V T P nRT T

32.

= Hence

+ . In an adiabatic process, = 0. = – = – 4.5 J. Hence the correct

33. Given T1 = 0ºC = 273 K, T2 = 400ºC = 673 K Work done W =

nR T2 1

T1

5 8.3 400 7 1 5

17.18 Comprehensive Physics—JEE Advanced

= 41500 J = 41.5 kJ

for a diatomic gas, atomic gas for which = 5/3. Therefore, the slope of the P-V curve is less for a diatomic gas than for a monoatomic gas. Hence curve 1 corresponds to diatomic gas and curve 2 to monoatomic gas. Thus

to be negative, i.e. W law of thermodynamics = + adiabatic process, = 0. Hence

. For an = – = implies that the internal energy increases. Hence the correct

38. Process of Q. 38)

34. For an adiabatic process TPn = 1

where n =

W Cp

,

Cv

Therefore,

and

W =W

gives

3 2

=

39. For an ideal gas, pV = nRT. Differentiating, we have since T is kept constant

1 = 3 or n

1

= 3, which

p

P)O2 =

1 mole RT V

P)He =

1 mole R 2T V

He

P

O2

1 p

–

p1

Given = 0 and < 0. Hence the change in internal energy < 0. Now, for an ideal gas, the internal energy can decrease only by decrease in

Since at any instant PV = constant,

P V

or

the slope of P-V curve is proportional to . Now,

p2 2

p2 = p1

2

7/5

7

or

Cv =

= 128

1

U = Cv T. Now Cp – Cv = R or

42. , i.e.

=

1

37. For an adiabatic process, PV = constant, Differentiating, we have V = 0 or

1

41. pV = constant. For a given mass of gas, V Hence p = constant

P)O2 =

+

1 V

=

of versus p. 40. Since is an isothermal process, the temperature of the gas remains constant between . Hence the P-T diagram must be perpendicular to the T-axis between and . Hence the correct choice

36.

–1

V p

=2

P)He

PV

+ V = 0 or

Hence

35. For a gas, PV = nRT. Hence

P

+W

or W

T –1/n

T 3. Hence

+W

or 5 = 10 + 0 + W

T

Given P

= P V2 – V1) = 10

Process C, occurs at constant volume. Hence W = 0. Given Q = 5 J, i.e. total work done is W = 5 J. Therefore, we have

is a constant.

Since n = constant for a given gas, P

=

1/ n

P=

occurs at constant pressure.

R 1

, where

=

Cp Cv

Cp Cv

–1=

. Hence

R Cv

.

17.19

U=

43. PV =

R T 8.3 8 = = 166 J 1 1 4 1)

m RT. Therefore, the density of the gas is M m PM PmN mP = V RT RT

to V1, P3 are as shown in Fig. 17.25. Let W1 and W2 be the work done in isothermal expansion and adiabatic compression respectively. Therefore, net work done is W 2) = W 1 – W 2 W = W1

44. Heat energy supplied = Cp . Change in internal energy = Cv . Therefore, work done = – Cp – Cv ) . =

Cp

Cv p

=1–

T

=1–

1

2 1 = 5 5/ 3 = 5/3 for a monoatomic gas)

45. Let L and r its radius. Since the mass of the gas remains unchanged and the pressures of the gas in both sides are equal, we have, from Charles’ law, V1 V = 2 T1 T2 L L r 2, V 2 = r 2, 5 5 2 2 T1 = 0°C = 273 K and T2 = 100°C = 373 K. Using

Given V1 =

L L 5 5 2 = 2 273 373 which gives L = 64.6 cm. 46. For an isothermal process: PV = constant and for an adiabatic process: PV = constant, where is the Cp /Cv) of the gas. When a gas is compressed from a volume V to a V – V), the increase in pressure is P)adia =

VP for an adiabatic compression V

VP for an isothermal compression. V Hence P3 will be greater than P1. Therefore, the P–V diagrams of isothermal expansion from V1, P1 to V2, P2 and adiabatic compression for V2, P2

Fig. 17.25

Now, the area under the adiabatic curve is more than that under the isothermal curve. Hence W2 > W1. Therefore, W < 0. Hence the correct . 47. Heat produced is given by 2

10

2

10 60 = 1200 J R 50 Since the container is rigid, the change in volume = 0. Hence work done = = 0. From the =

energy is

=

–

=

= 1200 J. Hence the

48. The two adiabatic paths and for the gas intersect the two isothermals and at temperatures a and lie T1 and T2 on the same adiabatic path, we have T1Va – 1) =T2V – 1) 1

Va V

or Since points path,

=

T2 T1

and c also lie on the same adiabatic T1 V )

– 1)

= T2 Vc

1

V Vc

or

P)iso =

Va V

= 1

=

T2 T1 V Vc

1

– 1)

17.20 Comprehensive Physics—JEE Advanced

= P V = nR T = nR T – T ) = 3 8.3

W

Va V = V Vc

or

49. Since the temperature T remains constant along the path , P will be inversely proportional to V along this path. Hence, as P increases, V must decrease in a nonlinear fashion. This is represented by the curve in Fig. 17.26. , the volume V is constant. Hence the graph of P against V is a straight line perpendicular to the V-axis. On a P – V diagram, the corresponding path is as shown in Fig. 17.26. For the path , V is directly proportional to T pressure remaining constant. The corresponding path is, therefore a straight line parallel to the V-axis. Thus the cyclic process on a P–V diagram

PV = nRT)

Process work done in this process W = 0. Since the whole process is cyclic, the change in internal energy in the complete cycle is zero, i.e. U = 0. Q = – 2510 J) Q= U+W=

U+W +W W +0

+W

or W The negative sign shows that the work is done 52.

oxygen is greater than 50 K. Therefore, between 50 K and 300 K, liquid oxygen undergoes a change of

53. The heat energy supplied is Q = I2

2

100 60) 3 10 J = 240 kJ

= 240

Since V W of thermodynamics, U = Q = 240 kJ. Thus the

Fig. 17.26

50. Work done on the gas is W = P V = P V – Vi) = 1 105 = 2000 J

10 –3

The internal energy is given by U =

PV 1

PV PVi ,U = . 1 1 Therefore, change in internal energy is P U = U – Ui = V – Vi) 1 Ui =

=

1 105

25 5 1.4 1

10

3

51. In process , the volume V increases linearly with temperature T. Hence process is isobaric

P0 and c = 3 P0. V0

These equations give m = – Now PV = nRT we have

P= T=

= 5000 J

energy supplied to the gas is Q = W + U = 2000 + 5000 = 7000 J

process is

54. The equation of straight line is P = mV + c where m is the slope and c is the intercept. For points and , we have 2 P0 = mV0 + c V 0) + c and P0 = m

nRT V

1 mV2 + cV nR

T will be maximum if

= 0 and

2 2

< 0.

V and putting = 0, we get

2 mV + c = 0 c which gives V=– 2m c2 1 Tmax = – =– 4nRm 4nR

3P0 2 = P0V0 P0 /V0 4nR

17.21

55. Mass of 2 litres of water = 2 kg. Heat energy needed to raise the temperature of 2 kg of water from 20°C to 75°C is Q=2 103) 55 = 4.62 105 J

P=

Differentiating we have

If is the time taken, heat energy supplied by the heater in time is Q1 time) = 1000

3nRT 2 T =

Heat energy lost in time is Q2 = 160 Heat energy available for heating water is Q = Q1 – Q2 = 840 J Equating Q = Q , we get 550 s. Thus the 56. Given PT 2 =

nRT . Hence V nRT 3 = V T

V

V V T

=

3nRT 2

3nRT 2

nRT and PT 2 = P 3 = . T

Using V = we get

PV = nRT, we have

II Multiple Choice Questions with One or More Choices Correct 1. Figure 17.27 is the P-V diagram for a Carnot cycle. In this diagram, represents isothermal process and adiabatic process Fig. 17.27 represents adiabatic process and isothermal process represents isothermal process and adiabatic process represents adiabatic process and isothermal process. 2. Figure 17.28 shows the P-V diagram of a cyclic process. If is the heat energy supplied to the system, is the internal energy of the system and is the work done by the system, then which of the following relations is/are correct = =0 Fig. 17.28

= =– 3. If Q represents the heat energy supplied to a system, U the increase in internal energy and W the work done by the system, then which of the following are correct? Q = W for an isothermal process U=– U=

W for an adiabatic process Q for an isochoric process

U = – Q for an isobaric process. 4. The initial state of n moles of an ideal gas is repP 2, V 2, resented by P1, V1, T1 T2. Wi is the work done by the gas in an isothermal T1 = T2 = T) and Wa in an adiabatic process, then V2 Wi = nRT loge V1 Wi = nRT loge Wa = Wa =

1 1 nR 1

P1 P2 P 1V 1 – P 2V 2) T 1 – T 2)

17.22 Comprehensive Physics—JEE Advanced

5.

P, volume V and temperature T is allowed to expand adiabatically until its volume becomes 5.66 V while its temperature falls to T/2. If is the number of degrees of freedom of gas molecules and W is the work done by the gas during the expansion, then =5 W=

6.

5PV 4

W=

3PV 2

namic process involving four steps. The amounts of heat involved in these steps are Q1 Q4 = 3645 J respecQ2 = – 5585 J, Q3 tively. The corresponding amounts of work done are W1 = 2200 J, W2 = – 825 J and W3 = – 1100 J and W4 . Then

Fig. 17.29

is 0.036 J. C is – 0.024 J. C is zero. is 0.06 J. 11. Figure 17.30 shows the P–V diagram for an ideal gas. From the graph, we conclude that

W4 = 275 J

W4

16% 7.

P, volume V and temperature T. The ratio CP/Cv = and U is the internal energy. If R is the gas constant, then R Cv = U = nCv T 1 U=

PV 1

zero.

U = nCpT

8. Fig. 17.30

of the source should be increased by 28.6°C. of the sink should be decreased by 28.6 °C.

9.

12. Figure 17.31 shows the P – V diagram for an ideal gas. From the graph we conclude that

800 J per cycle, the work output per cycle is 200 J. minutes. The room temperature is 27°C. The latent heat of fusion of ice is 80 cal/g. 105 J. to 10. Fig. 17.31

104 J. 10.

process

is adiabatic. P–V diagram of a cyclic

process.

17.23

to

, the temperature T of

decreases. in the process. 13. n moles of an ideal monoatomic gas is kept in a closed vessel of volume 0.0083 m3 at a temperature of 300 K and a pressure of 1.6 106 Pa. Heat 104 J is supplied to the gas. Given –1 –1 R = 8.3 J mol K . n=5 Cp = 20.75 J mol–1 K–1 106 Pa

R

v=

14. For an ideal gas

sure when the temperature of n moles of the gas changes by T is n Cv T.

adiabatic process is equal in magnitude to the work done by the gas. isothermal process. process. 15.

16. One mole of oxygen at 27 °C is enclosed in a vessel which is thermally insulated. The vessel is moved with a constant speed v and is then suddenly stopped. The process results in a rise of temperature of the gas by 1 °C. Then, if M = molecular mass of oxygen. 5 7 = Cp/Cv) = 3 5

P, volume V) to state P/2, Volume 2V) along a straight line in the P-V diagram as shown in Fig. 17.32. Then

M

1

v=

2R M

1

17. Cv and Cp of a gas at constant volume and constant pressure, respectively. Then Cp – Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas Cp + Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas Cp/Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas Cp Cv is larger for a diatomic ideal gas than for a monoatomic ideal gas 18. Figure 17.33 shows the P-V plot of an ideal gas taken through a cycle . The part is a semi-circle and is half of an ellipse. Then,

Fig. 17.33 Fig. 17.32

to exceeds the work that would be done by it if the system were taken from to along the isotherm. T-V diagram, the path becomes a part of a parabola. P-T diagram, the path becomes a part of a hyperbola.

is isother-

mal C zero

C is

17.24 Comprehensive Physics—JEE Advanced

ANSWERS AND SOLUTIONS 1. For adiabatic process, PV = constant. Differentiating w.r.t V we get V +P V

–1

or

=0 =

P V

For isothermal process, PV = constant. Hence P = V Now, / P-V) graph. Thus, P-V) graph for an adiabatic process is times that for an isothermal process. Hence curves and both represent adiabatic process and curves and both represent isothermal process. Thus the correct choices are 2. In a cyclic process, the system returns to its initial state. Hence the change in internal energy = 0. of thermodynamics, = + 3.

=

= 0)

W = P V and U = nCv T and = Cp/Cv Q = U + W. For an isothermal process, T = 0. Therefore U = 0. Hence Q = W adiabatic process Q = 0. So U = – W, which is W = 0, so Q = U Q= U+ W

4. 5. For an adiabatic change the relation between T and V is Cp = TV – 1) Cv 1 V T = V T T Given V = 5.66 V and T = . Therefore, 2 – 1) =2 Taking logarithm of both sides, we have

TV

or

– 1)

– 1)

=T V

=1+

l l

or

2) 0.3010 =1+ 5.66) 0.7528 = 1 + 0.4 = 1.4

Since = 1.4, the gas is diatomic. For a diatomic gas, the number of degrees of freedom of the molecules = 5. We know that the work done by the gas during adiabatic expansion is given by 1 W= PV – P V 1 where pressure P after expansion is obtained from the relation PV PV = T T V T or P =P V T T /2 P = T 11.32 P Putting = 1.4, V = 5.66 V and P = in 11.32 We have W= =

1 1 4 1) 1 0.4

PV

=P

V 5.66V

PV

P 11.32

1 PV 2

5.66V

= 1.25 PV

6. Since W2 and W3 are negative, it means that the work is done on the gas. Hence Q2 and Q3 are negative which implies that heat is evolved in processes 2 and 3. Since Q1 and Q4 are positive, heat is Q2 + Q3), the gas absorbs Q1 + Q4 a net amount of heat energy in a complete cycle, which is given by Q = Q1 + Q2 + Q3 + Q4

The net work done by the gas is W = W1 + W2 + W3 + W4 = 2200 – 825 – 1100 + W4 W4 Since the process is cyclic, the change in internal energy U namics, we have W= Q – U = Q or 275 + W4 = 1040 or W4 = 1040 – 275 = 765 J

17.25

Q = 800 net work done by the gas h= total heat absorbed by the gas

=

80 = 80,000 J = 80,000

275 765 3645

W = = Q1 Q4 1040

9.

=

= 0.1083 = 10.83%

7. The internal energy of n moles of an ideal gas at absolute temperature T is given by U = nCvT where Cv ume. We know that Cp Cv

–1=

R or – 1 = or Cv = Cv

R Cv

8.

PV RT

R 1

=

1

T=

T2 T1

=1–

0.25 = 1 –

300 which gives T1

= 0.25 + 0.05 = 0.30. The new temperature of the source should beT 1 so that T 300 0.30 = 1 – 2 = 1 – T1 T1

1 2

1 2

+

10–3 m3) 4 Nm–2 + 4 Nm–2 6 10–3m3 = 0.012 + 0.024 = 0.036 J

11. For point , PV = 5 10–3) = 10 10–3 J For point =2 10–3) = 10 10–3 J Since PV = constant, the process is isothermal. For an isothermal process, U = 0. Since the gas undergoes expansion W is positive and Q = W. 12. The P–V graph for an adiabatic process is not a P V = n R T and P V = n R T . Therefore PV T T 6 1 3 = = , PV T T 2 4 4 i.e. T > T . Hence the internal energy increases. Work done = area under upto the volume axis. Heat energy is absorbed in the process. Hence the

which gives T = 428.6 K. So, increase in temperature of the source = T1 – T1 = 428.6 – 400

0.30 = 1 –

104 J, which

W = – 0.024 J. The negative C sign shows that the work is done on the gas.

PV 1

T1

such that

3.36

WC = P V = 0 because V = 0. Work done in complete cycle = 0.036 – 0.024 + 0 = 0.012 J = 0.25.

T2

273 300 273

=

10. W

R

Q2 = mL = 103 4.2 = 3.36 105 J. So

Q1 = Q2 + W = 3.36 105 + 3.36 104 37 104 J Q 37 104 J Power = 1 = 20 kW. So choice 180 s

Now, the ideal gas equation for n nodes is PV PV = nRT or n = RT U=

=

T1 T2 is correct. Q 3.36 105 W= 2 = 10

11%

Cp – Cv = R or

T2

0.25 = 200 J. So

T2 of the sink should be T2 T =1– 2 T1 400

which gives T2 = 280 K = 7°C. Decrease in temperature of the sink is 27°C – 7°C = 20°C,

13.

PV = nRT =

n=

PV RT

1 6 106 ) 0.0083 16 = 8.3 300 3

17.26 Comprehensive Physics—JEE Advanced

Cp = = 20.75 J mol–1 K–1 Q = n Cv

5R 5 = 2 2

P0 and V0 be the intercepts on the P and V axes. The equation of straight line is P0 P= V – V 0) V0 P V P0 V0

8.3

3R 2

T Cv =

Q 2 Q 2 2. 104 = = 16 nCv 3nR 8.3 3 3 = 375 K Final temperature = 300 + 375 = 675 K = 402°C. T=

P2 P = 1 T2 T1 = 3.6 14.

15.

P2 =

P1

6

T2 T1

=

Since P =

P V P0 V 2 RT V =1 T= 0 R RV0 V P0 V0 which represents a parabola on the T-V graph.

1.6 106 675 300

V=

10 Pa

U = n Cv T Q = U + W. For an adiabatic process, Q = 0. Hence 0 = U + W or | U| = | W|. T = 0, Hence U = 0. Q = 0. to W1 = area of trapezium P 3 PV = P V= 2 2 =

3 RT 2

RT V

P P0

RT P

RT =1 PV0

which does not represent a hyperbola. So is correct. 16.

Therefore, Cv = 5 R/2 and Cp = 7 R/2. So = Cp/Cv = 7/5. The kinetic energy of oxygen mol1 M v2, where M = ecules with a velocity v0 = 2 molecular weight of oxygen. = Cv 1 = Cv Now heat energy = Cv Cp R Cp – Cv = R or –1= Cv Cv – 1) =

R or Cv = Cv

Therefore,

1 M v2 = 2

or

v=

If the process to work done would be W2 = RT loge Thus W1 > W2

V2 V1

were isothermal, the = RT loge

RT

R 1

R 1 2R

M

1

17. Cp + Cv) for diatomic gas = Fig. 17.34

V0 P 2 R P0

T = V0 P –

7R 2

Cp + Cv) for monoatomic gas = Cp Cv for diatomic gas =

5R =6R 2 5R 2

35 R 2 4

Cp Cv for monoatomic gas =

15 R 2 4

3R =4R 2

17.27

Cp

= =1+

2

U=

is smaller for diatomic gas than

Cv for monoatomic gas. Cp – Cv = R. 18.

PV

V 1

is negative because V < V .

From Q = U + W

Q is negative, C .

C the cyclic process is clockwise. Hence the correct

process increases with V rect. For the process C , volume is decreasW

III Multiple Choice Questions Based on Passage Questions 1 to 4 are based on the following passage Passage I

2. Which of the graphs shown in Fig. 17.36 represents the P – T diagram for the complete process?

7R at pressure P 2 and temperature T are isothermally expanded to twice the original volume. The gas is then compressed at constant pressure to its original volume. Finally the gas is heated at constant volume to its original pressure P . Three moles of an ideal gas

Cp

1. Which of the graphs shown in Fig. 17.35 represents the P – V diagram for the complete process?

Fig. 17.36

3. The net work done

W by the gas during the R is the gas constant) RT RT RT RT 4. The net heat energy supplied to the gas in the complete process is W W W

Fig. 17.35

SOLUTION 1 . During V C), P = constant and during ), V = constant. Hence the

1. During isothermal process

correct P – V

C

,P

2. During isothermal pro

), T = constant. C), P = constant and C ), P T. Hence the correct P – T diagram for the complete process

17.28 Comprehensive Physics—JEE Advanced

3. For process , P V = P V which gives P P = because V = 2V 2 V V = C . Since V = 2 V and For process C, T TC T PC = P VC = V and T = T , we get TC = 2 P . = 2 P P P . Since PC = For process C , C = T TC 2 T = 2TC. Work done is isothermal process W

= nRT loge

V V

is

= 3 RT loge

Questions 5 to 9 are based on the following passage Passage II Two moles of an ideal gas at volume V, pressure 2 P and temperature T undergo a cyclic process as shown in Fig. 17.37.

Work done in isobaric process C is P W = P VC – V ) = V – 2V ) 2 PV 3 =– =– RT . 2 2 Work done is isochoric process C is W Total work done 3 RT 2

W = 3 RT loge = 3 RT

RT

Thus th 4. Since the process is cyclic, U = 0. From Q = U + W, we get Q = W VC) of the gas in state C is

6. 8V 3 4V 3 7.

V 2V 3 V ) of the gas in state

V V

V W) in the complete cycle is RT loge

RT 4 RT 3

Fig. 17.37

9. V 3 V

V ) of the gas in state is 2V 3 4V 3

is

V

8.

5.

2 RT loge 3 Q) in the heat energy in the com-

plete process is

W W

W.

SOLUTION 5. For isobaric process V T

=

V T

V =

V T T

6. For isothermal process P V = PC VC

7. For isobaric process C

, =

V

4T / 3 4V = T 3

C,

PV 2 P 4V / 3 8V VC = = = , PC P 3

=0

VC V = T TC 8. W W

V =

V T TC

=P V –V )=2P = nRT loge

VC V

, =

8V / 3 T = 2V 4T / 3 4V 3

V

=

2 PV 3

17.29

=2

R

4T log e 3

W

= P C V – V C) = P

W

= nRT loge =2

R

Total work done is

8V / 3 8 = RT loge 4V / 3 3 8V 3

2V

=–

W=W

2 PV 3

+W

+W

2 PV 8 + RT loge 3 3

=

2 PV – 2 RT loge 3

2 RT loge 3 9. For a cyclic process, U = 0. Hence Q = W. So

V V

T loge

+W

=

V 2V

1 = – 2 RT loge 2

= 2 RT loge

Questions 10 to 12 are based on the following passage Passage III V 5 consists of one gram mole of a gas with Cp/Cv) = 3 7 at a certain temperature T. and another gas with = 5 The gram molecular weights of gases and are 4 and 32 respectively. The gases do not react with each other and are assumed to be ideal. The gaseous mixture follows the = constant in an adiabatic process. relation PV – 10. The number of moles of gas mixture is

11. The adiabatic compressibility of the mixture at pressure P is 3 5 5P 7P 13 7 P P 12. If the temperature T of the mixture is raised from 300 K to 301 K, the percentage change in the speed of sound in the gaseous mixture is 1 3

in the gaseous

1 % 6

1 2

SOLUTION 10. Let n and n be the number of moles of gases Cv) be their and C v) Since the gases do not react with each other and are assumed to be ideal, the internal molecular kinetic energy before and after the gases are mixed must be the same. Hence n +n Cv )m T n C v ) T + n C v) T Cv ) m ture at constant volume. Now R Cv ) = Cv ) = 1 R

C v) m = Therefore, n + 1

1

m

n 1

=

n m

n 1

5 7 , = and m is given by 3 5 the adiabatic relation for the mixture,

Given n = 1,

PV Given PV

5 1 3

R

m

= constant

= constant. Hence 1

-

1

=

+

m

=

13

. Using

n 1 n = 7 1 1 5 13

5n 13 1 n 3 + = 2 6 2 n = 13 + 13 n or 2 n = 4 or n = 2 Thus the mixture contains 2 moles of gas . or

17.30 Comprehensive Physics—JEE Advanced

of the mixture is

11. 1 . V We are given that PV where

m

=

13

m

= constant

. Partially differentiating, we have

PV

m

+

or

–

1 V

V m

V = P

V=0

m 1

PV

m

V =– P

or

1 m

V P

=–

=

m

PV

m 1

=–

12. We know that the speed of sound in a gas is proportional to the square root of absolute temperature, i.e. v = 1/2 = constant 1 Taking logarithm, log v = log + log T 2 v 1 T =0+ Differentiating, v 2 T 1 1K 1 = 2 300 K 600 Percentage change v 100 1 100 = = % v 600 6

V m

=

P

1 m

P

Questions 13 to 15 are based on the following passage Passage IV Two moles of a monoatomic ideal gas occupy a volume V at 27°C. The gas is expanded adiabatically to volume 2 2 V. Gas constant R = 8.3 JK–1 mol–1.

P

13 , P

=

2 14. The change in the internal energy of the gas in this process is

15. The work done by the gas during the process is

13. 150 2

SOLUTION 13. T1 = 300 K, V1 = V, V2 = 2 2 V. Let T2 obtained from the adiabatic relation T1 V1 or

– 1)

V1 V2

T2 = T1

For a monoatomic gas T2 = 300

= T2 V2

1 2 2

2/3

=

14. T2 is

– 1) 1

5 . Therefore, 3

U = nCv T2 – T1) For a monoatomic gas Cv =

3R . Therefore, 2

3 8.3 2 The negative sign implies that the internal energy U=2

15. For a adiabatic process, Q = 0. Hence W = – U

= 150 K

Questions 16 to 21 are based on the following passage Passage V is taken through the process and another sample of 2 kg of the same gas is taken through the process as

shown in Fig. 17.38. The molecular mass = 4 and R = 8.3 J K–1 mol–1. 16. The temperature of state

is

17.31

18. The temperature of state C is 19. The temperature of state

20. The work done in process 104 103 21. The heat supplied in process to 106 106

Fig. 17.38

17. The temperature state

is

is

is

104 J 103 J is nearly equal 106 J 106 J

SOLUTION 16. Number of moles of helium is 2000 mass in gram n= = = 500 4 molecular mass From equation state at , PV P V =nRT T = nR =

4.15 104 10 = 100 K 500 8.3

17. For isochoric process T T

=

P P

=

8.3 10

4

4.15 104

= 2.

19. For isobaric process T P 20 = = =2 T P 10 which gives T = 2 T = 200 K. So the correct 20. Work done in process W= W + W = 8.3

104

103 J

21. From U = n Cv T, heat energy in process for a monoatomic gas Cv = 3 R/2) Q

Thus T = 2 T 18. For isobaric process

is = 0 + P VC – V )

U)

W)

= n Cv TC – T ) + P V – V ) C,

= 500

TC V 20 = C = =2 T V 10 Thus TC = 2 T Questions 22 to 25 are based on the following passage Passage VI One mole of an ideal monoatomic gas is taken round the cyclic process

= 1.87

400 100 104

106

22. The work done by the gas is P 0V 0 P 0V 0

23.

P 0V 0 P0V0

is P0V0 3P0V0 2 2 5P0V0 P 0V 0 2 24. The heat energy absorbed by the gas in the process is 3P0V0 P0 V0 2 C

Fig. 17.39

3 83 2 + 4.15

17.32 Comprehensive Physics—JEE Advanced

5P0V0 2 25.

P0V0 2 5P0V0 2

P0 V0

in the process

C is

P0 V0 P0 V0

SOLUTION 22. Work done by the gas in the cyclic process W = area enclosed in the P–V diagram = area of triangle 1 = P 0 – P 0) 2

Using the ideal gas equation for points have P V = nRT or P0 V0 = nRT

is

1 2 V0 – V0) = P0V0, =

23. In the isobaric process by the gas is given by Q = nCp T = nCp T – TC Using ideal gas equation PV = nRT for points C, we have P V = nRT or P0 V0 = nRT and PC VC = nRTC or P0 V0) = nRTC

and

P V = nRT

and

T – TC = –

Q

=–

= 3 P0 V0

Q 25.

3R . Therefore 2 -

ergy absorbed by the gas is given by

where W = W = work done by the gas = P0V0. Since the process is cyclic, there is no change in the internal energy of the gas, i.e. U = 0. Hence Q = W = P 0V 0

C p P0 V0

If Q

R

is the heat absorbed in the process Q =Q

5R . Hence 2

5 P0 V0 2 24. The heat energy absorbed in the isochoric process is given by Q = nCv T = nCv T – T Q

= 2 Cv P0 V0 R

Q= U+ W

P0 V0 nR

Now, for a monoatomic gas Cp =

2 P0 V0 nR

Now, for a monoatomic gas, Cv =

P0V0 = nR TC – T ) or

P0)V0 = nRT

which give T – T =

Q

and , we

=–

Questions 26 to 30 are based on the following passage Passage VII Two moles of an ideal monoatomic gas, initially at pressure P1 = P and volume V1 = 2 2 V, undergo an adiabatic compression until its volume is V2 = V and the pressure is P2. Then the gas is given heat energy Q at constant volume V2.

+Q

+Q

P0V0 = 3P0V0 + Q or

Q

=

, then

–

5P0V0 2

P0V0 2

26. Which of the graphs shown in Fig. 17.40 represents the P–V diagram of the complete process? 27. Pressure P2 is 2P

2P

2 2P 28. The total work done by the gas is PV 2 PV

2 PV PV

17.33

29. The change in internal energy due to adiabatic compression is 2 PV PV 2 PV PV 30. The temperature T2 of the gas after it is adiabatiR is the gas constant). PV 2PV R R 2PV R

2 2PV R

Fig. 17.40

SOLUTION 26. 27. For an adiabatic change, P1V1 = P2 V2 . Therefore, P2 = P1 For a monoatomic gas, P2 = P

2 2V V

V1 V2 = 5/3. Hence

5/3

= P 2 2

5/3

. So the correct

28. In an adiabatic process, heat Q = 0. Hence from the W1 = – U1 = – n Cv T = – 2 Cv T2 – T1

n = 2)

Now for 2 moles P1 V1 = 2 RT1 and P2V2 = 2 R T2. Thus PV PV T1 = 1 1 and T2 = 2 2 2R 2R Using these, we have C W = – v P 2 V 2 – P 1 V 1) R C PV 1 1 =– v V2 PV 1 1 R V2 C =– v R

P 1V 1

V1 V2

1

1

=–

3 P1 V1 2

=–

3P 2 2V 2

V1 V2

2/3

1

2 2

2/3

1

=–3 2 The work done at contant volume = 0. Hence the 29. In adiabatic compression, Q = 0. Hence U = – W, 30. Since U = nCv T = 2 Cv T2 – T1) U we have T2 = T1 + 2Cv Using T2 = =

T1 =

P1 V1 3R and Cv = , we get 2R 2

P1 V1 3 P1 V1 + 2 2 3R / 2 2R P1 V1 2R =

V1 V2

V1 V2

2/3

P 2 2V 2 2PV = R R

2/3

1

17.34 Comprehensive Physics—JEE Advanced

Questions 31 to 35 are based on the following passage Passage VIII Two moles of an ideal mono-atomic gas is taken through a cycle as shown in the P–T During the process , pressure and temperature of the gas vary such that PT = , where is a constant. IIT, 2000

Fig. 17.41

31. Constant is given by PT 1 1 P 1T 1 2 P 1T 1 2 P 1T 1 32. The work done in process R is the gas constant) RT1 RT1 RT1 RT1 33. The heat energy released in process is RT1 RT1 RT1 RT1 34. The heat energy absorbed in process C is RT1 RT1 RT1 RT1 35. The heat energy absorbed in process C is RT1 RT1 e e RT1 RT1 e e

SOLUTION 31. Using the ideal gas equation PV = nRT, the volumes of the gas in states , and C are n RT n R 2T1 2 nRT1 V = = = P P1 P1 V =

n RT P

n R T1 2 P1

=

=

PV = nR

, we have

P=

V

1/ 2

nR V

V

=

nR

V

V

= 2 nR

W

V

= 2 2 nRT1

V

1 nRT1 2 P1

= 2 nR 2 PT 1 1 1 2

2nRT1 P1 = – 2 nRT1

2

= – 4 RT1. The negative sign indicates that heat is absorbed in 33. For a monoatomic gas Cv = 3R/2. The internal energy of the gas in the process is U) = n Cv T = n Cv T – T ) =2

or P V =

Q is given by

3R 2

U)

T 1 – 2 T 1) T1) = –3RT1

= 3R

1/ 2

nR V The work done in process or

=

2

P

V

= V

1 n RT1 2 P1

nRTC n R 2T1 nRT1 and VC = = = PC P1 2 P1 It is given that in the process , the pressure and temperature of the gas vary such that PT = where is a constant. Thus for point , we have = P T = P1 T 1) = 2 P 1T 1 32. For the process PV = nRT and PT = Eliminating T

V

W

+W

= – 3 RT1 – 4 RT1 = – 7 RT1 which

17.35

34. The process C takes place at a constant pressure P = 2 P1. Therefore, the work done in this process is W =P V P1 VC – V ) nRT1 P1

= 2P1

W

35. The process C takes place at a constant temperature T = 2 T1. Therefore, the work done in this P = nRT/V) V

W

V

= VC

1 nRT1 2 P1

VC

= nRT1 = 2 RT1

= nCv T1 = 2 Q

+W

U)

3R 2

V = 2. Therefore, VC

C is W

=2

2T1

R

= 4RT1loge

T1 = 3RT1

Questions 36 to 38 are based on the following passage Passage IX

log e T = 2 T 1)

= 0 as the temperature is constant.

U) Therefore Q

= 3 RT1 + 2 RT1 = 5 RT1

cyclic process starting from

V VC

= nRT log e

The change in the internal energy is process U) = nCv T = nCv TC – T ) T 1 – T 1) = nCv

nRT V

=

+W

U)

V V = 2 and V V R is the gas constant.

Given

= 4. The temperature T = 27°C.

as shown in Fig. 17.42.

IIT, 2001 36. The temperature of the gas at point

is

37. The total heat absorbed in the complete cycle is 38. The total work done by the gas in the complete cycle is

Fig. 17.42

SOLUTION 36. For process Hence V T

=

37. Process Hence Process

W

V T

V T = V

Q

U = 0. Hence Q = W.

, the plot of V versus T is linear.

T =2

300 K = 600 K

occurs at constant pressure =2

C

V

T.

5R T –T 2 C occurs at constant temperature. Q = U + W.

= nRT loge

VC V

= nRT loge

VD V

= nRT loge

VD V

=2 Q

C

R

V D = V C) V V

600 loge 4

= 1200 loge = 1200 loge

R R

1 2

17.36 Comprehensive Physics—JEE Advanced

Process C occurs at constant volume. Hence QC D = nCv T – T ) 3R =2 2 R Process Hence

= nRT loge

300

R

loge

= – 1200 loge

occurs at constant temperature. = WD

QD

=2

V VD

Questions 39 to 41 are based on the following paragraph Passage X 5 3 is trapped inside a liquid of density l

1 4

R

Total heat absorbed in the complete cycle is Q = 1500R + 1200 loge R R – 1200 loge R = 600R 38. In a cyclic process U = 0. Hence W = Q = 600R

40. When the gas bubble is at height its temperature is T0

with the liquid. The bubble contains n moles of gas. The temperature of the gas when the bubble is at the bottom is T0, the height of the liquid is H and the atmospheric pressure is P0 IIT, 2008

h

0

from the bottom,

2/ 5

0 2/ 5

0

T0

H

0

H

0

T0

3/ 5

0 3/ 5

0

T0

H

0

41. The buoyancy force acting on the gas bubble is R is the universal gas constant) Fig. 17.43

39.

ancy force the following forces are acting on it

l nR

-

0

the pressure of the liquid

0

pressure of the liquid and force due to viscosity of the liquid

7/5

0

0

2 /5

H

0

SOLUTION

41.

39. 1

40. For an adiabatic process TP For T0

= 5/3, 0

1 H

Which gives T

=– 2/5

T0

= constant.

2 . Therefore , 5

=T

2/5

0

8/ 5

0

0

H

3 /5

0

2/5

P0

F = weight of liquid displaced , V = volume of the bubble. nRT From PV = we have V = Therefore, P nR F= P =

nR

=

T0

0

2/5

0

3 /5

H

0

H

0

3/ 5

P0

l

to viscosity of the liquid

2/5

H

0

2/5

0

h

0

nR

= 0

H

2/5

P0

)]3 / 5

17.37

IV Matching 1.

P – V) graph of a cyclic process of an ideal gas is shown in Fig. 17.44. Match the process listed in column I with the consequences listed in Column II Column I

Column II

P

>0 <0 >0 <0 IIT, 2006

C C

SOLUTION

A D

B

C V

Fig. 17.44

In process , the volume remains constant. Hence = P T. Since the pressure decreases, temperature T decreases. Hence the gas loses heat energy, i.e., < 0. In process C, the pressure remains constant and volume increases. Now V T. Hence T increases which means that = is positive. In process C , the volume remains constant but pressure increases. Hence > 0 and = 0. In process , volume decreases and pressure increases. Hence < 0 and < 0. 2. Column I contains a list of processes involving expansion of an ideal gas. Match this with Column II describing the thermodynamic change during this process. Column I Column II a valve. Chamber I contains an ideal gas and the

Fig. 17.45

its original volume such that its pressure P where V is the volume of the gas. its original volume such that its pressure P where V is its volume

1 V2

.

1 V 4/3

or remains constant

.

pressure P and volume V follow the behaviour shown in the graph in Fig. 17.46.

Fig. 17.46

IIT, 2008

17.38 Comprehensive Physics—JEE Advanced

SOLUTION = 0. Since the system is completely insulated, no heat can = – = 0, i.e.there is no

enter or leave the sysytem, i.e.

nRT Given PV2 = constant. PV = n RT or P = n RT/V. Therefore, PV2 = V is increased, T will decrease, i.e. is negative. nR. . Hence For PV , the work done = 1 +

= nCv = n Cv

nCp

Cp = Cv + For monoatomic gas Cv =

+

V 2 = nRTV. Hence T

1 . Thus if V V

nR 1 nR 1

R 1

3R and for = 2 2 3R R 3R R + = –R= 2 1 2 2 2 is negative and Cp is positive, is negative, i.e. the gas loses heat. Hence the cor-

Cp = Now

= nCp

. Since

PV4/3 = constant, Then T Putting Cp = Cv +

1 V

1/ 3

. Thus if V is increased, T will decrease, i.e.

R 3R 3R = – 3R = – , which is negative. Now 1 4/3 2 2

ture of the gas increases. Therefore, = + . Since both and are positive,

= nCp

is negative.

. Since Cp and

are both negative,

is positive. Now will be positive, i.e. the gas gains heat. Hence the correct

3. One mole of a monatomic gas is taken through a cycle as shown in the P-V II gives the characteristics involved in the cycle. Match them with each of the processes given in Column I. IIT, 2011

Fig. 17.47

Column I C

Column II

17.39

C

SOLUTION is isobaric. Hence V T. Therefore T > T . U = nCv T = nCv T – T ). Since T < T , U is negative, i.e. internal energy decreases. Q = nCp T is also negative. Hence heat is lost. W = 3P V – V ) = – 6PV. Which is negative. Hence work is done on the gas. C is isochoric. Hence P internal energy decreases. W=P V V = 0)

T. Therefore T > TC. U = nCv T = nCv TC – T ) is negative, i.e. Q=

U + W),

Q=

U. Since

U is negative, heat is lost.

C is isobaric, i.e. V T. Hence TD > TC U = nCv TD – TC) is positive. Hence internal energy increases. Q = nCp TD – TC) is positive. Hence heat is gained by the gas. W = P V = P V – V) = 8 PV which is positive. So work is done by the gas.

gas is compressed, work is done on the gas, i.e. by the gas.

U = 0. Therefore Q = W. Since the W is negative. Hence Q is negative. Hence heat is lost

ANSWER

V Assertion-Reason Type Questions

following four choices out of which only one choice is correct. ment-2 is the correct explanation for Statement-1. ment-2 is the correct explanation for Statement-1.

Fig. 17.48

Statement-2 1. Statement-1 PV versus P graph for a cerFigure 17.48 shows T tain mass of oxygen gas at two temperatures T1 and T2. It follows from the graph that T1 > T2.

an ideal gas. 2. Statement-1 If two bodies of equal mass and made of the same material at different temperature T1 and T2 are

17.40 Comprehensive Physics—JEE Advanced

3.

4.

5.

6.

brought in thermal contact, the temperature of each T1 + T2)/2 when thermal equilibrium is attained. Statement-2 They have the same thermal capacity. Statement-1 Two vessels and of equal capacity are contains a gas at 0°C and 1 atmosphere pressure and vessel is evacuated. If the stopcock is suddenly and will be 0.5 atmosphere. Statement-2 If the temperature is kept constant, the pressure of a gas is inversely proportional to its volume. Statement-1 Two vessels and are connected to each other by a stopcock. Vessel contains a gas at 0°C and 1 atmosphere pressure and vessel is evaluated. The two vessels are thermally insulated from the surroundings. If the stopcock is suddenly opened, there will be no change in the internal energy of the gas. Statement-2 No transfer of heat energy takes place between the system and the surroundings. Statement-1 Two vessels and are connected to each other by a stopcock. Vessel contains a gas at 300 K and 1 atmosphere pressure and vessel is evacuated. The two vessels are thermally insulated from the surroundings. If the stopcock is suddenly opened, the expanding gas does no work. Statement-2 Since Q = 0 and U law of thermodynamics that W = 0. Statement-1 Heating system based on circulation of steam are

Statement-2 The latent heat of steam is high. 7. Statement-1 V-T graphs of a certain mass of an ideal gas at two pressures P1 and P2. It follows from the graphs that P1 is greater than P2.

Fig. 17.49

Statement-2 The slope of V-T graph for an ideal gas is directly proportional to pressure. 8. Statement-1 The curves and in Fig. 17.50 show P-V graphs for an isothermal and an adiabatic process for an ideal gas. The isothermal process is represented by curve .

Fig. 17.50

Statement-2 The slope of the P-V graph is less for an isothermal process than for an adiabatic process.

or circulation of hot water.

SOLUTION 1.

2.

is parallel to the P-axis. This means that PV/T is a constant, independent of pressure. Hence line corresponds to an ideal gas for which PV/T temperatures, a real gas behaves more like an ideal gas. Hence T1 is greater than T2. if the two bodies have the same thermal capacity which is equal to mass of the body heat capacity. Since the two bodies have the same

3.

4. 5.

mass and are made of the same material, they have the same thermal capacity. of equal capacity, the volume occupied by the gas is doubled when the stopclock is opened. Hence, pressure becomes half. mally insulated from the surroundings, no heat Q = 0. Since U W = 0.

17.41

6. 7. From PV = RT, we have PV R) T

an adiabatic process PV = constant. Differenting, we get + V =0 P V –1 P which gives =– = Cp/Cv V Since > 1, the slope of the P-V curve for an adiabatic process is greater than that for an isothermal process. Thus both the statements are true and statement-2 is the correct explanation for statement-1.

V R V 1 = , i.e. T P T P Hence Statement-1 is true but Statement-2 is false. 8. For an isothermal process PV = constant. DifferenP = – . For tiating we get + =0 V

VI Integer Answer Type 1. Two soap bubbles and are kept in a closed chamber where the air is maintained at pressure N/m2. The radii of bubbles and are 2 cm and 4 cm, respectively. Surface tension of the soap-water used to make bubbles is 0.04 N/m. Find the ratio

n /n , where n and n are the number of moles of air in bubbles and , respectively. [Neglect the effect of gravity.]

SOLUTION Using P V = n RT and P V = n RT, we have

1. If P0 is the external pressure, then P = P0 +

4 r

=8+

P = P0 +

4 r

=8+

4 0.04 2 10

2

4 0.04 4 10

2

= 16 N m–2 = 12 N m–2

n n

=

P P

V V

P P =

r r 12 16

3

V= 4 2

3

=6

4 r 3) 3

18

Kinetic Theory of Gases

Chapter

REVIEW OF BASIC CONCEPTS 18.1

18.3

PRESSURE EXERTED BY AN IDEAL GAS

The pressure of a gas in a container is a result of the continuous bombardment of the gas molecules against the walls of the container and is given by 2

2

1 m n vr ms 1 M vr ms 1 2 2U vr ms P= V 3 3 V 3 3V where m = mass of each molecule, n = number of molecules in the container, vrms = root mean square speed of molecules, V = volume of container, M = mass of gas in the container, = density of the gas and U = internal energy of the gas.

18.2

vrms =

v22

v32

vn2 )

1/ 2

where v1, v2, v3, vn are the speeds of the molecules 1, 2, 3, n respectively. In terms of P and , vrms is given by vrms =

3P

3k T m

The mean translational kinetic energy of a molecule of a gas is given by 1 mv2rms E= 2 In terms of E, the pressure of the gas is given by 2 nE 2 U 1 2 P= v rms = 3 V 3V 3 where U = nE is the total translational kinetic energy of all the n molecules of the gas. It is also called the internal energy of the gas. R Boltzman’s constant k = N0 where N0 is Avagdro number.

ROOT MEAN SQUARE SPEED 1 2 ( v1 n

MEAN TRANSLATIONAL KINETIC ENERGY

3 RT M

where m is the mass of each molecule, T the absolute temperature of the gas and k a constant called Boltzman’s constant. Its value is k = 1.38 10–23 J K–1 per molecule R is the universal gas constant and its value is R = 8.315 J K–1 mol–1

18.4

EQUATION OF STATE OF AN IDEAL GAS

The relationship between pressure P, volume V and absolute temperature T of an ideal gas is called the equation of state. For n moles of a gas, this relation is PV = nRT where R is the molar gas constant. From Avogadro’s law, it follows that one mole of all gases, at the same temperature and pressure, occupies gas occupies 22.4 litres at STP. Consequently, for on mole PV is constant for all gases. This constant is the the ratio T molar gas constant and can be evaluated as follows: At STP, V = 22.4 litre = 22.4 10–3 m3

18.2 Comprehensive Physics—JEE Advanced

P = 0.76 m Hg = 1.013 T = 273 K R=

18.5

PV T

105 Nm–2

1.013 105

22.4 10 273 = 8.315 J mol–1 K–1

According to Van der Waal the true pressure exerted by a gas is greater than P by an amount a/V 2 (where a is a constant) due to attractive forces between molecules and the true volume of the gas is less than V by an amount b (where b is another constant) because molecules themselves occupy a a

(V – b) = RT

V2 At high pressures, when the molecules are too many and too close together, the correction factors a and b both become important. But at low pressures, when they are not too many and not too close together, a gas behaves like an ideal gas and obeys the equation PV = RT.

18.6

DEGREES OF FREEDOM AND EQUIPARTITION OF ENERGY

The total number of coordinates or independent quantities tion of a dynamical system is called the degrees of freedom of the system. The molecules of a monoatomic gas consist of single atoms. Therefore, the molecules of a monoatomic gas have three degrees of freedom corresponding to translational motion. The molecules of a diatomic gas have degrees of freedom-three corresponding to translational motion and two for rotational motion. A polyatomic molecule has six degrees of freedom including one of vibrational motion. The law of equipartition of energy is stated as follows. In any dynamical system with a uniform absolute temperature T, the total energy is distributed equally among all the degrees of freedom and the average energy 1 kT, where per degree of freedom per molecule equals –23 –1 2 k = 1.38 10 J K .

If the molecules of a gas have f degrees of freedom, 1 kT. Therefore, then kinetic energy per molecule = f 2 kinetic energy per mole is 1 f U = Nf kT = RT 2 2 U f Now Cv = = R T 2

=

f 2

Cp

1 R

1

R

2 f

1

f R 2

Cv

f 2

f R+R= 2

Cp = Cv + R =

3

VAN DER WAAL’S EQUATION OF STATE

P

and

2 2 5 = for a monoatomic gas = 1 + 3 5 3 2 7 4 = for a diatomic gas = 1 + = for a triatomic 6 5 3 or polyatomic gas Thus

=1 +

Relation between Cp, Cv and R For a monoatomic gas;

Cv =

3R 5R and Cp = 2 2

For a diatomic gas;

Cv =

5R 7R and Cp = 2 2

For a polyatomic gas;

Cv = 3R and Cp = 4R

Relation between Cp, Cv and Cp – Cv = R =

Cp

Cp =

Cv

Cv – Cv = R

R

Cv =

Cp = Cv + R =

Cv 1

R 1

+R=

R 1

18.1 Calculate the root mean square speed of the molecules of hydrogen gas at S.T.P. Density of hydrogen at S.T.P. is 9 10–2 kg m–3. SOLUTION At S.T.P., pressure P = 1.01 = 9 10–2 kg m–3. vrms = =

105 Pa and density

3P 3 1.01 105 9 10

2

= 1840 ms–1

18.2 Calculate the temperature (in kelvin) at which the root mean square speed of a gas molecule is half its value at 0°C.

Kinetic Theory of Gases

SOLUTION

SOLUTION (a) Average translational K.E. of a molecule of an ideal gas is 3 E = kT, where T = temperature 2 in kelvin

3kT m

vrms = v rms = vrms

T T

1 = 2

T 273

T 273

T = 68.25 K

18.3 Find the mean translational kinetic energy of an oxygen molecule at 0°C. Given Avogadro number N = 6.03 1023 per mole and R = 8.3 JK–1 mol–1. SOLUTION 1 3 3 RT E = mv2rms = kT = 2 2 2 N E=

3 8.3 273 2

18.3

6.03 1023

= 5.64

10–21 J

18.4 Calculate the mean translational kinetic energy of 1 mole of hydrogen at S.T.P. Density of hydrogen at S.T.P. is 0.09 kg m–3. SOLUTION

At T = 0°C = 273 K, 3 E= (1.38 10–23) 2 = 5.65 10–21 J

273

At T = 100°C = 373 K, 3 E= (1.38 10–23) 2 = 7.72 10–21 J

373

(b) Number of molecules in 1 mole of a gas is N = 6.02 1023 K.E. of 1 mole at 273 K = (5.65 10–21) (6.02 1023) = 3.40 J K.E. of 1 mole at 373 K = (7.72 10–21) (6.02 1023) = 4.65 J 18.6 The speed of sound in a gas at S.T.P. is 330 ms–1 and the density of the gas is 1.3 kg m–3. Find the number of degrees of freedom of a molecule of the gas. SOLUTION

vrms =

3P

= 1.84 Mass of 1 mole is m = 22.4 = 2.016

3 1.01 105 0.09

=

103 ms–1 = 10

–3

m

0.09 kg m

–3

f = 10–3)

(1.84

v2 P

(330) 2

1.3

1.01 105

= 1.4

If f is the number of degrees of freedom, then

10–3 kg

1 K.E. = mv2rms 2 1 = (2.016 2 = 3.41

3

P

v =

103)2

103 J

18.5 Calculate (a) the average translational kinetic energy of the molecules of an ideal gas at 0°C and at 100°C and (b) the energy per mole of the gas at 0°C and 100°C. Given Avogadro’s number N = 6.02 1023 and Boltzmann’s constant k = 1.38 10–23 JK–1.

2 1

2 =5 (1.4 1)

18.7 The volume of 2 moles of a diatomic ideal gas at 300 K is doubled keeping its pressure constant. Find the change in the internal energy of the gas. Given R = 8.3 JK–1 mol–1. SOLUTION According to kinetic theory, there are no internal forces of interaction between the molecules of an ideal gas. This implies that the potential energy is

18.4 Comprehensive Physics—JEE Advanced

zero. Hence, for an ideal gas, the internal energy is only due to kinetic energy of the molecules. For n moles of an ideal gas at absolute temperature T, the internal energy is f nRT U= 2 where f is the number of degrees of freedom. For a diatomic gas f = 5. Therefore 5 U = nRT 2 Since pressure is kept constant, V T (Charles’ law), i.e. V1 V = 2 T1 T2 T2 =

V2 T1 = 2T1 = 600 K ( V1

U = U2 – U1 =

V2 = V1)

5 nR(T2 – T1) 2

5 2 8.3 2 = 1.245 104 J

=

(600 – 300)

Alternative Method U = nCv T 5R . Therefore 2 5R U= 2 (600 – 300) 2 = 1.245 104 J

For diatomic gas, Cv =

18.8 Hydrogen gas is contained in a vessel of volume 10 litres at 27°C. The gas pressure is 106 Pa. Find (a) total translational kinetic energy of hydrogen molecules, (b) mean (average) kinetic energy of the molecules and (c) total kinetic energy of the molecules. SOLUTION Volume of gas = 10 litres = 10

10–3 m3 = 10–2 m3) 3 3 106 (a) Total K.E. of molecules = PV = 2 2 10–2 = 1.5 104 J (b) Average K.E. of molecules 5 ( Hydrogen is diatomic) = kT 2

5 1.38 10–23 (273 + 27) 2 = 1.035 10–20 J 5 5 106 10–2 = 2.5 104 J (c) Total K.E. = PV = 2 2 =

18.9 n1 moles of a monoatomic gas are contained in a vessel A of volume V1 at pressure P1 and temperature T1. n2 moles of the same gas are contained in a vessel B of volume V2 at pressure P2 and temperature T2. The two vessels are now connected by a tube. Obtain the expression for (a) common temperature T and (b) common pressure P in the Vessels. SOLUTION If f = number of degrees of freedom, f f f n1RT1 + n2RT2 = (n1 + n2)RT 2 2 2 Here f = 3 (monoatomic gas) Equation (1) gives n T n2T2 T= 11 n1 n2 Final pressure is (n n2 ) RT PV P2V2 1 1 P= 1 V1 V2 V1 V2

18.7

IDEAL GAS LAWS

Boyle's Law At any given temperature, the volume of a given mass of a gas is inversely proportional to pressure, i.e. 1 V (T = constant and n = constant) P PV = constant P 1V 1 = P 2V 2 Charle's Law If the pressure of a gas is kept constant, the volume of a given mass of the gas is directly proportional to its absolute temperature, i.e. V T (P = constant and n = constant) V = constant T V1 V = 2 T1 T2

Kinetic Theory of Gases

If the volume of a gas is kept constant, the pressure of a given mass of the gas is directly proportional to its absolute temperature, i.e. P = constant (V = constant and n = constant) T P1 P = 2 T1 T2 Avogadro Law Equal volumes of all gasses at the same temperature and pressure contain an equal number of molecules. The number of molecules in one mole of any gas is N0 = 6.02 1023. N0 is called the Avogadro number. If volume V1 of one gas contains N1 molecules and volume V2 of another gas contains N2 molecules at the same temperature and pressure, then V1 V = 2 N1 N2 Equation of State in terms of Boltzmann constant (k) universal gas constant R Avogadro number N0 R PV = nRT = nN0 T = nN0kT N0 PV = NkT

k=

or where

N = nN0 = number of molecules

18.10 An electric bulb of volume 250 cm3 was sealed during manufacture at a pressure of 10–3 mm of Hg at 27°C. Find the number of air molecules in the bulb. SOLUTION Let N be the number of air molecules in the bulb. It is given that P = 10–3 mm of Hg = 10–4 cm of Hg, V = 250 cm3 and T = 273 + 27 = 300 K. Now PV = NkT (1) At STP, one mole of a gas occupies a volume V0 = 22400 cm3 and contains N0 = 6.02 1023 molecules, P0 = 76 cm of Hg and T0 = 273 K. (2) P0V0 = N0kT0 Dividing (1) by (2), we get N = N0 = (6.02 = 8.04

T0 T

P P0 1023)

V V0 273 300

1015 molecules

10 4 76

250 22400

18.5

18.11 A cylinder of volume 30 litres contains oxygen at a gauge pressure of 15 atm at 27°C. When some oxygen is ejected out from the cylinder, the gauge pressure falls to 11 atm and temperature falls to 17°C. Find the mass of oxygen ejected. R = 8.3 J K–1 mol–1 and molecular mass of oxygen = 32. SOLUTION Let N1 be the number of molecules, P1 be the pressure and T1 be the temperature of oxygen before some oxygen is ejected and N2, P2 and T2 be their values after the gas is ejected. Since the volume V remains unchanged (= volume of cylinder), (1) P1V = N1kT1 P2V = N2kT2

(2)

8.3

R N0

= 1.38 10–23 JK–1 per 6.02 1023 molecules is Boltzmann constant. Given, V = 30 litres = 30 10–3 m3, P1 = 15 atm = 15 1.01 105 Pa, T1 = 273 + 27 = 300 k, P2 = 11 atm = 11 1.01 105 Pa and T2 = 273 + 17 = 290 K. Substituting these values in Eqs. (1) and (2), we get N1 = 10.87 1024 and N2 = 8.25 1024 Number of molecules of oxygen ejected = N1 – N2 = 2.62 1024. The molecular mass of oxygen = 32, i.e. one mole of oxygen (which contains 6.02 1023 molecules) has a mass of 32 g = 0.032 kg. Hence 2.62 1024 molecules will have a mass of where k =

0.032

2.62 1024

6.02 1023

= 0.14 kg

18.12 A narrow glass tube of length 100 cm is closed at both ends. It lies horizontally with 20 cm of mercury column in the middle dividing the tube into two compartments I and II of equal length. The air in each compartment is at standard temperature and pressure. The tube is then turned to a vertical position. By what distance will the mercury column be displaced? SOLUTION Let A cm2 be the cross-sectional area of the tube. Then P0 = 76 cm of Hg and V0 = 40 A cm3 [Fig. 18.1(a)]. When the tube is turned to vertical position [Fig. 18.1(b)], Let x cm be the displacement of mercury column and let P1, V1 be the air pressure in compartment I and P2, V2 those in compartment II.

18.6 Comprehensive Physics—JEE Advanced

and

P1 =

P0V0 P 40 A = 0 V1 (40 x) A

40 P0 (40 x)

P2 =

P0V0 P 40 A = 0 V2 (40 x) A

40 P0 (40 x)

The mercury column will be in equilibrium if P2 (in cm of Hg) + 10 cm of Hg = P1(in cm of Hg), i.e. if 40 P0 40 P0 + 10 = 40 x (40 x) where P0 = 76 cm of Hg. Thus 40 76 40 76 + 10 = (40 x) (40 x)

Fig. 18.1

Using Boyle’s Law, we have

x2 + 608x – 1600 = 0 Solving for positive root, we get x = 2.6 cm

P0V0 = P1V1 and P0V0 = P2V2

I Multiple Choice Questions with Only One Choice Correct 1. In an adiabatic process, the root mean square speed of the molecules of a monoatomic gas becomes twice its initial value. The ratio of the initial volume (a) 2 (b) 23/2 (c) 4 (d) 8 2. The root mean square speed of hydrogen molecules at a certain temperature is v. If the temperature is doubled and the hydrogen gas dissociates into atomic hydrogen, the rms speed will become (a)

v 4

(b)

v 2

(c) 2v (d) 4v 3. A gas in a closed container has temperature T and pressure P. If the molecules of the gas undergo inelastic collisions with the walls of the container, then (a) both P and T will decrease (b) P decreases and T increases (c) P increases and T decreases (d) both P and T remain the same 4. Two moles of hydrogen are mixed with n moles of helium. The root mean square speed of the gas molecules in the mixture is 2 times the speed of sound in the mixture. The value of n is (a) 1 (b) 2 (c) 3 (d) 4

5. A vessel contains 4 moles of oxygen and 2 moles of argon at absolute temperature T. The total internal energy of the gas mixture is (a) 6 RT (b) 9 RT (c) 11 RT (d) 13 RT 6. The average translational kinetic energy of a molecule of a gas at absolute temperature T is proportional to 1 (b) T (a) T (c) T (d) T 2 7. The root mean square speed of the molecules of an enclosed gas is v. What will be the root mean square speed if the pressure is doubled, the temperature remaining the same? (a) v/2 (b) v (c) 2 v (d) 4 v 8. The mass of an oxygen molecule is about 16 times that of a hydrogen molecule. At room temperature the rms speed of oxygen molecules is v. The rms speed of the hydrogen molecule at the same temperature will be (a) v/16 (b) v/4 (c) 4 v (d) 16 v 9. The average kinetic energy of hydrogen molecules at 300 K is E. At the same temperature, the average kinetic energy of oxygen molecules will be

Kinetic Theory of Gases

(a) E/16 (b) E/4 (c) E (d) 4E 10. At room temperature (27°C) the rms speed of the molecules of a certain diatomic gas is found to be 1920 ms–1. The gas is (a) H2 (b) F2 (c) O2 (d) Cl2 IIT, 1984 11. The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K, the room mean square speed of the gas molecules is v, then at 480 K it will be (a) 4 v (b) 2 v v v (d) (c) 2 4 IIT, 1996 12. Three closed vessels A, B and C are at the same temperature. Vessel A contains only O2, B only N2 and C a mixture of equal quantities of O2 and N2. If the average speed of O2 molecules in vessel A is v1, that of N2 molecules in vessel B is v2, the average speed of O2 molecules is vessel C is 1 (v1 + v2) (b) v1 (a) 2 3k T (c) v1 v2 (d) M IIT, 1992 13. The average translational energy and the rms speed of molecules of a sample of oxygen gas at 300 K are 6.21 10–21 J and 484 ms–1 respectively. The corresponding values at 600 K are nearly (assuming ideal gas behaviour) (a) 12.42 10–21 J, 968 ms–1 (b) 8.78 10–21 J, 684 ms–1 (c) 6.21 10–21 J, 968 ms–1 (d) 12.42 10–21 J, 684 ms–1 IIT, 1997 14. A vessel contains 1 mole of O2 gas (molar mass 32) at a temperature T. The pressure of the gas is P. An identical vessel containing one mole of He gas (molar mass 4) at a temperature 2T has a pressure of P (b) P (a) 8 (c) 2P (d) 8P IIT, 1997 15. A vessel contains a mixture of 1 mole of oxygen and two moles of nitrogen at 300 K. The ratio of

18.7

the rotational kinetic energy per O2 molecule to that per N2 molecule is (a) 1 : 1 (b) 1 : 2 (c) 2 : 1 (d) depends on the moment of inertia of the two molecules. IIT, 1998 16.

with air at temperature (T1, T2), volume (V1, V2) and pressure (P1, P2) respectively. If the valve joining the two vessels is opened, the temperature inside the vessel at equilibrium will be (a) T1 + T2 (b) (T1 + T2)/2 (c) T1T2 (P1V1 + P2V2)/(P1V1T2 + P2V2 T1)

(d) T1T2 (P1V1 + P2V2)/(P1V1T1 + P2V2T2) 17. If the temperature of a gas is increased from 27°C to 927°C, the root mean square speed of its molecules (a) becomes half (b) is doubled (c) becomes 4 times (d) remains unchanged 18. At what temperature will oxygen molecules have the same root mean square speed as hydrogen molecules at 200 K? (a) 527°C (b) 1327°C (c) 2127°C (d) 2927°C 19. An enclosure of volume V contains a mixture of 8 g of oxygen, 14 g of nitrogen and 22 g of carbon dioxide at absolute temperature T. The pressure of the mixture of gases is (R is universal gas constant) 3RT RT (b) (a) 2V V 5RT 7 RT (c) (d) 4V 5V 20. At what absolute temperature T is the root mean square speed of a hydrogen molecule equal to its escape velocity from the surface of the moon? The radius of moon is R, g is the acceleration due to gravity on moon’s surface, m is the mass of a hydrogen molecule and k is the Boltzmann constant mgR 2mgR (b) (a) 2k k 3mgR 2mgR (c) (d) 2k 3k

18.8 Comprehensive Physics—JEE Advanced

23. 0.014 kg of nitrogen is enclosed in a vessel at a temperature of 27 °C. The amount of heat energy to be supplied to the gas to double the rms speed of its molecules is approximately equal to (a) 6350 J (b) 7350 J (c) 8350 J (d) 9350 J 24. A vessel of volume V contains an ideal gas at absolute temperature T and pressure P. The gas is allowed to leak till its pressure falls to P . Assuming that the temperature remains constant during leakage, the number of moles of the gas that have leaked is V V (P + P ) (b) (P + P ) (a) RT 2 RT

21. Two perfect gases at absolute temperatures T1 and T2 are mixed. There is no loss of energy in this process. If n1 and n2 are the respective number of molecules of the gases, the temperature of the mixture will be (a)

n1T1 n2T2 n1 n2

(c) T1 +

(b)

n2 T n1 2

n2T1 n1T2 n1 n2 n1 T n2 1

(d) T2 +

22. An insulated box containing a diatomic gas of molar mass M is moving with a velocity v. The box is suddenly stopped. The resulting change in temperature is (R is the gas constant) (a)

M v2 2R

(b)

M v2 3R

(c)

M v2 5R

(d)

2M v 2 5R

(c)

V (P – P ) RT

V (P – P ) 2 RT

(d)

IIT, 2003 ANSWERS

1. 7. 13. 19.

(d) (b) (d) (c)

2. 8. 14. 20.

(c) (c) (c) (d)

3. 9. 15. 21.

(d) (c) (a) (a)

4. 10. 16. 22.

(b) (a) (c) (c)

5. 11. 17. 23.

(d) (b) (b) (d)

6. 12. 18. 24.

(c) (b) (d) (c)

SOLUTIONS 1. Let T1 be the initial temperature. Since vrms T, T2 = 4 T1. For an adiabatic process, T1 V1( – 1) = T2 V2( – 1) V1 V2 V1 V2

1

5 1 3

T2 = =4 T1 =4

V1 = (4)3/2 = 8 V2 3 RT ; M = molecular mass. Now T is M doubled and M is halved. Hence the rms speed will become twice, which is choice (c). 3. Since the temperature of the molecules is the same as that of the walls, there is no exchange of

energy in a collision. Hence the molecules rebound with the same average speed whether the collision is elastic or inelastic. Therefore, both P and T will not change. 3 RT ; M = molecular mass of mixture M The speed of sound in the mixture is

4. vrms =

RT Cp ; = of the mixture M Cv Given vrms = 2 v. Therefore

v=

3 RT M

2. vrms =

= =

2

RT M

3 2

For the mixture, Cv =

n1 Cv

1

n1

n2 Cv n2

2

Kinetic Theory of Gases

and

Cp = Cp Cv

=

n1 C p

n2 C p

1

n1 n1 C p

n2 C p

2

(1)

n2 (C v ) 2

For hydrogen, n1 = 2, (Cv)1 = For helium, n2 = n, (Cv)2 =

n2

1

n1 (C v )1

2

3R 5R and (Cp)1 = 2 2

5R 7R and (Cp)2 = 2 2

12. For a given M, vav T only. Therefore, the correct choice is v1 as the temperatures of vessels A and C are the same. 13. E T and vrms T . Hence at 600 K, 600 E = 6.21 10–21 300 = 12.42 10 –21 J and

8. v = vh = Thus or

3 kT / m0 and

3kT / m . Therefore, v0 = 3kT / mh . v0 mh = = vh m0 v h = 4 v 0.

1 16

1 4

Hence the correct choice is (c). 9. The correct choice is (c) as explained above. 10. We know that vrms =

= 2 10 –3 kg = 2 g Since M = 2, the gas is hydrogen. 3RT or vrms M choice is (b).

11. vrms =

T . Hence the correct

684 ms–1

1 mole RT and V 1 mole R 2T (P)He = V P He = 2 or (P)He = 2 (P)O2 PO (P)O2 =

2

15. Since both the gases are diatomic, each has two degrees of freedom associated with rotational motion. According to the law of equipartition of energy, the rotational kinetic energy per degree of freedom is (1/2)kT. Since the temperatures of the two gases are equal, their rotational kinetic energies will be equal. Hence the correct choice is (a). 16. According to the kinetic theory, the average kinetic 3 kT. Let n1 energy (KE) per molecule of a gas = 2 and n2 be the number of moles of air in vessels 1 and 2 respectively. Before mixing, the total KE of molecules in the two vessels is 3 3 E1 = n1kT1 + n2 kT2 2 2 3 = k(n1T1 + n2T2) 2 After mixing, the total KE of molecules is

3 RT which gives M

3 RT 3 8.3 300 = M = v2 1920 2 r ms

600 300

14. For a gas, PV = nRT. Hence

U =

3 6. E = kT. So the correct choice is (c). 2 7. The rms speed = 3kT / m which is independent of pressure. Hence the correct choice is (b).

vrms = 484

which is choice (d).

Using these values and = 3/2 in Eq. (1) and solving, we get n = 2. 5. The internal energy of n moles of a gas temperature T is given by f (nRT) 2 where f = number of degrees of freedom. For oxygen f = 5 and for argon f = 3. Hence U = U1 + U2 5 3 = (4RT) + (2RT) 2 2 = 13RT

18.9

E2 =

3 (n1 + n2) kT 2

where T is the temperature when equilibrium is established. Since there is no loss of energy (because the vessels are insulated), E2 = E1 or 3 3 (n1 + n2)kT = k (n1T1 + n2T2) 2 2 or

T=

n1T1 n2T2 (n1 n2 )

(1)

18.10 Comprehensive Physics—JEE Advanced

Now P1V1 = n1RT1 and P2V2 = n2RT2 which give PV P2V2 1 1 and n2 = RT1 RT2 Using these in Eq. (1) and simplifying, we get

20. The root mean square speed is given by

n1 =

T1T2 ( PV P2V2 ) 1 1 T= ( PV P2V2T1 ) 1 1T2 17. The root mean square speed is given by 3kT m i.e. vrms T. Initial temperature T1 = 27 + 273 = 300 K. Final temperature T2 = 927 + 273 = 1200 K. Since temperature is increased by 4 times, the speed is doubled. Hence the correct choice is (b). vrms =

18. For oxygen : vrms = For hydrogen : vrms =

3 kT0 m0 3kTh mh

m0 Th = 16 200 = 3200 K = 2927°C mh Hence the correct choice is (d).

or T0 =

19. The pressure exerted by a gas is given by nRT P= V mass RT = molecular weight V 8 RT 1 RT = 32 V 4 V

14 RT 1 RT = 28 V 2 V Pressure exerted by carbon dioxide Pressure exerted by oxygen P2 =

P3 =

22 RT 1 RT = 44 V 2 V

From Dalton’s law of partial pressures, the total pressure exerted by the mixture is given by P = P1 + P2 + P3 1 RT 1 RT 1 RT + + 4 V 2 V 2 V 5RT = , which is choice (c). V =

3kT m

The escape velocity is given by ve =

2gR

For vrms = ve, we require 3kT m

2 gR or T =

2mg R , which is choice (d). 3k

21. Average kinetic energy per molecule of a perfect 3 gas = kT. 2 3 n1 kT1 2 Average KE of molecules of the second gas 3 = n2 kT2 2 =

Total KE of the molecules of the two gases before they are mixed is

The value of vrms will be the same if T0 Th = m0 mh

Pressure exerted by oxygen P1 =

vr ms =

3 3 n kT + n kT 2 1 1 2 2 2 3 = (n1 T1 + n2 T2)k (1) 2 If T is the temperature of the mixture, the kinetic energy of the molecules (n1 + n2) in the mixture is 3 K = (n1 + n2)kT (2) 2 Since there is no loss of energy K = K . Equating (1) and (2) we get n1T1 n2T2 n1 T1 + n2 T2 = (n1 + n2) T or T = n1 n2 Hence the correct choice is (a). 22. Let n be the number of moles of the gas in the box. 1 The kinetic energy of the gas = n M v 2 . When 2 the box is suddenly stopped, this energy is used up in changing the internal energy, as a result of which the temperature of the gas rises. The change in internal energy is given by 5 U = nCv T = n R T 2 5 For a diatomic gas Cv = R. 2 1 5R T=n Mv2 Hence n 2 2 K=

or

T=

M v2 , which is choice (c) 5R

Kinetic Theory of Gases

23. The root mean square speed is related to absolute temperature T as crms =

5R 2

U = 0.5

3kT m

18.11

(1200 – 300)

0.5 5 8.314 900 = 9353 J. 2 So the correct choice is (d) =

For a given gas, m crms T. Hence in order to double the root mean square speed, the absolute temperature must be increased to four times the initial value. Initial temperature T1 T2 = 4 T1 = 1200 K. V = 0. Hence the heat energy supplied to the gas does no work on the gas; it only increases the internal energy of the molecules. The increase in internal energy is U = n Cv T 5R Since nitrogen is diatomic Cv = . The number 2 of moles of nitrogen in the vessel is

PV . Let RT n be the number of moles of the gas that leaked till the pressure falls to P . Since volume V of the vessel cannot change and temperature T remains constant during leakage, we have

24. Number of moles present initially is n =

n =

PV RT

Number of moles that leaked is n=n–n =

V PV PV – = (P – P ) RT RT RT

So the correct choice is (c).

mass in kg 0.014 103 = = 0.5 n= molecular mass 28

II Multiple Choice Questions with One or More Choices Correct 1. The root mean square speed of the molecules of a gas depends upon (a) the pressure of the gas (b) the density of the gas (c) the temperature of the gas (d) the mass of a molecule of the gas 2. The average translational kinetic energy of a molecule of a gas at absolute temperature T is E and the root mean square speed of the molecules is v. Then (a) E T (b) E T (c) v T (d) v T 3. The root mean square speeds of the molecules of hydrogen, oxygen and carbon dioxide at the same temperature are vh, vo and vc respectively. Then (b) vh > vo (a) vh = vo = vc (c) vo > vc (d) vh < vo < vc 4. One mole of oxygen at 27 °C is enclosed in a vessel which is thermally insulated. The vessel is moved with a constant speed v and is then suddenly stopped. The process results in a rise of temperature

of the gas by 1 °C. Then, if M = molecular mass of oxygen. (a)

(= Cp /Cv) = R

(c) v =

M(

1)

5 3

(b)

=

(d) v =

7 5 2R M ( 1)

IIT, 1983, 1996 5. In a vessel a gas at temperature T has a pressure P. The density of the gas is and vrms is the average root-mean-square speed of a molecule. If N is the number of molecules per unit volume and m the mass of a molecule, then (k = Boltzmann constant) (a) P =

1 2 vrms 3

(b) P =

2 2 vrms 3

3kT (d) P = N kT m 6. N molecules of a gas, each of mass m, strike per A at an angle to the vertical and rebound with a speed v. (c) vrms =

18.12 Comprehensive Physics—JEE Advanced

The collisions are assumed to be elastic. (a) The magnitude of change in momentum is | p| = 2mv sin (b) | p| = 2mv cos | p| (c) The pressure exerted on the wall is P = NA N | p| (d) P = A 7. The pressure P of n moles of a gas varies with volume V as P = a – bV2

of the piston from the base of the cylinder is h1 = 20 cm as shown in Fig. 18.2. When the temperature is raised to T2 = 177 °C, the new height of the piston above the base becomes h2. The system is then insulated from the surroundings and the piston is brought back to its original height. The new equilibrium temperature becomes T3. Given (1.5)0.4 = 1.18. IIT, 2004

where a and b are positive constants. The highest absolute temperature to which the gas can be heated is Tmax. (a) Tmax = (b) Tmax =

8.

a3 / 2

;

nR 3b

R = gas constant

2a 3 / 2 3nR 3b 1/ 2

(c) At Tmax, V =

a b

(d) At Tmax, V =

a 3b

1/ 2

Fig. 18.2

gas at temperature T1 = 27 °C. Initially the height

(a) h2 = 30 cm (c) T3 = 81°C

(b) h2 = 24.5 cm (d) T3 = 258 °C

ANSWERS AND SOLUTIONS 1. vr ms =

3P

3kT . Hence all the four choices m

are correct. 3 3kT 2. E = kT and vr ms = . Hence the correct 2 m choices are (a) and (d). 3. The rms speed is the maximum for the gas which is the lightest. Hence the correct choices are (b) and (c). 4. Oxygen is diatomic; it has 5 degrees of freedom. Therefore, Cv = 5 R/2 and Cp = 7 R/2. So = Cp /Cv = 7/5. The kinetic energy of oxygen molecules with a 1 velocity v0 = M v2, where M = molecular weight of oxygen. 2 Now heat energy = Cv dT = Cv 1 = Cv Cp R But Cp – Cv = R or –1= Cv Cv or

R ( – 1) = or Cv = Cv

R 1

Therefore, or

1 M v2 = 2 v=

R 1 2R

M

1

So the correct choices are (b) and (d). 5. The average pressure of a gas is given by 1 v 2rms P= 3 where density = mN; here m is the mass of a molecule and N is the number of molecules per unit volume of the gas. Thus 1 2 mN vrms (1) P= 3 The average kinetic energy of a molecule is given by 3 1 mv 2r ms = kT (2) 2 2 Using (2) in (1), we get P = NkT Thus the correct choices are (a), (c) and (d).

Kinetic Theory of Gases

6. Refer to Fig. 18.3. Since the collision is elastic, the speed v of the molecule before collision = speed after rebound. Therefore, the magnitude of the change in momentum normal to the wall is | p| = mv cos – (– mv cos ) = 2 mv cos

or

a 3b

1/ 2

a a 3b

or or

b

1/ 2

a

a 3b

= nR Tmax

a 3

= nR Tmax

2a 3 / 2

Tmax =

18.13

(4)

3 3b nR d 2T

It is easy to check that

is indeed negative at dV 2 a = 3 bV . Hence Eq. (4) gives the maximum value of T. So the correct choices are (b) and (d). 8. Let A be the cross-sectional area of the base of the cylinder. Given T1 = 273 + 27 = 300 K and T2 = 273 + 177 = 450 K. At constant pressure, we have from Charles’ law, V1 V2 = T1 T2 2

Fig. 18.3

Number of collisions per second = N. Therefore, time interval between successive collisions is 1 N Force exerted on the wall by the molecules is = rate of change of momentum t=

F =

| p| t

F Pressure P = . Hence the correct choices are (b) A and (c). 7. For n moles of an ideal gas, PV = nRT Using the given relation between P and V, we have (a – bV2) V = nRT or aV – bV3 = nRT (1) Differentiating with respect to V, we get dT dV

a – 3 bV2 = nR

dT 1 = (a – 3 bV2) dV nR

or

a 3b

1/ 2

b

a 3b

3/ 2

= nR Tmax

or

Ah1 Ah2 = T1 T2 h2 =

T2 T1

h1

(2)

450 K × 20 cm 300 K = 30 cm

Since the system is completely insulated from the surroundings, it cannot take heat from or give heat to the surroundings. Hence the second process is adiabatic for which T 2V 2(

d 2T dT = 0 and is negative. T will be maximum if dV 2 dV dT = 0 in Eq. (2), we get Setting dV 1/ 2 a 2 (3) a = 3 bV or V = 3b Using (3) in (1) we get a

or

–1)

= T 3V 3(

–1)

where V2 = Ah2 and V3 = Ah1. Also, for a diatomic gas = 1.4. Therefore, ( 1) V2 T3 = T2 × V3 = 450 ×

30 20

(1.4 1)

= 450 × (1.5)0.4. = 450

1.18 = 531 K = 258 °C

So the correct choices are (a) and (d).

18.14 Comprehensive Physics—JEE Advanced

III Multiple Choice Questions Based on Passage Questions 1 to 5 are based on the following passage Passage I Kinetic Theory of Gases The molecules of a gas move in all directions with various speeds. The speeds of the molecules of a gas increase with rise in temperature. During its random motion, a fast molecule often strikes against the walls of the container of the gas. The collisions are assumed to be perfectly elastic, i.e. the molecule bounces back with the same speed with which it strikes the wall. Since the number of molecules is very large, billions of molecules strike against the walls of the container every second. These molecules exert a sizeable force on the wall. The force exerted per unit area is the pressure exerted by the gas on the walls. According to the kinetic theory, the pressure of a gas of density at absolute temperature T is given by 1 2 vr ms 3 where vr ms is the root mean square speed of the gas molecule and is given by P=

vr ms =

3kT m

where m is the mass of a molecule and k is Boltzmann constant. 1 2 1. From the relation P = vr ms , 3 dimensions of pressure are (a) ML–1T–1 (b) ML–1T–2 –2 –1 (c) ML T (d) ML–2T–2.

3kT , it follows that the m constant k should be expressed in units of (a) newton per metre per kelvin (b) newton per kelvin (c) joule per kelvin (d) joule per kilogram per kelvin 3. Choose the only correct statement from the following. (a) The pressure of a gas is equal to the total kinetic energy of its molecules per unit volume of the gas. (b) The product of pressure and volume of a gas is always constant. (c) The average kinetic energy of the molecules of a gas is proportional to its absolute temperature. (d) The root mean square speed of a molecule is proportional to the absolute temperature of the gas. 4. If the temperature of a gas is increased from 27°C to 927°C, the root mean square speed of its molecules (a) becomes half (b) is doubled (c) becomes four times (d) remains unchanged 5. The root mean square speed of oxygen gas molecule at T = 320 K is very nearly equal to (the molar mass of oxygen is M = 0.0320 kg per mole and gas constant R = 8.31 J mol–1 K–1) (a) 300 ms–1 (b) 500 ms–1 (c) 700 ms–1 (d) 900 ms–1 2. From the relation vr ms =

ANSWERS AND SOLUTIONS 1. Dimensions of P = dimensions of density sions of (velocity)2 = ML–3

dimen-

(LT–1)2 = ML–1T–2

Hence the correct choice is (b). 2. Squaring we have k=

2 mvrms unit of energy = 3T unit of temperature = joule per kelvin

3. The correct choice is (c) 4. The correct choice is (b) since vr ms is proportional to the square root of absolute temperature 5. vrms =

3kT = m

3RT = M

3 8.31 320 0.032

= 499.3 ms–1 Hence the correct choice is (b)

Kinetic Theory of Gases

Questions 6 to 12 are based on the following passage Passage II Van der Waals Equation of State The equation of state PV = nRT holds for an ideal gas. The behaviour of real gases shows departures from an ideal gas behaviour especially at high pressures. The model of an ideal gas is based on a number of assumptions. Van der PV = nRT by taking into account two of those assumptions which may not be valid. He argued that (i) the volume of the molecules may not be negligible compared to the volume V occupied by the gas and (ii) the attractive forces between the molecules may not be negligible. He said that pressure P in equation PV = nRT is less than the true pressure by an amount p because of attractive forces between the molecules. According to him, the pressure ‘defect’ p is inversely proportional to the square of volume, i.e. 1 p V2 a or p= 2 V where a is constant depending on the nature of the gas. a Thus the true pressure of the gas is P = P + p = P + 2 . V He further argued that V is not the true volume of the volume. According to him, the true volume of the gas is V = (V – b) where b is a factor depending on the actual volume of the molecules themselves. Thus Van der Waals’ equation for real gases is P V = nRT, i.e. a P V b = nRT V2

18.15

At high pressures, when the molecules are too many and too close together, the correction factors a and b become important. 6. In Van der Waals’ equation of state for real gases, the product PV has the same dimensions as those of a (a) bP (b) V ab (c) (d) nRT V2 7. The dimensions of a are the same as those of (a) PV (b) PV2 (c) P2V (d) P/V 8. The dimensions of b are the same as those of (a) P (b) V (c) PV (d) nRT a 9. The dimensions of are the same as those of b (a) work (b) force (c) pressure (d) power 10. The dimensional formula for ab is (a) ML2T–2 (b) ML4T–2 6 –2 (c) ML T (d) ML8T–2 11. The correction factors a and b depend upon (a) the pressure of the gas (b) the volume of the gas (c) the temperature of the gas (d) the nature of the gas 12. The equation of state PV = nRT holds if the gas has (a) low pressure and low density (b) low pressure and high density (c) high pressure and low density (d) high pressure and high density.

ANSWERS AND SOLUTIONS 6. From Van der Waals’ equation, we have a ab = nRT V V2 From the principle of homogeneity of dimensions, PV – bP +

a

= dimensions of P V2 Dimensions of a = dimensions of PV2. Hence the correct choice is (b).

7. Dimensions of

8. Dimensions of b = dimensions of V, which is choice (b)

dimensions of PV 2 a = dimensions of V b = dimensions of PV = dimensions of work Hence the correct choice is (a). 10. Dimensions of ab = dimensions of PV2 dimensions of V = dimensions of (PV3) 9. Dimensions of

= ML–1T–2 (L3)3 = ML8T–2 Hence the correct choice is (d). 11. The correct choice is (d). 12. The correct choice is (a).

18.16 Comprehensive Physics—JEE Advanced

Questions 13 and 14 are based on the following passage Passage III Mean Free Path During their random motion, the molecules of a gas often come close to each other. When the distance between two molecules is comparable with the diameter of a molecule, the forces between them become very strong. As a result, their individual momenta before and after the encounter are different. When this happens a ‘collision’ is said to have occurred. The average distance a molecule travels before it suffers a collision with another molecule is called the mean free path (lc), which can be estimated as follows. Suppose the average speed of a molecule of diameter d is v . In one second, this molecule sweeps out a volume d2 v this volume, it will suffer collisions with them. If n is the number density (i.e. number per unit volume) of the molecules, then the number of molecules in this volume = d2 v n. The number of collisions per second = c = d2 v n. Therefore, the average time between two collisions (called collision period Tc) is

Tc =

1

=

1

d 2 vn Hence the mean free path (i.e. the average distance the molecule travels between two successive collisions) is c

lc = v Tc =

1

d 2n For air at S.T.P. the value of lc 3 10–7 m. 13. The mean free path of a molecule of a gas depends (a) only on its diameter (d) (b) only on the number density (n) of the molecules (c) on both d and n (d) neither on d nor on n. 14. The average collision period in a gas (a) increases if the pressure is increased (b) decreases if the pressure is increased (c) increases if the temperature of the gas is increased. (d) decreases if the temperature of the gas is increased.

ANSWERS AND SOLUTIONS 13. The mean free path is given by lc =

1

d 2n Hence the correct choice is (c). 14. The collision period is given by 1 Tc = d 2 vn

If the pressure is increased, the volume of the gas decreases. Hence number density n increases. Therefore, Tc will decrease. If the temperature of the gas is increased, the average speed v of the molecules increases. Hence Tc will decrease. Thus the correct choices are (b) and (d).

19

Transmission of Heat

Chapter

REVIEW OF BASIC CONCEPTS 19.1

THERMAL CONDUCTIVITY

If a steady temperature difference (T1 – T2) is to be maintained between the ends of a rod, heat must be supplied at a steady rate at one end and the same must be taken out at the other end. Suppose an amount of heat Q t so that Q/t is the rate of (i) Q/t will be proportional to the area A of the crosssection of the rod. (ii) Q/t will be proportional to (T1 – T2). (iii) Q/t will be inversely proportional to l, the distance between ends of the rod.

H=

Q t

A T1 T2 l

or

k A T1 T2 Q = l t

where k is a constant of proportionality called the thermal conductivity of the substance. It is a measure of how stance. In the SI system, k is expressed in Js–1 m–1 °C–1 or J s–1 m–1 K–1 or Wm–1 K–1. –3 –1 K ] [k

19.2

CONDUCTION THROUGH A COMPOSITE SLAB

Fig. 19.1

temperatures T1 and T2 (T1 > T2 Q1 t Q2

and

t

= =

T2 )

(1)

L k 2 A 2 (T1

T2 )

(2)

L

(A1 + A2

L. If keq is the equivalent

the composite slab is k eq ( A1 Q = t Since

A2 ) (T1 L

T2 )

(3)

Q1 Q 2 Q = + t t t keq (A1 + A2) = k1 A1 + k2 A2

Case 1. Two slabs placed one on top of the other Suppose we have a composite slab made up of two different slabs of materials of thermal conductivities k1 and k2, and L cross-sectional areas A1 and A2

k 1 A1 (T1

keq = If A1 = A2, then keq =

k 1 A1 A1

k 2 A2 A2

1 (k1 + k2). 2

(4)

19.2 Comprehensive Physics—JEE Advanced

Case 2. Two slabs placed in contact one after the other L1 and L2 but of the same cross-sectional area A placed in contact as

T0 =

19.3

( k 1 T1

k 2 T2 )

(k 1

k 2)

THERMAL RESISTANCE

slab becomes much easier if we use the concept of thermal

R=

V I

R=

temperature difference rate of flow of heat

R=

(T1 T2 ) L k A(T1 T2 )

Fig. 19.2

temperatures T1 and T2 (T1 > T2). Let T0 be the temperature Q1 Q2 of the junction. In the steady state, = , i.e. t t k 1 A(T1

T0 )

L1 k1 L1

=

(T1 – T0) =

T0 =

k 2 A(T0

T2 )

potential difference , e

Q/t

posite slab is Req = R1 + R2

(T0 – T2)

L2

T2

L kA

L2 k2

T1

k 1 T1

k 2 T2

L1 k1

L2 k2

L1

L2

L1 L2 = L1 keq A k1 A (5) keq =

L2 k2 A

k 1k 2 ( L1 ( L1k2

L2 ) L2 k1 )

(b) If the two slabs are joined in parallel as shown in Q1 Q = t t

or

Q2

1 1 = R1 R

t

k 1 A(T1 T0 ) Q = L1 t

(6)

k 1k 2 A(T1 T2 ) Q = k 1 L2 k 2 L1 t

(7)

keq = If L1 = L2, keq =

k 1k 2 ( L1 k 1 L2

2 k 1k 2 (k 1

k 2)

and

L2 ) k 2 L1

A2 )

L

L1 + L2) but its cross-sectional area is A. If keq is the equivalent thermal conductivity of the composite slab, Q k A (T1 T2 ) = eq t ( L1 L2 )

keq ( A1

1 R2

(8)

keq =

( k 1 A1 ( A1

=

k1 A1

k2 A 2

L

L

k 2 A2 ) A2 )

19.1 is maintained at 100°C and the other end is immersed minutes. Calculate the thermal conductivity of metal. –1 . SOLUTION Given L = 0.3 m, A = r2 = (10–2)2 = T1 = 100°C and T2 = 0°C. Now

10–4 m2,

19.3 –1

Q = mLf = 31.4 = 31.4 Also

80 cal 80 4.2 J (

1 cal = 4.2 J)

k A (T1 T 2 ) Q = L t k= =

Fig. 19.3

SOLUTION

QL t A(T1 T2 ) (31.4 80 (5 60)

= 336 W m

0.3

10 4 )

( –1

4.2)

K

In the steady state,

(100

0)

Q1

k1 A1 (T1

–1

L1

19.2 A cylindrical metal boiler of radius 10 cm and on an electric heater. If the water boils at the rate –1 102 W m–1 K–1 and latent heat of vaporisation = 2.26 103 –1.

50

0.02

t

=

T0 )

Q2 t =

, i.e. k2 A 2 (T 0

T2 )

L2

(300 T0 ) 400 0.01 (T0 = 0.1 0.2 T0 = 100°C

0)

19.4 A steel rod (L1 = 10 cm, A1 = 0.02 m2, k1 = 50 W m–1 K–1) and a brass (L2 = 10 cm, A2 = 0.02 m2, k2 = 110 W m–1 K–1 ends of the composite rod are maintained at 403 K

SOLUTION Q = mLv Q t

–1

= 50

)

(2.26

2.26

103

–1

)

103 Js–1

Base area of boiler A = r2 = (0.1)2 = 10–2 m2 L = 3.14 cm = 3.14 10–2 m k A (T f T w ) Q = L t 3 50 2.26 10 = (1.13 102 )

(

10 2 )

3.14 10

(T f

Tw)

2

Tf – Tw = 1000°C Tf = 1000 + Tw = 1000 + 100 = 1100°C 19.3 A steel rod (L1 = 10 cm, A1 = 0.02 m2 and k1 = 50 J s–1 m–1 K–1) is welded to a silver rod (L2 = 20 cm, A2 = 0.01 m2, k2 = 400 J s–1 m–1 K–1 junction in the steady state.

Fig. 19.4

SOLUTION 1 1 (k1 + k2) = (50 + 110) 2 2 = 80 W m–1 K–1

keq =

k eq ( A1 A 2 ) (T1 T2 ) Q = 0.1 t 80 0.04 (403 273) = 0.1 = 4.16 103 Js–1 19.5 A metal cylinder of radius r and thermal conductivity k1 = 2k is surrounded by a cylindrical metallic shell of inner radius r and outer radius 2r conductivity k2 = k posite system are maintained at constant temperatures

19.4 Comprehensive Physics—JEE Advanced

T1 and T2 (T1 > T2). Find the equivalent thermal conductivity of the system.

Fig. 19.6 Fig. 19.5

SOLUTION L and cross-sectional area A1 = r2 of conductivity k1 and a rod of the same L and cross-sectional area A2 = [(2r)2 – r2] 2 = 3 r and conductivity k2 placed one on top of the other (in parallel). In the steady state Q1 Q2 Q = + t t t =

k1 A1 (T1

T2 )

k2 A 2 (T1

L

k1 ( r 2 ) (T1 Q = L t

Consider a small element of thickness dr at a distance r from the axis of the shell. Let dT be the temperature between the inner and outer surfaces of the element. If H 2 rL dr dr 2 kL dT = r H H=

r2

T2 )

r1

L T2 )

k2 (3 r 2 ) (T1

T2 )

L

ln

(i)

T2 )

(ii)

4 keq = k1 + 3 k2 keq =

k1

3k2 2k 3k 5k = = 4 4 4

SOLUTION

1

=

4 cm 2 cm

=

2 kL H

(T2 – T1)

2 3.14

.3 0.5 H

H = 3.14

19.4

(100 – 0)

.3 0.5 100 H 4 10 Js–1

CONVECTION

In convection heat is transferred by the physical movement of matter by molecules via collisions in their local regions.

19.5 19.6 A cylindrical metallic shell has inner radius r1 = 2 cm L = 50 and outer radius r2 T1 = 0°C and T2 –1 –1 K from the outer to the inner surface.

T

2 3.14

Cross-sectional area of the composite system is A = (2r)2 = 4 r2. If keq is the equivalent conductivity, then keq (4 r 2 ) (T1 Q = L t

dT

dr 2 kL 2 = dT r H T r2 r1

ln

k

RADIATION

All bodies emit heat from their surfaces at all temperaradiant heat or thermal radiation. (1) Black Body A perfect black body is one which

emitter of radiations. Consequently, a black body, when

19.5

black body radiation. (2) Emissive Power of a (3) Absorptive Power

e) of a body is area SI unit of e is Js–1 m–2 or Wm–2.

lo

e

T

T T0 ) T2 1

emissive power to the absorptive power is the same for all substances, i.e. e = constant a (5) Stefan’s Law by a unit area of a black body is proportional to the fourth power of its absolute temperature, i.e. E T 4 or E = T 4 where is a constant known as Stefan’s constant. Its value is = 5.735 10–8 Wm–2 K–4 When a black body at absolute temperature T is surround by another black body at a lower absolute temperature, T0,

=

t =K

T1 T2

e

T0 T0

ms k

K=

where by a completely absorbs all radiations, a = 1 for a black body. (4) Kirchhoff’s Law

T1 and T2, we have k t, ms

T1 to T2 when placed in a medium of temperature T0. An approximate formula is T1

T2

1 T1 T2 T0 t K 2 (7) Wien’s Displacement Law As the temperature of a black body increases, the maximum intensity of emission =

other words, mT

= b = constant

where m takes place at absolute temperature T constant b is b 10–3 mK (metre kelvin) 19.7 2

E=

4

(T –

T40)

a perfect black body, then E= (T 4 – T 40) where emissivity is always less than unity. power to that of the black body. (6) Newton’s Law of Cooling by a body is directly proportional to the excess of its temperature difference is small, i.e. dQ (T – T0) dt where T is the temperature of the body and T0 that of the m is the mass of the body, s heat and dT dt, then dQ = msdT dQ dT = ms = – k (T – T0) dt dt

If his skin temperature is 37°C, calculate the rate at 0.8 and Stefan’s constant

= 5.7

SOLUTION T = 273 + 37 = 310 K and T0 = 273 + 27 = 300 K (T4 – T04)

Rate of loss of heat = A = 0.8

2

10–8) [(310)4 – (300)4]

(5.7 102 W

= 1.03 19.8

constant b

10–3 mK.

SOLUTION

where – k

m

dT (T T0 ) =

k dt ms

10–8 W m–2 K–1.

T=

b m

=

T = b. 2.8

0

475.3 10

3

= 6080 K

19.6 Comprehensive Physics—JEE Advanced

If the temperature at the end of next 6 minutes is T, then

19.9 A body cools from 80°C to 50°C in 6 minutes in a room where the temperature is 20°C. What is the temperature of the body at the end of next 6 minutes?

e

50 20 T 20

=6K

30 T 20

e

(ii)

From (i) and (ii) SOLUTION e

80 50

20 20 e

e

=6K

(2)

30 = 2(T – 20)

(2) = 6 K

(i)

T = 35°C

I Multiple Choice Questions with Only One Choice Correct 1. their ends kept at the same temperatures 1 and 2 with 2 > 1. If A1 and A2 are their cross-sectional areas and k1 and k2 their thermal conductivities, same if A1 k1 = (a) A2 k2 (c)

A1 k1 = A2 k2

1

(b)

A1 k2 = A2 k1

(d)

A1 k2 = A2 k1

2

(a)

k1 = k2

1 s1

k1

3k2 4

(d)

k1 k2 k1 k2 3k1

2 s1

k1 1 = (d) k1 = k2 k2 2 4. A slab of stone of area 0.34 m2 and thickness 10 cm is exposed on the lower face to steam at 100°C. A block of ice at 0°C rests on the upmelted. Assume that the heat loss from the sides –1 . What is the thermal conductivity 3.4 105 of the stone in units of Js–1m–1°C–1?

(a) 1.0

(b) 1.5

(c) 2.0

(d) 2.5

5. face area A and

of the system is

(c)

1 s2

1

heat across the cylindrical surface and the system is

(b)

k1 = k2

(c)

2

2. A cylinder of radius R made of a material of thermal conductivity k, is surrounded by a cylindrical shell of inner radius R and outer radius 2R made of a material of thermal conductivity k2 two ends of the combined system are maintained

(a) k1 + k2

(b)

2 s2

k2 4

3. areas of cross-section have their ends kept at the same temperatures 1 and 2. If k1 and k2 are their thermal conductivities, 1 and 2 their densities and s1 and s2 of heat in the two rods will be the same if

(a) T = (c) T =

P

2

(b) T =

A P

P. If the emissivity is Stefan’s constant, the

1/ 2

P A P

1/ 4

(d) T = A A 6. What are the dimensions of Stefan’s constant? (a) ML–2 –2K –4 (b) ML–1 –2K–4 –3 –4 K (d) ML0 –3K–4 7. upon (a) the nature of its surface (b) the area of its surface

19.7

(c) the temperature of its surface (d) all the above factors

14.

8.

are joined end to end. But when they are joined depends upon (a) the nature of its surface (b) the area of its surface (c) the temperature of its surface (d) all the above factors

9. A composite slab consists of two slabs A and B of different materials but of the same thickness placed of A and B are k1 and k2 respectively. A steady temperature difference of 12°C is maintained across the composite slab. If k1 = k2/2, the temperature difference across slab A will be (a) 4°C (b) 8°C (c) 12°C (d) 16°C 10. l1 and l , radii r1 and r2 have thermal conductivities k1 and k2 at the same temperature difference. If l1 = 2l2 and r1 = r2 same if k1/k2 is (a) 1 (b) 2 (c) 4 (d) 8 11. A solid sphere and a hollow sphere of the same ature and allowed to cool in the same surround-

heat, in the same conditions, in (a) 24 s (b) 3 s

(b) 16 (d) 2

13. A body cools from 60°C to 50°C in 10 minutes. If of the body at the end of next 10 minutes will be (a) 38.5°C (b) 40°C (c) 42.85°C

(d) 45°C

(d) 48 s

(a) 2.5 s

(b) 10 s

(c) 20 s

(d) 5 s

15.

16. made from the same metal form the sides of an ABC B points A and B are maintained at temperatures T and 2 T respectively. In the steady state, the temperature of point C is TC heat conduction takes place, the ratio TC /T is 1 3 (a) (b) 2( 2 1) 2 1 1

(c)

3( 2

17.

T, then (a) the hollow sphere will cool at a faster rate for all values of T (b) the solid sphere will cool at a faster rate for all values of T (c) both spheres will cool at the same rate for all values of T (d) both spheres will cool at the same rate only for small values of T. 12. If the temperature of a black body increases from increases by a factor of 287 4 (a) 7 (c) 4

(c) 1.5 s

1)

(d)

1 2

1

S1 and S2 are made of the same S1 is three times that of S2. Both the spheres are heated

S1 to that of S2 is 1 (a) 3 (c)

3 1

(b) (d)

1 3 1 3

1/ 3

18. A spherical black body of radius 12 cm radiates 450 W power at 500 K. If the radius were halved and the temperature doubled, the power radiated in watt would be (a) 225 (b) 450

19.

19.8 Comprehensive Physics—JEE Advanced

U1, between

(a) 1 (c) 2/3

U2 and 1500 nm is U3 106 (a) U1 = 0 (c) U1 > U2

(b) 1/2 (d) 1/3

b = 2.88 (b) U3 = 0 (d) U2 > U1

20. black bodies at temperatures T1, T2 and T3 respecare such that Fig. 19.8

24. the same cross-section have been joined as shown Fig. 19.9 Fig. 19.7

(a) T1 > T2 > T3 (c) T2 > T3 > T1

the junction of the three rods will be (a) 45°C (b) 60°C (c) 30°C (d) 20°C

(b) T1 > T3 > T2 (d) T3 > T2 > T1

21. When the temperature of a black body increases, m to 0.13 body at the respective temperature is: 16 (a) 1 (c)

1 4

(c) 32

(b) it is the darkest body at all times

4 (b) 1 (d)

1 16

22. If the temperature of the sun were to increase from T to 2T and its radius from R to 2R, then the ratio was previously will be (a) 4

25. An ideal black-body at room temperature is thrown into a furnace. It is observed that (a) initially it is the darkest body and at later

(b) 16 (d) 64

23. K and 2K and thickness x and 4x, respectively, are T2 and T1 (T2 > T1 A T2 T1 K slab, in a steady state is f, with f x

(d) initially it is the darkest body and at later

26. fall of temperature (T ) of two bodies x and y same surface area, with time (t) due to emission of radiation. Find the correct relation between Fig. 19.10 emissive power (E) and absorptive power (a) of the two bodies. (a) Ex > Ey; ax < ay

(b) Ex < Ey; ax > ay

(c) Ex > Ey; ax > ay

(d) Ex < Ey; ax < ay

19.9

27.

temperature of the sun. Given Wien’s constant b = 2.88 10–3 mK. (a) 5000 K (b) 6000 K (c) 8000 K (b) 106 K

conductivity 420 W/m/K has one of its ends in area of cross-section is 10 cm2, the amount of ice that melts in 1 minute is

28. A body cools from 75°C to 65°C in 5 minutes in a ture of the body at the end of next 5 minutes will be (a) 55°C (b) 56°C (c) 57°C (b) 58°C 29. A liquid takes 6 minutes to cool from 80°C to 50°C.

35. the rate of 1400 Wm–2 the sun from the surface of the earth is 1.5 1011 m and the radius of the sun is 7.0 108 sun as a black body, it follows from the above data that the surface temperature of the sun is about (c) 6000 K

(d) 6100 K

36. rods are joined one after the other and this combi-

(a) 6 min. (b) 8 min. (c) 10 min. (d) 12 min. 30. A body initially at 80°C cools to 64°C in 5 minutes

rods are placed one on top of the other and connected to the same vessels. If q1 and q2 q1 is q2 2 (b) 1 1 (d) 8

the two cases, then the ratio (a) 15°C (b) 16°C (c) 20°C (d) 25°C 31. In Q. 30 above, the temperature of the body at the end of 15 minutes will be (a) 41°C (b) 43°C (c) 45°C (b) 47°C 32. of thermal expansion 1, 2 lii Y1, Y2

1 2 1 (c) 4

(a)

1 : 2 = 2 : 3, the thermal stresses developed in the two rods are equal provided Y1 : Y2 is equal to: (a) 2 : 3 (b) 1 : 1

33. A sphere, a cube and a thin circular plate have the same mass and are made of the same material. All

Fig. 19.11

37. (a) the for (b) the for (c) the the (d) the 34.

maximum for the sphere and minimum the plate. maximum for the sphere and minimum the cube. maximum for the plate and minimum for sphere. same for all the three.

A and B respectively are coated with carbon black on their tensity of emission of radiation are 300 nm and 500 them are in the ratio of (a)

5 3

(c)

5 3

(b) 2

(d)

5 3 5 3

4

19.10 Comprehensive Physics—JEE Advanced

38. transfer promarily due to radiation? 42. A layer of ice at 0 °C of thickness x1 pond of water. L, and k respectively are the latent heat of fusion of water, density of ice and thermal conductivity of ice. If the atmospheric temperature is – T °C, the time taken for the thickness of the layer of ice to increase from x1 to x2 L L (a) (x1 + x2)2 (b) (x2 – x1)2 2kT kT L 2 L 2 2 (c) (d) x2 x1 x2 x12 2kT kT 43. AB and CD L, cross-sectional area A and thermal conductivity k A, C and D are maintained at temperatures T1 = 20°C, T2 = 30°C and T3 temperature at B is (a) 32 °C (b) 33 °C (c) 34 °C (d) 35 °C

39. A spherical body of emissivity , placed inside a perfectly black body (emissivity = 1), is maintained at absolute temperature T a unit area of the body per second will be ( is Stefan’s constant) (b) T4 (a) T4 4 (c) (1 – ) T (d) (1 + ) T 4 40.

A and B made of the same

diameter of A is twice that of B, the ratio of rates of A and B will be (a) 1 : 1 (b) 2 : 1 (c) 4 : 1 (d) 1 : 4 41. A cubical ice box is made of thermocole of thick-

thermocole is 1.0 10–2 Wm–1 K–1 and latent heat –1 , the mass of of fusion of water is 3.35 105 ice left unmelted in 3 hours is very nearly equal to

Fig. 19.12

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43.

(b) (d) (c) (d) (a) (b) (c) (a)

2. 8. 14. 20. 26. 32. 38.

(c) (c) (d) (b) (c) (c) (d)

3. 9. 15. 21. 27. 33. 39.

(d) (b) (b) (d) (c) (c) (b)

4. 10. 16. 22. 28. 34. 40.

(a) (d) (b) (d) (c) (b) (c)

5. 11. 17. 23. 29. 35. 41.

(d) (c) (d) (c) (d) (a) (a)

6. 12. 18. 24. 30. 36. 42.

(d) (b) (d) (b) (b) (d) (c)

SOLUTION 1.

kA 2 Q 1 = l t Since ( 2 – 1) and l are the same for the two rods, Q/t will be the same if the product kA is the same for the two rods, i.e. if

k1A1 = k2A2 or 2.

A1 k2 = A2 k1

l and let its ends be maintained at temperatures 1 and 2. Area of

19.11

the cross-section of the inner cylinder = R2. Area of cross-section of outer cylinder = (2R)2 – R2 = 3 R 2. 2 Q1 = k1 R

1

2

k (4 R 2 ) l Q = Q1 + Q2

Q= Now

(i)

1

2

(iii) (iv) k = k1 + 3k2

k=

or

k1

where is the Stefan’s constant and T is the absolute temperature. E = 4 T Dimensions of =

l

k (3 R 2 ) 1 2 (ii) Q2 = 2 l Let the effective thermal conductivity of the compound cylinder be k across the compound cylinder is

3k2 4

dime

per unit area per second dimension of T 4

=

ML2

2

K

)

1

l is (d).

or Now

4.

=

=

mLd tA

2

or

–3

K–4

(

1

(

1

=

k2 A (

2)

l

– ) = k 2( – k2 = 2k1

2)

– ) = 2( – 2) 1 – 2 = 12°C or 2 = – ) = 2{ – (

3( 1 – ) = 24 or ce the correct choice is (b).

1 1

(i) 1 – 12 (ii)

– 12)} –

= 8°C

A and B are and

Q2 k r2 = 2 2 t l2

Q1 = Q2, if k1r12 k r2 k l = 2 2 or 1 = 1 l1 l2 k2 l2

perature T

r22 r12

=2

(2)2 = 8

11. upon its material, its surface area and its tempera-

A is P= A

1

Q1 k r2 = 1 1 t l1

T4

= ML0

4

10.

3.6 3.4 105 0.1 = 1 Js–1 m–1 °C–1 3600 0.34 100 0

=

k 1(

Also,

1

5.

1

9. Let 1 and 2 be the temperatures at the two faces of the composite slab and let be the temperature at the common face of the slab. If l of each slab and A the area of their face, then, in A= rate B, i.e. k1 A (

= mL, where m is the mass of ice melted and L its latent heat. Now time (t) = 1 hour = 3600 s and thickness d heat is mL Q= t Qd k= A 2 1

2

7. 8.

3.

6.

T4

E=

or T =

P A

1/ 4

at the same temperature, will cool at the same rate. 12. From Stefan’s law, E = T 4 E1 = (273 + 7)4

19.12 Comprehensive Physics—JEE Advanced

E2 =

E2 560 4 = E1 280 = (2)4 = 16

(273 + 287)4

13. Q = k(T – T0) t where k is a constant, T in time interval t and T0 is the temperature of the m is the mass of the body and s its is the temperature at the end of the next 10 minutes) ms 60 50 50 60 =k 25 and 10 min 2 ms 50 10 min

=k

Fig. 19.13

or

kA(TB TC ) l

2

25

14. Let Q be the heat transferred. If k is the thermal conductivity of each rod, their equivalent conductivity, when they are joined in series (end to end) is 2k. If t1 is time of transfer of heat, then 2k A l

3 2

tween the temperature of the body and that of the difference = 50 – 30 = 20° and in the second case the temperature difference = 40 – 30 = 10°. Since the temperature difference in the second case is half

TB > TA B to A and from B to C. In the steady state, rate B to C Q2 Q3 from C to A, i.e. t t

2 BC )

(

TB = T 2 )

d dt

r2 . Since ms

1

r3 , r

r 2)

T4

m1/3 m2 / 3 m 1

d dt

m1 / 3 d for S1 dt = d for S2 dt

1/ 3

m of S2 m of S1

1 3

=

1/ 3

18. P = (4 r2) T4 or P P2 = P1 or

r2 r1

2

P2 = 4P1 = 4

T2 T1

4

r2 T 4 1 2

2

2 1

4

=4

450 = 1800 W

19. From Wien’s displacement law

15.

16.

4 3

d = (4 dt

t1

equivalent conductivity is k/2. k A t2 2 Q2 = l Now Q1 = Q2 2k A t1 kA t2 = or t2 = 4t1 = 4 12 l 2l = 48 s

AC =

(

2l

17.

m=

Q1 =

T)

TC T

ms

50

kA(TC

m

T = b, the maxi-

b 2.88 106 nmK = 1000 nm T 2880 K U2 is the maximum. Since a black body U1 0 and

m =

U3 20. For black body radiations,

T = constant. It is > 2 3 > 1 follows that T1 > T3 > T2, which is choice (b). 21. T = constant, i.e. T2 0.26 1 =2 1 T1 = 2 T2 or T1 0.13 2 m

19.13

T2 = 2T1.

or

E1 = E1 E2

enters O

E = T4 T 41 and E2 = T 42 T14 T24

=

T1 T2

4

O, i.e. QA + QB = QC

4

1 2 (

or

1 = 16

kA

t

kA

l

l t) = t

t

T2 = 2T1)

or

t

kA t 0 l

3t = 180 or t = 60°C.

22. E=

T4

where R at absolute temperature T Q = 4 R2 T 4 If R and T are both doubled, we have Q = 4 (2R) 2 (2T ) 4 = 64 4 R2 T4 = 64 Q

Fig. 19.14

25.

23. Let A be the area of each slab. In the steady state, 26. T2

Q t

l1 K1 A

A (T2 T1 ) T1 = l1 l2 l2 K1 K 2 K2 A

(1)

Given l1 = x, l2 = 4x, K1 = K and K2 = 2K Q t

A (T2 T1 ) = x 4x K 2K

A (T2

T1 ) K x

1 3

1 3 24. Let A and l be the area of cross-section and the k conductivity and t°C the temperature of the junction O O from rods A and B kA t QA = l kA t and QB = l C is kA t 0 QC = l f=

of body x falls more rapidly with time than that of body y Ex > Ey ax > ay correct choice is (c). 27. Given k = 420 W/m/K, A = 10 cm2 = 10 10–4 m2 = 10–3 m2, 1 = 100°C, 2 = 0°C, t = 1 minute = 60 s and l k A( 1 2)t Q= l 2520 Q = 2520 J = = 600 cal. 4.2 m=

Q 600 cal = L 80 cal

28. We have e

or

e

T1 T0 T2 T0 75 25 65 25 5 e 4

= Kt =K = 5K

5 or

e

50 40

= 5K (1)

19.14 Comprehensive Physics—JEE Advanced

If the temperature at the end of next 5 minutes is T , we have 65 25 = 5K e T 25 or

e

e

or

T

40 25

= 5K

40 T 25

=

40 T 25

=

(2)

e

Stresses are equal for the two rods if Y1 1 t = Y2 2 t Y1 3 or = 2 = Y2 2 1

5 4

5 4

T = 57°C. 29. Given e

80 20 50 20

32. Let l heated so that the increase in temperature is t°C. lt where rod Strain = l t/l = t Stress = Y strain = Y t

= 6K

e

(2) = 6K

(1)

If the body takes t minutes to cool from 60°C to 30°C, then 60 20 = tK (2) e e (4) = tK 30 20

33. Since the material is the same the density is the same. Since the mass is the same and density is the same, all three have the same volume. For the same

heat by radiation is proportional to the surface area. 34.

m

T= lo t = lo 6 or

e e

4) 2)

lo 10 4) 0.602 lo 10 2) 0.301 t = 12 minutes.

b

2.88 10

T=b 3

= 6000K

480 10

m

2 35. Let R be the radius of the sun and let r be the radius of the earth’s orbit round the sun. If the surface temperature of the sun is T (in kelvin), the ener-

30. If T0 we have e

80 T0 64 T0

= 5K

(1)

4 R2 T 4, where is Stefan’s constant whose value is 5.67 10–8 Wm–2 K–4. Now the area of the spherical surface of radius r is 4 r2

e

64 T0 52 T0

= 5K

(2)

earth’s surface is 4 R2 T 4 4 r2

80 T0 64 T0 = 64 T0 52 T0

=

T4 =

R2 T 4 r2

= 1400

1400 r 2 R2

T0 = 16°C, which is choice (b). 31. If T is the temperature after 15 minutes, then 52 T0 = 5K e T T0 e

52 16 T 16

= 5K

=

1.5 1011

1400 5.67 10

= 1.1338 (3)

80 16 52 16 = T 16 64 16 T = 43°C, which is choice (b).

or

8

2

7 108

2

1015

T = 5803 K

36. If a steady temperature difference ( L and cross-sectional area A

1

–

2)

is

19.15

q=

k A(

1

2)

L

b4 A

=

4 m 2

m

where k

For a sphere of radius r, A = 4 r

the two rods are connected in series. If two rods L1 and L2 and conductivities k1 and k2 are joined in series, the equivalent conductivity ks L1

4

b

Q= A

Q= where k = 4

L2

L L2 = 1 (1) ks k1 k2 For two identical rods, L1 = L2 = L and k1 = k2 = k, in ks = k. Further, when two

and

Q1 = Q2

1 2 (2) 2L In the second case, the two rods are connected in

=

= k

r2

(3)

4 m

b4 r12 4 m 1

r22

Q2 = k

kA

k1 and k2 are joined in parallel, the equivalent condu-

4 m

Q1 = k

composite rod is (2L) but its cross-sectional area is A q1 =

b4 4 r 2

4 m 2

r1 r2

4 m 2 4 m 1

2

3 cm 5cm

2

500 nm 300 nm

4

=

5 3

2

38.

kp = k1 + k2 For two identical rods, k1 = k2 = k kp = (2k). Furthermore, the cross-sectional area of the composite rod is (2A q2 =

2k 2 A L

1

2

due to conduction or convection. 39. body of emissivity

(3)

q1 1 = . Now, the rate q2 8 37. black body at absolute temperature T (1) Q = AT 4 where A is the surface area of the body and is ment law, m T = b

b m

T 4, irrespective of the

40. proportional to the surface area. 41. 0.30 m. Since a cube has 6 faces, the total surface area of the cube exposed to air is A = 6 (0.3)2 = 0.54 m2. of thermacole (d) = 5.4 cm = 5.4 10–2 m of exposure (t) = 3 hours = 3 60 60 = 1.08 104 s –2 conductivity (k) = 1.0 10 J s–1 m–1°C–1

where m maximum emission of radiation and b is Wien’s T=

is

Ta) = 40°C Ti)= 0°C t is

(2) Q=

kA(Ta Ti )t d

19.16 Comprehensive Physics—JEE Advanced

=

1.0 10

2

0.54

5.4 10

= 4.32

0) 1.08 104

(40

kATdt L = A Ldx or dt = · xdx x kT

2

104 J

Now, heat of fusion of water, L = 3.35 –1 = 3.35 105

–1 102 6 10 J of

x

t

L 2 dt = xdx kT x

0

Q m= L

4.32 104

t=

or

3.35 105

L x2 kT 2

x2

L (x22 – x21 ), 2kT

= x1

which is choice (c). 43. Let TB be the temperature at B of heat from C towards B is Q1 k A (T2 TB ) = t L/2

42. When the temperature of the air is less than 0°C, the cold air near the surface of the pond takes heat

x be the thickness of the ice layer at a certain time. If the thickness increases by dx in time dt, then the dt

k A (T3 Q2 = t L/2

D towards B is TB ) B towards A is

Q3 k A (TB T1 ) = L t In the steady state, the rate at which heat enters B = rate at which heat leaves B, i.e. Q Q1 Q2 = 3 t t t

( T )]dt kATdt = (1) x x where A is the area of the layer of ice and – T °C is dm is the Q = dm L. But dm = A dx, where Q=

1

kA[0

or

2 k A (T2 L

2 k A (T3 L

TB )

TB )

k A (TB T1 ) L or 2(T2 – TB) + 2(T3 – TB) = TB – T1 =

TB = =

Fig. 19.15

Q = A Ldx

(2)

T1

2 T2 5

2 T3

20 2 30 2 40 5

= 32 °C, which is choice (a).

II Multiple Choice Questions with One or More Choices Correct 1. (a) All bodies emit thermal radiations at all temperatures

mirror a velocity of 3

108 ms–1

19.17

2. depends upon (a) the nature of its surface (b) the area of its surface (c) the temperature of its surface

body increases.

7. A composite block is made of slabs A, B, C, D and E

3. does not depend upon (a) the density of the body (b) the nature of its surface (c) the area of its surface (d) the temperature of its surface 4. mum emission are T1 and T2 respectively. If Wien’s constant is b 10–3 mK, then (b) T1 (a) T1 = 3718 K (c) T2 = 2318 K (d) T2 = 4677 K 5. A and B of equal surface area are placed one on top of the other to form a composite plate of A and B are

a constant K

L)

Q A and E slabs are same. E is maximum. (c) temperature difference across slab E is smallest. C B D.

the exposed surface of plate A is – 10 °C and that of the exposed surface of plate B the temperature of the contact surface is T1 if the plates A and B are made of the same material and T2 if their thermal conductivities are in the ratio 2 : 3 then (b) T1 = – 2°C (a) T1 = – 4°C (d) T2 = 0°C (c) T2 = – 3°C 6. Initially a black body at absolute temperature T is kept inside a closed chamber at absolute temperature T0 allow sun rays to enter. It is observed that temperatures T and T0 remains constant. Which of the

Fig. 19.16

8. Initially a black body at absolute temperature T is kept inside a closed chamber at absolute temperature T0 allow sun rays to enter. It is observed that temperatures T and T0 remains constant. Which of the

body remains the same. body increases. body increases.

body remains the same. body increases.

ANSWERS AND SOLUTIONS 1. 2. All the four choices are correct.

3. From Wien’s law, and m

m

T = b, where b is a constant m

temperature of the body.

depends only on T, the

19.18 Comprehensive Physics—JEE Advanced

4. From Wien’s law,

max

the thermal resistances are connected as shown

T=b 3

T1 =

7.8 10 .

T2 =

10

7

= 3718 K

3

= 4677 K 6.2 10 7 So the correct choices are (a) and (d). 5. Let T be the temperature of the contact surface. k A( 10 T ) k A(T 10) Q = A = B 4.0 6.0 t or

k A ( 10 T ) k (T 10) = B 4 6

(a) If kA = kB (b) If

(1)

T1 = – 2°C.

kA 2 = kB 3

Fig. 19.17

Reff 1 1 = Reff RB

T2 = 0°C.

6.

=

4

from a black body is proportional to (T – T0) . Since T and T0 remain constant, the rate of emission of

Reff =

rect choices are (a) and (d). 7. Let W L, area A (= LW) and thermal conductivity K L L = R= KA K ( LW ) A, B, C, D and E respectively are RA = where

L R = (2 K )(4 LW ) 8

3 4R

1 RC 2 R

1 RD 5 4 = 4R R

R 4 QA = (Q)eff = Q .

T , where T is the temperature R difference between the ends of the slab. Also the temperature difference between the ends of slabs B, C and D is the same = ( T)eff Q R ( T)A = QARA = A 8 Q R ( T)B = ( T)C = ( T)D = QAReff = A 4 Q R and ( T) = A ( Q = Q A) 24 E is the minimum. Q=

R=

L 1 = K ( LW ) KW

RB =

4L 4R = 3K ( LW ) 3

RC =

4L R = 4 K (2 LW ) 2

Now QB =

RD =

4L 4R = 5K ( LW ) 5

QC =

( T )C Q R/4 1 = A = QA R/2 RC 2

R =

L R = 6 K (4 LW ) 24

QD =

( T )D Q R/4 5 = A = QA RD 4R /5 16

Since slabs B, C and D are in parallel and slabs A and E are in series with this parallel combination,

( T )B Q R/4 3 = A = QA RB 4R /3 16

And Q = QA

B is

19.19

Now QB + QD =

T and T0 remain constant, the rate of emission of

3 5 1 QA + QA = QA which 16 16 2

-

is equal to QC the correct choices are (a), (c) and (d).

rect choices are (a) and (d).

8. 4

from a black body is proportional to (T – T0) . Since

III Multiple Choice Questions Based on Passage Questions 1 to 6 are based on the following passage Passage I Thermal Radiations

is transferred to us by a process which, unlike conduction or convection, does not require the help of a medium in the

1.

-

fan’s constant are (b) ML–1 –2K–4 (a) ML–2 –2K–4 –3 –4 K (d) ML0 –3K–4 2. What is the SI unit of Stefan’s constant? (b) W m–1K–4 (a) J s–1K–4 –2 –4 (c) W m K (d) J m–2K–4 3. thermal radiations lie? 4. thermal radiations? (a) Constant volume air thermometer (b) Platinum resistance thermometer

in terms of that emitted from a reference body (called the black body) at the same temperature. A black body E emitted by a unit area of a black body E = T4 where T is the absolute temperature of the body and is a constant known as Stefan’s constant. If the body is not a perfect black body, then E= T4 where is the emissivity of the body.

5. When a body A T1 is surrounded by another body B at a lower temperature T2, then the rate of loss of heat from body A will be proportional to (b) (T1 – T2)4 (a) T 41 (c) (T1 – T2) (d) (T 41 – T 42) 6. pends upon (a) the surface area of the body (b) the temperature of the body (c) the nature of the surface of the body (d) the emissivity of the surface of the body

ANSWERS AND SOLUTIONS 1.

E=

T4 where E

E = dimensions of 2

area

time

Dimensions of choice (d).

=

ML

2

–3

2

=

ML 3 K4

–3

K–4, which is

2. 3. 4. 5. 6. All the four choices are correct.

19.20 Comprehensive Physics—JEE Advanced

Questions 7 to 13 are based on the following passage

9. revealed that the atmospheres of stars contain

Passage II Stellar Spectra Like the solar spectrum, the spectra of stars show a continuous spectrum on which dark aborption lines are photosphere) of

10.

(c) uranium

(d) helium

(a) surface temperature (b) mass

the outer, relatively cooler, layer of the star, the radiations

(d) all the above factors 11. Wien’s displacement law tells us that an extremely hot star should look

in the outer layer of the star. T of a star can be estimated by m at which the intensity of

12. In Wien’s displacement law the SI unit of Wien’s constant b is (a) metre per kelvin (b) metre per kelvin squared (c) metre kelvin (d) metre kelvin squared 13. E of black body radiations where

displacement law which states that m T=b where b is a constant called Wien’s constant and the above relation is called Wien’s Displacement Law which states that as the temperature increases, the maximum intensity of emission shifts (or is displaced) towards the shorter b has been found 10–3 mK. 7. (a) continuous emission spectrum (b) emission line spectrum (c) emission band spectrum (d) absorption line spectrum 8.

the outer layers of the sun

(d) destructive interference between waves of Fig. 19.18

SOLUTION 7. 8. 9. 10. 11.

-

m

T = constant, if T

m

12. From m T = b, the SI unit of b = SI unit of m unit of T = metre kelvin, which is choice (c).

SI

13.

Questions 14 to 16 are based on the following passage Passage III A, B and C equal diameters are joined in series as shown in the folk, k and 0.5 k

Fig. 19.19

In the steady state, the free ends of rods A and C are at curved surfaces of rods.

19.21

14.

A and B is (a) 55.7°C (c) 75.7°C

16.

(b) 65.7°C (d) 85.7°C

15.

B and C is (a) 57.1°C (c) 37.1°C

nation is 7k (a) 3 (c)

(b) 47.1°C (d) 27.1°C

5k 3

(b)

2k 7

(d)

3k 5

SOLUTION 14.

T1 = 85.7°C. So the corsame for all rods. If T1 and T2 are the temperatures at the junction points between A and B and between and C respectively, then k A(100 T1 ) Q = A d t = Given

rect choice is (d). 15. 16.

T1 T2 = 57.1°C, which is choice (a). diameters, the equivalent thermal conductivity of

k A T2 0 k B A T1 T2 = C d d kA = 2k, kB = k and kC = 0.5k

1 1 = ke kA

1 kB

2(100 – T1)= (T1 – T2) = 0.5(T2 – 0) T1 = T1 – T2 and

ke =

(1)

T1 – T2 = 0.5 T2

1 1 = kC 2k

1 k

1 0.5k

2k . So the correct choice is (b). 7

(2)

Questions 17 to 19 are based on the following passage Passage IV

17. (a) 26.5°C (c) 24.5°C

(b) 25.5°C (d) 23.5°C

(a) 2.5°C (c) 1.5°C

(b) 2.0°C (d) 0.5°C

18. area 1 m2 and thickness 0.01 m separated by 0.05 m thick 19. –1

–1

K .

is nearly equal to (a) 1000 J s–1 (c) 3000 J s–1

(b) 2000 J s–1 (d) 4000 J s–1

SOLUTION 17.

T2 and T3 be the temperature k g A(T1 T2 ) dQ1 0.8 1 (300 T2 ) = = dt 0.01 dg

(1)

k A(T2 T3 ) 0.08 1 (T2 dQ2 = a = dt da 0.05

(2)

T3 )

k g A(T3 T4 ) dQ3 0.8 1(T3 273) = = dt 0.01 dg

(3)

In the steady, dQ3 dQ1 dQ2 = = dt dt dt Fig. 19.20

50

(300 – T2) = T2 – T3

(4)

19.22 Comprehensive Physics—JEE Advanced

19. (5) (300 – T2) = T3 – 273 26.5°C. T3 So the correct choice is (a). 18. T3 T2 = 273.5 K = 0.5°C, which is choice (d).

T2 = 273.5 K and T3 dQ1 dt

2000 J s–1, which is choice (b).

IV Assertion-Reason Type Questions Statement 2

correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1. (b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Statement-1. (c) Statement-1 is true, Statement-2 is false. (d) Statement-1is false, Statement-2 is true. 1. Statement 1 tion. Statement 2 e) and absorptive power (a temperature and for radiation of the same wave-

convection. 4. Statement 1 uid. Statement 2

5. Statement 1 netic waves. Statement 2 6. Statement 1 and 4 m and termperatures 4000 K and 1000 K

2. Statement 1

two spheres will be the same. Statement 2

same (room) temperature. Statement 2 tivity than plastic. 3. Statement 1 If the earth did not have an atmosphere, it would become extremely cold.

is directly proportional to its absolute temperature and inversely proportional to its surface area.

SOLUTION Conversely, if a body is a poor emitter of a radia-

1. law,

e

e = constant a a

2. -

temperature is lower than our body temperature. Since metal is a better conductor of heat than plastic, when we touch the metal cap and the plastic

19.23

4. 5. 6.

the metal cap much more quickly than to the plastic body. 3.

E = T 4A = T 4

4 R2; R = radius of sphere

is a poor conductor of heat, the atmosphere acts as a blanket for the earth and keeps the earth warn

V Integer Answer Type 1.

A is 3 times that of B. In the steady state, the temperature difference across the wall is 36°C. Find the temperature difference (in °C) across the layer A.

rod, end to end, with a second rod of a different material but of the same cross-section. At 25°C, the

10–5 per °C and that of the second rod is 10–5 per °C. Find the value of n.

= 1.7 =n

2. A wall has two layers A and B each made of different material. Both layers have the same thickness.

3.

A and B have thermal emissivities of of the two bodies are the same. If they emit total raratio TA/TB.

SOLUTIONS Given

1. = 30 (1 + 1.7 = 30.051 cm

10–5

T1 – T2 = 36

T2 = T1 – 36 T1 – T0

100)

100 = 7000

cm =2

cm = 10–5 per °C.

n = 2.

Fig. 19.21

2. layers A and B is the same, i.e. (a = area of crosskA a (T1 T0 ) k a(T0 T2 ) = B x x kA = 3 kB 3(T1 – T0) = T0 – T2

3. E1 = eA A E2 = eB Given E1 = E2 TA4,

TA e = B TB TA (1)

A TB4

1/4

=

0.81 0.01

1/4

=3

(2)

Electrostatic Field and Potential 20.1

20

Electrostatic Field and Potential

Chapter

REVIEW OF BASIC CONCEPTS 20.1

COULOMB’S LAW

On the basis of his measurements, Coulomb arrived at a law, known after his name as Coulomb’s law, which states that the magnitude of the electric force between two charges is directly proportional to the product of the magnitudes of charges and inversely proportional to the square of the distance between them, i.e. F or

q1 q2

F12 =

k=

4

0

0

r2

q1 q2 n

F21 = –

4

0

r2

9

109 Nm2 C–2

The force F is attractive for unlike charges (q1 q2 < 0) and repulsive for like charges (q1 q2 > 0). Coulomb’s Law in vector form

F21 =

20.2

4

0

r2

q1 q2 n 4

0

r2

RELATIVE PERMITTIVITY (OR DIELECTRIC CONSTANT)

the permittivity of the medium to permittivity of vacuum, i.e. r r

= 0

is also called the dielectric constant (K) of the medium.

Thus K =

or 0

Fig. 20.1

q1 q2 n

F12 = –

q1 q2

1

4

where n is a unit vector directed from q1 to q2. Force exerted by q1 on q2 is

r2 In the SI system, k is written as 1/4 0 where 0 is called the permittivity of vacuum and its value is 10–12 C 2 N–1 m–2 0 = 8.854 Then

q1 q2 n

Case (b): Like charges (q1 q2 > 0) [Fig. 20.1(b)]

r2

F= k

Case (a): Unlike charges (q1 q2 < 0) [Fig. 20.1(a)] Force exerted on q2 by q1 is

=K

0

K for air = 1. If

charges q1 and q2 are situated in a medium other than air or vacuum, the magnitude of force between them is given by q1 q2 q1 q2 F= 2 4 r 4 0 Kr 2

20.2 Comprehensive Physics—JEE Advanced

20.3

PRINCIPLE OF SUPERPOSITION

If many charges are present, the total force on a given charge is equal to the vector sum of the individual forces exerted on it by all other charges taken one at a time. 20.1 Two point charges q1 = +9 C and q2 = –1 C are held 10 cm apart. Where should at third charge +Q be placed from q2 on the line joining them so that charge Q does not experience any net force? SOLUTION Charge Q will not experience any net force if the forces exerted on it by charges q1 and q2 are equal and in opposite directions.

Fig. 20.2

It follows from Fig. 20.2 that charge Q will not experience forces in opposite direction if it lies at any point between AB. Let x be the distance of Q from q2. Then forces exerted on Q by q1 and q2 respectively are F1 =

and

q1 Q i 4

F2 = –

0

(0.1

q2 Q i 4

0

x

x)

2

=

4

0 6

1 10 2

4

6

9 10

0

9 10 4 9=

0

(0.1

(0.1

Qi x)

2

Fig. 20.3

Equilibrium of charge Q at O [See Fig. 20.3] Since Q is at the same distance from equal charges q1 and q2, it will be equilibrium for any positive or negative value, because it will experience equal and opposite forces. Equilibrium of charge q1 at A If Q is negative, it will repel q1. Also q2 is repel q1. Hence q1 cannot be in equilibrium if Q is negative. So Q must be positive. Force exerted on q1 by Q is F=

(0.1

x)

2

Qi x2

F =–

1 10 4

0

Qi x2

=0

x) 2

x2

0.1 x 3= x x = 0.05 m = 5 cm 20.2 Two charges, each equal to – 4 C, are held a certain distance apart. A charge Q is placed exactly mid-way between them. Find the magnitude and sign of Q so that the system of three charges is in equilibrium.

Qi

r 2

2

4 10

6

4 10

6

i

2

4 0r Net force on q1 will be zero if F + F = 0, i.e. if

4 –

0

6

Force exerted on q1 by q2 is

4 10

6

4 10 4

Qi

Net force on Q = F1 + F2 Net force on Q = 0 if F1 + F2 = 0 6

SOLUTION A system of charges is in equilibrium if no charge of the system experiences any net force.

0

6

Qi

r 2

2

–

Q= 1

4 10

6

4

4 10 0

r

2

6

i

=0

10–6 C = 1 C

It is easy to check that charge q2 will also be in equilibrium. Hence the system of three charges will be in equilibrium if Q = +1 C. 20.3 Four point charges, each equal to q = 4 C, are held at the corners of a square ABCD of side a = 10 cm. Find the magnitude and sign of a charge Q placed at the centre of the square so that the system of charges is in equilibrium. SOLUTION AC (= r) = 2a . Let us consider the equilibrium of charge q at A (Fig. 20.4)

Electrostatic Field and Potential 20.3

kq

i.e. if

a

q

q

2

2 2

2Q

=0

q Q = – (1 2 2 ) 4 =–

20.4

Force exerted on charge at A by charge at B is FAB =

kq a

i,

2

force F

1

where k =

4

FAD = FAC = = and

FOA =

=

kq

o

kq 2 (a 2 )2 kq 2 a2 2 2 kqQ a 2 2

(cos 45 i (i

sin 45 j)

(cos 45 i 2

sin 45 j)

k qQ

Fx = where

kq 2 a kq a2

=

kq

Fy =

j

2

kq 2

2 k qQ

2

a2

a 2 2 q

q 2 2

(1)

Electric r from a source charge q is given by E=

j)

( i j) a2 Net force on charge q at A in the x-direction is Fx =

F q0

If a charge q due to other charge or charges is E, then the charge q will experience a force F given by F = qE

j

a2

E is then given by E=

Similarly 2

ELECTRIC FIELD

positive point charge q0 at the point in space where the

Fig. 20.4

2

4 C (1 2 2 ) = – (1 2 2 ) C 4

2Q i

i i

1 4

q 0

r2

For a positive charge (+q), vector E is directed radially outwards from it and for a negative charge (–q), E is E is individual charges. (2) A pair of equal and opposite point charges separated by a certain distance is called an electric dipole. Let 2a be the separation between point charges –q and +q (Fig. 20.5).

q

2Q 2 2 a Similarly net force on charge q at A in the y-direction is 2

q

Fig. 20.5

Resultant force on charge q at A is F=

Fx2

Fy2

2

P due to +q and –q respectively are 2

Charge q will be in equilibrium if F = 0 i.e. if =0

2 E+ =

qi 4

0 (r

a)2

20.4 Comprehensive Physics—JEE Advanced

NOTE

qi

E– = –

0 (r

4

a)2

line of a dipole is along the dipole moment.

P is Ea = E+ + E–

torial line of a dipole is antiparallel to the dipole moment.

2q (2a)r

=

4

0 (r

2

a 2 )2

2p r

=

4

0 (r

2

a 2 )2

where p = q(2a) is the dipole moment and 2a is the vector distance between charges –q and +q. Dipole moment p is a vector quantity directed from –q to +q. For a very short dipole (a << r)

A conducting rod AB of negligible thickness and length L carries a charge a charge Q uniformly distributed P at a distance a from end B (Fig. 20.7), we consider a small element of length dx of the rod located at a distance x from P.

2p

Ea =

4

0r

3

Fig. 20.7

bisector (equatorial plane) of a dipole

Q dx = L

Charge of element is dq =

dx where

=

Q is L P due

to the element is dE =

dq 4

dx

0x

2

4

0x (L

Hence

=

Fig. 20.6

Q due to +q and –q are (see Fig.

20.6) E+ =

q 4

0x

and

2

E– =

4

0x

2

Ee = E+ cos + E– cos a and x = r 2 a 2 , we get x q ( 2a ) directed from Q to R Ee = 4 0 (r 2 a 2 )3 /2 In vector form p Ee = – 4 0 (r 2 a 2 )3 /2

For a very short dipole (a << r) Ee = –

p 4

0r

3

dE

4

E=

4

0

1 (L xa

0

a)

a

4 4

0

0

(L

dx x2

a)

1

E=

q Q is

Using cos =

E=

2

a)

1 a

L a( L a)

If Q is positive, E is directed from left to right. Consider a ring of radius R carrying a chare Q distributed uniformly on it. P on its axis at a distance x from the centre O, consider an element of length dl (Fig. 20.8). The charge of the element is Q dl dq = 2 R A is dE given by dq dE = 4 0r 2

Electrostatic Field and Potential 20.5

Fig. 20.9

Properties of Electric Field Lines Fig. 20.8

P due to element a diametrically opposite point B. The x components of y components cancel. Hence E=

E=

(iv) Field lines are closer together in the region where

dE cos on a charge is proportional to the magnitude of the charge.

dq cos

= =

(ii) Field lines originate from a positive charge and terminate on a negative charge.

4

0r

Q 2 R

2

1 4

1 4

x (R

0

2

2 3 /2

x )

dl

a closed loop. But if the electric is induced by a time-

Qx 0

(R2

x 2 )3 /2 dl

NOTE

2 R

20.6

ELECTRIC FLUX

The direction of E is from O to P if charge Q is positive.

20.5

ELECTRIC FIELD LINES

the tangent to which at a point gives the direction of the charge distributions.

through the surface. For a plane surface of surface area S E = E S = ES cos where S is called the area vector, its magnitude is S and its direction is normal to the surface and away from it. Angle is the angle between E and S. For a curved surface, =

E dS

(E n) dS

where n is a unit outward normal to the surface. dS is the surface area of an element of Fig. 20.10 the surface (Fig. 20.10). The SI –1 2 m or Vm (volt metre).

20.7

GAUSS’S LAW IN ELECTROSTATICS

surface S

E is equal to

the net charge enclosed in the surface and permittivity of vacuum.

q 0

0

, where q is is electrical

20.6 Comprehensive Physics—JEE Advanced

q

E dS =

(1)

0

S

Gauss’s law is used to obtain the expression for the electric

r from a spherical shell of radius R carrying a surface charge density (= q/4 R2) is given by

which are uniform and symmetric so that a proper and convenient closed surface (called the Gaussian surface) can be chosen to evaluate the surface integral in Eq. (1). Some Important Points about Gauss’s Law (1) Gauss’s law holds for any closed surface of any shape or size. (2) The surface that we choose to evaluate electric (1)] is called Gaussian surface. (3) If the Gaussian surface is so chosen that there are some charges outside and some inside the surface, then charge q on the R.H.S. of Eq. (1) is the net charge (taking into account the sign of charges) E on the charges both inside and outside the surface. (4) The exact location of charges inside Gaussian surface does not affect the value of the electric (5) If Coulomb’s law did not hold, Gauss’s law also would not hold.

20.8

APPLICATIONS OF GAUSS’S LAW

dicular distance r or wire carrying a uniform linear charge density given by E= where

is

n 2

0r

= q /L is the charge per unit length of the rod

and n is a unit vector pointing away from the rod if q is positive and towords it if q is negative. Electric r is given by E=

n 2

0

where = q / A is charge per unit area and n is a unit vector pointing away from the sheet if q is positive and towards it if q is negative. Notice that E is independent of r, the distance from the sheet.

E=

qn 4

0r

n

=

2

=

R2 n 0r

2

(for r > R)

(for r = R)

0

= zero (for r < R) where n is a unit vector pointing radially outwards if q is positive and inwards if q is negative.

20.9

ELECTRIC POTENTIAL

is the work per unit charge that is done to bring a small V = lim q0

W 0 q0

Electric potential at a point P charge is given by 1 q 4 0 r where r is the distance of the point P from the charge. This potential is spherically symmetric around the point, i.e. it depends only on r for a given charge q. Since potential is a scalar function, the spherical symmetry means that the potential at a point does not depend upon the direction of that point with respect to the point charge; it only depends on the distance of the point from the charge. Notice that the potential due to a positive charge (q > 0) is positive, it is negative in the neighbourhood of an isolated negative charge (q < 0). V=

to each charge, assuming that all other charges are absent, and then simply add these individual contributions. Since, addition here is the ordinary sum, not a vector sum. The potential at any point due to two point charges q1 and q2 is, therefore, simply given by 1

q1 q2 4 0 r1 r2 where r1 and r2 are the distances of the point in the question from charges q1 and q2 respectively. V=

Electrostatic Field and Potential 20.7

The potential at any point due to a system of N point charges in given by V = V1 + V2 +

20.10

+ VN =

1 4

N 0 n 1

qn rn

charge on an electron is 1.6 10–19 C, 1 eV = 1.6 10–19 J

20.13

POTENTIAL ENERGY OF AN ELECTRIC DIPOLE IN AN EXTERNAL ELECTRIC FIELD

RELATION BETWEEN E AND V

means that the potential decreases along the direction of E =–

20.11

dV dr

ELECTRIC POTENTIAL ENERGY

The electric potential energy of a system of point charges Fig. 20.12

We assume that the charges were at rest when kinetic energy. The electric potential energy of two point charges q1 and q2 separated by a distance r12 as shown in Fig. 20.11 (a) is given by 1 q1 q2 U12 = 4 0 r12

E, as shown in Fig. 20.12, it experiences a torque given by = p E sin where

is the angle between the line joining the two =p

The torque tends to rotate the dipole to a position where = 0, i.e, p is parallel to E. The electric potential energy of a dipole is U =– p E

20.14

Fig. 20.11

The electric potential energy of a system of three point charges as shown in Fig. 20.11 (b) is given by U = U12 + U23 + U31 =

1 4

0

q1 q2 r12

q2 q3 r23

E

ADDITIONAL USEFUL FORMULAE

(i) Charge q at each vertex of an equilateral triangle of side a (Fig. 20.13).

q1 q3 r13

This expression can be generalized for any number of charges.

20.12

THE ELECTRON-VOLT

The SI unit of potential energy is the joule. In atomic physics a more convenient unit called the electron-volt (written as eV) is used. An electron-volt is the potential energy gained or lost by an electron in moving through Since the magnitude of

Fig. 20.13

At centroid O, E0 = 0 and V0 = where r =

a 3

.

3q 4

0r

20.8 Comprehensive Physics—JEE Advanced

(ii) Charge q at each vertex of a square of side a (Fig. 20.14).

q

=

4

0r

1 4

1

2

1

q

=

4

1 16

0r

2

q

1

1 4

1 2

1 4

3

0r

2

Potential at O is q

V0 = Fig. 20.14

At center C, Ec = 0 and Vc =

4q 4

0r

; r

4 4

In the above two cases, if one of the charges is O and C is E = q/4 or2, directed towards the empty vertex.

1 0r

2

NOTE

1

q

=

a

0r

1

q 1 2

2

0r

(vi) A short electric dipole of dipole moment p (Fig. 20.17)

(iii) For Fig. 20.15,

Fig. 20.17 Fig. 20.15

Vc = 0

and

Ec =

2 2q 4

0r

2 p cos

At point A, Er =

4

2

0r

3

= Also tan = (

+ ).

p 4

3 0r

q, placed on the x-axis at x = r, x = 2r, x = 4r… and so O is q 1 1 1 E0 = 2 2 2 4 0r 1 2 42

2p 4

= 90°)

Electric potential at A is VA = At point B; VB =

p 4

At point C; VC = 0

0r

2

0r

3

0x

3

and EA is (

At point C on equatorial line, EC = (

4

E2

1 tan . Angle between 2

Fig. 20.16

p sin

1 1 /2

3 cos 2

At point B on axial line, EB =

Ec = 0

Er2

A is EA =

(iv) For Fig. 20.16,

Vc = 0

and E =

p cos 4

0r

2

= 0°) p 4

0y

3

Electrostatic Field and Potential 20.9

=

1

Q

At point P, EP =

Q ) L (i) Charge Q distributed uniformly on a rod of length L (Fig. 20.18) (Linear charge density

4

VP =

r (r

0

Q 4

0L

L) r

log e

L r

(ii) Charge Q distributed uniformly on a conducting sphere or shell of radius R (Fig. 20.19) (Surface Q charge density = ) 4 R2 Fig. 20.18

Fig. 20.19

At point C outside the sphere or shell (r > R), EC =

Q 4

0r

At centre O, V0 =

2

Q 4

0

R

4

0

At point B just outside the surface (r = R), EB =

Q 4

0

R2

At point A inside the sphere or shell (r < R), EA = 0 Potential at center O of sphere or shell, Q V0 = 4 0R Q At points inside (r < R), VA = = VB (at 4 0R surface) Q 4 0R (iii) Charge Q distributed uniformly on a semicircular wire of radius R (Fig. 20.20) (Linear charge Q density = ) R

Fig. 20.20

(iv) Charge Q distributed uniformly on a ring of radius R P on the axis is

At point C outside (r > R), VC =

At center O, E0 = 4 tion =

0R

along negative y-direc-

Q 4

2

0

R2

Fig. 20.21

EP =

Q 4

r 0

R2

r2

3 /2

20.10 Comprehensive Physics—JEE Advanced

R

E is maximum at r = Emax =

1 4

2

(ii) Charge Q kept at each vertex of a square of side a (Fig. 20.23)

and

2Q 0

3 3R 2

U=

Electric potential at point P is Q 1 VP = 4 0 R2 r 2

=

1 /2

(i) Charge Q kept at each vertex of an equilateral of side a. Potential energy is (Fig. 20.22). 3Q 2 4

0

a

Q2 4

0a

2Q 2 4 (4

0 (a

2)

2)

Fig. 20.23

(iii) An electric dipole of dipole moment p placed in E with angle between p and and potential energy U = – p.E. E. Torque = p The zero of potential energy is taken at = 90°. (a) When = 0°, = 0, U = minimum = –pE (stable equilibrium) (b) When = 90°, = maximum = pE, U = 0 (c) When = 180°, = 0, U = maximum = pE (unstable equilibrium) (d) Work done in turning a dipole from angle 1 to angle 2 is W = pE (cos 1 – cos 2)

V is maximum at r = 0 (i.e. at centre O) and Q Vmax = 4 0R

U=

4 Q2 4 0a

Fig. 20.22

If

= 0° and

1

2

= 180°, W = 2pE

I Multiple Choice Questions with Only One Choice Correct 1. Three point charges Q, – 2Q and – 2Q are placed at the vertices of an equilateral triangle of side r. The work done to increase their separation to 2 r is (a) zero (c)

(b)

2 Q2 4

0

r

(d)

Q2 4

0

(c) r

2 Q2 4 0r

2. A point charge Q is placed at point P at a distance R from the centre O of a metallic spherical shell of inner radius 2R and outer radius 2.5 R. The electric potential at the centre of the shell will be (a)

Q 4

(c) zero

0

R

(b) (d)

1 4

0

5Q 6R

0

9Q 10 R

1 4

(a)

3. Two concentric metallic shells of radii R and 2R are given charges Q and 2Q respectively. If the two shells are connected by a metallic wire, the change in electric potential on the outer shell is

Q 4 4

0

R

3Q 0 (2 R)

(b)

4

Q 0 (2 R)

(d) zero

4. A metal sphere of radius R carries a charge Q. The E and the electric potential is V. If R is doubled keeping Q the same, the new values of E and V will be (a)

E V and 4 2

(b)

E V and 2 4

(c) 4E and 2V (d) 2E and 4V 5. A metal sphere of radius R has surface charge density E and the electric potential is V. If R is halved, keeping the same, the new values of E and V will be (a) 4E and 2V (b) 2E and 4V (c) E and

V 2

(d)

E and V 2

Electrostatic Field and Potential 20.11

6.

at a point P on its axial line at a distance r from it is EP and at a point Q on its equatorial line at a distance 2 r from it is EQ . Then (a) EP = – 16 EQ

(b) EP = –8 EQ

(c) EP = 8 EQ

(d) EP = – EQ

7. A hollow metal sphere is charged such that the potential at its centre is V. The potential on the surface of the sphere is (a) zero (b) V (c) more than V (d) less than V 8. The work done in carrying a charge q once round a circle with a charge Q at the centre is W1. The work done is W2 if charge q is moved from one end of a diameter to the other. Then (a) W1 > W2 (b) W1 < W2 (c) W1 = W2 0 (d) W1 = W2 = 0 9. Three point charges 4q, Q and q are placed in a straight line of length l at points distant 0, l/2 and l respectively. If the net force on charge q is zero, the magnitude of the force on charge 4q is (a)

q2 0

(c)

l2

3 q2

(b)

2 q2 0

l2

4 q2

(d) 2 l2 0 l 10. 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 Q Q (b) – (a) – 2 4 Q Q (c) + (d) + 2 4 IIT, 1987 11. One thousand spherical water droplets, each of radius r and each carrying a charge q, coalesce to form a single spherical drop. If v is the electrical potential of each droplet and V that of the bigger drop, then V 1 V 1 = (b) = (a) v 1000 v 100 V V (c) = 100 (d) = 1000 v v 12. Two metallic identical spheres A and B carrying equal positive charge + q are a certain distance apart. The force of repulsion between them is F. A third uncharged sphere of the same size is brought in contact with sphere A and removed. It is then 0

brought in contact with sphere B and removed. What is the new force of repulsion between A and B? 3F (a) F (b) 8 F F (c) (d) 2 4 13. Two small identical balls P and Q, each of mass 3 /10 gram, carry identical charges and are suspended by threads of equal lengths. At equilibrium, they position themselves as shown in Fig. 20.24. 1 What is the charge on each ball. Given = 4 0 9 2 –2 –2 9 10 Nm C and take g = 10 ms . (b) 10–5 C (a) 10 –3 C –7 (d) 10–9 C (c) 10 C

Fig. 20.24

14. Two point charges q1 = 2 C and q2 = 1 C are placed at distances b = 1 cm and a = 2 cm from the origin on the y and x axes as shown in Fig. 20.25. P (a, b) will subtend an angle with the x-axis given by (a) tan = 1 (b) tan = 2 (c) tan = 3 (d) tan = 4

Fig. 20.25

15. An electric dipole placed with its axis in the direc(a) a force but no torque (b) a torque but no force (c) a force as well as a torque (d) neither a force nor a torque. 16. An electric dipole placed with its axis inclined at experiences

20.12 Comprehensive Physics—JEE Advanced

(a) a force but no torque (b) a torque but no force (c) a force as well as a torque (d) neither a force nor a torque 17. An electric dipole placed in a non-uniform electric (a) a force but no torque (b) a torque but no force (c) a force as well as a torque (d) neither a force nor a torque. 18. A cube of side b has a charge q at each of its vertices. What is the electric potential at the centre of the cube? 4q 3q (b) (a) 3 0b 0b 2q

(c)

4q 3

0b

2q

(c)

0b

3q

(b)

2

0b

2

(d) zero

2

(a)

20. Two point charges – q and + q are located at points (0, 0, – a) and (0, 0, a) respectively. What is the electric potential at point (0, 0, z)? qa q (b) (a) 2 4 4 0z 0a (c)

2qa 4

z2

0

a2

(d)

2qa 4

0

z2

a2

21. In Q. 20, how much work is done in moving a small test charge q0 from point (5, 0, 0) to a point (– 7, 0, 0) along the x-axis? (a) (c)

5 7 2 12

4

q0 q

4

(b) W1 < W2

0

a

q0 q 0

a

(b)

7 5

Fig. 20.26

(d) W1 = W2 = 0 25. Two concentric spheres of radii r1 and r2 carry charges q1 and q2 respectively. If the surface charge density ( ) is the same for both spheres, the electric potential at the common centre will be

19. (a)

24. A positive charge (+ q) is located at the centre of a circle as shown in Fig. 20.26. W1 is the work done in taking a unit positive charge from A to B and W2 is the work done in taking the same charge from A to C. Then (a) W1 > W2 (c) W1 = W2

(d) zero

0b

23. In a hydrogen atom, the electron and the proton are bound together at a separation of about 0.53 Å. nite separation of the electron from the proton, the potential energy of the electron-proton system is (a) – 54.4 eV (b) – 27.2 eV (c) – 13.6 eV (d) zero

4

q0 q 0

a

(d) zero

22. A neutral hydrogen molecule has two protons and two electrons. If one of the electrons is removed we get a hydrogen molecular ion (H +2). In the ground state of H +2 the two protons are separated by roughly 1.5 Å and the electron is roughly 1 Å from each proton. What is the potential energy of the system? (a) – 38.4 eV (b) – 19.2 eV (c) – 9.6 eV (d) zero

0

(c) 0

26.

r1 r2

(b)

(r1 – r2)

(d)

r2 r1

0

(r1 + r2)

0

a sphere of radius r having a uniform surface charge density is (b)

(a) 0

(c)

0r

(d)

2

0

2 0r

27. Two equal negative charges – q (0, a) and (0, – a). A positive charge + Q is released from rest at the point (2a, 0) on the x-axis. The charge Q will (a) execute SHM about the origin (b) move to the origin and remain at rest there. (d) execute oscillations but not SHM. IIT, 1985 28. Four charges q, 2q, 3q and 4q are placed at corners A, B, C and D of a square as shown in Fig. 20.27. P of the square has the direction along

Electrostatic Field and Potential 20.13

(a) zero q

(b)

4

2

(d) (a) AB (b) CB (c) AC (d) DB 29. Particle A has a charge + q and particle B has a charge + 4q, each having the same mass m. When allowed to fall from rest through the same potential difference, the ratio of their speeds vA/vB will be (a) 2 : 1 (b) 1 : 2 (c) 1 : 4 (d) 4 : 1 30. Four equal charges Q are placed at the four corners of a square of side a. The work done in removing a charge – Q is 2 Q2 (a) zero (b) 4 0a (c)

2 Q2 0a

(d)

Q2 2

a

31. A point charge + q is placed at the mid point of a cube of side L cube is q (b) zero (a)

(Q1

Q2 ) ( 2

1)

1 q Q1 24

0

Q2 R

IIT, 1992 34. An electron of mass me, initially at rest, moves in time t1. A proton of mass mp, also initially at rest, takes time t2 to move through an equal distance in gravity, the ratio t2/t1 is nearly equal to (a) 1

(b) (mp/me)1/2

(c) (me/mp)1/2

(d) 1836

IIT, 1997 35. A nonconducting ring of radius 0.5 m carries a total charge of 1.11 10–10 C distributed non-uniformly E everywhere in space. The value of the line integral l l

0

2R

q 2 (Q1 + Q2) 4 0R

(c)

Fig. 20.27

0

0

– E dl (l = 0 being centre of the ring) in

volts is (a) + 2 (c) – 2

(b) – 1 (d) zero

IIT, 1997 36. A metallic solid sphere is placed in a uniform

0

(c)

6 q L2 0

(d)

q 2

6L

0

32. A solid conducting sphere having a charge Q is surrounded by an uncharged concentric conducting hollow spherical shell. The potential difference between the surface of the solid sphere and the outer surface of the hollow shell is 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 IIT, 1989 33. Two identical thin rings, each of radius R are coaxially placed a distance R apart. If Q1 and Q2 are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to the centre of the other is

lines of force follow? (a) 1 (c) 3

(b) 2 (d) 4 IIT, 1996

Fig. 20.28

37. A charge + q x = 3 x0, x = 5x0

x = x 0, q is x = 2x0, x = 4x0, x = 6x0 x0 is a positive constant. The potential at the origin to this system of charges is (a) zero

(b)

4

0

q x0 ln 2

20.14 Comprehensive Physics—JEE Advanced

q ln 2 4 0 x0 IIT, 1998 38. Three charges Q, + q and + q are placed at the vertices of a right-angled isosceles triangle as shown in Fig. 20.29. The net electrostatic energy Q is equal to (a)

q 1

(b)

2

(c) – 2q

2q 2

2

(d) + q IIT, 2000

43. Two identical charges are placed at the two corners of an equilateral triangle. The potential energy of the system is U. The work done in bringing an (a) U (b) 2 U (c) 3 U (d) 4 U 44. The magnitude of electric intensity at a distance x from a charge q is E. An identical charge is placed at a distance 2x from it. Then the magnitude of the force it experiences is (a) qE (b) 2qE (c)

qE 2

(d)

qE 4

45. A particle carrying a charge q is shot with a speed v Q. It approaches Q up to a certain distance r and then returns as shown in Fig. 20.30.

Fig. 20.29

Fig. 20.30

39. Eight dipoles of charges of magnitude e are placed the cube will be 8e (a) 0

(c)

e

(b)

16 e

(c)

0

(d) zero

0

40. A point Q lies on the 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 Q is proportional to (b) p and r–2 (a) p–1 and r–1 2 –3 (c) p and r (d) p and r–3 41. A particle of mass m and charge q is released E. The kinetic energy attained by the particle after moving a distance x is (a) qEx2 (b) qE 2x (c) qEx (d) q2Ex 42.

3

Vm–1 along the y-axis. A body of mass 1 g and charge 10–6 C –1

positive x-axis with a velocity of 10 ms . Its speed (in ms–1) after 10 second will be (neglect gravitation) (a) 10 (c) 10 2

(b) 5 2 (d) 20

If charge q were moving with a speed 2v, the distance of the closest approach would be (a) r (b) 2r r 2

(d)

r 4

46. Four charges, each equal to – Q, are placed at the corners of a square and a charge + q is placed at its centre. If the system is in equilibrium, the value of q is (a)

Q 1 2 2 4

(b)

Q 1 2 2 4

(c)

Q 1 2 2 2

(d)

Q 1 2 2 2

47. A charge having magnitude Q is divided into two parts q and (Q – q) which are held a certain distance r apart. The force of repulsion between the two parts will be maximum if the ratio q/Q is (a)

1 2

(b)

1 3

1 1 (d) 4 5 48. A charge Q is given to a hollow metallic sphere of radius R. The electric potential at the surface of the sphere is (c)

Electrostatic Field and Potential 20.15

(a) zero

(b)

1

(c)

1 4

Q R

0

Q

(d) 4 0 Q/R 4 0 R2 49. In Q. 48, the potential at a distance r from the centre of the sphere where r < R is (a) zero (c)

(b)

1 4

Q 0

R

r

(d)

1 4 4 R

55. Three positive charges of equal value q, are placed at the vertices of an equilateral triangle. The resulting lines of force should be sketched as in (see Fig. 20.31).

Q R

0

r

0Q

r

50. The electric potential V at any point (x, y, z) in space is given by V = 4x2 volt where x, y and z are 2 m) in Vm–1 is (a) 8 along negative x-axis (b) 8 along positive x-axis (c) 16 along negative x-axis (d) 16 along positive x-axis

Fig. 20.31

IIT, 2001 IIT, 1992

51. Two spheres of radii r and R carry charges q and Q respectively. When they are connected by a wire, there will be no loss of energy of the system if (a) qr = QR (b) qR = Qr (c) qr2 = QR 2 (d) qR 2 = Qr 2 52. Two equal point charges of 1 C each are located at points ( i j k ) m and (2 i 3 j k ) m. What is the magnitude of electrostatic force between them? (a) 10–3 N (b) 10–6 N (c) 10–9 N (d) 10–12 N 53. Three equal point charges q are placed at the corners of an equilateral triangle. Another charge Q is placed at the centroid of the triangle. The system of charges will be in equilibrium if Q equals (a)

q 3

(b) –

q 3

q q (d) – 3 3 54. A metallic sphere A of radius a carries a charge Q. It is brought in contact with an uncharged sphere B of radius b. The charge on sphere A now will be (c)

aQ (a) b (c)

bQ a b

bQ (b) a (d)

aQ a b

V where 0 is the t permittivity of free space, L is a length, V is a potential difference and t is a time interval. The dimensional formula for X is the same as that of (a) resistance (b) charge (c) voltage (d) current IIT, 2001 57. 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 potentials at the points A, B and C satisfy: (a) VA < VB (b) VA > VB 56. A quantity X is given by

(c) VA < VC

0L

(d) VA > VC

IIT, 2001 58. Two equal poi x = – a and x = + a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x-axis, is approximately proportional to (a) x (b) x 2 (c) x3

(d) 1/x IIT, 2002

59. A metallic shell has a point charge q kept inside its circular cavity. The charge is not exactly at the centre of the cavity. Which of the diagrams in Fig. 20.32 correctly represents the electric lines of force?

20.16 Comprehensive Physics—JEE Advanced

q

(a) + q 3 alone

(b) + q1 and + q3

(c) + q1, + q3 and – q2

(d) + q1 and – q2 IIT, 2004

62.

charge densities 1 = – , 2 = + 2 and 3 = + 3 are placed parallel to the x–z plane at y = a, y = 3a and y = 4a at point P is

Fig. 20.32

IIT, 2003 60. Three negative point charges – q each, and three positive point charges + q, + q and + Q are placed at the vertices of a regular hexagon as shown in Fig. 20.33. For what value of Q O A, B, D, E and F be O due to charge Q at C alone? q (a) q (b) 2 (c) 2q

(d) 5q

Fig. 20.35

(a) zero

2

(b)

j 0

3

(c)

j

(d)

0

IIT, 2004

3

j 0

IIT, 2005 63. A metallic sphere of radius R is charged to a potential V distance r (> R) from the center of the sphere is V r

(b)

(c)

VR

(d) zero

r2 64. Two point charges q1 = 1 C and q2 = 2 C are placed at points A and B 6 cm apart as shown in Fig. 20.36. A third charge Q = 5 C is moved from C to D along the arc of a circle of radius 8 cm as shown. The change in the potential energy of the system is

Fig. 20.33

61. Figure 20.34 shows a spherical Gaussian surface and a charge distribution. When the Gaussian surface, be due to

(a)

Fig. 20.34

Vr R2

Fig. 20.36

(a) 3.0 J

(b) 3.6 J

(c) 5.0 J

(d) 7.2 J

Electrostatic Field and Potential 20.17

65. A partical of mass m and charge + q is midway

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

a charge + q and at a distance 2L apart. The middle charge is displaced slightly along the line joining oscillation is proportional to. (a) L1/2

(b) L

(c) L3/2

(d) L2

70.

66. Five point charges, each equal to + q, are placed L. The magnitude of the force on a point charge – q placed at the centre of the haxagon is (a) (c)

5q2 4

2 0L

q2 4

2 0L

(b)

3q 2 4

2 0L

(d) zero

IIT, 1992 67. Two isolated metal spheres of radii R and 2R are charged such that both have the same surface charge density . The spheres are located far away from each other. When they are connected by a thin conducting wire, the new surface charge density on the bigger sphere will be 2 3 (b) (a) 3 5 5 (c) (d) 6 2 IIT, 1996 68. A spherical portion has been removed from a solid sphere having a charge distributed uniformly in its volume as shown in Fig. 20.37. the emptied space is (a) zero everywhere (b) non-zero and uniform (c) non-uniform (d) zero only at its centre

71.

72.

Fig. 20.37

IIT, 2007 69. Positive and negative point charges of equal a a magnitude are kept at 0, 0, and 0, 0, 2 2

73.

IIT, 2007 A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinders are initially electrically neutral. (a) A potential difference appears between the two cylinders when a charge density is given to the inner cylinder (b) A potential difference appears between the two cylinders when a charge density is given to the outer cylinder (c) No potential difference appears between the two cylinders when a uniform line charge is kept along the axis of the cylinders (d) No potential difference appears between the two cylinders when same charge density is given to both the cylinders IIT, 2007 Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then, (a) negative and distributed uniformly over the surface of the sphere (b) negative and appears only at the point on the sphere closest to the point charge (c) negative and distributed non-uniformly over the entire surface of the sphere (d) zero IIT, 2008 Three concentric metallic spherical shells of radii R, 2R, 3R, are given charges Q1, Q2, Q3, respectively. It is found that the surface charge densities on the outer surfaces of the shells are equal. Then, the ratio of the charges given to the shells, Q1 : Q2 : Q3, is (a) 1 : 2 : 3 (b) 1 : 3 : 5 (c) 1 : 4 : 9 (d) 1 : 8 : 18 IIT, 2009 A disk of radius a/4 having a uniformly distributed charge 6C is placed in the x-y plane with its centre at (– a/2, 0, 0). A rod of length a carrying a uniformly distributed charge 8C is placed on the x-axis from x = a/4 to x = 5a/4. Two point charges – 7C and 3C are placed at (a/4, – a/4, 0) and (– 3a/4,3a/4, 0), respectively. Consider a cubical surface formed by

20.18 Comprehensive Physics—JEE Advanced

six surfaces x = ± a/2, y = ± a/2, z = ± a/2. The elec-

E = E0 x , where E0 is area (as

74.

20.38] 2C

(a)

(b)

0

0

10 C

(c)

2C

(d)

0

12 C 0

Fig. 20.39

Fig. 20.38

(a) 2E0a2

(b)

(c) E0a2

(d)

2 E 0a 2 E0 a 2 2

IIT, 2009 ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67. 73.

(a) (b) (c) (d) (d) (a) (d) (b) (a) (c) (c) (c) (a)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68. 74.

(d) (d) (b) (c) (a) (a) (b) (d) (a) (d) (c) (b) (c)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63. 69.

(d) (c) (d) (d) (d) (b) (d) (d) (b) (b) (c) (c)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64. 70.

(a) (d) (b) (b) (b) (b) (d) (a) (a) (b) (b) (a)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65. 71.

(c) (c) (c) (b) (b) (d) (c) (a) (b) (c) (c) (d)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66. 72.

(a) (b) (a) (d) (c) (d) (c) (b) (d) (b) (c) (b)

SOLUTIONS 1. or

W = Uf – Ui

Ui =

1 4

0

r

+

Q –Q

[(q) (–2q) + q(–2q) + (–2q) (–2q)]

2.5

=0 Uf =

2

1 4

0

(2 r )

[(q) (–2q) + q(–2q) + (–2q) (–2q)]

=0 W = 0. So the correct choice is (a). 2. Charge Q at point P will induce a charge – Q on the inner surface and a charge + Q on the outer surface of the shell (Fig. 20.40). The electric potential at O is

Fig. 20.40

V=

1 4

0

Q R

Q 2R

Q 2.5 R

Electrostatic Field and Potential 20.19

=

1 4

4q2 + 4qQ = 0

9Q 10 R

0

Hence Q = – q. The force on charge 4q is

3. Before connection, charge Q on the inner shell induces a charge – Q on the inner surface of the outer shell and a charge + Q on its outer surface. Therefore, the total charge on the outer surface of the outer shell = Q + 2Q = 3Q. When the two shells are connected by a conducting wire, the entire charge Q on the inner shell is conducted to the outer shell. Therefore, the charge on the outer shell now is Q + 2Q = 3Q, the same as before. Hence there will be no change in its electric potential. Q

4. E =

2

4 0R 4 0R Hence if R is doubled, E becomes E/4 and V becomes V/2. So the correct choice is (a). Q . In terms of , 5. = 4 R2 E=

and V = 0

R 0

If R is halved, E remains the same but V becomes V . So the correct choice is (c). 2

4q 2

4 qQ

F= 4

2

l 2

0

4

0

l2

Putting Q = – q, we get F =– | F |=

4q 2

16 q 2 4

0

3q 2 0

Q

and V =

4q (q + Q) = 0

l2

l2

4

0

l2

=–

3q 2 0

l2

, which is choice (c).

10. Refer to Fig. 20.42. The three charges will be in equilibrium if no net force acts on each charge. Charge q is in equilibrium because the forces acting on it by charge Q at A and charge Q at B are equal and opposite. Charge Q at A will be in equilibrium if Q2

qQ 4

0

l

2

4

0

(2 l )

2

=0

q=–

Q . 4

Similarly charge Q at B will be in equilibrium if Q . Hence the correct choice is (d). q=– 4

6. If p is the electric dipole moment, then EP =

2p 4

EQ = –

0

r3

Fig. 20.42

p 4

0

11. If R is the radius of the big drop, we have

( 2 r )3

EP = – 16 EQ, which is choice (a). 7. The potential inside a spherical conductor is constant including that on its surface. Hence the correct choice is (b). 8. The potential at every point on the circle due to charge Q is the same. Work done = q (potential difference). Hence the correct choice is (d). 9. Refer to Fig. 20.41.

Fig. 20.41

The net force on q will be zero if q (4 q) 4

0

l

2

qQ 4

0

l 2

2

=0

4 R3 4 r3 = 1000 3 3 which gives R = 10 r. The electrical potential of each droplet is q v = 4 0r and that of the big drop is V =

1000 q 4 0R

V 1000 r = = 100 ( R = 10r) v R Hence the correct choice is (c). 12. When two identical metallic spheres are brought in contact, the charges on them are equalized due to sphere C is brought in contact with sphere A having a charge + q, and then removed, the total charge q

20.20 Comprehensive Physics—JEE Advanced

is equally shared between the two so that the charge left on A is +q/2 and that developed on C is +q /2. The sphere C carrying a charge +q/2 is now brought in contact with sphere B which is already carrying a charge +q. The total charge is q / 2 + q = +3q/2 which must distribute equally on B and C. Thus when C is removed, B will have a charge of +3q/4 and C also has a charge of +3q/4. Hence when C is removed from both A and B, q New charge on A = + 2 3q New charge on B = + 4 Since force is proportional to the product of the charges, it follows that the new force of repulsion between A and B is 3/8 of the earlier force (F). Hence, the new force of repulsion between A and B is 3F/8. 13. Refer to Fig. 20.43. Let us consider forces on a ball, say, Q. Three forces act on it: (i) tension T in the thread, (ii) force mg due to gravity and (iii) force F due to Coulomb repulsion along + x-direction. For equilibrium, the sum of the x and y components of these forces must be zero, i.e. T cos 60° – F = 0 T sin 60° – mg = 0

and

1

E1 =

q1

4 0 a2 and is directed along + x E2 at (a, b) due to q2 has a magnitude 1 q2 E2 = 4 0 b2 and is directed along + y-axis. The angel E with the x-axis is given by

Fig. 20.44

tan

=

E2 q a2 = 2 2 E1 q1 b

=

1 2

2 1

2

=2

Hence the correct choice is (b). 15.

E exerts a force q E on charge + q and a force – qE on charge – q of the dipole. Since these forces are equal and opposite, they add upto zero. 16. The correct choice is (b). A torque acts on the 17. The correct choice is (c). In a non-uniform electric

Fig. 20.43

These equations give F = mg cot 60° = 10

1 3

= 10–3 N. Now F =

Putting 1 and 4 14.

F = 10 =9

0

–3

10 –3

V =

q2

1 4

3 10

translational motion and a torque which gives it a rotational motion. 18. The distance of a vertex from the the centre of the cube of side b is r = 3 b / 2. Now the potential due to charge q at the centre is q/4 0r. Hence the potential due to the arrangement of eight charges (each of magnitude q) at the centre is

r2

19.

N, r = 0.3 m

10 9, we get q = 10–7 coulomb.

0

due to q1 has a magnitude

E1 at (a, b)

8q 4

0r

=

4q 3

0b

torially. From the symmetry of the eight charges with respect to the centre of the cube, it is evident charges cancel in pairs (being equal and opposite). will be zero.

Electrostatic Field and Potential 20.21

20. Refer to Fig. 20.45. The distance of point P1 from charge + q is r1 = z – a and from charge – q is r2 = z + a.

Here q1 = q2 = q = + 1.6 10–19 C (proton), q3 = – q = – 1.6 10–19 C (electron), r12 = 1.5 Å = 1.5 10–10 m, r13 = r23 = 1 Å = 1 10–10 m and 1/4

=9

109 Nm2C–2. Thus

U= –

4 q 2 1010 joule 3 4 0

0

=– =–

=

1 4

0

q 4

0

q r1

q r2

=

4

0

q1 q2 r12

q1 q3 r13

1010

9 109

e2 joule 4 0 r 1 e =– eV 4 0 r

r2 r1 r1 r2

U= –

, 4 0 z 2 a2 which is choice (c). 21. Refer to Fig. 20.45 again. Any point on the perpendicular bisector passing through the centre of the dipole is at the same distance from the two charges. Hence the potentials at point P2(5, 0, 0) and at point P3(–7, 0, 0) are zero. Since P2 and P3 are at the same potential (zero), the potential difference between them is zero. Hence no work will be done in moving a charge from P2 to P3. The answer will not change if the path of the charge is changed because the work done is independent of the path taken. 22. Refer to Fig. 20.46. The total potential energy of the arrangement of charges is the sum of the energies of each pair of charges. The potential energy of the system comprising the three charges q1, q2 and q3 is U = W1 + W2 + W3 1

19

potential energy of the electron-proton system is

2qa

=

4 1.6 10

3 = – 19.2 eV 23. Charge on electron (– e) = – 1.6 10 –19 C, charge on proton (e) = 1.6 10–19 C, separation r = –10 0.53 Å = 0.53 10 m. If the zero of potential

Fig. 20.45

Potential at P1 =

4 q 1010 eV 3 4 0

=–

1

9 109

19

1.6 10

0.53 10 10 = – 27.2 eV Hence the correct choice is (b). 24. Points A, B and C are at the same distance from charge + q; hence electrical potential is the same at these points, i.e. there is no potential difference between A, B and C. Hence W1 = W2 = 0. 25. The electric potential at the common centre is q1 q2 V = + 4 0 r1 4 0 r2 Now

= V =

q1 4

r12

1

q2

4 r22

q1 r1 4

0

q2 q3 r23

=

=

r12

q2 r2

4 r22

(r1 + r2)

0

Hence the correct choice is (d). 26. If q surface is q E = 4 0 r2 Fig. 20.46

But

=

q 4 r2

. Therefore q = 4 r 2 .

20.22 Comprehensive Physics—JEE Advanced

Hence

E =

4 r2 4

0

r2

= 0

Thus the correct choice is (a). 27. Refer to Fig. 20.47. Forces exerted by charges – q at A and B on charge Q are F1 and F2 which are equal and have a magnitude F=

qQ 4

0

r2

Fig. 20.48

P due to charge 4 q at D is 1 4q E2 = along PB 4 0 r2 PB is E = E2 – E1 = q and 3q at A and C will be E =

1

; Fr is also proportional to (1/r2). r2 Hence charge Q will move towards the origin and because of its inertia it will overshoot the origin O. Thus, charge Q will oscillate about O but its motion is not simple harmonic because the force Fr is not proportional to its instantaneous distance from the origin as Fr 1/r2. Hence the correct choice is (d). 28. r from a point charge Q is given by Since F

E=

1 4

Q 0

r2

If Q from Q. Refer to Fig. 20.48. Let PA = PB = PC = PD = r P due to charge 2q at B is 1 2q along PD E1 = 4 0 r2

4

2q 0

r2

2q

4 0 r2 P due to charges

directed along PA.

Thus E = E , but they are mutually perpendicular to each other. therefore, their resultant will be along PQ (see Figure) which is parallel to CB. Hence the correct choice is (b)

Fig. 20.47

The resultant of these equal forces equally inclined with the x-axis is along the negative x-direction towards the origin. The magnitude of the resultant is given by F 2r = F 2 + F2 + 2 F2 cos = 2 F2 (1 + cos )

1

1

29.

E. The force on charge q is

FA = qE

qE m Similarly, the acceleration of charge 4q is 4q E aB = = 4 aA m Let s be the distance travelled by A and B to acquire speeds vA and vB. Then (since u = 0) v2A = 2 aA s and vB2 = 2 aB s

Therefore, its acceleration is aA =

v 2A vB2

=

aA aB

1 4

or vA/vB = 1/2. Hence the correct choice is (b). 30. Refer to Fig. 20.49. a AO = BO = CO = DO = r = 2 Since electric potential is a scalar and since the charges at corners are equal, the magnitude of the electric potential at point O due to the four charges = 4 times that due to a single charge. Thus 1 Q V=4 4 0 r

Electrostatic Field and Potential 20.23

1

=

Q a/ 2

0

=

34. Force F = qE. Therefore, acceleration a = qE/m. Hence the distance travelled by the particle in time t is 1 2 1 qE 2 at = t s= 2 2 m

2Q 0a

For electron,

1 2

se =

For proton,

qE me

1 2

sp =

t 21

qE mp

t 22

Given se = sp. Therefore t12 t2 = 2 me mp

Fig. 20.49

Work done = Q V = 31.

choice is (c). q

2Q 2 . Hence the correct 0a E

=

36.

1

V1 =

4 Potential at C2 is V2 =

0

1 4

0

Q1 R Q2 R

Q2

1

37. V =

4

+

0

1 4

1 4

q x0

0

q 3 x0

q 2 x0

0

=

q 5 x0

q 4 x0

1 2

q 1 x0

Q1 2R

gives the potential at the

to the surface and directed towards the centre of the sphere. Hence the correct choice is (d).

=

2R

me

surface of a conductor. On the surface of a metal-

0

32. When any additional negative charge is given to a hollow spherical shell, the potential on its surface falls, but the potential at each point within the shell also falls by the same amount. Hence the potential difference between the given surfaces remains unchanged. Thus, the correct choice is (a). 33. Refer to Fig. 20.50. Potential at C1 is

–E

1/ 2

centre of the ring, which is zero.

s

. Hence the correct choice is (a).

0

35. The integral

mp

t2 t1

or

1 3

q 6 x0

1 4

q 4

upto infinity upto infinity

1 5

1 6

Qq a

q log e 2 4 0 x0

loge (1 + 1) =

0 x0

38. The net electrostatic energy is ( of triangle = 2 a) U=

upto infinity

Qq

hypotenuse side

qq a

2a

For U = 0, we require Qq a

Fig. 20.50

Work done W = q (V1 – V2) = =

q 4

0

Q1 R

q 4

0

2R

Qq 2a

q2 = 0 or Q + Q + q = 0 a 2

which gives Q = – q Q2 2R

Q2 R

Q1 2R

(Q1 – Q2) ( 2 – 1)

2 2

1

=

2q 2

2

39. A dipole consists of two equal and opposite charges separated by a certain distance. Hence the total charge enclosed in the cube is zero. Therefore, the

20.24 Comprehensive Physics—JEE Advanced

40.

dicular bisector of a dipole is given by (for r >> a), here p is the dipole moment p

E=

4 0 r3 Hence the correct choice is (d). 41. Initial kinetic energy of the particle is zero. The gain in kinetic energy in distance x = decrease in to move the particle through a distance x = force distance = q Ex. Hence the correct choice is (c). 42. Given vx = 10 ms–1 directed along the y-axis, the acceleration of the body along the y-axis is ay =

qE m

10

6

10

103

45. Charge q will momentarily come to rest at a distance r from charge Q when all its KE is converted to PE, i.e. 1 qQ 1 mv2 = r 4 2 0 Therefore, the distance of closest approach is given by qQ 2 r= 4 0 mv 2 1 Thus r . Hence if v is doubled, r becomes v2 one-fourth. Thus the correct choice is (d). a 46. Let the side of the square be a. OA = OC = r = 2 (see Fig. 20.51).

= 1 ms–2

3

Therefore, the velocity of the body along the y-axis at time t = 10 s is vy = ayt = 1 × 10 = 10 ms–1 Resultant velocity v = (10) 2

=

vx2

(10)2

v 2y 10 2 ms

1

Hence the correct choice is (c). 43. Let q be the magnitude of each charge and a the length of each side of the triangle ABC. The potential energy of the system of two equal charges placed at vertices A and B is U (given). This means that U is the work done in bringing a charge q from B with the charge q at vertex A. Hence the work done in bringing an identical charge q C = work done to overcome the force of repulsion of q placed at A placed at a distance a from it + work done to overcome the force of repulsion of q placed at B at the same distance a from it = U + U = 2U, which is choice (b). q . Hence the magnitude of the 44. Given E = 4 0 x2 electric intensity at a distance 2x from charge q is E =

q 4

0 (2 x)

q 2

4

0x

2

1 4

E 4

Therefore, the force experienced by a similar charge q at a distance 2x is qE F = qE = 4 Hence the correct choice is (d).

Fig. 20.51

1. Stability of charge + q at the centre Charges – Q at corners A and C will attract charge + q with equal and opposite forces. Similarly charges – Q at corners B and D will attract charge + q with equal and opposite force. Hence no net force acts on charge – q. 2. Stability of charge –Q at any corner Q at corner A. This charge will experience four forces: (i) Force of repulsion F1 due to charge – Q at B (ii) Force of repulsion F2 due to charge – Q at D (iii) Force of repulsion F3 due to charge – Q at C (iv) Force of attraction F due to charge + q at O. Now

F1 = F2 =

and

F3 =

Q2 4

0a

2

Q2 4

0 ( 2r )

Q2 2

4

0 ( 2a

2

)

Electrostatic Field and Potential 20.25

and

q ( Q)

1

F =

2qQ

2

4 0 r 4 The resultant of F1 and F2 is given by F12

F=

F22

0a

2Q

2 F1 =

2

4 0a2 The forces F and F3 act along AP. Hence the net force acting on charge – Q at A due to charges – Q at B, C and D is F = F + F3 Q2

2 Q2

=

2qQ 4

0a

2

=

4

0 ( 2a

2

)

4

0 ( 2a

or

q r

dF For F to be maximum, = 0, i.e. dq q Q 0

r

q 2

=0

Since r d [q(Q – q)] = 0 dq q 1 which gives Q 2 Hence the correct choice is (a). 48. For points on the surface of the sphere or outside the sphere, a charged sphere behaves as if the charge is concentrated at its centre. Therefore, the potential at the surface of the sphere is given by 1

Q , which is choice (b). 4 0 R 49. At points inside a charged metallic sphere, i.e. for r < R, the potential is zero. Hence the correct choice is (a). dV 50. E = – i where i is a unit vector along the dx positive x-axis. Hence E at a point whose x-coordinate is x = 1 m is

3j

r=

q=

V=

4

0r

0R

Q . Hence the correct choice is (b). R

52. r = (2 i

)

d 1 dq 4

4

Q

k)

(i

k)

j

(i

2j

2 k ) m.

The magnitude of r is

Q 1 2 2 4 Hence the correct choice is (a). 47. The force of repulsion between the two parts is given by 1 q Q q F= 4 0 r2 or

q

V=

Q 2 (1 2 2 ) 2

8i Vm 1 .

8 xi

The negative sign shows that E is along the negative x-axis. Hence the correct choice is (a). 51. There will be no loss of energy if the potential of the spheres is the same i.e. if

Q 2 (1 2 2 )

4 0 a 2 4 0 ( 2a 2 ) For equilibrium, F = F , i.e.

d 4 x2 i dx

E=–

2

F= =

12

22

1

q1 q2

4

22

1 4

4 =3m

r2

0

9 109

6

10

10

6

= 10–3 N

2

3 Hence the correct choice is (a).

53. The system will be in equilibrium if the net force on charge q at one vertex due to charges q at the other two vertices is equal and opposite to the force due to charge Q at the centroid, i.e. (here a is the side of the triangle) 3 q2 4

0a

Qq

2

4

which gives Q = – is (b). 54.

q 3

0

a 3

2

. Hence the correct choice

A to B until their potentials become equal. If charge q A to B, then Q 4

q 0a

q 4

0b

a bQ q which gives q = . Hence b a b bQ aQ charge left on A = Q – q = Q – . a b a b Hence the correct choice is (d).

or Q – q =

55. Since all the three charges are of the same polarity, around them cannot be closed. Hence choices (a),

20.26 Comprehensive Physics—JEE Advanced

(b), the line in the middle is a closed circular loop. Thus, the only correct choice is (c). 56. The capacitance of a parallel plate capacitor is given by C = 0 A/d. Hence the dimensions of 0 L are the same as those of capacitance. V Dimensions of 0L t dimension of C dimensions of V = time dimension of Q = ( Q = CV) time charge = = current time Hence the correct choice is (d). 57.

59. Because the charge is not located at the centre of the cavity, inside the cavity the lines of force are skewed. Hence choice (a) and (b) are incorrect. Outside the shell, the lines of force are the same as if the charge were located at the centre of the cavity. Also there can be no line of force in the metallic body of the shell. Hence choice (d) also incorrect. Thus the correct pattern is shown in (c). 60. O due to charges – q at A and D are equal and opposite. Hence they cancel each other. Similarly charges + q at B and E do not contribute to electric at O. Due to charge – q at F O will be E1 =

Thus V decreases as dx increases in the direction VA VB , which is choice (b). 58. Potential energy of the system when charge Q is at O is

q 4

qQ a x

=

2qQ a

1

2qQ a

1

U = U – U0 = =

2qQ

Hence U

a

3

x2

0r

2

=

2Q 4

0r

a2

61. E.ds E is due to all the charges, both inside and outside the Gaussian surface. Hence the correct choice is (c). plane sheet carrying a uniform charge density given by E=

x2

1

a2 x2

(

a2 2qQ 1 a

x a

2 2

(x 2)

x 2 which is choice (b).

2

q . Hence the correct choice is (b). 2

qQ 2a

qQ a x

x

a)

O

directed from O to F 4 0 r2 E1 = 2E2, we require

For

62. Fig. 20.52

directed from O to F

2

Q

which gives Q =

When charge Q is shifted to position O , the potential energy will be (see Fig. 20.52).

U=

0r

E2 =

2qQ a

qQ a

4

O charges at A, B, D, E and F due to charge + Q at C is given by

dV E=– dx

qQ U0 = a

q

is

2 0 It is independent of the distance of point P from the sheet and is, therefore, uniform. The direction of dicular to it if is positive and is towards the sheet and perpendicular to it if is negative. Hence E1 =

2qQ a and

j along –ve y–direction

2

0

E2 =

2 2

0

E3 =

3 2

0

j along –ve y–direction j along – ve y–direction

Electrostatic Field and Potential 20.27

From the superposition principle, the net electric P is E = E1 + E2 + E3 2 = j 2 0 2 0 =–

3

3 2

j

F=

q 4

o

q (OD)2

=

q2 4

o

L2

j 0

j , which is choice (c). 0

63. Let the charge on the sphere be Q. Then Q V= 4 0R which gives Q = 4 0RV r is Q

E=

=

4

0 RV 2 0r

=

Fig. 20.53

RV

4 0r 2 r2 4 Thus the correct choice is (c). 64. If charge Q is moved from C to D along the arc, the potential energy between pairs (q1, Q) and (q1, q2) will not change as the distance between them remains unchanged ( AC = AD). The potential energy of the pair of chages q2 and Q will change. Now, distance BC = (8)2 (6) 2 = 10 cm and BD = 8 – 6 = 2 cm. Therefore, change in P.E. is

10–6)

= (2

U=

q2Q 1 4 0 BD

(5

10–6)

(9

1 BC 1 0.02

10–9)

1 0.1

65. If the middle charge is displaced by a distance x, the net force acting it, when it is released, is

=

q2

1 4

0

(L

q2

1

x)2

4

0

(L

x)2

4q 2 Lx 4

2 0 (L

For x << L, F = – where k = Now T = 2

q2

x 2 )2 q2 x 3 0L

and of the sphere of radius 2R is V2 =

Q2 4 (2 R)

When the spheres are connected by a thin wire,

= 3.6 J, which is choice (b).

F=

67. Initial charge on sphere of radius R is Q1 = 4 R2 and on sphere of radius 2R is Q2 = 4 (2R)2 = 16 R2 Total initial charge is Q = Q1 + Q2 = 20 R2 Initial potential of the sphere of radius R is Q1 V1 = 4 R

= – kx

3 0L

m k

So, the correct choice is (c). 66. As shown in Fig. 20.53, all forces cancel in pairs, except the forces F on charge – q due to charge + q at vertex D (i.e. the vertex opposite to the empty vertex). The magnitude of the net force on – q is

their potentials become equal. Let Q 1 and Q 2 be the new charges, then the potential of each sphere will be Q1 Q2 = V= R 4 4 (2 R) Q2 2 From conservation of charge, we have Q1 + Q2 = Q 1 + Q 2 which gives Q 1 =

20 R2

=

Q2 3Q 2 + Q2 = 2 2

40 R2 3 New surface charge density on the sphere of radius 2R is 40 R2 Q2 5 3 = = 2 = 2 4 (2 R) 6 16 R 68. Let O be the centre of the sphere and Q be the centre of the cavity. Let r be the separation between Q2 =

20.28 Comprehensive Physics—JEE Advanced

them. Let OP = b and QP = a. Here P is a point inside the cavity. Let be the charge density. From point P is given by [See Fig. 20.54] E = =

3

b 0

3

a = 0

r 3

3

b a 0

r

a

outer surface, so that the total charge on the outer surface of shell 2 is (Q1 + Q2). This charge induces a charge – (Q1 + Q2) on the inner surface of shell 3 and a charge + (Q1 + Q2) on its outer surface so that the total charge on the outer surface of shell 3 is (Q1 + Q2 + Q3) as shown in Fig. 20.56. Given 1= 2= 3 Q1

b

4

0

0

R

2

(Q1 4

Q2 ) 0 (2 R)

2

=

(Q1 4

Q2

Q3 )

0 (3R )

2

Fig. 20.54

choice is (b). 69. The two charges constitute an electric dipole. The points (– a, 0, 0) and (0, a, 0) are at the same distance from the positive and negative charges of equal magnitude. Hence the electric potential is zero at these points. Hence the work done is zero. 70. When a charge density is given to the inner cylinder, outer cylinders. Hence a potential difference appears between the two cylinders. If a charge density is given to the outer cylinder, the same potential appears on the inner and the outer cylinders. Hence, in this case, there is no potential difference between them. Thus the correct choice is (a). 71. When a positive charge + Q is placed outside a neutral conducting sphere, it will induce a negative charge – Q on the side of the sphere closer to it and an equal positive charge + Q on the opposite side of the sphere. Thus the net charge on the sphere will be zero, which is choice (d). [see Fig. 20.55]

Fig. 20.56

Solving we get Q1 = 3Q2 = 5Q3. Hence the correct choice is (b). q 73. E x through the cubical surface = where 0

q is the net charge enclosed in the surface. Since half the disc lies inside the surface, one-fourth of the rod lies inside the surface, point charge –7C lies inside the surface and point charge 3C lies outside the surface, the net charge enclosed in the surface is 6C 8C – 7C + 0 q= 2 4 = 3C + 2C – 7C = – 2C 2C . So the correct choice is (a). 0

74. Refer to Fig. 20.57.

Fig. 20.57 Fig. 20.55

72. Charge Q1 on shell 1 induces a charge – Q1 on the inner surface of shell 2 and a charge + Q1 on its

= E A Area vector A = a j

(a i

a k)

Electrostatic Field and Potential 20.29

= E0 i [a j

(a i

= E0 i [ a 2 k

= E0 a 2 ( i k

a k )]

i i)

= E0a2 (0 + 1) = E0a2

a2 i ]

II Multiple Choice Questions with One or More Choices Correct 1. Choose the correct statements from the following. potential must also be zero at that point. (b) If electric potential is constant in a given zero in that region. (c) Two different equipotential surfaces can never intersect. (d) Electrons move from a region of lower potential to a region of higher potential. 2. Figures 20.58 (a) and (b) show the lines of force q) and a negative charge (– q) respectively. Which of the following statements are correct? (a) Potential at P is greater than that at Q. (b) Potential at A is greater than that at B. (c) Potential energy at P is less than that at Q. (d) Potential energy at A is greater than that at B.

4. Which of the following statements are correct? nucleus in moving an electron around it in a complete orbit is zero irrespective of whether the orbit is circular or elliptical. (b) The equipotential surfaces corresponding to are concentric spheres with the point charge as the common centre. (c) If Coulomb’s law involved 1/r3 dependence instead of 1/r2, Gauss’s law would still hold good. (d) A single conductor cannot have any capacitance. 5. A pendulum bob of mass m carrying a charge q is at rest with its string making an angle with the E. The tension in the string is mg mg (b) (a) sin cos qE qE (d) sin cos 6. Four point charges + q, + q, – q and – q are placed respectively at corners A, B, C, and D of a square. Then (a) the potential at the centre O of the square is zero. O of the square is zero. (c) If E is the mid-point of side BC, the work done in carrying an electron from O to E is zero. (d) If F is the mid-point of side CD, the work done in carrying an electron from O to F is zero. 7. Six charges, each equal to + q, are placed at the corners of a regular hexagon of side a. The electric potential at the point where the diagonals intersect is V E. Then (c)

Fig. 20.58

3.

tive charge (+ q) and a negative charge (– q) are shown in Figs. 20.36 (a) and (b) above. Then (a) the work done in moving a small positive charge (+ q0) from Q to P will be positive (b) the work done in moving a small negative charge (– q0) from B to A will be positive (c) in going from Q to P, the a small negative charge (– (d) in going from B to A, the a small negative charge (–

kinetic energy of q0) increases kinetic energy of q0) decreases

20.30 Comprehensive Physics—JEE Advanced

(a) V = 0

(b) V =

(c) E = 0

(d) E =

Form the graph we conclude that

6q 4

0a

6q 4

0a

2

8. Two parallel plane sheets 1 and 2 carry uniform charge densities + and – as shown in Fig. 20.59. The electric I, II and III are E1, E2 and E3 respectively (a) E1 = 0 (b) E2 =

the surface of the sphere.

0

(c) E3 = 0 (d) E1 = E3 =

2

Fig. 20.59 0

9. Electric potential V due to a spherically symmetric charge system varies with distance r as shown in Fig. 20.60 Q Given V = 4 0 r0 for r < r 0

and V =

Q 4

the surface of the sphere. (b) for r > R, the entire charge Q may be assumed to be concentrated at the centre of the sphere. (c) the electric potential is maximum at centre of the sphere. (d) the electric potential is zero at centre of the sphere. 11. A sphere of raduis R is made of a conducting material and carries a charge Q. Choose the correct statements from the following. (a) Charge Q resides on the surface of the sphere.

0r

Fig. 20.60

for r > r0

(d) No work is needed to move a charge from one point to another on the surface of the sphere. 12. A proton and an electron are placed in a uniform enced by proton and electron are Fp and Fe and ap and ae are the respective magnitudes of their accelerations. Then (a) Fe > Fp (b) Fe = Fp (c) ae > ap (d) ae < ap 13. A metal ring of radius R carries a charge Q distributed uniformly on it. A point P lies on the axis of the ring at a distance r from its center. Then (a) The potential at the centre of the ring is zero. zero (c) For r >> R, the potential varies as 1/r. (d) For r >> R, r 2.

Which of the following statements are true? discontinuous at r = r0. (b) The net charge enclosed in a sphere of radius r = 2r0 is (c) No charge exists at any point in a spherical region of radius r < r0. (d) Electrostatic energy inside the sphere of radius r = r0 is zero. IIT, 2006 10. A sphere of radius R is made of a nonconducting material and carries a positive charge Q. Figure 20.61 shows the variation of E with distance r from the Fig. 20.61 centre of the sphere.

14.

ged. A point P lies at a perpendicular distance r from it Then r. r 2. istant straight lines prependicular to the wire.

15.

P lises at a perpendicular distance r from it. Then r. r 2. distant straight lines prependicular to the plane sheet sheet.

Electrostatic Field and Potential 20.31

16. In Gauss’s theorem

E ds s

17.

18.

19.

20.

q

. The surface

0

integral is evaluated by choosing a closed surface called the Gaussian surface. Here (a) the closed surface can be of any shape or size. (b) q is the net charge enclosed inside the Gaussian surface; charges outside the surface are not considered. (c) E both inside and outside the surface. (d) The exact location of the charges inside the surface does not affect the value of the integral. Two equal point charges q each are held at x = + a and x = – a. A third charge Q is placed at x = 0. The potential of the system will (a) decrease if Q is displaced by a small distance along the x-axis (b) increase if Q is displaced by a small distance along the x-axis (c) decrease if Q is displaced by a small distance along the y-axis (d) increase if Q is displaced by a small distance along the y-axis The electric potential at a point P at a distance x from a point charge is given by k V= r where k is a constant. (a) k is dimensionsless (b) the dimensions of k are [ML–3T–3A–1] k P= 2. r P = kr2. Two point charges q1= 4 C and q2 = –1 C are placed at x = 0 and x = 15 cm on the x-axis. (a) Electric potential is zero at x = 3 cm. (b) Electric potential is zero at x = 12 cm. x = 0 and x = 15 cm. x = ± 30cm. Choose the correct statement(s) from the following. (a) The electric potential due to a point charge at a distance r from it varies as 1/r (b) Electric potential at a distance r from the centre of a charged sphere varies as 1/r provided r is less than the radius of the sphere r from a point charge varies as 1/r2 IIT, 1980

21. Two small balls, each having a charge + Q are suspended by two insulating strings each of length L taken in a satellite orbiting the earth. In the satellite the angle between the strings is and the tension in each string is T. Then (a) = zero (b) = 180° (c) T =

Q2 16

2 0L

(d) T =

Q2 4

2 0L

IIT, 1986 22. A positively charged thin metal ring of radius R is x-y plane with its centre at origin O. A negatively charged particle P is released from rest at the point (0, 0, z) where z > 0. Then the motion of P is (a) periodic for all values of z satisfying 0
Fig. 20.62

A in the cavity = electric B in the cavity. (b) Charge density at A = charge density at B (c) Potential at A = potential at B

20.32 Comprehensive Physics—JEE Advanced

cavity is

q

.

0

IIT, 1999

25. A positive charge q dipole with dipole moment p is placed along the x-axis far away from the origin with p pointing along the positive x-axis. The kinetic energy when it reaches a distance x from the origin is K and the magnitude of the force experienced by charge q at this moment is F. Then (a) K varies as 1/x (b) K varies as 1/x2 2 (c) F varies as 1/x (d) F varies as 1/x3 IIT, 2003 26. Electric potential V due to a spherically symmetric charge system varies with distance r as shown in Fig. 20.63. Given

V=

and

V=

Q 4

0 r0

Q 4

0r

for r < r0 for r > r0

Which of the following statements are true?

27.

(d) Electrostatic energy inside the sphere of radius r = r0 is zero. IIT, 2005

+Q, a charge –q is moving around it in an elliptical orbit. Find out the correct statement(s). (a) The angular momentum of the charge – q is constant (b) The linear momentum of the charge – q is constant (c) The angular velocity of the charge – q is constant (d) The linear speed of the charge – q is constant IIT, 2009 28. A spherical metal shell A of radius RA and a solid metal sphere B of radius RB (< RA) are kept far apart and each is given charge ‘+Q’. Now they are connected by a thin metal wire. Then (a) EAinside = 0 (b) QA > QB (c) (d)

A

RB RA

B EAon surface

< EBon surface

IIT, 2011 29. Which of the following statement(s) is/are correct? varies as r–2.5 instead of r–2, then the Gauss law will still be valid. (b) The Guass law can be used to calculate the Fig. 20.63

discontinuous at r = r0. (b) The net charge enclosed in a sphere of radius r = 2r0 is (c) No charge exists at any point in a spherical region of radius r < r0.

is zero somewhere, then the sign of the two charges is the same. (d) The work done by the external force in moving a unit positive charge from point A at potential VA to point B at potential VB is (VB – VA). IIT, 2011

ANSWERS AND SOLUTIONS 1.

a point, electric potential is not necessarily zero at point exactly mid-way between two equal charges of the same sign, but the potential at this point is twice that due to a single charge. Statement (b) is correct because E = – V. If V is constant V = 0. Hence E = 0 in that region. Statement (c) is also correct. The

is normal to the surface at that point. If two different equipotential surfaces intersect, there would be two directions for the normals to the two surfaces at the point of intersection. Hence there will be two not possible. Statement (d) is correct. Since electron has a negative charge, it has less potential energy at a point where the potential is higher and vice versa.

Electrostatic Field and Potential 20.33

the region of lower potential to the region of higher potential. 2.

charge + q is given by 1 q 4 0 r where r is the distance of that point from the charge. Referring to Fig. 20.35 (a), the potentials at points P and Q are q 1 q 1 and VQ = VP = O Q 4 4 0 OP 0 V=

since OP < OQ, VP > VQ. Hence statement (a) is correct. charge – q is given by V=–

1 4

0

q r

Referring to Fig. 20.35(b), the potentials at points A and B are q q 1 1 and VB = – VA = – 4 0 OA 4 0 OB Since OA < OB, the potential at A is more negative than at B, i.e. VB > VA. Hence statement (b) is in-correct. We know that the potential energy of a charge q2 q1 at a distance r from it is given by 1 q1 q2 U= 4 0 r Referring to Fig. 20.21(a) the potential energy of a small negative charge (– q0 tive charge (+ q) at points P and Q is (setting q1 = q and q2 = – q0) U p= –

1 4

0

q q0 1 and UQ = – OP 4

0

q q0 OQ

Since OP < OQ, it is clear that UQ > UP. Hence statement (c) is correct. Referring to Fig. 20.21(b), the potential energy of a small negative charge (– q0 tive charge (– q) at points A and B is (setting q1 = – q and q2 = – q0) UA =

1 4

0

q q0 1 and UB = OA 4

0

q q0 OB

Since OA < OB, UA > UB. Hence statement (d) is correct.

3. The small positive charge (+ q0) will tend to move from P to Q due to the force of repulsion exerted on it by the charge (+ q charge (+ q) does positive work on charge q0 to move it from P to Q. Hence the work done by the q) in moving the charge (+ q0) from Q to P will be negative. Hence statement (a) is incorrect. The work done by the external agency to move a negative charge (– q0) from B to A is positive, since the external agency has to overcome the force of repulsion exerted by the negative charge (– q) on the small negative charge (– q0). Hence statement (b) is correct. In going from Q to P, a small negative charge is speeded up due to the force of attraction exerted on it by the positive charge (+ q). Hence statement (c) is correct. In going from B to A, the small negative charge (– q0) is slowed down due to the force of repulsion exerted on it by the negative charge (– q). Hence the kinetic energy of the small negative charge decreases. Hence statement (d) is correct. 4. E in moving a charge (– q) around a closed path of any shape (circular or elliptical) is given by W = q E d1 Now, we know that the line integral of an electroE d1 = 0 Hence the work done W = 0 irrespective of whether the path is circular or elliptical. Hence statement (a) is correct. We know that the potential at points equidistant from a point charge is the same. Hence, the equicharge are concentric spheres with the point charge as the common centre. Hence statement (b) is correct. The derivation of Gauss’s law assumes 1/r2 dependence of distance between charges in Coulomb’s law. Gauss’s law will not hold if Coulomb’s law involved 1/r3 or any other power of the distance r. Hence statement (c) is incorrect. A single conductor can have capacitance. It is a caa single spherical conductor of radius r has a capacitance C = 4 0r. Hence statement (d) is incorrect.

20.34 Comprehensive Physics—JEE Advanced

5. Refer to Fig. 20.64. Since the bob is in equilibrium For equilibrium, we have mg = T cos and qE = T sin mg qE Thus T = = . cos sin

Now, AF = BF = 5 a/2 and CF = DF = a/2. Putting these values, we get VF =

1

q

1 5 Work done in carrying a charge –e from O to F is W = – e (VF – V0) = – eVF 1 qe = 1 a 5 0

Hence the correct choices are (c) and (b).

0

a

Hence the correct choice is (a) and (c). 7. The distance of the point of intersection of diagonals = side of the hexagon = a. The potential 1 q . Thereat this point due to each charge = 4 0 a 1 6q fore, total potential = 4 0 a point due to equal charges at opposite corners will cancel each other in pairs. So the correct choices are (b) and (c). 8. Refer to Fig. 20.66. Since the charge on sheet 1 is

Fig. 20.64

6. Refer to Fig. 20.65. Potential at O is V0 =

1 4

q r

0

q r

q r

q r

=0

E = /2 0 which points away from it; to the left in region I and to the right in regions II and III. Sheet E = /2 0. Since the charge of sheet 2 is negative, the direction E is towards it; i.e. to the right in regions I and II and to the left in region III.

Fig. 20.65

Refer to Fig. 20.65 again. (AE)2 = a2 + =

5 a2 , giving AE = 4

and BE =

5a . Similarly DE = 2

a 2

2

5a 2

a a , CE = . Potential at E is 2 2

VE =

1 4

0

q AE

q BE

q DE

q CE

Fig. 20.66

=0

Work done in carrying a charge – e from O to E is W = – e (VE – VO) = – e (0 – 0) = 0 Potential at F is q q q q 1 VF = 4 0 AF BF DF C F

Region I Region II

: E1 = E – E = – : E2 = E + E =

2

2 0

2

0

2

=0 0

= 0

directed to the right. Region III : E3 = E – E = 0. Thus the correct choices are (a); (b) and (c).

0

Electrostatic Field and Potential 20.35

r > r0 is given by dV d Q E= – =– dr dr 4 0 r

9.

= For r

Q 4

r 0, E = –

0r

2

d Q dr 4 0 r0

=0 (

0

E

4 r2 =

Q 0

or

Q 4

0r

2

4 r2 =

Q 0

or Q = Q, which is independent of r as long as r is greater than r0. Hence statement (b) is also true. All the four choices (a), (b), (c) and (d) are correct 10. For r R, E =

Qr 4

0R

3

, i.e. E

and

r0 = constant) r = r 0.

Therefore, statement (a) is true. For r < r0, E = 0. Hence the charge resides only on the spherical surface of radius r = r0. No charge exists in the region for which r < r0. Therefore, statement (c) is also true. Electric energy density is given by 1 2 u= 0E 2 Since for r < r0, E = 0; u = 0 for r < r0. Hence statement (d) is true. Let Q be the net charge enclosed inside the spherical surface of radius r = 2r0. Then from Gauss’s theorem, we have Q E = or

12. Force on proton is Fp = eE, in the direction of E and force of electron is Fe = eE, opposite to the direction of E. Also ap = Fp /mp and ae = Fe /me. Since mp > me; ap < ae. So the correct choices are (b) and (c). 13. P are Q 1 V= 2 4 0 ( R r 2 )1 / 2 E=

1 4

Qr 0

(R

2

r 2 )3 / 2

,

which is zero for r = 0. Q Q and E = . So the For r >> R, V = 4 0r 4 0r 2 correct choices are (b), (c) and (d). P is given by

14. E=

n 2

0r

Where is the linear charge density and n is a unit vector perpendicular to the wire. So the correct choices are (a) and (d). 15. P is n

E=

2 0 where n is a unit vector perpendicular to the sheet. Since E is independent of r form. Hence the correct choices are (c) and (d). 16. All the four choices are correct. 17. Initial P.E. of the system [Fig. 20.67(a)] is U= =

1 4

0

q 4

0a

qQ a

qQ a

2Q

q 2

q2 2a (1)

r

For r > R, the entire charge Q may be assumed to be concentrated at the centre of the sphere. Hence for r > R, Q , i.e. E 1/r2 E = 4 0r 2 Since the sphere is made of a non-coducting meterial, charge Q is uniformly distributed over the entire volume of the sphere. For a unifrom distribution of charge, the potential is maximum at the dicular to the surface of a conductor. Hence the correct choices are (b) and (c). 11. For a conductor, all the four choices are correct.

Fig. 20.67

If charge Q is given a displacemet x along the x-axis the P.E. of the system becomes [Fig. 20.67 (b)]

20.36 Comprehensive Physics—JEE Advanced

U = =

1 4

qQ (a x)

0

q 4

2Q

0a

2

2

a (a

2

2

q 2

2

x )

state of weightlessness. The distance between the balls = 2 L. Hence

q2 2a

qQ (a x)

(2)

2

Since a > (a – x ); it follows from Eqs. (1) and (2) that U > If charge Q is given a displacement y along the y – axis, the P.E of the system will be [Fig. 20.67(c)] U = =

1 4

0

q 4

qQ r

2

qQ r

2aQ r 0a

q 2a

q 2

(3)

Since r > a, it follows from Eqs. (1) and (3) that U < U. So the correct choices are (b) and (c). work [ML2 T 2 ] . Therefore [V] = charge =[ML2T–3A–1] [k] = [V] [r]

18. V =

= [ML2T–3A–1] 3 –3

T=

Q2 (2 L) 2

4

So the correct choices are (b) and (c). 22. If Q1 is the charge on the ring and – Q2 is the charge on particle P, the force due to the ring on particle P when it is at (0, 0, z) is F =–

1

Q1 Q2 z

(R2

4

z 2 )3 / 2

When z > 0, F is in the – z direction. When z < 0, F is in the + z direction. So the motion of P will be periodic for 0 < z < . 1 Q1 Q2 z , i.e. F – z. Hence When z << R, F = 4 R2 for z << R, the motion of P will be simple harmonic. So the correct choices are (a) and (c). 23. E with distance r. [Fig. 20.68]

[L]

–1

= [ML T A ] dV d k k Now E = – =– = 2 dr dr r r Hence the correct choices are (b) and (c). 19. Let the electric potential be zero at a point P at a distance x from charge q1. Then 1 4

0

q1 x

q1 x (r

(r q2

x)

q2

Fig. 20.68

=0

x)

4 1 = x (r x)

4r 4 15 cm = =12 cm 5 5 E is zero at a value of x given by

which gives x =

2

q1 x = q2 (r x)2 which gives x = ± 30 cm. It is easy to check that E 0 between x = 0 and x = 15 cm. So the correct choices are (b), (c) and (d). 20. All the four statements (a), (b), (c) and (d) are correct. 21. The two charges will repel and will come to rest when = 180° because an orbiting satellite is in a

So the correct choices are (a) and (c). 24. The surface of a perfect conductor is an equipotential surface. Charge density at B is greater than at A because curvature at B is greater than that at A q closed surface = 0

face of the cavity is zero. Hence the correct choices are (c) and (d). 25. Potential energy at position x is U =–

1 4

qp x2

When the dipole is far away (x K = change in P.E. = 0 –

), U = 0. Hence 1

4

qp x2

Electrostatic Field and Potential 20.37

= F= 26.

1

qp

4

x

2

dU 2q p =– dx 4 x3

Hence the correct choices are (b) and (d). r > r0 is given by dV d Q =– dr dr 4 0 r

E =–

= For r

r 0, E = –

4

Q 4

0r

d Q dr 4 0 r0

=0 r0 = constant) r = r 0.

Therefore, statement (a) is true. For r < r0, E = 0. Hence the charge resides only on the spherical surface of radius r = r0. No charge exists in the region for which r < r0. Therefore, statement (c) is also true. Electric energy density is given by

1 2 u= 0E 2 Since for r < r0, E = 0; u = 0 for r < r0. Hence statement (d) is true. Let Q be the net charge enclosed inside the spherical surface of radius r = 2r0. Then from Gauss’s theorem, we have Q E = 0

or

4 r2 =

E

Q 0

or

Q 4

0r

2

4 r2 =

QA

QB

4

0 RA

QA QB

0 RB

RA RB

Since RB < RA, QA > QB. So choice (b) is correct. cal shell is zero. So choice (a) is correct.

2

(

are wrong because, for elliptical orbit, the speed of charge – q is the highest when it is closest to charge Q (as in planetary motion). Hence the only correct choice is (a). 28. Let QA and QB are the charges on metal shell A and metal sphere B after they are connected by a wire. Since their electric potentials will be equal, VA = VB

Q

Now

A

=

QA 4 A

and

RA2

=

B

QA QB

B

RB RA

QB

=

4 RB2 2

RA RB =

RB RA

2

RB RA

Hence choice (c) is also correct. are A

EA = And

0 B

EB =

0

EA = EB

A B

< 1, i.e. EA < EB.

So choice (d) is also correct. All the four choices are correct. 29. Gauss’s law is valid only if Coulomb's law holds, i.e. if E r–2. Hence choice (a) is wrong. Gauss’s

0

or Q = Q, which is independent of r as long as r is greater than r0. Hence statement (b) is also true. All the four choices (a), (b), (c) and (d) are correct 27. The torque of Coulomb force (which is radial) on dL charge – q is zero. Hence = 0 L = constant. dt Hence the angular momentum of charge – q is constant. So choice (a) is correct. All other choices

distribution around an electric dipole. So choice (b) is also wrong. Choice (c) is correct becausse the directions of similar charges. Work done WA

= q(VB – VA) = (VB – VA) ( q = +1 C) Hence choice (d) is correct. So the correct choices are (c) and (d). B

20.38 Comprehensive Physics—JEE Advanced

III Multiple Choice Questions Based on Passage Questions 1 to 4 are based on the following passage Passage I q, are placed along the x-axis at x = 1, x = 2, x = 4, x = 8 .... and so on. 1. The electric potential at the point x = 0 due to this set of charges is q q (b) (a) 2 0 0 q q (c) (d) 3 0 4 0

3. If the consecutive charges have opposite sign, the electric potential at x = 0 would be q

(a)

3

(a) (c)

q 3

(b) 0

q 5

(d) 0

q 4

0

0

q q (d) 5 0 6 0 4. If the consecutive charges have opposite sign, the x = 0 would be q

(a)

3

=

=

0

q 4

q 1 1

0

1

q 4

0

q 2 1 2

q 4 1 4

q 8 1 8

6

5

0

q

(d)

6

0

0

0

to to

1 2 1 1 2

(1 0) q = = , 1 4 0 2 0 2 which is choice (b). 2. Since the charges are placed along the same straight x = 0 will be directed along the x-axis and its magnitude is given by q

4

0

E=

1 4

q

q

q

2

2

2

82

1

0

q 4

q

=

=

2 1 4

1 0

Fig. 20.69

1

q

(b)

q

(c)

=

4

0

q

SOLUTION 1. The potential at x number of charges placed on the x-axis as shown in Fig. 20.69, is

V=

4

(c)

x = 0 is

2.

q

(b)

1

q 4

0

1 16

0

to

1 64

to

1 4 1 1 4

(1 0) q = 3 3 4

q 4

4

, which is choice (a). 0

3. If the consecutive charges have opposite sign, the potential at x = 0 is given by V= = –

1 4

0

q 4 1 2

q 1

q 2

q 4

1

1 4

1 16

0

1 8

1 32

to

q 8

q 16 to

q 32

to

Electrostatic Field and Potential 20.39

=

=

1

q 4

4 3

q 0

1 2

1

1 1 4

=

4 q = 3 6

–

1 1 4

0

4

1 2

0

Hence the correct choice is (d). 4. E =

1 4

q

q

2

2

2

1

0

q (16)

q

2 q

2

(32)

2

4

=

q 4

0

1 4

1 64

=

to

Question 5 to 9 are based on the following passage Passage II A rigid insulated wire frame in the form of a right-angled triangle ABC, is set in a vertical plane as shown in Fig. 20.70. Two beads of equal masses m each and Fig. 20.70 carrying charges q1 and q2 are conn-ected by a cord of length l and can slide without friction on the wires. The beads are stationary. IIT, 1978 5. The value of angle is (a) 30° (b) 45° (c) 60° (d) 75° 6. The tension in the chord is q1q2 (a) mg (b) 4 0l 2

1 256

1 1024

to

1

q 4

0

q (8)2

1 16

1

q 4

0

1 4

1 1 16 16 15

1 16 4 15

to

1

1 16

1

q 5

0

Hence the correct choice is (c).

(c) mg +

q1q2

(d) zero 4 0l 2 7. The normal reaction on bead P is (a)

2 mg

(b) 2 mg

(d) 3 mg (c) 3 mg 8. The normal reaction on bead Q is (a) mg

(b)

2 mg

(c) 2 mg

(d)

3 mg

9. If the cord is cut, the magnitude of the product q1 q2 of the charges for which the beads continue to remain stationary is (a)

mgl 2 4 0

(c)

3 (4

(b) )mgl2

(d) (4

3mgl 2 4 0 )mgl2

SOLUTION F =

1 4

q1 q2 0

l2

(iv) Normal reaction N1.

Fig. 20.71

Let us consider forces acting on bead P as shown in Fig. 20.71. These forces are: (i) Weight mg vertically downwards (ii) Tension T in the string (iii) Electric force between P and Q given by

The net force along the string is (T – F). Bead P will be in equilibrium, if the net force acting on it is zero. Resolving forces mg and (T – F) parallel and perpendicular to plane AB, we get, when the bead P is in equilibrium, mg cos 60° = (T – F) cos (1) and

N1 = mg cos 30° + (T – F) sin

(2)

For the bead at Q, we have mg sin 60° = (T – F) sin and N2 = mg cos 60° + (T – F) cos

(3) (4)

20.40 Comprehensive Physics—JEE Advanced

5. Dividing Eq. (3) by Eq. (1), we get tan = tan 60° or = 60°, which is choice (c). 6. Using = 60° in (3), we have mg sin 60° = (T – F) sin 60° 1

q1 q2

+ mg 4 0 l2 So the correct choice (c). 7. From Eq. (2) we have (since T – F = mg) N1 = mg cos 30° + mg sin 60° or T = F + mg =

= 2 mg cos 30° =

(5)

3 mg;

which is choice (c). 8. From Eq. (4) we have

N2 = mg cos 60° + mg cos 60° = mg Thus is the correct choice is (a). 9. When the string is cut, T = 0. Putting T = 0 in Eq. (5), we get 1 q1 q2 mg = – 4 0 l2 The right hand side of this equation should be positive which is possible if q1 and q2 have opposite signs. Thus, for equilibrium the beads must have unlike charges. The magnitude of the product of the charges is |q1 q2| = (4 0) mgl2, which is choice (d).

Questions 10 to 13 are based on the following passage (a) T = 2

Passage III Two charges, each equal to q, are kept at x = – a and x = a on the x-axis. A particle of mass m and charge q0 = q/2 is placed at the origin. 10. The charge q0 is given a small displaceplacement x (<< a) along the x-axis and then released. The restoring force acting on q0 is (k = 1/ (4 0)) kq 2

(a) – (c)

a3 kq 2 a2

2kq 2

(b) –

x

2kq 2

(d)

x

x

a3 a2

x

11. In Q.10 the time period of oscillation of the particle is

(c) T = 2

ma3

1/ 2

(b) T = 2

2kq 2 2kq 2

1/ 2

ma3

1/ 2

kq 2 kq 2

1/ 2

(d) T = 2 ma3 ma3 12. If charge q0 is given a small displacement y (<< a) along the y-axis, the net force acting on the particle is proportional to (a) y (b) – y 1 1 (d) – (c) y y 13. In Q. 12, the particle, (a) will execute simple harmonic motion. (b) will execute oscillatory but not simple harmonic motion. (c) will execute a non-periodic and non-oscillatory motion. (d) will never come back to x = 0.

SOLUTION 10. Refer to Fig. 20.72. Suppose the charge q0 = q/2 is given at a small displacement x from origin O. Then the force of repulsion on charge q0 due to charge q1 = q is F1 =

1 4

q1 q0 0

a

x

2

k

q

q/2

a

x

2

along positive x-direction The force of repulsion on charge q0 due to charge q2 = q is F2 =

1 4

q1 q0

kq 2

2( a a x along negative x-direction 0

2

x) 2

Fig. 20.72

Since F2 > F1, the net force on charge q0 is along the negative x-direction. Hence the restoring force on charge q0 is F = F1 – F2

Electrostatic Field and Potential 20.41

= k

q2 2

since x << a,

1 a

x

1 a

x

1 2

a

2

1

=

2

x x a

a2 1

2

x 2 a = a2 Expanding binomially and retaining only the terms x of order , we have a 2x 1 1 1 2x a = = 2 2 2 a a3 a a x 1

Similarly, Hence F= k

q2 2

1 a

2

x

=

1

2x

2

a3

a

2x

1

2x

2

3

2

a3

=–k

a

a

F = (F1)y + (F2)y = F1 cos = (F1 + F2) cos or cos

=

q2 y a2

y2

3/ 2

+ F2 cos

= 2F1cos

(

F 1 = F 2)

2

q cos

F= k

Now

F= k

1 a

F1 cos along positive y-direction. Similarly, force F2 can be resolved into two components (F2)x = F2 sin along negative x-direction and (F2)y = F2 cos along positive y-direction. Since F1 = F2, components (F1)x and (F2)x are equal and opposite and hence they cancel each other. The net force on charge q0 is along the positive y-direction and is given by

a2

y2

OC AC

y a2

y2

1/ 2

. Therefore,

along positive y-direction

2 q2 x a3

which is choice (b). 11. Since F (– x), the motion of the particle is simple harmonic whose time period is given by choice (a). 12. Refer to Fig. 20.73. Suppose the charge q0 = q/2 is given a small displacement OC = y along the y-axis, the force of repulsion on charge q0 due to the charge q1 = q is q q /2 q q0 k F1 = k = a2 y2 AC 2 The force of repulsion on charge q0 due to charge q2 = q is q q0 q q /2 = k 2 F2 = k 2 a y2 BC The directions of F1 and F2 are shown in Fig. 20.73. Force F1 can be resolved into two components (F1)x = F1 sin along positive x-direction and (F1)y =

Fig. 20.73

Since y << a, (a2 + y2)3/2 F= k

a3. Thus

q2 y a3

So the correct choice is (a). 13. Since the net force on charge q0 is along the postive y-direction, it will keep on moving in the positive y-direction away from the origin O and will never come back. So the correct choice is (d).

Questions 14 to 16 are based on the following passage Passage IV A point particle of mass M is attached to one end of a massless rigid non-conducting rod of length L. Another point particle of the same mass is attached to the other end of the rod. The two particles carry charges + q and – q. This E such that the rod makes a small angle (say of about 5°)

Fig. 20.74

IIT, 1989

20.42 Comprehensive Physics—JEE Advanced

14. The magnitude of the torque acting on the rod is (a) qEL sin (b) qEL cos (c) qEL (d) zero 15. When the rod is released, it will rotate with an angular frequency equal to (a)

qE ML

(c)

qE 2ML

1/ 2

1/ 2

(b)

2qE ML

(d)

1 qE 2 ML

16. The minimum time taken by the rod to align itself (a)

ML 2 2qE

1/ 2

(c) 2

1/ 2

2 ML qE

(d) 2

ML qE

(d) 2

ML 2qE

1/ 2

1/ 2

1/ 2

1/ 2

SOLUTION I= M

14. A non-conducting rigid rod having equal and opposite charges at the ends is an electric dipole.

L 2

=M

experiences a torque which tends to align it with forces F = qE each acting at A and B constitute a couple whose torque is given by = force perpendicular distance = F AC = F AB sin = qEL sin So the correct choice is (a).

(AO)2 + M

Thus

=

2

+M

(BO)2 L 2

2

=

M L2 2

M L2 2

(2)

Using Eq. (2) in Eq. (1), we get =– where

=

2q E ML

2q E ML 1/ 2

=–

2

, which is choice (b).

16. The time period of oscillation is T= Fig. 20.75

15. Since is small, sin radian. Thus = qEL Restoring torque

, where

is expressed in

= – qEL

(1)

If is the angular acceleration of the rotatory motion, =I where I is the moment of inertia of the two masses at A and B about an axis passing through the centre O and perpendicular to the rod. Since the rod is massless,

2

ML 2q E

=2

1/ 2

Rotating in the clockwise sense, the minimum time taken by the rod to align itself parallel to the elecof angular oscillation, i.e. tmin =

T 4

2

ML 2q E

1/ 2

So the correct choice is (a).

Questions 17 to 20 are based on the following passage

17. At the closest approach, speed v1 of particle A is

Passage V

(a) v1 = v (b) v1 = 2v (c) v1 = v/2 (d) v1 = v/ 2 18. At the closest approach, speed v2 of particle B is

Two identical particles A and B of mass m carry a charge Q each. Initially particle A is at rest on a smooth horizontal plane and the partical B is projected with a speed v along the horizontal plane from a large distance directly towards the first particle. The distance of closest approach is x.

(b) v2 = v/ 2 (a) v2 = v/2 (c) v2 = v (d) v2 = 2v 19. At the distance of closest approach, the total energy of the system is

Electrostatic Field and Potential 20.43

(a)

mv 2 2

(c) mv 2

(b) Q2 4

0x

mv 2 2

mv 2 (d) 4

Q2 4

20. The distance of closest approach is

0x

Q2

(a) x =

0 mv

Q2 4

0x

2

Q2

(c) x =

4

Q2

(b) x =

0 mv

2

0 mv

2Q 2

(d) x =

2

2

0 mv

2

SOLUTION 17. Initially particle A is at rest and particle B moves towards it with a velocity v from a large distance. Therefore, the initial momentum of the system is pi = mv + 0 = mv Since the initial distance is large, the particle B exerts move in the same direction as the second particle under the action of the force of repulsion. The distance between them, therefore, keeps decreasing until it attains a certain minimum value. Let v1 and v2 be the velocities of particles 1 and 2 and let t be the time taken by them to acquire these velocities. Then the distances travelled by them will be x1 = v1t and x2 = v2t. The separation between them at this time t is x = x1 – x2 = (v1 – v2)t This separation will be minimum if dx/dt = 0. Now dx = v1 – v2 dt For closest approach, v1 – v2 = 0 or v1 = v2.

18. The correct choice is (a). 19. If x energy of the system is Ef =

1 1 1 mv21 + mv22 + 2 2 4

1 m 2

or = or

Ef =

Ei

1 m 2

v 2

m v2 1 + 4 4

0

2

1

+

4

0

Q2 x

Q2 x

particle + KE of second particle + PE due to charge on each particle. 1 1 Q2 mv 2 + × =0 + 4 0 2

1 mv2 2 From the law of conservation of energy, Ei = Ef. Therefore or

Ei =

1 1 1 mv2 = mv2 + 2 4 4

From the law of conservation of momentum, pi = pf or mv = m (v1 + v2) or v = v1 + v2 Since the particles are identical, it follows that

0

Q2 x

which gives x=

v . 2 So the correct choice (c). v1 = v2 =

1 4

0

4 Q2

Q2

m v2

0

m v2

so the correct choice is (a).

Questions 21 to 24 are based on the following passage Passage VI Charge Q is uniformly distributed within a non-conducting sphere of radius R. The volume charge density is . IIT, 1992 21. The potential energy of the spherical distribution of charge is 3Q 2 20 0 R

+

0

which is choice (d). 20. The initial energy of the system is

closest approach is pf = mv1 + mv2 = m (v1 + v2)

(a)

2

v 2

Q2 x

(b)

2Q 2 5 0R

(c) 22.

Q2 4

0R

(d)

3Q 2 10 0 R

(> R ) from the centre of the sphere is (a) (c)

3 R3 0r

2

R3 3 0r 2

(b) (d)

3 R3 4 0r 2 3 R3 2 0r 2

r

20.44 Comprehensive Physics—JEE Advanced

23. If the same charge is given to a spherical conducting sphere of the same reduis R, the potential energy of the system will be (a) zero (c)

(b)

Q2 8

Q2 4

0R

(d) the same as in Q.21.

0R

24.

tance r (> R) from the centre of the sphere will be Q QR (b) (a) 4 0r 2 4 0r3 (c)

3Q 4

0r

3QR

(d)

2

-

4

0r

3

SOLUTION 21.

=

Q V

3Q

Q

(1)

3 4 R3 4 R 3 Refer to Fig. 20.76.

Hence, the total potential energy of the spherical distribution of charge is U=

2

4 3

R

0

r4dr =

0

4 3

2 0

R5 5

Using Eq. (1) in this we have 2

spherical shells. Consider one such spherical shell of radius r and thickness dr. The charge within the volume (4/3) r3 is

U=

4 r3 3 The charge in the spherical shell of thickness dr is dq = 4 r2dr

=

q=

Therefore, the potential energy due to charges q and dq is dU =

1 4

0

qd q r

=

4 4 3

2 0

4 r2 d r r

0

r4dr

4 R3 3

0

R5 5

3Q2 20 0 R

Q 4

0r

2

=

R3 3

0r

2

, which is choice (c).

23. Any charge given to a solid conductor, cannot stay in its body; it is immediately transferred to the surface of the conductor. The electrostatic energy of a charge Q on the surface of a conducting sphere of R is given by Q2 U= 2C where C = 4 0 R is the capacitance of the sphere. Thus

4 3 r 3

1

Q

So the correct choice is (a). 22. For points outside the sphere, the charge Q may be assumed to be concentrated at the centre. Hence (For r >R), E=

Fig. 20.76

=

4 3

U=

Q2 8

0

R

So the correct choice is (c). 24. For points outside the sphere, it behaves if the entire charge Q is concentrated at its centre. Hence the correct choice is (a).

Electrostatic Field and Potential 20.45

25. Take the electronic charge as ‘e’ and the permittivity as ‘ 0’. Use dimensional analysis to determine the correct expression for p.

Questions 25 to 26 are based on the following passage Passage VII A dense collection of equal number of electrons and positive ions is called neutral plasma. Certain solids containing treated as neutral plasma. Let ‘N’ be the number density of free electrons, each of mass ‘m’. When the electrons are becomes zero, the electrons begin to oscillate about the positive ions with a natural angular frequency ‘ p’ which is called the plasma frequency. To sustain the oscillations,

26.

an angular frequency

, where a part of the energy is approaches all the free electrons are set to resonance together and p

(a)

Ne m 0

(b)

m 0 Ne

(c)

Ne2 m 0

(d)

m

0

Ne2

tion will occur for a metal having the density of electrons N 4 1027 m–3. Taking 0 = 10–11 and m 10–30, where these quantities are in proper SI units. (a) 800 nm (b) 600 nm (c) 300 nm (d) 200 nm

IIT, 2011

SOLUTION 25. From F =

q1q2 4

= [ML3T–2]

0r

2

, dimensions of

26.

e2

= [Fr2]

p=

0

=

[N] = [L–3]. Hence Ne2 m 0 2 1/ 2

Ne m 0

= [L–3] –1

Ne2 m 0

4 1027 10

= 3.2

[ML3T–2]

[M–1] = [T–2]

= [T ] = dimension of

p

= =

p.

1/ 2

p

2 c

10

(1.6 10 30

15

10

)

11

rad s–1

. Since c = =

19 2 1/ 2

,

3 108

2 15

3.2 10

p

6

10–7 m = 600 nm

IV Assertion–Reason Type Questions In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each question has the following four choices out of which only one choice is correct. (a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1 (b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Statement-1.

(c) Statement-1 is true, Statement-2 is false. (d) Statement-1 is false, Statement-2 is true. 1. Statement-1 Figure 20.77 shows the tracks of two charged partices A and B two charged plates. The charge to mass ratio of B is greater than that of Neglect the effect of gravity.

20.46 Comprehensive Physics—JEE Advanced

7. Statement-1 In Q. 6 above, the work done to move an electron from P to Q and then back to P is zero. Statement-2 8. Statement-1 Fig. 20.77

Statement-2 The vertical acceleration of particle B is greater than that of particle 2. Statement-1 A positive charge + q is located at the centre of a circle as shown in Fig. 20.78. W1 is the work done in taking a small positive charge + q0 from A to B and W2 is the work done Fig. 20.78 in taking the same charge from A to C. Then W2 > W1 Statement-2 Work done = Charge potential difference.

9. Statement-1 Electrons move from a region of higher potential to a region of lower potential. Statement-2 An electron has less potential energy at a point where the potential is higher and vice versa.

3. Statement-1 A small test charge is initially at rest at a point in released, it will move along the line of force passing though that point. Statement-2 The tangent at a point on a line of force gives the 4. Statement-1

-

tial must also be zero at that point. Statement-2 potential.

5. Statement-1 If electric potential is constant in a certain region of Statement-2 potential. 6. Statement-1 If an electron is moved from P to Q, its potential energy increases (see Fig. 20.79) Statement-2 Potential at Q is less than that at P.

Fig. 20.79

moving an electron around it in a complete orbit is greater if the orbit is elliptical than if the orbit is circular. Statement-2

10. Statement-1 The equipotential surfaces corresponding to a x-direction are equidistant planes parallel to the y-z plane. Statement-2 Electric is normal to every point on an equipotential surface 11. Statement-1 is uniform. Statement-2 point charge is a sphere with the charge at its center. 12. Statement-1 If a metallic sphere A of radius r carrying a charge Q is brought in contact with an uncharged metallic sphere B of radius 2r, the charge on sphere A reduces to Q/3. Statement-2 A to B until their potentials are equalised. 13. Statement-1 A metal ring of radius R carries a charge + Q distributed uniformly. A point charge – q is placed on the axis of the ring at a distance x = 2R from the centre of the ring. If the charge is released from rest, it will execute a simple harmonic motion along the axis of the ring. Statement-2 In simple harmonic motion, the restoring force acting on the oscillator is proportional to (– x), where x is the displacement from the mean position. IIT, 1988

Electrostatic Field and Potential 20.47

charge Q uniformly distributed on the surface is Q . given by 4 0R

14. Statement-1 For practical purposes, the earth is used as a reference at zero potential in electrical circuits. Statement-2 The electrical potential of a sphere of radius R with

IIT, 2008

SOLUTION 1. The correct choice is (a). Let E be the electric

it from Q back to P an equal positive work is done

downwords. If q is the charge of the particle, it will experience a force F = qE. Hence its acceleration (in the vertical direction) is

So the correct choice is (a). 8. The correct choice is (d). The work done by the

F qE m m where m is the mass of the particle. If t is time spent by the particle between the plates (i.e in the region

path of any shape (circular or elliptical) is zero. 9. The correct choicce is (d). Since the electron has a negative charge, it has less energy at a point where the potenial is higher and vice versa. Hence in an

a=

lower potential to a region of higher potential. y=

1 2 qEt 2 at = 2m 2

q y. Since particle B m tion, it has a higher charge to mass ratio. The correct choice is (d). Points A, B and C are at the same distance from charge + q. Hence electric potential difference between points A, B and C is zero. Hence W1 = W2 = 0. The correct choice is (d). The test charge will move along the line of force if the line of force is straight (as in the case of a single charge). If the lines of force is curved, the charge will not move along the line of force. The reason is that the line of force does not give the direction of velocity, it gives the direction of the force which is along the tangent to the curve at that point. The correct choice is (d). Since E = – dV/dr, if E = 0, V = constant not necessarily equal to zero. The correct choice is (a). The correct choice is (c). Since charge of an electron is negative, P.E at P and Q is eq UP = – 4 0 (OP)

10.

Thus 2.

3.

4. 5. 6.

UQ = –

4

eq 0 (OQ )

Since OQ > OP, UQ is less negative than UP, i.e.UQ > UP. For the same reason, VQ > VP. 7. Since charge + q will attract the electron, work is done to move the electron from P to Q is negative

11.

tential surface. Therefore, for a constant electric x-direction, the equipotential surfaces are planes parallel to the y-z constant, the equipotential surfaces are equidistant from each other. The correct choice is (a). region around a point charge varies with distance E=

12.

4

0r

2

A to B until thier potentials become equal. If charge q A to B, then Q q q = 4 0r 4 0 ( 2r ) which gives Q – q = left on A = Q –

13.

Q

q 2

q=

2Q . Hence charge 3

2Q Q = . 3 3

x from its centre is given by E=

1

Qx

4

(R

2

x 2 )3 / 2

Force on charge – q is F=–qE=–

1 4

qQ x (R

2

x 2 )3 / 2

The motion will be simple harmonic if F – x which is true only if x << R and not when x = 2R.

20.48 Comprehensive Physics—JEE Advanced

Hence Statement-1 is false, but Statement-2 is true.

14. Both the statements are true but Statement-2 is not the correct explanation for Statement-1.

V Integer Answer Type 1. Two identical charged spheres are suspended by strings of equal length. The strings make an angle of 30° with each other. When suspended in a liquid of density 800 kg m–3, the angle remains the same. What is the dielectric constant of the liquid? The density of the material of the sphere is 1600 kg m–3. IIT, 1978 2. A circular ring of radius R with uniform positive charge density per unit length is located in the y-z plane with its centre at the origin O. A particle of mass m and positive charge q is projected from the point P (R 3 , 0, 0) on the positive x-axis directly towards O, with initial speed v. The smallest (nonzero) value of speed v such that the particle does not return to P is q , where n is an integer. Find the value v= n 0m of n. IIT, 1993

3. A solid sphere of radius R has a charge Q distributed in its volume with a charge density = kra, where k and a are constants and r is the distance R 1 r = is 2 8 times that at r = R a. IIT, 2009 4. Four point charges, each of +q of side ‘a is in equilibrium, and a = k

q2

1/ N

, where ‘k’ is

a constant. What is the value of N? IIT, 2011

SOLUTIONS If q is the charge on each sphere and r the separation between them, force F is given by

1. It follows from Fig. 20.80 that W = T cos and F = T sin F = tan W

(1)

F=

1

q2

4 0 r2 When the spheres are immersed in a liquid of dielectric constant K, the force of repulsion between them becomes F K Apparent weight is W = W – (weight of liquid displaced) F =

or Fig. 20.80

W = W – Vg

where is the density of the liquid. If is the density of the material of the sphere, W = Vg. Thus

Electrostatic Field and Potential 20.49

W =W 1

Vg W

=

=

As the angle Eq. (1) that tan Hence K 1

=

F W

Qq ( A P)

0

Now (AP)2 = (OA)2 + (OP)2 = R2+

W =W 1 F W

1 1 mv2 + 2 4

K 1

1 1 m v2 + 2 4

Therefore, EP =

remains unchanged, it follow from F F = W W 1 1

Given = 800 kg m–3 and Therefore, 1 K= 800 1 1600

2

=4R2

AP = 2R.

1

= 1 or K =

3R

=

= 1600 kg m–3. 1 1

1 2

=2

2. The positive charge on the ring is Q =2 R A particle with a positive charge q is projected towards the centre O of the ring with a certain velocity from a point P (lying on the x-axis) whose x-coordinate is 3 R. Due to repulsion of charge Q, the particle will keep slowing down. Now, at the centre O of a ring, carrying charge, the electric will be zero when it reaches O. It will then stop there and will not return to P. Let v be the smallest velocity of the particle when it is at P which is just O where it comes to rest. [see Fig. 20.81]

The total energy of the particle when it reaches O is (since its velocity at O is zero) E O = PE at O due to charge Q on the ring Qq 1 = 4 0 R From the law of conservation of energy, EP = EO. Thus Qq Qq 1 2 1 1 mv = 2 4 0 2R 4 0 R 1

which gives v =

4

1/ 2

Qq mR

0

1/ 2

=

q 2 0m

(

Q=2 R )

Thus the value of n = 2. 3. According to Gauss’s theorem 2

4 r =

E

q

dV

=

0

4 r 2 dr

=

0

0

where q is the charge enclosed inside a Gaussian r= sphere of radius r. If E1 R/2, then [see Fig. 20.82] E1

4

R 2

2

= =

1

R/2

0

0

4 r 2 dr

R/2

4 0

=

k r a r 2 dr

0 R/2

4 k 0

r (a

4 k r (a ( ) a 3 0

=

4 k 0 ( a 3)

E1 =

2)

dr

0

=

Fig. 20.81

The total energy of the particle at point P is EP = KE due to its motion + PE due to charge Q on the ring

0

Qq 2R

4 k 0 (a

3) R 2

3) R / 2 0

R 2

a 3

R 2

a 3

20.50 Comprehensive Physics—JEE Advanced

q2

F1 =

4

If E2 E2

4 R2 =

4 k 0

r

2)

(a

= dr (R)

0 (a

=

a+3

(R)a + 3

3) R 2

4

0 (a

3) R 2

R 2

2(a + 3) = 32

a 3

0a

2

+

4

0

2

2

q2 4

0a

4

02

2a 2

1 2

0a

2

2

1 2

=2 a q2

1/3

where k is a constant. Hence N = 3 =

k 8

0

(a 3) R 2

2(a + 3) = (2)5

4. Refer to Fig. 20.83.

Fig. 20.83

2a 2

2q 2

a= k

Hence 4k

4

q2

E1 1 = E2 8

Given

2

q2

Force on AB due to surface tension = 2 a. Hence

k

E2 =

2a

0

2q 2

0

4 k 0 ( a 3)

=

4

=

F = 2F1 + 2F2 cos 45°

r = R, then R

2

q2

F2 =

Fig. 20.82

0a

( R)

a 3

a=2

21

Capacitance and Capacitors

Chapter

REVIEW OF BASIC CONCEPTS 21.1

21.3

CAPACITANCE

When a conductor is given a charge Q, it acquires a potential V which is proportional to the charge given to it, i.e. Q V where C is a constant of proportionality and is called the Q

V or Q = CV or C =

Consider a spherical conductor of radius r having a charge Q i.e. the charge Q V=

.

4

Q r

Q C= V

coul volt

C= 4 -

r

capacitance.

21.4

–6

21.2

CAPACITANCE OF A SINGLE SPHERICAL CONDUCTOR

CAPACITORS

ENERGY OF A CHARGED CONDUCTOR

Q have a potential difference V Q = CV where C is the capacitance of the capacitor and its value

Q is given to a conductor of capacitance C, the potential energy in its

U =

Q = C

C Cm is a dielectric other than air, then the dielectric constant of

QV =

CV

Q = CV K=

Cm C

21.2 Comprehensive Physics—JEE Advanced

21.5

EXPRESSIONS FOR CAPACITANCE

Parallel Plate Capacitor C=

K

A

21.7

d where A is the area of each plate and d is the disK is dielectric constant of the

=

C

C

+

+

C

+

C3

ENERGY STORED IN A CAPACITOR

As in the case of a charged conductor, the energy stored Q = QV = CV C where Q = charge on each plate of the capacitor, V = C = capacitance U=

K Spherical Capacitor A spherical capacitor consists of a solid charged sphere of radius a low sphere of radius b K

C= 4

ab b a

3. Cylindrical Capacitor A cylindrical capacitor consists of two co-axial Kl b log e a where l is the length of each cylinder and a and b are the radii of the inner and outer cylinders.

21.8

LOSS OF ENERGY ON SHARING CHARGES

Q and Q and having capacitances C and C are connected with each

C=

V=

C=

21.6

d K

Q C

CV C

C C C C

E=

21.9

C V C

V –V

FORCE BETWEEN PLATES OF A PARALLEL PLATE CAPACITOR

CAPACITORS IN PARALLEL AND SERIES F=

capacitances.

V=

Q Q = C C

=

Q3 = C3

C = C + C + C3 +

Q = C V = C V = C3 V3 = V = V + V + V3 +

where

Q K A

F Q = A K A is the charge per unit area. f=

Q = Q + Q + Q3 + and

=

where V and V are the initial potentials of the charged

d and d having dielectric constants K and K , then A d K

Q C

21.1

=

K

Capacitance and Capacitors 21.3

V d

3

4

21.2 SOLUTION Q = CV

V d

–6

C

–6

CV =

U= E=

–6

–3

.

–3

Given C

C

C3

J

4

=4

Fig. 21.1

NOTE

SOLUTION

plates

-

C =

A d

C

C and C3 is C = C + C3 = 3

Fig. 21.2

C C

–6

and C Q V = C U =

CV

C

C

–6

=

–3

=4

=

C

C

C= C C = C C

J

+

=

6

Let Q , Q and Q3 Alternatively, U = E =

Q C V d

3

–3

=4

6

C , C , and C3 C and C are in series, charge on C = charge on C = Q . Let V and V differences across C and C C V Q = C V = CV V C 3 But V +V

J

4

=4

-

V +

3

V

V

V A d

C =

C

Q =CV –6

–6

Q =C V –4

C V

U = =

–6

–3

C

J

C

e C and C3 are in parallel, potential difference V across C = potential difference V3 across C3 = V Charge on capacitor C = C V C and Charge on capacitor C3 = C3V3 C C Q C, Q C and Q3

21.4 Comprehensive Physics—JEE Advanced

21.3

C 3 when S is closed.

S SOLUTION

SOLUTION

S –6

C

V U=

–6

CV =

=6

–3

J

Notice that CV +

U = U + U + U3 =

CV +

C 3V 3 .

21.4 C and C have a total a total capacitance 4.8 C and C .

Fig. 21.4

C and C3 are in series, charge on C = charge on C3 = Q tion of C and C3 C V = C 3V 3 Q = C V = C 3V 3 V = 3V3. Also V + V3

SOLUTION C +C

V3 S is closed, the circuit is as shown

CC C C C C C =

= 4.8

C C = 96

96 C

C

C

Fig. 21.5

C

C C =3+4=7

21.5 C C3 = 3

C4 = 4

C

C and C are in series, the charge on C = charge on C or CV =C V 3V = 7V 7V V = 3 7V Also V +V +V 3 V = 3V C3 and C4 are in parallel, potential difference across C3 = potential difference across C4 21.6

Fig. 21.3

S

S is closed.

Capacitance and Capacitors 21.5

SOLUTION A and B is Ca = C + C =

C C C C

C C C C

Fig. 21.7 6

=

=

C =

CC C C

C =

C3C4 C3 C4

Energy stored is Ua =

Ca V

.

4

J

4 8

CAB = C + C A and B is C C 3 7 Cb = C C 3 7

NOTE

6

=

Energy stored is Ub =

-

CbV

.

4

J

21.7 A and B C

C4

C3

found. Q

C C

Q

principle of charge conservation. Let Q ,Q the charges on the capacitors in the network and V , V Q = CV Q ,Q , ... etc. and V , V Q capacitance Ceq = , where E is the voltage of the E Fig. 21.6

21.8 A and B in

SOLUTION

C

C , C , C3 and C4 are such that the condition C3 C C C4 potential at P = potential at Q P and Q connected across A and B C is ineffective as no charge collects on its plates.

Fig. 21.8

SOLUTION

21.6 Comprehensive Physics—JEE Advanced

C and upper plate of C is

Fig. 21.10 Fig. 21.9

E A and B +Q

Q.

VA – VB

VA – VP

Q E= C CE Q Also

VA – VB

VA – VP

Q C CE = Q

q= – Q = – 4 C

VP – VB

Now the charge on the upper plate of C = C E =4 through A.

Q Q C Q–Q

Q +Q

VP – VR Q

E=

4 =4 C 3 When switch S is closed, the potential difference across C C B is q, then Q+q Q = CE =

Q

21.10

VR – VB

Q C Q = 4Q – Q

C Q –Q

capacitors C , C , C3, and C4 each of capacitance 4 E A and B.

Q Q= Ceq =

7CE Q E

7C

7 [

C Fig. 21.11

21.9 pacitors C

C =4

E points A and B when switch S is closed SOLUTION When switch S is open, the equivalent capacitance is C =

CC = C C

=

4 3

SOLUTION Capacitors C3 and C4 are in series, their equivalent capacitance is C3C4 4 4 C C3 C4 4 4 C is in parallel with C . is C =C +C

Fig. 21.12

Capacitance and Capacitors 21.7

But V is the p.d. across C and V across C , then

V +V =E 3V

C V =C V 4V = 6V

V =

3V

C3 and C4

21.11

V P and Q A and B

SOLUTION C

C

C3 = 3 A and B.

C , plate 4 of C C3 are connected to point A C , plate 3 of C and plate C , C and C3 6 of C3 are connected to point B capacitance Ceq A and B Ceq = C + C + C3

Fig. 21.13

Fig. 21.14

21.10

+V

CAPACITANCE OF DIELECTRIC FILLED CAPACITORS

Fig. 21.15

21.8 Comprehensive Physics—JEE Advanced

C =

K

A t

d

KK Kt K t

=

KK Kt K t

Keq =

KK t t Kt K t

d K eq

dielectric constant K C = KC =

=

K eq

Where

t

Capacitance of upper dielectric is K

C =

A/ d

=

K

K

K

where Keq = constant

K +K

K eq

A d

Where C =

A d

t

t K

K

d K eq

K A K3 A3 A A A3

d

d=t +t

K

,C = =

A d

K eq

A

d

KK K

A

d =

K

, Where Keq

K

K3

CHARGING AND DISCHARGING OF A CAPACITOR THROUGH A RESISTANCE

Consider a capacitor of capacitance C connected in series to a resistor R E -

C t d K

=

K eq

A

21.11 C=

A

Capacitors with K and K are in parallel and this K3

C = C + C + C3 = Where Keq =

CC C C

Keq =

C=

K A

A d

C = Where

K eq A A = d d

K

K

C =

A d C and C are connected -

together, C and C

K eq

=

A d

Capacitance of lower dielectric is C =

C=C +C =

KK A Kt K t

C =

A d

t = d, C = KC

Capacitance of the left capacitor is C =

K

Capacitance of the right capacitor is C =

A t K

A

C and C are in series, the capacitance of the CC C= C C

KK

= tt

A

Growth of Charge when t

A K t

Fig. 21.16

t

K t

and the potential difference q/C q is the charge at any t

Capacitance and Capacitors 21.9

t is given

of the capacitor rises till t E q = CE zero.

I=

dq = I e –t/RC dt

where I = E/R I

e –t/RC exponentially to a steady q as sho

q= q

Decay of Charge When the charge has attained a steady value EC K capacitor starts discharging through the resistor, i.e. the

Fig. 21.17

q = q e –t/RC = q e-t/

Time Constant RC C is in farad and R RC = q =q e –t/ t= , q= q

e

=q

CR will

t= ,q=q e

quickly d

q

.

q

t= ,

q t. Whether the charge decays slowly or RC.

of a CR

q . Whether the charge grows quickly or slowly depends R and C CR

Fig. 21.18

I Multiple Choice Questions with Only One Choice Correct 1. A spherical capacitor consists of an inner sphere of radius r = r and the outer sphere of radius r r C when the inner sphere is charged and the outer sphere is earthed and C when the inner sphere is earthed and the C /C is

C C

3 2. C =9

C =3 C is

Fig. 21.19

21.10 Comprehensive Physics—JEE Advanced

3.

C are conV. When key K is CV CV

CV CV 3

q

q r

r

4

q

q r

8

r

7.

r is charged to a voltage V V

q is the charge on the sphere and the

3V

V

V 8.

Fig. 21.20

V 3

C and C are connected in parallel and charged to a potential difference V capacitor of capacitance C K

4. state is reached, the energy stored in C is E and that stored in C is E E /E is

V K

K

3V K

3V K 3

V

9. A parallel plate capacitor of plate area A has a charge Q Q A

Q A

Q 10.

A n drops, each of capacitance C

o

Fig. 21.21

5.

n 3C C is given to the

6.

r r. A charge q is given to the inner sphere. When the inner sphere is con-

11.

12.

nC

n C n C n drops, each of capacitance C and charged to a potential V

n

n 4/3

n

n

Capacitance and Capacitors 21.11

17. has a capacitance C, the total capacitance of the C

-

C 9

C

C

13. 18. C C C C

the plate is

14. capacitors

connected ply. What is the charge

Fig. 21.22

19. points A and B

Fig. 21.24 Fig. 21.23

9 C C.

20. -

15. A capacitor of capacitance C

6

J kg K C

16.

plates and increase of potential difference across the plates

separation d plates can –6 d d –4 d –4 d

A A A A

–3

–4

21. A parallel plate capacitor of capacitance C is condifference V

change in the charge on the plates

A of the C N

C V.

capacitors are connected in parallel to each other no change in the charge on the plates

21.12 Comprehensive Physics—JEE Advanced

3

6 22.

9

CV

26. CV S S3 S S and V

CV

E in the annular region of a charged cylindrical capacitor

closed, V V closed, V = V and S closed V = V and S3 closed V

the inner cylinder r

where r

axis

Fig. 21.25

r

where r

the axis. 23.

resistor and a

27. A parallel plate capacitor of area A, plate separation d and capacitance C K ,K and K3 C in this capacitor, then its dielectric constant K is

e

24.

d is inserted in a parallel plate capacitor whose negative plate is at x and positive plate is at x = 3d charge. As x

Fig. 21.26

d,

K K

continuously

K=

= =

K

+

K

+

K +

K

K K K K

K3 K3

K3

K = K + K + K3

decreases and again increases. 28.

25. charges Q and Q

capacitances C and C

Q

pacitor with capacitance C, the potential difference Q

Q C

Q

Q

Q

Q C

Q C

ed in parallel is C /C is 3

Q C 3

C > C

-

6

4 3 6

Capacitance and Capacitors 21.13

29. capacitor A has a charge q on it whereas B is B

Fig. 21.29

Fig. 21.27

34.

q q

exists in the region

q -

30. plates is

C V and the other to V of the capacitors are connected together. When the positive ends are also

4 4

C V –V

4

C V –V

4

31.

35.

-

36. A capacitor of capacitance C

-

C V +V C V +V

-

and this capacitor is connected to a second uncharged capacitor of capacitance C 32. C = C = C3 = C4 = 4

C

of the total energy stored in the capacitors after

-

connection is

3 37. plate is A is d

Fig. 21.28

33.

A and B in

A d 3 A d

A d 4 A d

21.14 Comprehensive Physics—JEE Advanced

43.

Fig. 21.30

44.

38. plate is A d A d

3 A d

3 A d

3

4 A d

45.

A and B and C = 3

C

3 Fig. 21.31

39. A capacitor of capacitance C V pacitor of capacitance C V

Fig. 21.32

46. R ,R

,C

C =4

40. K and conductivity K K K K Fig. 21.33

41. plates of a parallel plate capacitor increases its d is the separation of the two plates of the capacitor, the thickness of the d 3 7d 9

d 9 d

47. A capacitior of capacitance C = C is charged to

capacitances C = C and C3 = C

42. CV 3 CV

CV 3

Capacitance and Capacitors 21.15

48.

R is increased

d initially. 3

constant K radius R

R /R is

3

speed V

t is

4 3 49. A perallel plate rapacitor of plate area A and plate separation d V. d is AV d

AV d

Fig. 21.35

6 R d Vt

AV

3AV d d 50. One plate of a parallel plate capacitor of plate area A and plate separation d is connected to the positive V

d d

dVt

R 9V t

6 R d Vt d

plates is

d AV d

9Vt dVt

R 9V t

52.

AV d

Vd A

9Vt

the switch S

Fig. 21.34

51. A parallel plate capacitor C with plates of unit area and separation d

Fig. 21.36

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49.

2. 8. 14. 20. 26. 32. 38. 44. 50.

3. 9. 15. 21. 27. 33. 39. 45. 51.

4. 10. 16. 22. 28. 34. 40. 46. 52.

5. 11. 17. 23. 29. 35. 41. 47.

6. 12. 18. 24. 30. 36. 42. 48.

21.16 Comprehensive Physics—JEE Advanced

SOLUTIONS P.D. across C C

rr

1. C = 4

r

r

C =4

r

r

C C

=

E = r r = = r r

Q = CV

Equivalent capacitance is CC C C

=

C is given to the positive plate, C. Let Q tive plate loses a charge Q and the negative plate gains a charge Q such that the total positive charge on the positive plate = total negative charge on the negative plate, i.e. Q Q

C

3. C CC C = C C C

K is open, the total C C = C 3

Q

Charge on capacitors is Q =

C C and

3 9 9 = 3 9 4

9 Q = CV = 4

capacitance =

V =V

=

5.

Q

C =

C V

E C = E C

2.

C

C V

E =

r

CV 3

V =

Q = C

C

6. C =4

r

C =4

r

r

Before connection, the total energy is q q = 8 C r After connection, the entire charge q of the inner ergy after connection is q q = U = C r U =

Fig. 21.37

When key K is closed, capacitor C is shortC. C is Q = CV Q –Q = CV –

CV CV = 3 3

=

4. . Current in the circuit is I =

U –U q

q

6 r r 7. Charge on the inner sphere is q = CV = 4 rV

6 V

=V –V

=

q r

Capacitance and Capacitors 21.17

= =

q r

4 q 4

.

r

=

=n

n =n Q = CV = C V

12. capacitors, each of capacitance C connected in par13.

KC C = KC

=

C V

E C V = E C V C Q = 3CV. When -

C = 3C

Q C

CV , E =

V

8. C=C capacitor C

11. E =

r

C K C. Q, the potential -

3CV 3V = K C K

-

9. method of virtual displacement. W d in the plate separation d to the resulting change U in the stored energy, i.e. W= U F plates, then the work W done to increase the plate d W= F d Now we know that the energy U of a parallel-plate capacitor of plate area A and capacitance C is U=

F =

Q

-

15. Energy stored in the capacitor is CV =

4

C

C

C

–4

16. Q on the plates does C Q = CV, the potential difference V

Q A r its radius, 4 r3 m= 3

n 4 R3 . 3 r. Now, the capacitance of a sphere C =n C

QV, a decrease in V results in a

stored energy =

17.

nm =

R=n

4

C

mc

d in d is,

d A

10.

R

their total capacitance is 6 capacitors in series each of capacitance 6

Q Q d = C A

where Q increase U in U due to an increase U=

14.

C

=

A

d When a dielectric of dielectric constant K is intro-

21.18 Comprehensive Physics—JEE Advanced

K

C =

A d

21. Q = CV and Q C V CV capacitors are connected in parallel such that the plates of opposite polarity are connected together,

C =K C

,

Q = C V and Q = CV

Now

C V = = C V K But V/V = 18.

V =

Q 4CV = C C

CV =V C

Equivalent capacitance C = C

K = 8.

8

Q C

t and dielectric constant K a parallel plate capacitor, the potential difference

U =

C V

=

C = 3C 3

V =

3C

CV

22. region of a charged cylindrical capacitor is given

d

V =E

t

K

E=

4 r where is the charge per unit length and r is the dis-

V, the separad

E d

=t

or K =

K

=

t t

23.

d

.

R = RC is zero.

.6

19.

r

24. 25.

Q and Q are E =

Q

and E =

A

Fig. 21.38

Q A

Q >Q

3 A and B 20.

C = K d = C A

K

E= E – E =

A 8.8

=

A

Q –Q Q

Q –Q

C

–3 6

.

Q C A d

26. When switch S3 is closed, the potential difference across C and C of V and V -

V E= d d

d

V = Ed =

d

A

=

V E

= x

V = dielectric strength

= 6.67

6

27. We have C =

A d

–6

C =

K

A

K d

= =

AK d AK d

,

Capacitance and Capacitors 21.19

AK AK = d d C and C are in parallel and their equivalent capacitance is A K +K C =C +C = d C3 capacitance is d d = + = + C3 A K K AK C C

U

C3 =

and

=

U Given

U

Q

Q Q

U U

get Q

Q

Q –Q

Given Q – Q

Q

Q. Q =

32.

d

=

K

A AK d

or C =

K

where

28. Given C + C =

K3

K

=

C C C C

CC

or

6C + 6C

CC

or

6C + 6C

CC

Let

C =xC 6C

+ 6x C

6x

x

which gives x = 29.

K

+

K

K3 Fig. 21.39

6

C +C

or

U

C C

CC

C , it is thus ineffective. Capacitors C and C4 are in series, hence equivalent capacitance of capacitors C and C3

xC

points A and B 3

or

C >C ,x=

3

3

is

33.

A to B Ui = U + U =

CV +

voltage is V=V –V As the capacitors C and C are in series, the

B

30.

C4 C3

CV

When they are connected, the potential across each is V =

V +V

Uf =

CV +

C CV = CV = C

V

V

or

C=

C

C

3

6

6

Decrease in energy = Ui – Uf C V +V

= =

31. U =

4

4

Q = CV C Potential difference across A and B = potential difference across capacitor C Q C C

C V +V

C V –V

Q Q and U = C C

34. V=E

d

21.20 Comprehensive Physics—JEE Advanced

A d

C=

is

A

A

or

.

C Uf =

K and thickness

Q C

Q C

t A

C = d

. C

t

K Now Q = CV = C V CV CV V = C C/

.

.

But

C

Ui = Uf

4

Ui

Q

Q C

C

Q C

Q C

=

37.

V

parallel, each of capacitance A C= d

35. A d Q is the charge on the plates, the potential difference is Q Qd V= C A Let d t and dielectric constant K is introduced, the new capacitance is is C =

capacitance is 3 A C = 3C = d

A

C = d

t

K

Q difference is V =

Fig. 21.40

Q C

Q d

t

38.

K A

Given V = V d=d –t

K

or d – d = t

Given d = d

K

capacitance C = A/d the equivalent capacitance is A C= C= d

t K

which gives K 36.

Q is the initial charge on capacitor C , the initial Ui = Q C When the two capacitors are connected together, on each capacitor is Q

Fig. 21.41

C =C

C = Q/V Q =Q =Q

C =

39. Q =C V

–6

–3

C

Capacitance and Capacitors 21.21

.

hold is Q =C V

–6

–3

C 44. Given

–3

pacitor cannot hold a charge of 8

C; it can

C

C

and

C3

C

C

3

C3 45.

–3 –3 6 capacitor is 3

V =

C

6

stand = V + V = 6 kilovolts + 3 kilovolts = Fig. 21.42

40. Let A

d the plate separation. RA d

= where R conductivity

d RA

C=

K

or

C=C +C

R=

d A

C=

A d = RC =

d A

K

CV , U =

U

=

C C

C C

4 8 = 3 9

3

R R

R = R =

=

3

4 3

s R R

=

=

3

C=C +C = RC = C

C

R R

For Circuit 3 R =

A d

thickness t A . Given C C = d t

s

K

d

C=

42. U =

A

= RC = 3 For Circuit 2 R =

capacitor is = RC =

-

For Circuit 1 R = R + R

is

=

41.

46. Let R and C tance of the circuit.

V =

3

6= 4 s

47. Charge on C is Q = C V = CV C and C3 is

CV

U

C =

U

CC C = C C

C = C3 = C C and C is

43. U = dU U

CV

U = CV V CV V CV

V V

V =

Q C

C

=

C

CV CV = C C C

C is Q =C V =

CV 3

= V 3

21.22 Comprehensive Physics—JEE Advanced

d 3

x

and the thickness of water col-

d =

d 3

x

is d =

wires is Q =Q =Q

= CV –

48.

CV CV = , which is 3 3

C =4 C =

4

capacitance is

R

Ceq =

RR R R

Given C = 3C 4

which gives

RR R R

=3

4 e R

R 3 = R

Now C =

W= 50.

Q C

x ,K

get Ceq =

Q 49. Energy stored initially is Ui = . if d C C C Q Uf = C Work needed is Q W = Uf – Ui = C

d 3

d =

Q C

CV

A d d k = RCeq Putting d = A

Vt

we

6 R d Vt

=

52. When switch S d is

V CV

U =

V =V

J

When switch S

Q = CV

A d

potential is V =

A V d

CV C C

q C

C

V

=

V

volt

drawn as shown in U = on the capacitor plates is A V Q = CV = d

C

C V V

= Percentage loss of energy is Fig. 21.43

51.

x ,

x = Vt

6 d

d 3

t

x

U

x = Vt.

U U

V

V =

V

II Multiple Choice Questions with One or More Choices Correct 1. A parallel plate capacitor is charged and the of insulating handles

V

J

Capacitance and Capacitors 21.23

QE

QE

6. When a capacitor of capacitance C is charge to a potential V , the energy stored in it is U . When this charged capacitor is connected to an uncharged capacitor of capacitance C is V U.

increases 2.

V V

point

U U

point 3. A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and

C C

C

V V

C

U U

C C

C C C C

7. V

constant K

-

the potential difference Q, E and

W

the charge on the plates is increased increased the energy stored in the capacitor increases. 8. A parallel plate capacitor is charged to a voltage V AV d

Q= E=

V Kd

Q= W=

V KA d AV d

K decreases

4. A parallel plate ai

-

and energy associated with this capacitor are givQ , V , E and U respectively. A dielectric

creases the energy stored in the capacitor decreases. 9. A parallel plate capacitor is charged to a voltage V.

Q, V, E and U are related to the Q>Q

V>V

E>E

U>U

5. A parallel plate capacitor of plate area A has a charge Q E Q

Q A

A

the charge on the plates is halved unchanged the energy stored in the capacitor is dou10.

V halved,

21.24 Comprehensive Physics—JEE Advanced

C C is 8

halved unchanged the energy stored in the capacitor is dou-

14.

C. C. A and B is C, are con-

V

11 S

-

Fig. 21.46

is C and the energy stored in the capacitors is U, then C C C C = Fig. 21.44

A

CV. B

U= 15.

V/3. CV . 3 CV . exists in the region

12.

4

CV

U=

CV

A and B of capacitances C and C 3 respectively are connected in series. When this supply, the potential difference across capacitor A A A and B are V and V C C 3

constant 4 is inserted parallel to the plates, the V thickness x

C C

V. V x

V x

13.

3

V

V

V

V

16.

Fig. 21.45

Fig. 21.47

Capacitance and Capacitors 21.25

A and B x and of the other x total capacitance –4

J.

17. attains a steady value, then

20.

A are arranged d apart. V Q is the charge on plate

fc is zero. capacitor is

3

Q4 on plate 4 then

rgy stored in the capacitor is nearly –8 J.

Fig. 21.50 Fig. 21.48

Q =

18. attains a steady value, then

Q4 = – C. –4

J.

AV d AV d

21. Capacitor C

AV d

Q = Q4 =

4

AV d C

separately allowed to discharge through equal t circuits is zero at t at t t Fig. 21.49

19.

C sooner then capacitor C initial charge.

ANSWERS AND SOLUTIONS 1.

voltage V across capacitor plates increases due to its capacitance C decreases. Now Q = CV charge Q

21.26 Comprehensive Physics—JEE Advanced

V increases and Q

5. on each plate of the capacitor is

QV also

F=

2.

A V E= d Q Qd Now V = = C A Q E= A

tween the plates it is E=

Q

Q 8.8 = A 8.8

3.

F= AV d

Q = CV =

V C = V C C

V Kd

A V

d A

d

U =

Q C

=

=

KA d

where C =

energy stored U = U>U are correct.

U =

CV

U U

K

=

+

CV

C C

C

V, the voltage of Q = CV, the charge on the plated C increases and V V/d V and d re

4.

Q

CV

7. Due to the introduction of dielectric the capacitance C

AV d

Work done W = U – U =

U = AV d

AV Kd

V=V increases capacitance C Q>Q unchanged, and V E = V/d

Q =C V . they are connected. Q = Q + Q . Using Q = CV, we have C V = C V + C V, which gives

K

Q U = = C

A d

QE.

6. V/d

E=

C

in the capacitor is U =

CV

C increases and V

d E=E CV increases

CV . U

C

8. Due to the introduction of dielectric, the capacitance C disconnected, the charge Q on the capacitor plates

Capacitance and Capacitors 21.27

V = Q/C C increases, V E = V/d

Q =

V 3

3C

energy

QV

stored in the capacitor is U =

V +

3C U –U=

Q

3

CV – CV =

3

=

3

CV

CV .

V decreases, U 12.

all the four choices are correct. 9. Charge Q capacitance C = A/d V = Q/C the capacitor is U = Q Q

V=E

d

d is C. U

C=

C

A = d

A .

A

or

K and thickness t A

C = d

10.

Q = CV E = V/d, E d is halved. Energy

CV

stored U =

K . C

=

unchanged equal to V, the capacitance C that Q V

t

.

.

CV C/

=

=

C

4

Now Q = CV = CV V =

C

and V

CV = C

V

=

11. When switch S closed, the potential differnce across capacitors A and B V the charges on capacitors A and B are Q = Q = CV. When the dielectric is introduced, the capacitance

x x

CV CV V = = volt C 3C 3

introduced, is U=

CV +

CV = CV

CV +

CV

-

pacitance is C =

C = KC = 3C. After the switch is opened, the potential difference across capacitor A V volt. Let V potential difference across capaciteo B. When the S charge on capacitor B Q. Q = CV = C V or V =

x . C

A

=

x

x

Now Q = C V = C V = Given V = V C =C . C C or = = or = x x which givens x 13. effective voltage is V=V –V As the capacitors C and C are in series, the effec-

introduced, is U =

C

=

C

=

C

=

3

6

21.28 Comprehensive Physics—JEE Advanced

or C =

17.

6

tors

cf

is Q = CV

C

Potential difference across A and B = potential difference across capacitor C Q C = = C Fig. 21.51

14. pacitance is C = C circuit is U=

CV =

4

15. Given V

CV

I=

V

V R

V = R

=

3

3

A

V is the potential difference across the capacitor, then applying Kirchhoff’s loop rule to loop abcfa Q, then we have

Q=C V =C V

– V – IR + V + V

C V = C V 3 When the dielectric of capacitor A

3

+V

V=

C =

volt 3 Charge on capacitor = CV

C

–6 –6

B C =C C C = C C 3 3 V and V are the new potential differences across A and B, then V C = V C 3 Also V +V V

CV

Energy stored is U = =

3 C.

–6

–8

3

J

18. BC and AD of the current is A

B

D

C

A through the

current in the circuit is 16.

I= capacitors each of capacitance 6

Now, p.d across A and C AB

is, therefore C 6

BC C.

Energy stored in the circuit = =

–6

CV

AD. Charge on each capacitor is Q = CV = C. Energy stored =

–4

J

CV + –4

J

CV = CV

–6

Capacitance and Capacitors 21.29

19.

A A ,C = d d

C =

Cb =

Ca = C + C =

A d

C x, then

C C =

A = Cb d

20.

x and of

A A and C = d x d x Ca = C

+ C

=

Q = CV =

A d d

AV d

x

C

A

=

x d Ca > Ca wrong. d

Cb =

Q4 = – C V = –

C C =

And

C +C =

AV d -

CC C C

A d

=

V ; V = voltage R

21. At t

Cb = C C C C Now

C.

e–t/ where = RC,

of

d

A x d

d

Ad x d x

x

III Multiple Choice Questions Based on Passage Questions 1 to 4 are based on the following passage Passage I

C=

A parallel plate capacitor C= each of area A, separated d. A dielectric

AK d t A t d K A

C=

area A and thickness t and dielectric constant K is introduced with its faces parallel to the plates as

d

t

A

C= d t

1. Fig. 21.52

K

K

21.30 Comprehensive Physics—JEE Advanced

2.

K = 3, for what value of t/d will the capacitance 3

3.

3

3 4

3 4 K = 3 and t/d

4

4.

capacitors is

SOLUTION 1. of two capacitor one of thickness t dielectric of dielectric constant K and the other of d–t tances respectively are KA C = t A and C = d t C CC = + or C = C C C C C

C = Ca

Given K = 3 and C/Ca t d 3 which gives

C= A d

t

3 A . d

Now Ca =

K

Ua =

2. Ca =

A d

t 3 = d 4

3. Putting K = 3 and t/d =

we get C=

t d K

A d

Ca C

Q Q and U = Ca C

4.

Questions 5 to 7 are based on the following passage Passage II C. 5.

A and B is

6.

A and B is Fig. 21.53

Q U = , which is Ua 3

Capacitance and Capacitors 21.31

7.

A and C is 8 3

3

SOLUTION 5.

As the 3 parallel, the potential difference across each is the

C of the parallel C C3

-

–6

=6

C3

–6

=8

=

C

C

4

or C = 3

C +6 C+8

3 C of the

C=8 C C and C =

C.

Potential difference across C =

C3 and C4

C 4 AB

=

C3 C or C A and B C and C = C + C

C4

6

3 7.

6.

C4

C = –6

=6 C

–6

Charge on 4

Now C and C capacitance C C

–6

Charge on 3

–6

C.

C3 and C4 is C C3 and

–6

A and C is =

C 3

6

=

6

Questions 8 to 12 are based on the following passage Passage III

8. through the circuit is C C

A and B with capacitances 3 9. capacitor A

capacitor C throught a switch S

C C 10. capacitor B C

C C

11. capacitor C C C Fig. 21.54

=

3

21.32 Comprehensive Physics—JEE Advanced

SOLUTION 8.

Q through the circuit. When the switch is pressed, the voltages developed on capacitors A, B and C are Q Q volt, VB = volt and VA = 6 6 VC =

Q

Q 6

6

A

volt.

3 Qf

A

Qi

6

–6

which gives Q 9.

Q

Q

6

Qi

Vi Ci

A

A

A is –Q

C,

10.

B is Qf

B

Qi

B

–Q

C C

11. Fig. 21.55

Applying Kirchhoff’s law to the loop, we have VA + VB – VC

E R =6 r

E C

A

Qf

C

Qf

C

C

Qf

A

C,

Questions 12 to 15 are based on the following passage Passage IV r

Qf

,R

12.

R is

13.

R is

C =4 14. 3 4 3 15.

C is C C

Fig. 21.56

SOLUTION 12. Current through R is I =

E

C C

14.

R r Potential difference across R is V = I R 13. Current throught R is E = =3A I = R r . . Potential difference across R is V = I R =3

V=V +V Effective capacitance C =

=

6

=

=

A

C

CC C C

4 3

15. Charge Q = CV =

4 3

–6

–6

C

C

Capacitance and Capacitors 21.33

Questions 16 to 18 are based on the following passage Passage V ra and rb + Q and the inner sphere is earthed. 16. Out of Q, a part Q which will appear on the outer surface of the outer sphere is Q rb ra Q rb ra rb ra Qrb rb ra 17.

Qrb ra

Qra rb

Q rb ra rb

Q rb ra ra

18. C=

Qra rb ra

rb

4 ra

ra ra

4 rb

C=

ra

4

C=

rb rb ra

4 rb

C=

Q which appears on the inner surface of the outer sphere is

SOLUTION 16. Out of charge Q, a part Q surface and another part Q surface of the outer shere such that Q+Q =Q

Q ra

Q rb

Q

A is earthed, VA Q or Q rb rb Q =Q–Q ra rb ra rb

Q–Q

Electric potential at shere B is

which given Q =

Q rb

Q rb ra rb

18. V = VB – VA = VB

Q Q Q rb rb rb Q = rb 4 Electric potential at the inner sphere A is Q Q Q VA = ra rb rb 4 VB =

4

V=

4

Q rb

Q rb ra rb

17. Q = Q – Q = Q –

Fig. 21.57

ra

Q = rb 4

C=

4 Q = rb V

=

Qra rb

A and B is VB Q rb

ra

rb rb ra

IV Assertion–Reason Type Questions following four choices out of which only one choice is correct.

-

21.34 Comprehensive Physics—JEE Advanced

Statement-2 1. Statement-1 4. Statement-1 source supplying a constant voltage V are kept connected to the source and the space charge on the plates will increase. Statement-2

Statement-2

on the shape and size of the conductor.

plates. 2. Statement-1

5. Statement-1

voltage V plates of a parallel plate capacitor charged to a

the energy stored in the capacitor will decrease. Statement-2

3. Statement-1

Statement-2 -

Fig. 21.58

in the capacitor will decrease.

SOLUTION 3.

1.

Q

constant voltage V capacitance C the plates is increased. Now, energy stored U = Q C is decreased, Q C U will increase.

V C increases due Q = CV, the charge Q on the capacitor plates will increase. 2.

4. 5.

Q on the

E charge q is F = qE

C increases due to the introduction of the dielectric. Now, energy Q stored U = Q C C increases, U will decrease.

V Integer Answer Type 1. -

E is constant, force F is the

Capacitance and Capacitors 21.35

2. R C

R

,

C =4

SOLUTION 1.

Given V = V

Q is the charge on the plates, the potential difference is Q V = C

d =d –t

Qd A

d –d =t

Let d t and dielectric constant K is introduced, the new capacitance is

d

or

K t

Given d – d

A

C =

K

K Which gives K = 3.

t

K

2.

Q difference is

R =

R R

R R

3

C =C +C V =

Q C

d

t

K A

= RC =

3

6=4 s

Electric Current and D.C. Circuits 22.1

22

Electric Current and D.C. Circuits

Chapter

REVIEW OF BASIC CONCEPTS 22.1

positive charge

ELECTRIC CURRENT

Conductor

negative charge

(+)

(–)

Fig. 22.2

current is said to be steady or constant

22.2

q I= t

DRIFT SPEED OF ELECTRONS IN A CONDUCTOR

q t.

vd E Fig. 22.1

vd =

I=

eE m

dq dt

Convention regarding direction of current -

Relations between drift speed and current I in a conductor as vd =

I ne A

n e A = cross-sectional area of conductor

22.2 Comprehensive Physics—JEE Advanced NOTE

=

R

R

R

R3

.

22.3

+…

OHM’S LAW

Ohm’s Law states that

22.6

EMF, TERMINAL VOLTAGE AND INTERNAL RESISTANCE OF A CELL

. I or V = R I

V

E

R

22.4

ELECTRICAL RESISTIVITY

V r

r

A R or

=

or A RA

R=

A

E connected to an external resistance R.

is

called the

Unit of

r

=

unit of R unit of A unit of

tre

Hence

Fig. 22.3

Total resistance of the circuit = R + r. The current in the circuit is I=

22.5

RESISTORS IN SERIES AND PARALLEL -

tance R

E R

r

Potential difference across r is v = Ir Potential difference across R is V = IR. V is called the v internal resistance. Thus E= V + v = IR + r

R = R + R + R3 + …

V = E – Ir

22.7

GROUPING OF CELLS

resistances r and r

E and E and internal

Electric Current and D.C. Circuits 22.3

nE R nr

I=

n

Fig. 22.4

E nE nR r

I= E

=E + E

r

=r + r

and

n m

For n cells in series E = E + E + … + En and

mnE nR m r

I=

rn = r + r + … + rn

22.1

SOLUTION N e Fig. 22.5

E r r

= =

E r r

E r

N=

It e

Ne . Thus t electrons

.

r

For n

22.2 E r

and

q t

q = Ne and I =

r

= =

E r

r

resistance is nr each cell.

E r

r

En rn

... ...

rn

E n nE and the total internal r is the internal resistance of

–7

free electrons

3

SOLUTION vd =

.

=

E internal resistance. 3. Cells should be connected in series if the external resistance R r. R < r. R r. n E

I en A 7

–3

t=

22.3

L vd

. 3

22.4 Comprehensive Physics—JEE Advanced

-

tance of the tube.

t

SOLUTION

SOLUTION I=

dq dt

dq = Idt

t q=

t

=

V R= I

t + 3t dt

to t

N + Ne e

I

t

.

t

dt 3 3 t 3

t

3

=

22.4

3

=

3

22.6

7

SOLUTION L RA

= L R= A I=

SOLUTION L increases and A

L V and I = A R

V A L L

But I = enAvd enAvd = vd =

V A L V Lne

R =

A

L A

7

=

R L = R L

–3

22.5 The current I t I t + 3t

AL = constant. Thus AL =AL AL L . Therefore A = L L R = A

R

A A

R

NOTE R is stretched to n n R.

Electric Current and D.C. Circuits 22.5

22.7

-

R L A = R L A AL = constant A A

L L

R = R

L L

A A

L L

SOLUTION R =

3

= 3 3=

nR

22.9

=

R

resistance and

R 22.8

SOLUTION AL = constant

A L =A L .

L =L +

A =A R =

L =L L L

A Fig. 22.6

L L and R = . Therefore A A R R

=

L L

A A

SOLUTION of R R3 and R is R R

R

=

R

R R=R –R

R R R

Simple Method

Fig. 22.7

Current in the circuit is

L L used.

I= L R= A R

R

E R

r

Potential difference across R is V = IR L

A

22.6 Comprehensive Physics—JEE Advanced

Potential difference across R is V = IR R R3 and R since these resisR3 is V I3 = R3

resistor. There-

v = Ir Fig. 22.9

V =E – v 22.10 Calculate the steady state current in the 3

Current I = Potential difference across R is V = IR

resistor in

resistor. Hence the . resistor is 3 22.11 and a resistance of the battery. SOLUTION Fig. 22.8

E =IR + r E

SOLUTION 3

and

and

is R =

r 22.12

SOLUTION Let R .

Fig. 22.10

r

E

r

r

r

r E

Electric Current and D.C. Circuits 22.7

PQ

be R

22.14 A and B S S is closed.

Fig. 22.11

RPQ =

R Fig. 22.14

R A and B is

R

RAB

R R

S

=R

R R

SOLUTION

R

R R

Current I =

Fig. 22.15

E R

r

RAB =

22.13

S

A and B has a resistance of 3

.

Fig. 22.16

RAB

Fig. 22.12

SOLUTION

22.15 I

Fig. 22.13

RAB

=

3

3

3

RAB

Fig. 22.17

22.8 Comprehensive Physics—JEE Advanced

SOLUTION

IR A and B is R =

.

Sign Convention for emfs and Voltage drops

=3

bcdeb and

C and D is R= Current I =

3 3 3 3 V R

. Fig. 22.20

Fig. 22.18

22.8

KIRCHHOFF'S LAWS that cell.

First Law or Junction Rule

IR junction A. junctions at b and e b or e I + I = I3 and bcdeb

Fig. 22.19

E – I 3R 3 – I R and

– E + I R + I 3R 3

I + I – I3 – I – I

E E R R and R3 I I and I3

I + I = I3 + I + I NOTE

Second Law or Loop Rule I and I

I3

Electric Current and D.C. Circuits 22.9

I3

I

BCDEB –I R + E – I3R3 – 3I I3

-

I +I

I 7I

22.16

and – 3I

I

I +I

I.

I

I I =

I =

and I3 = I + I = B and E B

E.

VB – V E = I 3R 3 = Fig. 22.21

22.18

SOLUTION P PQ Q

QR

as

A and B A and C

R I

-

E E nal resistances r and r are connected to an external resistor R

I R. 22.17 I I and I3 E

E

B and E R = 3 and

R

R3

Fig. 22.23

SOLUTION We choose any direction for the current I in the cirI I Fig. 22.22

SOLUTION B I + I = I3 ABEFA I 3R 3 – E + I R

Fig. 22.24

22.10 Comprehensive Physics—JEE Advanced

B and D are

ABCDEA – E + Ir + E + Ir + IR I

I

I

I I is

I I I

Fig. 22.25

Fig. 22.27

VA – VB = – E – Ir

Condition for balanced Wheatstone’s Bridge

A B. VB –VC = E – Ir B VR = IR

ABDA and BCDB C. IP+IG–IR I –I Q

22.19

G I

I

and

I +I S–IG I

IP–IR

I I

R P

IQ–IS

I I

S Q

P R = Q S

or

P Q = R S

Fig. 22.26

SOLUTION VA – VD

I – 3I

I

I But VA = VD I

22.9

A and D are earthed. I

WHEATSTONE’S BRIDGE P Q R and S A and C B and D

P Q R and S

Fig. 22.28

the resistor R =

AD = DC S

AC

Electric Current and D.C. Circuits 22.11

AD and DC and . For a balanced

22.21 A and B

P R = Q S 22.20 A and B R.

Fig. 22.31

Fig. 22.29

SOLUTION

SOLUTION

C and D because

C and D

rent I

A and B V across A and B

V

V = IR.

A and B R

R.

Fig. 22.32 Fig. 22.30

RAB =

R R

R =R R

and DEBFD V I–I

ABFGA ACDFA I–I +I

22.12 Comprehensive Physics—JEE Advanced

I +I I –I

I–I

I–I +I

3I – 3I

R R

I

I =V

I – 3I

and

I I =

I I =

I

7I

I

= R =

.

I.

22.10

=

3

3

3

R R =

3

THE POTENTIOMETER AB

I and I V 7 I

V

R R

3

I – 3I – I

R R

E is a

E RAB

22.22 resistances R and R

AC resistance is

A R

D such that

R and R . Fig. 22.34

AB and L

R = L/A

A V = IR or V = I L/A = KL V

R is the resistance

K = I /A

L

V = K or L

Fig. 22.33

difference v

SOLUTION

v

cell E or E = v =

R = R

3 V

R

R R

R R

=

R

=

Applications of potentiometer

R R

R =

3 3

E

and E and E E = E = E

and E =

v = . The E of the

Electric Current and D.C. Circuits 22.13

SOLUTION

E S

R=3

E=

AB = L AC =

and V

r

Fig. 22.36

AB I=

Fig. 22.35

S V of E V=

R is connected across the cell E is E

V

r

AB = I

22.24

v

=

resistance is 3

AB

.

v = V Ir = IR r= R R

and R

r is

22.23 Fig. 22.37

resistance of 3

E

SOLUTION AB

E.

r

E=

E = V + v = IR + Ir I E and r is its internal resistance. Thus v

R

AB is K =

E = V

V

V

I =

. 3

22.14 Comprehensive Physics—JEE Advanced

Potential difference across AB due to E is

Potential difference across AC cell is r VAC = IRAC

Potential difference across due to E . Vl = resistance due to E is .

I =

E is

tial difference across AC

VAC

resistance . due to E is

r

r

R connected across E

V V =V

I =

.

R

VR = I R = 22.25

R R

VR across AC VR = resistance of AC

current I

= R

R

R VR =

SOLUTION AC tance of AB.

AB AC

I

r be the resis-

22.11

R R

AMMETER

Fig. 22.38

AB =

r

AC is RAC = r

. Therer

Fig. 22.39

Electric Current and D.C. Circuits 22.15

I G I

V I R

S = shunt resistance G

S

I = I + Is

I G = I sS I–I S

IG S=

I I

G

I

22.26 Fig. 22.40

V = VG + VR =IG+IR V =G+R I SOLUTION I S=

I I I

I

G

G

.

R=

V –R I

22.27

. .

. .

RA =

GS G S

.

SOLUTION

resistance of the circuit and hence the current current.

22.12

G

I R=

V –G= I

VOLTMETER

V . .

RV = R + G tor of resistance R << RV

-

22.16 Comprehensive Physics—JEE Advanced

R and RV

RV >> R

R is con-

R

-

tion is

R.

R=

22.28 are connected resistance. sure the current in the circuit. Find the error in -

Fig. 22.42

SOLUTION

Total resistance of the circuit is R Current in the circuit is I =

. -

Fig. 22.41

V =

.

I=

. The current I =

.

V =I

.

I Multiple Choice Questions with Only One Choice Correct 1.

and +

are connected in series. +

2.

. The

Electric Current and D.C. Circuits 22.17

3.

resistor is Fig. 22.45

6. 3

Fig. 22.46 Fig. 22.43

4.

7. P and

3R

Q 3 3

A and B is

P to Q

R

Q to P

7R

P to Q

3

R Fig. 22.47

Q to P

3

8.

A and B is 3R R R R

5.

E is

is used

9.

Fig. 22.44

Fig. 22.48

I

across it is

22.18 Comprehensive Physics—JEE Advanced

10.

.

15.

16. 11.

L E E

-

L

L

L 3

L

L 3

R is

n n R R n

nR R n is bent into a circle. The

17. circle is

18.

12. tion. What is the resistance X

current I in the circuit is

Fig. 22.49

each are connected as D -

13. A and D

Fig. 22.51

is bent into the

19. r

3

3 3

20. Fig. 22.50

14.

A and B is

Electric Current and D.C. Circuits 22.19

22. R -

Fig. 22.52

21. t

I in the resistor AB t

I I I

K is -

23.

E and different internal resistances r and r are connected in series to an external resistance R R

t t

I

Fig. 22.54

Fig. 22.53

Fig. 22.55

24.

26. 25. the conductor is/are

r R= r

R=r +r

R=r –r

R=r =r

22.20 Comprehensive Physics—JEE Advanced

Fig. 22.56

27.

3 28.

resistances is S the total resistance is P n is

S = nP

Fig. 22.57

32. -

29.

in the ratio of

3

and

other. One cell has an and the other cell has

3

internal resistance difference across the X and Y is

3

30.

-

tance X X
33.

-

Y

X

Y 34.

31. A and B is

Fig. 22.58

current I in the circuit is

Electric Current and D.C. Circuits 22.21

R

R

R

R

38.

39. Fig. 22.59

35.

A

is connected in series

n R resistance of x R x and y

R is

xy

R= A

60 W

y

x

R=

y

xy

R

y–x

R

x+y

40. x E1

4V

R

1V

E2

y xy

Fig. 22.60

36.

x

xy y

y

y–x

C and D

x

xy

41. is

42. Fig. 22.61

37.

x is

A and B is

Fig. 22.62

Fig. 22.63

22.22 Comprehensive Physics—JEE Advanced

Fig. 22.64

43. R R R R R R R = R R3 R R3 = R R = R R

3

R

44.

Fig. 22.67

P and Q of

47. Rr R + r R R+r R+r r R R r

Fig. 22.65

45. V

V

V

V/3

Fig. 22.68

48. AC x Fig. 22.66

I is R . Then

46. R R R = R3 R R

x x

x x

AB is

Electric Current and D.C. Circuits 22.23

Fig. 22.71

51. The internal resistances of the cells in the circuit Fig. 22.69

the circuit is

49.

Fig. 22.72

A B A B

to to to to

D C B A.

52. ABCD

B and D

BD

A and C is

53.

Voltmeter

R

2W

Fig. 22.70 6V

50. Q S

Fig. 22.73

54.

A B and C and C R : R3

A and C

R R and R3 A and B B R :

22.24 Comprehensive Physics—JEE Advanced

Fig. 22.76 Fig. 22.74

55.

-

60.

RC circuit consists of a resistance R C F connected in series

-

e

resistor

56.

3

3 A and B E and of internal resistances L is an ideal inductor and C K is closed. When the

61. rA and rB R

A

Fig. 22.75

57.

sure is

Fig. 22.77

R = rA – rB if rA > rB. 58.

R= R=

59.

rA rB rA + rB R

A and B in

A 62.

RC circuit consists of a resistance R C F connected in series

Electric Current and D.C. Circuits 22.25

e

AB and BC

63.

L AB

BC is r. The current I

r

Fig. 22.78

-

65. X

BC

across AB.

BC AB. AB and BC AB and BC

X

64. S

Y to X is C

Fig. 22.79

C

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61.

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62.

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63.

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64.

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65.

6. 12. 18. 24. 30. 36. 42. 48. 54. 60.

SOLUTIONS 1.

A = cross-sec-

L R= R +R =

L A

d L L = A A

+

R=

L A

d

22.26 Comprehensive Physics—JEE Advanced

=

4.

+

R=

2. E = V + Ir = I R + r r is the internal resistance of the cell and the current in the circuit is I=

E R

r

I=

R V = IR and E= E–V

E E

V E

IR I R r R R

r

=

r R

r

r R

Fig. 22.81

r

= 3.

E. Points A B and C at A B and C

I +I =I APQDA

V

D D and E is V E and B is V = V V. Therefore V I = 3 V I = 3 V I3 =

I

I

I

V E and C =

I I =

I =

I =

3

I =

3

3

P I = I3 + I I3 = I – I =

3

3

=

P to Q 5. I = I +E I

I

I

6. Fig. 22.80

I = I + I3 V 3 Hence I3 =

=

V /3

V

=

V=

3 Fig. 22.82

E

3 I3

3

Electric Current and D.C. Circuits 22.27

A and B

.

V–V =

Current I =

I 3 = 3 3

10.

–3

-

7. 3

I=

–3

R –3

R

R 11.

E /L. Hence

Fig. 22.83

RAB =

=E

8.

L

E E = L

L

E=

R

x E=

xE L x=

L

.

12. =

Fig. 22.84

RAB =

X

R

7 X . Hence the correct choice is

13.

9. Current I =

=

A and D resistor is R

V=

or

=

+

R=3

R=

I =

= Fig. 22.85

V =

14.

I

=

3

22.28 Comprehensive Physics—JEE Advanced

I+

I=

I

R =

R

R

R I+R I

R

R =R

. Hence the correct

R

15. Let the three resistances be X X R = +

R

3

R

X and Y. Then X

or

=

R

+ R

R =

R

+

R

=

R

=

20.

X. 3 + X Y

Y

X X = 3. Hence Y

R circuit is

. Hence the

=

=

R V R

the

16.

R

=

R

+

R

+

n R

+n

resistor in

or R = R/n

A and B

17.

21. AB is

. Hence the 18.

t

I=

E. Hence no current

B

t as I= I e I=I

A and D is the

R or

=

+

R=3

=

–t/

t >>

t

I =

3 I

resistor

22.

19.

r R

r

R r r R =

R r

r=

R

r

r resistors connected in

Electric Current and D.C. Circuits 22.29

23.

A and B of the

26. E

AB I=

E r r

R

resistance r is E – I r I=

E=Ir

Ir r r

R

27.

R = r – r . Hence the correct choice

A and B Hence current in AC CD CD

.

sectional area A ity

-

-

24.

R=

-

A

V

V=

R

V

R Resistance R

R. The -

AB

resistances R bined resistance is =

. Current I

or R =

R

tional to

R

. Thus if

R

. a

R R

AB R R

R

R

R R of the R R R

Re =

R R

R R

3

R

R or

The roots are R . R

Fig. 22.87

28. Let x and y Fig. 22.86

25. of A

xy x y nxy x+y= x y

S = x + y and P = A

-

S = nP or

and

22.30 Comprehensive Physics—JEE Advanced

x y x=

xy + y

n

n

[n

y

or x =

x are

n

y

+

y

y

n

x

n

n n 29. = or

R = R

R A

R A

=

r

A = A =

3

R = R 3

and

=

r r r

3

Fig. 22.88

r r

=

E

R=

3

=

.

E

=

3

I R =I R or

and

A and B

. Hence

E Therefore

I I

3

R = R

3

E

R=

=

=

=7

3 32.

X 30. Y

= I= X = Y X = Y

X and Y

31.

is

33.

=

V = IR =

= or

E=

E ACDA CBDC and ADBGFA –

3

3

+ +

3

3

3

E

34. B

V

A =

I . = A

.

R

Electric Current and D.C. Circuits 22.31

CC C C

Ceff =

B.

3

F

F

3

Q is the

D and

R E =

R or R =

3

R

Q = VAB

3

Ceff

3 C AC is

VAC =

A and C

.

Q C

C and D is

VCD = VAD – VAC

CD is

37. R

R R by Fig. 22.89

R or

=

+

3

=

I= current due to E IR = E I R R

=

E R

=

R =

R

P Q

R S

R

R

R

38.

/

-

R

R = IR

and

. Thus R R

R R

36.

V

I=

7 35. Current due to E is

=

R

7

R=

R R + R and R + R R

R

. Hence the correct choice is I=

Potential difference across AB is VAB AD is VAD

. Hence the correct choice

39. For series connection x = nR R tion y = . Therefore xy = nR n 40.

R

R

Rn

Rn

R

R

Rn

y

n

R = R . Hence n

x

22.32 Comprehensive Physics—JEE Advanced

Subtracting (2) from (1), we have 1 1 = Rn x which gives Rn =

1 y

4 A, which is choice (b). 5

xy

which is choice (b). y x 41. Let m cells be connected in series and n such groups are connected in parallel. If the emf of each cell is E and internal resistance r, then the total emf of m cells in series in mE and the total internal resistance is mr. When n such groups are in parallel, the effective internal resistance is mr/n. Then the current through an external resistance R is I=

mE mr R n

mnE nR mr

(1)

mnE

= nR

mr

2

2 mnRr

Now, I will be maximum if the denominator is the minimum, i.e. if nR = mr. Given R = 3 and r = 1 . Using these values, we have 3n = m. But mn m m = 48 (given). Therefore = 48, which gives 3 m = 12. Thus n = 4. Hence the correct choice is (b). 42. The given circuit is a Wheatstone’s bridge. The current through the galvanometer will be zero if the bridge is balanced, i.e. if P R where P = 2 + 3 = 5 , Q = 10 Q S R = 4 . The value of S is given by

Fig. 22.90 where R1 =

R1

R2

I1 I

I2 Fig. 22.90

70 70 and R2 = 3 4

10 70 / 4

44. The emfs of cells connected in reverse polarity cancel each other. Hence cells marked 2, 3 and 4 together cancel the effect of cells marked 5, 6 and 7 and the circuit reduces to that shown in Fig. 22.91. Now cells Fig. 22.91 1 and 8 are in reverse polarity. Hence the voltmeter reading = 5 – 5 = 0 V. Hence the correct choice is (d). 45. containing the capacitor. Thus, the current, say I, R and 2R. Applying Kirchhoff’s second rule to the loop abcdefa, we have (see Fig. 22.92) V 2V – I(2R) – IR – V = 0 or I = 3R Potential drop across capacitor = 2V – V – I(2R) = V – =V–

2V 3

V 3R

2R

V , which is choice (c). 3

and

5 4 10 S or S = 8 . Thus the effective resistance of the parallel combination of 12 and x ohm must be 8 . Therefore 1 1 1 12 x 8 which gives x = 24 . Hence the correct choice is (d). 43. Since the seven resistances are in parallel, the effective resistance is R = 70/7 = 10 . Therefore, the current in the circuit is I = 14/10 = 7/5A. The given circuit can be redrawn as shown in

R 7 = R2 5

current I2 is given byI2= I

. The

Fig. 22.92

46.

R 6, resistances R1, R2, R3 and R4 constitute the four arms of a balanced Wheatestone’s bridge. Hence R1 R3 or R1 R4 = R2 R3 R2 R4 Thus the correct choice is (c). 47. Refer to Fig. 22.93. The branches ABPQ and PQCD are a balanced Wheatstone’s bridge. Therefore, resistances (each equal to 2R) between E and F and between F and G do not contribute and the circuit simplies to the Re between P and Q is given by

Electric Current and D.C. Circuits 22.33

Fig. 22.93

1 Re

1 4R

1 2r

1 4R

2Rr which gives Re = . Hence the correct choice R r is (a)

53. The correct choice is (b). 54. Let the value of each resistance be r. The network can be redrawn as shown in the Fig. 22.95.

48. is R R1 = AC R2 RCB where RAC and RCB are the resistances of the bridge wire of length AC and CB respectively. If the radius of the wire AB is doubled, the ratio RAC/RCB will remain unchanged. Hence the balance length will remain the same. Thus, the correct choice is (b). 49. The voltmeter must be connected in parallel with the resistor and the ammeter must be connected in series with the resistor. Hence the correct circuit is (a). 50. The potential at Q with respect to R is 15 V and R is at 25 V higher potential than S. Thus Q is 40 V higher than S. When Q is grounded, its potential becomes zero. thus, Vs = – 40 V. Hence the correct choice is (d). 51. Since the cells are in opposition, the effective emf = the positive to the negative terminal of the battery, a current of 4 = 0.5 A 5 3 B to C. Hence the correct choice is (b). 52. Resistance of each side of the square = 10 0.1 = 1 . As shown in Fig. 22.94, the square forms a W h e a t s t o n e ’s bridge which satisfies the balancing Fig. 22.94 condition. Thus, no the diagonal BD. Hence the correct choice is (a).

Fig. 22.95

(i) Net resistance R1 between points A and B r The series combination of resistances and 3 r which has an equivalent resistance r1 = 3 5 r = , is in parallel with resistance r. Hence 6 5r r r r1 6 = 5r R1 = 5r r r1 11 r 6 (ii) Net resistance R2 between points B and C The series combination of resistances r and

r , 2 r 2

r , 3

r = which has an equivalent resistance r2 = r + 3 4r r , is in parallel with resistance . Hence 3 2 r 4r r r2 2 3 = 4r R2 = 2 r r 4r 11 r 2 2 2 3 (iii) Net resistance R3 between points A and C r The series combination of resistances r and , 2

22.34 Comprehensive Physics—JEE Advanced

r = 2

which has an equivalent resistance r3 = r + 3r r , is in parallel with resistance . Hence 2 3 r 3 R3 = r 3

r3 r3

r 3 r 3

3r 2 = 3r 3r 11 2

5r 4r 3r Hence R1 : R2 : R3 = = 5 : 4 : 3. Thus : : 11 11 11 the correct choice is (c). 55. At instant of time t, the charge on the capacitor is given by q = q0 (1 – e–t /RC) and the potential drop across the capacitor is given by ( V = q/C) VC = V0 (1 – e–t/RC) where V0 is the voltage of the battery. The potential drop across the resistor is –t/RC

VR = V0 – VC = V0 – V0 (1 – e

–t/RC

) = V0 e

The resistance of wire of resistivity

is

2

l dl = kl2 (1) A m where k = d/m is a constant of the wire. Taking logarithm of both sides of (1) we have log R = log k + 2 log l Differentiating R=

R 2 l =0+ R l

R = 2 0.1% l R = 0.2%. Thus, the resistance of the increases by 0.2%. which is choice (b). 59. The resistance between points A and E is given by Given

l

2 l l

= 0.1%. Therefore,

1 1 = RAE 6

1 6

by giving RAE = 3 . The network reduces to that shown in Fig. 22.96 (a). Similarly the resistance

t / RC

VC 1 e = VR e t / RC Given

= e t /RC – 1

VC = 8. Therefore, VR 8 = et/RC – 1

et/RC = 9 = (3)2 t or = 2 loge (3) RC or t = RC 2 loge (3) = (5 106) (1 10–6) 2 1.1 = 11 s Hence the correct choice is (b). 56. The two sub circuits are closed loops. They cannot send any current through the 3 resistor. Hence the potential difference across the 3 resistor is zero, which is choice (a). 57. Given I = 1 mA = 10–3 A, G = 20 and R = 4980 . V Now I= R G or

or

V = I(R + G) = 10–3 = 5.0 V

(4980 + 20)

Hence the correct choice is (c). 58. The mass of a wire of length l, cross sectional area A and density d is given by m m = Ald or A = ld

Fig. 22.96

between points A and D in Fig. 22.96 (a) is RAD = 3 . The network reduces to that shown in Fig. 22.96 between points A and C is RAC = 3

. The network

effective resistance between points A and B is 2 So the correct choice is (b). 60. Let the internal resistance of each battery be r. Let R be the unknown resistance and G be the resistance of the galvanometer. Let E be the emf of each battery. When the batteries are connected in series, the total emf = 2E = 2 1.5 = 3 V and total internal resistance is 2r. The current in the circuit will be I=

R

3 G

2r

Given I = 1 A. Therefore

Electric Current and D.C. Circuits 22.35

1=

3 G

R

2r

or R + G = (3 – 2r) (1)

When the batteries are connected in parallel, the total emf = E = 1.5 V and the total internal resistance is r/2. Hence the current in the circuit will be r 2 Given I = 0.6 A. Therefore, R

15 . R

G

r 2

G

or R + G =

2.5

r 2

(2)

VC = 8. Therefore, VR 8 = et/RC – 1 et/RC = t = RC t= = =

or or or

r 3 – 2r = 2.5 – 2 1 which gives r = ohm, which is choice (d) 3 61. When the key K is inserted, the current starts growing and after some time it acquires a steady value. At

through the inductor (because an ideal inductor offers zero resistance to a steady current). Now, the network of resistors is a balanced Wheatstone’s bridge. Hence, no current flows through the resistance 2 R. Therefore, this resistance can be ignored. The net resistance between points X and Y = resistance of the parallel combination of 2R 2R = R. (2 R 2 R)

Hence the current in the circuit is I=

VC 1 e t / RC = = e t /RC – 1 VR e t / RC Given

From Eqs. (1) and (2), we have

2 R and 2 R =

VR = V0 – VC = V0 – V0 (1 – e–t/RC) = V0 e–t/RC

15 .

I =

0.6 =

VC = V0 (1 – e–t/RC) where V0 is the voltage of the battery. The potential drop across the resistor is

E

E

R rA

rB

=

2E R rA rB

Now, the terminal voltage of cell A is 2ErA VA = E – IrA = E – R rA rB

63. R =

9 = (3)2 2 loge (3) RC 2 loge (3) (5 106) (1 10–6) 11 s

1.1

l

. Since the two wires are made of the same r2 material, resistivity is the same for wires AB and BC. Since the wires have equal lengths, it follows that R 1/r2 . Hence

RAB 1 = , i.e RBC = 4RAB RBC 4 Since the current, is the same in the two wires, it follows from Ohm’s law (V = IR) that VBC = 4 VAB. Hence choice (a) is wrong. Now power dissipated is P = I2 R. Since I is the same, P R. Hence PBC PAB

=

RAB =4 RBC

Hence chioce (b) is correct. Choice (c) is wrong because current density (i.e. current per unit area) is different in wires AB and BC because their crosswire is E = V/l. Since the two wires have the same length (l), E is proportional to potential difference (V). Since VBC = 4 VAB, EBC = 4EAB. Hence choice (d) is also incorrect. 64. Refer to Fig. 22.97.

2ErA =0 R rA rB which gives 2 rA = R + rA + rB or R = rA – rB VA = 0, if E –

So the correct choice is (a). 62. At instant of time t, the charge on the capacitor is given by q = q0 (1 – e–t /RC) and the potential drop across the capacitor is given by ( V = q/C)

2

Fig. 22.97

22.36 Comprehensive Physics—JEE Advanced

Charge on capacitor C1 in Q1 = C1 V1 = 3 F 3V =9 C Charge on capacitor C2 is Q2 = C2 V2 = 6 F 6V = 36 C Now, when the switch is open, the initial charge Y to X = 18 – 18 = 0 C because the right plate of C1 has a charge – 18 C and left plate of C2 has a charge + 18 C. When the switch S is Y to X = – 9 + 36 = + 27 from Y to X when the switch is closed = 27 C – 0 = 27 C. 65. Corrected length L1 (= AJ) = 52 + 1 = 53 cm Corrected length L2 (BJ) = (100 – 52) + 2 = 50 cm For a balanced Wheatstone bridge, L X 53 = 1 L2 50 10

When the switch S is open, capacitors C1 and C2 are in series and their combined capacitance is 3 6 C1 C2 = =2 F 3 6 C1 C2 Charge on each capacitor = 2 F 9V = 18 C When the switch S is closed, in the steady state, C=

resistors R1 and R2 will be in series and their combined resistance is R = R1 + R2 = 3 + 6 = 9 . Therefore, the current in each resistor is 9V I= =1A 9 Potential difference across R1 is V1 = IR1 = 1 3 =3V Potential difference across R2 is V2 = IR2 = 1 6 =6V Since capacitor C1 is connected across R1, the potential difference across C1 is V1. Similarly, the potential difference across C2 is V2.

X = 10.6

II Multiple Choice Questions with One or More Choices Correct 1. A galvanometer has a resistance of 100 and fullscale range of 50 A. It can be used as a voltmeter or an ammeter, provided a resistance is connected to it. Pick the correct range and resistance combination (s): (a) 50 V range with 10 k resistance in series (b) 10 V range with 200 k resistance in series (c) 5 mA range with 1 resistance in parallel (d) 10 mA range with 2 resistance in parallel IIT, 1991 2. Choose the correct statements from the following. (a) A low voltage supply of, say, 6 V must have a very low internal resistance. (b) A high voltage supply of, say, 6000 V must have a very high internal resistance. (c) A wire carrying current stays electrically neutral. (d) A high resistance voltmeter is used to measure the emf of a cell. 3. The terminal voltage of a battery is (a) always equal to its emf (b) always less than its emf

(c) greater or less than its emf depending on the direction of the current through the battery (d) greater or less than its emf depending on the magnitude of its internal resistance 4. The internal resistance of the cell shown in Fig. 22.98 is negligible. On closing the key K, the ammeter reading changes from 0.25 A to 5/12 A, then (a) R1 = 10 (b) R1 = 15 (c) the power drawn from the cell increases (d) the current through R decreases by 40%

Fig. 22.98

5. A current I flows in the circuit shown in Fig. 22.99. Then

Electric Current and D.C. Circuits 22.37

(a) If a resistance R 2 = R is connected in parallel with R1 = R, the current through R1 will remain equal to I. (b) If a resistance R2 = 2R is connected in parallel with R1 = R, the current through R1 will remain equal to I. (c) If a resistance R2 = 2R is connected in parallel with R1 = R, the current through R1 will become I /3. (d) If a resistance R2 = 2R is connected in parallel with R1 = R, the current through R2 will be I/2.

(a) The potential at point P is 6 V. (b) The potential at point Q is – 0.5 V (c) If a voltmeter is connected across the 6 V battery, it will read 7 V. (d) If a voltmeter is connected across the 6 V battery, it will read 5 V. 8. In the circuit shown in Fig. 22.102, (a) the current through NP is 0.5 A (b) the value of R1 = 40 (c) the value of R = 14 (d) the potential difference across R = 49 V

Fig. 22.99

6. Two equal resistances R1 = R2 = R are connected with a 30 resistor and a battery of terminal voltage E. The currents in the two branches are 2.25 A and 1.5 A as shown in Fig. 22.100. Then (b) R2 = 60 (a) R2 = 15 (c) E = 36 V (d) E = 180 V

Fig. 22.100

7. Which of the following statements are correct about the circuit shown in Fig. 22.101 where 1 and 0.5 are the internal resistances of the 6 V and 12 V batteries respectively?

Fig. 22.101

Fig. 22.102

9. Choose the correct statements from the following. (a) If n identical cells, each of emf E and internal resistance r are connected in series, the emf of the combination is nE and the internal resistance of the combination is nr. (b) If n identical cells, each of emf E and internal resistance r are connected in parallel, the emf of the combination is E/n and the internal resistance of the combination is r/n. (c) Cells should be connected in series if the external resistance R is greater than internal resistance r. (d) Cells should be connected in parallel of R is smaller than r. 10. The resistance network shown in Fig. 22.103 is connected to a battery of emf 30 V and internal resistance of 1 . Then

Fig. 22.103

22.38 Comprehensive Physics—JEE Advanced

(a) (b) (c) (d)

the voltage drop across the 2 resistor is 12 V. the voltage drop across the 12 resitor is 12 V. the terminal voltage of the battery is 24 V. the voltage drop across the internal resistance of the battery is 6 V.

11. nonuniform cross-section. Which of the following quantities remain constant along the length of the conductor? (a) Current (b) drift speed

resistances r1 = 1 and r2 = 2 respectively, with polarities as shown in Fig. 22.105. IIT, 1997 5 V 6 (c) r = 1.5

12. A cell of emf E and internal resistance r supplies a current of 0.9 A through a 2 resistor and a current of 0.3 A through a 7 resistor. Then (a) r = 1.0 (b) r = 0.5 (c) E = 2.0 V (d) E = 2.25 V 13. A voltmeter graded as 6000 per volt reads 3 R is connected in series with it the reading of the

(b) E = 1.2 V

(a) E =

(d) r =

2 3

Fig. 22.105

16. In the circuit shown in Fig. 22.106, cells E1 and E2 have emfs 4 V and 8 V and internal resistances 0.5 and 1 respectively. IIT, 1978

resistance of the new voltmeter is R . Then (a) R = 5.4 104 (b) R = 3.6 104 V (c) R = 5.4 104 (d) R = 7.2 104 V 14. with 1 and 2 resistances as shown in Fig. 22.104. A 6 V battery of negligible internal resistance is connected between A and B. IIT, 1987 A 6V

1

1

2

1

2

1

2

2

To infinity

Fig. 22.106

(a) The potential difference across E1 is 4.25 V (b) The potential difference across E1 is 3.75 V (c) The potential difference across E2 is 8.5 V (d) The potential difference across E2 is 7.5 V. 17. In the circuit shown in Fig. 22.107, if the galvanometer resistance is 6 , then in the steady state B

B

Fig. 22.104

(a) The effective resistance between A and B is 3 . (b) The effective resistance between A and B is 2 . (c) The current in the 1 resistance closest to the battery is 3.0 A. (d) The current in the 2 resistance closest to the battery is 1.5 A. 15. A single battery of emf E and internal resistance r is equivalent to a parallel combination of two batteries of emfs E1 = 2 V and E2 = 1.5 V and internal

C1 = 10 m F

C2 = 20 m F

A

G=6 R1 = 8

R1 = 16 D

12 V

Fig. 22.107

C

Electric Current and D.C. Circuits 22.39

(b) the current through R2 is 4 A. (c) the charge on C1 is 80 C. (d) the charge on C2 is 80 C. 18. In the cirucit shown Fig. 22.108, the current through (a) the 3 resistor is 1.0 A 9 A (b) the 3 resistor is 15 (c) the 4 resistor is 0.50 A (d) the 4 resistor is 0.25 A

Fig. 22.108

IIT, 1998

ANSWERS AND SOLUTIONS 1. For voltmeter, the resistance R to be connected in series with the galvanometer is given by V –G R= Ig For 50 V range, R =

50 50 10

6

– 100

= 10 –6 – 100 106 1000 k . Hence choice (a) is incorrect. For 10 V range, R =

10

– 100 50 100 6 = 2 10 5 – 100 2

in the external circuit V < E. However, if the cur105

200 k . Hence choice (b) is correct. For ammeter, the shunt resistance is given by Ig G S= Is For 5 mA range, S =

50 10

5 10 Hence choice (c) is correct. For 10 mA range, S =

50 10

across the battery would show V > E. Hence the correct choice is (c). 4. Before closing the key, E E 0.25 = E = 2.5 V I1 = R 10 After closing the key,

6 3

ter one end of the wire as leave it from the other end. Statement (d) is incorrect. A voltmeter does not measure the emf; it measures only the potential difference because it draws some current from the cell. A potentiometer is used for emf measurement because at balance point, no current is drawn from the cell. The potential difference measured with a voltameter is always less than the emf. Thus the correct choices are (a), (b) and (c). 3. Terminal voltage V = E – I r, where E is the emf and r the internal resistance of the battery. If the current

100 = 1

Effective resistance is R2 =

6

100 = 0.5 10 10 3 Hence choice (d) is incorrect. Thus the correct choices are (b) and (c). 2. Statement (a) is correct. The current I drawn from a supply of voltage E and internal resistance r is given by I = E/r. So r must be very small so that a high current can be drawn. Statement (b) is also correct. At such a high voltage, the current drawn from the supply will become dangerously large if its internal resistance is small. Hence a high voltage supply must have a very high internal resistance so that the current drawn from it does not exceed the safe limit. Statement (c) is correct. A wire carrying current is not charged. It stays neutral because as many electrons en-

I2 =

RR1 10 R1 = . (10 R1 ) ( R R1 )

E R2

2.5 (10 R1 ) 5 = 10 R1 12 which gives R1 = 15 . Before closing the key, or

Power P1 = I12 R2 = (0.25)2 After closing the key,

10 = 0.625 W

5 2 10 15 = 1.04 W (10 15) 12 Hence power drawn from the cell increases. Before closing the key, the current through R is I1 = 0.25 A. After closing the key, the current through R is

Power P2 = I 22 R2 =

22.40 Comprehensive Physics—JEE Advanced

5 3 1 = = 0.25 A,which is equal to I1. 12 5 4 Hence the correct choices are (b) and (c). 5. When R2 = R is connected in parallel with R1 = R, the resistance of the combination is R/2. Therefore, the current in the circuit is V/R/2 = 2V/R = 2I, where I = V/R was the current in the circuit when R2 was not connected. Current 2 I divides equally among two equal parallel resistors. Hence the current through R1 will still be I. Thus choice (a) is correct. When R 2 = 2R is connected in parallel with R1 = R, the total resistance in the circuit is 2 R/3 and the current in the circuit is V 3V 3I = = 2R/3 2R 2 3I 2 Current through R1 = = I. Hence choice 2 3 (b) is also correct. The remaining current I through R2. Hence the correct choices are (a), (b), and (d). 6. Using Kirchhoff’s Ist law current through R2 is 2.25 – 1.5 = 0.75 A. Also since R2 is in parallel with the 30 resistance, R2 must be 60 since only half I1 =

through 30 resistor. Total resistance in the circuit becomes 60 + 20 = 80 Potential drop across the battery E = I R = 2.25 80 = 180 V. Hence the correct choices are (b) and (d). 7. Total resistance = 4 + 1 + 0.5 + 0.5 = 6 . Net volt6 age in the circuit is 6 V. Current I = = 1 A in the 6 anticlockwise direction VPR = 1

4=4V

Since R is connected to earth, VR = 0. Hence VP = 4 V VS Q = 0.5

69 yields R = 14 R 40 / 7 Hence the correct choices are (b) and (c). 9. The correct choices are (a), (c) and (d). In choice (b) the emf of the combination E and not E/n. 10. Total resistance of parallel combination is given by 1 1 1 1 = R1 12 6 4 or R1 = 2 . Total resistance in circuit is R = R1 + 2 = 4 . Therefore, current in the circuit is 3.5 =

=

upon the drift speed of free electrons. Resistivity depends only on the material of the conductor. Hence the correct choices are (a) and (c). 12. If E is the emf of the cell and r its internal resistance, then E E = 0.9 and = 0.3 2 r 7 r Dividing the two equations, we get 7 2

1 = 0.5 V. S is at a

higher potential than Q

E

30 =6A R r 4 1 Potential drop across 2 resistor = 2 6 = 12 V. The voltage drop across the 2 resistor is 12 V. Since the resistance of the parallel combination is also 2 , the voltage drop across this combination is also 12 V. Therefore, the total voltage drop across the network = 12 + 12 = 24 V. Thus the terminal voltage of the battery is 24 V. Voltage drop across the battery = 30 – 24 = 6 V. All the four choices are correct. 11. Current does not depend on the cross-section of the conductor. Drift speed is inversely proportional I =

or

r 9 = r 3 r = 0.5 .

VQ = – 0.5 V Current is being forced into the 6 V battery in the opposite direction. Hence V6 = E + I r = 6 + 1 1 = 7 V. Hence the correct choices are (b) and (c).

13. A voltmeter is graded according to its resistance

8. Potential difference across MP = p.d. across N O = p.d. across NP (see Fig. 22.104)

a voltmeter has a resistance R ohms and it reads V

Current across NP, INP

10 = 20

Across MP, 0.5 R1 = 20

or

1 or INP = 2 A R1 = 40

Total current = 2 + 0.5 + 1.0 = 3.5 A

Using r = 0.5

E

= 0.9, we get E = 2.25 V. (2 r ) Hence the correct choices are (b) and (d). in

as R/V ohm per volt. It is given that V = 3.0 V and the voltmeter is graded as 6000 /V. Hence the resistance of the voltmeter is R = 6000 /V 3.0 V = 1.8 10 4

Electric Current and D.C. Circuits 22.41

or (R + 1) (R – 2) = 0 voltmeter is V 3.0 = = 1.67 I= R 1.8 104

which gives R = – 1 or 2 . Since negative value of R is not admissible, R = 2 .

10 –4 A

In order to convert this instrument into a voltmeter tance R that must be connected in series with it is given by (here V = 12 V) R =

V 12 –R= I 1.67 10

= 5.4

4

– 1.8

10 4

10 4

The resistance of the new voltmeter = R + R = 1.8 104 + 5.6 104 = 7.2 104 . Hence the correct choice are (a) and (d). 14. Let circuit be broken as shown in Fig. 22.109(a). tance remains unaffected by removing one mesh nite network be R. The effective resistance of the remaining part of the circuit beyond CD is also R. The circuit can be recombined as shown in Fig. 22.109 (b). The resistance R and 2 are in parallel. Their combined resistance is 2R R = R 2 R is the series with remaining 1 total combined resistance is

Fig. 22.110

Applying Kirchhoff’s loop rule to the two meshes in Fig. 22.110, we have ( R = 2 ) 1 I + 2I = 6 (1) and 2(I – I ) – 2I = 0 which gives I = 2I . Therefore, using Eq. (1), we have 6 = 1.5 A. 2I + 2I = 6 or I = 4 So the correct choices are (b), (c) and (d). 15. Refer to Fig. 22.111. Applying Kirchhoff’s loop rule to loop abcda, we have – Ir2 + E2 + E1 – Ir 1 = 0 which gives I =

E1 r1

E2 2 1.5 7 = = A r2 1 2 6

resistance. The

2R +1 R 2 which must be equal to the total resistance of the Fig. 22.111

The emf across A and B is

Fig. 22.109

2R 3R 2 +1= R 2 R 2 2 or R + 2 R = 3 R + 2 or R2 – R – 2 = 0 or R2 – 2 R + R – 2 = 0 or R (R – 2) + (R – 2) = 0 R=

E = – E2 + Ir2 7 5 = – 1.5 + 2= V 6 6 This is the emf of the single battery. The internal resistance of the single battery is the resistance r of the parallel combination of internal resistances r1 and r2 which is r1 r2 1 2 2 r= = r1 r2 1 2 3 So the correct choices are (a) and (d). 16. Equivalent resistance of the parallel combination of 3 and 6 is R = 3 6/(3 + 6) = 2 . As the cells are in opposition, net emf E = E2 – E1 = 8 – 4 = 4 V.

22.42 Comprehensive Physics—JEE Advanced

= 2. Also V1 + V2 = 12. Therefore, V1 = 8 V and

Therefore, current is I=

R

E 4.5 r1

r2

=

4 2

4.5

0.5 1

V2 = 4V. Thus Q = C1V1 = 10 F 8 V = 80 C = C2. Thus the correct choices are (a), (c) and (d). 18. A and B of the AB is 4 . Therefore, total resistance = 3 + 4 + 2 = 9 . Current I = 9 V/9 = 1 A. The current is equally divided in the 8 A and B and the remainder 8 . Hence current in AC = 0.5 A. This current is equally CD CD. Therefore, current in the 4 resistor = 0.25 A. Hence the corrects choice are (a) and (d).

= 0.5 A

Potential difference across E1 is V1 = E1 + I r1 = 4 + 0.5 0.5 = 4.25 V Potential difference across E2 is V2 = E2 – Ir2 = 8 – 0.5 1 = 7.5 V Hence the correct choices are (a) and (d). 17. branch ABC ADC. So choice (a) is R2 R1 =I A and C = V C2 12 V. Therefore Q = V1C1 = V2C2. Thus 1 V2 C1

III Multiple Choice Questions Based on Passage Questions 1 to 4 are based on the following passage

R3 = 2R2 = 4 ohm C= 5 F

Passage I

IIT, 1988 1. The current in resistance R1 is (a) 0.5 A (b) (c) 1.5 A (d) 2. (a) 1.5 A (b) (c) 0.9 A (d) 3. (a) 0.3 A (b) (c) 0.2 A (d) 4. (b) (a) 4.8 10–6 J (d) (c) 1.44 10–5 J

Fig. 22.112

E1 = 3E2 = 2E3 = 6 volt R1 = 2R4 = 6 ohm

1.0 A zero R3 is 1.2 A 0.6 A R4 is 0.25 A zero 9.6 10–6 J 1.92 10–5 J

SOLUTION 1.

I2 R3 = E1 or I2 = R1 is zero,

2.

EBCDE

The correct choice is (a). 3. FDHGF

E1 = 1.5 A R3

Electric Current and D.C. Circuits 22.43

I 3 R 2 – I 2 R 3 + I 3 R 4= – E 2 – E 3

1 5 10–6 (2.4)2 = 14.4 2 Hence the correct choice is (c). =

or I3 (R2 + R4) = I2 R3 – E2 – E3 or

I3 (2 + 3) = 1.5

4–2–3

or 5I3 = 1 or I3 = 0.2 A So the correct choice is (c). 4. F and E is V= E2 + I3 R2 = 2 + 0.2 and A F and A 1 CV2 2

R1 = 6

A

E1 = 6V

B

C

I1

C=5 F

E

10 –6 J

F

I3

R2 = 2 E2 = 2V

I1 R3 = 4

E

I2 I3

G E3 = 3V

D

H

R4 = 3

Fig. 22.113

Questions 5 to 8 are based on the following passage Passage II E1 = 3 V, E2 = 2 V, E3 = 1 V and r1 = r2= r3= R = 1 .

Fig. 22.114

IIT, 1981

5. Current I1 (a) zero (c) 1.0 A 6. Current I2 (a) zero (c) 2.0 A 7. Current I3 (a) 1.0 A (c) 2.0 A 8. If r2 B (a) 1.0 A (c) 3.0 A

r1 is (b) 0.5 A (d) 1.5 A r2 is (b) 1.0 A (d) 3.0 A r3 is (b) 1.5 A (d) 2.5 A A is connected to R (b) 2.0 A (d) zero

SOLUTION 5.

abcda and

6

I2 choice (a).

abcdefa 7.

– I 1r 1 + E 1 – E 2 – I 2 r 2= 0 and – I1r1 + E1 – E3 – I3 r3= 0 E1 – I1r1=E2 + I2r2 = E3 + I3r3

(i) a

have I1 = I2 + I3

(ii)

E1 – (I2 + I3) r1= E3 + I3r3 or 2I3 + I2 = 2 (iii) Also E2 + I2r2= E3 + I3r3 or I3 – I2 = 1 (iv) I1 = 1 A. So the correct choice is (c). Questions 9 to 12 are based on the following passage Passage III

I3 = I1 – I2 = 1 – 0 = 1 A. So the correct choice is (a). 8. Since I2 a and d = emf E2 = 2 V and remains equal to 2 V r2 tial difference across a and d E1 and E2 do not the currents I1 and I3 A is conI1 = I3 B R

The direction of one of the currents in the branches is IIT, 1991 9. In branch ab, the current i1 is a to b

22.44 Comprehensive Physics—JEE Advanced b

b to a a to b b to a 10. In branch ad, the current i2 is a to d d to a a to d d to a 11. In branch bd, the current I3 is b to d d to b b to d d to b. 12. (a) 8.0 (b) 8.5 (c) 17 (d) 17.5

I4

I3

10 W

5W

I1

c 5

a

I5

I2 5W

I

10 W

I

d

f

10 W

e

8.5 V

Fig. 22.115

SOLUTION a, b and d I = I1 + I2, I4 = I1 – I3 and I5 = I2 + I3. abda, bcdb and adcefa 10 I1 + 5 I3 – 5 I2 = 0 or

I2 = 2 I1 + I3

(i)

5(I1 – I3) – 10(I2 + I3) – 5 I3 = 0 or and or

I1 – 2 I2 = 4 I3

I1 = 0.2 A, I2 = 0.3 A and I3 = – 0.1 A. Since I1 and I2 I3 indicates that the direction I3 should be from d to b and not from b to d. 9. The correct choice is (c). 10. The correct choice is (c). 11. The correct choice is (d). 12. Total current is I = I1 + I2 = 0.2 + 0.3 = 0.5 A. Hence

(ii)

5 I2 + 10(I2 + I3) – 8.5 + 10(I1 + I2) 2 I1 + 5 I2 + 2 I3 = 1.7

R= (iii)

So the correct choice is (c).

Questions 13 to 16 are based on the following passage Passage IV E, F, G and H of emfs 2 V, 1 V, 3 V and 1 V and internal resistances 2 , 1 , 3 and 1

13.

(+) E (–)

14.

A

V 8.5 = = 17 I 0.5

B

resistor is 1 A 7 1 (c) A 11 (a)

B and D is 2 V 7 2 (c) V 11 (a)

(+) F (–)

2W

(–) H (+)

15. D

(–)

G

(+)

cell G is (a) equal to 3 V (b) more than 3 V

C

Fig. 22.116

IIT, 1984

1 A 9 1 (d) A 13

(b)

2 V 9 2 (d) V 13 (b)

Electric Current and D.C. Circuits 22.45

16. cell H is (a) equal to 1 V (b) more than 2 V

SOLUTION 13. Let I1 and I2 be the currents in branches BAD and DCB I3 DB resistor. E

2V

I1

A

I1

I1

2W

5 6 A and I2 = A 13 13 1 A I3 = I1 – I2 = – 13 The current 2 BD assumed. So the correct choice is (d). 14. B and D = 2

H 1W

1V I3

D

I2

I2 3V

C

I2

G 3W

G = 3 – (6/13) choice is (d).

D have BADB and DCBD Questions 17 to 19 are based on the following passage Passage V are connected to a 60

resistor reads 30 V. IIT, 1980 Voltmeter

(1/13)

= 1 + 6/13 = 19/13 is (d).

1.6 V. So the correct

H 1.46. Thus the correct choice

17. The resistance of the voltmeter is (a) 600

(b) 800

(c) 1000

(d) 1200

18. The current in the circuit is 3 3 (a) A (b) A 32 16 3 3 A (d) A 8 4 19. When the same voltmeter is connected across the 300 (c)

400

300

3 = 21/13

16.

I1 = I2 + I3 or I3 = I1 – I2

connected across the 400

the DB as

15.

Fig. 22.117

and 400

(ii)

I1 =

1V

1W

I1

(i)

B I2

I3

2W

F

2I1 + I1 + 2(I1 – I2) = 2 – 1 = 1 or 5I1 – 2I2 = 1 and 3I2 + I2 – 2(I1 – I2) = 3 – 1 = 2 or 6I2 – 2I1 = 2

60 V

Fig. 22.118

(a) 40 V

(b) 22.5 V

(c) 37.5 V

(d) 25 V

SOLUTION 17. Potential difference across the 400

resistance

resistance, their combined resistance is

the 300 resistance = 60 – 30 V = 30 V. Let R be the resistance of the voltmeter. As the voltmeter is

R=

400 R 400 R

22.46 Comprehensive Physics—JEE Advanced

and 400 R should be equal to 300 . Thus

Total resistance in the circuit = 400 + 240 =640 Current in the circuit is

resistances,

So the correct choice is (a).

R = 1200 18. When the voltmeter is connected across the 300 resistance, their combined resistance is 300 R R = 300 R

300 1200 300 1200

19. 240

resistance

3 240 = 22.5 V 32 Thus the correct choice is (b). =

= 240

Questions 20 to 23 are based on the following passage Passage VI

23. 10– 4 J 10–4 J

(a) 2 (c) 6

state. The currents, the values of resistances and emfs of

10–4 J 10–4 J

(b) 4 (d) 8 1A

C = 4 F.

3

IIT, 1986 20. The value of current i1 is (a) 1 A (c) 3 A 21. The value of current i2 is (a) 1 A (c) 3 A 22. The value of current i3 is (a) 1 A (c) 3 A

60 V 3 = A 640 32

I=

400 R 300 = 400 R

4V

3

5

A

i1

E

2A

(b) 2 A (d) 4 A

4 F 3V

1

2A

(b) 2 A (d) 4 A

1

B

2

i2 D

i3

4

3 1A

(b) 2 A (d) 4 A

Fig. 22.119

SOLUTION 20.

A,

V = 5 i1 + 1 i1 + 2 i2 = 15 + 3 + 2 = 20 V

i1 21.

B, i2 + 1 = 2 or i2

C = 20 V

choice (a). 22.

D, i1 = i2 + i3

1 4 10 – 6 (20)2 = 8 2 Thus the correct choice is (d).

= 2 A. So the correct choice is (b). 23. AEDB

=

Questions 24 to 26 are based on the following passage Passage VII

A voltmeter of resistance 400 the 400 resistance.

1 CV2 2

i3 = i1 – i2 = 3 – 1

is connected across

IIT, 1996

10– 4 J

24. The value of current i1 is 1 1 (a) A (b) A 10 20 1 1 A (d) A (c) 30 40 25. The value of current i2 is 1 1 (a) A (b) A 30 15

Electric Current and D.C. Circuits 22.47

(c)

1 A 10

(d)

2 15

(a)

10 V 3

(b) 5 V

(c)

20 V 3

(d) 4 V

26.

SOLUTION

Fig. 22.120

24.

– 200 I3 + 200(I1 + I2 – I3) + 100(I2 – I3) = 0 Resistance R = 200 of the 400 resistor and the resistance of 400 the voltmeter.

or 2 I1 + 3 I2 – 5 I3 = 0 (3) Simultaneous solution of Eqs. (1), (2) and (3) yields 1 A I1 = I2 = I3 = 30 So the correct choice is (c). 25. The correct choice is (a). 26. R(= 200 ) 1 20 200 = V. = I3 R = 30 3 Hence the correct choice is (c).

of

– 100 I2 – 100(I2 – I3) + 100 I1 = 0 or

I1 – 2 I2 + I3 = 0

(1)

– 100 I1 – 200(I1 + I2 + I3) + 10 = 0 or

3 I1 + 2 I2 – 2 I3 – 0.1 = 0

(2)

Fig. 22.121

IV Assertion-Reason Type Questions correct.

22.48 Comprehensive Physics—JEE Advanced

(a) Statement-1 is True, Statement-2 is True; ment-1. (b) Statement-1 is True, Statement-2 is True; Statement-1. (c) Statement-1 is True, Statement-2 is False. (d) Statement-1 is False, Statement-2 is True. 1. Statement-1

5. Statement-1 E1 and E2 are the emfs of cells C1 and C2 E1 > E2. Cell C1 R is x x

-

AB is

R C1

-

Statement-2

D

x

A

C2

B

G

Fig. 22.123

2. Statement-1 Statement-2 –1

electron is also very small (= 1.6

10

– 19

C), yet

Statement-2 7

3. Statement-1

AD due to cell C1 = E2, the emf of cell C2. 6. Statement-1 AB tor R decrease. Statement-2

x

Statement-2 A and D due to cell C1 = emf E2 of cell C2. 7. Statement-1 Electrons in a metallic conductor have no motion if direction. 4. Statement-1

Statement-2 AC x. If the radius AB

IIT, 1982 8. Statement-1

becomes 4 x.

Statement-2 R1

R2 G

IIT, 1993 C

A

B

9. Statement-1 -

x

Fig. 22.122

Statement-2 the square of its radius.

of the standard resistance.

Electric Current and D.C. Circuits 22.49

Statement-2

IIT, 2008

SOLUTIONS 1. The correct choice is (c). The electrons suffer a

A and B due to cell C1 x. 6. The correct choice is (a). If the value of R is A and B due to cell C1 x. 7. Electrons in a conductor have random thermal motion. Statement-1 is false and Statement-2 is true. 8. Both the statements are true and Statement-2 is the

decelerated by collision. The net acceleration

2. The correct choice is (b). The current in a metal de-

1029 3.

3

9.

.

R is the standard resistance and X

rent stays neutral because as many electrons enter

4. The correct choice is (d). The condition for no R1 R2

RAC RCB

Fig. 22.124

RAC and RCB AC and CB AB is doubled, the ratio RAC/RCB 5.

X (100 l ) = R l The value of X

l the same, the value of R must be increased. So Statement-1 is false, Statement-2 is true.

AB

V Integer Answer Type Questions 1. emf of 5 V and an internal resistance of 0.2

2.

. Find

IIT, 1997 L internal resistances are connected in series. Due to

Fig. 22.125

22.50 Comprehensive Physics—JEE Advanced

T in a time t. A number N L

T in

the same time t. Find the value of N.

AJ = 60 cm. Find the value of X (in ohm). IIT, 2002 4.

IIT, 2001

AB in volts is

Fig. 22.126

3.

Fig. 22.127

AB resistance X and a resistance of 12

IIT, 2011 are connected

SOLUTIONS 1. All the cells are in series. Therefore, total emf = 40 V and total internal resistance = 1.6 . Therefore, current in the circuit is I=

40 = 25 A 1.6

3. In the balanced condition X BJ 40 = = 12 AJ 60

X=8

4. E – Ir = 5 – 25

0.2

= 5 – 5 = 0 volt. 2.

t is Q = V 2t/R, V Here Q = ms T

m R its resistance.

3V 2 t ms T = R m R second case, (2m) s T =

NV 2 t 2R

s Fig. 22.128

(1)

CDFEC 6–1

I–2

I – 3= 0 I= 1A

(2) VA – 6 + 1

I – VB = 0 VA – VB = 6 – 1

2

2=

N /2 or N 2 = 36 or N = 6, 9

1=5V

23

Current

Chapter

REVIEW OF BASIC CONCEPTS 23.1

23.2

HEATING EFFECT OF CURRENT

If a current I R V

t

POWER-VOLTAGE RATING OF ELECTRICAL APPLIANCES

(P – V P = VI

H I=

2

V t R J

H = VIt = I 2 Rt = H Electrical Power r R

V V V2 = = I P /V P 1. Power of Electrical Appliances Connected in Parallel R=

If E

Let R1 R2 R3 = VI = I2 R =

V

2

2

P

P V

E R V = R R r 2 [

Electrical Energy

P1 P2 P3

E = I(R + r A

V2 R1 = P1 P1 =

(60

V2 R1

V2 R2 = P2 P2 =

V2 R2

V2 R3 = P3 P3 =

V2 R3

R 1 1 1 1 = + + + ... R1 R2 R3 R

P=

= 1000 = 3.6

106

V2 = V2 R

1 R1

1 R2

1 R3

V2 V2 V2 + + + ... R1 R2 R3 P = P1 + P2 + P3 + ... =

...

23.2 Comprehensive Physics—JEE Advanced

2. Power of Electrical Appliances Connected in Series R1

R R = R1 + R2 + R3 + ...

R

R1 V

I=

P = I 2 R = I2 (R1 + R2 + R3

R

=

R1

300 1000 R1

P = P1 + P2 + P3 + ... 23.1

Fig. 23.1

R1 = IR1 =

SOLUTION P

V 2

V 200 200 = = 40 R= P 1000 R V 2 160 160 V = P = R 40 P P 100 P 1000 640 = 100 = 36% 1000

300 R1 = 100 1000 R1 23.3 A

300 R1 1000 R1

R1 = 500

B

23.2 SOLUTION A

RA =

V2 PA

= 200

200 = 1000 40

SOLUTION P

2

R=

P1 = P1 >> P

B

V

RB =

V2 200 200 = = 1000 R 40

V 12 300 300 = R 1000

=

V PB

200

200 60

= 666.7

I. A B

PA = I2 RA PB = I2 RB

PA I2 R R = 2 A = A PB RB I RB RA > RB; PA > PB

A B.

23.3

P2 = I 22 R = (5

V. A

PA =

V2 RA

B

PB =

V2 RB

R PA = B RA PB

105 2 10 = 2.5 P1 << P2

1012

23.6

RB < RA; PA < PB

E

B

r R

A. R=r

23.4

SOLUTION R = 0.2

103

P I=

V

P 10 103 = = 50 A V 200 2 0.2 I 2R

Fig. 23.2

SOLUTION ve

E

I=

R

r R

10 = 10.5 23.5

R

r

R=

E2R R

r

R

r

2

d 2P

dP dR

P

2

E

P = I 2R =

dR 2 R

10 dP = E2 dR

SOLUTION P V1

106

I1 = 2

8

R = 10

P 108 = = 5000 A V1 20 000

P1 = I 21 R 10 = 2.5 108 P 108 = =5 V2 200

dP dR R = r.

R E

R

r

105 A

2

2R

2

r

3

[(R + r

R+r d 2P dR 2

V2 I2 =

=

1

R

R=0

R=r

3

23.4 Comprehensive Physics—JEE Advanced

23.3

VARIATION OF RESISTANCE AND RESISTIVITY WITH TEMPERATURE

SOLUTION R2 = R1 [1 + (t2 – t1 t2 – t1 =

R2

117 100 R1 = = 1000°C R1 (1.7 10 4 100

t2 = 1000 + t1 = 1000 + 27 = 1027°C R2 = R1[1 +

(t2 – t1

R1

23.8

t1 R2 t2

2

=

1[1

at +

(t2 – t1

SOLUTION R2 R1 3.75 3.00 = = 0.0025 K–1 R1 t2 t1 3.00 (100 0 Rt R0 Rt = R0(1 + t = t R0 =

t=

23.7 at 27°C.

= 1.7

Rt

R0 3.15 3.00 = = 20°C R0 0.0025 3.00

10–4 K–1.

I Multiple Choice Questions with Only One Choice Correct 1.

R1 R2 (< R1 R1 – R2

3 4

8 9

1 (R1 – R2 2

R1 R2 R1 R2

R1 R2

2. tance

3.

1 3

1 2

Fig. 23.3

23.5

4.

4

3

2

V T P = k (T – T0

11.

k

T0

R T Y

2

RV V 2Y kR 5.

12.

kV 2Y R

kY

kYR V2 G

13.

S G S S= 1 G 3 3 S= G 2

S= 3 G 4 2 S= G 3

14.

6. n

15. P

n P

7.

P P

P 16.

8.

17. V

Q

t t 9.

Q 4 Q

10.

18.

E

r Q

r1

P1 P2 P2 /P1

Q 2 Q

t

23.6 Comprehensive Physics—JEE Advanced

r2 t r12

r

8 9

r22 r1

r2 1 (r1 + r2 2

25. R1

r1 r2

19.

R2

t1. 1 (R1 + R2 2 1 (R1R2 1/2 2

t2 1 (t1 + t2 2 t1 t2 t1 t2

t1 + t2

1 (R1 – R2 2 R 1R 2

1/2

26.

t1 t2

20.

t 1. t2.

1 (t1 + t2 2 t1 t2 t1 t2

t1 + t2 27. t1 t2

21.

4

. at

–1

–1

–1

–1

22.

23.

Fig. 23.5

28. –1 –1

–1 –1

R

24.

R = 0.5

R

Fig. 23.4

Fig. 23.6

23.7

33.

4 9 8 3

B1

W1 W2 W3 B1 B2 B3 W1 > W2 = W3 W1 > W2 > W3 W1 < W2 = W3 W1 < W2 < W3

29.

30. R R

1 5

Fig. 23.8

34.

Fig. 23.7

31.

32.

L

T

B2

B3

t

N L T t

N

Fig. 23.9

23.8 Comprehensive Physics—JEE Advanced

35.

A B

B A

42. 36.

43.

37.

AB

BC

L AB

10–1

–1

BC

K–1

r

I

BC AB. BC AB. AB

38. AB

Fig. 23.10

44. 2 39.

A

B

A

B A

B 1 4 40.

1 2 A

B Fig. 23.11

45. B A

1

41.

Fig. 23.12

BC BC

r

23.9

46. R1 R2

P1 > P2 > P3 P2 > P1 > P3

R3 R1 R2

P2

R3

P3 > P2 > P1 P1 = P2 = P3

P1

P3

Fig. 23.13

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43.

2. 8. 14. 20. 26. 32. 38. 44.

3. 9. 15. 21. 27. 33. 39. 45.

4. 10. 16. 22. 28. 34. 40. 46.

5. 11. 17. 23. 29. 35. 41.

6. 12. 18. 24. 30. 36. 42.

SOLUTIONS 1. If E

r R1 E

I

R1

r R1

Q1 = I R1 =

Q2 = Q1

R1

r

R1 Fig. 23.14

2

E R2

2

E

2

4 5

R2

r

Q2 r =

I = R1 R2 . 4 5

V 2 4 5

2. 1

6 =2 5

6 5

.

V1 =

= 2V 5

V 2

4 5

2V 5

23.10 Comprehensive Physics—JEE Advanced

V4 =

2V 5

V2 = V3 =

P1 =

V12 R1

3V 5

=

3V . 5

Y =

= 4V 25

2 V2 9V 2 P2 = 2 = = 4.5 V R2 25 2 25

V2Y kR

S = 5.

2 V2 9V 2 P3 = 3 = = 3V R3 25 3 25

S = stress

S Y

=

2

L L S

G

S G

S=V t

2 V2 4V 2 P4 = 4 = = V R4 25 4 25

H =

P2

V2t R

H

1 R

HG S = HS G

3. I1 =

4I 5

I2 =

3I 2I I3 = 5 5

I4 =

2 S = 3 G

I 5

P 1 = I 12 R 1 =

16 I 25

P 2 = I 22 R 2 =

9I2 25 4I 25

P 4 = I 42 R 4 =

I2 25

2 G 3

6. W = V × I

R = 12/2 = 6 2

2

1 =

16 I 25

2 =

18 I 2 25

2

P 3 = I 32 R 3 =

S=

3 = 4 =

12 I 25

I = W/V = 24/12

. If n

n 1.0 =2 6

n = 12

7. I =

2

4I2 25

500 =5A 100 100 R= = 20 5 R

P1 8 = P2 9

200 =5 R 20

4.

R = 20

8. V2 = k (T – T0 R L =

L T

k L V2 k = = L R

k T T=

L L

V I = 100

R I 2R = 900

23.11

R =

900 = 4.0 15 15

9.

V = IR = V–V =

10. P

10 6

3V 2 R P = 9 P. P = 16.

R 17. Let R

I1 =

100 106 = 5000 A 20 000

P1 = I 21 R =

2

R

t 2

Q =

V t 2R

R = 2.5 107 R R =R t

I2 =

100 106 =5 200

P2 =

I 22 R

105 A

Q =

V 2t 2V 2t = R R

Q = 4Q = 2.5

10

11

18.

R

2.5 1011 P2 = = 10 4 2.5 107 P1

I1 =

E r1

r

11.

2

P=I R

I

r2

2

P=

V R P 2 V = P V

V V

P=

r1

2V V R

r2 t

r

r1 r1 (r2 + r

r2

=

2

r

2

r2

r1 r2

R2

13.

R1 + R2 Q

14. 15. Let R

V

P =

V2 3R

2

r1 r2 (r1 – r2

r = 19. Let R1

2

r2 (r21 + 2r r1 + r2

r 2 (r1 – r2 2=

r

= r2 (r1 + r

r1 (r 22 + 2r r2 + r 2

P =–2 P

r 1t

r

2

E

Q2 = P P

I

12.

r1

P = 2I IR

P 2 I = P I

2

E

Q1 = I 2 r1 t =

Q= t=

V 2t R1 R2 Q R1

V = t1 + t2

R2 2

=

V

V 2 t1 V 2 t2 = R1 R2 =

Q R1 V

2

+

Q R2 V2

23.12 Comprehensive Physics—JEE Advanced

25.

20.

I 21 R1t = I 22 R2 t E I1 = R1 r

1 1 1 = + R1 R2 R 1 V2 = t1 Q R1 2 1 = V Q t

1 V2 = t2 Q R2 1 R1

1 R2

=

E2

1 1 + t1 t2

V2t R

t1 H

R2

R2

R1

R=

. r R2

2

r

R2 = (R2 + r

H=

t2 t1

R2 /R1

2

E

E2

R1 =

r=

26. H =

R2 = R0(1 + t2

R1 + r

R1

t1 t2 t = t1 t2 21. R1 = R0 (1 + t1 t2 R2 1 = R1 1

r

2

I2 =

2

R1

R2 l = A

l r2

V 2 t r2 l r l

27. Let I

t1

I 5

t2 22. 2

Q = V t R

2205 80 4.2 23. Let R

210

210

P1 = I2

–1

20

P2 = 1 P1 5

I 2

P2 =

2

P2 = P1 = 10 5 5

4 = I2 –1

28.

1

R1 = R + R + R = 3R R2 = R + R + 4R = 6R R1 R2 3R 6R R = = = 2R R1 R2 3R 6R

R R = R 4 V 2 = 4V 2 R R

4

= R = 2R R=2

–1

=4

29. 30.

24.

R 10 I=

R = 0.05

V2 P

100 10 R 10 10 R

I P = I 2R

2

100 100 = 1000

I = I

R 10

R

23.13

R

100 10 R 10 10 R

=

10 R R 10 10 R

= 1

R

10

5

R

5R R

W3 =

250 2 1042

34. Let R

10

R

25R 2 R

156.3 2 1042

W1 < W2 < W3

Q = I2 5

W2 =

R

R

R

2

R = 5 35.

V = 22

P

31. 32.

t V Q = ms T

Q = V 2t/R R=

m

P = VI V I= P

V R= I

s

220 2 V2 = = 484 100 P

R

V 2

ms T = 3V t R R

m

(2m s T =

2

2 = N /2 9

N2

P =

V R

2

=

110 2 484

36.

NV 2 t 2R

R = 20 P = V2/R V

N

P V2 = RP R

=

Pma

=

V=

RP

20 1000

= 100 2 V

33.

B1 B2

B3

are

R2 =

37.

250 2 = 625 100

V2 R1 = W1 2

250 60

= 1042

B3 B1

Q = KA 2 d t

= R3

V3

250 625 625 1042 V2

W1 =

V12 R1

A A=6

VR1 V1 = R1 R2

B2

1

93.7 2 625

=6

2

(0.5 m

1 1.5 Q = 1.68 10 1 10 3 t = 2.52 104 –1

100

23.14 Comprehensive Physics—JEE Advanced

H=

R=

V2 = 2.52 R 420 420 2.52 104

IB =

104 IA

IA

IB

IB

=7

V2 A V2 = l R

38.

V = 400 = 1 A RB 400

42.

200 2 = 80 500

R=

.

l V

A L

39. RA =

2L

RB =

r2 RA RB

2r

P=

2

RA = 2RB

Q 40.

l

43. R =

1 R

V2 50 2 = R 80

r2 AB

A V2 200 = RA = PA 40 RB =

BC

B are

1/r2

R

2

= 1000

RAB = 1 RBC 4

V2 200 2 = = 400 PB 100

V = IR P = I2 R

IA =

PA 40 = = 0.2 A V 200

IB =

PB 100 = = 0.5 A V 200

RBC = 4RAB

I

P

VBC = 4 VAB. R

PBC R = AB = 4 RBC PAB

AB

BC

R = RA + RB = 1000 + 400 = E = V/l l E VBC = 4 VAB EBC = 4EAB

1400 I=

V 400 = = 0.286 A R 1400

I

IA

(V IB

A 44.

B

2 ven V 2

41. V A IA =

B are

V 400 = = 0.4 A RA 1000

45. cur

V 2 = 30 3

V2

V

23.15 2

V2 R P 2 > P 1 > P 3.

V

P=

rr 46.

1 R

P R1

R1 = 1

R2 R3 = 2 .

R2 = 0.5

II Multiple Choice Questions with One or More Choices Correct 1. m 2m

I

5.

l

A

V 2.

V

A l

6. tance 0.5 R=8 3.

7. R

4. R = 0.5 E

8.

.

R

r P

P P

R = r. R = 2r. E2/2r. E2/4r.

23.16 Comprehensive Physics—JEE Advanced

9.

E&

A

B

r

A

P

B

I E I= r

A

2

P

=

B

E I= 2r

14. 2

E r

P

=

E 4r

10.

11. –1

. Fig. 23.16

BD

15. Fig. 23.15

–1

. A

12.

E

A

r sup

E r=1

D D

E r=2

13.

A B

A

B

Fig. 23.17

ANSWERS AND SOLUTIONS 1.

l A

d m = Ald A

d

I t H = I 2Rt

H H

R

23.17

R ence V

1/P

H 2 H = V t R

H

1 R

R1 R2 R

V 2 = 200 200 W 40 = 1000

2.

200

1/P R1

R2. If R V

I1 =

200

V R

R1

60 = 666.7

I2 =

200 200 100 = 400

V R

R2

R1 > R2 I2 > I1

= 1000 + 666.7 + 400 = 2066.7 I =

200 = 0.097 A 2066.7

V 2/R

3. R = If l

l = A r

2 V2 = V A l R

5.

l r2 R

6. (r

H = V 2/R

ce E = V l

0.5 = 3

E

E I=

4. Rb =

2 V2 = 200 = 400 Pb 100

E = 88 = 8 A R r 8 3 I2 (R + r

2 R h = 200 = 40 1000

7. V

I

r = 0.1

2

23.18 Comprehensive Physics—JEE Advanced

P = V I = 1.5 2

d 2P dR 2

2

Pc = I r

at R r

E2 8r 3

P

R

R=r

R=r

PR = P – Pc P

R 1.5 R R 0.10

2 = E 4r

9. R

R

=–

E

I

R

=

1.5 R 2.0 R 0.10

P = (E – E I E = Ir. P = EI – I2r

PR 1.5 R 2.0 = 2.6 R 0.10 R = 0.65 .

I dP = E – 2 Ir dI dP dI

P 8.

I= E

I =

E 2r d 2P

R r

dI 2

=–2r P

I = E/2r. I = E/2r

2

P = I 2R =

E R R r

E dP/dR

2

r

P

P d2P/dR2 R

E

10.

r dP = dR

E2 R r

2

2R R r

1

11.

dP/dR

d P dR

2

=

2E

R r

I2

P Q = I22 5 I1 4

3

3R R r

I2 C

I1

d2P/dR2 at R = r

2

I1

B

P R=r

R=r P 2

I

A

2R 1– =0 R r R=r

2 = E 4r

2

I 12

D

5

I1 = I 2 . 2 2 20 = I2 5

I2

2

–1

I1

4=1

4=4

23.19

14. A

B

E

P = I 2R =

R r

E 2 R1

P1 = P1 = P2

R1

r

R1 R2

R2 R1

E2

6

6 12

13. P1 = ce

100 R1

r r

R r

D

C

BD 2

2

3

E 2 R2 R2

r

2

15.

2

C CB B 6 + 2 = 8 A.

2

4 r 6 r

1.5 = 6 1.96 4

E2R

P2 =

2

D

I= 6 =6A 1

Fig. 23.18

12. I =

B

r 1

C

B

E

2

P2 =

200 R2

2

P 1 = P 2.

R1 = 1 R2 4

Fig. 23.19 2

current I

2

A V1 IR R = 1 = 1 = 1 V2 IR2 R2 4

D = (2 A

3

1

5

I 2 R1 P1 R = 2 = 1 = 1 I R2 P2 R2 4

III Multiple Choice Questions Based on Passage Questions 1 to 5 are based on the following passage passage I

B

A B

Joule’s Law

I q V

A

A

B

t

23.20 Comprehensive Physics—JEE Advanced

I= q t A

B V W = qV

3.

V R causes a current I

2 H = qV = I tV = I 2Rt = V t R P

1.

V = IR

(

2 P = H = IV = I2R = V t R A B

V R R

R V I R

V

4. 5. A B

P

B. A.

2.

P 4

P 2

P

P

ANSWERS AND EXPLANATIONS 1.

I2 (2R

I2 R = 2P

2V 2 V2 =2 = 2P 2R R 2.

4. 5. Let R

V P = V2/R

3.

2V R

2

=4

V2 = 4P. R

P V2 = 2 2R

R

V2 = P = 4P. R/4 2

23.21

Questions 6 to 9 are based on the following passage

7.

Passage II 8. 1/8 9. 6.

SOLUTION 6.

P

P1 + P2 = 16 + 80 = 96 7. 2

P1 2

8

1 8

8.

9. 2

P2 2

Questions 10 to 12 are based on the following passage

10.

I

Passage III 11.

12.

Fig. 23.20

SOLUTION 10.

3I 4

I2 = 15 I = 3I 20 4 I1 = 5 I = I 20 4 I 22

2

11.

5 = 45

I = 4 A.

I2

2 =

2

12.

I1 I = 4

4 6= 4

6

23.22 Comprehensive Physics—JEE Advanced

Questions 13 to 16 are based on the following passage

13.

R1

14.

R2

15.

R3

16.

R4

Passage IV R1 = 4

R2 = 6

R4 = 10

R3 = 12

S1

C = 500 S1 S1

S2

S2

Fig. 23.21

SOLUTION P = I 2R

S1

P1 P2 P3 = I 12 R1 I 22 R2 I 32 R3 U = 1 CV2 = 1 500 2 2 S1 R4 P4 = 0. Let I1 t R2 I3 I3 I 2R 2 = I 3R 3 I1 = I2 + I3 I1 = 2I3 + I3

I3 =

I1 I2 I3 = I1

2 I1 3

I1 3 I1 3

10–6

2

2

2

2

S2 R4

U R1 R2

I2

13. P1 = 3.6 3 6

R3

I1 = I2 + R2 R3 I2 = 12 I3 I2 = 2I3 I2 = 2 3

14. P2 = 3.6 2 6 3 . 6 1 15. P3 = 6

2 I1 3

16.

1 3

IV Assertion–Reason Type Questions

one

R3

R4 P4

12

23.23

Statement-2 1. Statement 1 3. Statement 1 (I – V T2 T 1.

T1 T2

Statement-2

4. Statement 1

Fig. 23.22

Statement-2

2. Statement 1 Statement-2

SOLUTIONS 1.

I1 =

I–V

2.

T2

R

R1

I2 =

V R

R2

R1 > R2 I2 > I1

T2 ture T1

V

T2 T 1. 4. R

1/P

R

P V

3.

R

R1 R2 R

V

V2/R

24

and Magnetism

Chapter

REVIEW OF BASIC CONCEPTS 24.1

BIOT-SAVART LAW

dB at a point whose position vector with respect to a current element dl is r is given by I (dI r ) (1) dB = 0 4 r3 where

0

24.2

10–7 Hm–1

=4

AMPERE'S CIRCUITAL LAW

Fig. 24.1

a closed curve is proportional to the current threading or passing through the closed circuit i.e. B·dI = where

24.3

0

B=

0I

directed into the page if I is clockwise 2r and outside the page if I is anticlockwise.

0I

is the permeability of free space.

MAGNETIC FIELD DUE TO CURRENT CARRYING CONDUCTORS

For a coil of N turns. NI B= 0 2r (iii) (Fig. 24.3). B=

(i) I (Fig. 24.1) B=

0I

directed into the page (away from the

2 r reader) if the I is upwards and towards the reader if I is downwards. At points Q or S, B = 0 (ii) radius r (Fig. 24.2)

Fig. 24.2

0I

2r

2

directed into the page. For a semi-circular element ( = ) I B= 0 4r

Fig. 24.3

24.2 Comprehensive Physics—JEE Advanced

straight conductor are parallel. Therefore, dl = 0. Hence B = 0 at point .

(iv) (Fig. 24.4) B=

0I

(sin

4 r directed into the page Special Cases (a) If the conductor XY

and point lies near the centre of the conductor (as in Fig. 24.1), = = 90° so B= =

0I

4 r

+ sin )

r

(v) (Fig. 24.7) O due to straight portions and ST is zero and due to semicircular part is I B = 0 directed into the page. 4r

Fig. 24.4

(sin 90° + sin 90°)

Fig. 24.7

0I

If the current in anticlockwise, B is directed towards the reader.

2 r (vi)

(Fig. 24.8)

(b) If the conductor XY is of

a 2 b2 directed into the page ab For a square coil (b = a) B=

lies near the end X or Y as shown in Fig. 24.5, then = 90° and = 0°, Fig. 24.5 then I B = 0 (sin 90° + sin 0°) 4 r =

2

0I

B=

0I

2 2 a

0I

4 r

Note: near one of its ends. (c) If the conductor XY L and point lies on the right bisector of the conductor, as shown in Fig. 24.6, then = and L/2 sin = sin = x L/2 = L 2 r2 2 L = 2 4r L2

so B =

0I

2 r

Fig. 24.8

llow metal pipe of (Fig. 24.9)

(vii) At point

,B=

0I 2 r

L 4r

2

L2

Fig. 24.6

(d) If the point lies on the straight conductor or on its axis, then dl and r for each element of the

Fig. 24.9

Magnetic Effect of Current and Magnetism 24.3

(Fig. 24.9) Inside the pipe at a point at a distance r from the axis, Ir B= 0 2 2 Outside the pipe at a point at a distance r from the axis I B= 0 2 r (viii) (Fig. 24.10) At point

4

, B =

0nI,

where n =

N N ; N = total no. of turns. Outside the L 2 solenoid, B = 0. 24.1 bent wires and STU lying in the x- plane and each carrying a current I as shown. Find the magniO. Given OQ = OT = a

2M

0

,B=

For a toroid of radius

(

2

2 3/2

) where M = IA = I is the magnetic moment. If current I is anticlockwise B is directed from O to . For clockwise current B is form to O. 2

Fig. 24.10

(ix) revolving in a circular orbit of radius r with speed r with speed v (Fig. 24.11) e ev = en = Current along orbit I = T 2 r The direction of I is opposite to the direction of motion of the electron. O is I B= 0 4r ev and 2 r is directed into the page.

where I =

SOLUTION As point O is along the line segments and ST, the O due to and ST is zero. The O due to wires and TU respectively are I (k ) I (k ) and B2 = 0 B1 = 0 4 (OQ) 4 (OT ) both directed along the positive z-axis. The resultant t O is ( OQ = OT = a)

Fig. 24.11

e vr r2 = 2

Magnetic moment M = IA = I (x) solenoid In the middle region B = per unit length.

Fig. 24.12

0nI;n

At the ends of solenoid, B =

= no. of turns

0 nI

2

B = B1 + B2 = 2

0I

4 a

(k )

0I

2 a

k

24.2 I in the opposite direction are placed perpendicular to the x- plane. One wire is located at point (0, a, 0) and the other wire at Q (0, –a, 0). Find the magnitude and direction of at point A(x, 0, 0).

24.4 Comprehensive Physics—JEE Advanced

SOLUTION Refer to Fig. 24.13. Wire 1 carries a current I along the positive z-direction and wire 2 carries a current I along the negative z-direction. = OQ = a, OA = x,

= QA = r.

SOLUTION (a) (i) Radius r of wire A when it is bent into a circle is given by 2 r=L

L 2

r=

loop is B1 =

0I 2r

0I

(1)

L

O is (OT = a)

due to wire B

= =

Fig. 24.13

=

A due to wire 1 is

M

0I

B1 =

2 (

0I (sin 45° + sin 45°) 4 a 0I

2 2 a 4

0I

a

2 L

L 8

0I 2

)

2 a x2 According to Biot-Savart law, B1 is perpendicular to both and wire 1 and therefore in the x- plane. A due to wire 2 is 0I 2

B2 =

2 a x2 The -components of B1 and B2 cancel each other but the x-components add up. These components are B1 cos and B2 cos both along the positive x-direcA is B = B1 + B2 = (B1 cos + B2 cos ) i = (a

0I 2

a 2

x )

a cos

=

(a

0I 2

a 2

x )

2

x2 a r

Since centre O of the square is at the same distance from each side of the square and each arm carries the same current, the mag-

(i ) a a

2

Fig. 24.14

x

2

(i)

24.3 Two wires A and B have the same length L and carry equal currents I. Wire A is bent into a circle and wire B is bent into a square. (a) Obtain expression for the i) the centre of the circular loop and (ii) the centre of the square. Which

of the same and in the same direction. Hence O is B2 = 4B

=

16

0I

2 L

0I

8 2 L

(2)

(b) Dividing (2) by (1) we get B2 B1

8 2 2

8 1.41 (3.14) 2

= 1.16

Hence B2 > B1 centre due to the square loop will be greater than that due to the circular loop.

Magnetic Effect of Current and Magnetism 24.5

24.4 Figure 24.15 shows a wire loop ABCDEA carrying a current I as shown. Given AE = ED = a and AB = CD = a/2. Find the magnitude and direction of the magnetic F where BF = CF = a/2.

24.5 A wire ABCDE is bent as shown in Fig. 24.16. The wire carries a current I and the radius of the bent coil BCD is r. Find the magnitude and direction of the O. SOLUTION The straight line segments AB and DE are collinear with O AB and DE at O is zero. Angle subtended at O by arc BCD = 2 – . BCD at O is B =

0I 2 2r 2

Fig. 24.15

SOLUTION F is B = BAB + BBC + BCD + BDE + BEA Since point F lies in line with current elements AB and CD, BAB = BCD = 0 Also BDE = BEA =

0I

0I

(sin 0° + sin 45°) = 4 a 4 2 a directed out of the page and towards the reader. BBC =

0I

(sin 45° + sin 45°)

4 BC/2

directed into the page and away from the reader. Now BF 2

BC =

0I

BBC =

1

a

4

2

a 2

FC 2

2

a 2

2

a 2

0I

1

a

2

0I

0I

a

24.6 A wire ABCEF is bent as shown in Fig. 24.17 and caries a current I. The radius of the smaller arc ABC is r1 = r and that of the bigger arc is r2 = 2r. Find the magnitude of the Fig. 24.17

SOLUTION ABC at O is

directed out of the page.

2 2 a Since BBC > BDE + BEA the page and has a magnitude B =

The current through BCD is clockwise. Therefore, the O is into the page and away from the reader.

O.

2 2 directed into the page. Now BDE + BEA =

Fig. 24.16

0I

2 2 a

B is directed into 0I

a

1

B1 =

1 2 2

B2 =

0I 2 2r1 2 DEF at O is 0I

2r2 2

24.6 Comprehensive Physics—JEE Advanced

since B1 and B2 are both directed into the page, the O is B = B1 + B2 =

0I

2r1

B

2 r=

0I

2r

1

0 Ir 2

=

2

0I

2 2

2r2 2

B=

Putting r1 = r and r2 = 2r, we get B =

0i

2

2

r

NOTE

4

24.7 A long straight cylinder of radius carries a current I which is uniformly distributed across its cross-secr from the axis of the cylinder in cases (a) r > (b) r <

0I

(1) For r < ; B

1 . Figure 24.19 r

r and for r > ; B

shows the variation of B with r.

and

SOLUTION Figure 24.18 shows the cross-sectional view of the cylinder.

Fig. 24.19

walls. Therefore, in the case r < [Fig. 24.18 (b)], no current threads the Amperian loop. Hence B = 0 for points inside a hollow cylinder.

Fig. 24.18

Case (a) r > . For this case, the Amperian loop is a circle of radius r concentric with the cross-section [Fig. 24.18 (a)]. For this loop, L = 2 r and the current threading the loop is i = I. From Ampere’s circuital law. I B 2 r = 0I B= 0 BL = 0i 2 r Case (b) r < . For this case, L = 2 r [Fig. 24.18 (b)] and the current threading the loop is i = current per unit cross-sectional area of the cylinder cross-sectional area of the Amperian loop =

I 2

r2

Ir

2 2

From Ampere’s law,

24.8 The current density J (current per unit area) in a solid cylinder of radius varies with distance r from its axis as J = kr where k is a constant. Find the magnetic where (a) r > and (b) r < . SOLUTION Current I =

JdA

kr

(2 rdr )

Case (a) We take the Amperian loop of radius r > . Since the loop is outside the cylinder, the current through the loop is I=

kr

(2 rdr )

2 k r 2 dr

0

B

2 r =

3

0

0I

=

2

3

2

3

3

B

0

r 3 Case (b) For r < , the current through the Amperian loop is

Magnetic Effect of Current and Magnetism 24.7

r

I =

kr (2 rdr ) 0

B

24.4

2 r =

0I

=

2 kr 3 3 0

(vi) Frequency of revolution is v =

2 kr 3 3

B

0 kr

2

3

FORCE ON A MOVING CHARGE IN A MAGNETIC FIELD

The force on a charge q moving with a velocity v in a B is given by F = q(v B) The direction of F is perpendicular to both v and B. The magnitude F of vector F is given by F = qvB sin where is the angle between vectors v and B. (1) F = 0 if v = 0, i.e. a charge at rest does not experience any magnetic force. (2) F = 0 if = 0 or 180°, i.e. the magnetic force vanishes if v is either parallel or antiparallel to the direction of B. = 90°, i.e. if v is (3) F is maximum = Fmax if perpendicular to B, the magnetic force has a maximum value given by Fmax = qvB The direction of the force when v B is given by Fleming’s left hand rule. (4) If v is perpendicular to both E and B and E is E . perpendicular to B, then F = 0 if v = B

24.5

MOTION OF A CHARGED PARTICLE IN A MAGNETIC FIELD

Case (a): If v is perpendicular to B, the particle describes a circle in the region of the magn F v. (i) The speed along the circular path is constant. (ii) The kinetic energy is constant. (iii) Velocity and momentum continually change. mv 2 (iv) The radius r of the circular path is given by r qvB r=

mv qB

qB which is indepen2 m

dent of both v and r.

Case (b): If v is inclined to B at an angle , the particle mv sin moves in a helical path. The radius of helix is r = , qB time period T =

2 m and pitch of the helix = v cos qB

Applications (i)

-

region of magnetic

describes a semi-circle of radius. mv r= qB

Fig. 24.20

T m 2 qB (ii) If the particle enters the t=

as shown in Fig. 24.21, then

and

r=

mv qB

t=

2 m qB

Fig. 24.21

where is in radian. (iii) In Fig. 24.22, the particle wall not be able to hit the wall if d > r, i.e. d>

mv qB

B

mv qd

2mK qB

where m = mass of particle and K = kinetic energy. If the particle is accelerated through a potential difference V, then K = qV. (v) Time period of revolution is T =

2 m qB

T

Fig. 24.22

Fig. 24.23

24.8 Comprehensive Physics—JEE Advanced

(iv)

SOLUTION Refer to Fig. 24.24. d qBd r mv Linear defection x = r(1 – cos ) sin

=

24.9 through a potential difference of 18 kV and enters a certain initial velocity. What is the trajectory of the Fig. 24.24

initial velocity and (b) makes an angle of 30° with the 10–31 kg.

A is

SOLUTION V = 18 103 V 1 mv2 = eV 2 =

2

B= 2eV m

v=

(1.6 10

19

1/2

8 107 ms

31

9 10

1

(a) Since v is to B, = 90°, the trajectory of the electron is circular having a radius r =

mv eB

(9 10

31

)

(1.6 10 = 4.5

(8 107 ) 19

)

0.1

–3

10 m = 4.5 mm

(b) The trajectory of the electron is helical. The radius of heix is r=

mv eB

mv eB

sin

= 4.5 mm

2 r

4

10 2

7

10 0.05

4 10

5

T

directed inwards along the negative z-direction (a) = 90°. Therefore, force on proton is F = qvB sin

1/2

(18 103 )

)

0I

sin 30° = 2.25 mm

EXAM 24.10 A long straight wire lying along the -axis carries a current of 10 A along the positive -direction. A proton moving with a velocity of 107 ms–1 is at a distance 5 cm from the wire at a certain instant. Find the magnitude and direction of the force acting on the proton at that instant if its velocity is directed (a) along the negative x-direction (b) along the positive -direction and (c) along the positive z-direction

= (1.6 10–19) 107 (4 10–5) sin 90° = 6.4 10–17 N According to Fleming's L.H. rule, the direction of the force is parallel to the wire and opposite to the direction of current I, i.e. F is along the negative -direction (b) = 90°, F = qvB = 6.4 10–17 N. The force is directed towards the wire, i.e. along negative x-direction (c) = 180°. F = qvB sin 180º = 0 24.11 A proton and an -particle move perpendicular to a -particle is four times that of a proton and its charge is twice that of a proton. Find the ratio of radii of the circular path followed by them if both (a) have equal velocities, (b) have equal linear momenta, (c) have equal kinetic energies and (d) are accelerated through the same potential difference. SOLUTION Given

m =4 mp

and

q =2 qp

Magnetic Effect of Current and Magnetism 24.9

mv qB

(a) r =

mpv

rp

rp

mp

r

m

qp B q qp

m v q B

and r

1 4

1 2

2

mv qB

(b) r =

Kinetic energy K = r=

m qB

2K m

q qp

mp

rp r mv qB

(c) r =

rp r

m

1 2 mv 2

2K m

v

Fig. 24.27

1 2 mK qB 1 4

2

(iv) Force per unit length between two long straight 1 and I2 and

1

p qB q qp

f =

2 r attractive if I1 and I2 are in the same direction and repulsive if I1 and I2 are in opposite directions. Force on a segment of length l of either wire is F = f l. (v) 1 placed at a

2

(d) K = qV. Therefore 1 1 2mV r= 2mqV qB B q

24.6

rp

mp

r

m

q qp

1 4

0 I1 I 2

2

current I2 as shown in Fig. 24.28. II L F = 0 1 2 loge 1 vertically upwards. 2 r x

1 2

FORCE ON A CURRENT CARRYING CONDUCTOR IN A MAGNETIC FIELD

(i) Force on a straight conductor placed perpendicuF = BIL upwards if current I is from left to right and downwards if I is from right to left (given by Fleming’s left hand rule) Fig. 24.28

(vi)

1

2

Fig. 24.25

as shown in Fig. 24.29.

Fig. 24.26

(ii) (Fig. 24.26) F = BI(2 ) = 2 vertically upward for clockwise current and downward for anticlockwise current (iii) Net force F = F1 – F2 = 0

Fig. 24.29

24.10 Comprehensive Physics—JEE Advanced

Force F1 and F2 being equal and opposite cancel and F3 and F4 are given by expression above. Net force on coil is F = F3 – F4 =

0 I1 I 2 a

0 I1 I 2 a

2 x

2 (x

L)

0 I1 I 2 aL 2 x( x L) directed towards the wire (attractive)

F=

5

20 5 10

2

2 5 10 2 = 2 10–5 N (repulsive since I1 and I2 are in opposite directions) Force exerted by on CD is F2 =

24.12 The battery of a car is connected to the motor by 50 cm long wires which are 1.0 cm apart. If the current

10 7 )

= (4

=

0 I1 I 2

2 r2 (4

CD

10 7 ) 2

5

20 5 10

2 10

2

2

= 5 10–5 N (attractive since I1 and I2 are in the same direction)

Is the force attractive or repulsive. SOLUTION Force per unit length is f=

0 I1 I 2

10 7 )

(4

2 r

2

200

200 2

(1.0 10 )

0.8 Nm

1

F = f l = 0.8 0.5 = 0.4 N Since the currents in the wires are in opposite direction, the force is repulsive. 24.13 A small rectangular loop ABCD of sides 5 cm and 3 cm carries a current of 5 A. It is placed with its longer side parallel to a long straight conductor of length 5 m at a distance of 2 cm from it as shown in Fig. 24.30. If the current in force on the loop. Is the loop attracted towards or

to current in is directed outwards (towards the reader) and perpendicular to the plane of the coil. Therefore, forces F3 and F4 on BC and AD are equal and opposite and hence cancel each other. Therefore, the net force on coil ABCD is F = F2 – F1 = 5 10–5 – 2 10–5 = 3 10–5 N (attractive). Hence coil is attracted towards . 24.14 A particle of charge q and mass m moving in region I with a velocity v enters normally a region II of width d where a uniform B (directed inwards) exists as shown in Fig. 24.31. There is no and III. (a) What is the maximum speed (vmax) of the particle so that it returns back Fig. 24.31

(b) What will happen if v = 2 vmax SOLUTION

Fig. 24.30

SOLUTION Force exerted by F1 =

0 I1 I 2 2 r1

(a) Refer to Fig. 24.32(a). The particle describes a circular path of radius r = mv/qB in region II. It will return to region I if it describes a semicircle in region II. This happens if r
on AB is AB

vmax =

qBd mv < d or v < m qB qBd m

Magnetic Effect of Current and Magnetism 24.11

(a) Tension in the wires will be zero if Fm = Fg BIl = mg B=

mg Il

(60 10 3 ) 10

= 0.24 T 5 (50 10 2 ) (b) If current I is reversed, force Fm acts downwards. Hence Tension T = BIl + mg = 0.24 5 0.5 + 60 10–3 10 = 1.2 N Fig. 24.32

(b) Refer to Fig. 24.32 (b). If v > vmax, the particle is cross over to region III after describing a circular trajectory in rejoin II with O as the centre. In region III, te particle is move along the tangent at Q. The particle will suffer a deviation . In triangle sin

=

sin

=

If v =

OQ

d r

qBd mv

vmax v

2 vmax, then sin

24.7

TORQUE ON A CURRENT CARRYING COIL IN A MAGNETIC FIELD

The torque on a coil of N turns, area A carrying a current I B is given by =M B Magnitude of torque is = MB sin = NIAB sin where M = NIA is the magnetic moment and is the angle The magnitude of torque on a coil in radial magnetic =k where k is the restoring couple per unit twist and

=

1 2

= 45°

Then k or I NAB Current sensitivity of the galvanometer is NAB Cs = I k NIAB = k

24.15 A straight horizontal conducting rod of length 50 cm and mass 60 g is suspended by two vertical wires at its ends. A current of 5 A set up in the rod. to the conductor in order that the tension in (b) What will be the tension in the wires if the direction of the current is reversed, keeping Ignore the mass of the wires and take g = 10 m/s–2. SOLUTION Refer to Fig. 24.33.

Fig. 24.33

is the = 90°.

24.8

I=

TORQUE ON A BAR MAGNET IN A MAGNETIC FIELD

The magnetic dipole moment of a bar magnet of pole strength q and length (2a M = q(2a) It is a vector pointing from the south to the north pole of a magnet. Force on north pole N of magnet = qB (in the direction of B) Force on south pole S of the magnet = –qB (opposite to B) opposite forces on the magnet. The two forces, therefore, constitute a coupe which tends to rotate the magnet in the clockwise direction. The arm of the couple is 2 a. The torque is given by = arm of the coupe force = 2a qB = q(2a) B or =M B

24.12 Comprehensive Physics—JEE Advanced

where M = q(2a) is called the magnetic moment of the bar magnet. The direction of is perpendicular to both M and B. If M and B are both in the plane of the paper then the torque will be perpendicular to the plane of the paper and directed into it away from the reader. The magnitude of the torque is = MB sin where is the angle between M and B. The SI unit of M is Nm T–1 or JT–1 (joule per tesla).

24.9

POTENTIAL ENERGY OF A MAGNETIC DIPOLE

Fig. 24.34

The magnetic potential energy of a magnetic dipole in any orientation B is the dipole from its zero energy position ( = 90°) to the given position. . U = –MB cos In vector notation, U = – (M·B) For stable equilibrium U is minimum. Hence = 0 and = 0. For unstable equilibrium, U is maximum i.e. = 180°. Hence = 0

24.10

SOME USEFUL TIPS

1. Magnetic dipole moment of a bar magnet is M = m l, where m is pole strength and l is the length of the magnet. The value of M depends on the volume of the magnet. (a) If a magnet is cut into two equal parts by cutting it by a plane along its length, its volume is halved, Hence the magnetic dipole moment of a piece is halved = M/2. The pole strength M is also halved as length l remains m = l the same. (b) If a magnet is cut into two equal parts by cutting it by a plane transverse to its length, the volume and length are both halved. Hence the magnetic moment becomes M/2 but pole strength m remains the same. (c) If a wire of magnetic dipole moment M and length l is bent as shown in Fig. 24.34, the 1 and distance between the pole becomes 2 magnetic moment becomes l M M =m 2 2 If the wire is bent as shown in Fig. 24.35, the magnetic moment becomes

M =m

Fig. 24.35

l 2

M 2

NOTE Magnetic dipole moment M is a vector quantity directed from south pole to north pole.

experiences no net force but experiences a torque = M B. The magnitude of is = MB sin where is the angle between M and B. (a) when = 90°, max = MB (b) when = 0°, min = 0 (stable equilibrium) (c) when = 180°, min = 0 (unstable equilibrium) Potential energy is U = –M.B = –MB cos . When = 0°, P.E is minimum Umin = –MB. U is max = Umax = MB when = 180° (d) Work done in rotating the magnet from 1 to 2 is W = MB (cos 1 – cos 2) experiences a force as well as a torque. 3. The time period of a bar magnet oscillating in a I , where I T=2 MB ml 2 , 12 m = mass of magnet and l = length of magnet. (a) If a bar magnet is cut into two equal parts by cutting along its length, then each part has M = M/2 and I = I/2. Hence T = T. (b) If a bar magnet is cut into two equal parts by cutting perpendicular to its length, then each part has M = M/2 and I = I/8. Hence T = T/2. (c) If two bar magnets of magnetic moments M1 and M2 are placed one on top of the other as shown n Fig. 24.36, then time period is given by (since I = I1 + I2 and M = M1 + M2)

is the moment of inertia of the bar magnet =

Magnetic Effect of Current and Magnetism 24.13

24.16

( I1 I 2 ) ( M1 M 2 ) B

T1 = 2

A closely wound solenoid of 1000 turns and area of cross-section 5 cm2 carries a current of 3 A. It is suspended through its centre (a) what is the magnetic 10–2 T is set up at an angle of 30° with the axis of the

Fig. 24.36

If the magnets are placed as shown in Fig. 24.37, then I = I1 + I2 but M = M1 – M2 and ( I1 I 2 ) T2 = 2 ( M1 M 2 ) B M T1 and T2 are related as 1 M2

(T22

T12 )

(T22

T12 )

SOLUTION M = e r2 = 3.14 (1.6 10–10)2 = 1.26 10–23 Am2

e to a bar magnet (a) At a point on axial line (Fig. 24.38) 2Mr

0

(r 2

4

l 2 )2

parallel to M = m

For a very short magnet (l << r), Ba =

acting on the solenoid is zero Torque = MB sin = 1.5 (8 10–2) = 6 10–2 J

3

(5

sin 30°

24.17 In a hydrogen atom, the electron moves in a circular orbit of radius 0.5 Å making 1016 revolutions per second. Calculate the magnetic moment associated with the orbital motion of electron.

Fig. 24.37

Ba =

SOLUTION (a) Magnetic moment M = NIA = 1000 10–4) = 1.5 JT–1 or Am2

0

4

2l.

10–19)

1016

(0.5

24.18 A bar magnet is suspended at a place where it is acted

2M r3

a magnitude

2

10–2 T. The magnet attains stable

SOLUTION B1 exerts anticlockwise torque 1 to orient M B2 exerts a stockwise torque 2 to orient M along itself (Fig. 24.40). For equilibrium,

Fig. 24.38

(b) At a point Q on the equatorial line (Fig. 24.39) Be =

M

0

4

(r

2

l 2 )3/2

antiparallel to M For l << r, Be =

0M 3

4 r

Fig. 24.39

Fig. 24.40

24.14 Comprehensive Physics—JEE Advanced

1

MB1 sin B2 =

=

2

1 = MB2 sin

2

2 10 2 sin 30 =10–2 T sin 45

B1 sin 1 sin 2

24.19 A uniform wire is bent into the shape of an equilateral triangle of side a. It is suspended from a vertex at B exists parallel to its plane. Find the magnitude of the torque acting on the coil when a current I is passed through it.

24.20 A wire loop ABCD carrying a current I2 is placed on a frictionless horizontal table as shown in Fig. 24.42. A long straight wire carrying a current I1 is placed at a distance a from side AB = l. Find the work done by Q

to position

SOLUTION Area of the coil is (AB = a, BD = a/2)

Fig. 24.42

SOLUTION

Fig. 24.41

A= 2

area of triangle ABD 1 2

=2

1 2

=2

Fig. 24.43

AD

BD

3 a 2

a 2

3 2 a 4 Magnetic moment of the loop is =

3 2 a 4 Since the current is clockwise the direction of vector M is perpendicular to the plane of the coil directed inwards as shown in Fig. 24.41. Hence = 90°. The magnitude of the torque acting on the coil is M = IA = I

= MB sin

=

3 2 3 2 Ia B sin 90° = Ia B 4 4

Divide the loop into a large number of elements each of a very small width dx. Consider one such element at a distance x from [see Fig. 24.43].

B=

0 I1

2 x

directed inwards

Magnetic moment of the element is dM = I2 area of element = I2

ldx

Since the current in the coil is anticlockwise, dM is directed outwards. Hence angle between vectors B and dM is = 180°. Potential energy when the wire is a distance x from the elements is dU = –dM·B = –dM B cos

Magnetic Effect of Current and Magnetism 24.15

0 I1

= –I2 ldx

2 x

By symmetry, the potential energy of the system when the wire is shifted to position P Q is UP Q = –UPQ

cos 180°

0 I1 I 2 l

dx 2 x Potential energy of the system when the wire is at position PQ is =

UPQ =

0 I1 I 2 l

2

b

0 I1 I 2 l

dx x a

2

ln

Work done in shifting the wire from position PQ to P Q is W = –(UP Q – UPQ)

b a

= 2UPQ =

0 I1 I 2 l

b a

ln

I Multiple Choice Questions with Only One Choice Correct 1. The dimensions of

B 2 RC

(where B is magnetic

0

R is resistance, C is capacitance and 0 is permeability of free space) are the same as those of (a) impulse (b) angular momentum (c) energy (d) viscosity 2. A current carrying metal wire of diameter 2 mm 2 10–3 T. The current in the wire is (a) 10 A (b) 20 A (c) 40 A (d) 40 2 A 3. with velocity vector making an angle of 30° with

Fig. 24.44

5. A straight wire PQ of length L carries at current I. R is [see Fig. 24.45] (a) zero (b)

0I

2 2 L

trajectory of pitch x. The radius of the helix is x (a) 2 (c)

(b) x

(d)

x

(c)

2 2 3x 2

2 3 4. A current carrying wire AB is placed near another CD as shown in Fig. 24.44. Wire CD wire AB is free to move. When a current is passed through wire AB, it will have (a) only translational motion (b) only rotational motion (c) both translational as well as rotational motions (d) neither translational nor rotational motion.

(d)

0I

4 2 L 0I

2 L Fig. 24.45

6. A metal rod of length L and mass M is slipping down a smooth inclined plane of inclination 30° with a constant velocity v. The rod carries a current I directed inwards perpendicular to the plane upwards perpendicular to the length of the rod ex-

(a)

Mg IL

(b)

Mg 2 IL

24.16 Comprehensive Physics—JEE Advanced

(c)

3 Mg 2 IL

(d)

Mg 3 IL

7. Equal current I ACB and ADB of equal radius r as shown in Fig. 24.46. If O is (a) (b) (c) (d)

0I

r

(c)

R2 R1

(c)

R1 R2

1/ 2

(b)

R2 R1

(d)

R1 R2

2

IIT, 1988 11. Two identical coils carry equal currents and have a common centre, but their planes are at right angles to each other. What is the magnitude of the resultant

0I

coil alone is B? (a) zero

2r 0I

Fig. 24.46

0I

0

I

4 0

I

2

1 R1

1 R2

(b)

1 R1

1 R2

(d)

0

I

1 R1

1 R2

I

1 R1

1 R2

4 0

2

(b) B/ 2

(c) 2 B (d) 2B 12. When a charged particle moves perpendicular to a

3r

4r 8. The wire loop PQRSP formed by joining two semi– circular wires to radii R1 and R2 carries a current I as shown in Fig. 24.47. The magnitude of the magnetic induction at centre C is (a)

(a)

(a) energy and momentum both change (b) energy changes but momentum remains unchanged (c) momentum changes but energy remains unchanged (d) energy and momentum both do not change. 13. A proton with kinetic energy K describes a circle of radius r –particle with kinetic energy K moving in the same magnetic r 2 (c) 2 r (a)

IIT, 1983

14. An

(b) r (d) 4 r

–particle moving with a velocity v in a

at frequency called the cyclotron frequency. The cyclotron frequency of a proton moving with a speed 2v (a)

Fig. 24.47

9.

(c) diagonals of a square loop of side L carrying a current I is I 2 0I (b) (a) 0 L L (c)

2

0

L

I

(d)

2 2

0

I

L

10. Two particles X and Y having equal charges, after being accelerated through the same potential and describe circular paths of radii R1 and R2 respectively. The ratio of mass of X to the mass of Y is

15.

4

(b)

2 (d) 2

ated through a potential difference of 2.88 kV and

of an electron = 9 10–31 kg. The trajectory of the electron is a (a) circle of radius 1.6 mm (b) circle of radius 0.9 mm (c) helix of radius 0.9 mm (d) helix of radius 1.6 mm 16. In the region around a charge at rest, there is

Magnetic Effect of Current and Magnetism 24.17

l2 I l2 I (d) 2 4 23. Two circular current carrying coils of radii 3 cm and 6 cm are each equivalent to a magnetic dipole having equal magnetic moments. The currents through the coils are in the ratio of (c)

17. In the region around a moving charge, there is

18. An electron is accelerated to a high speed down the axis of a cathode ray tube by the application of a potential difference of V volts between the cathode and the anode. The particle then passes through experiences a force F. If the potential difference between the anode and the cathode is increased to 2 V, the electron will now experience a force (a) F/ 2

(b) F/2

(c) 2 F (d) 2 F 19. A magnetic needle is kept in a non–uniform (a) a force as well as a torque (b) a force but no torque (c) a torque but no force (d) neither a force nor a torque. 20. A conducting circular loop of radius r carries a constant i B such that B is perpendicular to the plane of the loop. The magnetic force acting on the loop is (a) i rB (b) 2 i r B (c) zero (d) irB IIT, 1983 21. A rectangular loop carrying a current i is situated near a long straight wire such that the wire is parallel to one of the sides of the loop. If a steady current I is established in the wire, as shown in Fig. 24.48, the loop will (a) rotate about an axis parallel to the wire (b) move away from the wire Fig. 24.48 (c) move towards the wire (d) remain stationary. 22. A wire of length l metres carrying a current I amperes is bent in the form of a circle. The magnitude of the magnetic moment is (a)

l I2 2

(b)

l I2 4

(a)

2 :1

(b) 2 : 1

(c) 2 2 : 1 (d) 4 : 1 24. An electron of charge e moves in a circular orbit of radius r due to orbital motion of the electron at the site of the nucleus is B. The angular velocity of the electron is 2 0 eB eB (b) = 0 (a) = 4 r r 4 rB 2 rB (c) = (d) = e 0 0e 25. Three long, straight and parallel wires C, D and G carrying currents are arranged as shown in Fig. 24.49. The force experienced by a 25 cm length of wire C is (a) 0.4 N (b) 0.04 N (d) 4 10–4 N (c) 4 10–3 N

Fig. 24.49

26. A potential difference of 600 V is applied across the plates of a parallel plate capacitor. The separation between the plates is 3 mm. An electron projected vertically, parallel to the plates, with a velocity of 2 106 ms–1 between the capacitor plates? (a) 0.1 T (b) 0.2 T (c) 0.3 T (d) 0.4 T 27. A current I long, straight, thin walled pipe. Then is the same but not zero.

24.18 Comprehensive Physics—JEE Advanced

+ y direction enters a region of uniform electric is zero. the pipe. points inside the pipe. IIT, 1993 28. Two straight and long conductors AOB and COD are perpendicular to each other and carry currents of I1 and I2 a point P at a distance a from point O in a direction perpendicular to the plane ABCD is (a)

(c)

0

2 a 0

2 a

(I1 + I2) (I

2 1

+

I22)1/2

(b)

(d)

0

2 a 0

2 a

(I1 – I2) I1 I 2 I1 I 2

29. If a magnetic dipole of dipole moment M is rotated through an angle with respect to the direction of H, then the work done is (a) MH sin (b) MH (1 – sin ) (c) MH cos (d) MH (1 – cos ) 30. A 2 MeV proton is moving perpendicular to a proton is (b) 8 10–11 N (a) 2.5 10–10 N –11 (d) 8 10–12 N (c) 2.5 10 N 31. A straight section PQ of a circuit lies along the a a to x = and carries a current x-axis from x = – 2 2 I PQ at point x = + a will be (a) proportional to a (b) proportional to a2 1 (c) proportional to (d) equal to zero. a 32. Two charged particles M and N enter a region of Fig. 24.50. The possible reason is (a) The charge of M is greater than that of N (b) The momentum of M is greater than that of N (c) The charge to mass ratio of M is greater than that of N (d) The speed of M is greater than that of N. 33. The monoenergetic beam of electrons moving along Fig. 24.50

B and E are directed respectively along (a) – y axis and – z axis (b) + z axis and + x axis (c) + x axis and + z axis (d) – x axis and – y axis 34. A proton of mass 1.67 10–27 kg and charge 1.6 10–19 C is projected with a speed of 2 106 ms–1 at an angle of 60° to the x-axis. If a uniform y-axis, the path of the proton is (a) a circle of radius 0.1 m 2 10–7 s (b) a circle of radius 0.2 m 10–7 s (c) a helix of radius 0.1 m 2 10–7 s (d) a helix of radius 0.2 m 4 10–7 s

and time period and time period and time period and time period

IIT, 1995 35. A proton, a deuteron and an alpha particle having the same kinetic energy are moving in circular trarp, rd and r denote respectively the radii of trajectories of these particles, then (b) r > rd > rp (a) r = rp < rd (c) r = rd > rp (d) rp = rd = r IIT, 1997 36. Two particles each of mass m and charge q, are attached to the two ends of a light rigid rod of length 2l. The rod is rotated at a constant angular speed about a perpendicular axis passing through its centre. The ratio of the magnitudes of the magnetic moment of the system and its angular momentum about the centre of the rod is q q (b) (a) 2m m 2q q (c) (d) m m IIT, 1998 37. Two very long straight parallel wires carry steady currents I and – I. The distance between the wires is d. At a certain instant of time, a point charge q is at a point equidistant from the two wires, in the plane of the wires. Its instantaneous velocity v is perpendicular to this plane. The magnitude of the at this instant is (a)

I qv 2 d 0

(b)

0

I qv d

Magnetic Effect of Current and Magnetism 24.19

(c)

2

0

I qv d

42. An electron revolves in a circle of radius 0.4 Å with a speed of 106 m/s in Hydrogen atom. The magnetic

(d) zero

IIT, 1998 38. A charged particle is released from rest in a region which are parallel to each other. The particle will move in a (a) straight line (b) circle (c) helix (d) cycloid IIT, 1999 39. An ionized gas contains both positive and negative ions. If it is subjected simultaneously to an electric x along the + z direction, then y direction and negative ions towards – y direction y direction y direction y direction and negative ions towards + y direction IIT, 2000 40. Two long parallel wires are at a distance 2d apart. plane of the paper, as shown in Fig. 24.51. The B along the line XX is given by IIT, 2000

10–7 motion of the electron in tesla is [ 0 = 4 –19 H/m; Charge on the electron = 1.6 10 C] (a) 0.1 (b) 1.0 (c) 10 (d) 100 43. A proton of velocity (3 i 2 j) ms–1

of

magnetic induction (2 j 3k) tesla. The acceleration produced in the proton is (charge to mass ratio of proton = 0.96 108 C kg–1) (a) 2.8 (b) 2.88 (c) 2.8 (d) 2.88

108 (2 i 3 j) 108 (2 i 3 j 2k) 108 (2 i 3k) 108 ( i 3 j 2k)

44. A long wire carries a steady current. It is bent into at the centre of the loop is B. The wire is then bent into a circular coil of n turns and the same current of coil will be (a) nB (c) 2nB

(b) n2 B (d) 2n2 B

45. circular loop of radius 3 cm at a point on its axis at a distance of 4 cm from the centre is 54 T. The magT) at the centre of the loop will be (a) 250 (b) 150 (c) 125 (d) 72 46. A wire ABCDEF (with each side of length L) bent as shown in Fig. 24.52 and carrying a current I is B parallel to the positive y–direction. What is the magnitude and direction of the force experienced by the wire?

Fig. 24.51

41. An electron moves with a speed of 2 along the positive x

105 ms–1 B

= (i 4 j 3k ) tesla. The magnitude of the force (in newton) experienced by the electron is (the charge on electron = 1.6 10–19 C) (a) 1.18 10–13 (b) 1.28 10–13 (c) 1.6 10–13 (d) 1.72 10–13

Fig. 24.52

(a) BIL along positive z-direction

24.20 Comprehensive Physics—JEE Advanced

(b) BI 2/L along positive z-direction (c) BIL along negative z-direction (d) BL/I along negative z-direction 47. long bent wires are placed in the x-y plane as shown in Fig. 24.53. The wires carry a current I = 10 A each as shown. The segments RL and SM are along the x-axis. The segments PR and QS are along the y-axis, such that OS = OR = 0.02 m. What is the magnitude and direction of the magnetic induction at the origin O? (a) 100 Wb m–2 vertically upward (b) 10–4 Wb m–2 vertically downward (c) 10–4 Wb m–2 vertically upward (d) 10–2 Wb m–2 vertically downward IIT, 1989

(a) 1 MeV (b) 2 MeV (c) 4 MeV (d) 0.5 MeV 51. A loosely wound helix made of stiff wire is mounted vertically with the lower end just touching a dish of mercury. When a current from the battery is started in the coil through the mercury (a) the wire oscillates (b) the wire continues making contact (c) the wire breaks contact just when the current is passed (d) the mercury will expand by heating due to passage of current. IIT, 1981 52. A non-planar loop of conducting wire carrying a current is placed as shown in Fig. 24.54. Each of the straight sections of the loop is of length 2a. The P(a, 0, a) points in the direction (a) (b) (c)

Fig. 24.53

(d)

48. A particle of charge q and mass m moves in a circular orbit of radius r with angular speed . The ratio of the magnitude of its magnetic moment to that of its angular momentum depends on (a) and q (b) , q and m (c) q and m (d) and m IIT, 2000 49. Two long parallel wires P and Q are held perpendicular to the plane of the paper at a separation of 5 m. If P and Q carry currents of 2.5 A and 5 A respectively in the same direction, then the magnetic P and Q is (a)

0

(b)

3

0

3 0 (c) (d) 2 2 50. A proton of mass m and charge +e is moving in 0

MeV. What should be the energy of an -particle (mass 4 m and charge + 2e) so that it revolves in a circular orbit of the same radius in the same mag-

1 2 1 3 1 3

i

1 2

j

k

j

k j

i

i k

k

Fig. 24.54

IIT, 2001 53. Two particles A and B of masses mA and mB respectively and having the same charge are moving in a plane. A uniform magnetic this plane. The speeds of the particles are vA and vB respectively and the trajectories are as shown in Fig. 24.55. Then (a) vA < mB vB (b) mA vA > mB vB (c) mA < mB and vA < vB (d) mA = mB and vA = vB

Fig. 23.55

IIT, 2003 54. A coil having N turns is wound tightly in the form of a spiral with inner and outer radii a and b respectively. When a current I passes through the coil, the

(a)

0 NI

b

(b)

2

0 NI

a

Magnetic Effect of Current and Magnetism 24.21

(c)

0 NI

2 b

a

ln

b a

(d)

0 NI

2 b

a

ln

a b

(b) E = a i ; B = c k (c) E = 0; B = c j

IIT, 2001 55. A particle of mass m and charge q moves with a constant velocity v along the positive x direction. It B directed along the negative z direction, extending from x = a to x = b. The minimum value of v required so that the particle can just enter the region x > b is (a) qbB/m (b) q(b – a)B/m (c) qaB/m (d) q(b + a)B/2m IIT, 2002 56. Which pattern shown in Fig. 24.56 correctly repre-

(d) E = a i ; B = c k

bi bk bj

IIT, 2003 58. A magnetized wire of magnetic moment M is bent into an arc of a circle that subtends an angle of 60° at the centre. The equivalent magnetic moment is (a) (c)

M

(b)

3M

(d)

2M 4M

59. are held 0.1 m apart and carry a current of 5 A each in the same direction. The magnitude of the magis (a) 10–5 T

(b)

2

10–5 T

10–5 T (d) 2 10–5 T (c) 3 60. A proton moving with a speed u along the positive x-axis enters at y = 0 a region of uniform magnetic B = B0 k which exists to the right of y-axis as shown in Fig. 24.58. The proton leaves the region after some time with a speed v at co-ordinate y. Then (a) v > u, y < 0 (b) v = u, y > 0 (c) v > u, y > 0 (d) v = u, y < 0 Fig. 24.56

57. An electron is moving in the x-y plane along the positive x-axis. There is a sudden change in its path due to the presence of electric and/or magP as shown in Fig. 24.57. The curved path lies in the x-y plane and is found to be non-circular. Which of the following combinations is possible?

Fig. 24.58

IIT, 2004 61. A particle of charge q moves with a velocity v = ai B = b j + c k where a, b and c are constants. The magnitude of the force experienced by the particle is (a) zero (b) qa(b + c) 2 2 1/2 (c) qa(b – c ) (d) qa(b2 + c2)1/2 62. A current I lateral triangle of side a. The magnitude of the mag-

Fig. 24.57

(a) E = 0; B = b i

ck

(a)

3 0I 2 a

(b)

9 0I 2 a

24.22 Comprehensive Physics—JEE Advanced

(c)

3 3 0I 2 a

(d) zero

63. A particle of mass m and charge q, accelerated by a potential difference V enters a region of a uniform B. If d is the thickness of the region of B, the angle through which the particle deviates from the initial direction on leaving the region is given by 1/2

(a) sin

q = Bd 2mV = Bd

q 2mV

1/2

(b) cos

q 2mV

1/2

(c) tan

= Bd

67.

q 1/2 2mV 64. A metal wire of mass m slides without friction on two rails spaced at a distance d apart. The track B. A constant current I and back down the other rail. If the wire is initially at rest, the time taken by it to move through a distance x along the track is (d) cot

Fig. 24.59

(a)

(b)

(c)

(d)

= Bd

(a) t =

BId 2 xm

(b) t =

2xm BId

(c) t =

BIdm 2x

(d) t =

2dm BIx

65. A particle of charge q and mass m is released from

Fig. 24.60

IIT, 2011 68. A long insulated copper wire is closely wound as a spiral of ‘N’ turns. The spiral has inner radius ‘a’ and outer radius ‘b’. The spiral lies in the XY plane and a steady current ‘I Z the spiral is [see Fig. 24.61] (a)

the origin with a velocity v = a i in a uniform magB = b k . The particle will cross the y-axis at a point whose y-coordinate is ma 2ma (b) y = (a) y = qb qb (c) y = –

ma qb

(d) y = –

(c)

0 NI

2 b a 0 NI

2b

ln

ln

b a

b a

(d)

0 NI

2 b a 0 NI

2b

ln

ln

b a b a

b a b a IIT, 2011

2ma qb

66. A thin wire loop carrying a current I is placed in a B pointing out of the plane of the coil as shown in Fig. 24.59. The loop will tend to (a) move towards positive x-direction (b) move towards negative y-direction (c) contract (d) expand

(b)

Fig. 24.61

Magnetic Effect of Current and Magnetism 24.23

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67.

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68.

(d) (c) (b) (a) (d) (d) (d) (b) (c) (b) (d) (c)

(a) (a) (d) (c) (a) (c) (a) (b) (a) (d) (b) (a)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63.

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64.

(c) (d) (c) (c) (b) (c) (c) (a) (a) (b) (a)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65.

(c) (c) (a) (d) (c) (c) (b) (a) (d) (c) (b)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66.

(c) (c) (d) (d) (d) (a) (c) (c) (b) (c) (d)

(d) (c) (c) (c) (d) (a) (c) (c) (c) (d) (c)

SOLUTIONS 1.

B2 = magnetic energy density. Therefore 2 0 B2

energy [ML2 T 2 ] = = [ML–1 T–2] 3 volume [L ] 0 RC = time constant = [T] B 2 RC

=

= [ML–1 T–2] [T] = [ML–1 T–1], which

0

Fig. 24.62

are the dimensions of viscosity. 2.

In Fig. (b), r = L,

of the wire and is given by I B= 0 (1) 2 R R = 1 mm = 1 10–3 m, 0 = 4 10–7 Hm–1 and –3 B = 2 10 T. Using these values in Eq. (1), we get I = 10 A. 2 mv cos 30 2 mv 3 = (1) 3. Pitch x = qB qB 2 Radius r =

mv sin 60° mv = qB 2qB

x From Eqs. (1) and (2), we get r = 2 3 choice (c).

= 0° and

6. Figure 24.63 shows the cross-sectional view of the rod. From Fleming left-hand rule, the magnetic force F acting on the rod is directed to the right and is given by F = B I L. Since the rod moves with a constant velocity, no net force acts on it. Hence the components of Mg and F parallel to the inclined plane must balance, i.e.

(2) , which is

4.

CD is nonuniform, wire AB will experience as force as well as a torque. Hence it will have both translational as well as rotational motions. 5. Refer to the Fig. 24.62. R in Fig. (a) is given by I B = 0 (sin + sin ) (1) 4 r

= 45°. Using these

Fig. 24.63

or or

F cos

= Mg sin

B I L cos

= Mg sin B=

Mg tan IL

24.24 Comprehensive Physics—JEE Advanced

Mg

Putting = 30°, we get B =

3 IL

, which is choice

=

I (cos 45° – cos 90°) 4 r

=

I 0I (cos 45° – 0) = 4 r 4 2 r

(d). 7. centre is B=

0

0

0I

2r

45

ADB at centre O is B1 =

90

0I

2r

2

L

directed out of the page.

E

r

O

ACB at centre O is B2 =

0I 2r

2

directed into the page

2 O is

Fig. 24.64

B = B2 – B1 = Putting

0I

( – ) directed into the page.

2 r

= 60° =

3

, we get B =

O due to current element DE is the same as that due to AE. Hence, O due to one side AD is

0I . 3r

8. The magnitude of magnetic induction due to a circular loop of radius R carrying a current I is I given by B = 0 2R

BAD =

2

0

I

4 2 r

2 0I 4 r

=

Since the centre of the square is equidistant from the ends A, B, C and D of each side of the square and each side produces at the centre O the same that due to one side. Hence (because r = L/2)

B=

I 4R

B = 4BAD =

0

The direction of B is normal to the plane of the loop. Since the current in the bigger loop is clockwise and that in the smaller loop is anticlockwise centre C are in opposite directions. Therefore, the magnetic induction at the centre is given by 0

B = B1 – B2 =

I

1 R1

4

1 R2

The wires PQ and S R the centre C. Hence the correct choice is (a). L 9. Refer to Fig. 24.64. Here r = OE = . Referring to 2 O due to the current element AE is given by BAE = –

I 4 r

45

0

sin 90

d

=

I 45 | cos |90 4 r 0

2

0

r

I

=

2 2

0

L

I

.

Hence the correct choice is (d). 10. Let the masses of X and Y be m1 and m2 and let their velocities after being accelerated be v 1 and v2 respectively. Since the particles have equal charges and have been accelerated through the same potential difference, their kinetic energies are equal, i.e. 1 1 m1v 21 = m2v 22 2 2 B, the radii of the circular paths are given by m1v12 mv = q B v 1 or q B = 1 1 R1 R1 and

m2 v22 mv = q B v 2 or q B = 2 2 R2 R2

Therefore

m1v1 mv m2v2 = 2 2 or 1 2 1 R1 R2 R1

m22 v22 R22

Magnetic Effect of Current and Magnetism 24.25

But

m1v 21 = m2v 22. Therefore, we have m1 = m2

R1 R2

2

helix whose radius is r=

Hence the correct choice is (c). 11.

tre due to each coil is B. Since the planes of the coils are at right angles to each other, the directions Br =

B

2

B

2

= 2 B Hence the correct choice is (c). pendicular to the direction of motion of the particle, the speed of the particle cannot change but its velocity changes. Hence the correct choice is (c). 13. The radius of the circular orbit is given by 2m K r= qB The charge of an –particle is twice that of a proton and its mass is four times the mass of a proton. Therefore m /q is the same for both. Hence r will the same for both particles. Thus the correct choice is (b). 14. The cyclotron frequency is given by qB = 2 m It is independent of the speed of the particle and the radius of its circular path. Now q /m. The charge of a proton is half that of an -particle and the mass of a proton is one-fourth. Therefore, will be doubled. Hence the correct choice is (d). 15. V = 2.88 10 3 V. The velocity of the electron is given by 1 m v2 = e V 2 v=

=

=

2eV m

3.2 107

1.6 10 =9

10

–4

9 10 2 1.6 10

19

9 10

sin 30

0.1

Hence the correct choice is (c). 16. The correct choice is (a). 17. The correct choice is (d). 18. The velocity when the potential difference is V is 2 eV m and force F = e v B When the potential difference is doubled, i.e. V = 2V, the velocity is v =

2 eV m

=

2e

2V m

2.88 103 2.88 103

I 1/2

31

10 7 ms–1

with the velocity v, then v = v sin and v11 = v cos . The electron has two motions: a linear motion parallel to magnetic

2 v

21. Referring to Fig. 24.65, the forces acting on arms BC and AD are equal and opposite. The force on arm AB is given by Ii F1 = 0 2 a which is directed towards the wire. The force on arm CD is given by 0I i F2 = 2 a b which is directed away from the wire. Since F1 > F2, the loop will move towards the wire. Hence the correct choice is (c).

1/2

31

=

Force F = e v B = 2 evB = 2 F. Hence the correct choice is (c). 19. The correct choice is (a). 20. produced by the current in the loop and the external

i 19

19

m = 0.9 mm

1/2

2 1.6 10

= 3.2

31

9 10

v=

12.

or

m v sin eB

Fig. 24.65

24.26 Comprehensive Physics—JEE Advanced

22. Magnetic moment m = AI = r 2I, where r is the radius of the circular loop. Now, the circumference of the circle = length of the wire, i.e. l2

2

2 r = l or r =

l2 I

r 2I =

Therefore, m =

2

4

4

l2 I , 4

=

2

26. Electric intensity E =

V d

where V is the potential difference between the plates and d, the separation between them. d = 3 mm = 3 10 –3 m E=

V 600 = d 3 10

=2

3

10 5 V m–1

which is choice (d). 23. Magnetic moment m = IA = I r 2. m = I1 r 21 = I2 r 22 I1 r2 = 22 = I2 r1

6 3

2

B e v = e E or B =

=4

Hence the correct choice is (d). 24. An electron moving in a circular orbit is equivalent to a current carrying loop. As explained above, the current is e I= e = T where T is the time period of the motion of the electron around the nucleus. If v is the speed of the electron, 2 r T= v I=

B=

ev 2 r

e 2

v =r )

(

I e = 0 2r 4 r 0

or

=

4 rB 0e

Hence the correct choice is (c). 25.

2I 10 7 2 30 = =2 r 0.03

0

4

10–4 T

which is directed into the page. G at C is 10

B = BD – BG = 2 = 1.6

10 –4 T

and is directed into the page. The force on 25 cm of wire C is F = B I l sin 90° = 1.6 10–4 10 =4

10 –4 T

10 –4 – 0.4

10 –4 N

0.25

B dl =

0I

Since no current exists in the medium (air) inside the pipe I = 0. Hence B = 0. Hence the correct choice is (b). 28. P due to current I1 in conductor AOB is I B1 = 0 1 2 a P due to current I2 in conductor COD is I B2 = 0 2 2 a Since the two conductors are perpendicular to each B1 and B2 will be perpendicular to each P is

C is 10–4

(I21 + I22)1/2

0

2 a

Hence the correct choice is (c). 29. The work done is given by M H sin

W=

7

2 20 = 0.4 0.1 which is directed out of the page. BG =

27. From Ampere’s law, we have

B = (B21 + B22)1/2 =

D at wire C is BD =

E 2 105 = = 0.1 T v 2 106

d

= MH |– cos | 0

0

= MH (1 – cos ) Hence the correct choice is (d). 30. The kinetic energy of proton is K = 2 MeV = 2 106 eV =2

106

1.6

10–19 J = 3.2

1 mv 2 = 3.2 10–13 2 Now, mass of proton is m = 1.67 fore,

10–13 J

10–27 kg. There-

Magnetic Effect of Current and Magnetism 24.27

v2 =

2 3.2 10 1.67 10

v = 1.96

or

13

27

= 3.83

m

1014

It is given that the charge q is moving in a circular path of radius 2l. Therefore, the time period = 2 (2l)/v. Hence qv (2l)2 = qvl m = 2 (2l )

107 ms–1.

Now force on proton is F = evB = 1.6 10–19 1.96 = 7.84 10–12 N

107

mv 2 = evB or r = r

The angular momentum L = mv(2l). Therefore, qvl q m = L mv(2 l ) 2m

2.5

Hence the closest choice is (d). 31. The point x = + a lies along the line of the straight section PQ at point x = a is zero. 32. The radius of the circular path of a particle of mass m, charge e moving with a speed v perpendicular to B is given by m v e B

e , the m charge to mass ratio. Hence the correct choice is (c). 33. The total Lorentz force on the electron is F = – e (E + v B) Thus, r is inversely proportional to

37. point equidistant from the two wires will be equal is zero. Hence the force on a charge at this point is zero. 38. E, the force on a particle of charge q is F = qE in the direction of the electric E is parallel to B, the velocity v of the particle is parallel to B. Hence B will not affect the motion of the particle since v B = 0. Thus the correct choice is (a). 39. E on a particle of charge q is to impart to it a velocity v which is proportional to qE, i.e.

v B. If E is along + z-direction, the force – e E will be along – z-direction. If B is along + x direction, force – e (v B) will be along + z direction. When eE = evB, the electron moves along + y-direction

v

r=

mv 2 = qB

2m K 1 , where, K = mv 2. qB 2

m since K and B are the same for q the three particles. If mp is the mass of a proton and qp its charge, then md = 2mp and qd = qp and m = 4 mp and q = 2qp. From these it follows that r = rp < rd. 36. According to Ampere’s Law, the magnetic moment of a current I cross-section A is given by Thus r

qE i

qE

B on a charge moving with a velocity v is to exert on it a force F = q(v

for an electron moving along + y direction, the elecz direction and magnetic x direction, then the electron will be 34. Refer to the solution of Q. 15. The correct choice is (c). 35. The radius of the circular path is given by

= IA

B)

q (qE i )

Bk

Thus

F

q2EB ( j )

Since F be

j k j) q2, both positive and negative ions will y direction.

k

q2EB i

40. and 2 at a point between them act in opposite directions. But at a point to the left of wire 1 and to the B along the line XX is as shown in choice (b). 41. Given v = (2 given by

105 i ) ms–1. The force vector is

F = q(v

B)

= q {2 =2

105

105 i

(i 4 j

q ( 4k

3k ) }

3 j)

Therefore, the y and z components of the force are

24.28 Comprehensive Physics—JEE Advanced

Fy = 6 and

105

Fz = –8

10

q 5

q Fy2

Magnitude of force = =q

2R coil of n turns and radius r is

Fz2

B =

105

10

= 1.6

10–19

105

= 1.6

10–13 N, which is choice (c).

10

2 r 2 = v

10 7 )

(4

2

2 0.4 10

2

=

10

j + 9i

= q (6i

= 10 tesla

(2j + 3k)

k + 4j

j + 6j

k)

= 3q (2i – 3j + 2k) newton

=3

F 3q = (2 i 3 j 2 k ) m m

(0.96

= 2.88

(1)

2(r 2 x 2 )3 / 2

(2)

Hence the correct choice is (a).

= q (6k – 9j + 0 + 6i )

Acceleration =

r2

Substituting the values of r and x, we get B0 125 or B 27 125 125 B0 = B = 54 T = 250 T 27 27

43. Given v = (3i + 2j) ms–1 and B = (2j + 3 k ) tesla. Force experienced by the proton is B) = q (3i + 2j)

oI

oI 2r Dividing (1) by (2), we get

10–3 A

Hence the correct choice is (c).

F = q(v

2

B0 r3 = B (r 2 x 2 )3/2

3

10

0 In

B0 =

10–17 s

=8

19 17

B=

10

106

e 1.6 10 = t 8 10

Current I =

I B= 0 = 2r

0.4 10

0 nI

is given by

42. Given r = 0.4 Å = 0.4 10–10 m, v = 106 ms–1 Speed of electron in the orbit is 2 r v= ; here t is time taken by the t electron to complete one revolution. Thus t=

0 nI

= n 2B 2r 2R / n 2R Hence the correct choice is (b). 45. Given x = 4 cm = 0.04 m and r = 3 cm = 0.03 m.

(6 105 ) 2 ( 8 105 ) 2

= q

0I

Now B =

46. Wires AB and EF experience no forces since curforces on BC and DE are equal in magnitude but are directed in opposite directions. Hence their resultant is zero. Only force acting is on CD. Hence the correct choice is (a). 47. Since point O lies along the segments LR and SM, at point O. As point O is close to R and S, the net O due to segments PR and QS is B = BP + BQ =

108) (2i – 3j + 2k)

108 (2i – 3j + 2k) ms–2

Hence the correct choice is (b). 44. Let L be the length of the wire and let R be the radius of the circle when the wire is bent into one circular turn and r be the radius of the coil of n turns. Then R L = 2 R = 2 nr or R = nr or r = n

=

0

4

2I d

0I

4 RO

= 10–7

0I

4 SO

0I

4

1 d

1 d

2 10 = 10–4 Wb m–2 0.02

outside the plane of the paper. Hence the correct choice is (c). 48. Magnetic moment M = current area charge area = time period

Magnetic Effect of Current and Magnetism 24.29

=

r2

q T

Angular momentum L = m r2 1 q r2 M 2 q . Hence the correct choice 2 L m 2 m r is (c). 49.

B=

0

4

·

2 I1 r1

0

4

·

2I 2 r2

Given I1 = 2.5 A, I2 = 5 A and r1 = r2 = 2.5 m. Using 0 . The magnitude of B 2 is 0/2 . Hence the correct choice is (c). mv 50. For proton: r = eB m v 4mv 2mv For -particle r = eB 2eB eB v Given r = r . Hence v = . 2 1 2 Energy of proton E = mv . Energy of -particle 2 is

these values, we get B = –

E =

1 mv 2

2

=

1 2

4m

v 2

mAv A mB vB

2 T

1 q r2 2

2

=

1 mv 2 = E 2

Hence E = 1 MeV which is choice (a). 51. When a current is passed through the helix, the neighbouring coils of the helix attract each other due to which it contracts. As a result the contact is broken and the coils will recover their original state is made again and the process continues. Thus the wire oscillates. Hence the correct choice is (a). 52.

in the y-z plane, which is along the x-direction and (ii) the other due to the planar loop in the x-y plane, which is along the z-direction. The direction of the

( i k ) . Hence the correct choice is (d). 53. The radius r of the circular path of a particle of mass m and charge q moving with velocity v B is given by 2 mv = qvB r or mv = qrB. Hence mA v B = qrA B and mB v B = q rB B.

rA rB

rA > rB. Hence mA v A > mB v B. Thus the correct choice is (b). 54. The correct choice is (c). For derivation of the expression, refer to a Textbook of Physics. 55. The radius r of the circular path is given by (see Fig. 24.66) mv 2 = qvB r qB or v= (r) m qB vmin= (rmin) m qB (b – a), m which is choice (b). =

Fig. 24.66

56. The correct choice is (d) because the lines of force are continuous inside the magnet. 57. Since the path of the particle beyond P is non-circular, both E and B P. Hence choices (a) and (c) are incorrect. Since the curved path lies in the x-y must be in the x-z plane. Hence choice (d) is also incorrect. Thus, the correct choice is (b). 58. Let r be the radius of the circle. The length of the 60 r arc = (2 r) = . Therefore, the length 2l 360 3 of the magnet is r 6l 2l = or r = 3 If m is the pole strength of each pole of the magnet, 6l the magnetic moment of the arc = m r = m =

3 (2 ml )

=

3M

Hence the correct choice is (c). 59. Wires A and B carry current I = 5 A each coming out of the plane of the page as shown in Fig. 24.66. The P due wire A is equal to that due to wire B, i.e. 2I BA = BB = 0 4 a =

10

7

2 5 01 .

24.30 Comprehensive Physics—JEE Advanced

= 10–5 T BA is perpendicular to PA and BB is perpendicular to PB. Therefore, = 60°. The P is given by BR2 = B 2A + B 2B + 2BA BB cos which gives BR = 2BA cos

The correct choice is (d). 62. Refer to Fig. 24.68. Let AB = BC = AC = a. Let OD = r.

2

3 = 2 Hence the correct choice is (c). 10–5

=2

Magnitude of F = [(q ab)2 + (q ac)2]1/2 = qa(b2 + c2)1/2

3

10–5 T

Fig. 24.68

O due to current I AB of the triangle is given by BAB = It is clear that OD = r =

BAB= Fig. 24.67

60. When the proton enters the region of the magnetic F given by F = q (u B) where q is the charge of the proton. The force F is perpendicular to both u and B. Since the force is perpendicular to the velocity of the particle, it does not do any work. Hence the magnitude of the velocity of the particle will remain unchanged; only the direction of the velocity changes. Hence v = u. Since u is perpendicular to B, the proton moves in a circular path. Since the charge of proton is positive, u is along posotive x-axis and B is directed out of the page, the proton will move in a circle in the x-y plane in the clockwise direction. Hence its y coordinate will be negative, when it leaves the region. Thus the correct choice is (d). 61.

F = q(v

B) = q{a i

= q(ab i

j + ac i

(bj + c k)} k)

= q(ab k – acj) = qa(b k – cj)

=

0I

4 r =

(sin

+ sin )

= 60° and

AD a/ 2 = tan tan 60

0I 4 r

2 3 a

=

a/2 3

=

a 2 3

(sin 60° + sin 60°)

3 0I 2 a

sides BC and AC = that due to side AB. Hence, the O due to the current in the three sides of triangle ABC is B = BAB + BBC + BCA = 3BAB Hence the correct choice is (b). 63. Refer to Fig. 24.69. Let v be the velocity of the particle. Its kinetic energy is 1 2 2qV 1 / 2 mv = qV or v = (1) 2 m The particle follows a circular path from A to B of radius r which is given by mv 2 mv = qvB or r = r qB

(2)

Using (1) and (2), we have r=

m 2qV qB m

1/2

1 2mV B q

1/2

Magnetic Effect of Current and Magnetism 24.31

B = b in the positive z-direction, and the charge of the particle is positive, the path of the particle is a circle as shown in the circular path is r=

mv ma = qB qb

Thus y = – 2 Fig. 24.69

In triangle BCD, sin

=

BD d = . Therefore, BC r

q 1/ 2 , which is choice (a). 2mV 64. Refer to Fig. 24.70. Wire PQ of length d, the spacing between rails carries a current I vertically sin

= Bd

the reader and perpendicular to the length PQ of the wire. Thus angle between I and B is 90°. The force exerted on the wire of length d by the F = BId sin 90° = BId Using Fleming’s left hand rule, the direction of the force is to the left. The acceleration of the wire is force F BId = = a= mass m m 1 2 at 2 choice is (b).

Now x =

t =

2x . Hence the correct a

r=–

Fig. 24.71

2ma qb

So the correct choice is (d). 66. A circular metal loop carries a current. Hence charge, say, q moves along the circle with a velocity, say v which is tangential to the circle at every point (Fig. 24.59 on page 24.22). The force experienced by the charge is F = q(v B). Since v is along the tangent and B is directed out of the x – y plane, the direction of the force is towards the centre O of the loop. Hence the force tends to contract the loop. Further, since F is perpendicular to v, no work is done on the loop. Hence it cannot have any translational motion. Thus the correct choice is (c). 67.

Hence the correct choice is (c). 68. 0I . The direc2r z-direction if the current is

radius r and carrying a current I =

anticlockwise. Consider a small element of width dr. The current through the element is total current in spiral dI = width of element total width of spiral =

Idr b a b

B= a

Fig. 24.70

65. Refer to Fig. 24.71. Since the velocity of the particle is v = a along the positive x-axis and the direction of

=

b

0 NdI

=

2r 0 NI

b

0 NI

dr 2 b a r a

dr b 0 NI = ln 2 b a a r 2 b a a

24.32 Comprehensive Physics—JEE Advanced

II Multiple Choice Questions with One or More Choice Correct 1. A straight wire carrying current is parallel to the y–axis as shown in Fig. 24.72. The P is parallel to the x– axis z–axis (c) magnetic lines are concentric circles with the wire passing through their common centre wire are oppositely directed.

(a) E = 0, B = 0 (c) E 0, B = 0

(b) E = 0, B (d) E 0, B

0 0 IIT, 1985 6. The force F experienced by a particle of charge q moving with a velocity v B is given by F = q (v B). Which pairs of vectors are at right angles to each other? (a) F and v (b) F and B (c) B and v (d) F and (v B) 7. Choose the correct statements from the following. in magnitude from point to point but has a constant direction, is set up in a region of space. A charged particle enters that region

Fig. 24.72

2.

dal solenoid (a) is independent of the radius of the solenoid (b) depends on the number of turns and the current in the solenoid (c) is constant in magnitude inside the solenoid (d) is always radial inside the solenoid.

3. (a) is independent of the radius of the solenoid (b) depends on the number of turns and current in the solenoid (c) is uniform throughout the solenoid (d) is perpendicular to the axis of the solenoid. 4. A charged particle of mass m and charge q enters a B with a velocity v at an angle with B. (a) The kinetic energy of the particle will not change if 0. (b) The momentum of the particle will not change if = 0. (c) The particle moves in a circle of radius mv/qB if = 90°. (d) The frequency of circular motion is independent of the speed of the particle if = 90°. 5. A proton moving with a constant velocity passes through a region of space without any change in its velocity. If E and B represent the electric and may have

a constant speed. The initial velocity of the particle is either along the direction of the

from point to point in magnitude and direction is set up in a certain region of space. A charged particle enters the region with a certain initial velocity. The direction of the the initial velocity. (c) A electron travelling in the positive x–direction enters a region of space having an y– direction. positive z–direction, the electron will travel (d) A charged particle moves in a uniform magof the particle will change during this time. 8. Which of the following statements are correct? (a) A current carrying coil, free to rotate, when itself such that its plane becomes perpendicu(b) The trajectory of a charged particle moving -

between two long parallel wires carrying equal currents in the same direction is zero.

Magnetic Effect of Current and Magnetism 24.33

(d) An electron is moving in the anticlockwise sense in a horizontal circular orbit. The

(a) M =

of the orbit will be vertically upward.

(c) B =

9.

us r carries a current I. It is placed in a uniform B. The tension in the loop will be doubled if (a) I is doubled (b) B is doubled (c) r is doubled (d) Both B and I are doubled. 10. A particle having a mass of 0.5 g carries a charge of 2.5 10 –8 C. The particle is given an initial horizontal velocity of 6 10 4 ms –1. To keep the particle moving in a horizontal direction the direction of the velocity tion of the velocity of 3.27 T 11. A wire is bent into a circular loop of radius R and carries a current I of the loop is B. The same wire is bent into a double loop. If both loops carry the same current in the the double loop is B1. If they carry the same current I centre is B2. Then (b) B1 = 4B (a) B1 = 0 B (c) B2 = 0 (d) B2 = 4 12. A straight horizontal conducting rod of mass m and lenght l is suspended by two vertical wires (of negligible mass) at its ends. A current I is set up in the B normal to the conductor is required to keep the tension in the wires equal to zero. If the direction of the current is reversed B, a tension T is developed in the wires. Then mg 2mg (b) B = (a) B = Il Il (c) T = BIl + mg (d) T = BIl – mg 13. In the hydrogen atom the electron (charge e) moves around the proton with a speed v in a circular orbit of radius r. The magnetic dipole moment of the circulating electron is M the site of the proton (i.e. at the centre of the orbit) is B. Then

e vr 4 0 ev 2

4 r

e vr 2

(b) M = (d) B =

0 ev 2

2 r

14. a circular coil of radius r, having n turns and carrying a current I can be doubled by (a) changing I to 2I, keeping n unchanged (b) changing n to 2n, keeping I unchanged (c) changing I to 2I and n to 2n (d) changing n to n , keeping I unchaged. 2 15. The force experienced by a charged particle moving B with a velocity v is zero if (a) v = 0 (b) v is parallel to B (c) v is perpendicular to B (d) v is antiparallel to B 16. pends upon (a) the number of turns in the solenoid (b) the current in the solenoid (c) the radius of the toroid (d) the permeability of the core of the solenoid. 17. The plates of a parallel plate capacitor are in the y – z plane. The separation between the plates is 3 mm and a potential difference of 600 V is applied across the plates. An electron is projected between the plates with a velocity of 2 106 ms– 1 along the positive y-direction. The electron moves unde105 V m–1. region between the plates is 0.1 T. the positive z-direction. the negative z IIT, 1981 18. An alpha particle and a deuteron, after being accelerated through the same potential difference, enter dicular to their velocities. If r and rd are the radii of the circular paths of the alpha particle and the deuteron respectively and and d their respective frequency of revolution, then (b) r = rd (a) r = 2 rd (c)

=

d

2

(d)

=

d

IIT, 2004

24.34 Comprehensive Physics—JEE Advanced

19.

rent I in the same direction are in the x-y plane in a gravity free space (see Fig. 24.73). The central wire is at x = 0 while the other two wires are at x = d. IIT, 1997

(b) the magnitude of the angular momentum about the centre is doubled. (c) the radius r becomes half (d) the frequency remains unchanged. 22. A long, thin and hollow cylindrical metal pipe of radius R carries a current I along its length. For such a pipe, the pipe and increases as we go towards the wall. 0I . (c) On the surface of the pipe, B = 2 R (d) at a point outside the pipe at distance r from r 2. 23. A particle of charge q moving with a velocity B = B0 ( i

v = v0 i

j k) .

(a) The particle describes a circle in the magnetic

Fig. 24.73

(a) cannot be zero beyond B (i.e. x > d) (b) cannot be zero beyond A (i.e. – x > – d) (c) will be zero between x = 0 and x = d. (d) will be zero between x = 0 and x = – d. 20. Two charged particles 1 and 2 of masses m1 and m2, charges q1 and q2 enter a uniform magnetic v1 and v2 shown in Fig. 24.74.

(b) The magnitude of the force acting on the particle is qv0B0 (c) The magnitude of the force on the particle is 2 qv0B0. (d) The force vector lies in the y-z plane. 24. A thin rod AB of length l carries a current I1. It is PQ carrying a current I2 as shown in Fig. 24.75.

Fig. 24.75

(a) The force experienced by the rod is F = Fig. 24.74

(a) If q1 = q2, m1v1 > m2v2 (b) If q1 = q2, m1v1 < m2v2 (c) If v1 = v2,

m1 m2 > q1 q2

(d) If m1 = m2, v1q1 < v2q2. 21. A charged particle is accelerated through a potential difference V. It then enters a region of uniform r and its frequency of revolution is . If V is doubled. (a) the kinetic energy is doubled

0 I1 I 2

4

loge 1

l 2a

(b) The force experienced by the rod is F = l 0 I1 I 2 loge 1 4 a (c) The rod experiences no torque (d) The rod experiences a force as well as a torque. 25. Choose the correct statements from the following. is [ML0 T–2 A–1] (b) 0 is dimensionless.

Magnetic Effect of Current and Magnetism 24.35

(c) The dimensions of 0 0 are the same as those of speed. E are the same as those (d) The dimensions of B of speed. 26. An annular wire loop ABCD carries a current I1 as shown in Fig. 24.76. O is the common centre of the curved parts AB and CD of the loop. A straight wire passing through O and perpendicular to the plane of the loop carries a current I2 directed towards the reader. Then

(d) The proton follows track D and electron follows track A. IIT, 1984 28. A particle of charge + q and mass m moving under E i and B k follows a trajectory from P to Q as shown in Fig. 24.78. The velocity at P is v i and at Q is – 2v j . Which of the following statements is/are correct ?

Fig. 24.76

(a) the net force on the loop is zero. (b) the net torque on the loop is zero. (c) As seen from O the loop will rotate in clockwise sense about axis OP. (d) As seen from O the loop will rotate in anticlockwise sense about axis OP. IIT, 2006 27. A neutron, a proton, an electron and an alpha par-

inward normal to the plane of the paper. The tracks of the particles are labeled as shown in Fig. 24.77. (a) The electron follows track D and alpha particle follows track B (b) The proton follows track A and alpha particle follows track B

Fig. 24.78 2

(a) E =

3mv 4qa

P 3

is

3mv . 4a P

is zero. Q is zero. IIT, 1991 29. A particle of mass m and charge q moving with velocity v enters Region II normal to the boundary as shown in Fig. 24.79. Region II has a uniform B perpendicular to the plane of the paper. The length of the Region II is l. Choose the correct choice(s)

Fig. 24.79

(a) The particle entres Region III only if its qlB velocity v > m Fig. 24.77

(c) The electron follows track A and neutron follows tracks C

(b) The particle enters Region III only if its qlB velocity v < m

24.36 Comprehensive Physics—JEE Advanced

(c) Path length of the particle in Region II is qlB maximum when velocity v = m (d) Time spent is Region II is the same for any velocity v as long as the particle returns to Region I IIT, 2008 30. H+, He+, and O++ all having the same kinetic energy pass through a region in which there is a uniform masses of H+, He+ and O++ are lu, 4u and 16u respectively. Then (a) H+

(b) O++ (c) He+ and O++ IIT, 1994 31. A proton and an electron moving with the same velocity v B which is perpendicular to their velocity. If rp and re are the radii of their circular trajectories and tp and te the time after which each particle comes out (a) rq < re (c) tp < te

(b) rp > re (d) tp > te IIT, 2011

ANSWERS AND SOLUTIONS 1. to the plane containing the point P and the current element I, which is the z-direction. Hence choice (a) the left and to the right of the current element are oppositely directed. Hence choice (d) is correct. concentric circles with their common centre at the wire. Hence choice (c) is also correct. 2. The correct choices are (a), (b) and (c). The mag3. uniform at the edges of the solenoid and is parallel to the axis. 4. perpendicular to the velocity of the particle; the speed of the particle cannot change; only the direction of motion, i.e. velocity (and hence momentum) will change. The correct choices are (a), (c) and (d). 5. The force on a charge q moving with a velocity v is given by F = q (E + v B) There will be no change in velocity if F = 0. This can happen in three cases. (i) Both E and B are zero which is choice (a). (ii) E = 0 and v and B are parallel so that v B = 0 in which case B 0 which is choice (b). (iii) The electric force qE is equal and opposite to magnetic force q (v B) in which case the net force is zero which is choice (d). Hence the correct choices are (a), (b) and (d). 6. The pairs F and v and F and B are always at right angles to each other, because F is always perpendicular to the plane containing B and v. Vectors B and v may have any angle between them.

7. Statement (a) is correct. Since the direction of the velocity of the particle remains unchanged, no magnetic force acts on the particle. The force experienced by a particle of charge q moving with a velocity v B is F = q (v B) Since F = 0, (v B) = 0, i.e. v and B are parallel to each other. Thus, the initial velocity of the particle is either Statement (b) is also correct. Since the magnetic force is always perpendicular to particle velocity, it cannot change the magnitude of the velocity (i.e. speed); it can only change the direction of is equal to its initial speed. The direction of the initial velocity. y-direction, the electron y-direction. It will y-direction. Since the magnetic force is perpendicular to the magnetic

the negative z-direction (use Fleming’s left-hand rule). Statement (d) is also incorrect. No work is done Hence its kinetic energy remains constant. 8. Statement (a) is correct. The loop will rotate until equilibrium state is attained. It will then come to rest because the torque acting on it becomes zero. We know that the torque is given by = BIA sin

Magnetic Effect of Current and Magnetism 24.37

where

is the angle between the direction of magB and the normal to the plane of the loop. It is clear that = 0 when = 0. Thus, in the equilibrium state, the orientation of the loop is such

of the loop. Statement (b) is incorrect. If v is parallel to B, the particle does not experience any force. Hence its trajectory will be a straight line. Statement (c) is correct. Since currents are equal and in the same point midway between the wires will be equal and opposite and hence they will cancel each other. Statement (d) is incorrect. The direction of the hand rule, will be vertically downward. Note that the charge of an electron is negative. 9. The force acting on the loop is given by F = m B sin where m = r2I. Force will be doubled if I or B are doubled. Hence the correct choices are (a) and (b). 10. experience gravitational force mg. As a result the particle will not continue moving in the horizontal direction but will describe a parabolic path. So must be perpendicular to the direction of the velocity. The magnetic force experienced by the particle is given by F = q (v B) The magnitude of the force is F = q v B sin . If the particles is to move in the horizontal direction, this force must balance the force of gravity, i.e. mg = q v b sin The minimum value of B corresponds to sin = 1 or = 90°. Thus mg = v B mg 0.5 10 3 9.8 = 3.27 T or B = qv 2.5 10 8 6 10 4 Hence the correct choices are (a) and (c). 11. The radius of the double loop r = R/2. Now I B= 0 2R r at the centre of the loop is I I ( r = 1/2) B1 = 0 = 0 = 2 B 2r 2R Similarly for the second loop of the double loop, B2 = 2 B Since the currents in the two loops are in the same B1 + B2 = 4B.

Since the currents in the two loops are in opposite B1 and B2 are equal and opposite. the double loop = B1 – B2 = 0. Hence the correct choices are (b) and (c). 12. In order that the tension in the supporting wires is zero the downward gravitational force mg on the rod must be balanced by an upward force BIl due BIl = mg mg or B= Il If the current is reversed, the direction of the force due to B becomes downwards, in the direction of the gravitational force. Hence the tension in the string is T = BIl + mg The correct choices are (a) and (c). 2 r e ev . Current I = = . 13. Time period T = v T 2 r Magnetic moment M = current area of orbit ev e vr =I r2 = r2 2 r 2

B=

0I

=

0 ev 2

2r 4 r So the correct choices are (b) and (c). 14. the coil is given by 0 nI 2r Hence the correct choices are (a) and (b). 15. F = q (v B). The magnitude of the force is F = q v B sin Thus F = 0 if v = 0 or = 0° or 180°. Hence the correct choices are (a), (b) and (d). 16. The correct statements are (a), (b) and (d). The

B=

B = μnI where is the permeability of the core of the solenoid. B is independent of the radius of the toroid. 17. plates is E =

V 600 = d 3 10

3

Fig. 24.80

=2

105 V m–1.

24.38 Comprehensive Physics—JEE Advanced

The electron will experience a force qE in the negative x ed in the region between the plates, the magnetic force experienced by it must be along the positive x-direction and its magnitude must be equal to qE. Since magnetic force = q(v B), it follows that the v (as well as E) and directed along the negative z-direction. The magnitude of the magnetic force =qvB sin 90° = qvB. Thus 2 10 5 E = = 0.1 T qvB = qE or B = v 2 10 6 So the correct choices are (a), (b) and (d). 18. If a particle of mass m and charge q is accelerated through a potential difference V, it acquires kinetic energy = qV. Hence 1 mv2 = qV 2 2qV or v= m where v is the velocity acquired by the particle. The radius of the circular path is given by r=

mv m 2qV = = qB qB m

q qd

q m

q . Hence m

For given V and B, r r = rd

2V B

md m

2q p

2m p

qp

4m p

=1

The frequency of revolution is given by qB q = 2 m m Hence

III are along the positive z-axis. Hence the net magB cannot be zero beyond A, (i.e. – x > – d). For Points between O and B: The directions of the along the negative z-axis but the direction of the the positive z B will be zero for some value of x lying between x = 0 and x = d. For Points between O and A: The directions of the along the positive z-axis but that due to current in wire III is along the negative z-axis. Hence the net B will be zero for some value of x lying between x = 0 and x = – d. So all the four choices are correct. 20. The radius r of the circular path of a particle of mass m and charge q moving with a velocity v perpenB is given by 2 mv = qvB mv = qrB (1) r mv r If q1 = q2, then 1 1 = 1 m2 v2 r2 It follows from Fig. 24.66 that r1 > r2. Hence m1v1 > m 2v 2. If v1 = v2, then from Eq. (1), we have m1 qr = 11 m2 q2 r2

(1)

Now, alpha particle has 2 protons and 2 neutrons and a deuteron has 1 proton and 1 neutron. Therefore, m = 4 mp, md = 2 mp, q = 2 qd and qd = qp. Using these in (1), we get r = rd

For Points on the x-axis beyond A: The directions of

=1 d

Thus the correct choices are (b) and (d). 19. By using the right hand rule, the following conclusions can be drawn: For Points on the x-axis beyond B: The directions III are along the negative z-axis. Hence the net magB cannot be zero beyond B (i.e. x > d).

r1 m m > 1, m1q2 > m2q1 or 1 > 2 . If m1 = m2, r2 q1 q2 v1q2 > v2q1. So the correct choices are (a) and (c). 21. Kinetic energy = qV. Hence if V is doubled, K.E. is also doubled. The magnitude of the angular momentum about the centre is L = mvr. Now r = mv/qB. Therefore, m2 v 2 2m 1 2 2m = (K.E.) L= mv = qB qB 2 qB Thus if V is doubled, K.E. is doubled and L is also doubled. The radius of the circular path is given by m2 v 2 2m 1 2 = 2m qV r2 = 2 2 = 2 2 mv q B q B q2 B 2 2 Since

=

2mV qB 2

Magnetic Effect of Current and Magnetism 24.39

Thus r

V . The frequency of revolution is

qB 2 m Hence the correct choices are (a), (b) and (d). 22. Consider a point P at a distance r from the axis of the pipe. For all points inside the pipe, r < R P, we choose an Amperean loop to be a cylindrical loop of radius R as shown in Fig. 24.81. Since the current enclosed in the Amperean surface is zero, it follows from Ampere’s =

24. Consider a small element of length dx of the rod at a distance x from wire PQ as shown in Fig. 24.82. PQ at the element is 0 I2 B= 2 x Therefore, force exerted on the element is

points inside the pipe. For points outside the pipe, r > R, we have from Ampere’s law,

Fig. 24.82 0 I1 I 2 dx

dF = BI1 dx =

2 x Total force experienced by the rod is F=

Fig. 24.81

B .dl =

0I

B

dl =

0I

B

2 r=

0I 0I

B=

0 I1 I 2

= (B is || dl)

2

, i.e. B

0I

2 R Hence the correct choices are (a) and (c). 23. Since v is not perpendicular to B, the particle will not describe a circle. The force acting on the particle is F = q(v B) = q[v0 i

= qv0B0 k j Thus the magnitude of the force is F = qv0B0 = qv0B0(12 + 12)1/2 = 2 qv0B0 and it lies in the y-z plane. Hence the correct choices are (c) and (d).

dx x

a l 2 a B is not uniform; it decreases as we go from and A to end B of the rod. Hence the rod will also experience a torque. Thus the correct choices are (b) and (d). 25. From F = qvB, it is easy to see that the dimensional formula of B is [ML0 T–2 A–1], which is choice (a). Choice (b) is wrong. From = 0/(2 ), the dimensions of 0 are B r [ 0] = I =

B0 i j k ]

= qv0B0 i i i j i k

x (a l )

x a

0 I1 I 2

=

1 2 r r At the surface of the pipe (r = R) B=

dF

ML0 T 2 A A

From F = [ 0] = [

0

r2

0

q1q2 2

L

q1q2

1 4

1

=

log e

= [MLT–2 A–2] , the dimensions of

A2T2 2

2

0

are

= M–1 L–3 T4 A2

MLT L F r –2 –2 –1 –3 4 2 [M L T A ] = [L–2 T2] 0] = [MLT A ]

0 0

= [L–1 T], which is reciprocal to speed.

Hence choice (c) is incorrect.

24.40 Comprehensive Physics—JEE Advanced

E = v. Hence choice B

From qE = qvB (d) is correct. 26.

I2 is tangential to the curved parts AB and CD of the loop. Hence every current element dl of parts AB and CD is parallel or antiparallel to B. The magnetic force on AB or CD is zero since = 0° or 180° in the expression dF = Bldl sin . The magnetic force on straight parts AD and BC is not zero. The magnetic force on AD is directed towards the reader which is equal and opposite to the force on BC which is directed away from the reader. These equal and opposite forces cancel each other. Therefore, the net force on the loop ABCD is zero. Since these equal and opposite forces do not act at the same point, they will exert a net torque on the loop which will rotate it in the clockwise sense when viewed from O. Hence the correct choices are (a) and (c). 27. The radius of the circular track is given by mv r= qB For a neutron q = 0. For an electron q = –e and for a proton q = +e. For an alpha particle q = 2e and m = mass of four protons. Hence the correct choices are (a) and (b). 28. In going from P to Q, the change in the kinetic 1 1 3 energy of the particle = m(2v)2 – mv2 = 2 2 2 mv2 ing the particle from P to Q = F·d where d is the displacement in the direction of E. So work done W = qE i 2a i 2qaE i i = change in K.E. Hence 2qaE =

2qaE . Now work done

3 2 mv 2

3mv 2 4qa

E

Thus choice (a) is correct. P is F vp

qE i v i

qE v

q

3mv 2 4qa

r=

mv qB

Therefore, the particle can enter region III if r > l, qBl . i.e. if v > m In region II, the maximum path length is r = l, qBl which gives v = m The time period of the circular motion is 2 r 2 mv 2 m T= v v qB qB The particle will return to region I if the time spent T m by it in region II is , which is independent 2 qB of the velocity. Hence the correct choices are (a), (c) and (d). 1 1 1 30. mHv2H = mHev2He = mOv2O. Given mH = 1u, 2 2 2 1 mHe = 4u and mO = 16u. Hence vHe = vH and 2 1 v O = v H. 4 The radius of the circular path is given by mv ; q = charge R= qB Now charge of H+ = e, charge of He+ = e and charge of O++ = 2e. It is easy to check that rHe = rO and 1 rH = rHe 2 Smaller the value of r Thus the correct choices are (a) and (c). 31. the electron and proton in the region of magnetic Since

r =

mv , rp > re qB

v

3mv3 4a So choice (b) is also correct. Q is zero 0 . Hence the correct

because i j 0 and i k choices are (a), (b) and (d). 29. In region II, the particle follows a circular path of radius

Fig. 24.83

Since v B, the charged particle describes a semicircular trajectory. Hence the time after which period, i.e.

Magnetic Effect of Current and Magnetism 24.41

t=

T 2

Since mp > me and q and B are the same, tp > te. Hence the correct choices are (b) and (d).

m qB

III Multiple Choice Questions based on Passage Questions 1 to 3 are based on the following passage Passage I Two long parallel wires carrying currents 2.5 amperes and I ampere in the same direction (directed into the plane of the paper) are held at P and Q respectively such that they are perpendicular to the plane of the paper. The points P and Q are located at a distance of 5 m and 2 m, respectively, from a collinear point R (see Fig. 24.84).

Fig. 24.84

An electron moving with a velocity of 4 105 m/s along the positive x-direction experiences a force of magnitude 3.2 10–20 N at the point R. IIT, 1990 1. R is (b) 5.0 10–7 T (a) 2.5 10–7 T (d) 2.5 10–6 T (c) 5.0 10–6 T 2. R due to current I = 2.5 A in wire P is (b) 2 10–7 T (a) 1 10–7 T –7 (d) 4 10–7 T (b) 3 10 T 3. The current I in wire Q is (a) 1 A (b) 2 A (c) 3 A (d) 4 A

SOLUTION 1. The magnitude of the force experienced by a particle of charge q moving with a velocity v in a F = qvB sin where is the angle between v and B. Given F = 3.2 10–20 N, v = 4 105 ms–1 and = 90°. For electron q = 1.6 10–19 C. Using these value we get B = 5 10–7 T, which is choice (b) 2. R due to currect I in wire P is 7 I 2.5 = 1 10–7 T B1 = 0 = 4 10 2 r1 2 5 The correct choice is (a).

3.

R due to currect I in wire Q is B2 =

0I

2 r2

=

4

10 2

7

2

I

=I

10–7 T

B1 and B2 will be in the downwar direction, parallel and colliner. Hence the resultant R is B = B1 + B2 = (1 + I) 10–7 T Now B = 5 10–7 T. Therefore (1 + I) 10–7 = 5 10–7 or 1 + I = 5 or I = 4 A. So the correct choice is (d).

Questions 4 to 7 are based on the following passage

4. The force experienced by the charged particle in the

Passage II

(a) along the positive y-direction (b) along the negative y-direction (c) in the x-y plane (d) in the y-z plane. 5. If the particle emerges from the region of magnetic

The region between x = 0 and x = L B0 k . A particle of mass m, positive charge and velocity v0 i travels along x-axis and enters IIT, 1999

initial velocity, the value of L is

24.42 Comprehensive Physics—JEE Advanced

(a)

2mv0 qB0

(b)

(c)

mv0 2qB0

(d)

mv0 qB0

(c) v0 j (d) – v0 j 7. In Q. 6, the time spent by the particle in the mag-

3mv0 2qB0

6.

(a) t =

2 m qB0

(b) t =

(c) t =

3 m 2qB0

(d) t =

x = 2.1 L,

(a) v0 i

(b) – v0 i

2 m qB0 m qB0

SOLUTION 4. The force experienced by the charged particle is given by F = q(v

B) = (v0 i )

= qv0B0( i

mv0 (2) qB0 Since the particle emerges from the region of the r=

(B0 k )

k)

= qv0B0(– j )

angle of 30° with the initial vector, it follows from triangle ABC that AB = AC sin 30° or L = r sin 30° m v0 m v0 sin 30 = (3) = 2qB0 qB0

(1)

The force is along the negative y-direction, which is choice (b). 5. Refer to Fig. 24.85.

Thus the correct choice is (c). 6. will continue to move in a circular path till it completes half the circular path and emerges out of the v0 i moving along the negative x-axis as shown in Fig. 24.39. 7. Distance travelled by the particle in the magnetic . Therefore, time t=

Fig. 24.85

The particle describes a circle of radius Questions 8 to 11 are based on the following passage

9.

(c)

2 3 r

(d)

4 r

m qB0

[Use Eq. (2)]

B2 at O due to the circular arc ACB is (a) 5 0 I 12r

A wire loop consists of a straight segment AB and a circular arc ACB of radius r. The segment AB subtends an angle of 60° at the centre O of the circular arc. The wire loop carries a current I in the clockwise direction (Fig. 24.86). 8. B1 at O due to the straight segment AB is 0I 0I (b) (a) 2 r 2 2 r 0I

v0

=

So the correct choice is (d).

Passage III

0I

r

0I

(b)

2r (c) 3

0I

8r 7 0I (d) 18r 10.

Fig. 24.86

B at O due to the whole wire loop is

Magnetic Effect of Current and Magnetism 24.43

(a) B = B1 + B2 B12

(b) B = B2 – B1

(d) B = B22 11. B is (a) parallel to the plane of the coil (c) B =

B22

(b) perpendicular to the plane of the coil and directed out of the page (c) perpendicular to the plane of the coil and directed into the page. (d) inclined at an angle of 60° with the plane of the coil.

B12

SOLUTION 8.

O due to current I in AB is given by (use Biot-Savart law)

9. (n = 1 turn) circular loop of radius r and carrying a current I is 0 nI

B=

Here loop ACB is a fraction of a circle i.e. n < 1. Since ACB subtends an angle (360° – 60°) = 300° at O, hence the fraction n is

Fig. 24.87

BAB =

0I

4 x

(sin

+ sin

=

0I

4 0I

2 3 r

3r/ 2

(sin 30

(0.5 + 0.5) =

sin 30 ) 0I

2 3 r

n=

)

3r Here = = 30°. Also x = r cos = r cos 30° = . 2 Therefore, B1 =

2r

,

which is choice (c). plane of the paper directed into the page. Questions 12 to 15 are based on the following passage Passage IV A moving coil galvanometer consists of a coil of N turns and area suspended by a thin phosphor bronze strip in B. The moment of inertia of the coil about the axis of rotation is I and C is the torsional constant of the phosphor bronze strip. When a current is (in radian). IIT, 2004 12. Choose the correct statement from the following. The magnitude of the torque experienced by the coil is independent of (a) N (b) B (c) i (d) I 13. The current sensitivity of the galvanometer is increased if (a) N, A and B are increased and C is decreased.

300 360

=

5 6 ACB is

5 I 5 0I 6 = B2 = 12r 2r As the current in ACB is clockwise, the direction of 0

the paper and directed into the page. The correct choice is (a). 10. Since B1 and B2 are in the same direction, the net B = (B1 + B2), which is choice (a). 11. The correct choice is (c). (b) N and A are increased and B and C are decreased (c) N, B and C are increased and A is decreased (d) N, A, B and C are all increased. 14. When a charge is passed almost instantly through the coil, the angular speed acquired by the coil is NAB BAQ (b) (a) QI NI NABQ NAQI (c) (d) I B 15. radian) of the coil is I 1 (b) max = (a) max = I C C (c)

max

= I

C

(d)

max

=

IC

24.44 Comprehensive Physics—JEE Advanced

SOLUTION 12. The magnitude of torque experienced by the coil is given by = iNAB sin where is the angle which the normal to the plane of the coil makes with the direction of the magnetic coil is always parallel to the direction of the magnet= 90°. Hence = iNAB = Ki = where K = NAB So the correct choice is (d). 13. Let current i is passed through the coil. Then, restoring torque = C . When the coil is in equilibrium, iNAB = NAB i C Hence the correct choice is (a). Current sensitivity is

=

Questions 16 to 18 are based on the following passage Passage V 10–19 C E along the direction shown in Fig. 24.88. The speed of the particle is 107 ms–1 the inward normal to the plane of the paper. The particle F. IIT, 1984 16. The value of angle is (a) 30° (b) 45° (c) 60° (d) 75° 17. The radius of the circular path of the particle in the A particle of mass 1.6

(a) 0.1 m (c) 0.3 m

14. If is the angular speed acquired by the coil when a charge is passed through it for very short time t, then angular momentum = = I timeinterval t Q or = t = = KQ i t NABQ or I = NABQ or = , which is choice I (c). 15. From the principle of conservation of energy, we have 1 2

=

1 2

which gives

2 max

max

=

I , which is choice (a) C

18. The distance EF is (a)

2m

(b)

(c)

2 m 10

(d)

10–27 kg and charge 1.6

(b) 0.2 m (d) 0.4 m

2

Fig. 24.88

SOLUTION 16. Let O be the centre of the circular path. It is obF such that BF is tangent to the circle. OE and OF are normals, which meet at O (see Fig. 24.89). Therefore, angle OFE = angle OEF. Hence = 45°. mv . Substituting the given values and solving 17. r = qB we get r = 0.1 m 2 m 18. EF = 2 r cos 45° = 2 r = 2 0.1 10

Fig. 24.89

2 m 5 1 m 2

Magnetic Effect of Current and Magnetism 24.45

IV Matrix Matching Type 1. In Column I are listed some charged bodies and current carrying conductors. Match them with the effects they produce listed in column II Column I Column II

with a constant angular velocity (c) A coil carrying a current I = I0 sin

t

(r) Magnetic moment IIT, 2006

ANSWER (a) (c)

(p) (p), (r), (s)

(b) (d)

(p), (q), (r) (q)

2. Two wires each carrying a steady current I Column I. Some of the resulting effects are described in Column II. Match the statemens in Column I with those in Column II. Column I Column II (a) Point P B) at P due to the currents in the wires are in the same direction. (b) Point P joining the centers of the circular wires which have same radii.

B) at P due to the current in the wires are in opposite directions.

(c) Point P joining the centers of the circular wires, which have same radii

(d) Point P is situated at the common centre of the wires.

P.

(s) The wires repel each other.

IIT, 2007

24.46 Comprehensive Physics—JEE Advanced

SOLUTION (a) At point P the currents are equal but in opposite directions. Hence the correct choices are (q) and (r). (b) The currents in the two coils are in the same sense (anticlockwise). Hence, at point, P the current are the same direction, which is choice (p). P due to the currents are in the opposite directions. Since point P is the same distance from the centres of the coils, and their radii are point P is zero. Hence the correct choices are (q) and (r). P of the coils are not equal (because their radii are different). Further, direction. If the currents are in opposite directions in coils (or wires), they repel each other. Hence the correct choices are (q) and (s). (a) (q), (r) (b) (p) (c) (q), (r) (d) (q), (s) 3. Six point charges, each of the same magnitude q, are arranged in different manners as shown in Column II. In each case, a point M and a line PQ passing through M are shown. Let E V be the electric potential at M Now, the whole system is set into rotation with a constant angular velocity about the line PQ. Let B the magnetic M and μ be the magnetic moment of the system in this condition. Assume each rotating charge to be equivalent to a steady current. IIT, 2007 Column I Column II

(a) E = 0 (p)

(p) Charges are at the corners of a regular hexagon. M is at the centre of the hexagon. PQ is perpendicular to the plane of the hexagon.

(b) V

0 (q)

(q) Charges are on a line perpendicular to PQ at equal intervals. M is the mid-point between the two innermost charges.

(c) B = 0 (r)

(r) Charges are placed on two coplanar insulating rings at equal intervals. M is the common centre of the rings. PQ is perpendicular to the plane of the rings.

(d) μ

(s) Charges are placed at the corners of a rectangle of sides a and 2a and at the mid point of the longer sides. M is at the centre of the rectangle. PQ is parralled to the longer sides.

(t)

0 (s)

(t) Charges are placed on two coplanar, identical insulating rings at equal intervals. M is the mid-point between the centres of the rings. PQ is perpendicular to the line joining the centres coplanar to

Magnetic Effect of Current and Magnetism 24.47

SOLUTION M due to charge at A and B =

(p)

2q 4

0r

2

where r = AM = BM directed from M to B. The electric

M due to other charges are net V = 0. Since the total charge of the system is zero, on rotation net current = 0. Hence B= 0 and μ = 0

(q) The net charge on the right of M is 2q – q = q and on the left of M is –q M and directed to the left of M. But net V = 0. Net charge = 0. On rotation, current = 0. Hence B = 0 and μ = 0.

not zero. Hence B

M 0 and μ

V

0. On rotating, net current is

0.

E = 0. But net V 0. On rotating, net current 0. Hence B 0 and μ 0. E 0 and is directed to the right but net V = 0. On rotation, net current = 0. Hence B = 0 and μ = 0.

ANSWERS

(a) (c)

(p), (r), (s) (p), (q), (t)

(b) (d)

(r), (s) (r), (s)

24.48 Comprehensive Physics—JEE Advanced

V Assertion-Reason Type Questions In the following questions, Statement-1 (Assertion) is followed by statement-2 (Reason). Each question has the following four options out of which only one choice is correct. (a) Statement-1 is True, Statement-2 is True and State-ment-2 is a correct explanation for Statement-1. (b) Statement-1 is True, Statement-2 is True; but 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. 1. Statement 1 nitude from point to point but has a constant direction, is set up in a region of space. If a charged particle enters the region in the direction of the uniform rate in the region. Statement 2 The force F experienced by a particle of charge B is moving with a velocity v given by F = ( v B ). 2. Statement 1 A charged particle moves in a uniform magnetic energy of the particle cannot change but its momentum can change. Statement 2 The magnetic force is always perpendicular to the velocity of the particle. IIT, 1993 3. Statement 2 A current carrying loop is free to rotate. It is placed Statement 2 The torque on the coil is zero when its plane is 4. Statement 1 An electron moving in the positive -direction enters a region where uniform electric and magnetic

Statement 2 If a charged particle moves in a direction perpenacting on it is given by Fleming’s left-hand rule. 5. Statement 1 If a charged particle is released from rest in a region each other, it will move in a straight line. Statement 2 6. Statement 1 A proton and an alpha particle having the same kinetic energy are moving in circular paths in a paths will be equal. Statement 2 Any two charged particles having equal kinetic energies and entering a region of uniform magnetB in a direction perpendicular to B , will ic describe circular trajectories of equal radii. 7. Statement 1 Two particles having equal charges and masses m1 and m2, after being accelerated by the same potential difference (V), enter a region of uniform r1 and r2 respectively. Then m1 = m2

r1 r2

Statement 2 Gain in kinetic energy = work done to accelerate the charged particle through potential difference . 8. Statement-1 No net force acts on a rectangular coil carrying a steady current when a suspended freely in a uniStatement-2 The magnitude of force experienced by a straight conductor of length L carrying a current I and placed B is BIL. IIT, 1981

Magnetic Effect of Current and Magnetism 24.49

9. Statement-I There is no change in the energy of a charged partiforce is acting on it. Statement-2 The magnetic force acting on a moving charged particle is always perpendicular to its velocity. IIT, 1983 10. Statement-I A charged particle enters a region of uniform magpath of the particle is a circle. Statement-2 allel and equidistant. IIT, 1983 11. Statement-I An electron and a proton are moving with the same kinetic energy along the same direction. When to their direction of motion, they describe circular path of the same radius. Statement-2 The radius of the circular pathof a particle of charge q, mass m and moving with velocity v perpendicuB is given by

r=

mv qB IIT, 1985

12. Statement-1 tude from point to point but has a constant direction, is set up in a region of space. If a charged particle enters the region in the direction of the magnetic the region Statement-2 The force F experienced by a particle of charged B q moving with a velocity v is given by F = q( v B ) IIT, 1989 13. Statement-I The sensitivity of a moving coil galvanometer is increased by placing a suitable magnetic material as a core inside the coil. Statement-2 Soft iron has a high magnetic permeability and cannot be easily magnetized or demagnetized. IIT, 2008

SOLUTIONS 1. The correct choice is (d). If v is parallel to B , F = 0. Hence the particle does not experience any force and is, therefore, not accelerated in the region. It

plane of the coil is perpendicular to the magnetic 4. The correct choice is (a). Because electron has a

2. The correct choice is (a). Since the magnetic force is always perpendicular to the velocity, no work particle. Hence magnetic force cannot change the magnitude of velocity (i.e. speed); it can only change the direction of velocity. Hence kinetic en1 mv 2 remains unchanged but momentum ergy 2 p = v will change. 3. The correct choice is (a). The loop will rotate and come to rest when the torque acting on it becomes zero. The magnitude to torque acting on a loop of area A and carrying a current I B is given by = BIA sin where

the magnetic force is perpendicular to the magnetic ing’s left-hand rule) should be along the negative -direction. 5. force is F = E in the direction of E . Since E is parallel to B , the particle velocity v (acquired due to force F ) is parallel to B . Hence B will not exert any force since v B = 0 and the motion of the particle is not affected by B . 6. The correct choice is (c). The radius of the circular path is given by

is the angle between the direction of the =

coil. It is clear that

= 0 when

= 0, i.e. when the

mv = qB

2mK 1 ; where = v2 qB 2

24.50 Comprehensive Physics—JEE Advanced

These forces do not exert any net force on the coil. The force on arms AD and BC are F2 = B I b sin . These forces merely compress the coil and are resisted by its rigidity. These forces also exert no net force on the coil. Thus, both the statements are true but Statement-2 is not the correct explanation for statement-1. 9. F = q(v B). Since F is perpendicular to v, power p = F v = 0. Hence both the statements are true and statement-2 is the correct explanation for Statement-1. 10. The path of the Particle is a helix; it is a circle if

m . q Now, the charge of an alpha particle is twice that of a proton and its mass is four times the mass of a proton, m / q will be the same for both particles. Hence will be the same for both particles. 7. The correct choice is (d). Kinetic energy K = qV. Since and are the same for the two particles, r

Therefore r1 = r1 Hence = r2

2m1qV and r2 = qB m1 m2

m1 = m2

2m2 qV qB r1 r2

2

.

8. A coil ABCD B. Its planes makes an angle with B and it carries a current I as shown in the Fig. 24.90. The forces F1 = B Il on AB and CD are equal and opposite but constitute a couple which tends to rotate the coil.

is true. mv 11. r = qB

2m K , where K qB

1 mv 2 2

Since K and q are the same, r m . Hence the electron will describe a circle of a circle of a smaller radius. So. Statement-1 is false but Statement-2 is true. 12. If v is parallel to B, F = 0. Hence the particle does not experience any force and is, therefore, not with a constant speed. Statement-1 is false and Statement-2 is true. 13. Placing a soft iron core inside the coil makes the

in the coil. The sensitivity increases because, for Fig. 24.90

when a core is inserted the coil. Statement-1 is true and Statement-2 is false.

VI Integer Answer Type 1. A potential difference of 600 V is applied across the plates of a parallel plate capacitor. The separation between the plates is 3 mm. An electron projected vertically, parallel to the plates with a velocity of 2 105 ms-1 in the region between the capacitor plates. Neglect edge effects. IIT, 1981

2. Two long straight parallel wires are 2 m apart, perpendicular to the plane of the paper (Fig. 24.91) Wire A carries a current of 9.6 A directed into the plane of the paper. Wire B carries a current P at a distance of 10/11 m from the wire is zero. Find the magnitude of the current (in ampere) in wire B. 3. A steady current current I goes through a wire loop PQR having shape of a right angle triangle with

Magnetic Effect of Current and Magnetism 24.51

PQ = 3x. PR = 4x and QR = 5x. If the magnitude of 0I , P due to this loop is k 48 x k. IIT, 1990

4. A long circular tube of length 10 m and radius 0.3 m carries a current I along its curved surface as shown. A wire-loop of resistance 0.005 ohm and of radius 0.1 m is placed inside the tube with its axis coinciding with the axis of the tube. The current varies as I = I 0cos(300 t) where I0 is constant. If the magnetic moment of the loop is Nμ0I0sin (300 t), then ‘N’ is IIT, 2011

Fig. 24.92

Fig. 24.91

SOLUTION V 1. E = d

600 3 10

3

=2

5

10 Vm

–1

directed to the right in the plane of the paper. Its magnitude is given by (see Fig. 24.94 on page 24.52) B1 =

24.93] e v B = eE

E B= v

2 105 2 105

= 1T

or Given

Fig. 24.93

2 R1

P B2 due to I2 is equal and opposite to B1. Therefore, the current in wire B should be normal to it and directed out the plane of the paper. Therefore B1 = B2 or

2. Let the currents in A and B be I1 and l2 respectively. Let AP= R1, BP = R2. According to Ampere’s P due to I1 is

0 I1

Thus

0 I1

2 R1

=

I2 =

0 I2

2 R2 I1 R2 2 R1

10 m and 11 10 32 R1 = AB + BP = 2 + m. 11 11 I1 = 9.6 A, R2 = BP =

I2 =

9.6 10/11 32/11

9.6 10 = 3.0 A 32

24.52 Comprehensive Physics—JEE Advanced

25 r 2

1 =

sin

144 x

r=

2

3 and cos 5

=

0I

3 4 r 5 k =7

B(at P) =

=

12 x 5

4 5

7 0I 20 r

4 5

7 0I 48 x

4. inside the tube is B= =

P is (Fig. 24.95) B (at P) = =

0I

4 r 0I

4 r

[sin

L

; L = length of tube

0 I0

cos (300 t ) L

If r through it is

Fig. 24.94

3.

0I

0 I0

= BA =

+ sin 90° – ]

cos (300 t ) r 2 L

Magnitude of induced emf is (sin

+ cos )

|e| =

r2

0 I 0 (300)

d dt

sin (300 t )

L

The induced current through the loop is i=

0 I 0 (300)

e R

r2

sin (300 t )

LR

Magnetic moment of the loop is M=i

Fig. 24.95

Now and

sin cos

r 4x r = 3x

=

These equations give sin2

+ cos2

=

r2

r2

16 x 2

9 x2

r2 =

300

2 4

r RL

0 I0

sin (300t)

Given M = Nμ0I0 sin (300t)

(i)

Comparing (i) and (ii) we get

(ii)

N =

300 2 r 4 RL

300 = 5.92

(3.14) 2 (0.1)4 0.005 10 6

Electromagnetic Induction and A.C. Circuits

25

Electromagnetic Induction and A.C. Circuits

Chapter

REVIEW OF BASIC CONCEPTS 25.1

e

B is

A =B

A = B A cos

where

d dt

Lenz’s Law

MAGNETIC FLUX

Thus d dt

e= – k where k

N turns, the

k = 1 and one

can write e= –

= N B A cos

d dt e

=

25.2

25.1

B dA

N turns

FARADAY’S LAWS OF ELECTROMAGNETIC INDUCTION

The

d dt

e

N

d dt

i=

1 d induced em = total resistance circuit R dt

The direct law.

Faraday’s Law of Electromagnetic Induction

Flow of Induced Charge

-

circuit. q = idt =

q 1 d 1 dt = d R dt R

=

change in lu Resistance

25.2 Comprehensive Physics—JEE Advanced

Heat Dissipation d idt = i d dt = induced current

H = eidt =

Fig. 25.3

Fleming’s Right Hand Rule

I1 and I2

25.3

I1 and I2 will decrease. I1 and I2

EXPRESSION FOR INDUCED EMF

(B). B increases with time, the induced current i is

Applications of Lenz’s Law

B

towards the magnet, the induced current i is

Fig. 25.4

Fig. 25.1

d d BA dt dt B decreases with time, I B dR increases at a rate , then dt

i

R2

e

e Fig. 25.2

B

d dt

R2

B

dB dt

2 R

PQRS

I

or I increases with time. or I decreases with time. I1 and I2

v

Fig. 25.5

dR . dt

v, the

B

-

The current I

-

Electromagnetic Induction and A.C. Circuits

e Blv where l = PS = QR Induced current i R is the resistance i= Force F

e = Blv v

Bl v R

e R

v is F = Bil =

25.3

v

v

v

B 2l 2 v R

Power needed is P = F v =

B 2l 2 v2 R

Fig. 25.7

e = Blv sin R is B as

v NOTE -

ends P and Q 2BvR

e = Bv

R

v

XY sliding on metallic rails PQRS to the right as shown in Fig. Fig. 25.8

PQ

l

B

v

e=

1 B l2 2

Fig. 25.6

Change in

) N turns is rotated , the

A B e = e0 sin where e0 = NBA

25.4

= e0 sin

t

Fig. 25.9

MOTIONAL EMF v

l B as shown in Fig. P and Q Fig. 25.10

P is

25.4 Comprehensive Physics—JEE Advanced

R centre O

25.5

A solen

P

e=

25.2

B as

1 B R2 2

-

ELECTRIC MOTOR SOLUTION I in a solenoid

V

n e

i=

V

B=

R is

V

0nI

e

R Vi and heat loss = i2R = Vi – i2R = ei. ei e = Vi V Some important points about a d.c. motor e and hence current i V V and e= 2 and initial current = V/R

ing the coil is = NAB cos

= NAB cos ° = NAB = N r2 0nI r

= 0.1 m. e=0

I1 = 0 to

I2 = 2 A is = N r2

0n

I2 – I1 0.12

= 100

i

is minimum.

= 0°. N the -

A

2

10 e=

25.1

10

t

, 100 turns and side

=

2

10

0 3

2

EXAMPLE 25.3 -

1 2

t

SOLUTION R=2

T

, N = 100, A = 0.1

0.1 = 10–2 m2,

B = 0.1 T and t

NBA 2 – cos 1 = 100 0.1 10 –2 1 = 0.1 0 2 change in lu 0.0 In e= = 0.0 time e 1 Induced Current i = R 2

t t2 Find the induced current in the coil at t = 10 s. SOLUTION e| =

d d = dt dt

t2

t

10– 3

At t = 10 s, e Induced current at t = 10 s is I=

e = R

t

10

3

= 10–2A

Electromagnetic Induction and A.C. Circuits

25.5

25.4

EX

10 T.

Fig. 25.12

SOLUTION 100

=

3

l = 0. =2 1 2 Bl 2 1 = 2

e=

3

-

2

10

10

3

10

its diameter. Assume that the normal to the = 0 with t = 0. SOLUTION coil changes with time.

Fig. 25.11

the coil does not change with time, hence no

10

coil does not change with time, hence the magnetic

25.5 –1

in the eastward direction at a 10 T. Find the

Fig. 25.13

a diameter, as shown in

SOLUTION –1

v e = Blv

= 300 ms–1 10 A =

25.6 r

N turns is B

r2

Fig. 25.14

then at time t, = t this time is = NBA cos = NBA cos

t

25.6 Comprehensive Physics—JEE Advanced

Q acquires a negad d =– dt dt

e= – =

NBA sin

=

NB

NBA cos

sin

dF = qvB = qv T

t t

25.7 A metal rod PQ

dE = v

CD at distances a and b

Force on the element is

t

r2 sin

= Nr2B

t

P and Q

Now

2 x

qv 0 I Iv dF = = 0 q 2 x 2 xq

dE = –

I as

0I

dV dx

dV = – dE ere dV

dx = –

0Iv dx 2 x

PQ is 0 Iv

V

2

EXAMPLE 25.8 A metal rod PQ

b

dx = x a

l AB and CD

straight wire XY

0 Iv

2

ln

b a

v I as shown in Fig.

R PQ

Fig. 25.15

SOLUTION dx. Consider one such element CD

at distance x

Fig. 25.17

resistance R. on rod PQ

v.

SOLUTION -

Fig. 25.16

I in

wire CD is B=

ACPQ with time.

0I

2 x

The directi v

PQ

P to Q P to Q. Hence end P

Fig. 25.18

Electromagnetic Induction and A.C. Circuits

dx at a distance x

where r is th

SOLUTION

PQ is 0 Ir

d = BdA = Brdx =

2 x

a

= Induecd

e= =

a l

0 Ir

2

a

dx Ir a l = 0 ln x 2 a

0Iv

ln

a l a

v=

dr dt

e=

d dt

v=

i R 2kx Bl

dx = Blv dt R ACPQ = R + 2kx induced in the circuit is e Bl v = i= R 2kx R 2kx

I a l dr d = 0 ln dt 2 a dt

2 ent induced in R is i=

t. 0 Ir dx 2 x

= d =

ACPQ when the rod is at a R is = BA = Blx

distance x

dx

PQ ACPQ is a l

25.7

= Bl

25.10 A metal rod PQ

AB and CD

m l

Iv a l e = 0 ln R 2 R a

R -

width dx

ACPQ dF = Bidx =

0I

2 x PQ is

F F= = = 25.9 A metal rod PQ AB and CD B

dF =

idx

0 Ii

2

a

0I v

0I

2

a l

2 R

ln

v 0I a l ln R 2 a

u

distance x dx x a l

a l a

a

dx x

2

Fig. 25.20

SOLUTION

l k R is connected

F PQ so that it is accelerated

Induced currnet i =

Force F = Bil = – ma = – m

dv dt

The ne m

v PQ when it is at a distance x R.

Bvl R

dv =–B dt =–

Fig. 25.19

B 2 vl 2 R

Bvl R

l

25.8 Comprehensive Physics—JEE Advanced

dv B 2l 2 =– dt = –kdt v mR k=

k dt 0

v u

ln

dqR = d d dq = R

t

dv v u

d dt

dq d R= dt dt

B 2l 2 mR

v

Intergrating

iR =

– kt

Q=

v = e–kt u

R

=

2 NBA 2 NB = R R

25.12 Two circular coils A and B b>a

v = ue–kt dx = ue –kt dt The rod comes to rest when t = Integrating

r2

a and b

-

3b

x= sient current I

B

A is R

A

x

dx = u e 0

kt

dt

0

x=– x=

u e k

kt

0

u k

=–

u k

umR B 2l 2

25.11 r has N turns and a resis-

tance R

Fig. 25.21

B Q

SOLUTION A due to current

I in coil B is SOLUTION A

r

BAB =

2

NBA cos 0°

= NBA

2b

2

0 Ib

2

x

2 3/2

=

Sin

= NBA

NBA

NBA

= BAB Induc

d e| = dt e Induced current is i = R

d dt

1 R

b

x = 3b

A, hence A is

NBA

– NBA

0I

area o e

A d dt

d dt IRdt = d IR =

cos 0° =

A, = 0°. a2

0I

b

Electromagnetic Induction and A.C. Circuits

or

where V

1 d = R R

Idt = Q=

R

a2

0I

=

d d =– BA dt dt d =– kt x2 dt

V= –

bR

25.13 R and mass M

Q

E= B = kt

k is a

kx 2

Force acting on the element is dF = dq

t

E= =

Q R2 kQ 2

R Torque acting on the disc is

t.

R

SOLUTION

=

R2

xdF = 0

Q dx at a distance xdx

Q R2

2 xdx

kQR 2

kx 2

x2dx

x3 dx =

kQR 2

0

is the angu-

lar acceleration I=

dq =

R

kQ R2

2 xdx

= I , where I

R2 x

kx2

2 x = – kx2

–E

constant and t

25.9

1 MR2 and 2

=

1 MR2 2

d =

kQ dt 2M

=

kQ dt 2M 0

=

kQt 2M

=

d . Hence dt

d dt

t

d 0

25.6

i = Mi where M i is changing with time, the

Fig. 25.22

E. Since E= –

dV dl

dV = – V= – E

MUTUAL INDUCTANCE

di dt Expressions of M in some situations l N2 turns. e= – M

Edl = – E 2 x

2 x

N1 wound

25.10 Comprehensive Physics—JEE Advanced

M=

0

R >> r

N1 N 2 A ; A = common cross-sectional area l R and r r2 2R

0

M=

25.7

SELF INDUCTANCE = Li,

i where L di e= L . dt

N turns, cross-

sectional area A and length l 0N

L=

2

A

l

Fig. 25.23

r a and b with a, b >> r Fig. 25.26

M=

2

0

r

2

a ab

2

b

2

U=

1 2 Li . 2

L = L 1 + L2 M

Fig. 25.27

Fig. 25.24

a and b

tance x

L = L1 + L2 + 2M

-

L = L1 + L2 – 2M M=

0

2

a

x

log e

b x

1 L

1 L

1 L2

M

M = k L1 L2 ; k

25.8

Fig. 25.25

GROWTH AND DECAY OF CURRENT IN A D.C L—R CIRCUIT (FIG. 25.28) S1 is closed at t = 0, with switch S2

NOTE =

0

r,

-

0

where

r

t, i = io

e–t/

L = is the time constant. R

Electromagnetic Induction and A.C. Circuits

25.11

SOLUTION = LI

L=

I

=

0 1 10

3

–3

25.15 Fig. 25.28

io = E/R At

t =

t = , i = io 1

1 e

SOLUTION

i 0.

dI = = 20 As–1 dt 0.1

1

e

L

–3

=L

dI dt H

L

25.16 Fig. 25.29

Decay of current: At time t = 0, let i0 = E/R S1 S2 i = i0 e–t/ At

t= , i =

i0 e

SOLUTION

i 0.

Mutual inductance M = 0

=

Fig. 25.30

25.9

M

dI = dt

2

H –

0.01

–2

25.17

ENERGY STORED IN AN INDUCTOR

are connected in series to a

L is increased I U=

l

0. 1.0 –

e

0 AN1 N 2

through a switch. The switch is closed at time t = 0

1 LI2 2

current.

25.14 -

tor at t

e–2

-

25.12 Comprehensive Physics—JEE Advanced

25.19

SOLUTION i = i0 – e t/

i0

i

, i = i0 =

t

L R

Time constant

100

e–t/

i0 = i0 1 2

E = R 100

SOLUTION L = 100 mH = 0.1 H, R 2 E i0 = = R 0.1 L = = = 0.002 s R i = e–t/ i0

– e–t/

e–t/ = 1–

1 1 = 2 2 t

e t/ = 2

1 = e–t/ e–t/ 2 t = ln2 = 0.002

t VL = – L

di d =–L [i0 dt dt 1

= – Li0 Li0

e–t/

25.10

e–t/

,E

ln2 =

t 10–3 s

TRANSFORMER

e–t/ .

e–2

e

primary secondary

25.18

EX

-

SOLUTION i0 =

12 E 3 = = R

1 2 1 Li 0 = 2 2 3 P = i 0E = U=

or

P = i 02 R =

10

3

2 s p

=

Ns or Np

s

=

Ns Np

p

where Ns 3

Np

2

ep

d p/dt

es

d s/dt

Electromagnetic Induction and A.C. Circuits

Thus d Ns d t Np

es = or

SOLUTION Pi = epIp =

Ns d p Np d t

p

Ip =

Ns ep Np

es =

Ns > Np, then es > ep N s < N p, step-down

step-up

then es < ep

=

and es

, ep =100

Pi = 100 ep Po Pi

Po =

Is =

Po 3000 = es 200

=

25.13

Pi =

25.22

I2R and that

SOLUTION Np Ns = 2000, Po es Is = ep Ip

or

Ip

es = ep Is

Ns Np

ep

Np

es

Ns

=

25.20

cienc

ep =

Po Pi

=

es =

Ns

1000

2000

Po ep I p 12000 200 = A . 200 3

Ip =

Po = ep

Is =

Po 12000 = = 12 A es 1000 2

Pow Ip Pi ep Ip = 220 Po es Is = 10

Np

= 200

where Ip and Is

SOLUTION ep es

es

Rp =

Ip

Is

2

Is

Po = 1100 Pi

Rs

200 3

2

2

25.23

25.21 -

25.14 Comprehensive Physics—JEE Advanced

SOLUTION Ns =

es ep

Is =

Po Pi = = es es

Np =

1000 = 20,000

220 .

= 1.2 A

Simila I = I0

t

= where

2 T

=

Vo2 1 T 02

=

V02 1 dt T 20

=

V02 T T 2

1 2

2 t 2

=

V02 T T 2

0

V02 2

t dt T

1 20

2 t dt T 0

Roo

=2 –1

T

V0

V2 =

Vrms =

T I = I0 sin

t dt

T

as

reads

1 V02 T T

t

sinusoidally varying current whose magnitude changes continuously with time and whose direction reverses periodically is called an alternating current T

=0

I =0

V2 =

V = V0 sin t resistance R, the current I in the circuit is V V I= = 0 sin t = I0 sin R R where I0 = V0 /R, is the maximum or peak value I

0

t

V = V0

ALTERNATING CURRENT

T

V0 2

=

Is is called th 25.11

V0 2 t cos T T

=–

Simila ing current I = I0

2 t T

Irms =

Root Mean Square Voltage and Current

2

I0

I2 =

2

1 2

=

V -

t 1 2

=

I

X t T

EXA

25.24 V

T

X t dt X =

0

T

=

T

dt

t

1 X t dt T0

V

t

0

V = V0

V0 sin

t

SOLUTION

2 t T

V = V0

T

1 V = V0 T0 V0 sin T 0

V0

t dt

Vrms =

T

=

t dt = –

V0 co T

t

t

T 0

and or

2

V0 2

=

100 2

Electromagnetic Induction and A.C. Circuits

25.25 A 100

mum.

Z=R

25.15

-

Special Cases (a) A.C. circuit containing only a pure resistor (Fig. 25.32) SOLUTION

Vrms V0 =

2 Vrms =

Irms = I0 =

25.12

2

200 Vrms = =2A 100 R Fig. 25.32

V0 . = 100 R

SERIES LCR CIRCUIT V VR

C

VC

VL

L

R

VR = V0 sin

t

I = I0 sin

t

V0 R e across R current in the circuit. (b) A.C. circuit containing only an ideal inductor (Fig. 25.33) where

I0 =

Fig. 25.31

V = VR2

Z=

R

VL VC

2

XL

2

XC

2

=

R

2

L

The current in the circuit is t– I = I0 tan

=

L

1 , i.e. C

L <

1 , i.e. C

1 L= , i.e. C

Fig. 25.33

2

VL = V0 sin I = I0 sin

1 C

L>

1 C

where 1 , then tan LC

>

< =

1 LC 1 , then LC

I0 =

t t

2

V0 V = 0 XL L

XL = L is call across the inductance leads the current in the circuit /2. (c) A.C. circuit containing only an ideal capacitor (Fig. 25. 34) VC = V0 sin

=0

I = I0 sin

t t

2

25.16 Comprehensive Physics—JEE Advanced

where I0 =

V0 = XC

is cal

edance. I = I0

CV0

t+

1

where = tan–1

R C

Fig. 25.34

XC = age a

C

i

Fig. 25.36

/2. (d) A.C. circuit containing an ideal inductor and a pure resistor (Fig. 25.35) V0 = I0 Z V0 =

VR2 VL2

25.13

POWER IN LCR CIRCUIT

In a series LCR V = V0 sin t, the current in the circuit is t± I = I0 XL < XC or XL > XC.

VR = IR, VL = IXL V and Z = 0 = I0 nce.

R

2

X L2

=

R

2

L

2

I0 =

is called and

Vo ; Z = R2 Z

= tan

Insta source is

2

R

VI = V0 sin

Pt

1 C

1 C

L

–1

L

= V0 I0 sin

t

I0

t t+

t

T

Fig. 25.35

I = I0

t–

L is t where = tan R nd current in the circuit. (e) A.C. circuit containing an ideal capacitor and a pure resistor ( Fig. 25.36) V0 = I0 Z –1

where

V0 = VR2 VC2 VR = IR, VC = IXC

V and Z = 0 = I0

R2

1 P t dt T0

P

X C2 =

R2

1 C

2

= =

1 T

T

sin t

V0 I0

V0 I 0 cos T

t cos

cos

t sin

dt

0 T

T

sin 2

t dt

sin f sin

0

t

0

t dt VI = 0 0 cos T

T 2

0

V0 V0 I 0 cos = 2 2 or P = Vrms Irms cos =

I0 2

cos

Electromagnetic Induction and A.C. Circuits

Power Factor of an A.C. Circuit on Vrms and Irms

is

tan and

cos

=

V0 R

I0

I0 = R2 1/2

1 tan 2

I0

R

R

ma

2 1/2

1 C

L

1 Figure 2

.

I0

1 1 C

L

1

2

R R

= R2

L

1 C

2 1/2

R Z

Fig. 25.37

resis tance im e

P

and i.e. when 1

Special Cases For an A.C. circuit containing only a resistor, R Z = R and cos = = 1 = 0 and R P =V I

when I0 =

2

I0 R2

ma

R 1 C

L

rms rms

For an A.C. circuit containing only an inductor or a capacitor, P =0 LCR circuit. Hence P =

R2 +

2

R – L

2=

Bandwidth and Quality Factor of LCR Circuit For an LCR V = V0 sin t

2 r

2

–

1 R = LC L

= 0, where

Case 2: L –

R + 1 2L

r

=

1 LC

V0 2

L

1 C

2 1/2

2 2 1/2 rL 2

R

1 = –R C 2

R

2

= R2

1 =+R C –

R

Th

P = Vrms Irms

V0 Z

2

ma

1 =±R C

L–

Wattless Current

I0

=

2 1/2

1 C

L

Case 1: L –

I0 =

V0

, I0

I0

R

=

=

In ter

1 C

L

=

=

I0

25.17

+

R – L

R + 1 1 = – 2L

2 r

=0 2 2 1/2 rL 2

R

I0

ma

2

,

25.18 Comprehensive Physics—JEE Advanced

= Qualit

2

–

1

R L

=

Q Q=

qu width

1

=

t1 =

LCR

resonant

=

r

=

t1 =

2

2 t–

I

1 L L = R R C

LC

t=+1

V

t2 =

2

2

t2 –

– -

Q i t = t1 – t2 =

Q.

P

Now tan

–

=

100 L = 100 R . °=

=

t=

rad

.

=

= 3.2

100

10–3 s

25.27

Fig. 25.38

Q. The resonance Q, the

25.26 -

tance 100

SOLUTION Vrms L

tan

V0 2

R L = R

2 Vrms =

2 2 1/2

,

L

100

100

2 1/2

, tan

=

1 CR

1/2 2

120 10

t=+1

V

t1 =

2

2 t+

t2 + .

, Vrms =

–1

2

I is

2 2

1/2

1

2

t1 =

2

F, R

2 110

I0 =

t–

1 C2

R2

rad s–1,

=2 =2 R = 100 . V = V0 sin t

I0 =

I0 =

V0

I0 =

I = I0

V0 =

SOLUTION C = 100 F = 100 10 =2 =2 V = V0 sin t t+ I = I0

=

2

t2 =

2

– -

Electromagnetic Induction and A.C. Circuits

t = t1 – t2 = Now tan

VR

1 = CR 120

=

1

33.

=

VL,C

=

33.

10–3 s

120

rms

230

R=

= Irms

1 rC

rL

= Vrms

Irms

= Vrms

Irms

=

25.28 A series LCR circuit with L R a.c. source.

= Irms

=

rad P

t=

rms

25.19

F and

C

cos

2 230 230 Vrms = R

25.29 A A and B. L, C

and R at resonance.

A and B. SOLUTION

L and C at resonance.

A is a resistor and B Irms Vrms

-

=

r

=

1

r

2

1

=

LC

Z= R

=

L

VL

rms

Im is 1 C

VC

rms

R2

Irms =

1 L= C

25.14

X C2 =

2

2

Vrms 220 = Z .3

LC OSCILLATIONS L and

XL = Irms

C

rL

230

= Irms =

Z=

2 1/2

2 230

= Irms =

220 Vrms = I rms . A and B are connected in series

Reactance o B is XC =

2

2

V0 = Z

–1

0 10

=R I0 =

220 Vrms = I rms .

A is R =

SOLUTION r

B, again the -

230

XC = Irms 1 10

1 LC

1 rC

and tim =

q = q0

T= 2 t+

LC I = I0

t+

25.20 Comprehensive Physics—JEE Advanced

I Multiple Choice Questions with Only One Choice Correct 1. to

v v1/2 v2

v v3/2

t = t1 t1 and t2.

g less than g g

B

t = t2

PQR

a is at B at t = 0 as

v and its edge R t = t0

-

4.

induced current i with time t

Fig. 25.39

2.

Fig. 25.41 (a)

the other coil the current will decrease. 3. t = t2

t = t1

Fig. 25.41 (b)

5. Fig. 25.40

g g

t t

PQ and RS CD CD

v as

Electromagnetic Induction and A.C. Circuits

25.21

9. v

Fig. 25.42

6.

-

I. The centre O x >> a

v

I = 3 sin

2

2

2

10. A series LCR tance L tance R

A A L, C and R, the

C

L C R R or L or C.

v

1 x 1 x

t

2

x x2

t

3

a tance x

t as

11.

2

Fig. 25.43

7. as shown. The ratio I1 / I2 L1 L2

t is L2 L1

L1 L2

12. In a series LCR

L2 L1

-

2 13. Fig. 25.44

V = 200 2

I2 = 3A in the

8.

where V the ammeter is

resistor is

14.

Fig. 25.45

t+ t

25.22 Comprehensive Physics—JEE Advanced

F F -

15. line is 1

16. at t = 0

t

I in the circuit shown K

Fig. 25.48

18.

19.

f is connected

to an LCR

I in f

Fig. 25.46

Fig. 25.49

20. Fig. 25.47

17. V across the inductor L with time t K t = 0, in the circuit shown in 21.

-

LC 2

22. L, C and R

= 10.

Electromagnetic Induction and A.C. Circuits

28. RC L/R

25.23

inductance and a resistance connected in series

LC C/L

23. V = 200 2 is connected to 1

t

29.

V = V0sin t a circuit. As a result a current I = I0

t– V 0I 0

24.

V 0I 0

30.

V 0I 0 L and resistance v B,

R

A and B is

Fig. 25.50

25.

V and

current I V and

t

I = 100 sin 100 t

3

mA Fig. 25.51

BLv/R BLv/R BLv/R

26.

31.

27.

A

B. A

connected across the ends such that the total resisR BR A

AB R B2 A

ABR

R2

32. B

R is MNQ the

25.24 Comprehensive Physics—JEE Advanced

across the ring is

P at a distance r region

v

1 Bv R2 and M 2 RBv and Q RBv and Q

1 r 1

. r2 IIT, 2000

r

Fig. 25.53 Fig. 25.52

37.

33.

l

B r

L L >> l

-

R

34.

2

l L

l L

L l

L2 l

in the x-y

r2 B 2R

I, lies

R

38.

r B 2R

r2 B R 2

r

2

B

2

2

R –1

x-y is 0.2 × 10

I R R

35.

is

36.

Bt a and is directed

39.

F is charged to a

40.

r B.

Electromagnetic Induction and A.C. Circuits

25.25

t Br 2 1 10 t Br 2 1 t

Br 2 1 t Br 2 1 t l, mass m and resistance

41. R

Fig. 25.55

IIT, 2000

height h 45. m 2 gR 2

mgR 2 Bl mgR

2B l 2

m 2 gR 2 is induced AD BC

2B l

42.

r1 and r2 r22

0

2r1

0

v

2 2

B 3l 3

0

ABCD

r1 > r2

BC

r12

2r2

r1 r2 2r1

2

0

r1 r2 2r2

2

v in B = 2T as shown in Fig.

43. tance 1

BC AD AD nor in

Fig. 25.56

AD and BC IIT, 2001 P and Q

46.

-

S Ip IQ1 are:

P Q S

E IQ1 and IQ2

IQ2

Q. E

equal to 3 –1

–1

–1

–1

Fig. 25.57

Fig. 25.54

44. Two IIT, 2002

25.26 Comprehensive Physics—JEE Advanced

47.

51. An LCR series circuit with R = 100

2

E = E0

48.

IIT, 2002 t

is connected to

3 A 2

3 52. v

E and the current I in the circuit

current I its ends at distances r1 and r2

R-C or R-L or L-C

r2 > r1 0 Iv

R R R R

, , , ,

2

C = 10 F C=1 F L = 10 H L=1H

0 Iv

2 53. IIT, 2003

loge

r1 r2

0 Iv

r2 r1

log e 1

increasing according to the equation I

r1 r2 t2

54. s. The Fig. 25.58

49. An air ms–1 is 2

a and b

55. B = kx where k

10 1 10

2 10

3 10

1

coil as shown in Fig.

kab2 1 kab2 2

L and resistance R is con-

50. 1 L loge R

L loge R

L loge R

L loge R

-

56.

2 kab2

Fig. 25.59

x as

Electromagnetic Induction and A.C. Circuits

57.

58. -

resented as

along the x

25.27

x x x x direction

59.

I1 and I2 are the currents in the segments AB and CD. Then, I1 > I2 I1 < I2 I1 is in the direction BA and I2 is in the direction CD I1 is in the direction AB and I2 is in the direction DC

Fig 25.61 Fig. 25.60

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49. 55.

2. 8. 14. 20. 26. 32. 38. 44. 50. 6.

3. 9. 15. 21. 27. 33. 39. 45. 51. 57.

4. 10. 16. 22. 28. 34. 40. 46. 52. 58.

5. 11. 17. 3. 29. 35. 41. 47. 53. 59.

6. 12. 18. 24. 30. 36. 42. 48. 54.

SOLUTIONS 1.

l e

Bl v

R

i=

Bl v R

25.28 Comprehensive Physics—JEE Advanced

F = Bil

B2 l 2 v = R Power needed is P = F v =

i=

B 2 l 2 v2 . So the R

Bva R

i against t

Bva R

-

2.

B v2 a t Rb

B v2 a . Rb

5. As the rod CD in each coil will decrease.

C to D C to D and, as a result, end C

3.

CD D

t1 and t2 no current is ing g. 4.

t

right a distance CD = v t

charge. 6. = BA = B a I Now B = 0 2 x

RD = b.

=

I a2 2 x

=

Fig. 25.62

7.

Now

across L2

RC CT = RD QD

I a 2 dx d = 0 2 dt dt 2 x

vt

b b

=

Bv

2

x

2

1 x2 L1 t. Hence.

Integrating, we get L1 I1 = L2 I2

CT =

e

I a2 v

dI1 dI 2 = L2 dt dt L1 dI1 = L2 dI2

CT a/2

a b vt 2b a ST = 2CT = b

0

L1

RD CD CT = RD QD

8.

I1 L = 2. I2 L1

b – vt

ST = Bv a

a vt b

R i=

e Bv = a R R

a vt b

A = a2

0

e=

Hence e

2

Fig. 25.63

v

dx dt

Electromagnetic Induction and A.C. Circuits

I2 tance = 3

I1 =

resis-

12

VL = VC

= 2 A. There-

VR2

V=

I = I1 + I2

VL

2

VC

= VR

Current in the circuit is I=

9. 2

I0 = Irms =

25.29

I0 2

2R

2

Hence =

2

VL = IXL =

A 1 > C

XC > XL or

10. L R = Z

R

Irms = Vrms

2

=

L

C

100

2

10 -

tor, i.e. 1 1 LI2 CV 2 = LI 2 or C = 2 2 V2

1 1

200 2

14.

2

1 C

= 100 rad s–1.

V0 = 200 2

1/ 2

=

XL = R

R

13.

2

1 C

R2

2R 2

1/ 2

=

V = Z

L

=

1

2

2

10

2

R

F

15. Current in the transmission line is I=

R, C and L are 11.

nance, i.e. when Z = R, where R Vrms R=

I= 2= 12.

E R 2

r

LR or

R = XL = XC XC = 0 and the X L2 =

–R t /L

I = I0

e

I = I0

e –t /

where = L/R is the time constant. At t = 0, I = 0. At t >> t, I = I0.

r = 1.0

R2

2

16.

;r

VR = VL = VC

= I2R

= 100 A

10000

Vrms 12 = =2 I rms

r

Z=

Power loss

-

=

ltage

R2

R2 = 2 R R = XL

17. V=–L

dI d = – LI0 dt dt = – LI0

R –Rt /L e L

e –Rt/ L

25.30 Comprehensive Physics—JEE Advanced

V0 e –t / , where V0 = I0R is the initial V = V0 at t t >> . Hence

V

22.

L R 100 tan = 2 1000 =

18. R tan

L=

10

ltage current

RC 23.

2

=2 cL

V0 = 200

L

I=

. Thus I increases with increase in = c 1

=

cC

or

3

LC > c, I decreases

25.

V0

I P=

1 = or e t/ = 2 or t = 2

=

=

2

=

LC circuit is 1

21. 2

LC

2

100

1

10

10 –3 A = 20 mA

10 –3

=

-

3

V0 I 0 cos 2 100 100 10 2

3

27.

L 10

For

10

H.

due to which it contracts. As a result the contact is

3

1 103 10

2

current is induced in the coil. 2

where L

2

200 2

1 C

26.

1

C=

C=

I0 = 100 mA = 100 I and V is

I0 / 2 is

Zc

24.

I = I0 e –t/

or

C=

= 20

10 –3 s

=2

Current at time t

e –t/

2

C

inductance L 1 1 1 1 3 3 = + + = = =1 L L L L L 3 or L

= 2 millisecond.

where I0

V0

=

20. L 100 10 = R

= 100 rad s –1

2

Vrms = Vrms Zc

1

=

c

t t,

2 1/ 2

1 C

when I with increase in =

V = 200 2 equation with V = V0 sin

V R

where

charge charge = = time ltage charge/time

–3

19. The current in an LCR I=

charge /

RC

2

10 28.

–12

C

wire oscillates. the circuit is

Electromagnetic Induction and A.C. Circuits

V R2 + V 2L

V

1/2

l l << L

1/2

V and current I is

29.

0

I l2

L

Mutual inductance 12

M12 = no current is induced. 31. Induced charge q =

i

= BA

change resistance

q = BA/R.

i

R

l2 L

l2 L

i.e. M12

.

0

34.

x-y

Rv

d d =– dt dt

BA

B

dA = 2RBv. dt -

rent I B 0

=

f

2 2

=

f

tion. Hence Q

=

=

I

v the decrease in

area with time is dA dt e=–

33.

2 2

2 12 = Bl =

30.

32.

25.31

2

0

I L/2

+ cos Fig. 25.65

I

35.

L 2 2

0I

L

I= =

V R L = R

=

1.0 = or or – or 36. Fig. 25.64

12

e –t

=1–

t = loge . ms

e –t/

= L/R.

. mH

e –t 1 1 = 2 2 1 2

t -

25.32 Comprehensive Physics—JEE Advanced

distance r radius a

39. B

cos

a2 d B E= 2r d t At r = a, E a dB/dt

t=

Q Q0

10–3 H and C = 2

P=

-

V02 sin 2 R

10

t

40.

r2B 2

e= =

B L2 2 /

=

1 2

rad s–1.

sin 10 2 3

t

sin

3

ld change in area time

2

A = r2 –

R

tim

F. Using

2 3

2 r

2

L2

BA = B 2

.

B A t

e= L2

12

ma

e

t>

1 2R

r B

3

t

t = CV0

r B

2

L A=

sin

Q0

2

1 V02 2 R

=

I=

I

circular coil. Hence

2

= V0 < sin2 R

=

dQ d Q0 cos dt dt = – Q0 sin t

Now

r2

V2 R

t=

10 10

=

t and

1 V0 = 2

1 or 2

LC

L

V0 = AB 1 A= 2

12

1

=

37. t V = V0 sin

V V0

t. Also

Now E r > a, E

decreases as 1/r

38.

Q = CV and Q = Q0 cos

Q0 = CV0.

41.

v=

t

1 B L2 2

. There-

= r2 1

1

e = Blv = Bl 2gh .

Bl 2 gh R

Force F = BIl = 2

2

2gh

I=

n

r2

B

r 2

r, r

B 2l 2 2 gh R

equals mg B 2l 2 2 gh = mg R

Electromagnetic Induction and A.C. Circuits

h=

t = 0. The current t0

46.

m 2 gR 2

P

Ip

2B l

25.33

Ip Q IQ

42. B=

0I

1

-

2r1

where I the inner coil is

IQ r 22 =

=B = MI M=

43.

0I

S

r22

2r1

IQ r22

0

Induced current I = I=1 Hence

P and

R=

. Total resistance R = 3 + 1 = e = Blv = 2 0.1 v =

0.2 v.

0.2v

e R

.

n l r

10–3 =

v=2

0.2v

–2

–1

l

or R

r2

Power P =

V2 R

P2 = P1

n2 n1

10 ms = 2 cm s , which is

m = r 21 l1 d = or

we get Fig. 25.66

is changing with time. Hence the correct choice is

n2 l /r2 2

nr l

or P r2 r1

48.

r 22 l2 d

l1 = l2

r2 r1

P2 = P1

n2 n1

n2 n1 P2 P1

AD and BC,

. Here

2

2

l1 l2

l1 and radius r1 is stretched to a length l2 such that its radius reduced to r2, then

–1

44. Th

45.

2

10–3 A 1

2

i.e. V

PQRS

should IP

IP

47.

2r1

R

1

E = E0 E = E0 sin t

2

we get 2

r2 r1 r2 r1

2

d

1 2

1 2

= 1, which is t

= 100 rad s–1

-

R-C R-L circuit. Hence the circuit does not contain an

25.34 Comprehensive Physics—JEE Advanced

For R-C I

or

E and

tan

=

L or R

= 100 rad s–1. Using these

=

tan tan

L loge R

51.

1 RC

=

t=

=

1 100 RC

or

1 100

RC =

L = R tan

R and C tan

49. Now BV = BH tan

10

=2 3

–2

10

BV lv = 2 3

e

1 RC

1 = R tan C LCR

BV =2

=

10

1 C

Z=

R2

L

=

R2

100 an

20

3 10

2

100 ta

2

= R = 100

50.

LR I = I0 I – e–Rt/L

I=

where I0 at time t is

52. U=

1 2 LI 2 -

U= U0 =

U0

r I in the wire at a

where

1 LI 02 2

1 2 1 1 2 LI = LI 0 2 2 I0 = I0 2

or

dr distance r

when

i.e.

V 200 = R 100

e–Rt/L =

1 2

or I =

e–Rt/L or

–

I0 2

1 = 1 – e–Rt/L 2 Rt 1 = loge L 2

Fig. 25.67

Electromagnetic Induction and A.C. Circuits b

I 2 r 0

B=

= ka 0

dr is

0

or 53. W =

Iv

2 0

e=

r1

Iv

2

Iv dr 2 r dr = r

0

Iv

2

1 2

log e r

r2 r1

r2 r1

loge

1 L I 02 , where I0 2 W=

I = 2 A. Thus

Fig. 25.68

2

2.0

56.

1 2 1 LI 0 = CV2 2 2 joule heating as R

I t

54. e = – M or

r2

M=–

I0 = V e I =– I

000

10 10 57.

= 0.1 H 55.

x width dx and length a at a distance x

Bd A =

C L

2 10

e = – d /dt

0

3

=1A

-

58. O, as shown I1 and

59.

b

=

1 kab2 2

0

de = Bvdr =

e=

xd x =

25.35

I2

kx adx 0

II Multiple Choice Questions with One or More Choices Correct 1.

-

25.36 Comprehensive Physics—JEE Advanced

2.

V

ing.

t

2 6.

current.

is 2 2 A. 7. F and R

C

.

LCR circuit connectL

.

3.

Fig. 25.69

8. L, C and R 1 RC

R L

4. Two circuits A and B are connected to identical dc A tance L1 = 10 H and circuit B L2

R

1

LC

LC

9.

A and B is 1. A and B A and B

Fig. 25.70

A and B 5.

t

At a certain instant the current I 3 As–1. At that instant, the

Electromagnetic Induction and A.C. Circuits

L L

A and B A and B

25.37

14.

10.

I t I

t Fig. 25.71

15. t=1s

0.2 I P. Then

t=2s 11.

L and resistance R V

16. and the I. The sole-

is now is I . Then =2 I =2I 12. A reatangular coil 20 cm

I = 100 A P A1L and B1M R B

I P tance w wire CD

-

v. Neglect the

= I

I 10–3

Fig. 25.72

2 Bwv . R

Bv 2 . R CD tends to

R. the wire CD

Bwv 2/R. IIT, 2001

13.

10

17. An LR 100 L= mH and a resistance R = 10 . A sinusoidal V = V0 Vrms state is 20 A.

t

25.38 Comprehensive Physics—JEE Advanced

/2. t = 0, the current in the circuit is – 10 2 A. T t = , T , ..., where T = 0.02 s. 18. A metal rod PQ v

20. Two metallic rings A and B, identical in

A

and

B,

identical solenoids as

I noids in identical manner, the rings A and B hA > hB. The to heights hA and hB

B

their masses mA and mB A > B and mA = A < B and mA = A > B and mA > A < B and mA <

the rod

Fig. 25.74

Fig. 25.73

21. A series R-C

P to Q.

mB mB mB mB C IR through the resistor and

VC

19.

I RA

IR

VCA VC

2

I RA

IR

VCA VC IIT, 2011

ANSWERS AND SOLUTIONS 1.

v region.

B. F=e v

2.

B v

B and v B

-

-

when an iron core is inserted in the coil.

tance L is inserted in the coil the inductance L increases due creases, causing a decrease in current in the circuit. As

-

-

3. -

Electromagnetic Induction and A.C. Circuits

25.39

7. C C is decreases, the reactance will increase and as a re-

r

1

=

LC 1

= r

4. E

L1 = 10 H, L2 = 10 10–3 H and R I0 = E/R

the current I0 is E= For circuit 1, E1 =

=

r

2

=

2

. Z= R

12

E R

I0 =

=

2

L

1 C

1 2 LI 0 2

I0 =

1 1 L1I 02 = 2 2

=

10–1 10–3

10

V0 V = 0 = Z R

2

10

2 Vrms R 2

2

Irms =

Vrms 200 = R L is

= Irms

L

r

I0 is I20 R. Since I0 and

C is

R

V

1

t

V0 with V = V0 sin t = 100 rad s–1 or 2 = 100 or V 200 = 100 2 Also Vrms = 0 = 2 2

10 R is = Irms 8.

2 100 2

1 rC

= Irms

2

P = I02R

=

200

The rms current in the circuit is

10

E1 = 1000 E2

6. Vrms

2 1/ 2

= r = 1/ LC L = 1/ C Z=R

For circuit 2, 1 1 E2 = L2 I02 = 2 2

5.

R

1 C =2 .

L and resistance, where

2 A. 9.

Irms =

–1

/2

10

2 2 2

= 2 A. VA – I R + E – L

dI = VB dt

2 A.

25.40 Comprehensive Physics—JEE Advanced

or

VB – VA = – IR + E – L

dI dt

dI Since I is decreasing with t, dt Hence 10–3

VB – VA

I

Ip =

ep

Is =

3

= 10. e

L

dI dt

=

10 000 200

es 1000

14. 10

d dt

–3

10

t 10–3

Power P = eI

t –3

Power at t

10

Power at t

10–3

-

15.

L R

11. Time constant =

–3

I=

V R

I=

L

220 210 0.2

R CD is made to slide on wires A1L and

16. L =

L

and

R =

12. A = 20 cm N = 100, B

R

=

L L/ = R R/

I =

V V = R R

The new time constant

B 1M

10 cm = 200 cm2 = 200 10–3

L = . R

=

=

m2. 1200

rad s–1. The

=2

d d = dt dt

dA dt dx is time dt, then

BA

CD A = wdx

I.

10

e

w = CD d wdx e B dt The induced current is e B wv = I= R R

B

Bw

dx = Bwv dt

CD. e = e0 sin where E0 = NA B and

13.

es =

Ns Np

I

= angle which the normal

ep =

2000

CD the wire CD w F = BIw = B

200

CD B wv R

distance dx in time dt is dW = Fdx

w=

B 2 w2 v R

CD through a small

Electromagnetic Induction and A.C. Circuits

P -

dW dx =F = Fv dt dt

P=

the direction Q to P. Thus the correct choices are 19. From

B 2 w2 v 2 P= R reactance XL =

17.

100

=2

3

10

= LI e

L= 2

25.41

L

= 10

-

L dI dt

L

From U =

L is

1 2 LI 2

L is

2

ance is Z

XL2 + R2

1/2

V0 Z

Vrms

I0 =

2

+ 102

2

=

= 10 2

200

2

= 20 A

10 2

Z

= tan–1

1/2

XL R

tan

1

10 10

hA > hB A is greater than that on B. Hence, current induced in A is greater than that in B, i.e. e

tan

1

.

or

I = 20 sin

where T = At

1

=

2

t = 0, I = 20 sin

21. Case A: I RA =

t VCA =

=

T

,

,

Q to P. As a result a current P to Q. Also end Q

VC =

2

V R C

2

I

C. Hence V R2

T

1 C

R2

I RA = C

IR =

10 2 A.

V

V = ZA

Case B: CB = KC

= 0.02 s.

I = 0 at t = 18.

RB > RA

A . Hence A < B. Mass mA l than, equal to or greater than mB. Hence the correct

2 t T 1

e RB

Now R =

t I0 = 20 A and = t as

I = I0 sin

e RA

IA > IB

In an LR V = V0 current I

20. Since the rings are identical and the solenoids are B = 0nI and hence mag=B A A

IR = C

1 C

2

V R C

I RA < I R and VCA > VC

2

25.42 Comprehensive Physics—JEE Advanced

III Multiple Choice Questions Based on Passage

The Alternating Current Genetator -

E = E0 sin

t

R is connected across the terminals, a

current I

E E = 0 sin t = I0 sin t R R where I0 = E0/R. Such a current is called a.c. or alternating current. 1. A N B through the coil is NAB I=

4. In an a.c. generator

is used.

5.

1 NAB 2 1

Fig. 25.75

E = 100 sin 100 t

3 t in seconds.

where E

NAB

2

2.

t t

3.

E with time t

SOLUTION 1.

= NAB cos where

Electromagnetic Induction and A.C. Circuits

E = E0

2.

3. 4. 5.

E = 0. Since

E = 0 at t

=

t+ E = E0 = 100 or 2

t,

t = 0, E = E0 sin

3

= 100 or , i.e.

magnet is used. . 6. B vd

d R

MN

m

2 x R

B vd

B vd

1 B vd 2

B vd R

B vd 2 x

F

2 B vd R 2 x

MN through R.

2 x

B vd R 2 x

8. B2d 2 2 x 1 2 m R B2d 2 1 R

R 2 x

R B2d 2 log e 1 2 x 2 m B2d 2 2 x log e 1 2 m R

Fig. 25.76

SOLUTION R to MN x MN and R is xd and the magnetic Bxd

6.

e where v 7.

R

7.

induction B

d Bxd dt

Bd

dx = Bvd dt

MN.

F = IBd = m or

R and MN

is R + 2 x I=

e Bvd = R 2 x R 2 x

8. Force acting on the rod,

2 2 dv dx = B d dt R 2 x dt

dv =

B2d 2 m

dv =

B2d 2 m

v 0

or

B2d 2 v R 2 x

v=

2 2

dx R 2 x x 0

dx R 2 x

B d R 2 x loge 2 m R

25.43

=

3

=

25.44 Comprehensive Physics—JEE Advanced

9.

m

downward direction is

B 2 L2 v mR

g L. A conducting massless rod R

g m, tied to the other

B 2 L2 v mR

B 2 L2 v mR

g

10.

B

g

gR mg R BL

mg R B 2 L2

11.

m g 2

g g 3

g

Fig. 25.77

SOLUTION 9.

x-direction. Since the rod is

v x

T i.e. T = F = BIL. a downward direction, then

B y induced in the rod is e = B Lv induced current is e BLv = I= R R L

or

F m

a =g –

F = BIL a =g –

Bl L B BL2 v =g– m mR =g –

10.

Fig. 25.78

B 2 L2 v mR

a = 0. Putting a = 0 and v = v t 0= g –

11.

m m = mg – T = mg – F

ma I in magnetic

x-direction,

v t when

B 2 L2 vt mgR or v t = 2 2 mR B L

Electromagnetic Induction and A.C. Circuits

v=

a=g–

vt mgR = , 2 2 B 2 L2

=g–

B 2 L2 2mR

mgR B L

B 2 L2 vt / 2 mR 0M

0M

M v in the x-direction. a and y-z x

centre at x

2

0M

14.

0M

a x=2a 3

a 0M v 2

a

x on

12.

1

0M v 2

3 32

0M v 2

1

0M v 2

a

0M

0M 2 x

15.

2 x2

0M 3

a

a

x=2a 3

3

0Mv

0Mv

32 R

0M

2 x 13.

g g = 2 2

=g–

2 2

25.45

2 x

R 3

0Mv

0Mv

2R

x=2a

R

SOLUTION 12.

distance x

l

14.

x=2a

M B=

2

Mx

0

Since x >> l

e=

x2 B=

l2

d dt

dx d dt dx

2 0

=–

0M 3

2 x

v

Ma 2 v d 1 2 d x x3

d dx 3 2

0Ma

15.

Fig. 25.79

0M 3

2 x

0Ma

2

v

x R I a2

M0 = I

13. Due to B = BA = B a2

3 2

e R

a2

0Ma 3

2x

=

2

Putting x = 2 a

3 2

0Ma

x R

v

x

3 0 Mv , which is 32 a 2

Putting x = 2 a, we get e =

I=

2

v

25.46 Comprehensive Physics—JEE Advanced 0 IR

x=

2I and I2 = I 16.

R and r = R/100 are I1 =

3 R

0 IR

0

due to current I1 = 2 I 0I R

2

3R

R

2I M and m

19.

0I

10

2 I

I 0I R

10

IR

0 IR

10

18.

0I

17.

3

10

M/m

is

is

10 102

2

SOLUTION 16.

=B

due to current I1

r2

=B

R 100

=B = Fig. 25.80

B=

0 I1 R 2 2 3/2

B=

M=

x

Putting I1 = 2 I and x =

19.

3 R

0I

x, the magnetic and

B.

20. An LCR series circuit with 100

10

R

0 IR

=

10

resistance is connected

I2

=

I

magnetic moment = current

R

17. Since r

R2

0I

18.

2

2R

2

M = I1

R2 = 2

I R2

m = I2

r2 = I

R 100

I1 = 2I 2

=

LCR circuit is

21. The current in the circuit is 2 2 2 22.

IR 2 10

2 2

Electromagnetic Induction and A.C. Circuits

25.47

SOLUTION 20.

R2

Z= tan

R2

=

L or L = R tan = R

100 tan

1 C

2

100 tan

2

21. I=

1 RC

=

1 C

= R = 100 -

tan

L

V R

200 =2A 100

22.

= R tan

P = I 2R

LCR 23.

wheel is

QR 2

QR2 radius R

2

QR 2 2

Q

QR2

24.

.

x x y y

IIT, 2003

25. 1 QR2 B 2 QR2 B

1 QRL B 2 BRL B

26. x y z z

Fig. 25.81

SOLUTION 23.

T x Q I= T m=I

24.

R2 =

Q 2

Q 2

25. Torque R2 =

=

B = mi

Bj

= mB k

1 QR2 2

= 26.

1 B Q R2 2

k

25.48 Comprehensive Physics—JEE Advanced

q 0. An LCR

t

q = q0 cos

S1 and S2

LC t

q = q0 cos q = – LC 1

q=–

LC dt 2 d 2q

LC dt 2 t

and S2 is closed. Then t

27. At time t = 0 switch S1 is closed and S2 is q0 and Then

RC circuit.

t = q = q0/2. t = 2 , q = q0 t = 2 , q = q0

2

d 2q

29. Fig. 25.82

2

V, switch S1

-

t t in one direction.

–2

e e–1

C V. L

circuit is 28. At time t q, switch S1

2

is closed.

SOLUTION 27. In an RC at a time t

q= – q = q0

where

e

28.

e–2t/

q = q0 cos t,

1

where d 2q dt 2

t as t+

2

q0

=

1 LC

dq d q0 cos t q0sin t dt dt which is alternating and not unidirectional. The I

is not equal to t, we get =–

where

I=

LC

Now q 0 at t wrong. The charge q q = q0

dt 2

t as

q

is closed and S1 oscillates in the LC

d 2q

t

e–2

q0

dt 2

= – LC

29. At t

2

=

2

–t/

= RC is the time constant. At t = 2 , we q = q0

1 d 2q

t+

2

=

q0 = =

q

1 LC C V L

CV

q0 = CV

Electromagnetic Induction and A.C. Circuits

30.

25.49

to

31.

,

32.

train ,

SOLUTION 32.

30. 31.

IV Assertion-Reason Type Questions

correct.

ment-1.

tor.

one -

-

with time.

changes at a constant rate.

s law.

-

25.50 Comprehensive Physics—JEE Advanced

9. than g

10.

Three identical coils A, B and C

A

A and C

B, with coils B and C B

-

tance decreases.

Fig. 25.84

11.

-

A to B 12. AB

xB

Fig. 25.83

Fig. 25.85

Electromagnetic Induction and A.C. Circuits

25.51

z A

A to B.

y

rod goes through a conducting ring as shown in Fig.

13.

rent is induced in the ring

than g

Fig. 25.86

SOLUTIONS 1.

v

to B

F =e v = B

8. B . Since the 9.

v

2. 3.

changes.

-

4.

10. Since coil C rent in C B A B

region. 11.

ment B

-

12.

7.

B, then according -

-

5.

6.

B, the cur-

inserted in the coil.

13.

in the coil, its inductance L increases. Hence its reactance L increases, causing a decrease in the 14. C C is decreased, the reactance will increase and as a result the current in the circuit is decreased causing a decrease in the Statement-1.

25.52 Comprehensive Physics—JEE Advanced

V Integer Answer Type 1.

each other, and the ground, are connected to a mil–1

along

10 that 2.

when

line AD into two regions.

the

R1 and R2

B

Find the terminal L –1

ACD is a semicircular conductr = 10 cm with centre at O, the

5. Fig. 25.87

-

–1

through O

R=

3. An LCR series circuit with 100

resistance is

Fig. 25.88

F and the resulting LC circuit is set oscillating at its Q denote the instantaneous I the current in the charge Q is 200

I 10–3 m2

6.

-

LCR circuit.

through the coil.

4.

IIT, 2000 AC

7. A series R-C R1 and R2 as shown Fig. L

R-C circuit is R 1.2 , the

SOLUTIONS 1.

v

–1

–1

. The induced

e = Blv = 0.2

10

1.0

10–3

Electromagnetic Induction and A.C. Circuits

2.

1 Br2 2

e

I= =

e R

Br 2R

.

0.1 2 2 0.

L or R

=

P1

P2 . 1.2

I1 I2

2

3.

tan

-

25.53

30

=2A

L = R tan -

tan

1 RC

=

1 C

= R tan Fig. 25.89

LCR R2

Z=

= R

2

L

1 C

100 tan

2

I = I1 + I2. Thus 100 tan

= I1 +

2

= R = 100 I=

V R

4.

I1 =

30I1

or

I=

mg Bd

0.2 .

Current I I1 and I2 R2 I = I1 + I2 e and R2

and

. 1

R1 and R1 and R2, then R1

e = Bvt d or vt = 5. The charge Q the current I =

e = I 1R 1 = I 2R 2 P1 = eI1 P2 = eI2 P1 P2

I1 I2

R1 and R2 are

=

e Bd

= 2 A. . /

P1 I1

vt is induced in it is

A

I1

=

30 e=

BId = mg

A

A. Hence

I2 = I – I1 =

200 =2A 100

I=

. .

1

= 1 ms–1

1 LC 1

. 10 2.0 10 –1 = 10 rad s Charge Q . Since at t = 0, Q = Q0

1/2

t at angular

25.54 Comprehensive Physics—JEE Advanced

Q = Q0 cos

t

dQ d = Q0 Current I = dt dt = – Q0 sin I dB 6. dt

. 0.2

=

Q0 = 10

7.

t

Z=

Z2 = R 2

t 10 1.2 R

–1

= NAB d dB = NA e= dt dt 10–3

= 100 Induced current I =

e R

2 .

2

= R2 +

R= Time constant =

1

R2

C

2

1 C2

2

1 C2

2

2 C

= RC =

2

=

2

10–3s

26

Ray Optics and Optical Instruments

Chapter

REVIEW OF BASIC CONCEPTS 26.1

26.2

REFLECTION OF LIGHT

REFLECTION AT A SPHERICAL SURFACE (CONCAVE MIRROR OR CONVEX MIRROR)

(i) -

(ii)

( 1 v

(in

1

1 ;f f

R

v 360 360

1 if 1

if

360

f

f h h0

360 v

f f

360

if

f

R

ge

360

-

v

f

26.2 Comprehensive Physics—JEE Advanced

26.3

SNELL’S LAW OF REFRACTION

1

1

26.4

REFRACTIVE INDEX AND SPEED OF LIGHT

v Fig. 26.2

v1 v

1

26.5

OA h OI h

REFRACTION OF LIGHT THROUGH A PLANE SLAB

26.7

AI h

h

h

h

h h

TOTAL INTERNAL REFLECTION ) 1

(<

1 1

1

1

Fig. 26.3

Fig. 26.1 1

)

x

26.8

REFRACTION THROUGH A PRISM

26.6 REAL AND APPARENT DEPTH (OR HEIGHT)

OA

d

AI

d d

-

d 1

OI

d

d

d

d

d 1

1

;

A

+

-

1

A

Fig. 26.4

26.3

A

1

+ 1

v

1

A

n

R

Application: Image formed by a transparent sphere O

1

A

n

R I

O

P 1

1

1

v

1

R

(

A

n

R

I

1

Q

A

n

1 v

A

1

I

A

Q

1

v

R R

(

O

26.9

REFRACTION AT A SPHERICAL SURFACE

1

v v

v

R

R

Fig. 26.6

h1 h0

v

1 v

1

1

R O

(

R 1 v

1 1

R R

1

v

1

1 R R

(

v

v Fig. 26.5

R

26.4 Comprehensive Physics—JEE Advanced

1

1 f v

1 R

1 R

Applications of lens maker’s formula Fig. 26.7

R1

26.10

REFRACTION THROUGH A LENS v

1 v

1 f

f

1 R

1 R

R

f v

h h

v

f

R1

f

f

R R

1 f f

-

f

R1

-

1 R1

1

R R

f

1 R

R1

R1

R1

R) R

1 R

1 R

R1

R) R

R1 R

1 f

1

1 f

R

Fig. 26.9

1 f

1

R R

R1 R

1 R

1 f

R

Fig. 26.10

;f

1 R

;f

-

1 f

3 R1

R Fig. 26.11

R1 1 f R1 1 f

Fig. 26.8

P

1 f (in

f e)

R R

R

1 R R 1

1

(

1) R

1 R

1)

( R

26.5

1

P

f (in

e) D)

P

P

f (in

Fig. 26.12

R1 1 f

R R

R1 1 f

R

f 1 R

1

(

1)

26.12

CO-AXIAL COMBINATION OF LENSES

R

R 1 1 R

(

f1

1)

F 1 F

-

1 f

1 f P

R1 R

1 f

1 R

d 1 F

1 R Fig. 26.13

R1 R

1 f R > R1 f

P1 + P

R

R > R1 f

R1

f

R

f

R1

e)

P

26.13

R 1 R

1 R

1 f

d f f

1 f

P1 + P

P1 P d

EFFECT OF SILVERING ONE OF THE REFRACTING FACES OF A LENS

Fig. 26.14

f

F

AB f

Fig. 26.16

1 f

1 F

Fig. 26.15

CD

f

f

26.11

POWER OF A LENS 1 F

1 f

1 f

1 f

f

f

26.6 Comprehensive Physics—JEE Advanced

f

f

26.1

R R 1

1 F

f

f

1 f

1 R

SOLUTION

1 R

R

f 1 F

1 R

1 R

R R1

1 F

( R

R 1)

R

R

R

F

(R1

R

(

R R

1)

Fig. 26.19

)

AI

1

1

1

1

1 F f

f

f

(

1) Fig. 26.17

R f

AB f

F

R (

)

R1 1 F 1 f

f

f

(

1)

F

R

R

R R

f 1 F

26.2

Fig. 26.18

( R

Fig. 26.20

) R

R

R

SOLUTION AC

CB

26.7

1 n

SOLUTION

1 n

1

26.3

> 26.5

SOLUTION

a v

a 1 v 1

or

a

1

1 f

1 a

1 30

a SOLUTION P1

P P

-

P1 + P

a

F F 26.6

v 1

1

d 1 30

-

d SOLUTION 1 1 + v

26.4

1 f

(i) -

1 1

>

> <

(b)

1 1

<

< >

v or

(

f f

or

1 f

f

f 1

or

(ii)

v 1+ or Fig. 26.21

+v

v v d

v or 1 + f f (1 +

)

v f (iii)

26.8 Comprehensive Physics—JEE Advanced

1

d f

+ f (1 +

)

1

26.7 f f Fig. 26.23

SOLUTION

1 v

1 f

1

26.9 3

v

f 1

or

or

A

v

1

-

A

f

A

SOLUTION

f

A

3 26.8

n

PQR

A

-

A

n

PR

A

n

A A

n

n

nA A

PQ A

PR

A A

n A

A

3

A

or

A 3

A

26.10 3

Fig. 26.22

SOLUTION AB PR

BC

-

SOLUTION n

PBC

A n

3 n n

1

n 30 3

n 30

A

A

26.9

30

26.14

26.11

3

3

COMPOUND MICROSCOPE

SOLUTION f M 1 v

e

0

1

1

1

1

f

|M

v0

1

0

1 v

D fe

1 10

1

v

-

0

v0

v u

fe D

|M

26.15

v0 D 0 fe D x

M

TELESCOPE

x x D M |M

f0 fe

-

f0

fe f0 + fe

L

D v

|M

f0 1 fe

fe D

D 1 v

f 1

1

26.10 Comprehensive Physics—JEE Advanced

1 u

1 10

1

26.14

or

-

x D v / D x

M

SOLUTION D f

1

M 1

fo

fe

10

M

o

e

D fe

1

o

26.12

1

o

-

o

-

o e

Position of object AB AB AB

SOLUTION fo

v

fe

1 v

1

1 v

1

v

o

1 1 v

o

fo 1

1

1

1 0

1

1

v AB M

v

-

D fe

1

1

6 -

26.13

AB v

v

-

SOLUTION M fe M

o

e

o o

1

D D fe

o

1

v

Fig. 26.24

26.11

Positioning of eyepiece

M A B

e

v

AB v

D

D

v

o

fo fe

M 1 v

1

1

fo fe

or

e

1 6

1

1

26.16

D

-

fe D

DISPERSION OF LIGHT

6 AB OE

v+|

OB + B E

-

n n

26.15 -

SOLUTION

fo + fe fo + fe fo fe

M

fo fe

fe

(ii)

Angular Dispersion and Dispersion Power

fo

V

R

A

26.16 V

V

V

R

A

R

V

R)A

R

A

V

R

SOLUTION fo

A fe

1

(

V

+

R

26.12 Comprehensive Physics—JEE Advanced

(

R )A

V

(

V

1) A

R

1

Dispersion without Deviation and Deviation without Dispersion

f

f1

d

f1

f

(ii) Chromatic aberration V

+( V A+(

R)A

V

R)A

+(

-

A

A+(

( (

R)A

A

fR > fY > fV R)A

V

Fig. 26.25 Fig. 26.27

26.17

ABERRATIONS

(i) Spherical aberration

f f Fig. 26.28

d Fig. 26.26

I Multiple Choice Questions with Only One Choice Correct 1.

x y

x i

j

+x (b) +x

1 (f + f

Ray Optics and Optical Instruments 26.13

(c) 4 ms–1 along the – x axis (d) 4 ms–1 along the + x axis 2. A plane mirror is made of glass of thickness 3 cm and refractive index 1.5 by silvering one of its faces. A point object is placed a distance of 6 cm in front of the unsilvered face. The distance of the image from the silvered face is (a) 6 cm (b) 7 cm (c) 12 cm (d) 15 cm 3. The image distance (v) is plotted against the object distance (u) for a concave mirror of focal length f. Which of the graphs shown in Fig. 26.29 represents the variation of v versus u as u is varied from zero v

v

v

v

Fig. 26.29

4. A hollow thin convex lens made of glass is placed in air. It will behave like a (a) convex lens (b) concave lens (c) prism (d) glass plate 5. Two thin equi-convex lenses each of focal length 3 are placed in conf and made of glass g 2 4 . The focal length of the combination is 3 f 2f (a) (b) 2 3 3f 4f (c) (d) 5 7 6. Two lenses A and B of focal lengths 30 cm and 20 cm are placed co-axially a distance d apart. A ray of

light parallel to the common principal axis is incident on lens A as shown in Fig. 26.30. What should be the value of d so that the ray emerges from lens B (a) 50 cm (b) 40 cm (c) 25 cm (d) 10 cm

Fig. 26.30

7. A convex lens has a diameter d and focal length f. A point object is placed on its principal axis at a distance 3f from it. The eye of an observer is placed at a distance of 3f on the other side of the lens and at a distance h below the principal axis. The maximum value of h so that the observer can see the image of the object is 2d (a) d (b) 3 d d (c) (d) 2 3 8. A glass prism ABC of refractive index 1.5 is immersed in water of refractive index 4/3 as shown in Fig. 26.31. A ray of light incident normally on face AB AC if (a) sin 8/9 (b) sin 2/3 (c) sin = 3 / 2 (d) 2 / 3 < sin < 8 / 9 IIT, 1981

Fig. 26.31

9. What is the relation between refractive indices , 1 and 2 if the behaviour of light rays is as shown in Fig. 26.32.

26.14 Comprehensive Physics—JEE Advanced

(a)

>

2

(c)

<

2;

>

(b)

1

=

(d)

1

< 2

<

2

<

1;

1

=

2

Fig. 26.32

10. A lens forms a sharp image on a screen. On inserting a parallel sided glass slab between the lens and the screen, it is found necessary to move the screen a distance d away from the lens in order to focus the image sharply. If the refractive index of glass relative to air is , the thickness of the glass slab is given by (a) d/ (b) d 1 d (c) (d) 1 d 1 1 11. A convex lens is placed between an object and a age obtained on the screen is m1. When the lens is moved by a distance d image obtained on the same screen is m2. The focal length of the lens is (m1 > m2). d d (a) (b) m1 m2 m1 m2 (c) d

m1 m2

(d) d

m2 m1

12. A ray of light, travelling in a medium of refractive index , is incident at an angle i on a composite transparent plate consisting of three plates of refractive indices 1, 2 and 3. The ray emerges from the composite plate into a medium of refractive index , at angle x. Then (a) sin x = sin i sin i

(b) sin x = (c) sin x = (d) sin x =

sin i 1

3

2

2

sin i

13. The principal section of a glass prism is an isosceles triangle ABC with AB = AC. The face AC is silvered. A ray incident normally on face AB, after BC in a direction perpendicular to it. What is the BAC of (a) 30° (b) 36° (c) 60° (d) 72° 14. An object is placed at a distance of 10 cm from a co-axial combination of two lenses A and B in contact. The combination forms a real image three times the size of the object. If lens B is concave with a focal length of 30 cm, what is the nature and focal length of lens A (a) Convex, 12 cm (b) Concave, 12 cm (c) Convex, 6 cm (d) Convex, 18 cm 15. A plano-convex lens acts like a concave mirror of 28 cm focal length when its plane surface is silvered and like a concave mirror of 10 cm focal length when its curved surface is silvered. What is (a) 1.50 (b) 1.55 (c) 1.60 (d) 1.65 16. A pin is placed 10 cm in front of a convex lens of focal length 20 cm and refractive index 1.5. The surface of the lens farther away from the pin is silvered and has a radius of curvature of 22 cm. (a) 10 cm (b) 11 cm (c) 12 cm (d) 13 cm 17. Two glasses have dispersive powers in the ratio of 2 : 3. These glasses are used in the manufacture of an achromatic objective of focal length 20 cm. What are the focal lengths of the two lenses of the (a) 6.67 cm, – 10 cm (b) 7.5 cm, – 12.5 cm (c) 9.67 cm, – 15 cm (d) 12.5 cm, – 20 cm 18. A giant telescope in an observatory has an objective of focal length 19 m and an eye-piece of focal length 1.0 cm. In normal adjustment, the telescope is used to view the moon. What is the diameter of The diameter of the moon is 3.5 106 m and the radius of the lunar orbit round the earth is 3.8 108 m. (a) 10 cm (b) 12.5 cm (c) 15 cm (d) 17.5 cm 19. Two convex lenses of focal lengths f1 and f2 are separated co-axially by a distance d. The power of the combination will be zero if (a) d =

f1

2

f2

(b) d =

f1

2

f2

Ray Optics and Optical Instruments 26.15

(c) d = f1 + f2

(d) d =

f1 f 2

20. A thin lens of focal length f has a aperture of diameter d. It forms an image of intensity I. Now, the central part of the aperture upto diameter d/2 is blocked by opaque paper. The focal length and the image intensity will change to f I I (a) , (b) f, 2 2 4 3f I 3I , (d) f , 4 2 4 21. The diameter of a plano-convex lens is 6 cm and the thickness at the centre is 3 mm. If the speed of light in the material of the lens is 2 108 ms–1, the focal length of the lens is (a) 15 cm (b) 20 cm (c) 30 cm (d) 10 cm 22. A thin prism P1 with angle 4° and made from glass of refractive index 1.54 is combined with another thin prism P2 made from glass of refractive index 1.72 to produce dispersion without deviation. The angle of prism P2 is (a) 5.33° (b) 4° (c) 3° (d) 2.6° IIT, 1990 23. A ray is incident at an angle of incidence i on one face of a prism of a small angle A and emerges normally from the opposite face. If the refractive index of the prism is , the angle of incidence i is nearly equal to (c)

(a)

A

(b)

A 2

A 2 24. A converging lens is used to form an image on a screen. When the upper half of the lens is covered by an opaque screen, (a) half the image will disappear (b) complete image will be formed (c) intensity of the image will increase (d) none of these IIT, 1986 25. The angle of a prism is A and one of its refracting surfaces is silvered. Light rays falling at an angle of incidence 2 A (c)

A

(d)

the second (silvered) surface. The refractive of the material of the prism is (a) 2 sin A (b) 2 cos A

1 cos A (d) tan A 2 26. When a ray of light enters a glass slab from air, (a) its wavelength decreases (b) its wavelength increases (c) its frequency increases (d) neither wavelength nor frequency changes IIT, 1980 27. A beam of light consisting or red, green and blue colours is incident on a right-angled prism ABC as shown in Fig. 26.33. The refractive indices of the material of the prism for red, green and blue wavelengths, respectively are 1.39, 1.44 and 1.47. The prism will (c)

Fig. 26.33

(a) separate part of the red colour from green and blue colours (b) separate part of the blue colour from red and green colours (c) separate part of three colours from one another (d) not separate even partially any colour from the other two colours. IIT, 1989 28. Spherical aberration in a thin lens can be reduced by (a) using a monochromasstic light (b) using a doublet combination (c) using a circular annular mask over the lens (d) increasing the size of the lens. IIT, 1994 29. A real image of a distant object is formed by a plano-convex lens on its principal axis. The spherical aberration (a) is absent (b) is smaller if the curved surfaces of the lens faces the object (c) is smaller if the plane surface of the lens faces the object

26.16 Comprehensive Physics—JEE Advanced

(d) is the same whichever side of the lens faces the object IIT, 1998 30. A concave mirror is placed on a horizontal table, with its axis directed vertically upwards. Let O be the pole of the mirror and C its centre of curvature. A point object is placed at C. It has a real image, also located at C water, the image will be (a) real and will remain at C (b) real and located at a point between C and

31.

32.

33.

34.

(c) virtual and located at a point between C and O (d) real and located at a point between C and O. IIT, 1998 A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at point O and PO = OQ. The distance PO is equal to (a) 5R (b) 3R (c) 2R (d) 1.5R IIT, 1998 An eye specialist prescribes spectacles having a combination of a convex lens of focal length 40 cm in contact with a concave lens of focal length 25 cm. The power of this lens combination is (a) + 1.5 D (b) – 1.5 D (c) + 6.67 D (d) – 6.67 D IIT, 1997 A concave lens of glass, refractive index 1.5, has both surfaces of the same radius of curvature R. On immersion in a medium of refractive index 1.75, it will behave as a (a) convergent lens of focal length 3.5 R (b) convergent lens of focal length 3.0 R (c) divergent lens of focal length 3.5 R (d) divergent lens of focal length 3.0 R IIT, 1999 In a compound microscope, the intermediate image is

of two liquids L1 or L2 having refractive indices n1 and n2 respectively (n2 > n1 > 1). The lens will (a) (b) (c) (d)

air and placed in air air and immersed in L1 L1 and immersed in L2 L2 and immersed in L1.

IIT, 2000 36. A diverging beam of light from a point source S having divergence angle , falls symmetrically on a glass slab as shown in Fig. 26.34. The angles of incidence for the two extreme rays are equal. If the thickness of the slab is t and refractive index is n, then the divergence angle of the emergent beam is (a) zero (b) 1 –1 1 (d) 2 sin–1 (c) sin n n IIT, 2000

Fig. 26.34

37. A rectangular glass slab ABCD of refractive index n1 is immersed in water of refractive index n2 (n1 > n2). A ray of light is incident at the face AB of the slab as shown in Fig. 26.35. The maximum value of the angle of incidence max, such that the ray comes out only from the other side CD is given by (a) sin–1

(b) sin–1 n1 cos sin (c) sin–1

n1 n2

(d) sin–1

n2 n1

(d) virtual, erect and reduced.

IIT, 2000 35. A hollow double concave lens is made of a very thin

n1 cos sin n2

1

1

n2 n1 1 n2

IIT, 2000

Ray Optics and Optical Instruments 26.17

41. A plano convex lens of refractive index 1.5 and radius of curvature 30 cm is silvered at the curved surface. Now this lens has been used to form the image of an object. At what distance from this lens an object be placed in order to have a real image of

Fig. 26.35

38. When a glass prism of refracting angle 60° is immersed in a liquid, its angle of minimum deviation is 30°. The critical angle of glass with respect to the liquid medium is (a) 42° (b) 45° (c) 50° (d) 52° 39. In the visible region, the dispersive powers and the prism are and and d and d respectively. When the two prisms are combined, the condition of zero dispersion by the combination is (a)

d

(b)

d+

d =0

(c)

d+

d =0

d

2

(d) ( d) + (

=0

d )2 = 0

40. A ray of light is incident normally on one face of a glass-air interface as shown in Fig. 26.36. If the of the prism is n, then 1 (a) n < 2 (c) n >

1 2

(b) n >

2

(d) n <

2

(a) 20 cm (b) 30 cm (c) 60 cm (d) 80 cm 42. The two lenses of an achromatic doublet should have (a) equal powers (b) equal dispersive powers (c) equal ratio of their power and dispersive power (d) equal product of their power and dispersive power. 43. Rays from the sun subtend an angle (in radians) at the pole of a concave mirror of focal length f. If the diameter of the sun is D, the diameter of the image of the sun formed by the mirror is (a) D (b) 2 D (c) f (d) 2 f 44. A point source of light S is placed at a distance L in front of the centre of a plane mirror PQ of width d hung vertically on a wall as shown in Fig. 26.37. A man walks in front of the mirror along a line parallel to the mirror at a distance 2L from it as shown. The greatest distance over which he can see the image of the light source in the mirror is d (b) d (a) 2 (c) 2d (d) 3d IIT, 2000

Fig. 26.37

45°

45°

Fig. 26.36

45. A thin glass prism of refractive index 1.5 produces a deviation of 4° of a ray incident at a small angle. What will be the deviation of the same incident ray by the same prism if it is immersed in water of (a) 1° (b) 2° (c) 8° (d) 16° 46. An air bubble in a glass slab ( = 1.5) is 5 cm deep when viewed through one face and 2 cm deep when

26.18 Comprehensive Physics—JEE Advanced

viewed through the opposite face. What is the thick(a) 7.0 cm (b) 7.5 cm (c) 10.0 cm (d) 10.5 cm 47. The refracting angle of a prism is A and the refractive index is cot (A/2). The angle of minimum deviation is (a) 180° – A (b) 180° – 2A (c) 180° – 3A (d) 180° – 4A 48. A ray of light passes through four transparent media with refractive indices 1, 2, 3 and 4 as shown in Fig. 26.38. The surfaces of all media are parallel. If the emergent ray CD is parallel to the incident ray AB, we must have

Fig. 26.40

51. Which one of the spherical lenses shown in

Fig. 26.38

(a) (c)

1 3

= =

(b) (d)

2 4

2 4

= =

3 1

IIT, 2001 49. A given ray of light suffers minimum deviation in an equilateral prism P. Additional prisms Q and R of identical shape and of the same material as P are now added as shown in Fig. 26.39. The ray will now suffer

curvature of the surfaces of the lenses are as given in the diagrams. IIT, 2002

Fig. 26.39

(a) greater deviation (b) no deviation (c) same deviation as before 50. An observer can see through a pin-hole the top end of a thin rod of height h, placed as shown in Fig. 26.40. The beaker height is 3h and its radius h. 2h, he can see the lower end of the rod. Then the refractive index of the liquid is (a) 5/2 (b)

5/ 2

(c) 3/ 2 (d) 3/2

IIT, 2002

Fig. 26.41

52. Two plane mirrors A and B are aligned parallel to each other, as shown in Fig. 26.42. A light ray is incident at an angle of 30° at a point just inside one end of A. The plane of incidence coincides with the The maximum number of times the ray undergoes es out is (a) 28 (b) 30 (c) 32 (d) 34 IIT, 2002

Ray Optics and Optical Instruments 26.19

f

(a)

u

53. The size of the image of an object, which is at in30 cm is 1.6 cm. If a concave lens of focal length 20 cm is placed between the convex lens and the image at a distance at 26 cm from the convex lens, (a) 0.8 cm (c) 2.0 cm

(b) 1.2 cm (d) 2.4 cm

54. A ray of light is incident at the glass-water interface at an angle i face of water as shown in the Fig. 26.43. The value of g would be

(c)

4 3

sin i

(b)

2

1 sin i

3 sin i

IIT, 2003

of focal length 20 cm. A second motor car 2.8 m of 15 ms–1. The speed of the image of the second 1 ms 1 10 (c) 10 ms–1

(a)

(b) 1 ms 1 15 (d) 15 ms–1

56. A short linear object of length b lies on the axis of a concave mirror of focal length f at a distance u from the pole. The lenght of the image will be

u

f

b

(d)

u

(b) 3r

f f f

b 2

b

f

2r

(d) 2r

58. Light is incident at an angle on one planar end of a transparent cylindrical rod of refractive index n. The least value of n so that the light entering the rod does not emerge from the curved suface of the rod for any value of is 4 3

(c) 1.5

Fig. 26.43

55.

(c)

(a)

(d) 1.5

(b)

57. A point source of light S is placed at the bottom of a vessal containing a liquid of refractive index 2 . A person is viewing the source from above the surface. There is an opaque disc of radius r on the surface. The centre of the disc is vertically above the source S. The liquid is gradually drained out from the vessel through a tap. The maximum height of the liquid for which the source cannot be seen from above is (a) r

IIT, 2003

(a)

u

b 2

f

(c)

Fig. 26.42

f

(b)

2

(d)

3

59. A ray of light is incident at an angle of 60° on one face of a prism of refracting angle 30°. The ray emerges out of the prism making an angle of 30° with the incident ray. The refractive index of the material of the prism is (a)

2

(b) 1.5

(c)

3

(d)

1 1 2

3

60. A square wire of side 3.0 m is placed 25 cm from a concave mirror of focal length 10 cm. The area enclosed by the image of the wire is (b) 4 cm2 (a) 1 cm2 (c) 16 cm2 (d) 25 cm2 61. A ray of light PQ is incident on an isosceles glass prims placed on a horizontal table. If the prism is in the minimum deviation position for the ray PQ 26.44). IIT, 2004 (a) (c)

= <

(b) (d)

> +

= 90°

26.20 Comprehensive Physics—JEE Advanced

Fig. 26.44

62. White light is incident on face AB of a glass prism. The path of the green component is shown in the AC as shown in Fig. 26.45 the light emerging from face AC will contain (a) yellow, orange and red colours (b) violet, indigo and blue colours (c) all colours (d) all colours except green. IIT, 2004

Fig. 26.46

65. A convex lens and a concave lens are placed in contact. The ratio of the magnitude of the power of the convex lens to that of the concave lens is 4 : 3. If the focal length of the convex lens is 12 cm, the focal length of the combination will be (a) 16 cm (b) 24 cm (c) 32 cm (d) 48 cm IIT, 2005 66. A right-angled prism is to be made by selecting a proper material and angles A and B (B A), as shown in Fig. 26.47. It is desired that a ray of light incident on face AB emerges parallel to the incident be the minimum refractive index n for this to be IIT, 1987

Fig. 26.45

63. A point object is placed at the centre of a glass sphere of diameter 12 cm and refractive index 1.5. What is the distance of the virtual image from the (a) 4 cm (c) 9 cm 64. A beaker contains water (

Fig. 26.47

(b) 6 cm (d) 12 cm IIT, 2004

surface of water as shown in Fig. 26.46. An object is placed at the bottom of the beaker and its image is formed 14 cm below the surface of water. The focal length of the mirror is (a) 8 cm (b) 12 cm (c) 16 cm (d) 20 cm IIT, 2005

(a) nmin =

1 sin A

(b) nmin =

1 sin B

(c) nmin =

sin A sin B

(d) nmin =

sin A sin B

67. A plano-convex lens has thickness 4 cm. When placed on a horizontal table with the curved face in

Ray Optics and Optical Instruments 26.21

contact with it, the apparent depth of the bottommost point of the lens is found to be 3 cm. If the lens is inverted such that the plane face is in contact with the table, the apparent depth of the centre of 25 cm. The the plane face of the lens is found to be 8 focal length of the lens is IIT, 1984 (a) 25 cm (b) 50 cm (c) 75 cm (d) 100 cm 68. Figure 26.48 shows a lens having radii of curvature R1 and R2 and 1 < 2 < 3. If the thickness of the lens is negligible and R1 = R2 = R, the focal length of the lens will be

Fig. 26.50. A ray of light is incident on the prism at a small height h. The focal length of this crude converging lens is h (a) f = (b) f = (c) f =

Fig. 26.50

h (

1) h

(d) f =

h (

1)

IIT, 2005 71. A ray of light is incident at an angle of 45° on a square slab of a transparent material. What should be the refractive index of the material of the slab face AB Fig. 26.48

(a) f = (c) f =

( (

3R 3 1R 3

1)

2)

(b) f = (d) f =

2R

(

3

1)

(

2

1)R

(

3

1)

IIT, 2003 69. A parallel sides slab ABCD of refractive index 2 is sandwiched between two slabs of refractive indices 2 and 3 as shown in the Fig. 26.49. The minimum value of angle such that the ray PQ AB and CD is (a) 30° (c) 60°

(b) 45° (d) 75°

IIT, 2005

Fig. 26.51

(a)

5 3 72. Parallel rays from a distant object fall on a solid transparent sphere of radius R and refractive index . The distance of the image from the sphere is (c) 1.5

(a) (c) Fig. 26.49

70. Two identical thin isosceles prisms of refracting angle (in radian) and refractive index are placed with their bases touching each other as shown in

3 2

(b)

2

(d)

R(2 2(

) 1)

R (

1)

(b)

R (

1)

(d) R ( – 1)

73. An object O is placed at a distance of 20 cm from a thin plano-convex lens of focal length 15 cm. The plane surface of the lens is silvered as shown in Fig. 26.52. The image is formed at a distance of

26.22 Comprehensive Physics—JEE Advanced

distance x from the pole P. The student looks at the pin and its inverted image from a distance keeping his/her eye in line with PA. When the student shifts his/her eye towards left, the image appears to the right of the object pin. Then, (a) x < f (b) f < x < 2f (c) x = 2f (d) x > 2f IIT, 2007 Fig. 26.52

(a) (b) (c) (d)

60 30 24 12

cm cm cm cm

to to to to

the the the the

right of the lens left of the lens right of the lens left of the lens

IIT, 2006 74. Figure 26.53 shows the graph between the image distance v (in cm) and the object distance u (in cm) for a thin convex lens. The focal length of the lens is

77. A ray of light travelling in water is incident on its surface open to air. The angle of incidence is , which is less than the critical angle. Then there will be (Fig. 26.54)

between them would be less than 180° – 2 between them would be greater than 180° – 2 IIT, 2007

Fig. 26.54 Fig. 26.53

(a) (5.00 ± 0.05) cm (c) (10.0 ± 0.10) cm

(b) (5.00 ± 0.10) cm (d) (0.50 ± 0.05) cm IIT, 2006 75. Image of the sun is formed by a biconvex lens of focal length f. The image is a circular patch of radius r and is formed on the focal plane of the lens. Choose the correct statement from the following. (a) The area of the image is r 2 and it is directly proportional to f. (b) The area of the image is r 2 and it is directly proportional to f 2. (c) The intensity of the image will increase if f is increased. (d) If the lower half of the lens is covered with black paper, the area of the image will become half. IIT, 2006 76. In an experiment to determine the focal length (f) of a concave mirror by the u – v method, a student places the object pin A on the principal axis at a

78. Two beams of red and violet colours are made to pass separately through a prism (angle of the prism is 60°). In the position of minimum deviation, the angle of refraction will be (a) 30° for both the colours (b) greater for the violet colour (c) greater for the red colour (d) equal but not 30° for both the colours IIT, 2008 79. A light beam is traveling from Region I to Region IV (Refer to Fig. 26.55). The refractive indices in n n n Regions I, II, III and IV are n0, 0 , 0 and 0 , 2 6 8 respectively. The angle of incidence for which the beam just misses entering region IV is (a) sin–1 (c) sin–1

3 4 1 4

(b) sin–1

1 8

(d) sin–1

1 3

Ray Optics and Optical Instruments 26.23

(c)

Fig. 26.55

IIT, 2008 80. A light ray travelling in glass medium is incident on glass-air interface at an angle of incidence . The R) and transmitted (T) intensities, both as function of , are plotted. The correct sketch is [refer to Fig. 26.56]

(d)

(a) Fig. 26.56

IIT, 2011

(b)

ANSWERS

1. 7. 13 19. 25. 31. 37. 43. 49 55. 61. 67. 73. 79.

(c) (c) (b) (c) (b) (a) (a) (d) (c) (b) (a) (c) (d) (b)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68. 74. 80.

(b) (a) (c) (d) (a) (b) (b) (d) (b) (c) (a) (a) (b) (c)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63. 69. 75.

(b) (c) (b) (c) (a) (a) (c) (a) (c) (a) (b) (c) (b)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64. 70. 76.

(b) (c) (b) (c) (c) (a) (b) (d) (b) (b) (b) (d) (b)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65. 71. 77.

(d) (a) (a) (c) (b) (d) (a) (b) (c) (c) (d) (b) (c)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66. 72. 78.

(a) (b) (d) (b) (d) (b) (d) (d) (b) (b) (b) (a) (a)

26.24 Comprehensive Physics—JEE Advanced

SOLUTIONS 1. Refer to Fig. 26.57. v v

1 1 1 = . For a concave f v u mirror u = – u and f = – f. Hence

3. For a spherical mirror

v

1 1 = v u

v

–v

Fig. 26.57

1 f

uf f u

v=

If u < f, the image is virtual. Hence v is positive for u lying between zero and f. If u > f, the image is real. Hence v is negative for u lying between f and u ,v – f. When u f, v ± . Hence the correct graph is (b). 4. Refer to Fig. 26.59.

Velocity of the object is vo = (2 i + 2 j ) ms–1 Speed of object is vo =

22

22 = 2 2 ms–1

= speed of the image (vi). The velocity vi of the image will be as shown in Fig. (a). The relative velocity of the image with respect to the object is v = vi – vo = vi + (– vo) The magnitude of v is given by [see Fig. (b)] v = [v02 + v2i – 2vo vi cos 90°]1/2 =

2 2

2

2 2

2 1/ 2

= 4 ms–1 along – x axis.

Fig. 26.59

The hollow lens can be considered to be a combination of two lens A and B. For a lens,

2. Refer to Fig. 26.58.

1 = ( – 1) f

1 R1

1 R2

For lens A, R1 = + R1 and R2 = + R2. Hence 1 = ( – 1) fA

1 R1

1 R2

For lens B, R1 = – R1 and R2 = – R2. Hence 1 = ( – 1) fB

Fig. 26.58

I1 is the image of O due to refraction at face I AI1 = (OA) = 1.5 6 = 9 cm

I2 A

=

15 = 10 cm 1.5

Distance of I3 from B = 10 – 3 = 7 cm.

1 R2

=

1 fA

Thus fB = – fA. The focal length of the hollow lens is given by

I2 is the image of I1 I1 B = 9 + 3 = 12 cm, I2 B = 12 cm. I3 is the image of I2 due to refraction at Face I again. AI3 =

1 R1

1 1 = fh fA

1 =0 fB

(

f B = – f A)

or fh = . Hence a hollow lens behaves like a glass slab. 5.

1 = f

3 2

1

2 R

=

1 R

(

R1 = – R2 = R)

Ray Optics and Optical Instruments 26.25

which gives f = R. When the space between the lens of = 4/3 surrounded by a medium of 3/2. Therefore, for the water lens, 1 = f =

3 4 2 3 4/3 1 4R

2 R

(

g

=

R1 = – R and R2 = + R)

1 4f

Fig. 26.61

The focal length of the combination of the three lenses is given by 1 1 = f F =

2 f

1 f

1 f

1 4f

7 4f

4f F= , which is choice (d). 7 6. Refer to Fig. 26.60.

8. The ray falling normally on face AB is refracted undeviated into the prism and is incident on face AC at an angle of incidence i = face AC 4 8 w sin i or sin or sin 3 1.5 9 g Hence the correct choice is (a). 9. The ray does not suffer any deviation on entering the lens. Hence 1 = . The ray leaves the second surface of the lens bending towards the normal. Hence 2 > . Thus the correct choice is (c). 10. When a glass plate of thickness t is introduced, the 1 image shifts by an amount t 1 . Hence 1

d=t 1

d

or t =

1

1 Thus the correct choice is (c).

11.

Fig. 26.60

For no deviation, the ray must emerge from lens B parallel to the principal axis. For this to happen, point F must be at the second focus of lens A and B. Hence d = 30 + 20 = 50 cm, which is choice (a). 7. u = – 3f. The distance v of the image I is given by 1 v

1 1 = 3f f

m1 – m2 = =

v = 1.5 f

The distance of the eye from I is = 3f – 1.5f = 1.5f (see Fig. 26.61) Triangles ABI and CDI are similar. Hence CD DI = AB BI

v . For the u conjugate (second) position, since u and v are u interchanged, we have m2 = . v Therefore m1 =

h d /2

1.5 f 1.5 f

d or h = . So the correct choice is (c). 2

But

f =

Hence m1 – m2 =

v u v2 u 2 – = u v uv (v

u) v uv

uv and v – u = d. u v d f

d . m1 m2 Thus the correct choice is (a). or

f=

u

26.26 Comprehensive Physics—JEE Advanced

12. Refer to Fig. 26.62.

If F is the focal length of the combination, we have 1 1 1 1 15 = – = + or F = cm F v u 30 10 2 Focal length of the concave lens B = f1 = – 30 cm. If f2 is the focal length of lens A, we have 1 1 1 + = or f1 f2 F

1 1 1 = – f2 f1 F =

2 1 + 15 30

which gives f2 = 6 cm. Since f2 is positive, the lens is convex. Hence the correct choice is (c). 15. fm = ) Fig. 26.62

Applying Snell’s law to successive refractions, we have sin i = sin r1

1

sin r1 = sin r2

;

sin r3 sin r2 = 3; = sin r3 sin x 2 Multiplying these equations we get sin i = sin x

or sin x =

2 1 2 1 = + = f1 f1 28 fm In the second case we have 1 2 1 = + f1 10 fm

2 1

3

(ii)

where fm is the focal length of the curved silvered R surface. Hence fm = where R is the radius of 2 curvature of the curved surface. Subtracting (i) from (ii) we get

sin i

13. The path of the ray is shown in Fig. 26.63.

(i)

1 1 1 140 = – or fm = cm. fm 10 28 9 280 cm 9 From (i) we have f1 = 28 2 = 56 cm. Now 1 1 = ( – 1) f1 R Therefore R = 2 fm =

R 280 = = 0.55 or = 1.55 f1 9 56 16. The value of the effective focal length F is given by 1 1 1 1 2 1 F = f1 + f m + f = f + f or

–1=

1

Fig. 26.63

It is clear that + Hence,

v =–3 ( the image is inverted) u v= – 3 u u = – 10 cm, therefore v = + 30 cm.

14. Given m = or Now

= 180° and =2 = 36°, which is choice (b).

1

m

2 2 + 20 22 110 or |F| = cm 21 Since the convex lens with a silvered surface behaves as a concave mirror of effective focal length F, we have 110 F=– cm and u = – 10 cm 21 =

Ray Optics and Optical Instruments 26.27

Substituting these values in the mirror formula 1 1 1 + = v u F 1 21 1 we have =– + or v = – 11 cm. v 110 10 The negative sign shows that the image is in front of the effective mirror and hence is real. 17. For an achromatic combination f 1 1 =– =– f f f 2 . Therefore, 3 1 2 =– f 3f The focal length of the combination is 1 1 1 = + F f f where F = + 20 cm. Therefore where

Hence the intensity of the image reduces by a factor of 4. Thus the intensity becomes I – I/4 = 3I/4. Hence the correct choice is (d). 21. Refer to Fig. 26.64. Here AB = 6 cm. Therefore, a = AE = 3 cm. Let C be the centre of curvature of the lens. The radius of curvature of the lens is R = AC = BC = CD

=

1 1 1 = + 20 f f From Eqs (i) and (ii) we get

(i)

Fig. 26.64

(ii)

3f = – 10 cm f = 6.67 cm and f = – 2 18. Since u >> f0, v = f0 = 19 m. Now u = – 3.8 108m. ctive is v 19 =– = – 0.5 10 –7 8 u 3.8 10 Diameter of the image of the moon is 3.5 10 6 0.5 10 –7 = 0.175 m = 17.5 cm Hence the correct choice is (d). 19. The focal length F of the combination is given m0 =

1 1 1 d = + – f1 f2 F f1 f 2 In terms of powers we have P = P1 + P2 – d P1P2 for P = 0, 1 1 P1 P2 = + = f 2 + f 1. P2 P1 P1 P2 Hence the correct choice is (c). 20. The focal length of a lens does not change if a part of it is blcoked. If the central part of the aperture upto d/2 is blocked, the exposed area of the aperture reduces by one-fourth the earlier area because d=

d 2 1 2 = 2 4 d

Now ED = 3 mm = t, say. In triangle AEC, we have (AC)2 = (AE)2 + (CE)2 or R2 = a2 + (R – t)2 = a2 + R2 – 2Rt + t2 t = a2 2R Since t << R, the term t/2 R can be neglected compared to 1. Hence 2 Rt = a2 a2 3 cm 3 cm or R= = = 15 cm 2t 2 0 3 cm or

2Rt 1

c 3 108 ms 1 v 2 108 ms 1 = 1.5. Therefore, the focal length of the planoconvex lens is given by 1 1 1 1 = ( – 1) = (1.5 – 1) f R 15 30 Now the refractive index

=

or f = 30 cm. Hence the correct choice is (c). 22. For a prism with a very small refracting angle A, the deviation is given by (Fig. 26.65) = ( – 1) A 1 = ( 1 – 1) A1 and that produced by the second prism is 2 = ( 2 – 1) A2

The total deviation will be zero if 1 + 2 = 0. The emergent ray will then the parallel to the incident ray (see Fig. 26.65). Thus

26.28 Comprehensive Physics—JEE Advanced

(

2

– 1) A2 = – (

1

– 1) A1

The negative sign shows that the refracting angles of the two prisms are in opposite directions. Thus

27.

Fig. 26.65

| A2 | =

1 2

1 A1 1

=

154 . 1 4 1.72 1

= 3°

28.

23. Deviation produced by a prism having a small refracting angle is given by = ( – 1) A Also A + = i + e = i ( e = 0, since the ray emerges normally from the opposite face. Thus = i – A or ( – 1) A = i – A or i = A 24. A complete image will be formed but the intensity of the image will decrease due to decrease in aperture. Hence the correct choice is (b). 25. Refer to Fig. 26.66. The refracted ray BC will retrace its path if it falls normally on the silvered face PQ of the prism, i.e. PCB = 90°. Therefore, angle in triangle PBC is = 90° – A. Hence r = A. Now sin i = sin r =

sin 2 A sin A

29.

30.

v= or = v/ , wavelength decreases because speed v decreases. As the beam falls normally on face AB, it goes through undeviated. The critical angles for red, green and blue colours respectively are 1 1 = 46°, ig = sin–1 = 44° ir = sin–1 139 . 1.44 1 and ib = sin–1 = 43° 1.47 The angle of incidence at face AC is i = 45°. Since i is greater than ig and ib but less than ir, the red colour will be refracted out from face AC but green AC towards the base BC. Hence the correct choice is (a). Using monochromatic light eliminates chromatic aberration. Using a doublet combination minimizes chromatic aberration. Increasing the size of the lens increases its resolving power. To reduce spherical aberration, the aperture (i.e. exposed portion of the lens) must be decreased. Hence the correct choice is (c). Spherical aberration is reduced if the total deviation is distributed over the two surfaces of the lens. If the plane surface of the lens faces the object, all the deviation takes place at the curved surface. Hence spherical aberration is not reduced. Hence the correct choice is (b). Figure 26.67 shows the ray diagram for the image with water, the image is real and located at C which is between O and C. Hence the correct choice is (d).

2 sin A cos A = 2 cos A sin A

Hence the correct choice is (b).

Fig. 26.67

31. Using the formula for a spherical surface a

Fig. 26.66

26. Since the refractive index of glass is greater than that of air, the speed of light is less in glass than in air. The frequency of light never changes due to

u 1.0 we have u which gives u =

g

g

a

= v R 15 . 15 . 1.0 = (since v = u) u R 5R.

Ray Optics and Optical Instruments 26.29

32. Power of the lens combination is 1 1 P = P1 + P2 = f1 in m f 2 in m =

1 + 0.40 m

sin ic =

1 0.25 m

= – 1.5 m–1 = – 1.5 D 33. The focal length f of a lens of refractive index 2 surrounded by a medium of refractive index 1 is given by 1 = f

2

1 1

1 R1

1 R2

Now, for a concave lens, f = – f. Given R1 = R and R2 = – R1 = – R. Also 2 = 1.5 and 1 = 1.75. Hence, we have –

1 f

=

15 . 1.75 1.75

1 R

1 R

n2 n1

=

0.25 1.75

Fig. 26.68

Thus angle i must at least be equal to ic. Therefore ic = 90° – r or r = 90° – ic. n n sin max = 1 sin (90° – i) = 1 cos ic n2 n2 n1 cos n2

=

2 R

which gives f = 3.5 R. Since the focal length is positive, the lens acts like a convergent lens. Hence the correct choice is (a). 34. In a compound microscope, the object is placed just beyond the focus of the objective. Hence the image formed by the object is real, inverted and highly 35. Refer to the solution of Q. 33 above. Since the lens is concave f = – f. Also R2 = – R1. Therefore, we have n2 n1 1 1 1 = – n1 R1 R2 f where n2 the lens and n1 that of of the liquid in which the lens is immersed. It will act as a divergent lens (i.e. f will remain negative) if n2 > n1. Hence the correct choice is (d). 36. When a ray of light passes through a glass slab with parallel faces, it does not suffer any deviation; it is only displaced parallel to itself. Therefore, the direction of the beam remains unchanged after passing through the glass slab. However, the rays are displaced slightly towards the outer side. Hence the divergence angle of the emergent beam will be the same as that of the incident beam. 37. The ray will emerge from side CD of the slab if the ray refracted in the slab suffers repeated total AD and BC of the slab as shown in Fig. 26.68. From Snell’s law, we have n n sin r = 2 or sin max = 1 sin r n1 n2 sin max The critical angle ic is given by

sin

1

n2 n1 n2 n1

sin ic or

n1 cos sin n2

= sin–1

max

1

n2 n1

38. The refractive index of the prism with respect to the liquid in which it is immersed is given by =

=

1 A m 2 1 sin ( A) 2

sin

sin

60

30

2 60 sin 2

=

sin 45 sin 30

2

The critical angle ic is given by 1 sin ic = , which give ic = 45°. Hence the correct 2 choice is (b). 39. Mean angular deviations produced by crown and d = ( – 1) A and d = ( Their dispersive powers are =

v

1

r

and

=

– 1) A v

Their angular dispersions respectively are D = ( v – r ) A and D = ( v –

r

1

r)

A

When the prisms are combined, the dispersion by the combination will be zero if D+D =0

26.30 Comprehensive Physics—JEE Advanced

or ( or or

(

v

–

r)

A+(

r)

v

v

–

r)

( – 1) A +

A =0 (

v

r)

( – 1)A = 0 ( ( 1) 1) d+ d = 0. which is choice (c).

40. given by 1 n

sin ic = If n =

2 , sin ic =

1 2

or ic = 45°. For total internal greater

fore, n is greater than is (b).

i = 45°. Hence the critical 2 . Thus the correct choice

Fig. 26.69

45. = (

41. 1 F

R 2 or F = 2 R

– 1)A

g

in water is

30 2 1.5

=

15 . 4/3

g w

=(

u = 2F = 2

1

g

(ii)

9 1 8 . 1 15

1 1

=

2

9 8

– 1)A

42. f1 =– f2

(i)

4° = 1°

Hence the correct choice is (a). P2 =– P1 or P1

1

= – P2

46. Real thickness =

1 2

apparent thickness

= 1.5

2

43.

angle in sin = R = 2f

(

R = 2f )

44.

AB = 2OB = 2

3d = 3d, 2

d 2

d

=

-

tance is AB. Now, since PR = RD = L angles PRT PBD we have d =d BD = 2RT = 2RS = 2 2 d OD = . Therefore, OB = OD + BD 2 =

47.

or 3d 2

or

A

m

2 A sin 2

. Given

sin

cot

A 2

=

cos

A 2

= sin

A 2

= sin

sin 90

= cot A

A . Thus 2

m

2 A sin 2 A

m

2 A

m

2

A 2 2 or m = (180° – 2A) which is choice (b). which gives

A

m

= 90° –

26.31

48. sin i = sin r1 = sin r2 =

1 2 3

2 3 4

1

sin r1 sin r2 sin e

or

=

sin i sin r

QG / QD sin 45

h / 5h 1/ 2

2 5

5 , which is choice (b). 2

=

51.

correct choice is (c). 52. AB x = AC tan 30°

x = BC which is given by 3 = 0.2/ 3 3

Fig. 26.70

n=

(1) 1 sin i = 4 sin e Ray CD will be parallel to ray AB, if e = i. Hence 1= 4

2 3

= 30 0.2/ Hence the correct choice is (b).

49. P

P is such that the reP P, Q

Q

R P suffers P Q R will sufP.

R

53. Refer to the Fig. 26.73. AB

Thus the correct choice is (c). 50.

h QD travelling

Q

Fig. 26.72

DB Fig. 26.71 that D ABPR. Hence DE = PE = h. Also 45° since DPE

PB of BDF =

plane. It serves as the virtual object for the concave lens. A B f u v = ? Now

Fig. 26.73

1 v

1 u

1 1 v 4 which gives v or

Fig. 26.71

1 f 1 20

26.32 Comprehensive Physics—JEE Advanced

AB v = AB u AB 5c = 1.6 c 4c

or

2

f

| v| = b

, which is choice (c). u f 57. Referring to Fig. 26.74, the source S cannot be seen

or A B 54. For refraction at glass-water interface, we have sin i = sin r For refraction at water-air interface, we have sin r = a sin 90° = a = 1 g

x so that rays such as SA SB suffer internal relection. The critical angle ic is given by

(i) (ii)

Using (ii) in (i), we get g

sin i = 1 or

g

=

1 , sin i

which is choice (b). 55. Using f

1 1 1 = , f v u 1 1 1 = , v u f

u

we get v = +

56 3

we have –

1 dv v2 d t

1 du u2 d t

0

dv dt

or

v2 d u u2 d t

v2

Fig. 26.74

sin ic =

1

1

, which gives ic = 45°. 2 In triangle OAS, we have r r r = tan ic or x = = = r. tan ic x tan 45 Hence the correct choice is (a). 58. Refer to Fig. 26.75.

u2 correct choice is (b). 56. 1 v

1 1 = u f

(1) f is

v

u

v2

u2

u 1 v

u f

0

v=–

or

u, we get v f u u f

or

v=–

(2)

f

sin sin

n= (3)

or

sin

The ray AB

sin n B

(1)

). This

For this to happen (90° – or b

=

n at point B

2

f u

u

u

u f Given u = b. Therefore

at the planar face be the angle of refrac-

2

2

f

v=–

v u

Fig. 26.75

Ray OA

(4)

sin (90° – ) 2

-

B. ic, the critical angle sin ic or cos

sin ic

1/2

or (1 – sin ) sin ic 2 2 or 1 – sin sin ic The critical angle is given by

(2)

26.33

1 n

sin ic =

or

1–

sin

1

2

2

or n2 – sin2

1 2

follows that

A = r1 = 90° – 2 Hence = , which is choice (a). But

2

n n (1 + sin2 )

or n2

A = 90° – 2

or

(3)

n2

2

or

n

= + 1, it

2

AB to suffer total internal BC will be CD suffering another total D -

B

Fig. 26.77

59. Given i = 60°, A= 30°. Using = i + e – A, we get e AC of the

62.

1 2

Now the wavelength of violet light is the least

Now r1 = i – = 60° – 30° = 30°. Hence

Fig. 26.76

sin i sin 60 = = 3. sin r1 sin 30 Hence the correct choice is (c). =

60. u

f

1 v

1 f

1 u

AC,

1 10

1 25

or

v=–

3 50

sin ic =

G

2

Area of the object wire is 3.0 area of age area of object

2

. The

2

= 4 9

1

v2

50 3 25

2

4 9

u

2

2

, which is choise (b).

61.

PQ RS the refracting faces AB AC i1 Fig. 26.77). Therefore, r1 = r2 = r. Also A r1 + r2 = A, which gives r1 = r2 = . 2 In triangle ABC, A + + = 180° or

A + 2 = 180°

Fig. 26.78

i2 (see

where

G

ic is the critical angle for green

26.34 Comprehensive Physics—JEE Advanced

AC at angles greater than their respec-

u = OI

Now

v = OI

AC back into the AC at angles less than their respective critical angles. Hence these colours will AC AC. Thus the correct choice is (a). 63. O fall

O. Hence the virtual O -

1 1 = f v

1 u

we have 1 = f f

or

1 20

1 12

Hence the correct choice is (b). | P1 | | P2 |

65. Given f1 f2 is negative,

the correct choice is (b).

1 30

f2 f1

4 f1 = 3 The focal length F 1 f2

4 3

4 . Hence 3 4 3

f2 =

1 1 = f1 F

|f | 4 or 2 | f1 | 3

1 12

1 16

1 48

which gives F 66. Referring to Fig. 26.81, the path of the ray is PQRS Q R. It is clear er than the critical angle given by 1 sin ic = n

Fig. 26.79

64. 26.80) d=

32 4/3

Thus I I

Fig. 26.81

Also angle A = 1 n

sin

67. Refer to Fig. 26.82. Fig. 26.80

B= A n is given by n

=

1 sin

=

1 sin B

B

26.35

point O I1 at the plane surface as shown in Fig. 26.82 (a).

2

1

v 2

or

v

=

2

=

2

1

R1 1

(i)

R1 =v–t

v(

t

is negligible), v = + v1

Fig. 26.82

Hence 4 3 When the plane face of the lens is in contact with O of the plane I2 =

refraction at this face, we have 1

v where 1 = 1, 25 – 8

2

=

3

1 = f

2

v = OI2=

4

=

3

=

2

(ii)

R2

2

f

1

R1 2

3

=

2

R2 1 R1

1 3

3

1 R2

2 3

(iii)

R1 = R2 we get [put R1 = R2 = R in (iii)] 1 = f

8 25 Putting

1

=

2

v

the ray is parallel to the principal v1 = f using v1 = f (i) in (ii), we get

or

R = , u = OP 2 u

3

v1

R = + R2. We have

3

1 R

1 3

1 69. Refer to Fig. 26.84.

R

4 3

R

by

or

1 1 = ( – 1) R f f

4 3

1

1 25

1 75

68. Refer to Fig. 26.83. Fig. 26.84

AB, angle i1 given by sin i1 =

2 1

2 2

1 2

which gives i1 = 45º CD, angle i2 given by Fig. 26.83

sin i2 = u=– ,v=+v

R = + R1. We have

3 1

which gives i2 = 60º.

3 2

26.36 Comprehensive Physics—JEE Advanced

surfaces AB CD = 60º. Thus the correct choice is (c). 70.

ABC DBC their bases BC touching each other as shown in Fig. 26.85, which shows the path of parallel F. Distance OF = f tan

=

h f

1

sin ic =

(1)

Angle r is given by =

sin 45 sin r

1

or sin r =

(2)

2

ic + r = 90° or ic = 90° – r 1 which gives cos r = sin r = (1 – cos2r)1/2 1

= 1

1

1

2

1

( 2–1)1/2 =

1/ 2

2

which gives

(

2

– 1)1/2

=

3 , 2

which the correct choice is (b). 72. Refer to Fig. 26.87. For refraction at face I, 1 u1

Fig. 26.85

h PQ

v1

h f

h is given by (here angle = ( – 1) h = ( – 1) Thus f

R , we have 1

=

v1 =

or

or f =

1

=

u1 =

then by where =

v1

R R 1)

(

(1)

h 1

71. at E than the critical angle ic given by Fig. 26.87

For refraction at face II, u2 = – (v1 – 2R) = 2R – v1. u2 = 2R –

(

R 1)

R( (

2) 1)

(2)

v2 is given by u2 Fig. 26.86

1 = v2

1 R

(3)

26.37

v2 =

R (2 2(

) . 1)

The correct choice is (a). 73. given by 1 1 2 2 1 2 = + = + = , which gives f F f 15 15 m F=

Fig. 26.88

15 2

-

76. x
1 1 1 + = , we have v u F u

x = 2f

1 1 = v 20 which gives v

15 2

an F

20 2 15

false. If x lies between f f to the eye than the object as shown in Fig. 26.89.

-

74.

u v 1 1 1 1 = – = – f v u 10

C

1 , which gives f 10 f

Object pin

Image

Eye

F x

f is

f

P

2x

given by f f

2

=

where u

v

u

v u v are the least counts of u 2

u= v

=

v

u

f=

Fig. 26.89

2

2

(10)2

77.

f2

v u 0.1 0.1 2

v

ray is (5)

(10)2

2

= 180° – ( + > , choice is (c).

5 = 100

78.

Hence f 75. Refer to the Fig. 26.88. = r/f tan , where we have r = f

< 180° – 2 . Hence the correct

r = A/2 = 60°/2 = 30° for lights of all wavelengths. However the corre-

colours. The correct choice is (a). 79.

or

r2 = which is choice (b).

r=f 2

f 2. Thus area

)

f 2,

n0 sin

=

n0 sin 2

1

26.38 Comprehensive Physics—JEE Advanced I

II

III

IV q2

q1 q

n0 sin 6 1 which gives sin = . 8 80. If I =I =

q3 = 90°

q2

q1

tain critical angle

c. For

2

=

n0 sin 90° 8

R+T

<

c

a part of the inci-

> c Hence for > c, the value of R is 100 %. Hence the only correct option is (c).

Fig. 26.90

II Multiple Choice Questions with One or More Choices Correct 4.

1.

x

y respectively. If v. Then

–1

(a) x (b) y (c) v = v

1 21

–1

5.

–1

(a) violet light than for yellow light

2.

-

6.

3. 7. The focal lenght of a lens is (a) greater for violet light than for yellow light (c) greater for green light than for blue light

26.39

8.

(a) R (c) f

D -

R f

13.

jugate positions A B d (see Fig. 26.91). The respective linear m2 for posim1 tions A B of the lens. Then (a) d d 7 3 7 5 m1 = , m2 = (c) m1 = , m2 = 3 7 5 7

R

(a) in case (i) the lens will behave as a convergent lens of focal length 2.5R lens of focal length 3.0R (c) in case (ii) the lens will behave as a convergent lens of focal length 3.0R lens of focal length 3.5R 14. A short linear object of length b f u

a. If the Vo Vi. Then f

(a) a = b

u

f

Fig. 26.91

9.

(c) Vi = Vo slab

(b) a = b

f u

u

f 2

f

Vi = Vo

f

2

f

u

f

IIT, 1987 15.

AB of length f f, such that its AB m. Then

. (a) A B = 10. When light travelling in air enters a glass slab,

f 3

(b) A B =

(c) m

f 2

m = 3/2 IIT, 1991

16. . 11.

R point object P mR shown in Fig. 26.92. The value of m for which a ray P

Its focal length is f1 (a) f1 (c) f2

f2. Then f1 f2

12. R. When the plane surface f. Then

Fig. 26.92

26.40 Comprehensive Physics—JEE Advanced

(a) either

4 3 or 3 2

(b) neither

4 3

4 3 nor 3 2

is . Then (a) i0 = 45° (c)

3 2

(b) i0 = 60° = 60° IIT, 2005

IIT, 1999 ABC DCE,

17. 3 in Fig. 26.93. A light ray PQ AB at an angle i DCE

of i ABC

C C Fig. 26.93

i0. The angle through which C so that

DCE

ANSWERS AND SOLUTIONS 1. 2. All the four choices are correct. 3. 4. 5.

Putting f d

D m= =

Thus | m | = 6. 7.

v u

D d D d

100 40 7 =– . 100 40 3

7 3 . For lens at L2, | m | = 3 7

9.

-

8. Refer to Fig. 26.94. If O

I L2 are the two L1 conjugate positions of the lens, then

v =

. Fig. 26.94

10. D d For lens at L1, u = – x = – 2 Using these in

D d v= . 2

v=

1 1 1 = – , we get f v u f=

D2 d 2 4D

11.

1 = ( – 1) f1

1 = (1.5 – 1) R

1 10

f1

26.41

is given by 1 2( 1) 2 (1.5 1) = = f2 f2 R 10 Thus the only correct choice is (c). 12.

2

v f =– t u f

= 1.5, f1 = 2 R

R 1)

=

1 1 = f u

(1) f is

v v2

=–

u u2

2

f u

f

V0

30 2(1.5 1)

v=–

= 0 or

v u

2

u

(2)

2

f

Vi = V0 1 v

u = V0 t

v =– t

Vi = 2(

f

u . Given t

-

13. 14.

–

u

(4) by t, we have

R = f1

f=

2

f

a=b

u

f

15. Refer to Fig. 26.95. A B AB. B B B B. Location of B x B or B O u = – x, v =–x f = – f. Using these in the spherical

u, we get u u v f + 1= or = v f u u f

1 v

(3) We have

2

f

v= –

u

u f Given u = b. Therefore v= –

1 x

1 1 =– x f

which gives x = 2f. Thus OB = OB = 2f Location of A : A

2

f u

–

1 1 = u f

f

b

(4) -

AO = y = 2f –

Fig. 26.95

f 5f = . 3 3

Thus for point A, u = – y = – tance A O

A

5f 3

AB =

f , 3

-

26.42 Comprehensive Physics—JEE Advanced

1 v

1 1 1 = or u f v

( ray BC Therefore, we have

1 3 = f 5f

2

5f 5f which gives v = – . Thus A O = as shown 2 2

u

5f f – 2f = 2 2

AB =

2

=

or

1

R

which gives (1.5 m + 1)

f 3

m=

0.5 = 1.5

2 4 = . 15 . 3

The only correct choice is (c).

f /2 AB 3 = = . f /3 AB 2

17.

16. Q P surface. Ray PA refracts along AB Q

1

v

15 . 15 . 1 –0= R 15 . m 1 R

or AB =

–

R = – R.

ABC, the ray QR is parallel to base BC. Hence r1 = r2 r1 + r2 = 60º

P -

ity, we have 2

v Here

1

= 1,

– 2

15 . – v

1

u

=

2

1

Fig. 26.97

= 1.5, u = – mR. Therefore, 1 mR

= 0 or v = – 1.5 mR

which give r1 = r2 = 30º sin i =

sin r1

=

3

=

3 2

sin 30º

which gives i = 60º e Fig. 26.97 that the ray RS RS

Fig. 26.96

BC Q serves

– R (1.5m

DC

u = – (v + R) = – (1.5 mR + R) = v = –m

RS on face DC DCE C through an angle of 60º in the clockwise or anti-

26.43

III Multiple Choice Questions Based on Passage Questions 1 to 5 are based on the following passage focal lengths. Passage I The Compound Microscope

focal lengths. eyepiece has a long focal length eyepiece has a short focal length. 4.

1. (c) the objective of a higher focal length is

5.

2.

(a) its resolving power will increase

3. is high if

SOLUTION 1. The correct choice is (a). 2. 3. given by L D M= fo fe where L eyepiece, D

where 2 = angle of the cone of light rays entering the objective, -

AB fo

(for D >> fo ) or fe fo fe = focal length

of the eyepiece. 4. (see Fig. 26.98) R.P. =

2 sin Fig. 26.98

26.44 Comprehensive Physics—JEE Advanced

5. value of increases. Therefore, the light gathering capacity of the objective increases. As a result, the Questions 6 to 10 are based on the following passage Passage II

8.

The Astronomical Telescope

focal lengths focal lengths eyepiece has a long focal length focal plane of the objective. The position of the eyepiece

eyepiece has a short focal length 9.

(c) the objective of a higher focal length is 6.

-

10. If the aperture of the objective of a telescope is (a) its resolving power will increase

7.

SOLUTION 9. The resolving power of a telescope is given by R.P. = d 1.22 where d = wave-

6. The correct choice is (b). 7. 8. M=

fo D fe fe D

of fo or fe. Hence the correct choice is (a). 10.

where D Questions 11 to 13 are based on the following passage Passage III

t A n(y) given by n(y) = (ky

3/2

+ 1)

1/2

where k = 1.0.

–3/2

. The refractive

of air is

IIT, 1995 11. The relation between the slope of the trajectory of the ray at the point B(x, y i at that point is (a)

dy = sin i dx

(b)

dy = cos i dx

26.45

dy = tan i dx

(c)

dy = cot i dx

12. (a) y = (x)4 x 3

(c) y =

4

4

x 4 4 x1 of point P where the ray intery=

13. Fig. 26.99

x 2

(b) y =

(a) x1

x1

(c) x1

x1

SOLUTION 11. Refer to Fig. 26.100. The variation of refractive n(y) = (ky3/2 + 1)1/2

(1)

because na A = 90°. or

ia at

sin i =

1 n 1

cot i =

1/ 2

1 n2 1 n

= (n2 –1)12

dy = (n2 – 1)1/2 or dx

dy dx

Fig. 26.100

y = 0 to y = t. Therefore, the ray SA entering the slab at A i

dy = tan dx

= tan (90° – i) = cot i (2)

12. If i of refract n have n sin i = na sin ia = 1

sin 90° = 1

= n2 – 1

(4)

= ky3/2 + 1 – 1 = ky3/2 dy

dy = k1/2 y3/4 or dx Integrating, we get or

y3/ 4

B(x, y) is i, then the slope CD of the trajectory at B is

2

2

dy ABP.

dy dx

(3)

= k1/2

y

3/ 4

= k1/2 dx

dx or 4y1/4 = k1/2 x

Given k = 1.0. Therefore, we have y1/4 = 13. For point P, y get x y1 x1 is (a).

1 x or y = 4

x 4

4

(5)

y P are

26.46 Comprehensive Physics—JEE Advanced

Questions 14 to 16 are based on the following passage Passage IV

15. (a) 30°

The x-z z

2

z

3 . A ray of

(b) 45°

16. where a

A=6 3 i +8 3

b are

Ar = a i + b j + c k

(a) a = 6 3 , b = 8 3

j –10 k

(b) a = 3 3 , b = 4 3

IIT, 1999 14. The angle between vector A z (a) 90° (b) 120°

(c) a = 8 3 , b = 6 3 a = 4 3, b = 3 3

SOLUTION 14. Refer to Fig. 26.101. Ray PQ in x-z plane travelz i on the z QR in z r. be the angle between vector A k is the unit vector along the tive z positive z A.k cos = A =

(6 3 i + 8 3 j 10k ) (k ) (6 3 ) 2 + (8 3)2 =

which gives 15.

10

20

=–

( 10) 2

1/ 2

1 2

Fig. 26.101

( i . k = j . k = 0) = 120°. which is choice (b). i = 180° – = 180° –

sin i = sin r or

which gives r = 45°. so the correct choice is (b).

sin r =

16. x-y plane), the Ar = a i + b j + c k with a = 6 3

2 1 1

sin i =

2

2 3

sin 60° =

This the corrcet choice is (a).

2 18.

19.

–1

IIT, 2004 17. The focal length of the lens is

Ar = 6 3 i + 8 3 j + c k

1

Questions 17 to 19 are based on the following passage Passage V

b=8 3 A. Thus

–1

–1

–1

–1

26.47

SOLUTION 17.

1 = ( – 1) f

1 = (1.5 – 1) 1 gives f R 10

18.

u 1 v or

1 v

The

(1)

or

v

1

=

u

v

2

1

=

(30) 2

du dt

u2

(2)

v du u 2 dt

dm = dt

1 du =0 u 2 dt dv = dt

,

t, we have

1 1 = which gives v 30 20

2

–1

v u

m=

1 1 = u f

2

19.

f = + 20

t, we get 1 dv v 2 dt

60 30

=

1 dv u dt du dt

u

dv dt

(–60 × 1 – 30 × 4)

dm dt

v2 u2

IV Assertion-Reason Type Questions 3. Statement-1 four choices out of which only one choice is correct. R. The focal length of the lens R. Statement-2 4. Statement-1 1. Statement-1

at an angle i face of water as shown in Fig. 26.102. Then

focal length will increase. Statement-2 2. Statement-1 length increases. Statement-2 Fig. 26.102

-

26.48 Comprehensive Physics—JEE Advanced

g

=

Statement-2

4 3

sin i

SOLUTIONS 3. The correct choice is (c). The focal length f is given by 1 1 1 2 ( = (1.45 – 1) 1) f R1 R2 R R which gives f = , i.e. f is greater than R. 0.9 4.

1. curvature. 2. The correct choice is (a). The focal length of a lens

f R1

2

R2 is

g

1

2

1 R1

1 1

It is easy to see that f 1 = 1 for air).

sin r

(i)

For refraction at water-air interface, we have

given by 1 = f

sin i = sin r =

1 R2

a

sin 90° =

a

=1

(ii)

Using (ii) in (i), we get g

1

sin i = 1 or

g

=

1 sin i

V Integer Answer Type 1.

IIT, 1980 2. Fig. 26.103

100 tance of n value of n.

4. -

IIT, 1981 3. 5. =

a

a. IIT, 1987

IIT, 1988 4 ) in a tank is 18 3 7 lies on water 4

26.49

R

S R = 6 cm

m = 1.0

x

m = 7/4

x IIT, 2011

m = 4/3

Fig. 26.104

SOLUTION 1. For the objective,

1 v0

1 u0

f0

– 200

m0 =

v0 u0

A

1 . Putting u0 = f0 200 v0 = 3

200 / 3 200

d i

e r1

1 3

r2

B

C

have 1 ve

Fig. 26.105

1 1 = fe ue

Therefore, r1 = i –

where ve values, we get ue = eye-piece is v me = e ue

fe

=

25 6

sin 60 sin 30

26.106]

1 3

v where Thus 1 v1

6=–2

2.

1

1 = u = 4/3,

4/3

=

2

1

R u=

1

I1

v For refraction at the concave surface 1 1 = , where R v u R 1.5 1 1.5 1 100 = u= 15 u 50 9 Hence n = 9. 3. Given i = 60°, A = 30°. We have =i+e–A

(1)

e – 30° or e = 0. = 30°. [see Fig. 26.105]

R

1 4/3 2

which gives v1

Here also i

3

Hence the value of a = 3.

25 =6 25 / 6 me =

sin i sin r1

4. 2

m = m0 | m |= 2

= 60° – 30° = 30°. Hence

Fig. 26.106

26.50 Comprehensive Physics—JEE Advanced

I1

2

1

v2

u

=

2

1

R

(i)

For this refraction, we use 1

2

=

1

7/4 1 7/4 1 – = v ( 24) 6

2

v2 u R where u R 4/3 1 4/3 1 = ( 2) v2 10 which gives v2 I2

v u 4/3 7/4 – =0 v 21 v

5. 1 2,

then

VI Matrix Match Type 1.

Column I Column II. Match all the Column I.

Column II Column I (a)

(b)

(c)

Column II

S

S

S

S

IIT, 2009

SOLUTION -

1 =( f

– 1)

1 R1

1 R2

26.51

where R1 R2 R1 R2 the lens is a converging lens for which all the four choices are correct.

R2 > R1, f is positive, i.e.

ANSWER (a) (c) 2.

1

3

2 1,

2

3

Column I (a)

1

<

2

Column II (p)

m3

(b)

1

>

2

m3

(c)

2

=

>

3

m1

m2

(r)

3

m3

2

m1

m2

m1

m2

(s)

m3

m1

m2

(t) m3

m2

m1

IIT, 2010

ANSWERS (a) (c)

(p), (r)

(b)

26.52 Comprehensive Physics—JEE Advanced

Explanation: II

(p)

Ray AB Hence Hence

> 2= 2

1.

I

C

Ray BC

3.

m1

m2

m3

I

II

Ray AB 2 < 1. The ray BC face II. Hence 3 < 2.

A

B

B

C

A m1

m2

m3

II

(r)

Ray AB 2 > 1. Ray BC 3 = 2.

C

Ray AB 2 < 1. Ray BC II. Hence 3 < 2.

m3

B

C

A I

m3

m3

m1

m2

B C

Ray AB 2 < 1. Ray BC 3 = 2.

m1

m2

II

(t)

A

B

II

(s)

I

m2

A I m1

27

Wave Optics

Chapter

REVIEW OF BASIC CONCEPTS 27.1

where x1 and x2 are paths of the waves up to point P where =

WAVE NATURE OF LIGHT

Light is an electromagnetic wave which does not require a material medium for propagation. The electric and magnetic

=

where v

of a wave at a point x and at time t is given

=(

1

–

2)t

– (k1 x1 – k2 x2)

2) 1

–

2)t

–2

x1

x2

1

2

or

Phase difference =

2 2

1

(x2 – x1) (path difference)

Intensity proportional to the square of its amplitude at that point.

REFLECTION AND REFRACTION OF LIGHT

t – kx undergo a change and the wave is said to suffer refraction.

Phase Difference Suppose two waves meeting at a point P are represented

and

–

=

27.2

representing the wave, i.e. =

1

i.e., the phase difference is independent of time and x2 – x1). This holds

k

Phase The phase

=(

1. If the two waves have different frequencies, i.e., depends on time t. 2 and 2 then 1 2. If 1 = 2, then 1 = 2. In this case

E = A sin ( t – kx)

v=

2

=2 (

of an electromagnetic wave even in free space. aries in space and time as

which represents a wave travelling along the + x direction. –1 A = amplitude, = 2 ( 2 and k= ; = wavelength. Also

–

1

E1 = A1 sin (

1t

– k 1x 1)

E2 = A2 sin (

2t

– k 2x 2)

v1 in the medium in which the incident wave propagates and v2 refracted wave propagates, then 1 2

27.2 Comprehensive Physics—JEE Advanced

1

where

1

and

2

2

=

v1 v2

1

mum if cos

2

are the wavelengths of the same wave and Then

remains the same as that of the incident wave.

I is mini-

= – 1, i.e. = (2 n – 1) =

1 2

n

; n = 1, 2, 3, .....etc.

Imin = I1 + I2 – 2I1I2 suffers a phase change of 180° (or radians) in relation to that of the incident wave. No phase change occurs if a

=

I1

I2

2

minima constitute the bright and dark fringes.

27.3

INTERFERENCE OF LIGHT

27.4

When two or more light waves meet (or superpose) at a principle of superposition

which states that

more waves is known as If two waves of intensities I1 and I2, differing in phase I = I1 + I2 + 2 I1 I 2 cos I

= + 1, i.e.

-

= 2n ; where n = 0, 1, 2, 3, etc. is an integer or

2

or

depends on the phase difference ( ) between the two interfering waves. This phase difference depends upon two factors—(1) the initial phase difference between difference resulting from the path difference for that point. The initial phase difference depends upon the time –8 to 10–10

E = E1 + E2 +

imum if cos

COHERENT LIGHT SOURCES

= 2n

with time that, due to persistence of vision, we are unable to see the interference pattern. Thus, non-coherent sources cannot produce sustained interference effects. We con1. The sources must be coherent. 2. The wavelength of the interfering waves must be the same. Thus,

27.5

YOUNG’S DOUBLE SLIT EXPERIMENT

Monochromatic light from a source slit S illuminates two slits S1 and S2 equidistant from S (Fig. 27.1).

=n

where is the path difference between the interfering waves. Then I = I1 + I2 + 2 I1 I 2 =

I1

I2

2

The interference is said to be constructive. If the two interfering waves have equal intensities I1 = I2 = I0, then I

= 4 I0

Fig. 27.1

S1 and S2 interfere giving rise to bright and dark fringes on the screen. There is bright fringe at centre P0 of the screen.

Wave Optics 27.3

(i) The distance of the nth bright fringe from the centre yn =

n D

; n = 0, 1, 2, ... etc.

where = wavelength of light used, = seperation between slits S1 and S2 and D = distance between the screen and the plane of the two slits. (ii) The distance of the n th dark fringe from the centre 1 2

y*n = n

D

D

(iv) Angular separation between nth bright fringe and the central fringe is yn n ; n = n is in radian. D (v) Angular separation between nth dark fringe and the central fringe is 1 * n n = 2

27.6

DISPLACEMENT OF FRINGES

If a transparent plate of thickness t is introduced in the path of one of the interfering waves, tD

y0 = ( – 1) Number of fringes shifted =

27.7

2f , where a f the slit.

27.8

SOME IMPORTANT POINTS ABOUT INTERFERENCE OF LIGHT

; n = 1, 2, 3, ... etc

(iii) The separation between two consecutive bright or dark fringes is called fringe width ( ) which is =

2 a

(

1)t

will be white; all other fringes will be coloured. White light consists of colours between violet and rad (VIBGYOR). Wavelength is the shortest for violet light and longest for red light. At the central

the central fringe will be violet colour. ness t , then (a) the path difference at the centre of the screen – 1)t. amount y0 =

(

1)tD

.

(c) at the centre of the screen there will be a bright fringe if ( – 1) t = n ; n = 1, 2, 3, ... etc. (d) At the centre of the screen there will be a dark fringe if 1 n = 1, 2, 3, ... etc. ( – 1)t = n 2 (e) the fringe width will remain the same.

DIFFRACTION AT A SLIT

When a parallel beam of monochromatic light falls (because the two interfering beam do not now If is the wavelength of light and a is the width of the slit, then 3 5 , ... etc (i) For bright fringes : sin = 0, , 2a 2a (ii) For dark fringes : sin

=

2 3 ... etc , a a a ,

will become less distinct. is closed (or covered with black paper), the interpattern which has a bright central fringe bordered in still water rather than in air, the fringe width

27.4 Comprehensive Physics—JEE Advanced

is greater than that of the air, the speed of light in water (v) will be less than that in air (c). Since the v = c 1 and a = which give = = . c a Now

< a. Fringe width in air.

in water <

sources have intensities in the ratio n : 1, then in the interference pattern I ma I ma

=

=

I1

I2

2 I1 I 2

I1

I2

2 I1 I 2

I1

I2

I1

I2

I1 I2 I1 I2

in

=

2 n (n 1)

. I1 = n and I2

n : 1, then

I ma = I min

light sources have intensities in the ratio n : 1, i.e. I1 =n I2 and minima in the interference pattern is =

in

-

light has two wavelengths 1 and 2 will coincide if n1 1 = n2 2, where n1 and n2 are integers.

I ma I ma

I I

n 1 n 1

x is the width of the source slit S and X its distance from the plane of the slits, the interference fringes will not be seen (because the interference pattern becomes indistinct) if the condition x < X

2 2

27.9

RESOLVING POWER

2

1

n 1

2

close together. (a)

n 1

1

R.P. =

ent light sources, if the ratio of the intensities of

where 2

2 sin 1.22 = refrac-

I ma = n, then ratio of the intensities of n : 1, i.e. I min the coherent sources is I1 = I2

I ma = Lmin

n =

2

n 1

= wavelength of light used to

2

Objective

because

n 1

Object

2

I1 I2

1

I1 I2

1

I1 I2

1

I1 I2

1

O

2q

Fig. 27.2

I1 I2

n 1 n 1

2

(b) R.P. =

D 1.22

Wave Optics 27.5

where D length of light

27.10

= wave-

POLARIZATION OF LIGHT

diffraction are common on both transverse and longitudinal waves, mechanical as well as electromagnetic. The be

.

all the possible orientations in a plane perpendicular to the direction of propagation. When this light is passed

27.1 light at a point on the screen where the path difference is is K a point where the path difference is /3; being the wavelength of light used? SOLUTION Path difference 2 = 2 = =

or

= . Therefore, phase difference

= 2 is = I1 + I2 + 2 I1 I 2 cos = I1 + I2 + 2 I1 I 2 cos 2 = I1 + I2 + 2 I1 I 2

Polarization by Reflection: Brewster’s Angle

particular angle ip on the surface of a transparent medium,

= I + I + 2I

4I = K units ( I1 = I2 = I)

i.e. I = K difference is = =

3

or

2

=

2 3

=

2 is 3

I = I + I + 2I cos = 2I – I = I = Fig. 27.3

ip is Brewster’s angle

2 3

K units 4

27.2

angle ip at the surface of glass. Brewster discovered that when i = ip 90° apart. The angle ip when this happens is called the or the . If rp is the corresponding ip + rp = 90° or rp = 90° – ip

n= =

sin i p sin rp sin i p cos i p

=

sin i p sin(90

ip )

I = I1 + I2 + 2 I1 I 2 = 100 + 1 + 2 100 1 = 121

= tan ip

n = tan ip This equation is called the

SOLUTION Given I1/I2 = 100, i.e. I1 = 100 units and I2 = 1 unit.

I = I1 + I2 – 2 I1 I 2 = 81 and the special .

I 121 = = 1.49 I 81

27.6 Comprehensive Physics—JEE Advanced

27.3 where

and the central bright fringe is measured to be 1.2 periment?

4 D

x 1 x 1

=

7 3

x 1 x 1

I1 25 = x2 = I2 4

or

= (y4 – y0)

27.5 the intensities at points P and Q on the screen where the path difference between the interfering waves is

(i)

4D It is given that (y4 – y0) = 1.2 cm = 1.2 10–2 m, D = 1.4 m and = 0.28 mm = 0.28 m. Substituting these values in (i) and solving, we get 7 m(or 600 nm or 6000 Å). =6

(a) light used.

4

, where

is the wavelength of

SOLUTION I = I1 + I2 + 2 I1 I 2 cos

27.4 where

in an interference pattern is 49 : 9. What is the ratio of

=

2

;

= path difference

I1 = I2 = I0. Therefore, I = I0 + I0 + 2I0 cos = 2I0 (1 + cos )

SOLUTION Given

(a) I ma 49 = I min 9 I

and

2

I1 . I2

x=

n D

For the central bright fringe (n = 0), y0 = 0. For the fourth bright fringe (n = 4), y4 = 4 D/ . Therefore y4 – y0 =

2

I2 I2

49 x 1 2 9 x 1 5 which gives x = . Therefore 2

SOLUTION The position of the nth bright fringe with respect to yn =

I1 I1

=

= 0,

°

I1 = 2I0 (1 + cos 0°) = 4I0 (b)

= I1 + I2 + 2 I1 I 2

Imin = I1 + I2 – 2 I1 I 2 I ma I = 1 I min I1

For

I2

2 I1 I 2

I2

2 I1 I 2

For

=

4

,

=

2

I2 = 2I0 1 cos

4

= 2I0

2

I1 =2 I2

I Multiple Choice Questions with Only One Choice Correct 1. Monochromatic light of wavelength in air is re(a) (b) . The (d) (c) wavelength of light in glass is

2

2

Wave Optics 27.7

2.

air and then in a liquid. It is observed that the 10th

(a) 1 : 1

(b) 1 : 2

(c) 2 : 1

(d)

2 :1

8. (a)

3 2

(b)

4 3

(c)

5 3

(d)

20 17

3.

y when a glass plate of re-

than in air? (a) Fewer fringes will be visible (b) Fringes will be broader (c) Fringes will be narrower (d) No fringes will be observed. 9.

another plate of the same thickness but of refractive

4.

(a)

3y 2

(b)

2y 3

(c)

7y 6

(d)

6y 7

monochromatic light of wavelength . The separation between the slits is = 50 and the distance between the screen and the slits is D = 1000 . If I0 the slits is I (a) 0 2 (c) 2I0

(b) I0 (d) 4I0

5. light from the slit? (a) The fringe width will decrease (b) The fringes will become more distinct (c) The bright fringes will become less bright and

10. What is the effect on the interference fringes in moved closer to the double slit plane? (a) The fringe width increases (b) The fringe width decreases (c) The fringes become more distinct (d) The fringes become less distinct. 11. What is the effect on the interference fringes in source slit is increased? (a) The fringe width increases (b) The fringe width decreases (c) The fringes become more distinct (d) The fringes become less distinct. 12. What is the effect on the interference fringes in two slits are increased? (a) The fringe width increases (b) The fringe width decreases

(d) No fringes will be observed 6.

are in the ratio of 1 : 2, the ratio of the intensities at (a) 1 : 2 (c) 1 : 4

7.

(b) 1 : 3 (d) 1 : 9

as coherent sources of equal amplitude A and of wavelength set up, the two slits are sources of equal amplitude A and wavelength but are incoherent. The ratio

ment affected if one of the slits is covered with black opaque paper? (a) The bright fringes become fainter (b) The fringe width decreases (c) There will be uniform illumination all over the screen (d) There will be a bright central fringe bor-

(d) The bright fringes are no longer 13.

0.5 mm apart and interference is observed on a screen placed at a distance of 100 cm from the slits. It is found that the 9th bright fringe is at a distance of 9.0 mm from the second dark fringe from the centre of the fringe pattern. What is the wavelength of light used.

27.8 Comprehensive Physics—JEE Advanced

(a) 2000 Å

(b) 4000 Å

(c) 6000 Å

(d) 8000 Å

20.

14. A screen is placed at a certain distance from a -

scope?

I. If the width of each slit is doubled I 2 (c) 2 I

21.

imum of wavelength

(a)

and if

is the width of the slit, what is

(b) 2 sin –1 ( / ) (a) sin –1 ( / ) –1 (d) sin –1 ( /2 ) (c) sin (2 / ) 16. A single-slit diffraction pattern is obtained using a beam of red light. What happens if the red light is (a) There is no change in the diffraction pattern. (b) Diffraction fringes become narrower and crowded together. (c) Diffraction fringes become broader and farther apart. (d) The diffraction pattern disappears. 17. A parallel beam of light of wavelength 6000 Å is

the two slit interference pattern within the central (a) 0.1 mm (b) 0.2 mm (c) 0.3 mm (d) 0.4 mm 19. Monochromatic light is refracted from air into glass . The ratio of the wavelengths of the incident and refracted waves is (a) 1 : 1 (b) 1 : (c) :1 (d) 2 : 1

(b)

2

2 1

1

2

(a)

(d)

2

=

(c)

2

2 1

=

2

2

1 2

1

(b)

=

(d)

=

2

2

1 D 1 D

2

-

33.

centre of the screen is I. What will be the intenclosed? (a) I

(b) 2 mm (d) 4 mm

between the slits S1 and S2 is 1.0 mm. What should the

1

wavelengths are missing. The missing wavelengths are (here = 0, 1, 2, is an integer)

length 50 cm. If the lens is placed close to the slit, the distance between the minima on both sides of

18.

is at a distance y2 from its central y1/ y2 will be

22. White light is used to illuminate the two slits in tween the slits is and the distance between the screen and the slit is D (>> ). At a point on the

diffraction pattern is observed on a screen which is

(a) 1 mm (c) 3 mm

2

2

1

is at a distance y1 from

2

(c) above is

(d) 4 I

wavelength

observed. (c) One bright fringe is observed at the centre of the screen. (d) A bright central fringe bordered on both

15.

(b) I

(a)

(c) 24.

I 4

(b)

I 2

(d) none of these

and blue light of wavelength 4000 Å is incident

(a) red (b) blue (c) violet (d) green 25. Light of wavelength is incident on a slit of width . The resulting diffraction pattern is observed on a screen at a distance D. The linear width of the if D equals

Wave Optics 27.9

(a)

(b) 2

(c)

(d)

2

pattern will reveal 2

2

(b) more number of fringes (c) less number of fringes (d) no diffraction pattern

2 26. Two waves of intensities I and 4I superpose, then (a) 5I, 3I (c) 9I, 3I 27.

(a)

n

IIT, 1999

(b) 9I, I (d) 5I, I

32.

IIT, 1982 n, wavev in air, enters

length

,

(c) n, ,

,

v

v

(b) n, (d)

n

, ,

v

wavelength of light used is 6000 Å. If the path difference between waves reaching a point P on the screen is 1.5 microns, then at that point P: (a) Second bright band occurs (b) Second dark band occurs (c) Third dark band occurs (d) Third bright band occurs 33. The difference in the number of wavelengths, when propagates through air and vacuum columns of the

, v

28. A beam of light of wavelength 600 nm from a distant source falls on a single slit 1.0 mm wide and the resulting diffraction pattern is observed fringe is (a) 1.2 cm (c) 2.4 cm

X

is 1.0003, the thickness of the air column is (a) 1.8 mm (b) 2 mm (c) 2 cm (d) 2.2 cm 34.

-

(b) 1.2 mm (d) 2.4 mm

IIT, 1994 29. A parallel beam of monochromatic light is incident

= 3.0

35. r

formed on a screen placed perpendicular to the the diffraction pattern, the phase difference between

(c)

= 4.0. Then

remains unchanged. becomes half unchanged

2 (d) 2 IIT, 1998 –2

30. A string of length 0.4 m and mass 10 clamped at its ends. The tension in the string is 1.6 N. Identical wave pulses are produced at one end at equal intervals of time t. The minimum value of t which allows constructive interference between successive pulses is (a) 0.05 s (b) 0.10 s (c) 0.20 s (d) 0.40 s IIT, 1998 31. Yellow light is used in a single slit diffraction

unchanged 36. Two beams of light having intensities I and 4I interfere to produce a fringe pattern on a screen. The phase difference between the beams is /2 at point A and at point B. Then the difference between the resultant intensities at A and B is (a) 2 I (b) 4I (c) 5I (d) 7I IIT, 2001 37. A double slit apparatus is immersed in a liquid

27.10 Comprehensive Physics—JEE Advanced

1 mm and distance between the plane of slits and

(c) sec

beam of light whose wavelength in air is 6300 Å. What is the fringe width? 0.63 mm (a) (1.33 0.63) mm (b) 133 . 0.63 mm (d) 0.63 mm (c) 133 . 2 IIT, 1996 38.

40.

(b) 0.30 (d) 0.53

3

2 3

=

IIT, 2003

Fig. 27.4

43.

observed to be formed in a certain region of the screen when light of wavelength 600 nm is used. If the light of wavelength 400 nm is used, the number of fringes observed in the same region of the screen will be (a) 12 (b) 18 (c) 24 (d) 8 IIT, 2000 fringes are obtained on a screen placed at some distance of 5 10–2 m towards the slits, the change in the fringe width is 3 10–5 m. If the separation between the slits is 10–3 m, the wavelength of light used is (b) 6 10–7 m (a) 5 10–7 m –7 (d) 6 10–6 m (c) 7 10 m

41.

(d) sec

to be 0.4 mm. If the whole apparatus is immersed in in mm is (a) 0.25 (c) 0.40

39.

=

t is introduced in the path of one of the interfering beams (wavelength

remains unchanged. The minimum thickness of the glass-plate is (a) 2 (b) 2 /3 (c) /3 (d) IIT, 2002 42. In Fig. 27.4, PQ represents a plane wavefront and AO and BP monochromatic light of wavelength . The value of angle BP OP (a) cos

=

2

(b) cos

=

4

is 2.0 mm, The separation between the 9th bright fringe and the second dark fringe from the centre (a) 5.0 mm (c) 15 mm

44.

(b) 10 mm (d) 20 mm

is covered with a transparent sheet of thickness 3.6 10–3 cm the central fringe shifts to a position

(a) 1.50 (b) 1.55 (c) 1.60 (d) 1.65 45. A beam of light, consisting of two wavelengths 6500 Å and 5200 Å is used to obtain interference separation between the slits is 2.6 mm and the distance between the plane of the slits and the screen is 1.0 m. The least distance from the central wavelengths coincide is (a) 1.0 mm (c) 2.0 mm

(b) 1.5 mm (d) 2.5 mm n are

46. interference pattern is n 1 (a) n 1 (c)

n 1 n 1

(b) (d)

n 1 n 1

2

n 1

2

n 1

47. pattern is n. The ratio of the intensities of the two coherent sources is

Wave Optics 27.11

(c) 48.

2

n 1

(a)

n 1 n

2

1

n2 1

(b)

n 1 n 1

(d)

n 1 n 1

2

interference pattern is n 1 n

(c) 49.

n 1 n

(d)

n 1

2

of a fringe formed on a distant screen is 0.1°. If the wavelength of light used is 628 nm, the spacing between the slits is (b) 1.8 10–4 m (a) 0.9 10–4 m (d) 7.2 10–4 m (c) 3.6 10–4 m

n (n 1)

(c)

n n 1

(b)

2

55.

2 n n 1

2

51. Interference pattern is obtained with two coherent light sources of intensities I and 4I a point where the phase difference is /2 is (a) I (b) 2I (c) 3I (d) 5I 52. light of wavelength at this wavelength. The smallest thickness of the sheet to bring (a)

2(

1)

(b)

(

(d)

1)

(c)

(d) 2 53. Monochromatic light of wavelength 500 nm is inof 5 10–4 m. The interference pattern is obtained on a screen at a distance of 1.0 m from the slits. I0. When

(b) 4 mm (d) 2 mm

mum is observed at a point on the screen when light of wavelength 480 nm is used. If this light is

(a) 16 (b) 14 (c) 12 (d) 10 56. A beam of light consisting of two wavelengths 4500 Å and 7500 Å is used to obtain interference distance between the slits is 1 mm and the distance between the plane of the slits and the screen is 120 cm. What is the minimum distance between two successive regions of complete darkness on the screen? (a) 4.5 mm (b) 5.4 mm (c) 2.7 mm (d) 1.2 mm IIT, 2004

2 n (n 1)

(d)

I0 3

54. In Q. 53 above, the lateral shift of the central

50. Interference pattern is obtained with two coherent n. In the interference I ma I in will be pattern, the ratio I ma I in (a)

(b)

(a) 1.5 mm (c) 3 mm

n 1

2

n 1

I0 2 I0 (c) 4

n

(b)

10–6

(a)

have widths in the ratio n:1. The ratio of the

(a)

thickness 1.5

57.

a point P

-

of light used is and is the distance between the slits, the angular separation between point P and the centre of the screen is (a) sin–1 (c) sin–1

(b) sin–1

3

(d) sin–1

2 4

IIT, 2005 58. A parallel beam of fast moving electrons is inciat a large distance from the slit. If the speed of the electrons is increased, which of the following statements is correct? (a) Diffraction pattern is not observed on the screen in the case of electrons of the diffraction pattern will increase will decrease

27.12 Comprehensive Physics—JEE Advanced

(a) (b) (c) (d)

will remain the same. IIT, 2007 -

59.

62.

certain distance on the screen when a transparent sheet of thickness t is introduced in the path of one of the interfering waves. The sheet is then removed and the distance between the screen and the slits is doubled. It is found that the distance

= ( – 1)t

(b)

=

1 ( – 1)t 2

(c)

= ( + 1)t

(d)

=

1 ( + 1)t 2

composed of two wavelengths 1 and 2 close to each other (with 2 greater than 1) is used. The order n up to which the fringes can be seen on the 2

(a) n =

same as the observed fringe shift when the sheet was introduced. The wavelength of light used is IIT, 1983 (a)

3 4 5 much greater than 5.

(c) n =

2

2(

(b) n =

1 2 2

(d) n =

1)

The wavelength of light used is 6000 Å. What will mental arrangement is immersed in water? Refrac4 . 3 (a) 0.15° (b) 0.18° (c) 0.2° (d) 0.27° 61. A coherent parallel beam of microwaves of wave-

D 3

(b)

(c)

D

(d) 2

Fig. 27.5

ANSWERS 2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62.

(b) (c) (d) (c) (b) (c) (b) (a) (b) (c) (d)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63.

(a) (d) (b) (a) (b) (b) (b) (a) (d) (d) (a)

2(

2 D 3

(a)

slit apparatus. The separation between the slits is 1.0 mm and the screen is placed at a distance of 1.0 m from the slits. The number of minima in the interference pattern observed on the screen is IIT, 1998

(c) (c) (c) (c) (c) (a) (c) (c) (c) (a) (b)

2

1 1 2

1)

63. Monochromatic light of wavelength emerging from slit S illuminates slits S1 and S2 which are placed with respect to S as shown in Fig. 27.5. The distances x and D are large compared to the separation between the slits. If x = D/2, the minimum value of so that there is a dark fringe at the centre P of the screen is

60.

1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61.

1

4. 10. 16. 22. 28. 34. 40. 46. 52. 58.

(c) (d) (b) (a) (d) (c) (a) (d) (a) (c)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59.

(c) (d) (c) (c) (d) (c) (d) (a) (d) (b)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60.

(d) (d) (b) (c) (b) (b) (a) (c) (a) (a)

D 3

Wave Optics 27.13

SOLUTIONS 1.

of light remains the same, we

have

c=

and

v=

6. The amplitudes of the two coherent waves will be A1 = 2A and A2 = A. Therefore I ma = I min

where c = speed of light in air and v = speed of light in glass. Therefore c v

=

=

.

10 D

1 2

and y8 = 8

y10 10 = y8 7.5

D

4 3

=

Given y10 = y8

( =

y = (1.5 – 1)

For point P,

2 =

Phase difference P is

=

50 2

3y 2

= 25

25 50 1000

=

2

2

=

I = 4I0 cos2 = 4I0 cos2 5.

y =

D

For the point P of the slits, y=

tD

tD

y 0.75 3 = = y 0.5 2 =

)

tD

y = (1.75 – 1)

4. Path difference

=

5 4 5 5 = 4 2

2 5 4

I2

2

=

2A 2A

A =9 A

= 2I0

from the covered slit will decrease resulting in a difference in the intensities of the two virtual sourc-

(A 2 + A2) or I2 = k

2A2

I1 4 k A2 = =2 I2 2 k A2

4 , which is choice (b). 3

3. Displacement of fringes = ( – 1)

A2 A2

7. If the two sources are coherent, the resultant amplitude at the midpoint of the screen due to interference = A + A = 2 A I1 (2 A) 2 or I1 = k 4 A 2 where k is a constant of

So the correct choice is (c). 2. y10 =

A1 A1

8.

periment will be less in water than in air. Since the

it follows from the relation = v / that the wavelength in water is less than in air. Since fringe width , the value of the correct choice is (c). 9. We then obtain a single slit diffraction pattern on 10. and 11. Let x be the width of the source slit and X the distance between the source slit and the plane of the two slits. For interference fringes to be disx /X < / should x is too large (i.e. the source slit is too wide) or if X is too small (X is the distance between the source slit and the two slits) the requirement x / X < / es will no longer be distinct. The reason is that the interference patterns due to various parts of

remains visible, a change in x or X has no effect on the fringe width . 12. The single slit diffraction effects at the two slits becomes important and as a result, the interference

13. The distance of the central fringe is

th bright fringe from the

27.14 Comprehensive Physics—JEE Advanced

y = where

since is small, sin . Therefore, = 3 10 –3 rad. If the lens is placed close to the slit then

=

x = f tan

= D / is the fringe width. y9 = 9

The distance of the th dark fringe from the central fringe is 1 1 D = y = 2 2 (ii)

From Eqs. (i) and (ii), we get y9 – y 2 = 9 –

3 2

2 x = 2f = 2

)

10–3

3

10 –3 m = 3 mm.

18. Let the width of each slit be a. The linear separation between bright fringes in the double slit y = Since y << D, the angular separation between bright fringes will be

9.0 2 = 1.2 mm 15 = 1.2 10 –3 m

14.

0.5 =3

=

=

Now =

tan

between two minima on both sides of the central

15 = 2

It is given that y9 – y 2

is small,

where x

(i)

3 y2 = 2

(

f

y = D

For 10 bright fringes we have

/ D. Substituting for , and D, we get = 6 10 –7 m = 6000 Å

10

=

10

(i)

in the diffraction pattern due to a slit of width a is

a bright central fringe bordered on both sides with

2 a Equating Eqs. (i) and (ii), we get 2

15. sin where

=

10

is the angular separation between the

of it so that 2 is the angular width of the central = sin –1

2 = 2 sin –1

<< , then sin , where is in radians. In that case, the =2 / . 16. The wavelength of blue light is less than that of the will decrease which means that the fringes become narrower and crowded together. 17. The angular separation of the minima on both sides where sin

1

=

=

6000 10 0.2 10

10 3

=3

10 –3

=

=

2 a

or a =

5

=

(ii)

1.0 mm 5

= 0.2 mm. 19.

n of the light does not change as light travels from air into glass, we have va = n a and v = n Therefore

20. I = I

a

=

va = v

= I1 + I2 + 2 I1 I 2

When the width of each slit is doubled, I1 becomes 2 I1 and I2 becomes 2I2. Therefore, I =I

= 2 I1 + 2I2 + 2 2 I1

= 2 (I1 + I2 + 2 I1 I 2 ) = 2 I

2 I2 = 2I.

Wave Optics 27.15

21. We know that y = length

1,

y1 =

. Therefore, for wave-

10

1

D

5

2

D

y2 =

22. The distance of the

and for wavelength y1 2 = y2

,

1

where

2

or

is the width of the slit. The linear width of

for a value of D 2 D

D 26. Given

of the slits, y =

2

25. If D >> , the linear width of the central principal 2 D , distance D =

th minimum from the centre 1 2

y=

2,

1 2 2

2

or D =

2

= I1 + I2 + 2 I1 I 2 = I + 4I + 2 4 I 2 = 9I

D and

2

=

=

I1 = I and I 2 = 4I. Now I

=

= 3867 Å

which is close to violet light ( the correct choice is (c).

I min = I1 + I 2 – 2 I1 I 2 = 5I – 4I = I

1

-

27. 23. Let I0

to

=

glass is v =n

I = I0 + I0 + 2 I 0 I 0 = 4 I0 I 4 24. When a wave of wavelength falls at an angle of incidence i and thickness t, then the condition for constructive or

I0 =

2

1 2

t cos r =

;

= 0, 1, 2, 3,

where r normal incidence i = 0, hence r = 0. Therefore ( = 1 for air) 1 2

2t = or Now

Wave of wave of

=

2

,

3 5 , 2 2

etc.

4t 4t = 4t, , , etc. 1 3 5 2 t = 0.00029 mm = 2.9 10 –5 cm = 2900 Å. Therefore, = 11600 Å, 3867 Å, 2320 Å, etc. =

2t

= 11600 Å is in the infrared region and = 2320 Å is in the ultraviolet. These

=

n

=

v

28. The angular separation between the th dark fringe a sin

= = 1. Therefore

a sin

1

=

or sin 1 = /a. Since << a, sin 1 ~ 1 1 = /a. This is also the angular separation between

either side of the central bright fringe = 2 1 = 2 /a . Therefore, their separation at distance = 2 m is 2 a

=

2

= 2.4

(600 10

9

1.0 10

3

)

2

10–3 m = 2.4 mm

29. If follows from Fig. 27.6 that the path difference at the A and B is = BC = a sin 1 But sin 1 = /a. Therefore, = a /a = . A path difference of corresponds to a phase difference of 2 .

27.16 Comprehensive Physics—JEE Advanced

10 – 6 =

or1.5

1 2

n

10 – 7 )

(6

1 5 or n 2 2 difference of 1.5 10–6 m corresponds to the third dark fringe. Thus the correct choice is (c).

which gives n –

33. Let

a

and

va and c be their respective speeds in air and vacuum. Since the have =

Fig. 27.6

= 10 kg/0.4 m = 2.5

–2

–1

10 kg m .

Speed of each pulse is v =

c

=

a

30. Mass per unit length of the string is –2

va

c = va

or

T a

a

. But c = va

=

or

a

=

(1)

a

1.6

=

2.5 10

2

= 8 ms

–1

To have a constructive interference, the pulses must arrive at the same end of the string in the same L is the length of the string, the minimum value of t is 2L 2 0.4 = = 0.1 s ( t)min = v 8 31.

10

–10

Å m. Now, the angular width of the central

2 D a where D is the distance of the screen from the slit. Since 0

=

32. Given = 6000 Å = 6000 10–10 m = 6 10–7 m and = 1.5 microns = 1.5 10–6 m. For bright fringes: = n ; where n is an integer. n=

=

1.5 10 6 10

6 7

=

5 , which is not an integer. 2 10–6 m does not

correspond to a bright fringe. For dark fringes, we have 1 = n 2

a

a

Now, if t is the thickness of each column, then the number of wavelengths in the two media are t t na = and n = .Given (na – n) = a

l=

t

t

=

t

a

1

(2)

a

Using (1) in (2), we have t ( a – 1) 1=

(3)

Given a = 1.0003 and = 6000 Å = 6000 cm. Using these values in (3), we have t l= (1.0003 – 1) 6000 10 8 or

t=

6000 10 0.0003

8

= 0.2 cm = 2 mm

34. Referring to Fig. 27.7, the direction we have the n sin = n sin or n= If

10–8

along which

= 2 , we have n = 2 sin

is 90 , n = 2 sin 90 = 2. Thus, there are two interference (which corresponds to which is choice (c).

Wave Optics 27.17

l

D

=

=

l

6300 10 133 .

10

133 . 3

1 10

10–3 m = 0.63 mm

= 0.63

0.4 mm = 0.3 mm, which is choice (b). nl 4/3 width of region L or n = . 39. Number of fringes = fringe width 38.

l

=

a

Now, fringe width Fig. 27.7

35.

n=

r

=

4 =2

If c is the speed of the electromagnetic wave in vacuum and v in the dielectric medium, then c c c or 2 = or v = n= v v 2 When a wave passes from one medium into 0 and are the wavelengths of the wave in vacuum and v= v c

and c = or

=

0

0

0,

c /2 c

which give 1 or 2

=

2

IA = I + 4I + 0 = 5I IB = I + 4I + 2 I

2 4 I cos

37. The fringe width in liquid is

where

a

=

=

5 10

10–5 =

3 which gives

=6

10

. Given

is the fringe width in air. Thus

D =

= 10–3 m. Using 2

3

10–7 m which is choice (b).

41. When a transparent plate is introduced in the path of one of the interfering beams, the entire fringe y = ( – 1)t, where t is the thickness of the

or ( – 1)t =

or

3 2

1 t=

or t = 2 .

42. Since P and Q are points on the same wavefront, ference at point P BP and the reOP is = QO + OP

(i)

PR . Also cos cos in triangle QOP, QO = OP sin (90° – 2 ) = OP cos 2

Now, in triangle POR, OP =

= OP cos 2 + OP = OP (cos 2 + 1) = 2OP cos2

a

nl D

5 10–2 m, = 3 10–5 m and these values, we have

y=

= 5I – 4I = I IA – IB = 5I – I = 4I, which is choice (b).

l

D

=

-

I = I1 + I2 + 2 I1 I 2 cos . Given I1 = I and I2 = 4I.

and

which is choice (b). D . Therefore, 40. =

0

(c). 36.

600 = 18, 400

= 12

n =n

n of a dielectric medium is given

is proportional to wavelength.

=2

cos

cos2

= 2 cos

27.18 Comprehensive Physics—JEE Advanced

Since there is a sudden path change of re ence at P is

or

3 = , , 4 4

cos

or

where

=

etc.

y9 = 9

(1) th dark fringe from the 1 2

y =

D

=

15 2

3 2

=

y0 = 30 But 30

y0 =

D

D

D

=

and n

Substituting the values of , , D and , we get ymin = 1.0 mm, which is choice (a). I 46. Given 1 = n. Therefore, the amplitude ratio is I2

(2)

A1 = n A2 I = (A1 + A2)2 and Imin = (A1 – A2)2 ( A1 I ma = ( A1 I min

A2 )

2

A2 ) 2

=

15 2

D

2

A1 A2

1

A1 A2

1

2

n 1

=

2.0 mm = 15 mm

n 1

2 2

the correct choice is (d). I 47. Given ma = n. I min ( A1 ( A1

D

( A1 ( A1

or

( – 1)t

which gives

( – 1)t

30 (6000 10 10 ) 30 = = 0.5 (3.6 10 5 ) t = 1.5 45. Let the nth bright fringe of wavelength n and the th bright fringe of wavelength coincide at a distance y or

5 4

ymin =

Now

44. y30 = 30

6500 5200

i.e., the 5th bright fringe of wavelength 5200 Å coincides with the 4th bright fringe of wavelength 6500 Å. The smallest value of y at which this happens is

1 2

3 y2= 2 From Eqs. (1) and (2) we get y9 – y 2 = 9 –

n

= 5 and n = 4

is the fringe width.

The distance of the central fringe is

n

=

D

The least integral values of the above condition are

43. The distance of the th bright fringe from the central fringe is D y = = D

n

y=

3 = , etc. 2 2 3 = , , etc. 2 2

2 cos

or

2

due to

A2 ) 2 A2 ) 2

=n

A2 ) = A2 ) A1 = A2

n n 1 n 1

I1 n 1 = I2 n 1 the correct choice is (a).

( – 1) =

48.

2

proportional to its width. Since the amplitude is we have

Wave Optics 27.19

A1 A2

I1 I2

n 1

Ir = I + I + 2 I = 2I – 2I = 0

n

As shown in solution of Q. 46, the correct choice is (c).

I cos 3 (

cos 3 = – 1)

is a dark fringe at the centre.

49. or

=

=

(1)

Given = 628 nm = 628 10–9 m and = 0.1° 0.1 = rad. Using these values in Eq. (1), we 180 = 3.6 10–4 is (c). 50. The correct choice is (b). Refer to the solution of Q. 46. 51. The correct choice is (d). Use I = I1 + I2 + 2 I1 I 2 cos where

I1 = I, I2 = 4I

. 2 52. If t is the thickness of the glass sheet, the fringes are y = ( – 1)

and

=

54. The correct choice is (a). The lateral shift is given y = ( – 1) 55. The position of the n n D yn = For a given point yn D and are n2 n = constant, i.e. n1 1 = n2 2 20 480 1 = n1 = = 16, which is choice (a). 600 2 56. Let the nth dark fringe of wavelength n and the th dark fringe of wavelength coincide at a distance y from the centre of the screen, then

the central bright fringe), the fringes must be

or

D

, which is choice (a). 2( 1) 53. When a transparent plate of thickness t and refracis introduced in one of the interfering waves, the path difference at the centre of the screen is tD = ( – 1) Phase difference =

2

=

2

( – 1)

1 2 1 2

= n

(1)

= 1.5, Given = 5000 Å = 5 10–7 m, –6 t = 1.5 10 m, D = 1 m and = 5 10–4 m. Using these values in Eq. (1). we get = 3 . If I is

1 2

D

(1)

7500Å 5 = 4500Å 3

=

regions of darkness ar

and tD

=

6n 2 3n 1 = (2) 10 5 Integral values of n and n = 3, = 2; n = 8, = 5; and so on. Let n1 = 3 and n2 = 8, then from Eq. (1) the distances from which gives

t=

nD

At this position there is complete darkness on the screen. Eq. (1) gives n

1 = y= 2 2 tD 1 D ( – 1) = 2

1 2

y= n

tD -

tD

y1 = n1

1 2

nD

y2 = n2

1 2

nD nD

y = y2 – y1 = (n2 – n1) = (8 – 3) = 2.7

4500 10

10 3

1 10 10 –3 m = 2.7 mm,

1 2

27.20 Comprehensive Physics—JEE Advanced

where

57. If is the phase difference between the interfering waves at point P P is I= I Given

cos2

sin

a where a is the width of the slit. Thus, if decreases, and hence 2 will decrease. Therefore, the correct statement is (c). 59. When a sheet of thickness t is introduced in one of the interfering waves, the distance y0 through which the fringes shift is

2

I ma 2 1 = , which gives 2 2

I= cos2

2

or

4

=

2 The angular separation between points P and O = y/D. Since tan ~ sin y (1) sin = D If is the fringe width, then y /2 1 = (2) 2 4 This is so because the phase difference . Now

y0 = ( – 1)t

D

4

or

= sin–1

4

D

=

When the distance D between the slits and the screen is doubled, the new fringe width becomes

between D = .

D

=2

1 4

y = D 4 Using Eq. (1) in Eq. (3), we have =

(1)

(2)

It is given that y0 = . Equating Eqs. (1) and (2) we get D D 2 = ( – 1) t

or

sin

D

The fringe width , i.e., the distance between suc-

Using this in Eq. (2), we get =

=

=

(3) 60.

1 ( – 1)t 2

which is choice (b)

, which is choice (d).

= In air:

a

=

a

= =

or

a

= Fig. 27.8

58. De Broglie wavelength of electron is h h p v If speed v of electron is increased, momentum p(= v will decrease. Now, the angular width of the central =

=

(1)

a

speed of light in air speed of light in water

a

va v

where unchanged. Using Eq (2) in Eq (1) we have 0.2 = 0.15° 4/3 So the correct choice is (a). =

a

a

a

(2)

Wave Optics 27.21

61. Refer to Fig. 27.9. When the incident beam falls S1 and S2, the path difference at the central point P0 P 0.

or

n

=

1

1 2

n

which gives n=

2

2

or 2n

2

= (2n – 1)

1

2

, which is choice (d).

2

1

63. Refer to Fig. 27.10. To reach point P, wave 1 has to travel a path (SS2 + S2P) while wave 2 has to travel a path (SS1 + S1P). Therefore, when the waves arrive at P, the path difference is Fig. 27.9

Let the minima appear along directions with respect to the incident direction. Coherent waves from S1 and S2 along this direction are brought to a focus at P. It is clear that the path difference between the waves from S1 and S2 on reaching P is = sin The interference minima will appear on the screen if 1 = 2 1 or sin = 2 where is an integer. Thus the directions of mini1 2 = 1.0 mm and = 0.5 mm. Therefore

sin Given

1 0.5 1 1 2 1.0 2 2 The allowed values of are those integers for which sin is not more than + 1 or less than – 1. These values are minima will be observed. The correct choice is (b) sin

62.

=

=

wavelengths close to each other, two interference patterns corresponding to the two wavelengths are remains distinct upto a point on the screen where the nth 1 = 5890 Å falls on the nth order minimum of the other wavelength 2 = 5895 Å. Thus, interference pattern can be seen upto a distance yn from the centre of the screen if n 1D ; (n yn = =

n

1 2

2D

; (nth minimum) (2)

Fig. 27.10

= (SS2 + S2P) – (SS1 + S1P) Now, in triangle SS2S1, we have SS2 = (x2 +

2 1/2

)

=x 1

2 1/ 2

x2

2

= 1

(1)

(

2 x2 2

2 1/2

S2P = (D +

)

<< x) 2

=D 1

2 D2 ( << D) Also (SS1 + S1P) = x + D. Using these in Eq. (1), we have 2

=x 1 =x+ or

2 x2 2

+D+

2x 2

=

1 x

2

2

+D 1 2

2D

–x–D

1 D

In order to have a dark fringe at P, 2

=

or D 2 choice is (a).

Putting x =

2

2 =

– (x + D)

2 D2

1 x

1 D

xD x D

=

2 (1)

1/ 2

(2)

27.22 Comprehensive Physics—JEE Advanced

II Multiple Choice Questions with One or More Choices Correct 1.

(c) the 5th dark fringe is at a distance of

is no change in its (c) wavelength

the central fringe.

(d) speed

(d) the 5th dark fringe is at a distance of

2. When a light wave travels from air into glass, which of the following will change? (c) wavelength 3.

with light of wavelength 1 is 1 and with light of wavelength 2 is 2. Using light of wavelength 1, the fringe width becomes 3 if the entire apperatus is immersed in a transparent liquid of refractive . Then (a) (c)

4.

from the central fringe. 7.

(d) speed

2

=

3

=

1 1

1 2

(b) (d)

2

=

3

=

fringe of wavelength central fringe. Then

1

9x 20

is at a distance x from the

(a) the 5th bright fringe of wavelength 2 will be x at a distance of 1 from the central fringe. 2 2 (b) the 5th bright fringe of wavelength 2 will x be at a distance of 2 from the central 2 1 fringe. (c) the 5th dark fringe of wavelength 2 will be at 9 2x a distance of from the central fringe. 20 1

2

1

x from 2

1

1

of equal widths, one slit is made twice as wide as the other. Then in the interference pattern, the in-

(d) the 5th dark fringe of wavelength 2 will be at 9 1x a distance of from the central fringe. 20 2 8.

(c) of minima will increase (d) of minima will decrease IIT, 2000 5.

of the slit is made double its original width. Then become (a) narrower (c) broader

6.

(b) fainter (d) brighter

fringe is at a distance x from the central fringe. Then 19 x (a) the 10th dark fringe is at a distance of 20 from the central fringe. 21x (b) the 10th dark fringe is at a distance of 20 from the central fringe.

is observed on a screen placed at a distance of 100 cm from the slits. The wavelength of light used is 600 nm. (a) the distance of the 4th bright fringe from the central fringe is 8 mm. (b) the distance of the 4th dark fringe from the central fringe is 7 mm. (c) the distance between the 9th dark fringe and the second bright fringe on the same side of the central fringe is 15 mm. second bright fringe on the opposite sides of the central fringe is 5 mm. 9. A narrow slit of width 1.3 10–6 m is illuminated

is 30°.

Wave Optics 27.23

is 45°. (d) the angular separation between the central 3 . 4 10. A parallel beam of monochromatic light of waveis sin–1

width 0.3 mm. The diffraction pattern is observed on a screen which is placed at the focal plane of

x1 and x2 (a) x1 = 1 mm (b) x1 = 2 mm (c) x2 = 3 mm (d) x2 = 6 mm 11. A beam of light consisting of two wavelengths 750 nm and 450 nm is used to obtain interference fringes between the slits is 1 mm and the distance between the plane of the slits and the screen is 100 cm. The bright fringes due to both the wavelengths coincide is ymin and y min is the corresponding distance where the dark fringes due to both the wavelengths coincide. Then IIT, 2004 (b) ymin = 2.0 mm (a) ymin = 2.25 mm (c) y min = 4.5 mm 12.

(d) y min = 0.1 mm

covered with a thin glass sheet of thickness t and 1 while the lower slit is covered with another glass plate of the same thickness t but 2 (> 1). Interference pattern is observed using light of wavelength . It is observed that the point P on the screen where the n = 0) fell before the plates were

2 3 5 (d) 3

(a)

is 60°. (c) the angular separation between central

(b)

3 4 (c) 3 13.

between the slits is 2 mm and the distance of the screen from the plane of the slits is 2.5 m. A light of wavelengths in the range 200 nm to 800 nm is allowed to fall on the slits. The wavelengths in the visible region that will be present on the screen at

(a) 400 nm (b) 500 nm (c) 600 nm (d) 700 nm 14. A parallel beam of light containing two wavelengths and slit. Fraunhofer diffraction pattern is obtained on a screen placed at the focal plane of a lens of focal length 0.5 m. It is observed that the second minimum of and the third minimum of overlap at the same point on the screen 2.5 mm from the centre of the screen. If = 6000 Å, (a) = 4000 Å (b) = 3000 Å (c) Slit width = 0.20 mm (d) Slit width = 0.24 mm 15 between the two slits is and the wavelength of the light is Choose the correct choice(s). (a) If = (b) If

<

< 2

served on the screen

increased so that it becomes equal to that of slit 1, the intesities of the observed dark and bright frings will increase.

SOLUTIONS

3.

1

=

1D

,

1

=

2D

, Therefore,

-

increased so that it becomes equal to that of slit 2, the intensities of the observed dark and bright fringes will increase.

phase difference between the interfering waves at point P now is

1. The correct choices are (b), (c) and (d). 2. The correct choices are (a), (c) and (d).

IIT, 1997

2

2

=

1

Now

, which is choice (b).

1

l

=

a

=

1

. Therefore,

27.24 Comprehensive Physics—JEE Advanced

3

=

1 l

1

1

4. In the case when the slits are of equal width, the

and

= I0 + I0 + 2 I 0 I 0 = 4I0

Imin = I0 + I0 – 2 I 0 I 0 = 0

I1 = 2I0 but I2 = I0

S1 is made twice as wide as the S1 is doubled, i.e.

= I1 + I2 + 2 I 1 I 2

I

= wavelength and a = width of the slit. 2, where 2

3 2a 10. The correct choices are (b) and (d). The linear 2f and a 3f . a 11. Let nth bright fringe of wavelength n and the th bright fringe of wavelength coincide at a distance y from the centre of the screen. Then sin

I0. Then I

where

, which is choice (c).

= 2I0 + I0 + 2 2 I 0 I 0 = 3I0 + 2 2 I and

0

y =

= 5.83 I0

or

Imin = I1 + I2 – 2 I 1 I 2

or

= 2I0 + I0 – 2 2 I 0 I 0 = 3I0 – 2 2 I Thus I

>I

0

or

= 0.17I0

and I min > Imin,

xn =

n D

The distance of the nth dark fringe from the central 1 D 2 7. The correct choices are (b) and (c). 8. All the four choices are correct. 9. The correct choices are (b) and (d). The angular 1 where 1 is x*n = n

sin

1=

a

D

=

n

=

n

n

750 = 450 n

5 . The minimum integral values of and n 3 n = 5 and n = 3. Therefore, the minimum value of y is

decreases to half its earlier value. This implies that thermore if a -

6. The correct choices are (a), (c) and (d). The distance of the nth bright fringe from the central fringe

n

or

so the correct choice are (a) and (c). 5. The angular width of the /a where a is the width of the slit. If the value of a is

The correct choice are (a) and (d).

n

1=

ymin =

n

nD

3 750 10 10

9

1

3

= 2.25 × 10–3 m = 2.25 mm wrong. 12.

periment, slits S1 and S2 are equidistant from the source S light emerging from S1 and S2 has the same intenI0. Before the plates are introduced, the path difference (and hence the phase difference ) at the central point P between the interfering waves P is I = I0 + I0 + 2 I 0 I 0 cos 0° = 4I0

Fig. 27.11

Wave Optics 27.25

When the two plates are introduced the path difference at the central point P between the interfering waves becomes = ( 2 – 1)t – ( 1 – 1)t = ( 2 – 1) t Phase difference

=

2

2

(

2

–

2 a The third minimum of

1)t

3

P becomes

Given have

I = I0 + I0 + 2 I 0 I 0 cos = 2I0 + 2I0 cos 3I 4 I 0 = 3I0. Thus 4 4 3I0 = 2I0 + 2I0 cos 1 or cos = 2 5 7 or = , etc. , , 3 3 3 Thus the correct choices are (a) and (d). 13. The distance of the n n D

Given Thus

=2 = =

or

=

10

n 1 n

8

2 10 2.5

n= 1 ;

1

For

n= 2 ;

2

For

n= 3 ;

3

For

n= 4 ;

4

3 a

3

(3)

2 3 = a a =

2

=

2f a

or a =

2f y

(4)

Given f = 0.5 m, = 6000 Å = 4 10–7 m and y = 2.5 mm = 2.5 10–3 m. Using these values in Eq. (4) we get a = 2.4 10–4 m = 0.24 mm. So the correct choices are (a) and (d).

n

3

15. Path difference

=

sin

=n

n = 0, 1, 2,

n = 0),

sin 0 = 0 which gives 0 mum appears at point P0. (n = 1), sin 1= . For = , we have sin

8000 Å n

n is an integer having values 1, 2, 3, For

lies along a direction n = 3 in Eq. (1).

3

y= f

nD

10–7 m =

=

or

10–3 m, D = 2.5 m and yn = 10–3 m. 3

2

=

(2)

2 2 6000 Å = 4000 Å 3 3 If y is the distance of the second minimum of (or third minimum of ), then, using Eq. (1), we have

Given that I =

yn =

=

2

etc.

1

= 8000 Å

(a) is correct.

= 4000 Å 8000 Å = 3 = 2000 Å

If 2667 Å

Wavelength 8000 Å is in the infrared region, wave-length 4000 Å is in the visible region and wavelengths 2667 Å and 2000 Å are both in the ultraviolet region of the electromagnetic spectrum. 14. The nth minimum lies along a direction n a sin n = n ; n = 1, 2, 3, etc. ( or a n = n n is small) n or (1) n = a The second minimum of lies along a direction 2 n = 2 in Eq. (1).

sin

1

lies between

lies choice (b) is also correct.

1

= 1 or -

1 and 1, i.e. 2

1

-

27.26 Comprehensive Physics—JEE Advanced

If I1 = I = I2, than I

Ir = I1 + I2 + 2 I1 I 2 cos I and

= I1 + I2 + 2 I1 I 2

If I1 = 4I = I2, then I

Imin = I1 + I2 – 2 I1 I 2

For I1 = 4I and I2, = I, we have I

= 9I and Imin = I.

= 4I and Imin = 0. So the in= 16 I and Imin

incorrect. Choice (a) and (b) are correct.

III Multiple Choice Questions Based on Passage Questions 1 to 3 are based on the following passage Passage I romatic and parallel beam of light of wavelength 6000 10 Wm–2 circular apertures A and B of radii 0.001 m and 0.002

1. The ratio of the powers received at aperture A to that at aperture B is (a) 1 : 2

(b) 1 : 4

(c) 1 : 8

(d) 1 : 16

2. The phase difference between the interfering waves at point F is (a)

is placed in front of aperture A (Fig. 27.12). IIT, 1989

(c) 3.

Fig. 27.12

(b)

6

(d)

4

3 2

in the original direction, the resultant power at point F will be (a) 5 W

(b) 6 W

(c) 7 W

(d) 8 W

SOLUTION I) =

1

10

Wm–2

=

Power received at aperture A = I cross-sectional area of A 10 = (0.001)2 = 10–5 W Power received at aperture B = So the correct choice is (b). 2. The phase difference at F is = ( – 1)

t

2

10

=4

(0.002)2 10–5 W

(15 . 1)

8

(2000 10 (6000 10

8

)

2

)

3

rad

The correct choice is (b). 3. Since 10% the power received at each aperture goes in the original direction, the power at point F PA = 10% of 10–5 W = 10–6 W PB = 10% of 4

10–5 W = 4

10–6 W

the square of the amplitude. If A1 and A2 are the amplitudes at F due to the two sources, we have PA = kA21 and PB = kA22, where k is the proportional-

Wave Optics 27.27

PA and A2 = k Resulting amplitude A=

A12

A22

7 10 6 k Resultant power at F = kA2

PB k

A1 =

A=

2 A1 A2 cos

7 10 k

=k

Substituting the values of A1, A2 and , we get

6

= 7 10–6 W

Questions 4 to 6 are based on the following passage Passage II

IIT, 1990

4. (a)

3I 16

(b)

3I 32

9I 9I (d) 32 64 5. The ratio of the intensities of beams 1 and 2 is 16 4 (a) (b) 9 3 25 5 (c) (d) 16 4 6. in the interference pattern is (a) 16 : 1 (b) 25 : 1 (c) 36 : 1 (d) 49 : 1 (c)

Fig. 27.13

I is incident on a glass plate A as shown in Fig. 27.13. Another identical glass plate B is kept close to A and incident on it and transmits the remaining. Interference

SOLUTION I is incident on plate I

4. A I1 = I

25 100

3I 4

I 4

25 3I = 100 16 I/16 falls on plate A which

of beam 2 is I2 =

3I 16

75 100

9I 64

So the correct choice is (d). I /4 16 = , which is choice (a). 9 I / 64 9 a 16 4 . Thus . 6. The ratio of amplitudes is 1 a2 9 3 5.

I1 I2

a1 = 4 units

Fig. 27.14

I/4 falls on plate B which B is

and

I ma = I min

a1 a1

a2 = 3 units. a2 a2

2

Thus the correct choice is (d).

4 4

3 3

2

= 49

27.28 Comprehensive Physics—JEE Advanced

Questions 7 to 9 are based on the following passage Passage III Two parallel beams of light P and Q (separation ) each containing radiations of wavelengths 4000 Å and 5000

where

is a positive constant. The value of

the face AC IIT, 1991 7.

= 1.20 +

is in Å and

2

( = 4000 Å, = 5000 Å) (a) greater for beam than for beam (b) less for beam than for beam (c) equal for both the wavelengths

8. The value of is (given sin (a) 2 105 (Å)2 (c) 8 9.

Fig. 27.15

= 0.8) (b) 4

105 (Å)2

(d) 1

104 (Å)2

105 (Å)2

-

length is (a) 1.5 (c) greater than 1.25 (d) less than 1.25.

2

(b)

SOLUTION 7. Given

= 4000 Å and = 1.20 +

2

= 5000 Å. Also (1)

It follows from this relation that the value of is greater for than for . The angle of deviation is greater for greater value of choice is (a). 8. and critical angle ic is sin ic =

Fig. 27.16

1

Thus, the smaller the value of , the greater is the angle ic. It follows from relation (1) that for is greater than for ic for is less than i c for . It follows from Fig. 27.16 that the angle of incidence at face AC is the same (= ) for both beams. It is given that the condition of total interAC of the wavelengths. Since ic for (= 4000 Å) is less than i c for (= 5000 Å), it is obvious that the radiation of wavelength = 5000 Å is transmitted through the face AC (see Fig. 27.16)

=

1 sin ic

1 sin [

Substituting we have

= 1.25 and

1.25 = 1.20 +

1 = 1.25 0.8 sin

= 0.8 (given)]

= 4000 Å in relation (1),

(4000)2

which gives = 8 105 (Å)2, which is choice (c). 9. Using = 8 105 (Å)2 and = 5000 Å in Eq. (1), we get 8 105 = 1.20 + = 1.232 (5000) 2 So the correct choice is (d).

Wave Optics 27.29

Questions 10 to 13 are based on the following passage

11. If

Passage IV 3=

t

of wavelength

1.5 is coated with a 2 = 1.8. Light

= 648 nm, the least value of t for which the

(a) 90 nm (c) 108 nm 12. (a) t =

10.

n 2

IIT, 2000

n (a) t = 2

(b) t = 2

n (c) t =

2

1 2

n 2( n

(d) t =

2

3)

2

2

1 2

(b) t = 2

n (c) t =

(n is an integer)

(b) 180 nm (d) 216 nm

13. If

2

1 2

n 2

3

n (d) t =

2

2

1 2 3

= 648 nm, the least value of t for which the

(a) 90 nm

(b) 180 nm

(c) 108 nm

(d) 216 nm

3

SOLUTION 10.

( 1 thickness t

= 1.8) of P as wave meeting

2

the surface of the glass plate ( 3 at point Q and travels along QP.

path for wave 2 from P to Q and from Q to P in the 2(PQ)

2

= 2 2t = 2 2t Optical path difference between waves 1 and 2 at point p is =

2

–

1

= 2 2t –

2 Now, for constructive interference, 1, 2, 2 2t –

or

Fig. 27.17

Waves 1 and 2 meet at point P We know that when a wave is travelling in a rarer denser medium, it undergoes a phase change of or a path change of /2. Thus wave 1 has an optical path of 1 = /2. Wave 2 travelling from P to Q in the at Q from

t=

= n or 2 2t =

2

n

1 2

1 2

n or

n ; n = 0,

2

2

So the correct choice is (c). 11. The minimum value of t corresponds to n = 0. tmin =

=

4

2

648 nm = 90 nm. 4 18 .

So the correct choice is (a) 12. For destructive interference 2

2t

–

2

= n

= n 1 2

1 2

27.30 Comprehensive Physics—JEE Advanced

which gives t =

n 2

, which is choice (a).

tmin =

2

13. The minimum value of t corresponds to n = 1. Questions 14 to 16 are based on the following passage Passage V Figure 27.18 shows a surface XY separating two transparent media, medium-1 and medium-2. The lines and represent wavefronts of a light wave traveling in medium-1 and incident on The lines of ef and represent wavefronts of the light wave in medium-2 after refraction.

Fig. 27.18

IIT, 2007

2

= 2

648nm = 180 nm 2 1.8

Thus the correct choice is (b). 14. Light travels as a (a) parallel beam in each medium (b) convergent beam in each medium (c) divergent beam in each medium (d) divergent beam in one medium and convergent beam in the other medium 15. The phases of the light wave at and f are , , and c e f c, f. (a) c cannot be equal to (b) can be equal to e (c) ( – f) is equal to ( c – e) (d) ( – c) is not equal to ( f – e) 16. Speed of light is (a) (b) (c) (d)

the same in medium-1 and medium-2 larger in medium-1 than in medium-2 larger in medium-2 than in medium-1 different at and

SOLUTION 14.

front. Since the wavefronts in both the media are plane and parallel, the corresponding beam of light in each medium will be parallel. Therefore, the correct choice is (a).

smaller in medium-2 than in medium-1. Thus the correct choice is (b).

15. All points on a wavefronts are in the same and phase of oscillation. Therefore c = f= e ( 16.

– f) = (

c

–

e)

wavefront. It is clear from the diagram that the angle r is less than angle i) as the beam travels from medium-1 into medium-2. Therefore medium-2 is

Fig. 27.19

Wave Optics 27.31

IV Assertion-Reason Type Questions In the following questions, Statement-1 (Assertion) is

3. Statement-1

correct. (a) Statement-1 is True, Statement-2 is True; State(b) Statement-1 is True, Statement-2 is True;

become sharper and brighter. Statement-2 4. Statement-1 When light travels from a rarer to a denser medium, its speed decreases. Statement-2

Statement-1. (c) Statement-1 is True, Statement-2 is False. (d) Statement-1 is False, Statement-2 is True. 1. Statement-1 Red light travels faster in glass than green light. Statement-2

5. Statement-1 When a light wave travels from one medium to

than for green light. 2. Statement-1

Statement-2 The speed of the wave undergoes a change. 6. Statement-1

of the source slit is increased, the fringe pattern becomes indistinct. Statement-2

undergoes a phase change of . Statement-2 The direction of the propagation of light is changed

increases if the width of the source slit is increased.

SOLUTIONS -

1. =

speed of light in vacuum speed of light in the medium

longer wavelength. The wavelength of red light is R < G which implies that the speed of red light is more than that of green light in glass. 2. The correct choice is (c). If the source slit is wide, the interference pattern becomes indistinct because the interference patterns due to various parts of the source

3. The correct choice is (a). 4. 5.

not depend on the speed of the wave.

does not depend on its speed or wavelength; it deduces that wave. 6. The correct choice is (c). The phase change is due tion from the mirror.

27.32 Comprehensive Physics—JEE Advanced

V Matrix Match Type 1. Column I from the slits S1 and S2. In each of these cases S1P0 = S2P0, S1P1 – S2P1 = /4 and S1P2 – S2P2 = /3, where is and thickness t is pasted on slit S2. The thicknesses of the sheets are different in different cases. The phase difference between the light waves reaching a point P (P I(P). Match each situation given in Column I with the statement(s) in Column II valid for that situation. Column I Column II (a)

(p)

(P0) = 0

(b) ( – 1) t = /4

(q)

(P1) = 0

(c) ( – 1) t = /2

(r) I (P1) = 0

(d) ( – 1) t – 3 /4

(s) I (P0) > I (P1)

(t) I (P2) > I (P1)

IIT, 2009

ANSWERS (a) (c)

(p), (s) (t)

(b) (d)

(q) (r), (s), (t)

Explanation: I (P) = 4I0 cos2

where I0 2 each interfering beam and is the phase difference =

(a) At P0, 4I0. At P1,

2

4

=

2

(P0) = 0 and I(P0) = . Therefore

4

= 2I0

I (P0) > I (P1). At P2,

(P)

(P0) = 0, therefore (P1) =

I (P1) = 4I0 cos2

(P2) =

I (P2) = 4I0 cos2

3

=

2 . Therefore 3

= I0 3 I (P2) < I (P1) So the correct choices are (p) and (s).

Wave Optics 27.33

(b) At P0, (P0) =

=

4

At P1,

(P1) = 0. Therefore, I (P1) = 4I0

At P2,

(P2) =

3

4

2

2

=

12

Therefore I (P2) = 4 I0 cos

Therefore I (P2) = 4I0 cos2

. Therefore, I (P0) = 2I0

2

12

6

(d) At P0,

(P0) =

3 4

At P1,

(P1) =

3 4

At P2,

(P2) =

3 4

3.7 I0

12

So I (P1) > I (P2 (c) At P0,

(P0) =

At P1,

(P1) =

At P2,

(P2) =

4

. Therefore I (P0) = 0

2

4

4

2

3

6

=

. Therefore I (P1) = 2 I0 =

3

2

5 12

= 3I0

6

I (P0) = 2 I0 I (P1) = 0

4

2

3

5 12

5 6

Therefore, I (P2) = 4I0 cos2

5 12

= 0.27 I0

0

So the correct choices are (r), (s) and (t).

.

VI Integer Answer Type 1.

2. A coherent parallel beam of microwaves of wavea parallel beam of light of wavelength 600 nm. The width of the slit (in mm).

slit apparatus. The separation between the slits is 1.0 mm and the screen is placed at a distance of 1.0 m from the slits. Find the number of minima in the interference pattern observed on the screen. IIT, 1998

10–3 m. Find the IIT, 1997

SOLUTION 1. For a slit of width a, the angular separation between the n

The linear separation y =

n sin n = a

a= 1=

n << a, sin

1

2

=

1

1

=

(in radian) and

2 a

1

=

a

a

=5

. Since

. Therefore,

2f

2.

y

= 10

–3

2f a

2 0.25 600 109 6 10 m = 5 mm

3

S1 and S2, the path difference at the central point P0 P0. (Fig. 27.20) Let the minima appear along directions with respect to the incident direction. Coherent waves from S1 and S2 along this direction are brought

27.34 Comprehensive Physics—JEE Advanced

to a focus at P. It is clear that the path difference between the waves from S1 and S2 on reaching P is = sin The interference minima will appear on the screen if

or where

sin

1 2

=

1 2

=

1 2

0.5 1.0

=

1 2

= 1.0 mm and

= 0.5 mm. Therefore

1 2

1 2

The allowed values of are those integers for which sin is not more than + 1 or less than – 1. These values are minima will be observed.

is an integer. Thus the directions of minisin

Given

=

sin

Fig. 27.20

Atomic Physics 28.1

28

Atomic Physics

Chapter

REVIEW OF BASIC CONCEPTS 28.1

PHOTOELECTRIC EFFECT

When electromagnetic radiation of appropriate frequency falls on a metal, electrons are emitted. This phenomenon is called photoelectric effect and the emitted electrons are called photoelectrons because they are liberated by means of light. Einstein’s Photoelectric Equation The classical electromagnetic wave theory of light, which successfully explained interference, diffraction, and polarization of light, could not account for the observations related to photoelectric effect. In 1900 Planck postulated that light waves consist of tiny bundles of energy called photons or quanta. The energy of a light wave of frequency is given by E = h , where h is Planck’s constant. Photon is simply a light wave of energy E. Following Planck’s idea, Einstein proposed a theory for photoelectric effect. According to him, when a photon of light falls on a metal, it is absorbed, resulting in the emission of a photoelectron. The maximum kinetic energy Kmax = 1/2 mv 2max of the emitted electron is given by 1 mv 2max = h – W0 (1) 2 This is the famous Einstein’s photoelectric equation. The term h represents the total energy of the photon incident on the metal surface. The photon penetrates a distance of about 10–8 m before it is completely absorbed. In disappearing, the photon imparts all its energy to a single electron. Part of this energy is used up by the electron in freeing itself from the atoms of the metal. This energy designated by W0 in Eq. (1) is called the work-function of the metal and is a characteristic of it. The rest of the energy is used up in giving the electron kinetic energy.

The work function W0, i.e. the energy required to pull an electron away from the surface of the metal, is large for heavier elements like platinum whereas for other elements like alkali metals, W0 is quite small. The minimum, or threshold, energy which a photon must have to free the electron from the surface of the metal should be equal to its work function. If the threshold frequency is 0 the threshold energy is h 0. Thus W0 = h 0 Einstein’s photoelectric equation therefore becomes 1 m v 2max = h ( – 0) (2) 2 It is evident that when < 0, no electron is emitted for any intensity of light. When > 0, the energy of the electron increases linearly with the frequency of light. Since intensity of light is a measure of the number of photons and since each photon emits a photoelectron on absorption, the intensity of photoelectrons is proportional to the intensity of light. Below a certain negative voltage V0, no photoelectrons are emitted no matter what the intensity of light is. This voltage is called the cut-off or stopping potential. Since there is no photoelectric emission at potentials less than V0, the maximum velocity v max acquired by the photoelectrons is given by 1 2 Kmax = m v max = eV0 2 where Kmax is the maximum kinetic energy. V0 is given by eV0 = h ( – 0 ) or

V0 =

h ( – e

0)

(3)

28.2 Comprehensive Physics—JEE Advanced

Laws and graphs of photoelectric effect (1) For a given emitter illuminated by radiation of a given frequency, the photoelectric current is proportional to the intensity of radiation (Fig. 28.1)

(5) Graph of photoelectric current (i) versus voltage (V) for radiations of different intensities (I1 > I2) but of the same frequency (Fig. 28.4). 1 2

–

0

Fig. 28.4

Fig. 28.1

(2) The maximum kinetic energy (Kmax) of photoelectrons is proportional to the frequency ( ) of the incident radiation and is independent of intensity of the radiation (Fig. 28.2).

(6) Graph of photoelectric current (i) versus voltage (V) for radiations of different frequencies ( 1 > 2) but of the same intensity (Fig. 28.5)

Fig. 28.5

(7) Threshold wavelength is tric emission < 0. Fig. 28.2

Kmax = h( –

0)

Slope of graph = h(Plank’s constant). Kmax = 0 when (threshold frequency) ( 0) below which no photoelectrons are emitter no matter what the intensity of radiation is. (4) Graph of stopping potential (V0) versus frequency of incident radiation (Fig. 28.3).

0

=

c

. For photoelec-

0

28.1 A metal of work function 3.3 eV is illuminated by light of wavelength 300 nm. Find (a) the threshold frequency of photoelectric emission, (b) the maximum kinetic energy of photoelectrons and (c) the stopping potential. Take h = 6.6 10–34 Js. SOLUTION 10–19 J

(a) W0 = 3.3 eV = 3.3 1.6 Threshold frequency is 0

=

W0 h

3.3 1.6 10 6.6 10

19

34

= 8 1014 Hz (b) Frequency of incident radiation is =

3 108

c

300 10

Fig. 28.3

h V0 = ( – 0) e h Slope of graph = , which is the same for all e metals.

= 10 Kmax = h( – = (6.6 = 1.32

9

1014 Hz 0) –34

10 ) 10–19 J

(10

1014 – 8

1014)

Atomic Physics 28.3

(c) Kmax = eV0

K max e

V0 =

1.32 10 1.6 10

19 9

= 0.825 V 28.2 Photoelectric emission from a metal begins at a frequency of 6 1014 Hz. The emitted electrons are fully stopped by a retarding potential of 3.3 V. Find the wavelength (in nm) of the incident radiation. Take h = 6.6 10–34 Js. SOLUTION

SOLUTION eV0 = h( – (1.6

10–19)

3 108

c

1.4 10

( –6

1014)

1015 Hz 10–7 m = 214 nm

= 2.14

15

eV0 = h( –

0)

10–34)

3.3 = (6.6 = 1.4

=

Fig. 28.6

28.3 Light of wavelength 300 nm is incident on two metals A and B whose work functions are respectively 4 eV and 2 eV. Which of the two metals will emit photoelectrons?

E=h =

hc

=

(6.6 10

=

34

)

300 10

=6

(3 108 )

1.6 10

19 19

= 3.75 eV

Photoelectrons will be emitted from the metal if E is greater than the work function of the metal. Hence metal B will emit photoelectrons but A will not. 28.4 For photoelectric affect in a metal, the graph of stopping potential V0 (in volt) versus frequency (in Hz) of the incident radiation is shown in Fig. 28.6. From (a) threshold frequency (b) Planck’s constant and (c) work function of the metal.

h = e slope = (1.6 10–19) (4.125 10–15) = 6.6 10–34 Js W0 = h 0 = (6.6 10–34) (4 1015) = 26.4 10–19 J = 16.5 eV

9

10–19 J 6 10

h (1) ( 0) e (a) It follows from Eq. (1) that = 0 if V0 = 0. Hence 0 = 4 1015 Hz h (b) The slope of V0 versus graph = e 16.5 0 Now slope = = 4.125 10–15 15 (8 4) 10 V0 =

(c)

SOLUTION Energy of incident radiation is

0)

28.5 Calculate the number of photons emitted per second by a transmitter of power 10 kW sending radiowaves of frequency 6 105 Hz. Take h = 6.63 10–34 Js. SOLUTION Let N be the number of photons emitted in time t. Energy of 1 photon = h . Therefore, energy of N photons = Nh . Therefore, the power is Nh = nh ; t n = number of photons emitted per second. P=

or

n=

P 10 103 = h (6.63 10 34 ) 6 105

= 2.5

1031 photons per second

28.4 Comprehensive Physics—JEE Advanced

28.6 The stopping potential of a metal is 3 V when it is illuminated by light of wavelength 500 nm. What will be the stopping potential of the metal when the cy and work function of the metal. Take h = 6.6 10–34 Js.

SOLUTION =

3 108

c

250 10

eV0 = h( –

= 6.6 10–34

1

0)

(1)

eV0 = h(

2

–

0)

(2)

Subtracting, we get e(V1 – V2) = h(

1

–

2)

h V1 – V2 = ( e = = =

h c e 1 hc e

2

28.2

1 1 2 3

3 108

1.6 10

19 7

5 10

6 5 10

7

14

V2 = V1 – 0.4 = 3 – 0.4 = 2.6 V From Eq. (1),

=

–

c 1

eV1 h

3 10 6 10

= 4.27 W0 = h

0

= 6.6 = 2.82

6.6 10

c

3 108

y

600 10

34

= 0.9

1015 Hz

9

= 0.5

1014 Hz

1.6 10

7

10

19

6.6 10

10

BOHR’S THEORY OF HYDROGEN LIKE ATOM

(a) Bohr’s quantization condition: The magnitude of angular momentum of the electron in a circular orbit is nh nh L= mvn rn 2 2 where m = mass of electron, rn = radius of nth circular orbit, vn = orbital speed of electron in the nth orbit, h = Planck’s constant and n is an integer called the principal quantum number. (b) Speed of electron in nth orbit is vn =

eV1 h 8

=

19

6 10

2

6.6 10

1

W0 h

Since y < 0, the photocell will not work with yellow light as no photoelectrons will be emitted.

c

= 0.4 V

=

=

y =

2)

1

0

1.2

Frequency of yellow light is

6 10

0

Now

1.2 1015 – 1.6 10–19

10–19 J = 3.75 eV

=6 –

= h – W0

0)

W0 = h – eV0

SOLUTION eV1 = h(

1015 Hz

= 1.2

9

14

–34

3 34

Hz 4.27

28.7 Ultraviolet light of wavelength 250 nm falls on the metal emitter of a photocell. If the stopping potential

Z 2 0h n

where Z = atomic number of atom. For hydrogen 1 Z = 1. For a given atom vn . Substituting the n known values of e, 0 and h we get 106 ms–1)

vn = (2.2

1014

10–19 J = 1.76 eV

e2

(c) Radius of nth orbit is rn =

0

h2

me

2

n2 = (0.53 Z = (0.53Å)

photocell work if yellow light of wavelength 600 nm is used? Take h = 6.6 10–34 Js.

Z n

For a given atom rn

n 2.

10–10 m) n2 Z

n2 Z

Atomic Physics 28.5

(d) Total energy of electron in nth orbit K.E. =

Ze

2

0 rn

8

Ze

; P.E. =

2

0 rn

4

1

P.E. = –2K.E. Total energy of electron in nth object is En = K.E. + P.E. = K.E. – 2K.E. = –K.E. Z e2 8 0 rn Putting the value of rn, we get En = –

m e4

En =

2 0

8

h

2

1

n

2

Z

10–19 J)

= (–13.6eV)

Z

n2

2

(f) Frequency of revolution of the electron in nth orbit is 1 = (6.6 Tn

Z2

1015 Hz)

n3

(g) Wavelength of emitted radiation: When an electron jumps from a higher energy state n = n2 to a lower state n = n1, a photon of energy h of radiation is emitted. h = En2 En1 hc

1

1

where RH =

m e4

=

=

2 0

8

h

2

m e4 8

2 0

= R HZ 2

Z2

3

Z2

1

1

n12

n22

h c

1

1

n22

n12

1

1

n12

n22

= 1.097

107 m–1 is called

8 02 h3c Rydberg constant. (h) Main Series of Hydrogen Spectrum (Z = 1) (1) Lyman series: n1 = 1, n2 = 2, 3, 4,… 1

min

= RH

= RH

1

1

1

12

n22

= 91.2 nm,

max

1

2

n22

2

1

1

2

n22

3

= RH 1 42

1 n22

min = 1458 nm, max = 4050 nm These spectral lines also lie in the infrared region. (i) The energy of electron in hydrogen atom in the ground state is –13.6 eV. (j) The ionization potential of hydrogen atom in the ground state is 13.6 V. Ionization potential of a 13.6 Z 2 volt. hydrogen like atom in nth state = n2

28.8 The energy required to excite a hydrogen atom from n = 1 to n = 2 energy state 10.2 eV. What is the wavelength of the radiation emitted by the atom when it goes back to its ground state? SOLUTION Given E2 – E1 = 10.2 eV = 10.2 1.6 10–19 J. Therefore, frequency the emitted radiation is =

m e4

1

min = 820.1 nm, max = 1874.6 nm Spectral lines in Paschen series lie in the infrared region. (4) Brackett series: n1 = 4, n2 = 5, 6, 7,…

2

n2 (e) Time period of revolution of electron in nth orbit is 2 rn 4 02 h3 n3 n3 –16 = = (1.51 10 s) Tn = vn Z2 m e4 Z2

=

= RH

min = 364.5 nm, max = 656.1 nm Spectral lines in Balmer series lie in the visible region. (3) Paschen series: n1 = 3, n2 = 4, 5, 6,…

Z2

= (–21.76

n

Spectral lines in Lyman series lie in the ultraviolet region. (2) Balmer series: n1 = 2, n2 = 3, 4, 5,…

E2

E1 h

and wavelength is = =

c

ch E2

3 108

E1 6.6 10

10.2 1.6 10 = 1.22 10–7 m = 121.6 nm

34

19

= 1220 Å = 122 nm

28.6 Comprehensive Physics—JEE Advanced

28.9 The ionization potential of the hydrogen atom is 13.6 V. Find the energy of the atom in n = 2 energy state.

28.12 The wavelength of the second line of Balmer series of Lyman series?

SOLUTION Energy of hydrogen atom in the ground state is ( ionization potential is 13.6 V) E1 = – 13.6 eV 1 Since En , the energy in the n = 2 state is n2 E1 13.6 E2 = = – 3.4 eV 2 4 ( 2) 28.10 The innermost orbit of hydrogen atom has a diameter of 1.06 Å. What is the diameter of the 10th orbit? SOLUTION Given d1 = 1.06 Å. We know that dn d10 = (10)2 1.06 Å

SOLUTION Wavelengths in Balmer series for hydrogen are given by 1 1 1 = RH 2 n2 2 1 4

= RH

1 n2

n = 2) in Lyman series is 1

n2d1. Hence

= RH 1

1

1 2

1

= RH 1

1

=

l

l s

4 3RH

=

4 3RH

=

4 3RH

=

4 3

1

=

1

1

RH

1 RH

= RH 1 l

1

=

2

s

and

1

or

n2 1 where n = 2, 3, 4, ... . The shortest wavelength ( s) corresponds to n = and the longest wavelength ( l) corresponds to n = 2.

s

or

2

SOLUTION For Lyman series

1

RH 1

2

=

28.11 Find the ratio of longest and shortest wavelength in the Lyman series of hydrogen atom.

= RH

1 4

3RH 4

3, 4, 5 ...

The second line in Balmer series corresponds to n = 4. Hence 3RH 1 1 16 1 or 2 = RH 4 16 16 3RH 2

= 106 Å

1

; n

28.3

2

4

3RH 16

1 4

3RH 4

1 4

486.4 4 = 121.6 nm

X-RAYS

X-rays are produced when energetic electrons fall on a suitable target. The apparatus used for the production of X-rays is called the Coolidge tube in which electrons are produced by thermionic emission. X-rays are electromagnetic waves of wavelength of the order of 1Å or 0.001 nm. 1. Duane-Hunt Law The shortest X-ray wavelength emitted when electrons incident on the target are accelerated through a potential V volts is given by hc 1239.6 nm min = eV V This is called the Duane-Hunt Law.

Atomic Physics 28.7

2. Diffraction of X-rays A crystal is a natural grating for diffraction of X-rays, since the spacing d between the crystal planes is of the order of the wavelength of X-rays. The angle of diffraction of X-rays is given by the relation 2d sin = n where n is an integer having values 1, 2, 3, etc. and is the wavelength of X-rays incident on the crystal. This relation is called Bragg’s equation.

5. The characteristic X-rays consist of K and L series. The K X-rays are produced when electrons jump from n = 2 to n = 1 orbit. They are called L when electrons jump from n = 3 to n = 2 orbit. The wavelength of the K line of the X-ray spectrum is given by K series limit Kg

n = 4 (N shell) Lb

Ma

n = 3 (M shell) Kb

3. Absorption of X-rays X-rays are absorbed by materials following the exponential relation I = I0 e– x

n=¥

L series limit

La

n = 2 (L shell) Ka

where I0 is the initial intensity of X-rays, I their intensity after they have traversed a thickness x and is the

n = 1 (K shell)

Fig. 28.7

4. X-ray Spectra spectrum and characteristic spectrum. X-ray spectrum consists of a series of discrete spectral lines superimposed on a continuous luminous background. The background spectrum consists of all sorts of wavelengths and is called the continuous spectrum which is the same for all target materials. The discrete line spectrum is a characteristic of the target metal and is, therefore, called the characteristic spectrum. The spectral lines in this spectrum are due to the transitions of electrons from the outer orbits to the inner K, L, M shells, etc. The emitted radiations are called K , K , K , radiations. These transitions are shown in Fig. 28.7. The frequency ( ) of the characteristic X-rays is related to the atomic number (Z) of the target metal by the relation = a (Z – b) where a and b are constants. This relation is known as Moseley’s law. Some facts about X-rays can be summerized as follows: 1. X-rays are electromagnetic waves of wavelength of the order of 1 Å or 0 001 nm. 2. 3. X-rays travel in vacuum at the speed of light, c = 3 108 ms 1. 4. If electrons are accelerated through a potential difference V and then are made to fall on a target, X-rays of wavelength greater than hc/eV are produced. The shortest wavelength emitted is min = hc/eV.

1

= RH(Z – 1)2

1

1

1 22 where R is the Rydberg constant and Z is the atomic number. The wavelength of L line is 1

= RH(Z

7 4)2

2

1

1

2

2 32 6. The frequency of the characteristics X-rays is proportional to (Z b)2 where b is a constant. 7. X-rays are absorbed by materials according to the relation I = I0 e x where I0 = incident intensity, = constant for the given material and x = distance penetrated. 28.13 What is the maximum frequency of X-rays emitted from an X-ray tube operating at 50 kV? SOLUTION eV = h max. Hence eV h

(1.6 10

= 1.2

1019 Hz

max =

19

)

(50 103 )

(6.63 10

34

)

28.14 An X-ray tube produces a continuous spectrum of radiation with its shortest wavelength end at 0.66 Å. What is the maximum energy of a photon of this radiation? Take h = 6.6 10–34 Js

28.8 Comprehensive Physics—JEE Advanced

SOLUTION Emax = h =

max

=

(6.6 10

=3

hc min 34

)

(0.66 10 10–15 J

(

c=

)

10

28.4

I 3.2 10 3 = =2 e 1.6 10 19

1016 electrons per second

WAVE NATURE OF MATTER

150 V

=

)

SOLUTION

2 m qV

For an electron,

(3 108 )

28.15 The potential difference applied to an X-ray tube is 5 kV and the current through it is 3.2 mA. Find the number of electrons striking the target per second.

n=

h

=

2 1/ 2

v c2 = m0 v 2. In terms of kinetic energy K, de Broglie wavelength is given by h = 2m K h 1

3. If a particle of charge q is accelerated through a potential difference V, its de Broglie wavelength is given by

Å

4. For a gas molecule of mass m at temperature T kelvin, the de Broglie wavelength is given by h = 3m k T where k is the Boltzmann constant. 28.16 Calculate the wavelength associated with a dust particle of mass 1 g moving with a velocity of 106 ms–1. Given h = 6.6 10–34 Js. SOLUTION m = 1 g = 10–6 g = 10–9 kg

In 1924, Louis de Broglie, a French theoretical physicist, derived an equation which predicted that all atomic wavelength. Under certain circumstances, a beam of electrons or atoms will behave like a group of waves. On the basis of theoretical considerations, de Broglie predicted that the wavelength of these waves is given by h h = = mv p where h is Planck’s constant and p is the momentum of the particles. This equation is known as de Broglie’s wave equation. For an electron moving at a high speed, the momentum is large and the wavelength is small. The faster the electron, the shorter is the wavelength. Notice that the particle need not have a charge to have an associated wave. This is why de Broglie waves are sometimes referred to as matter waves. 1. If the rest mass of a particle is m0, its de Broglie wavelength is given by

1/ 2

=

h 6.6 10 34 = mv 10 9 106 = 6.6

10–31 m

28.17 Calculate de Broglie wavelength of an electron having kinetic energy of 1 BeV. Given mass of electron (m) = 9.1 10–31 kg, h = 6.6 10–34 Js and e = 1.6 10–19 C. SOLUTION K = 1 BeV = 109 eV = 109 =

h 2mK

1.6 10–19 = 1.6

6.6 10

=

2 9.1 10 = 3.87

31

10–10 J

34

1.6 10

10

10–14 m

28.18 Calculate de Broglie wavelength associated with an electron accelerated through a potential difference of 200 V. Given m = 9.1 10–31 kg and h = 6.6 10–34 Js. SOLUTION K = 200 eV = 200 =

h 2mK

1.6

10–19 = 3.2 6.6 10

=

2 9.1 10 = 0.86

31

10–17 J

34

3.2 10

10–10 m = 0.86 Å

17

Atomic Physics 28.9

28.21 Ultraviolet light of wavelength 99 nm falls on a metal plate of work function 1.0 eV. Find the wavelength of the fastest photoelectron emitted. Mass of electron (m) = 9.1 10–31 kg and h = 6.6 10–34 Js.

28.19 Calculate the wavelength of de Broglie waves associated with a neutron at room temperature of 27°C. Given mass of neutron (m) = 1.67 10–27 kg, Boltzman constant (k) = 1.38 10–23 JK–1 and h = 6.63 10–34 Js.

SOLUTION SOLUTION Kinetic energy of neutron due to thermal speed is 3 K = kT, where T = 273 + 27 = 300 K. 2 h h h = 2mK 3 3mkT 2m kT 2 Substituting the values of h, m, k and T, we get = 1.45 10–10 m = 1.45 Å

=

(6.6 10

1

=

hc

(6.6 10

)

6.6 10 (9.1 10

hc 2

(6.6 10

34

)

30 10

31

34

) 1.84 10

18

10–10 m = 0.36 nm

28.22 A proton and an electron have equal kinetic energy. Which of the two has a greater de Broglie wavelength?

8

9

= 10 10–18 J Energy of photon after collision is Ef =

10–19

2mK max

2

(3 10 )

19.8 10

1

10–19

h

= 3.6

Ei = h

1.6

10–18 J

=

SOLUTION Energy of photon before collision is

(3 108 )

)

10–18 – 1.6

= 1.84 =

34

W0

99 10 9 – 1.0

=2

28.20 A photon of wavelength 19.8 nm collides with an electron at rest. After the collision, the wavelength of the photon is found to be 30 nm. Is the collision elastic or inelastic? Calculate the energy of the scattered electron. Given h = 6.6 10–34 Js.

34

hc

Kmax = h – W0 =

SOLUTION

(3 108 )

p

9

= 6.6 10–18 J Since there is a loss of kinetic energy ( Ef < Ei), the collision is inelastic. The energy of the scattered electron = Ei – Ef = 3.4 10–18 J.

e p

=

=

h 2m p K mp me

,

>1

h e

2me K (

m p > m e)

Hence the electron has the greater wavelength

I Multiple Choice Questions with Only One Choice Correct 1. Figure 28.8 shows the variation of photoelectric current (i) with voltage (V) between the electrodes in a photo-cell for two different radiations.

If Ia and Ib are the intensities of the incident radiations and a and b are the respective frequencies, then

28.10 Comprehensive Physics—JEE Advanced

(a) Ia > Ib,

b

<

a

(b) Ia < Ib,

b

>

a

(c) Ia > Ib,

a

=

b

(d) Ia < Ib,

b

<

a

i

b a

V

0

Fig. 28.8

2. When a radiation of energy 5 eV falls on a surface, the emitted photoelectrons have a maximum kinetic energy of 3 eV. The stopping potential is (a) 2 V (b) 3 V (c) 5 V (d) 8 V 3. When a radiation of wavelength 1 falls on a surface, the maximum kinetic energy of the emitted photoelectrons is K1. For a radiation of wavelength 2, the maximum kinetic energy is K2. If 1 = 2 /2, then (a) K1 = 2K2 (b) K2 = 2K1 (c) K1 > 2K2 (d) K1 < 2K2 4. Figure 28.9 represents the observed intensity (I) of X-rays emitted by an X-ray tube as a function of wavelength ( ). (a) Peaks A and B represent K lines (b) Peaks A and B represents K lines (c) Peak A represents K line and peak B represents K line (d) Peak A represents K line and peak B represents K line I

B A

l

Fig. 28.9

5. A particle at rest disintegrates into two fragments of masses in the ratio of 1 : 2 having non-zero velo-

cities. The ratio of the de Broglie wavelength of the lighter particle to that of the heavier particle is 1 (b) 1 (a) 2 1 (c) 2 (d) 4 6. A potential difference V is applied across an X-ray tube. If e/m is the charge to mass ratio of an electron and c the speed of light in vacuum, the ratio of the de Broglie wavelength of the incident electrons to the shortest wavelength of X-rays produced is given by (a)

1 c

eV 2m

(b)

1 c

eV m

(c)

1 c

2 eV m

(d)

2 c

eV m

7. The de Broglie wavelength of the electron in the nth energy state of a hydrogen atom is proportional to 1 1 (b) 2 (a) n n (c) n (d) n2 8. The magnitude of the magnetic moment of the electron in the nth energy state of a hydrogen atom is proportional to 1 (b) n (a) n (c) n (d) n2 9. The shortest wavelength in the Brackett series of a hydrogen like atom of atomic number Z is equal to the shortest wavelength in the Balmer series of hydrogen atom. The value of Z is (a) 2 (b) 3 (c) 4 (d) 5 10. The difference between (n + 1)th Bohr radius and nth Bohr radius is equal to the (n – 1)th Bohr radius. The value of n is (a) 1 (b) 2 (c) 3 (d) 4 11. atom due to the motion of an electron in the nth orbit is inversely proportional to (b) n3 (a) n2 4 (c) n (d) n5 12. The magnitude of the angular momentum of an electron revolving in a circular orbit of radius r in a hydrogen atom is proportional to (b) r (a) r1/2 (c) r3/2 (d) r2

Atomic Physics 28.11

13. The wavelength in air associated with a photon of energy E is (c is the speed of light in air and h is the Planck’s constant) hc h (b) (a) E cE c (c) (d) hcE hE 14. An image of the sun is formed by a lens of focal length 30 cm on the metal surface of a photoelectric cell and it produces a current I. The lens forming the image is then replaced by another lens of the same diameter but of focal length 15 cm. The photoelectric current in this case will be (a) I / 2 (b) 2 I (c) I (d) 4 I 15. The momentum of a photon of wavelength is h (b) h c (a) c (c) h

(d)

h

16. Violet light can cause photoelectric emission from a metal but blue light cannot. If sodium light is incident on the metal, then (a) the photoelectric current decreases (b) the number of photoelectrons ejected per second increases (c) the velocity of photoelectrons increases (d) no photoelectric emission occurs. 17. 1020 photons of wavelength 660 nm are emitted per second from a lamp. What is the wattage of the lamp. Planck’s constant = 6 6 10 34 Js (a) 30 W (b) 60 W (c) 100 W (d) 500 W 18. When a certain photosensitive surface is illuminated with monochromatic light of frequency , the stopping potential for photoelectric current is V0/ 2. When the same surface is illuminated by monochromatic light of frequency /2, the stopping potential is V0. The threshold frequency for photoelectric emission is 2 3 (a) (b) 3 2 3 5 (d) 5 3 19. When a certain photosensitive surface is illuminated with monochromatic light of wavelength , the stopping potential for photoelectric current is 2V0. When the same surface is illuminated with (c)

monochromatic light of wavelength 2 , the stopping potential is V0/2. The threshold wavelength for photoelectric emission is (a) 1.5 (b) 2.0 (c) 2.5 (d) 3 20. The threshold frequency for a certain photosensitive metal is 0. When it is illuminated by light of frequency = 2 0, the stopping potential for photoelectric current is V0. What will be the stopping potential when the same metal is illuminated by light of frequency = 3 0? (b) 2 V0 (a) 1 5 V0 (c) 2 5 V0 (d) 3 V0 21. The threshold frequency for a certain photosensitive metal is 0. When it is illuminated by light of frequency = 2 0, the maximum velocity of photoelectrons is v0. What will be the maximum velocity of the photoelectrons when the same metal is illuminated by light of frequency = 5 0? (b) 2 v0 (a) 2 v0 (c) 2 2 v0 (d) 4 v0 22. The threshold wavelength for a metal whose work function is W0 is 0. What is the threshold wavelength for a metal whose work function is W0/2? (a) (c) 2

0

(b)

4 0

(d) 4

0

2 0

23. The minimum wavelength of X-rays emitted from an X-ray tube operating at a voltage of 104 volts is roughly equal to (a) 1 Å (b) 1 5 Å (c) 2 Å (d) 2 5 Å 24. The energy in monochromatic X-rays of wavelength 1 Å is roughly equal to (a) 2 10 15 J (b) 2 10 16 J (d) 2 10 18 J (c) 2 10 17 J 25. When the accelerating voltage applied on the electrons, in an X-ray tube, is increased beyond a critical value (a) the spectrum of white radiation is unaffected (b) only the intensities of various wavelengths are increased (c) only the wavelength of characteristic radiation is affected (d) the intensities of characteristic lines relative to the white spectrum are increased but there is no change in their wavelength.

28.12 Comprehensive Physics—JEE Advanced

26. In the continuous part of the spectrum of X-rays the limiting frequency is (a) inversely proportional to the potential through which electrons have been accelerated (b) is directly proportional to the accelerating potential (c) not dependent upon the accelerating potential (d) is dependent upon the nature of the target material. 27. The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation, (a) the intensity increases (b) the minimum wavelength increases (c) the intensity decreases (d) the minimum wavelength decreases. IIT, 1988 28. In an X-ray tube, electrons accelerated through a very high potential difference strike a metal target. If the potential difference is increased, the speed of the emitted X-rays (a) increases (b) decreases (c) remains unchanged (d) is always equal to 3 108 ms 1. 29. An X-ray tube is operated at 66 kV. Then, in the continuous spectrum of the emitted X-rays (a) wavelengths 0.01 nm and 0.02 nm will both be present (b) wavelengths 0.01 nm and 0.02 nm will both be absent (c) wavelengths 0.01 nm will be present but wavelength 0.02 nm will be absent (d) wavelength 0.01 nm will be absent but wavelength 0.02 nm will be present. 30. The minimum wavelength of X-rays produced in an X-ray tube is when the operating voltage is V. What is the minimum wavelength of the X-rays when the operating voltage is V/2? (b) 2 (c) 2 (d) 4 31. The maximum frequency of X-rays produced in an X-ray tube is when the operating voltage is V. What is the maximum frequency of the X-rays when the operating voltage is V/2? (a)

(a)

2 (c) 2

(b) (d) 4

32. X-rays are incident normally on a crystal of lattice constant 0 diffraction from the crystal occurs at an angle of 30°. What is the wavelength of X-rays used? (a) 0 3 nm (b) 0 6 nm (c) 1 2 nm (d) 2 4 nm 33. X-rays of wavelength are incident normally on a tion from the crystal is observed at an angle of 45°. The lattice constant of the crystal is (a)

34.

35.

36.

37.

38.

39.

2

(b)

2

(c) (d) 2 The frequency of K line of a source of atomic number Z is proportional to (a) Z2 (b) (Z 1)2 (c) 1/Z (d) Z The wavelength of K line from an element of atomic number 41 is . Then the wavelength of K line of an element of atomic number 21 is (a) 4 (b) /4 (c) 3 08 (d) 0 26 IIT, 2005 The continuous X-ray spectrum is produced due to (a) acceleration of electrons towards the nuclei of the target atoms (b) retardation of energetic electrons when they approach the nuclei of the target atoms (c) fall of the electrons of the target atoms from higher energy level to lower energy levels (d) knocking out of the electrons from the target atoms by the fast moving incident electrons An X-ray photon of wavelength and frequency collides with an electron and bounces off. If and are respectively the wavelength and frequency of the scattered photon, then (a) = ; = (b) < ; > (c) > ; > (d) > ; < The binding energy of the innermost electron in tungsten is 40 keV. To produce characteristic X-rays using a tungsten target in an X-ray tube, the accelerating voltage should be greater than (a) 4 kV (b) 40 kV (c) 400 kV (d) 4000 kV The shortest wavelength of X-rays emitted from an X-ray tube depends upon (a) the current in the tube (b) the voltage applied to the tube

Atomic Physics 28.13

40.

41.

42.

43.

(c) the nature of the gas in the tube (d) the atomic number of the target material IIT, 1982 Which one of the following parameters of the emitted X-rays increases when the potential difference between the electrodes of an X-ray tube is increased? (a) intensity (b) frequency (c) wavelength (d) speed A proton, when accelerated through a potential difference of V volts, has a wavelength associated with it. If an alpha particle is to have the same wavelength , it must be accelerated through a potential difference of (a) V/ 8 volts (b) V/4 volts (c) 4 V volts (d) 8 V volts Two particles of masses m and 2m have equal kinetic energies. Their de Broglie wavelengths are in the ratio of (a) 1 : 1 (b) 1 : 2 (d) 2 : 1 (c) 1 : 2 The de Broglie wavelength of a neutron at 927°C is . What will be its wavelength at 27°C?

(b) 2 (c) 2 (d) 4 44. The de Broglie wavelength of a neutron when its kinetic energy is K is . What will be its wavelength when its kinetic energy is 4K? (a)

(a)

4 (c) 2

(b)

2 (d) 4

45. What is the de Broglie wavelength of an electron of energy 180 eV? Mass of electron = 9 10 31 kg and Planck’s constant = 6 6 10 34 Js. (a) 0 5 Å (b) 0 9 Å (c) 1 3 Å (d) 1 8 Å 46. Moving with the same velocity, which of the following has the longest de Broglie wavelength? (a) -particle (b) -particle (c) proton (d) neutron 47. The de Broglie wavelength of an electron moving with a velocity 1 5 108 ms 1 is equal to that of a photon. The ratio of the kinetic energy of the electron to that of the photon is (a) 2 (b) 4 1 1 (d) (c) 2 4 IIT, 2004

48. The shortest wavelength in the Lyman series is 911 6 Å. Then the longest wavelength in the Lyman series is (a) 1215 Å (b) (c) 2430 Å (d) 600 Å 49. If electron orbits with principal quantum numbers n > 3 were not allowed, the number of possible elements would be (a) 28 (b) 90 (c) 32 (d) 64 50. Pauli’s exclusion principle states that no two electrons in an atom can have identical values for (a) one of the four quantum numbers (b) two of the four quantum numbers (c) three of the four quantum numbers (d) all four quantum numbers 51. Which energy state of the triply ionized beryllium (Be+++) has the same electron orbital radius as that of the ground state of hydrogen? Given Z for beryllium = 4. (a) n = 1 (b) n = 2 (c) n = 3 (d) n = 4 52. In Q. 51, what is the ratio of the energy state of beryllium and that of hydrogen? (a) 1 (b) 2 (c) 3 (d) 4 53. Which energy state of doubly ionized lithium (Li++) has the same energy as that of the ground state of hydrogen? Given Z for lithium = 3. (a) n = 1 (b) n = 2 (c) n = 3 (d) n = 4 54. In Q. 53, what is the ratio of the electron orbital radius of Li++ to that of hydrogen? (a) 1 (b) 2 (c) 3 (d) 4 55. state of hydrogen is 3 4 eV. What is the kinetic energy of the electron in this state? (a) + 1 7 eV (b) + 3 4 eV (c) + 6 8 eV (d) + 13 4 eV 56. In Q. 55, the potential energy of the electron is (a) – 1 7 eV (b) – 3 4 eV (c) – 6 8 eV (d) – 13 4 eV 57. the hydrogen spectrum is . What is the wavelength of the second line. (a)

20 27

(b)

3 16

(c)

5 36

(d)

3 4

28.14 Comprehensive Physics—JEE Advanced

58.

59.

60.

61.

62.

63.

64.

65.

the hydrogen spectrum is . What is the frequency of the corresponding line in the spectrum of doubly ionized Lithium? (a) (b) 3 (c) 9 (d) 27 The energy difference between the first two levels of hydrogen atom is 10 2 eV. What is the corresponding energy difference for a singly ionized helium atom? (a) 10 2 eV (b) 20 4 eV (c) 40 8 eV (d) 81 6 eV The ionization energy of hydrogen atom is 13.6 eV. What is the ionization energy of helium atom? (a) 3.4 eV (b) 13.6 eV (c) 54.4 eV (d) 108.8 eV The ionization energy of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by electromagnetic radiation of energy 12.1 eV. How many spectral lines will be emitted by the hydrogen atom? (a) one (b) two (c) three (d) four If an orbital electron of the hydrogen atom jumps from the ground state to a higher energy state, its orbital speed reduces to half its initial value. If the radius of the electron orbit in the ground state is r, then the radius of the new orbit would be (a) 2r (b) 4r (c) 8r (d) 16r The orbital speed of the electron in the ground state of hydrogen is v. What will be its orbital speed when it is excited to the energy state 3.4 eV? v (a) 2 v (b) 2 v v (c) (d) 4 8 In the Bohr model of the hydrogen atom, the ratio of the kinetic energy to the total energy of the electron in a quantum state n is (a) – 1 (b) + 1 1 1 (d) 2 (c) n n The ratio of the wavelengths of the longest wavelength lines in the Lyman and Balmer series of hydrogen spectrum is 3 5 (b) (a) 23 27 7 9 (c) (d) 29 31

66. If a hydrogen atom at rest, emits a photon of wavelength , the recoil speed of the atom of mass m is given by (a)

h m

(b)

mh

(c) mh (d) none of these 67. If elements with principal quantum number n > 4 were not allowed in nature, the number of possible elements would be (a) 60 (b) 32 (c) 4 (d) 64 IIT, 1983 68. When a monochromatic source of light is at a distance of 0.2 m from a photoelectric cell, the cut-off voltage and the saturation current are respectively 0.6 V and 18 mA. If the same source is placed 0.6 m away from the cell, then (a) the stopping potential will be 0.2 V (b) the stopping potential will be 1.8 V (c) the saturation current will be 6.0 mA (d) the saturation current will be 2.0 mA IIT, 1992 69. The energy of a photon of frequency is E = h and the momentum of a photon of wavelength is p = h / . From this statement one may conclude that the wave velocity of light is equal to E (b) (a) 3 10 8 ms–1 p 2

E (c) E p (d) p 70. When a centimetre thick surface is illuminated with light of wavelength , the stopping potential is V. When the same surface is illuminated by light of wavelength 2 , the stopping potential is V/3. The threshold wavelength for the surface is 4 (b) 4 (a) 3 8 (c) 6 (d) 3 71. Energy levels A, B and C of a certain atom correspond to increasing values of energy, i.e. EA < EB < EC. If 1, 2 and 3 are the wavelengths of radiations corresponding to transitions C to B, B to A and C to A respectively, which of the following relations is correct? (a)

3

=

1

+

2

(b)

3

(d)

2 3

1

=

2

1

(c)

1

+

2

+

3

=0

=

2 1

2

+

2 2

Atomic Physics 28.15

72. Figure 28.10 represents the observed intensity (I) of X-rays emitted by an X-ray tube, as a function of wavelength ( ). The sharp peaks A and B denote (a) band spectrum (b) continuous spectrum (c) characteristic radiations (d) white radiations.

Fig. 28.10

73. In a photo-emissive cell, with exciting wavelength , the fastest electron has a speed v. If the exciting wavelength is changed to 3 / 4, the speed of the fastest emitted electron will be (a) v

3 4

75.

76.

77.

4 3

4 4 (d) more than v 3 3 An energy of 24.6 eV is required to remove one of the electrons from the neutral helium atom. The energy (in eV) required to remove both the electron from a neutral helium atom is (a) 38.2 (b) 49.2 (c) 51.8 (d) 79.0 IIT, 1995 As per Bohr model, the minimum energy (in eV) required to remove an electron from the ground state of doubly ionized Li atom (Z = 3) is (a) 1.51 (b) 13.6 (c) 40.8 (d) 122.4 IIT, 1997 The K X-ray emission line of tungsten occurs at = 0.021 nm. The energy difference between K and L levels in this atom is about (a) 0.51 MeV (b) 1.2 MeV (c) 59 keV (d) 136 eV IIT, 1997 The maximum kinetic energy of photoelectrons emitted from a surface when photons of energy 6 eV fall on it is 4 eV. The stopping potential is (a) 2V (b) 4V (c) less than v

74.

(b) v

(c) 6V

(d) 10V

IIT, 1997 78. The electron in a hydrogen atom makes a transition n2 where n1 and n2 are the principal quantum n1 numbers of the two states. Assume the Bohr model to be valid. The time period of the electron in the possible values of n1 and n2 are (b) n1 = 8, n2 = 2 (a) n1 = 4, n2 = 1 (d) n1 = 6, n2 = 3 (c) n1 = 8, n2 = 1 IIT, 1998 79. X-rays are produced in an X-ray tube operating at a given accelerating voltage. The wavelength of the continuous X-rays has values from (a) 0 to (b) min to where min > 0 (c) 0 to max where max < (d) min to max where 0 < min < max < . IIT, 1998 80. The work function of a substance is 4.0 eV. The longest wavelength of light that can cause photoelectron emission from this substance is approximately (a) 540 nm (b) 400 nm (c) 310 nm (d) 220 nm IIT, 1998 81. A particle of mass M at rest decays into two particles of masses m1 and m2, having non zero velocities. The ratio of the de Broglie wavelengths of the particles, 1/ 2, is m1 m2

(b)

(c) 1.0

(d)

(a)

m2 m1 m2 m1

IIT, 1998 82. Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atom model and consider all possible excited level. The longest wavelength photon that will be emitted has wavelength (given in terms of the Rydberg constant R for the hydrogen atom) equal to 9 36 (b) (a) 5R 5R (c)

18 5R

(d)

4 R IIT, 2000

28.16 Comprehensive Physics—JEE Advanced

83. Electrons with energy 80 keV are incident on the tungsten target of an X-ray tube. K shell electrons of tungsten have – 72.5 keV energy. X-rays emitted by the tube contain (a) a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of about 0.155Å (b) a continuous X-ray spectrum (Bremsstrahlung) with all wavelengths (c) the characteristic X-ray spectrum of tungsten (d) a continuous X-ray spectrum (Bremsstrahlung) with a minimum wavelength of about 0.155 Å and the characteristics X-ray spectrum of tungsten. IIT, 2000 84. Photoelectric emission is observed from a metallic surface for frequencies v1 and v2 of the incident light (v1 > v2). If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio 1 : n, then the threshold frequency of the metallic surface is (a)

v1 v2 n 1

(b)

(c)

nv2 v1 n 1

(d)

nv1 v2 n 1 v1

v2 n

85. If 0 is the de Broglie wavelength for a proton accelerated through a potential difference of 100 V, the de Broglie wavelength for -particle accelerated through the same potential difference is (a) 2 2 (c)

0

0

2 2

(b)

0

2 0

(d)

2

86. The de-Broglie wavelength of a particle moving with a velocity 2.25 108 m/s is equal to the wavelength of a photon. The ratio of kinetic energy of the particle to the energy of the photon is 1 3 (b) (a) 8 8 (c)

5 8

(d)

7 8

87. A radiation of energy E falls normally on a perfectly the surface is E (a) c

(b)

2E c

(c) Ec

(d)

E

c2 88. The kinetic energy of the most energetic photoelectrons emitted from a metal surface is doubled when the wavelength of the incident radiation is reduced from 1 to 2. The work function of the metal is hc hc (a) (2 2 – 1) (b) (2 1 – 2) 1

(c)

2

hc 1

1

(

1

+

2)

(d)

2

2

hc 1

(

1

–

2)

2

89. The slope of the graph of the frequency of incident light versus the stopping potential for a given metallic surface is h (a) h (b) e e (c) (d) eh h 90. Lights of two different frequencies, whose photons have energies 2 eV and 10 eV respectively, successively illuminate a metal of work function 1 eV. The ratio of the maximum speeds of the emitted electrons will be (a) 1 : 5 (b) 3 : 11 (c) 1 : 9 (d) 1 : 3 91. The mass of a photon of wavelength is given by h (a) h c (b) c hc h (c) (d) c 92. The de-Broglie wavelength of an electron moving in the nth Bohr orbit of radius r is given by 2 r (b) n r (a) n nr nr (d) (c) 2 93. The momentum of a particle of mass m and charge q is equal to that of a photon of wavelength . The speed of the particle is given by h h (b) (a) m qm mh (c) qh (d) 94. The kinetic energies of photoelectrons emitted from a metal are K1 and K2 when it is irradiated with lights of wavelength 1 and 2 respectively. The work function of the metal is (a)

K1

1

K2

2

1

2

(b)

K1

1

K2

2

1

2

Atomic Physics 28.17

(c)

K1

2

K2

2

1

1

(d)

K1

2

K2

2

1

1

95. When the energy of the incident radiation is increased by 20%, the kinetic energy of the photoelectrons emitted from a metal increased from 0.5 eV to 0.8 eV. The work function of the metal is (a) 0.65 eV (b) 1.0 eV (c) 1.3 eV (d) 1.5 eV 96. An electron of mass m is moving such that its momentum is equal to that of a photon of wavelength . The velocity of the electron is (h is the Planck’s constant) h mh (a) (b) m (c)

2h m

(d) mh

97. The radius of hydrogen atom in the ground state is 0.53 Å. After collision with an electron, it is found to have a radius of 2.12 Å. What is the principal quantum number n the atom? (a) n = 1 (b) n = 2 (c) n = 3 (d) n = 4 IIT, 2003 98. Figure 28.11 shows the variation of photoelectric current (i) with anode potential (V) for a photosensitive surface for two radiations of intensities Ia and Ib and frequencies a and b for the curves a and b respectively. It follows from the graph that (b) a = b, Ib > Ia (a) a = b, Ib < Ia (c) a < b, Ib > Ia (d) a < b, Ib = Ia ( IIT, 2004 i

b a

-VO

O

V

Fig. 28.11

99. A photon of energy 10.2 eV undergoes an inelastic collision with a hydrogen atom in the ground state. After a few microseconds, another photon of energy 14.6 eV collides inelastically with the same hydrogen atom. In these processes

(a) a photon of energy 3.4 eV and an electron of energy 1.0 eV are released. (b) a photon of energy 10.2 eV and an electron of energy 1.0 eV are released (c) two photons of energy 10.2 eV are released (d) two photons of energy 3.4 eV and 1.0 eV are released. IIT, 2005 100. Ultraviolet light of wavelengths 1 and 2 (with 2 > 1) when allowed to fall on hydrogen atoms in their ground state is found to liberate electrons with kinetic energies E1 and E2 respectively. The value of the Planck’s constant can be found from the relation IIT, 1983 1 (a) h = ( 2 – 1)(E1 – E2) c (b) h =

1 ( c

(c) h =

( E1 E2 ) 1 c( 2 1)

(d) h =

( E1 E2 ) 1 c( 2 1)

2

+

1)(E1

+ E 2)

2

2

101. The wavelength of the characteristic X-ray K line emitted by a hydrogen like element is 0.32 Å. The wavelength of K line emitted by the same element will be IIT, 1990 (a) 0.21 Å (b) 0.27 Å (c) 0.34 Å (d) 0.40 Å 102. The potential energy U of a moving particle of mass m varies U with x as shown in E Fig. 28.12. The deBroglie wavelengths of the particle in the 0 1 x regions 0 x 1 and x > 1 are 1 and 2 Fig. 28.12 respectively. If the total energy of the particle is nE, the ratio 1/ 2 is IIT, 2005 (a)

(c)

n (n 1) n2 (n 2 1)

(b)

(d)

n (n 1) n2 (n 2 1)

28.18 Comprehensive Physics—JEE Advanced

103. The wavelength of K line from an element of atomic number 41 is . Then the wavelength of K line of an element of atomic number 21 is (a) 4 (b) /4 (c) 3.08 (d) 0.26 IIT, 2005 104. Electrons with de-Broglie wavelength fall on the target in an X- ray tube. The cut-off wavelength of the emitted X-ray is (a) (c)

0

= 2mc h

0

=

2

2m 2 c 2 h2

(b)

0

=

(d)

0

=

107. Photoelectric effect experiments are performed using three different metal plates p, q and r having work functions p = 2.0 eV. q = 2.5 eV and r = 3.0 eV, respectively. A light beam containing wavelengths of 550 nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct I-V graph for the experiment is (see Fig. 28.13) [Take hc = 1240 eV nm]

2h mc

2

IIT, 2007 105. The largest wavelength in the ultraviolet region of the hydrogen spectrum is 122 nm. The smallest wavelength in the infrared region of the hydrogen spectrum (to the nearest integer) is (a) 802 nm (b) 823 nm (c) 1882 nm (d) 1648 nm IIT, 2007 106. Which one of the following statements is Wrong in the context of X-rays generated from a X-ray tube? (a) Wavelength of characteristic X-ray decreases when the atomic number of the target increases. (b) Cut-off wavelength of the continuous X-rays depends on the atomic number of the target (c) Intensity of the characteristic X-rays depends on the electric power given to the X-rays tube (d) Cut-off wavelength of the continuous X-rays depends on the energy of the electrons in the X-ray tube. IIT, 2008

Fig. 28.13

IIT, 2009 108. Balmer series of hydrogen atom is 6561Å. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is (a) 1215 Å (b) 1640 Å (c) 2430 Å (d) 4687 Å IIT, 2011

ANSWERS

1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67.

(b) (c) (a) (d) (d) (a) (d) (c) (a) (b) (c) (a)

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68.

(b) (c) (c) (b) (b) (b) (b) (b) (d) (c) (b) (d)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63. 69.

(d) (d) (d) (b) (d) (b) (b) (b) (b) (a) (b) (b)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64. 70.

(d) (d) (d) (c) (d) (b) (b) (a) (d) (c) (a) (b)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65. 71.

(b) (d) (a) (a) (d) (b) (a) (d) (c) (c) (b) (b)

6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66. 72.

(a) (a) (b) (a) (c) (b) (d) (a) (c) (c) (a) (c)

Atomic Physics 28.19

73. 79. 85. 91. 97. 103.

74. 80. 86. 92. 98. 104.

(d) (b) (c) (b) (b) (b)

(d) (c) (b) (a) (b) (a)

75. 81. 87. 93. 99. 105.

76. 82. 88. 94. 100. 106.

(d) (c) (b) (a) (b) (b)

77. 83. 89. 95. 101. 107.

(c) (c) (a) (a) (c) (a)

78. 84. 90. 96. 102. 108.

(b) (d) (c) (b) (b) (a)

(d) (b) (d) (a) (b) (a)

SOLUTIONS 1. The saturation current is proportional to the intensity of incident radiation. The slopping potential (the magnitude of V when i = 0) increases with increase in frequency. Hence Ib > Ia and b > a. So the correct choice is (b). 2. Stopping potential is the negative potential needed to stop the fastest moving electrons. Kmax = eV0. V0 = 3 V. Thus 3 eV = eV0 hc 3. K1 = –W (1)

The shortest X-ray wavelength ( 2) corresponds to continuous X-ray spectrum and is given by hc , which gives E=h eV = 2

hc 2 = eV From Eqs. (1) and (2), we get 1 2

1

K2 =

hc

–W

(2)

2

Putting

=

1

=

2 /2

in Eq. (1) and using Eq. (2), 2 hc K1 = –W

eV 2m

1 c

7. Speed of the electron in the nth orbit is e2 v= 2 0 hn Momentum p = mv =

2

= 2 (K2 + W) – W Hence K1 > 2K2, which is choice (c). 4. K X-rays are produced when electrons jump from n = 2 to n = 1 state and K X-rays are produced in a transition n = 3 to n = 1. Hence the energy of K rays is greater than that of K rays. Since E =

hc

;

the wavelength K rays is less than that of K rays. Hence peak A represents K line and peak B represents K line. 5. From the conservation of linear momentum, the two fragments will have equal and opposite momenta. Now de Broglie wavelength = h/p. Hence p 1 = 2 p1 2 Since p1 = p2,

1

= 1.

2

6. de Broglie wavelength of electrons is h h = 1 = p 2 meV

m e2 2 0 hn

de Broglie wavelength =

K1 = 2K2 + W

(1)

(2)

2 h2 0 h = n. Hence p m e2

n, which is choice (c). 8. Magnitude of magnetic moment of an electron moving in an orbit of radius r with angular frequency is charge area M= time period =

r2

e T

=

1 e r2 2

Magnitude of angular momentum is L = m r2 M e = L 2m eL M= 2m nh From Bohr’s postulate L = . Hence 2 eh M= n 4 m So M

n, which is choice (c).

T

2

28.20 Comprehensive Physics—JEE Advanced

9. The wavelengths in Brackett series are given by 1 1 1 = RH Z2 ; n = 5, 6, 7, … 2 n2 4 where RH is the Rydberg constant. The shortest wavelength corresponds to n = . Thus 1 1 = RH Z2 16 The shortest wavelength in Balmer series of hydrogen (Z = 1) is given by 1

13. Energy of a photon is E = h . Now E= 14.

15.

1 = RH 4

16 Given = . Therefore Z2 = =4 Z = 2. So 4 the correct choice is (a). 10. rn n2 or rn = k n2 where k is a constant. Hence rn + 1 = k (n + 1)2 rn = k n2 rn – 1 = k (n – 1)2

16.

Given rn

+ 1

– rn = rn

2

– 1

17.

2

k (n + 1) – k n = k (n – 1)2 2

2

(n + 1) – n = (n – 1)

2

which gives n = 4. 11. Current due to circulating electron is e vn e e = = I= 2 rn 2 rn /vn Tn

0

B=

I

2 rn 0

Now rn =

0

=

(1)

2 rn2

h2 n2

and vn =

m e2

e2 2 0 hn B

correct choice is (d). 2

12. r =

0

n h

2

m e2 L=

nh 2

r=

0

L= Thus L

1 n

5

. So the

or

=

c

. Hence

hc E

Hence the correct choice is (a). Lenses of equal diameters collect the same amount of lighter so that the intensity remains the same; hence the photoelectric current also remains the same. Energy of photon h = m c 2. Therefore, h h m= 2 = ( c= ) c c h . Hence the Momentum of photon = mc = correct choice is (d). The wavelength of blue light is longer than that of violet light and the wavelength of sodium light (yellow light) is longer than that of blue light. Since no photoelectric emission occurs for blue light, it follows that the wavelength threshold ( min) is less than the wavelength of blue light and hence less than that of yellow light. Hence the correct choice is (d). Energy of photons of frequency ( ) = h . If the lamp emits n photons per second, then the power of the lamp is P = nh . Now, = c / . Therefore nhc P =

=

e vn

hc

=

1020

6.6 10

34

3 108

= 30 W 660 10 9 18. h = h 0 + eV, where V is the stopping potential. For frequency , we have eV0 (i) h =h 0+ 2 and for frequency / 2, we have h = h 0 + eV0 2

(ii)

From (i) and (ii) on eliminating V0, we get 0 = 3 / 2. Hence the correct choice is (b). hc hc 19. h = h 0 + eV. Since = c / , we have = 0

nh = 2 L

(2 L) 2 m e2

m e2 4 0

+ eV, where 0 is the threshold wavelength. For wavelength , we have hc hc = + 2 eV0 (i) 0

and for wavelength 2 , we have

1/ 2

r1/2

r1/2, which is choice (a)

eV0 hc hc = + 2 2 0

(ii)

Atomic Physics 28.21

Eliminating V0 from (i) and (ii) we get Hence the correct choice is (d). 20. For light of frequency 2 0, we have 2 h 0 = h 0 + eV0 or h 0 = eV0

0

=3 .

For light of frequency 3 0, we have 3 h 0 = h 0 + eV or 2 h 0 = eV

min

=

c

=

max

(i) (ii)

From (i) and (ii) we get V = 2V0. Hence the correct choice is (b). 1 mv 2max = h ( – 0). For light of frequency 21. 2 = 2 0 we have 1 mv 20 = h (2 0 – 0) = h 0 (i) 2 For light of frequency = 5 0, we have 1 mv 2 = h (5 0 – 0) = 4 h 0 2

27.

(ii)

Dividing (ii) by (i) we get v 2 = 4v20 or v = 2v0. Hence the correct choice is (b). hc . For a metal of work function 22. W0 = h 0 =

Thus 23. Now

= max

0.

=

eV . Since h min

=

c

=

=

max

=

3 108

c

=

1.6 10 1

19

34

104

24. E = h =

hc

30.

min =

34 3 108 = 6.6 10 10 10

= 1.98 10 –15 J 2 10 –15 J Hence the correct choice is (a). 25. The correct choice is (d) because the emission of characteristic X-rays starts only when the incident electrons are accelerated beyond a certain critical value. 26. The maximum frequency is given by eV , i.e. max V max = h Hence the correct choice is (b).

3 108

hc , i.e. eV

1 . Hence the correct choice V

min

is (c). eV . Thus h choice is (a). max

=

V. Hence the correct

max

32. We use Bragg’s equation 2 d sin = n . Here order n = 1, = 30° and d = 0.6 nm. Therefore,

10 –10 m = 1 Å

Hence the correct choice is (a).

34

Since wavelength 0.01 nm is less than min, it will be absent from the continuous spectrum. Hence the correct choice is (d).

31.

6.6 10

6.6 10

1.6 10 19 66, 000 = 1.87 10 –11 m = 0.0187 nm

, we have

ch eV

. Thus if V is

min

29. V = 66 kV = 66,000 V. Now hc min = eV

is given by

Hence the correct choice is (c).

1

increased, min decreased. Hence the correct choice is (d). 28. X-rays, being electromagnetic waves, always travel with the speed of light i.e., 3 10 8 ms –1 in vacuum. It is only the energy of the incident electrons (and, therefore, the energy of the emitted X-rays) that depends on the accelerating p.d. The speed of X -rays has nothing to do with this p.d., it is a characteristic only of the medium of propagation. Unlike the case of visible light, the speed of X-rays changes but little with a change in medium (the refractive index of different media for X-rays is very close to unity)

0

W0 / 2, the threshold wavelength W0 hc = 2

ch . Thus V eV

=2

0.6

sin 30° = 0.6 nm

33. Here order n = 2 and tion, we have 2 d sin 45° = 2 or

d =

= 45°. From Bragg’s equa-

2

34. According to Moseley’s law = a (Z – b). Hence (Z – b)2. For K line, b = 1. Hence the correct choice is (b). 35. From Moseley’s law, we have 41

=

21

21

=

41 1 21 1 41

4

=

2

= 4. Hence

4

28.22 Comprehensive Physics—JEE Advanced

36. The correct choice is (b). 37. Due to collision, the energy of the scattered photon will be less than that of the incident photon as some hc energy is lost in the collision. Now E = h = . Thus if E decreases decreases and increases. Hence < and > . 38. To produce characteristic X-rays, the energy of the incident electrons must be greater than the binding energy of the innermost electron. Thus the accelerating voltage must be greater than 40 kV, then the accelerated electrons will have energy greater than 40 keV. Hence the correct choice is (b). hc . Hence the correct choice is (b). 39. min = eV 40. If V is increased, the energy of X-rays E = eV also increases. But E = h . Hence frequency is increased. 41. The wavelength associated with a particle of charge q, mass m and accelerated through a potential difference V is given by h = 2 m qV or

V =

for proton:

V =

For

-particle :V =

=

T1 = T2

h

45. K = 180 eV = 180 Now

=

=

2

46.

2m q qp q

2

=

1 4

1 1 = 2 8

42. The de Broglie wavelength of a particle of mass m moving with a speed v is given by, h = mv 1 Kinetic energy K = mv 2 or m v 2 = 2 K or m2v 2 2 2 K m . Therefore h

2K m Since K is the same for both particles, we have

. Hence

T

927 273 =2 27 273

10 –19 J = 2.88

1.6

10–17 J.

h 2m K 6.6 10 2 9 10

2

2

= 2 . Hence the correct choice is (c). 1 44. In terms of kinetic energy K = mv2, de Broglie 2 wavelength is h = 2m K 1 Thus . If K is increased by a factor of 4, K is decreased by a factor of 2. Hence the correct choice is (b).

h2 2 mp qp

2m = m 1

43. We have seen above that

2

2mq

m2 = m1

Hence the correct choice is (d).

=

( m = 4 mp and q = 2qp) Thus V = V/8. Hence the correct choice is (a).

=

=

2

h2

mp V = m V

= 2 Km or mv =

1

6.6 10

34

7.2 10

24

31

34

2.88 10

= 0.9

17 1/ 2

10 –10 m

-particle is an electron. Since electron has the smallest mass, its de Broglie wavelength is the longest as the velocity of all particles is the same.

47. The speed of photon = c = 3 10 8 ms –1. Wavelength of photon is, say, . h Wavelength of electron = = . Now mv 1 m v2 KE of electron 1 m v2 2 = = KE of photon 2 hc h =

m v2 h 2hc mv

=

v 1 = 2c 4

48. For Lyman series, we have 4 911.6 = 1215 Å. l= 3

=

l s

h mv

(

v = c / 2).

=

4 . Hence 3

Atomic Physics 28.23

49. The maximum number of electrons that can be accommodated in orbits with n = 3 is 2 12 + 2 22 + 2 23 = 28. 50. The correct choice is (d). 51. For an atom of atomic number Z, the radius of the nth orbit is given by [see Eq. (4)] K n2 (i) rn = Z 2 0h is a constant. For the ground m e2

where K =

K 32 = 3K 3 r3 3K = = 3. r1 K Hence the correct choice is (c). r3 =

55. The kinetic and potential energies of an electron in the nth excited state are given by

state of hydrogen (Z = 1), n = 1 so that

and

r1 = K Let n be the energy state of Be+++ for which the orbital radius is r1. Putting Z = 4 and rn = r1 = K in Eq. (i) we get n2 or n 2 = 4 or n = 2 K=K 4 Hence the correct choice is (b). 52. The energy in the nth state is given by [see Eq. (5)] En = – C=–

where

me 2 8

0

Z2

h2 n2

=

C Z2 n2

me2

is a constant. For the ground 8 0 h2 state n = 1 of hydrogen and for n = 2 state of Be+++, we have E1 = C and E 2 = E2 4C = = 4. E1 C Hence the correct choice is (d).

53. En =

CZ

C 4 2

En =

0

e2 rn

(ii)

1 8

0

e2 rn

(iii)

n = 3. Therefore, 1

= RH

1 4

32

= RH

1 4

1 9

1

1

=

1 5 RH 36

36 . We have seen above that 5 RH

2

=

16 . 3 RH

Hence

9C

K n2 . Therefore, for hydrogen (n = 1 Z state), we have ( Z = 1) r1 = K

54. Now rn =

Li++ (n = 3 state) we have (

4

57.

Z = 3)

16 3 RH 1 20 2 = 27 2

. For n = 1 state of hydrogen, we have

2

1

(i)

Notice from (i) and (iii) that E = – KE. Given E = – 3.4 eV. Hence KE = – E = – (– 3.4) = + 3.4 eV. Thus the correct choice is (b). PE 56. E= or PE = 2E 2 = 2 – 3.4 = – 6.8 eV. Hence the correct choice is (c).

or

= 2 . For E1 = En, we require n2 n 9C C = 2 which gives n = 3. Hence the correct n choice is (c).

and for

E = KE + PE = –

= 4C

n E1 = C and for the nth state of Li++, we have C 3

0

e2 rn

Adding (i) and (ii) we get the total energy E which is

2

2

8

PE = –

2

2

1

KE =

or

=

1

20 5 RH = 27 36 20 = . 27

58. Now Z 2. For doubly ionized lithium Z = 3. Hence the correct choice is (c). 59. Energy difference E Z2. For a singly ionized helium atom Z = 2. Hence E = 10.2 (2)2 = 40.8 eV. 60. Ionization energy E Z2. For helium Z = 2. Hence E for helium = 13.6 (2)2 = 54.4 eV, which is choice (c). 61. When an electron in the ground state receives 12.1 eV of energy, it jumps to a level where its

28.24 Comprehensive Physics—JEE Advanced

62.

63.

64.

65.

energy = 13.6 – 12.1 = 1.5 eV. This corresponds to the third excited state corresponding to n = 3. It can have three transitions, namely from n = 3 to n = 2, from n = 2 to n = 1 and from n = 3 to n = 1. Hence three spectral lines will be emitted. 1 . Therefore, n = 2. Now Orbital speed v0 n n 2. Hence the radius of the new orbit rn 2 = (2) r = 4r. Energy state – 3.4 eV corresponds to a level n given by – 13.6 eV / n 2 = – 3.4 eV which gives 1 n = 2. Now, orbital speed v0 . Hence the n orbital speed in the excited state is v/2. The total energy of an electron bound to an atom is negative and is the sum of its P.E. and K.E. The magnitude of the P.E. is twice that of the K.E. (as per the Bohr model) but since the P.E. is – ve, we have Total Energy = (– 2K) + (K ) = – K (K = Kinetic energy). Hence the ratio of total energy and kinetic energy = – 1 : 1 = – 1. The longest wavelengths in the two series are given by 1 1 1 3 = RH = RH 2 2 4 1 2 L 1

and

B

= RH

1

1

2

2

2

3

= RH

5 36

3 36 27 or L : B = 5 : 27 4 5 5 L h 66. Since the momentum of photon is , from the law B

=

of conservation of momentum, the recoil speed v of an atom of mass m is given by h h or v = mv = m 67. The maximum number of electrons allowed in an orbit is 2 n2. Hence the number of possible elements is 2 (12 + 22 + 32 + 42) = 60 68. The stopping potential depends on the frequency (or wavelength) of the incident electromagnetic wave and is independent of the distance of the source from photocell. Hence the stopping potential will still be 0.6 V. However, the saturation current varies as 1/r 2, where r is the distance of the source from the photocell. Since r is increased by a factor of 3, the saturation current will decrease by a factor of (3)2 = 9, i.e. it will be 18 mA/ 9 = 2 mA at r = 0.6 m. Hence the correct choice is (d).

E and 69. We know that c = . Now = h Therefore, E h E = c= = h p p Hence the correct choice is (b).

=

h . p

70. h = eV + W0 or eV = h – W0 = h – h 0. Now = c / and 0 = c / 0. Thus, for wavelength , we have 1 1 eV = h c (i) 0

and for wavelength eV = hc 3

1 2

1 0

1 2

eV = 3 hc

or

V = V/3)

= 2 , we have (

1

(ii)

0

From (i) and (ii) we have 1

1

1 2

=3

0

which gives (b).

0

1 0

= 4 . Hence the correct choice is

71. When an electron falls from energy level E2 to energy level E1, the frequency of the emitted radiation is given by h = E2 – E1 hc or = E2 – E1 1

1 (E2 – E1) = k (E2 – E1) hc where k = 1 / h c. For energy levels A, B and C, we have 1 = k (EC – EB) (i) or

=

1

1

= k (EB – EA)

(ii)

= k (EC – EA)

(iii)

2

1

and

3

Adding (i), (ii) and (iii), we get 1

1

1

1

2

3

= k {(EC – EB) + (EB – EA) + (EC – EA)} 2 = k (2EC – 2EA) = 3

or

1

1

1

2

=

1

or

3

Hence the correct choice is (b).

3

1

= 1

2 2

Atomic Physics 28.25

72. Peaks A and B represent characteristic X-rays; they are different for different emitters. Hence the correct choice is (c). 73. For wavelength we have 1 m v2 = h – W0 2 =

hc

hc

hc

0

0

For wavelength 1 mv 2

(i) 0

, we have

2

=

hc

80.

0

(ii) 0

From (i) and (ii) we get v

2

v2 Now

=

0

=

0

/ 4. Hence

Therefore,

v

2

v

2

v

=

4 3

3 /4

0 0

2

4 4 . Hence v > v . 3 3

is greater than

v 74. Energy required to remove one electron is E1 = 24.6 eV. The energy required to remove the second electron is 2

E2 = Z2 RHhc

1

1

( )2 1 = (2)2 13.6 eV = 54.4 eV Total energy required = E1 + E2 = 24.6 + 54.4 = 79.0 eV. 75. E = Z RHhc 2

76.

E=

hc

1

1

12

( )2

=

2

= (3) 34

6.63 10

2

9

= 59

103 eV

Also

r=

0 2

me

2

n ,

T=

4h 3 me 4

hc 6.626 10 34 3 108 = W0 4.0 1.6 10 19

= 3.10 10–7 m = 310 nm 81. Let v1 and v2 be the respective speeds of the two particles. The law of conservation of linear momentum gives m2 v2 m1v1 + m2v2 = 0 or = 1.0 m1 v1 Since de Broglie wavelength = h/(m v ), we will have m v 1 = 2 2 = 1.0 m1 v1 2 82. The energy levels of the hypothetical particle of double the mass of the electron but having the same charge as the electron are given by 2R hc En = – H2 ( Z = 2) n For the longest wavelength photon emission leav-

2 0

n

3

1

1

2

2

= 2 RHhc

5 36

2 3 5 = RHhc 18 The wavelength corresponding to this energy is =

2 r and m v r = nh/2 . Therefore, v 2 r 4 2m 2 T= = r n h / (2 m r ) nh h2

=

| E| = 2 RHhc

= 59 keV. 77. The stopping potential in volts = kinetic energy of the emitted photoelectrons in eV. Hence the correct choice is (b). 78. T =

max

transition from n = 3 state to n = 2 state. Hence

13.6 eV = 122.4 eV

3 108

0.021 10

3

T1 n1 = . Given T1 = T2 n2 8T2. Hence n1/n2 = 2. Hence the correct choice is (d). 79. For a given accelerating voltage, the wavelength of the continuous X-rays will vary from a minimum value min min is greater than zero. Hence the correct choice is (b). For the two orbits

hc 18 = , which is choice (c). E 5RH

83. The energy of incident electrons is E = 80 keV = 80 103 eV = 80 103 1.6 10–19 J = 1.28 10–14 J The minimum wavelength of the continuous X-ray spectrum is hc 6.6 10 34 3 108 = = min E 1.28 10 14 = 0.155 10–10 m = 0.155 Å Since the energy of the incident electrons is more than that of the K shell electrons in tungsten, the

28.26 Comprehensive Physics—JEE Advanced

characteristic X-ray spectrum of tungsten will appear as peaks on the background of the continuous X-ray spectrum. Hence the correct choice is (d). 84. E1 = h ( we get

1

–

0)

and E2 = h(

E2 = E1 Given

2

0

1

0

2

–

0).

Dividing them,

2

0

1

0

2E Momentum transferred = p – (– p) = 2p = . c Hence the correct choice is (b). hc – W0 (i) 88. Given Kmax = 1

Dividing, we get

1

=

0

=

0

2 2

m0 m1

(i)

, which is choice (c).

Kinetic energy of the particle is 1 E = mv2 2 From (i) and (ii), we have 1 mv 2 E v 2.25 108 2 = E m vc 2c 2 3 108 2.25 6 Hence the correct choice is (b).

1

E c

(ii)

2

–

1)

which is choice

=

e V0. Hence the slope of h

e which is choice (c). h

1 mv 2 = h 0 – W0. Now E1 = 2 – 1 = 1 eV and 2 E2 = 10 – 1 = 9 eV. Therefore E1/E2 = 1/9, i.e. 1 mv 2 2 1 = 1 1 9 mv22 2 v1 1 or . Hence the correct choice is (d). v2 3 hc 91. h = mc 2 or = mc2 ( c = ). Hence m=

h which is choice (b). c

nh 92. For nth Bohr orbit, mvr = . The de-Broglie 2 wavelength is h = mv nh But mv = . Therefore, 2 r =h

is

(2 2

90. E =

3 8

87. The momentum of a photon of wavelength h

which gives W0 = (a).

versus V0 graph is

e0 e1

hc is the de Broglie wavelength. The energy mv of a photon of this wavelength is hc hc E= h = = = mvc (i) h / mv

h

(ii)

1

hc

2m1 e1 V

=

p=

– W0

Dividing (ii) by (i), we get hc W0 2 2= hc W0

89. eV0 = h . Therefore,

h

Now, -particle has twice the charge and 4 times the mass of a proton, i.e. m1 = 4 m0 and e1 = 2 e0. Using these in Eq. (i), we get 1

hc 2

85. Let m0 and e0 be the mass and charge of a proton and m1 and e1 those of -particle. Then h and 0 = 2m0 e0 V =

2Kmax =

and

n 2 which gives 0 = 1 , which is choice (b). (n 1)

1

). Since the photon is

(– p).

E2 = nE1. Hence, we have n=

86.

(because E = h and c =

93. Given mv = 94. Given

hc 1

h

2 r nh

2 r which is choice (a). n

. Hence the correct choice is (a).

– W0 = K1

(i)

Atomic Physics 28.27

and

hc

– W0 = K2

(ii)

2

Eliminate hc from (i) and (ii). The correct choice is (a). 95. h = E + W0 = 0.5 eV + W0 (i) When the energy of the incident photon is increased by 20%, we have 6 h = E + W = 0.8 eV + W (ii) 0 0 5 Subtracting (ii) from (i), we get h = 1.5 eV. Hence W0 = h – 0.5 eV = 1.5 eV – 0.5 eV = 1.0 eV. Thus the correct choice is (b). h 96. Momentum of photon is p = . Momentum of an h electron moving with velocity v is mv. Given = h mv or v = , which is choice (a). m rn . If r1 and r2 are 97. We know that rn n2 or n

100. The energy of a photon of wavelength by hc E=

The energies of radiations of wavelengths 1 and are hc hc E1 = and E2 = respectively 1

1

1

1

2

The value of h is given by choice (c) 101. For a hydrogen like element, we have 1

= Z 2R H 1

For K -line:

1

For K -line:

1

1

n12

n22

= Z 2R H = Z2 RH

n=

1

1

12

22

1

1

12

32

3 Z 2 RH 4 =

8 Z 2 RH 9

Dividing, we have

98.

=

potential (V0) is the same for the two radiations. We know that eV0 = Emax and Emax = h – W0. Since V0 is the same, Emax and hence is the same for radiations a and b. Hence a = b. Since the saturation current is greater for radiation b than for radiation a, the intensity Ib is greater than Ia. Hence the correct choice is (b). 99. The energy required to excite a hydrogen atom from state (n = 1 state) to the n = 2 state is 10.2 eV. So when a photon of energy 10.2 eV undergoes an inelastic collision with a hydrogen atom in the ground state, the electron of the hydrogen atom jumps from n = 1 state to n = 2 state. The electron spends a time between 10–10 s to 10–8 s in the excited state before falling back to the ground state.In this process, a photon of energy 10.2 eV is released. After a few microseconds, when a another photon of energy 14.6 eV collides with the same

3 4

9 27 = 8 32

27 27 = 0.32 Å = 0.27 Å. 32 32 The correct choice is (b). 102. Total energy = KE + PE = K + U =

In the region 0 x 1; U1 = E. Therefore, kinetic energy is K1 = total energy – U1 = nE – E = (n – 1) E 1

=

h

h

2mK1

2 m (n 1) E

(1)

In the region x > 1; U2 = 0. Therefore, kinetic energy is K2 = nE – 0 = nE 2

=

h

h

2mK 2

2 mnE

Dividing (1) by (2), we get state. Now, the ionization energy of hydrogen atom is 13.6 eV. The part 13.6 eV of the energy of the incident photon is used up in ionizing the atom, i.e. in knocking the electron from the atom and remaining energy = 14.6 – 13.6 = 1.0 eV is used up in imparting kinetic energy to the released electron. Hence the correct choice is (b).

2

2

E1 – E2 = hc

then r2 2 .12 = =2 r1 0 . 53 Hence the correct choice is (b).

is given

1 2

=

n (n 1)

. So the correct choice is (b).

103. From Moseley’s law, we have 41 21

=

41 1 21 1

2

= 4. Hence

(2)

28.28 Comprehensive Physics—JEE Advanced

21

41

=

107. For photoelectric emission, the wavelength of the incident radiation must be less than the cut-off wavelength of the metal given by 0 = hc/ 0. For metals p, q and r, the cut-off wavelengths are hc 1240 eV nm = = 620 nm p = 2.0 eV p

=

4 4 104. de-Broglie wavelength is h = (1) 2mE where E is the kinetic energy of the electrons. The out-off wavelength is hc 0 = E h2 2m

=

2mc h

105.

1

2

. Hence 2

1 1 n22 n12 The largest wavelength in the ultraviolet region of the hydrogen spectrum corresponds to the transition n1 = 2 to n2 = 1 (Lyman series). Thus = RH

1 1 = RH 1 122 nm 4 RH =

which gives

=

108.

1

= RH Z 2

= RH =

106.

1

1 n12

1 n22

1 = RH 6561 RH =

R = H 9

9 9 3 122nm = RH 4

=

For Balmer series n1 = 2.

4 3 122 nm

1 32

1240 = 496 nm 2.5

For singly ionized helium atom, Z = 2. For hydrogen atom Z = 1.

3RH 4

The smallest wavelength in the infrared region of the hydrogen spectrum corresponds to n1 = and n2 = 3 (Paschen series). Therefore 1

=

1240 = 413.3 nm 3.0 Hence metal plate p emits photoelectrons for all the three given radiations, metal plate q emits photoelectrons for radiation of wavelengths 450 nm and 350 nm and metal plate r emits photoelectrons only for wavelength 350 nm. Therefore, photoelectric current is maximum for metal p and minimum for r, i.e. Ip > Iq > Ir. So the correct choice is (a). r

From Eq. (1) E =

0

q

(1)2

1 22

=

5R 36

36 5 6561

For second spectral line of helium, 1 1 3RH 1 = RH (2)2 2 = 2 2 4 4

823 nm

hc , which is independent. Hence choice (a) eV is wrong.

=

min =

3 4

36 5 6561

= 1215 Å

II Multiple Choice Questions with One or More Choices Correct 1. The threshold frequency for photoelectric emission from a material is 4.5 1014 Hz. Photoelectrons will be emitted when this material is illuminated with monochromatic light from a

1 32

(a) (b) (c) (d)

50 watt infrared lamp 100 watt red neon lamp 60 watt sodium lamp 5 watt ultraviolet lamp

Atomic Physics 28.29

2. When monochromatic light from a bulb falls on a photosensitive surface, the number of photoelectrons emitted per second is n and their maximum kinetic energy is Kmax. If the distance of the lamp from the surface is halved, then (a) n is doubled (b) n becomes 4 times (c) Kmax is doubled (d) Kmax remains unchanged 3. The maximum kinetic energy of photoelectrons in a photocell depends upon (a) the frequency of the incident radiation (b) the work function of the photosensitive material used in the cell (c) the intensity of the incident radiation (d) all the above parameters. 4. When ultraviolet light is incident on a photocell, its stopping potential is V0 and the maximum kinetic energy of the photoelectrons is Kmax. When X-rays are incident on the same cell, then (a) V0 will increase (b) K max will increase (c) V0 will decrease (d) K max will decrease 5. The work function of metal A is greater than that for metal B. The two metals are illuminated with appropriate radiation of frequency so as to cause photoelectric emission in both metals. If 0 is the threshold frequency and Kmax, the maximum kinetic energy of photoelectrons, then (a) 0 for metal A is greater than that for metal B. (b) 0 for metal A is less than that for metal B. (c) Kmax for metal A is greater than that for metal B. (d) Kmax for metal A is less than that for metal B. 6. X-rays are used to cause photoelectric emission from sodium and copper. Then (a) the stopping potential is more for copper than for sodium. (b) the stopping potential is less for copper than for sodium. (c) the threshold frequency is more for copper than for sodium. (d) the threshold frequency is less for copper than for sodium. 7. When a monochromatic point source of light is at a distance of 0 2 m from a photoelectric cell,

the cut-off voltage and the saturation current are respectively 0 6 volt and 18 0 mA. If the same source is placed 0 6 m away from the photoelectric cell, then (a) the stopping potential will be 0 2 volt (b) the stopping potential will be 0 6 volt (c) the saturation current will be 6 0 mA (d) the saturation current will be 2 0 mA. 8. When a point light source, of power W emitting monochromatic light of wavelength is kept at a distance a from a photosensitive surface of work function , and area S, we will have (a) number of photons striking the surface per W S unit time as 4 h c a2 (b) the maximum energy of the emitted photo1 electrons as (hc ) (c) the stopping potential needed to stop the most energetic emitted photoelectrons as e (hc ). (d) photoemission occurs only if range 0 h c/

lies in the

9. Which of the following statements are correct about photons? (a) The rest mass of a photon is zero (b) The energy of a photon of frequency is h (c) The momentum of a photon of frequency h is c (d) Photons do not exert any pressure on a surface on which they are incident. 10. Figure 28.14 shows the stopping potential V0 versus frequency for photoelectric emission from two metals A and B. Choose the correct statement(s) from the following.

Fig. 28.14

(a) Work function of A is greater than that of B. (b) Work function of B is greater than that of A.

28.30 Comprehensive Physics—JEE Advanced

(c) Threshold frequency of A is greater than that of B. (d) Threshold frequency of B is greater than that of A 11. The intensity of X-rays from a Coolidge tube is plotted against wavelength as shown in Fig. 28.15. The minimum wavelength found is C and the wavelength of K line is k. If the accelerating voltage is increased

(d) Its kinetic, potential and tatal energies decrease. IIT, 2000 14. Figure 28.16 shows graphs between cut-off voltage 1 V0 and for three metals 1, 2 and 3, where is the wavelength of the incident radiation in nm. If W1, W2 and W3 are the work functions of metals 1, 2 and 3 respectively, then (a) W1 : W2 : W3 = 1 : 2 : 4 (b) W1 : W2 : W3 = 4 : 2 : 1 (c) The graphs for metals 1, 2 and 3 are parallel to each other and the slope of each graph is hc/e, where h = Planck’s contant, c = speed of light and e = charge of an electron. (d) Ultraviolet light will eject photoelectrons from metals 1 and 2 and not from metal 3.

Fig. 28.15

(a) C decreases (b) K increases (c) ( K – C ) increases (d) C and K both decrease but ( K – C ) remains unchanged 12. A hydrogen atom and a Li2+ ion are both in the second excited state. If lH and lLi are their respective electronic angular momenta, and EH and ELi their respective energies, then (a) lH = lLi (b) lH > lLi (c) EH < ELi (b) EH > ELi 13. The electron in a hydrogen atom makes a transition from an excited state to the ground state. Which of the following statements are not true? (a) Its kinetic energy increases and its potential and total energies decrease. (b) Its kinetic energy decreases, potential energy increases and its total energy remains the same (c) Its kinetic and total energies decrease and its potential energy increases.

Fig. 28.16

IIT, 2006 15. In a hydrogen-like atom of atomic number Z = 11, an electron makes a transition from the nth orbit and emits a radiation in the Lyman series. What is the value of n if the de Broglie wavelength of the electron in the nth orbit is equal to the wavelength of the emitted radiation? IIT, 2006 (a) n < 10 (b) n between 11 and 24 (c) n = 25 (d) n > 30

SOLUTIONS 1. Electrons will be emitted if the frequency of incident light is greater than 4.5 1014 Hz. Wavelength of infrared light 10,000 Å, its frequency is

(infrared) =

3 108

10, 000 10 10 = 3 1014 Hz. Wavelength of red light is about 7800 Å, its frequency is about 3.8 1014 Hz. Frequency of

sodium light is about 5 1014 Hz and the frequency of ultraviolet light is about 15 1014 Hz. Hence the correct choices are (c) and (d). 2. The value of n is proportional to the intensity of incident light. If the distance of the lamp is halved, intensity becomes four times. But Kmax is independent of the intensity of light. Hence the correct choices are (b) and (d). 3. The correct choices are (a) and (b)

Atomic Physics 28.31

4. The frequency of X-rays is higher than that of ultraviolet light. Now Kmax = h ( – 0). Hence Kmax increases as is increased. Also Kmax = eV0, where V0 is the stopping potential. Hence V0 also increases with frequency. Hence the correct choices are (a) and (b). 5. Work function W0 = h 0 and Kmax = h( – 0). So the correct choices are (a) and (d). 6. The work function of sodium is smaller than that of copper. Since W0 = h 0, the threshold frequency for sodium is less than that for copper. So choice (c) is correct and choice (d) is incorrect. Since the work function of sodium is lower than that of copper, it is easier to extract electrons from sodium than from copper. Therefore, the electrons ejected from sodium will have a greater kinetic energy and will hence need a greater stopping potential. So choice (a) is incorrect and choice (b) is correct. 7. The cut off potential remains the same as long as the frequency of the incident light remains unchanged. The saturation current is proportion to the intensity of light. Since the distance has become 3 times 1 1 its previous value, the intensity is only 2 or 9 3 of its previous value. Hence the new value of the 18.0 saturation current is mA = 2.0 mA. Hence the 9 correct choices are (b) and (d). hc so that the num8. The energy of each photon is hc . ber of photons released per unit time is W ÷ These photons are spread out in all directions over an area 4 a 2 so that the ‘share’ of an area S is a fraction S/4 a 2 of the total number of photons emitted. The maximum energy of the emitted photoelectrons is hc 1 – = (h c – ). Emax = h – = The stopping potential is given by eVS = Emax. 1 1 Hence VS = Emax = (hc – ). e e Hence choice (c) is incorrect. For photoemission to be possible, we must have hc h . Hence or h c/ Thus the permitted range of values of is 0 hc / . Hence the correct choices are (a), (b) and (d). 9. The correct statements are (a), (b) and (c).

h h h – ( – 0). 0 = e e e Thus V0 = 0 if = 0. If follow from graphs in Fig. 28.14 that the intercept 0 on -axis is less for metal A than for metal B. Hence choice (d) is correct. Also since W0 = h 0, the work function of metal A is less than that of metal B. Hence choice (b) is also correct. 11. The minimum wavelength is given by hc C= eV As V increases, C decreases. Since the wavelength of K line is due to transition n = 2 to n = 1 in the element of the target in the tube, wavelength K remains unchanged as V is increased. Hence the difference ( K – C ) increases with increase in the accelerating voltage. Thus the correct choices are (a) and (c). 12. For a hydrogen-like atom, the energy in the nth excited state is Z2 E n2 2+ Since Z for Li is greater than Z for H+, |ELi| > |EH|. h Also l = n(n 1) . Hence lLi = lH. Thus, the 2 correct choices are (a) and (c).

10. h = h

0

+ eV0 or V0 =

13. Potential energy (PE) = –

Kinetic energy (KE) =

1 2

1 4

0

Ze2 r

1 4

0

Ze2 r

Total energy E = PE + KE Ze2 1 1 2 4 0 r When an electron makes a transition from an excited state to the ground state, the value of r decreases. From the above expressions it follows that the kinetic energy increases, while the potential energy and the total energy both decrease as they become more negative. Hence the only correct choice is (a). hc , where 0 is the 14. Work function W = h 0 = =–

threshold wavelength. Hence W1 : W2 : W3 =

(

0

hc hc hc : : ) ( ) ( 0 1 0 2 0 )3

28.32 Comprehensive Physics—JEE Advanced

=

1

:

1

:

where rn is the radius of the nth orbit. The deBroglie wavelength of the electron in the nth orbit is 2 rn h = = n p

1

( 0 )1 ( 0 ) 2 ( 0 )3 = 0.001 : 0.002 : 0.004 =1:2:4 Hence choice (a) is correct. In photoelectric emission, the relation between V0 and is given by hc eV0 = h – W = –W V0 =

or

hc 1 e

Now rn =

1 0 )2

2 r1n (1) Z The wavelength of the transition n = n to n = 1 in Lyman series is given by 1 1 = RZ2 1 2 n 2 n which gives = (2) 2 2 RZ (n 1) where R = 1.097 107 m–1. Equating (1) and (2), we have n2 2 r1n = RH Z 2 (n 2 1) Z n or n2 – 1 = 2 r1 RH Z n = 10 2 3.14 (0.53 10 ) (1.097 107 ) 11 =

= 0.002 nm–1, therefore ( 0)2 = 500 nm = 5,000 Å

1

= 0.004 nm–1, therefore ( 0) = 250 nm = 2,500 Å For photoelectric emission, the wavelength of the incident radiation must be less than the threshold wavelength. Since the wavelength of ultraviolet light is about 1200 Å, it will eject photoelectrons from all the three metals. Hence the correct choices are (a) and (d). 15. According to Bohr’s hypothesis, the momentum of the electron in the nth orbit is given by nh p= 2 rn (

10–10 m is the

Therefore,

W e

1 Hence the slope of the graph between V0 and hc is which is the same for all metals. Therefore, e choice (c) is correct. The threshold wavelength for the three metals are 1 = 0.001 nm–1, therefore ( 0)1 = 1000 nm ( 0 )1 = 10,000 Å (

r1n 2 , where r1 = 0.53 Z

0 )3

= 25 n or n – 25n – 1 = 0 The two roots of n are 1 1 n= (25 ± 25) 25 629 2 2 i.e. n = 25 or n = 0. Now n = 0 is not possible. Hence the value of n = 25. The only correct choice is (c). 2

III Multiple Choice Questions Based on Passage Questions 1 to 3 are based on the following passage Passage I A beam of light has three wavelengths 440 nm, 495 nm and 660 nm with a total intensity of 3.24 10–3 Wm–2 equally distributed amongst the three wavelengths. The beam falls normally on an area of 1.0 cm2 of a clean metallic surface of work function 2.2 eV. Assume that there

capable photon ejects one electron and take h = 6.6 10–34 Js. IIT, 1989 1. Photoelectric emission is caused by (a) light of wavelength 440 nm alone (b) light of wavelength 660 nm alone

Atomic Physics 28.33

(c) (d) 2. The is (a)

lights of wavelengths 440 nm and 495 nm lights of wavelengths 495 nm and 660 nm incident energy (in Js–1) of each wavelength 10–7

3.24

(b) 1.62

10–7

(c) 1.08 10–7 (d) 0.81 10–7 3. The total number of photoelectrons liberated per second is (b) 5.1 1011 (a) 4.9 1011 11 (c) 5.3 10 (d) 5.5 1011

SOLUTION 1. The threshold wavelength is hc (6.6 10 34 ) (3 108 ) = W0 2.2 1.6 10 19 = 6 10–7 m = 600 nm Out of the three given wavelengths, two wavelengths 1 = 440 nm and 2 = 495 nm will cause photoelectric emission as these wavelengths are less than 0. Thus the correct choice is (c). 1 3.24 2. Intensity of each wavelength is I = 3 10–3 = 1.08 10–3 W m–2. Area of metal surface is A = 1 cm2 = 1 10–4 m2. Therefore, energy of each wavelength is E = I A = 1.08 10–7 J s–1, which is choice (c). 0

=

Questions 4 to 6 are based on the following passage Passage II In a photoelectric effect set-up, a point source of light of power 3.2 10–3 W emits monoenergetic photons of energy 5.0 eV. The source is located at a dsitance of 0.8 m from the centre of a stationary metallic sphere of work function 3.0 eV and of radius 8.0 10–3 of photoelectric emission is one for every 106 incident photons. Assume that the sphere is isolated and initially neutral, and that photoelectrons are instantly swept away after emission. IIT, 1995

3. Let n1 be the number of photons of wavelength 1 incident per second. The energy of one photon = hc/ 1. Hence n hc E = 1 1

or

n1 =

E 1 (1.08 10 7 ) (440 10 9 ) = hc (6.6 10 34 ) (3 108 )

= 2.4 1011 Similarly n2 = 2.7 1011 Total number n = n1 + n2 = 5.1 choice (b).

1011, which is

4. The number of photoelectrons emitted per second is (a) 105 (b) 107 (c) 109 (d) 1011 5. The kinetic energy of the fastest electron is (a) 8 eV (b) 5 eV (c) 3 eV (d) 2 eV 6. The photoelectric emission stops when the sphere acquires a potential of (a) 2 V (b) 3 V (c) 5 V (d) 8 V

SOLUTION 4. Refer to Fig. 28.17.

If r is the radius of the metallic sphere and R its distance from the source S, the power received at the sphere is r2 P r2 = P =P 4 R2 4 R2 =

Fig. 28.17

Power of the source of light, P = 3.2 10–3 W. Energy of the emitted photon, E = 5.0 eV = 5.0 1.6 10–19 = 8.0 10–19 J.

3.2 10

3

(8.0 10 3 )2

4 (0.8) 2 = 8 10–8 W (Js–1) Number of photons striking the sphere per second is P 8 10 8 Js 1 = = 1011 s–1 n = E 8.0 10 19 J

28.34 Comprehensive Physics—JEE Advanced

Since one photoelectron is emitted for every 106 incident photons, the number of photoelectrons emitted per second is n 1011 = 105 per second n= 6 = 6 10 10 The correct choice is (a). 5. Kinetic energy of the fastest electron is

attractive force exerted by the positive charge of the sphere on the electrons. The photoelectric emission will stop when the sphere acquires a positive potential equal to the stopping potential. In other words, the work function of the sphere keeps on increasing with time till it becomes equal to 5.0 eV (which is the energy of the incident photon). At this time, the emission of photoelectrons stops. Increase in work function = 5.0 eV – 3.0 eV = 2.0 eV. This implies that the photoelectric emission will stop when the sphere has acquired a potential of 2.0 V due to accumulation of charge. Hence the correct choice is (a).

Emax = 5.0 – 3.0 = 2.0 eV, which is choice (d). 6. Due to the emission of photoelectrons, the metallic sphere acquires a positive charge and it will oppose the ejection of photoelectrons due to Questions 7 to 10 are based on the following passage Passage III A single electron orbits around a stationary nucleus of charge + Ze, where Z is a constant and e is the magnitude of electronic charge. It requires 47.2 eV to excite the electron from the second Bohr orbit to third Bohr orbit. The ionization energy of hydrogen atom = 13.6 eV. IIT, 1981 7. The value of Z is (a) 3 (b) 4 (c) 5 (d) 6 8. The ionization energy of the atom is

(a) 340 eV (c) 122.4 eV

(b) 217.6 eV (d) 13.6 eV

orbit is (a) – 680 eV (c) – 170 eV

(b) – 340 eV (d) – 85 eV

9.

10. Bohr orbit is (h = Planck’s constant) (a)

h 2

(b)

(c)

3h 2

(d)

h 5h 2

SOLUTION 7. The energy of the electron in the nth Bohr orbit is En = –

Z 2 Rhc

n2 Given, ionization energy of hydrogen atom = Rhc = 13.6 eV. Therefore, En = –

(13.6 eV) Z 2 n2

The energy required to excite the electron from n = 2 to n = 3, Bohr orbit is 1 1 E3 – E2 = – (13.6 eV) Z2 2 3 22 =

(13.6 eV) 5Z 2 36

Questions 11 to 14 are based on the following passage Passage IV Light from a discharge tube containing hydrogen atoms falls on the surface of a piece of sodium. The kinetic

Given E3 – E2 = 47.2 eV. Hence (13.6 eV) 5Z 2 = 47.2 eV 36 which gives Z = 5. So the correct choice is (c). 8. Ionization energy of the atom is 1 1 Z2 E – E1 = – (13.6 eV) 12 = (13.6 eV) (5)2 = 340 eV The correct choice is (a). 9. orbit is E1 = – 340 eV. The potential energy in this orbit = 2E1 = – 680 eV, which is choice (a). nh h 10. Angular momentum = = ( n = 1). So 2 2 the correct choice is (a). energy of the fastest photoelectrons emitted from sodium is 0.73 eV. The work function for sodium is 1.82 eV. Ionization potential of hydrogen is 13.6 V and the mass of hydrogen atom = 1.67 10–27 kg. IIT, 1992

Atomic Physics 28.35

11. The energy of the photons causing the photoelectric emission is (b) 2.912 10–19 J (a) 1.168 10–19 J –19 (d) 4.08 10–19 J (c) 1.744 10 J 12. The quantum numbers of the two levels in the emission of the photons are (a) n = 1, n = 4 (b) n = 1, n = 3 (c) n = 2, n = 4 (d) n = 3, n = 4 13. In the transition in Q. 12, the change in the angular momentum of the electron is (h = Planck’s constant)

(a)

h 2

(b)

h

3h 2 14. The recoil speed of the emitting atom (assuming that it is at rest before the transition) is of the order of (b) 1 ms–1 (a) 1 cm s–1 2 –1 (c) 10 ms (d) 104 ms–1 (c)

2h

(d)

SOLUTION 11. Given, Emax = 0.73 eV, W = 1.82 eV, ionization potential of hydrogen atom = 13.6 V and mass of hydrogen atom, m = 1.67 10–27 kg. (a) From Einstein’s photoelectric equation,

E4 – E2 = – (0.85) – (– 3.4) = 2.55 eV So the correct choice is (c). 13. According to Bohr’s theory, the angular momentum of the electron in the nth energy state is

h = W + Emax = 1.82 + 0.73

Ln =

–19

= 2.55 eV or 4.08 10 J The correct choice is (d). 12. We are given that the ionization potential of hydrogen atom = 13.6 V. Therefore, the ionization energy = 13.6 eV. The energy levels of hydrogen atom are given by RH hc

13.6

For

nh 4h 2h . For n = 4, L4 = = 2 2

2h h = 2 Change in angular momentum is 2h h h L = L4 – L2 = = n = 2, L2 =

= – 0.85 eV 42 The quantum numbers for which the energy difference is 2.55 eV are n = 2 and n = 4 since

Thus the correct choice is (b). 14. The recoil speed v of the emitting atom of mass m can be found by using the principle of conservation of linear momentum. The momentum p of a photon of wavelength is h h h = = p= c/ c where is the frequency of the emitted radiation. Hence h h mv = p = or v = c mc We have seen above that h = 4.08 10–19 J. Therefore, 4.08 10 19 = 0.81 ms–1 v= 27 8 1.67 10 3 10 The correct choice is (b).

Questions 15 to 17 are based on the following passage Passage V Assume that the de Broglie wave associated with an electron can form a standing wave between the atoms arranged in a one dimensional array with nodes at each of the atomic sites. It is found that one such standing wave is formed if the distance d between the atoms of the array is 2 Å.

A similar standing wave is again formed if d is increased to 2.5 Å but not for any intermediate value of d. IIT, 1997 15. The wavelength of the de Broglie wave associated with the electron is (a) 0.5 Å (b) 1.0 Å (c) 2 Å (d) 2.5 Å

En = –

eV n2 n Hence the values of energy the electron in n = 1, 2, 3 and 4 energy levels are E1 = E2 = E3 = E4 =

2

13.6 12 13.6 22 13.6 32

=–

= – 13.6 eV = – 3.4 eV = – 1.51 eV and

13.6

28.36 Comprehensive Physics—JEE Advanced

16. The minimum value of d for the formation of the standing wave is (a) 2.25 Å (b) 1.5 Å (c) 0.75 Å (d) 0.5 Å

17. The energy of the electron (in eV) is (a) 151 (b) 15.1 (c) 1.51 (d) 0.51

SOLUTION 15. From the condition of standing wave formation, we have n

2

Hence

2

=

h2 p2

= 2Å E=

= 2.5 Å 2 where n is an integer. These equations on subtracand (n + 1)

tion give

= 2.5 – 2 = 0.5 Å

secutive nodes = =

2

h2 2 mE h2

2m

2

6.63 10

=

= 1.0 Å

2

2 16. The minimun value of d = distance between con-

17.

=

= 2.42

9.1 10

h h 1 = and E = mv2 and p mv 2

=

2.42 10

2mE .

Questions 18 to 20 are based on the following passage Passage VI A hydrogen-like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of energy 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. The ground state energy of hydrogen atom = – 13.6 eV. IIT, 2000

31

10

10 2

10–17 J

= 0.5 Å =

34 2

1.6 10

17

= 151 eV

19

18. The atomic number Z of the atom is (a) 1 (b) 2 (c) 3 (d) 4 19. The quantum number n is (a) 1 (b) 2 (c) 3 (d) 4 20. The ground state energy (in eV) of the atom is (a) – 217.6 (b) – 108.8 (c) – 54.4 (d) – 13.6

SOLUTION 18,19.The atom will emit maximum energy for a transition n1 = 2n to n2 = 1, we have E2n – E1 = 204 eV (1) where E1 is the ground state energy of the atom, Also (2) E2n – En = 408 eV Now

E2n = – (13.6 eV)

Z2 2n

2

,

Questions 21 to 23 are based on the following passage Passage VII When a beam of 10.6 eV photons of intensity 2.0 W/m2 falls on a platinum surface of area 1.0 10–4 m2 and work function 5.6 eV, 0.53% of the incident photons eject photoelectrons. IIT, 2001 21. The maximum kinetic energy of the photoelectrons is

Z2

En = – (13.6 eV) and

E1 = – (13.6 eV)

n2 Z2

12 Using these in Equations (1) and (2) and solving, we get Z = 4 and n = 2 20. E1 = – (13.6 eV) (4)2 = – 217.6 eV (a) 10.6 eV (b) 8.1 eV (c) 5.0 eV (d) 0.53 eV 22. The minimum kinetic energy of the photoelectrons is (a) zero (b) 0.53 eV (c) 1.0 eV (d) 5.0 eV 23. The number of photoelectrons emitted per second is (a) 6.25 (c) 6.25

1019 1011

(b) 6.25 (d) 6.25

1016 106

Atomic Physics 28.37

SOLUTION 21. From Einstain’s photolectric equation, Kmax = h – W

N=

Intensity × area energy of photon

= 10.6 eV – 5.6 eV = 5.0 eV 22. The emitted photoelectrons have kinetic energies ranging from zero to a maximum value. 23. Number of photoelectrons emitted per second is Questions 24 to 26 are based on the following passage Passage VIII A hydrogen-like atom (described by the Bohr model) is observed to emit six wavelengths, originating from all possible transitions between a group of levels. These levels have energies between – 0.85 eV and – 0.544 eV (including both these values). Given hc = 1240 eV-nm and ground state energy of hydrogen atom = – 13.6 eV. IIT, 2002

=

2.0 10.6

1.0 10

4

1.6 10

19

0.53 100 0.53 100

1011

= 6.25

24. The quantum number of the lowest energy level is (a) 1 (b) 6 (c) 12 (d) 15 25. The atomic number of the atom is (a) 1 (b) 2 (c) 3 (d) 4 26. The smallest wavelength emitted in the possible transitions is very nearly equal to (a) 4 nm (b) 40 nm (c) 400 nm (d) 4000 nm

SOLUTION 24,25. A total of four energy levels will involve six elecronic transitions as shown in Fig. 28.18 It is given that En = – 0.85 eV En+3 = – 0.544 eV

The given value is En 0.85 = En 3 0.544 n

Hence,

Also, form Bohr's theory we know that

3

2

n n

=

0.85 0.544

= 1.5625

3

=1.25 or n = 12 n The atomic number of the atom may be computed as follows.

or

Z2 n2

(– 13.6 eV) = – 0.85 eV

0.85 0.85 (n 2 ) = (122 ) = 9 13.6 13.6 which gives Z = 3 26. The smallest wavelength corresponds to maximum energy difference. ( E)max = Emax – En = – 0.544 – (– 0.85) Z2 =

Fig. 28.18

En =

Z2 n2

En+3 = – Hence,

(– 13.6 eV)

= 0.306 eV

Z2 n

n 3 En = n2 En 3

3

2

(– 13.6 eV)

2

Questions 27 to 29 are based on the following passage Passage IX In a mixture of H–He+ gas (He+ is singly inoized He atom). H atoms and He+ ions excited to their respective

min

=

hc 1240 eV nm = E max 0.306 eV

= 4052 nm

total excitation energy to He+ ions (by collisions). Assume that the Bohr model of atom is exactly valid. IIT, 2007

28.38 Comprehensive Physics—JEE Advanced

27. The quantum number n lated in He+ ions is (a) 2 (b) 3 (c) 4 (d) 5 28. The wavelength of light emitted in the visible region by He+ lons after collisions with H atoms is (a) 6.5 10–7m (b) 5.6 10–7m

(d) 6.5 10–7m (c) 4.8 10–7m 29. The ratio of the kinetic energy of the n = 2 electron for the H atom to that of He+ ion is 1 1 (a) (b) 4 2 (c) 1 (d) 2

SOLUTION 27. En = – (13.6 eV)

Z2

hc hc = =– (13.6 eV)Z2 E

n2 For hydrogen Z = 1 and for helium Z = 2. Hydrogen

Helium

=4

– 0.85 eV

– 3.4 eV

=3

– 1.51 eV

– 6.04 eV

=2

– 3.4 eV

– 13.6 eV

– 13.6 eV

=1

– 54.4 eV

We notice that the hydrogen atom in n = 2 state has the same energy as the helium atom in the n = 4 state. If all the excitation energy of hydrogen in n = 2 state is transferred to helium, the energy transfer is 10.2 eV. Due to this the helium atom gets excited to n = 4 state. 1 1 28. E = – (13.6 eV) Z2 n12 n22 Questions 30 to 32 are based on the following passage Passage X When a particle is restricted to move along x-axis between x = 0 and x = a, where a is of nanometre dimension, its energies of the particle moving in such a restricted region, correspond to the formation of standing waves with nodes at its ends x = 0 and x = a. The wavelength of this standing wave is related to the linear momentum p of the particle according to the de Broglie relation. The energy of the particle of mass m is related to its linear momentum as E = p2 . Thus, the energy of the particle can be denoted by a 2m quantum number ‘n’ taking values 1, 2, 3, ... (n = 1, called the ground state) corresponding to the number of loops in the standing wave.

1

1

n12

n22

1

Given h = 4.1 10–15 eVs and c = 3 108 ms–1. For He, Z = 2, using these values, we get =

(4.1 10

15

eVs ) (3 108 ms 1 )

(13.6 eV) (2)2 = (– 0.226

–7

10 m)

1

1

n12

n22

1

1

1

1

n12 n22 The wavelength in the visible region corresponds to transitions n1 = 4 to n2 = 3. Thus 10–7m)

= (– 0.226 = 4.65

10–7m

1 16

1 9

1

The closest choice is (c). 29. For a given state n, kinetic energy is proportional to Z2 . Hence the correct choice is (a). Use the model described above to answer the following three questions for a particle moving in the line x = 0 to x = a. Take h = 6.6 10–34 J s and e = 1.6 10–19 C. IIT, 2009 30. The allowed energy for the particle for a particular value of n is proportional to (b) a–3/2 (a) a–2 –1 (c) a (d) a2 31. If the mass of the particle is m = 1.0 10–30 kg and a = 6.6 nm, the energy of the particle in its ground state is closest to (a) 0.8 meV (b) 8 meV (c) 80 meV (d) 800 meV 32. The speed of the particle, that can take discrete values, is proportional to (b) n–1 (a) n–3/2 1/2 (c) n (d) n

SOLUTION 30.

=

h p

p=

h

E=

p2 h2 = 2m 2m

2

(1)

Atomic Physics 28.39

For a standing wave, the distance between consecutive node = /2. If there are n nodes between x = 0 and x = a, then n 2a a= = (2) 2 n Using (2) in (1), we get

6.6 10 = 8 = 1.25 =

2 2

h n

E=

8ma i.e. E a–2, which is choice (a). 31. From Eq. (3), setting n = 1, we have 8 m a2

Questions 33 to 35 are based on the following passage Passage XI The key feature of Bohr’s theory of spectrum of hydrogen atom is the quantization of angular momentum when an electron is revolving around a proton. We will extend this energy of a diatomic molecule assuming it to be rigid. The rule to be applied is Bohr’s quantization condition. IIT, 2010 33. A diatomic molecule has moment of inertia I. By Bohr’s quantization condition its rotational energy in the nth level (n = 0 is not allowed) is (a)

h2

1 n2

(c) n

2

8

(b)

I

h2 8

2

1 n

(d) n2

I

6.6 10

9 2

1.25 10

21 19

= 7.8

10–3 eV

The closest choice is (b). 1 2E 2h 2 n 2 mv2 v2 = = 32. E = 2 m 8m 2 a 2 Thus v n. So the correct choice is (d).

h2

E1 =

30

10–21 J

1.6 10 = 7.8 meV

(3)

2

1.0 10

34 2

h2 2

8

I

h2 2

8

34. It is found that the excitation frequency from ground 4 1011 Hz. Then the moment of ecule is close to inertia of CO molecule about its centre of mass is close to (Take h = 2 10–34 Js) (a) (b) (c) (d)

10–46 10–46 10–47 10–47

2.76 1.87 4.67 1.17

kg kg kg kg

m2 m2 m2 m2

35. In a CO molecule, the distance between C (mass = 12 a.m.u) and O (mass = 16 a.m.u.), where 1 a.m.u. 5 = 10–27 kg, is close to 3 (a) 2.4 10–10 m (b) 1.9 10–10 m –10 (d) 4.4 10–11 m (c) 1.3 10 m

I

SOLUTION 33. Bohr’s quantization condition is nh nh L= I = 2 2 1 Rotational K.E. = I 2 =

2

1 = I 2

This is the excitation energy which is equal to h . Hence 3h 2

nh 2 I

n2 h2

( K.E.)rot =

h2 8

2

I

(22 – 12) =

3h 2 8

2

I

=h

I=

3h

8 2v Substituting the given values of h and , we get I = 1.87 10–46 kgm2. So the correct choice is (b). 8

2

, which is choice (d). 8 2I 34. We have shown in Q.33 that n2 h2 Rotational K.E. = (1) 8 2I For ground state, n n = 2. Putting n = 2 and n = 1 in Eq. (1), we have

2

I

35.

r1 =

m2 r m1 r and r2 = m1 m2 m1 m2

I = m1r12 + m2r22

28.40 Comprehensive Physics—JEE Advanced

= m1 =

r=

m2 r m1 m2

2

+ m2

m1 r m1 m2

2

1/ 2

(12 16) 5 12 16 10 27 3 = 1.3 10–10 m, which is choice (c). =

m1 m2 r 2 (m1 m2 ) I (m1 m2 ) m1 m2

1.87 10

46

1/ 2

IV Assertion-Reason Type Questions In the following questions, Statement-1 (Assertion) is followd by Statement-2 (Reason). Each question has the following four choices out of which only one choice is correct. (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. 1. Statement-1 A particle of mass M at rest decays into two particles of masses m1 and m2 which move with velocities v1 and v2 respectively. Their respective de Broglie wavelengths are 1 and 2. If m1 > m2, then 1 > 2. Statement-2 The de Broglie wavelength of a particle having momentum p is = h/p. 2. Statement-1 The de-Broglie wavelength of an electron of mass m moving in the nth Bohr orbit of radius r is 2 r/n. Statement-2 According to Bohr’s theory, the magnitude of the angular momentum of an electron moving with velocity v in the nth orbit is L = nh/2 . 3. Statement-1 Figure 28.19 shows the graphs of Kmax (maximum kinetic energy) of the emitted photoelectrons versus the frequency of the incident light for two different metals A and B. The lines for metals A and B are always parallel to each other. Statement-2 In photoelectric emission Kmax = h – W0, where W0 is the work function of the metal.

Fig. 28.19

4. Statement-1 From the graph in Q.3 above it follows that the work function of metal B is greater than that of metal A. Statement-2 The work function does not depend upon the slope of the graph. 5. Statement-1 When an electron in a hydrogen atom makes a transition from an excited state to the ground state, its kinetic energy increases, its potential energy decreases and its total energy remains constant. Statement-2 In a given orbit, the total energy of the electron consists of its kinetic energy and the electrostatic potential energy of the electron and the proton in the hydrogen atom. 6. Statement-1 When monochromatic light falls on a photosensitive material, the number of photoelectrons emitted per second is n and their maximum kinetic energy is Kmax. If the intensity I of the incident light is doubled, n is doubled but Kmax remains the same. Statement-2 The value of n is directly proportional to I but Kmax is independent of I.

Atomic Physics 28.41

7. Statement-1 When ultraviolet light is incident on a photocell, its stopping potential is V0 and the maximum kinetic energy of the photoelectrons is Kmax. When the ultraviolet light is replaced by X-rays, Kmax increases but V0 decreases. Statement-2 Photoelectrons are emitted with speeds ranging from zero to a maximum value. Below a certain

negative voltage V0, no photoelectrons are emitted in a photocell. 8. Statement-1 If the voltage of an X-ray tube is increased, the minimum wavelength of the emitted radiation decreases. Statement-2 The maximum frequency of the radiation in an X-ray tube is directly proportional to the voltage.

SOLUTIONS 1. The correct choice is (d). The law of conservation of linear momentum gives mv m1v1 + m2v2 = 0 or 2 2 = 1.0 m1v1

for metals A and B but 0 for B is greater than that for A. 5. The correct choice is (d). 1 e2 Potential energy (P.E.) = – 4 0 rn

Since de Broglie wavelength = h1/mv, we will have m v 1 = 2 2 = 1.0 m1v1 2 2. The correct choice is (a). For nth Bohr orbit, mvr = nh . The de-Broglie wavelength is 2 h = mv nh But mv = . Therefore, 2 r 2 r 2 r =h = nh n 3. The correct choice is (a). The slope of each graph = h, the Planck’s constant. 4. The correct choice is (c). The intercept of the line on the -axis gives the threshold frequency 0 and work function W0 = h 0. Thus work function = slope intercept. The value of slope is the same

Kinetic energy (K.E.) =

1 8

0

e2 rn

6. The correct choice is (a). 7. The correct choice is (d). The frequency of X-rays is higher than that of ultraviolet light. Since Kmax = h – W0, Kmax will increase if is increased. Also, Kmax = eV0. Hence V0 will also increase if is increased. 8. The correct choice is (a). The maximum frequency is given by eV . Thus max V max = h Hence

min

=

c max

Thus

min

=

ch eV

1 V

V Integer Answer Type Questions 1. Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength 975 Å. How many different lines are possible in the resulting spectrum? The ionization energy of hydrogen atom is 13.6 eV. IIT, 1982 2. A hydrogen-like atom of atomic number Z is in an excited state of quantum number n = 6. This

state by successively emitting two photons of energies 10.20 eV and 17.00 eV respectively. Find the value of Z. The ionization energy of hydrogen is 13.6 eV. IIT, 1994 3. The electric potential between a proton and an electron is given by

28.42 Comprehensive Physics—JEE Advanced

r r0 where V0 and r0 are constants and r is the radius of the electron orbit around the proton. Assuming Bohr’s model to be applicable, it is found that r is proportional to nx, where n is the principal quantum number. Find the value of x. IIT, 2003 4. The de Broglie wavelength of an electron moving with a velocity of 1.5 108 ms–1 is equal to that of a photon. Find the ratio of the kinetic energy of the photon to that of the electron. IIT, 2004 5. An element of atomic number 9 emits K X-ray of wavelength . Find the atomic number of the element which emits K X-ray of wavelength 4 . IIT, 2004 V = V0 ln

6. An -particle and a proton are accelerated from rest by a potential difference of 100 V. After this, their and p respectively. de Broglie wavelengths are The ratio

p

, to the nearest integer, is

7. A silver sphere of radius 1 cm and work function 4.7 eV is suspended from an insulating thread in free-space. It is under continuous illumination of 200 nm wavelength light. As photoelectrons are emitted, the sphere gets charged and acquires a potential. The maximum number of photoelectrons emitted from the sphere is A 10z (where 1 < A < 10). The value of ‘Z’ is IIT, 2011

SOLUTION 1. Energy of monochromatic radiation is E =h =

hc

=

6.63 10

34

975 10

8

3 10 10

V mv 2 = 0 mv2 = V0 r r Bohr’s quantum condition is

10–19 J = 12.75 eV

= 20.4

Let n be the number of the possible spectral lines. Then 1 1 12.75 = 13.6 2 n2 1

mvr = v=

n

2

0.85 13.6

=

2.

r2 = 1 16

n

4

n = 2. Hence – 13.6 eV

1

1

2

2

6

1 36 Z2 = 9

2

Z 2 = 10.20 + 17.00 = 27.20 eV

1 2 27.20 Z = 4 13.6 or

Z = 3.

r . Therefore, r0 V dV = 0 F= – r dr If v is the orbital speed of the electron and m its mass

3. V = V0 ln

nh 2 nh 2 mr

(2)

Using (2) in the (1) we get

1 12.75 =1 – 2 n 13.6 1

(1)

r

n.

h2 4

2

m V0

n2

Hence x = 1

4. Speed of photon (c) = 3 108 ms–1. Let be the wavelength of the photon. The de Broglie waveh length of the electron = . mv Given

=

h . Now mv

K.E. of photon h = 1 2 K.E. of electron mv 2 =

=

2hc mv

2c v 2 3 108 1.5 108

(

2

(

4

=

=

c

)

h ) mv

Atomic Physics 28.43

5. For K X-ray, (Z – 1)2 (9 – 1)

2

= constant. Hence 2

= (Z – 1) (4 )

64 16 4 Z – 1= 4 or

(Z – 1)2 =

6.

=

= Z=5

h

6.63 10

34

hc

3 108

200 10

9

10–17 J

= 9.945 = 6.2 eV

2 mqV

For proton m = mp, q = qp = e For -particle m = m = 4 mp and q = q = 2e p

7. Energy of incident radiation = hv =

=

m mp

q qp

Work function W0 = 4.7 eV stopping potential is V0 = 6.2 – 4.7 = 1.5 eV q Now V0 = 4 0r 1.5 =

=

4

2

8

2.83

The integer nearest to 2.83 is 3.

ne 4

0r

n = 1.04

=

n 1.6 10

19

1 10 2 107. Hence Z = 7.

9 109

29

Nuclear Physics

Chapter

REVIEW OF BASIC CONCEPTS 29.1

RUTHERFORD’S EXPERIMENT

- PARTICLE SCATTERING

A beam of fast -particles was made to fall on a thin metal foil. Rutherford observed that most of the -particles The distance of the closest approach of an given by r0 =

1 4

0

-particle is

2 Z e2 Ek

1 where Ek = mv2 is the initial kinetic energy of particle 2 and Z is the atomic number of the nucleus. The impact parameter is given by b = where

29.2

2 Z e2 cot ( / 2)

1 4

0

mv

is the scattering angle.

COMPOSITION OF THE NUCLEUS

The nucleus of an atom contains protons and neutrons which are collectively called nucleons. The total number of nucleons is called the mass number and it is denoted by A. The number of protons in a nucleus is called its atomic number Z. The number of neutrons is denoted by N, so that A=Z+N

29.3

ISOTOPES, ISOBARS AND ISOTONES

Nuclei with the same Z but different A are called isotopes, such 238 236 206 207 208 as 168O, 178O, 188O; 235 92U, 92U, 92U ; 82Pb, 82Pb, 82Pb.

Nuclei with the same A but different Z are called isobars, 40 such as 31H, 32He ; 73Li, 74Be ; 40 18Ar, 20Ca. Nuclei with the same (A Z) = N but different Z are called isotones such as 31H, 32H ; 31H, 42He ; 177N, 188O, 199F

29.4

THE ATOMIC NUCLEUS

and negative charges within the atom was given by Ernest Rutherford. His -particle scattering experiments suggest that most of the mass of an atom is located in a very small volume of radius of the order of 10–14 m, called the nucleus. The electrons reside outside the nucleus. The nucleus of every atom carries a positive charge. The radius of an atom is of the order of 10–10 m. The radius R of a nucleus of mass number A is given by the relation R = R0 A1/3 where R0 is a quantity which varies slightly from one nucleus to another.

29.5

MASS DEFECT AND BINDING ENERGY

The mass of a nucleus which contains Z protons and (A –Z) neutrons is always less than the sum of the masses of these particles in the free state. The difference is called the mass defect of the given nucleus and is given by m = Z mp + (A – Z) mn – m where m = mass of the nucleus, mp = mass of a proton and mn = mass of a neutron. The binding energy of the nucleus is given by BE = ( m)c2 = [Z mp + (A – Z) mn – m]c2 where c is the speed of light in free space. Since A is the total number of nucleons, Z m p ( A Z ) mn m c 2 BE per nucleon = A

29.2 Comprehensive Physics—JEE Advanced

29.6

NUCLEAR FISSION

The splitting of a heavy nucleus into two or more fragments of moderate and comparable sizes is called by the reaction. 235 92 U

141 56Ba

1 + 92 36Kr + 2 0n + energy

much more than the energy released in the usual nuclear reactions. This makes the a particularly suitable source of energy.

release of 2 to 3 neutrons—the very particles that initiated in a bulk sample of uranium increase in geometric ratio. The rate of energy release also increases similarly . of excess neutrons, the chain reaction is a one and is used in nuclear reactors. When there is no such control on the number of released neutrons, we have an uncontrolled chain reaction and this is the source of energy in the atom bomb. known as the moderator. The role of the moderator is to may be easily absorbed by another 23592 which contain nuclei of masses comparable to the neutron

29.7

NUCLEAR FUSION

The process of nuclear fusion consists in the ‘combination’ of two light nuclei to form a stable nucleus of mass less than the total initial mass. It is believed to be the main source of energy for the sun and the stars. The fusion reaction in stars is believed to occur either the cycle or the . The proton-proton cycle is as follows: 1 1 2 + 1H + 1H 1H + e + n 1 2 3 1H + 1H 2He + 3 3 4 1 1 2He + 2He 2He + 1H + 1H The energy released in this sequence works out to be Nuclear fusion occurs at very high temperatures of about 107 K and under extremely high pressures.

29.8

RADIOACTIVITY

The phenomenon of self-emission of radiations from a nucleus is called radioactivity and substances which emit

these radiations are called radioactive substances. The radiations emitted from a radioactive element are of three types. 1. : These rays consist of -particles. An alpha particle is a helium nucleus having two protons and two neutrons. It has a positive charge equal to the charge of two protons. It has an initial speed of about 107 ms–1. They have very little penetrating power. 2. : These rays consist of electrons. Their speed is very nearly equal to the speed of light. They have more penetrating power than alpha particles. 3. : These are high frequency electromagnetic waves having a high penetrating power. Alpha Decay The process of emission of an alpha particle from a nucleus is called alpha decay. When a nucleus emits an alpha particle 42 He , it loses two protons and two neutrons which means that the daughter nucleus has its mass number reduced by 4 and its atomic number reduced by 2. When a nucleus AZ X of mass number A and atomic number Z emits an -particle 42 He , it is transformed into a nucleus

A Z

4 2Y

whose mass number is (A – 4) and

atomic number is (Z – 2). Alpha decay is represented by A ZX

min us 42 He

A Z

4 2Y

+ energy

Beta Decay The process of the emission of an electron from a nucleus is called . In this process also, the nucleus achieves greater stability by emitting an electron. A neutron inside the nucleus is changed into a proton by the emission of an electron. Because the mass of an electron is negligibly small, the mass number of the resulting nucleus remains unaltered but its atomic number is increased by one. For example, when a radium nucleus 228 88 Ra emits a -particle, the resulting element is an isotope of actinium 228 89 Ac. Thus in -decay also, a new element is formed. The transformation of a nucleus AZ X into the nucleus A Z + 1 Y by -decay is represented by an equation A ZX

minus e Z

A 1Y

Gamma Decay Gamma rays are high frequency electromagnetic radiations (i.e. photons) which do not carry any charge. Hence in -decay, the mass number and atomic number of the nucleus remain unchanged so that no new element is formed.

Nuclear Physics 29.3

Radioactive Decay Law If N is the number of radioactive nuclei present in a sample at a given instant of time, then the rate of decay at that instant is proportional to N, i.e.

The distance of closest approach is r0 =

N

=–

The proportionality constant is called the disintegration constant. If N0 is the number of radioactive nuclei at time = 0, then the number of radioactive nuclei at a later time is given by N = N0 e– : The half life of a radioactive element is the time in which half the number of nuclei decay. It is given by ln 2 0.693 T = : The average life of a radioactive sample is the reciprocal to its disintegration constant, i.e.

=

EXAMP

=–

N or | R| =

N=

0.693 N T

As N decreases exponentially with time, R will also decrease exponentially with time. The SI unit of the decay rate R is called 1010 disintegrations per second, i.e. 1 Ci (curie) = 3.7 1010 disintegrations/s

3.7

1 mCi (millicurie) = 3.7

107 disintegrations/s

1

104 disintegrations/s

Ci (microcurie) = 3.7

9 109

1.6 10

8 10

19 2

13

10–14 m

Calculate the ratio of the radii of two nuclei of mass numbers 1 and 27.

SOLUTION 1/3 R1 = R0A1/3 1 and R2 = R0A2

R1 = R2

1/3

A1 A2

=

1 27

1/3

=

1 3

29.3 16 g of radioactive radon is kept in a container. How much radon will disintegrate in 19 days? Half life of radon is 3.8 days. SOLUTION Number of half lives in 19 days is 19 days =5 T 3.8 days Number of atoms left undecayed after 5 half lives is n=

=

N = N0

29.1 Calculate the distance of closest approach when a

79

29.2

: It is useful to use the concept of the decay rate R of radioactive disintegrations taking place in a sample per second, which is given by

Z e2 K

1 4 E0

= 2.3

1

Tav =

R=

10–13 J.

=8

1 2

n

= N0

1 2

5

=

N0 32

s of radon left undecayed after 19 days =

number of gold is Z = 79.

16 g = 0.5 g 32 ntegrated = 16 – 0.5 = 15.5 g

SOLUTION

29.4

Kinetic energy of porton is 106

K =5

106

1.6

10–19 J

The half life of a radioactive substance is 30 days. What is the time taken for 3/4 of its original mass to disintegrate?

29.4 Comprehensive Physics—JEE Advanced

Number of atoms of B after 2 hrs (i.e. after 1 half life of B) is

SOLUTION T = 30 days 3N 0 N = 0 4 4 n 1 1 1 = 2 4 2

NB =

N = N0 – N = N0

0.693 N . Therefore. T 0.693 N A 0.693 N B and RB = RA = TA TB

|R| = N =

Now n

n=2

Time taken = 2 half lives =2

1 The activity of a radioactive element reduces to th 16 of its original value in 30 years. Find the half life and the decay constant of the element. SOLUTION |R| = N, i.e. the activity of proportional to the number atoms present in the sample. N0 Given N= 16 From

N 1 = N0 2

n

, we have

1 1 n = n=4 16 2 i.e. there are 4 half lives in 30 years. Therefore the half life of the element is 30 years = 7.5 years T= 4 Decay constant

RA N = A RB NB

30 = 60 days.

29.5

0.693 0.693 = = T 7.5 years = 0.0924 per year.

29.6 Two radioactive substances A and B initially contain equal number of atoms. The half lives of A and B are 1 hour and 2 hours respectively. Find the ratio of their rates of disintegration at the end of 2 hours.

No 2

N /4 TB = 0 TA N0 / 2

2 =1 1

29.7 A sample contains 2.3 g of radioactive 230 90Th of half life 2.4 1011 seconds. How many disintegrations per second occur in the sample? Take Avogadro number = 6 1023 atoms per mole. SOLUTION 1 mole of a substance has a mass equal to its atomic mass expressed in grams. Hence, number of moles in 2.3 g of 230 90Th is 2.3 g = 10–2 mole 230 g/mole Also Avogdro number = number of atoms in 1 mole of the substance. Hence number of atoms in 10–2 mole is N = 6 1023 10–2 = 6 1021 atoms |R| =

0.693 6 1021 0.693N = T 2.4 1011

= 1.73

1010 disintegrations per second.

29.8 A radioactive substance of half life of 69.3 days is kept in a container. After a certain lapse of time, it was found that 20% of the substance is left undecayed. Find the time elapsed. Given ln (5) = 1.61. SOLUTION T = 69.3 days

SOLUTION Given TA = 1 hour and TB = 2 hours. At = 0, number of atoms of A = number of atoms of B = N0. Number of atoms of A after 2 hours (i.e. after 2 half lives of A) is NA = N0

1 2

2

=

N0 4

=

0.693 0.693 = 69.3 days T = 10–2 per day

N = 20% of N0 = 0.2 N0. Therefore N0 =5 N

Nuclear Physics 29.5

N = e– N0

N = N0 e–

Now

N0 =e N =

=

N0 N

ln

ln N 0 / N

1.61 2

SOLUTION (a) In -decay, the mass number remains unchanged and the atomic number increases by 1. The resulting isotope has a mass number 137 and atomic number 56. This corresponds to the isotope 137 56 Ba.

= ln 5

=

2

(b) The equation of -decay is

per day

days = 161 days

137 55 Cs

0 1e

137 56 Ba

Q

where is an antineutrino and Q is the energy released in -decay. Initial activity is R0 = 1.0 mCi Since there are 2 half lives in 60 years and since the activity is proportional to the number of atoms, the activity at the end of 60 years is R 1 2 1.0 mCi = 0 = R = R0 4 2 4 = 0.25 mCi

29.9 1 gram of caesuim

-decay

137 55 Cs

decays by -emission with

a half life of 30 years. (a) Name the resulting isotope. (b) Write the equation of decay. (c) If the initial activity of caesuim is 1.0 millicurie (mCi), what is its activity after 60 years.

I Multiple Choice Questions with Only One Choice Correct 1. The distance of the closest approach of an alpha K is r0. The distance of the closest approach when kinetic energy 2K will be (b) 4r0 (a) 2r0 r0 2

r0 4 2. The distance of the closest approach of an alpha p is r0. The distance of the closest approach when the (c)

(d)

momentum 2p will be (a) 2r0 (c)

r0 2

(b) 4r0 (d)

r0 4

3. When high energy alpha-particles ( 42He) pass through nitrogen gas, an isotope of oxygen is formed with the emission of particles named x. The nuclear reaction is 14 7N

+ 42He

What is the name of x?

17 8O

+x

(a) electron (c) neutron

(b) proton (d) positron

4. What is particle x in the following nuclear reaction? 9 4Be

+ 42He

12 6C

+x

(a) electron (b) proton (c) neutron (d) photon 5. When aluminium is bombarded with fast neutrons, it changes into sodium with emission of particle x according to the equation 27 13Al

What is x? (a) electron (c) neutron 6. In the equation

+ 10n

24 11Na

+x

(b) proton (d) alpha-particle

27 13Al

30 + 42He 15P + X, The correct symbol for X is

(a) (c)

0 1e 4 2He

(b) 11H (d)

1 0n

7. ond required to produce 3.2 W of power is

29.6 Comprehensive Physics—JEE Advanced

(a) 107 (c) 10

(b) 1010

15

(d) 10

17

8. 236 92U

117 46X

1 + 117 46Y + 2 0n given that the binding energy per nucleon of X and 236 92

9. The binding energy of deuteron (21H) is 1 per nucleon and an alpha particle (42He) has a binding energy of 7 reaction 2 2 4 1H + 1H 2He + Q the energy Q released is (c) 23 10. If M is the mass of a nucleus and A its atomic mass, then the packing fraction is M A M A (a) (b) M+A M (c)

M

A

M+A (d) M A

A

11. The half life of a certain radio isotope is 10 minutes. The number of radioactive nuclei at a given instant of time is 108. Then the number of radioactive nuclei left 5 minutes later would be (a) (c)

108 2 2

(b) 104 107

(d)

108 2

13. The radioactivity of a sample is X at a time 1 and Y at a time 2. If the mean life of the specimen is , the number of atoms that have disintegrated in the time interval ( 2 1) is (a) X 1 Y 2 (b) X Y Y)/

(d) (X

Y)

14. The decay constant of a radioactive sample is . The half-life and mean-life of the sample are (respectively) given by: (a) 1/ and (ln 2)/ (b) (ln 2)/ and 1/ (c) 1/ and

(ln 2)

(d)

(c)

A 1 ZY

(ln 2) and 1/

(d)

A Z

4 2Y

17. The radioactive decay of uranium into thorium is represented by the equation 238 234 92U 90Th + x What is x? (a) an electron (b) a proton (c) an alpha particle (d) a neutron 18. A carbon nucleus emits a particle x and changes into nitrogen according to the equation 14 6C

14 7N

+x

What is x? (a) an electron (b) a proton (c) an alpha particle (d) a photon 19. The radioactive decay of an element X to elements Y and K is represented by the equation A ZX

12. The half life of Pa 218 is 3 minutes. What mass of a 16 g sample of Pa 218 will remain after 15 minutes? (a) 3.2 g (b) 2.0 g (c) 1.6 g (d) 0.5 g

(c) (X

15. The ionising power and the penetration range of radioactive radiations increase in the order (a) , , and , , respectively (b) , , and , , respectively (c) , , and , , respectively (d) , , and , , respectively IIT, 1994 16. A radioactive element X has atomic number Z and atomic mass number A. It decays by the emission of an alpha particle and a gamma ray. The new element is (b) ZA 42 Y (a) ZA 12 Y

A Z+1 Y

A 4 Z 1

K

A 4 Z 1

K

The sequence of the emitted radiations is (a) , , (b) , , (c) , , (d) , , 20. The half-life of a radioactive substance is 10 days. This means that (a) the substance completely disintegrates in 20 days (b) the substance completely disintegrates in 40 days (c) 1/8 part of the mass of the substance will be left intact at the end of 40 days (d) 7/8 part of the mass of the substance disintegrates in 30 days. 21. A rate-meter measures the number of disintegrations per second from a radioactive source. It gives a count of 320 counts per second. Ninety minutes later, it gives 40 counts per second. What is the half-life of the source? (a) 30 minutes (b) 45 minutes (c) 60 minutes (d) 75 minutes

Nuclear Physics 29.7

22. Beta rays emitted by a radioactive material are (a) electromagnetic radiations (b) the electrons orbiting around the nucleus (c) charged particles emitted by the nucleus (d) neutral particles IIT, 1983

(a) 1.0 Ci (b) 0.5 Ci (c) 4 Ci (d) 8 Ci 28. If the binding energy per nucleon in 7 Li and 4 He the energy of the reaction 7

Li + p

2 42He is

23. The equation 411H+ (a)

4 + 2He

+ 2e+

-decay

(b) -decay

24. What is the number of and particles emitted in the following radioactive decay? 200 90X

168 80Y

(a) 8 and 6 (c) 8 and 8

(b) 6 and 8 (d) 6 and 6

25. A freshly prepared radioactive source of half life 2 hours 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 the source is (a) 6 hours (b) 12 hours (c) 24 hours (d) 128 hours IIT, 1988 26. A beam of fast-moving alpha particles was directed A , B and C ing to the incident parts A, B and C of the beam are shown in the Fig. 29.1. The number of alpha particles in (a) B will be minimum and in C maximum (b) A will be maximum and in B minimum (c) A will be minimum and in B maximum (d) C will be minimum and in B maximum

29. In nuclear reactions, there is conservation of (a) mass only (b) energy only (c) momentum only (d) mass, energy and momentum. 30. If the half life of a radioactive atom is 2.3 days, its decay constant would be (a) 0.1 (b) 0.2 (c) 0.3 (d) 2.3 1 of initial 64 value in 60 seconds. The half life of the substance is (a) 5 s (b) 10 s (c) 30 s (d) 20 s

31. A radioactive substance disintegrates

32. The atomic weight of boron is 10.81 and it has two isotopes 105 B and 115 B. The ratio of 105 B : 115 B in nature would be (a) 19 : 81 (b) 10 : 11 (c) 15 : 16 (d) 81 : 19 33. The mass number of a nucleus is (a) always less than its atomic number (b) always more than its atomic number (c) always equal to its atomic number (d) sometimes more and sometimes equal to its atomic number. IIT, 1986 34. An closest approach is of the order of (a) 1 Å (b) 10 –10 cm –12 (c) 10 cm (d) 10–15 cm IIT, 1981 35. A star initially has 10 deuterons. It produces energy via the processes 2 2 3 1H + 1H 1H + p 40

Fig. 29.1

27. A radioactive sample with a half life of 1 month carries a label “Activity = 2 microcuries on 1.8.1991”. What was the activity two months later?

and

2 1H

+ 31H

4 2He

+n

where the masses of the nuclei are : m ( 2H) = 2.014 amu, m (p) = 1.007 amu, m (n) = 1.008 amu and m (4He) = 4.001 amu. If the average power radiated

29.8 Comprehensive Physics—JEE Advanced

by the star is 1016 W, the deuteron supply of the star is exhausted in a time of the order of (a) 10 6 s (b) 10 8 s (c) 1012 s (d) 1016 s IIT, 1993 36. A nucleus ruptures into two nuclear parts which have their velocities in the ratio of 2 : 1. What will be the ratio of their nuclear sizes (radii)? (a) 2 1/3 : 1 (b) 1 : 2 1/3 (c) 3 1/2 : 1 (d) 1 : 31/2

43.

22

Ne nucleus, after absorbing energy, decays into two -particles and an unknown nucleus. The unknown nucleus is (a) nitrogen (b) carbon (c) boron (d) oxygen IIT, 1999

44. Binding energy per nucleon versus mass number curve for nuclei is shown in Fig. 29.2. W, X, Y and Z are four nuclei indicated on the curve. The process that would release energy is

37. A free neutron decays into a proton, an electron and (a) a neutrino (b) an antineutrino (c) an -particle (d) a -particle 38. A radioactive element 238 90X decays into number of -particles emitted is (a) 4 (b) 6 (c) 2 (d) 1

222 83Y.

The

39. Fast neutrons can easily be slowed down by (a) the use of lead shielding (b) passing them through water (c) elastic collisions with heavy nuclei IIT, 1994 40.

64 29

64 30

of two isobars Cu and Zn are 63.9298 u and 63.9292 u respectively. It can be concluded from these data that (a) both the isobars are stable (b) 64Zn is radioactive, decaying to 64Cu through -decay (c)

64

64

Zn through

(d)

64

64

Zn through

Cu is radioactive, decaying to -decay Cu is radioactive, decaying to -decay.

Fig. 29.2

(a) Y (c) W

2Z 2Y

(b) W (d) X

X+Z Y+Z IIT, 1999

45. Two radioactive materials X1 and X2 have decay constants 10 and respectively. If initially they have the same number of nuclei, then the ratio of the number of nuclei of X1 to that of X2 will be 1/e after a time 1 1 (a) (b) 10 11 (c)

11 10

(d)

1 9

IIT, 2000 46. A heavy nucleus at rest breaks into two fragments

IIT, 1997 41. The half-life of I is 8 days. Given a sample of 131I at time = 0, we can assert that (a) no nucleus will decay before = 4 days (b) no nucleus will decay before = 8 days (c) all nuclei will decay before = 16 days (d) a given nucleus may decay any time after =0. IIT, 1998 131

42. The order of magnitude of density of uranium nucleus is, (mp = 1.67 10–27 kg) (a) 1020 kg m–3 (c) 1014 kg m–3

(b) 1017 kg m–3 (d) 1011 kg m–3 IIT, 1999

The ratio of the radii of the fragments (assumed spherical) is (a) 1 : 2 (b) 1 : 4 (c) 4 : 1 (d) 2 : 1 47. A nucleus at rest splits into two nuclear parts having radii in the ratio 1 : 2. Their velocities are in the ratio (a) 8 : 1 (b) 6 : 1 (c) 4 : 1 (d) 2 : 1 48. If the radius of a nucleus 256X is 8 fermi, the radius of 4He nucleus will be (a) 1 fermi (b) 2 fermi (c) 3 fermi (d) 4 fermi

Nuclear Physics 29.9

49. The binding energy per nucleon of C-12 is

C-13 is very nearly equal to (a) 0.2

(b) 3.7

(c) 3.9

(d) 5

50. A radioactive element of mass number 208 at rest disintegrates by emitting an -particle. If E is the energy of the emitted -particle, the energy of disintegration is 52 E 51 (c) 52 E (a)

56. A radioactive sample consists of two distinct species having equal number of atoms initially. The mean life time of one species is and that of the other is 5 . The decay products in both cases are stable. A plot is made of the total number of radioactive nuclei as a function of time. Which of the plot? (see Fig. 29.3).

51 E 52 (d) E

(b)

51. moving in opposite directions. They annihilate each other by emitting two photons. If the rest mass eneach photon is (a) 5.1 10–3 Å

(b) 10.2

10–3 Å

(c) 8.2 10–3 Å (d) 6.2 10–3 Å 52. A gamma ray photon creates an electron-positron pair. If the total kinetic energy of the electronma ray photon is (given the rest mass energy of

53. A neutral -meson at rest disintegrates to form two identical photons. The mass of -meson is 264.2 m0, where m0 is the rest mass of an electron. equal to (a) 67.4

(b) 132.1

(c) 200

(d) 931

54. The half life of a substance is 20 minutes. What is the time interval between 33% decay and 67% decay? (a) 40 min

(b) 20 min

(c) 30 min

(d) 25 min

55. The electron emitted in beta radiation originates from (a) inner orbits of atoms (b) free electrons existing in nuclei (c) decay of a neutron in a nucleus (d) photon escaping from the nucleus IIT, 2001

Fig. 29.3

57. The half-life of 215At is 100 s. The time taken for the radioactivity of a sample of 215At to decay to 1/16th of its initial value is (a) 400 s

(b) 6.3 s

(c) 40 s

(d) 300 s

IIT, 2002 58. Which of the following processes represents a gamma-decay? (a)

A Z

X

(b)

A Z

X

(c)

A ZX

(d)

A ZX

A Z 1 A 3 Z 2

1 0n

A ZX

+

– 1e

b X

c

+ A Z–1 X

+ IIT, 2002

59. For a certain radioactive substance, it is observed that after 4 hours, only 6 25% of the original sample is left undecayed. Choose the only wrong statement from the following. (a) the half life of the sample is 1 hour (b) the mean life of the sample 1/2 hour (c) the decay constant of the sample is (ln 2) hour 1 (d) after a further 4 hours, the amount of the substance left over would be only 0.39% of the original amount.

29.10 Comprehensive Physics—JEE Advanced

60. If the end A of a wire is irradiated with alpha rays and the end B is irradiated with beta rays, then (a) there will be no current in the wire A to B B to A midpoint of the wire. 61. During a negative beta decay: (a) an atomic electron is ejected (b) an electron which is already present within the nucleus is ejected (c) a neutron in the nucleus decays emitting an electron (d) a part of the binding energy of the nucleus is converted into an electron. 62. The nucleus of

230 90 Th

decays to

226 88 Ra

and

4 2 He

with the emission of energy. If the original nucleus was at rest, the ratio of kinetic energies of He and Ra nuclei will be very nearly equal to (a) 22 (b) 44 113 (d) 113 (c) 2 63. A nucleus X, initially at rest, undergoes -decay according to the equation, A 92 X

2 28 Z Y

+

The values of A and Z are (a) 232 and 90 (b) 234 and 94 (c) 230 and 88 (d) 232 and 92 64. At a given instant there are 25% undecayed radioactive nuclei in a sample. After 69.3 s, the number of undecayed nuclei reduces to 12.5%. The mean life of the sample is (a) 1 s (b) 10 s (c) 100 s

(d) 1000 s

65. In Q. 64 above, the time in which the number of undecayed nuclei will further reduce to 6.25% of the reduced number will be (a) 6.93 s (b) 69.3 s (c) 693 s

(d) 277.2 s

66. A uranium nucleus 238 92 U emits two alpha particles and two beta particles and transforms into a thorium nucleus. The mass number and atomic number of thorium nucleus so formed are (a) 230 and 90

(b) 232 and 90

(c) 234 and 92

(d) 234 and 88

67. A radioactive substance of half life 69.3 days is kept in a container. The time in which 80% of the substance will disintegrate will be [take ln(5) = 1.61] (a) 1.61 days (b) 16.1 days (c) 161 days (d) 1610 days 68. The sequence of decays of a radioactive nucleus is D

D1

D2

D3

D4

If the mass number and atomic number of D2 are 176 and 71 respectively, the corresponding values of D will be (a) 180, 72 (b) 176, 70 (c) 172, 69 (d) 168, 67 69. The mass m of a uranium nucleus varies with its volume V as 1 (a) m V (b) m V (c) m V (d) m V2 IIT, 2003 70. A nucleus of mass number 220, initially at rest, emits an -particle. If the Q value of the reaction -particle will be

IIT, 2003 71. After 24 hours; the activity of a radioactive sample is 2000 dps (disintegrations per second). After another 12 hours, the activity reduces to 1000 dps. The initial activity of the sample in dps is (a) 1000 (b) 2000 (c) 4000 (d) 8000 IIT, 2004 72. If all the Helium nuclei in the core of a star get converted into oxygen nuclei, then the energy released per oxygen nucleus is (mass of 42 He-nucleus = 4.0026 amu, mass of 168O-nucleus = 15.9994 amu) IIT, 2005

73.

221 87Ra

undergoes radioactive decay with a half life of 4 days. The probability that a Ra nucleus will disintegrate in 8 days is 1 (a) 1 (b) 2 1 3 (c) (d) 4 4 IIT, 2006

Nuclear Physics 29.11

74. The half life of a radioactive sample is 6.93 days. After how many days will only one-twentieth of the sample be left over? Take loge (20) = 3.0. IIT, 1981 (a) 20 days (b) 27 days

236 92 U

>E

137 53 I

+E

97 39 Y

+ 2E (n)

(b) E

236 92 U

<E

137 53 I

+E

97 39 Y

+ 2E (n)

236 92 U

<E

140 56 Ba

+E

94 36 Kr

+ 2E (n)

(d) E

236 92 U

=E

140 56 Ba

+E

94 36 Kr

+ 2E (n)

IIT, 2007 76. A radioactive sample S1 having an activity of 5 Ci has twice the number of nuclei as another sample S2 which has an activily 10 Ci. The half lives of S1 and S2 can be (a) 20 years and 5 years, respectively (b) 20 years and 10 years, respectively (c) 10 years each (d) 5 years each IIT, 2008

(c) 30 days (d) 35 days 75. In the options given below, let E denote the rest mass energy of a nucleus and n a neutron. The correct option is (a) E

(c) E

ANSWERS 1. 7. 13. 19. 25. 31. 37. 43. 49. 55. 61. 67. 73.

2. 8. 14. 20. 26. 32. 38. 44. 50. 56. 62. 68. 74.

(c) (d) (b) (b) (b) (b) (b) (b) (d) (c) (c) (c) (d)

3. 9. 15. 21. 27. 33. 39. 45. 51. 57. 63. 69. 75.

(d) (c) (b) (d) (b) (a) (d) (c) (a) (d) (c) (a) (c)

(b) (c) (b) (a) (b) (d) (b) (d) (c) (a) (a) (c) (a)

4. 10. 16. 22. 28. 34. 40. 46. 52. 58. 64. 70. 76.

(c) (c) (b) (c) (d) (c) (d) (a) (d) (c) (c) (b) (a)

5. 11. 17. 23. 29. 35. 41. 47. 53. 59. 65. 71.

6. 12. 18. 24. 30. 36. 42. 48. 54. 60. 66. 72.

(d) (d) (c) (c) (d) (c) (d) (a) (a) (b) (d) (d)

(d) (d) (a) (a) (c) (b) (b) (b) (b) (b) (a) (c)

SOLUTIONS 1. The distance of the closest approach is given by r0 =

2 Z e2 K

1 4

0

1 1 m v 2. Thus r0 . When K is dou2 K bled, r0 becomes half. Hence the correct choice is (c).

where K =

2. K =

1 1 m v2 = 2m 2 r0 =

Thus r0

1

mv

=

p2 . Therefore, 2m

4 m Z e2

1 4

2

0

p2

. When p is doubled, r0 becomes onep2 fourth. Hence the correct choice is (d).

3. The charge number and the mass number must be the same on both sides of a nuclear reaction. Particle x must have charge number + 1 and mass number 1. Hence it is a proton. 4. Particle x must have zero charge and mass number 1. Hence it is a neutron. 5. The mass number of x = 27 + 1 – 24 = 4 and its atomic number = 13 + 0 – 11 = 2. Hence particle x is helium nucleus, which is called alpha particle. 6. The mass number of X is A = 27 + 4 – 30 = 1 and its atomic number Z = 13 + 2 – 15 = 0. Hence particle X is a neutron and its symbol is AZ n which is 10 n. 7. Energ 200 106 number is

1.6

10

–19

10 6 J. Therefore, the required

3.2 200 10

6

1.6 10

19

= 10 17

29.12 Comprehensive Physics—JEE Advanced

8. The energy released is of the order of (8.5 (117 + 117) 9. Binding energy of 21H = 1.15 = 1.15 4 2 He

number of nucelons -

= 7.1

number of nucleons = 7.1 Q the correct choice is (c).

4 = 28.4

20. In 30 days (i.e. 3 half-lives) 1/ 23 = 1/8 of the sustance is left. Hence the correct choice is (d).

10. The correct choice is (c). 11. We have T1/2 = 10 minutes. Therefore 5 minutes = 1 of half-life. It follows that the number of nuclei 2 left after 5 minutes will be 1 2 1/ 2

=

1 2

of the original number =

the same. Hence an electron ( -particle) is emitted. A–4 In transition ZA + 1Y Z–1K, the mass number decreases by 4 and charge number decreases by 2. Hence an -particle is emitted. In the third transition, mass and charge numbers do not change. Hence a -ray is emitted. Hence the correct choice is (b).

108 2

Hence the correct choice is (d).

21. The count decreases by a factor of 320/40 = 8 in 90 minutes. Now (2)3 = 8. Hence 90 minutes = 3 halflives. Hence the half life of the radioactive source is 30 minutes. 22. The correct choice is (c). 23. It represents a fusion of four 11 H nuclei with the emission a huge amount of energy. The correct choice is (c).

12. Since 15 minutes = 5

3 minutes = 5 half lives, the 1 1 number of nuclei left after 15 minutes = 5 32 2 of the original number. Therefore, the mass of 16 g 16 sample left after 15 minutes = = 0.5 g. Hence 32 the correct choice is (d).

24. The mass number reduces by 200 – 168 = 32. Hence 8 -particles are emitted. The emission of 8 -particles reduces the atomic number by 16. But the atomic number reduces by 90 – 80 = 10. Hence the number of –particle emitted = 16 – 10 = 6. Hence the correct choice is (a).

13. There were X atoms at time = 1 and we are left with Y atoms at time = 2. Therefore, the number of atoms that have disintegrated in the time interval ( 2 – 1) is X-Y. Hence the correct choice is (b). 14. For a radioactive material of disintegration constant , 1n 2 , and Half-life = 1 / 2 =

25. Since the half life is 2 hours, the intensity of the radiation falls by a factor of 2 every 2 hours. In 12 hours it will fall by a factor of (2)6 = 64. Thus, in 12 hours the intensity attains the safe level. Hence the correct choice is (b).

mean-life =

=

1

.

Hence the correct choice is (b). 15. Out of the three radioactive radiations, the -rays are the most ionising but least penetrative while the -rays are the least ionising but most penetrative. 16. Alpha particle has mass number 4 and atomic number 2. Thus A decreases to A – 4 and Z decreases to Z – 2. Hence the correct choice is (b). 17. The mass number of x must be 238 – 234 = 4 and atomic number of x must be 92 – 90 = 2. Hence the correct choice is (c). 18. The charge of x is opposite to that of a proton. Hence x is an electron. A 19. In transition AZ X Z + 1Y, the atomic (or charge) number increases by unity; mass number remaining

26. The probability of a beam of

-particles passing

number of particles in the beam. Thus beam A has the maximum number of particles and B has the minimum number of particles. Hence the correct choice is (b). 27. Two months = 2 half lives. The activity of the sam1 ple will become 2 , i.e. one-fourth in 2 months. 2 Hence the correct choice is (b). 28. Binding energy of 7Li = (binding energy per nucleons) (no. of nucleons) = 5.60 of 4He = 7.06 energy of two 4He nuclei = 28.24 Hence, the energy of reaction is (56.48 – 39.20) 29. The correct choice is (d). 30. The decay constant

is given by

Nuclear Physics 29.13

log e 2 = T

0.693 = 2.3

Now power P = 10 16 W. Therefore, the number of 1016 = 2.58 10 27 deutrons per second is 12 . 3 87 10 per second 1040 Deutron supply will exhaust in 2.58 1027

0.3

Hence the correct choice is (c). 60s = 10 s. 6 32. Let 1 and 2 be the respective abundance of isotopes 10B and 11B of mass number Z1 and Z2 of boron of mass number Z, then

31. Now 64 = (2)6. Therefore, half life =

1Z 1

or

+

=(

1

+

2)Z

1

+

2)

10

1

+ 11

2

=(

or 10.81

1

– 10

1

= 11

0.81

1

= 0.19

or which gives

1 2

33.

2Z 2

=

2

10.81

– 10.81

So the correct choice is (c). 36. Let m1 and m2 be the masses and v1 and v2 the velocities of the two parts. From the principle of conservation of momentum, we have m1 v = 2 m2 v1 Now, the radius of a nucleus R is the atomic mass. Since A m, nucleus; R m1/3.

2

m1v1 = m2 v2 or

2

19 , which is choice (a). 81

Z is greater than atomic number A for all nuclei; the single exception being the hydrogen nucleus for which A = Z. Hence the correct choice is (d).

34. The distance of closest approach is given by 1 2 Z e2 r0 = 4 0 Ek Here Z = 92 for uranium and E k 5 10 6 1.6 10–19 J. Also e = 1.6 10 –19 C and 1 = 9 10 9 Nm2 C –2. Using these values, we 4 0 get r0

10 –12 cm, which is choice (c).

35. Let the mass of 31H be x amu. Then the mass defect ( m)1 = m (21H) + m (21H) – m (31H) – m (p) = 2.014 + 2.014 – x – 1.007 = (3.021 – x) amu ( m)2 = m (21H) + m (31H) – m (42He) – m (n) = 2.014 + x – 4.001 – 1.008 = (x – 2.995) amu Total mass defect is m = ( m)1 + ( m)2 = (3.021 – x) + (x –2.995) = 0.026 amu 106 1.6 10–19 J. Hence, energy released is given by E = 0.026 931 106 1.6 10–19 = 3.87 10 –12 J

10 12 seconds

= 3.87

R Thus 1 = R2

m1 m2

1/ 3

1 2

1/ 3

1 (given) 2 A 1/3, where A the mass of the

1 1/ 3

2

Hence the correct choice is (b). 37. The conservation of spin requires that the third particle is an antinuetrino ( ). The decay reaction is 1 0 represented as 01 n 1 p + –1e + 38.

-particles are electrons. When a nucleus emits an electron, its atomic number increases by 1 but its mass number remains unchanged. Here the mass number decreases by 238 – 222 = 16. The mass number of a helium nucleus is 4. Hence 16 / 4 = 4 -particles are emitted. An -particle is 42He. Hence the atomic number decreases by 8 by the emission of 4 -particles. But the atomic number decreases by 90 – 83 = 7. This is possible if only one -particle is emitted. 39. The correct choice is (b). 40. The atomic mass of stable Cu is smaller than that of Zn. Since the given mass of Cu is greater than that of Zn, 64Cu will be unstable. In -decay, the atomic number is increased by one while the mass number remains unchanged. Thus 64 29 Cu

-decay

64 30 Zn

+ –10 e

Hence the correct choice is (d). 41. A radioactive nucleus may decay any time after = 0. The number of nuclei at any time is given 0.693 by N = N0 e– , where = T where T is the half-life. Hence the correct choice is (d).

29.14 Comprehensive Physics—JEE Advanced

42. If A is the atomic number, the mass of uranium nucleus is m = (1.67 10 –27)A kg and its volume is V=

4 3

r3 =

4 3 8.2

( m Density = V

10

Am

or

3

1/3

10 –15 m)

r = r0 A ; r0 = 1.25 27

1.67 10 8.2 10 2

45

m1 = m2

10–15 m A1/3]3

{1.25 –45

Now m1 = A0 (R1)1/3 and m2 = A0 (R2)1/3. Hence

A m3

2

X+Z

120

7.5

(c) W

2Y

120

7.5

(d) X

Y+Z

90

8.0

(90

30

2 (60

60

v1 m = 2 v2 m1

5.0)

R2 = R1

5.0)

45. N1 = N0 e– 1 , N2 = N0 e – 2 . Therefore N1 e 1 = = e – ( 1 – 2) 2 N2 e = e – (10 – ) = e– 9 N1 Given, = e–1. N2 1 Hence – 9 = – 1 or = . 9 46. Let m1 and m2 be the atomic masses of the fragments and v1 and v2 their velocities. From the principle of conservation of linear momentum, we have m 1v 1 + m 2v 2 = 0

or

m1 m2

v2 v1

1 8

3

1 8

8 , which is choice (a). 1

A2 A1

1/ 3

4 254

1/ 3

1 4

R1 8 fermi = 2 fermi 4 4 Hence the correct choice is (b).

or The binding energy of the products in reaction (c) is greater than that of the reactant. Hence reaction (c) releases energy.

1 2

48. R = R0 (A)1/3. Therefore

8.5

8.5 + 30

3

m1 1 = m2 8 From the principle of conservation of momentum, the magnitude of p1 = magnitude of p2 or m1 v1 = m2 v2, which gives

5.0

8.0 + 30

R1 R2

Hence the ratio of their masses is

44. The binding energies of the reactant and the products in the given nuclear reactions are as follows:

(b) W

1 2

1/ 3

A1 A2

or

Since the atomic number is conserved, we have 10 = 2 + 2 + Z or Z = 6. The carbon nucleus has atomic number 6.

8.5

R1 R2

or

Dividing, we get R1 A1 = R2 A2

43. The given nuclear reaction is given by the equation 22 4 4 X 10 Ne 2 He + 2 He + Z

60

1/ 3

47. Let A1 and A2 be the mass numbers of the two nuclear parts. Their radii are given by R1 = R0 (A1)1/3 and R2 = (A2)1/3

A kg

Hence the correct choice is (b).

2Z

1/ 3

Thus, the correct choice is (a).

1017 kg m –3

(a) Y

R1 R2

1 = 8

R1 R2

R2 =

49. Energy required = (7.48 50.

13 – 12

M) = 208 – 4 = 204 Now, total energy of disintegration = energy of daughter nucleus + energy of -particle p2 P2 2m 2m Since momentum is conserved, p = P. Hence or

E =

E =

p2 2

1 m

Energy of -particle =

1 M

p2 2

m M mM

p2 = E. Hence 2m

Nuclear Physics 29.15

E =

m

M

1 2

E

M

204 52 E E 204 51 Hence the correct choice is (a). =

which gives 1 =

Hence energy of each photon is E according to Duane-Hunt law, the wavelength of a photon of energy E 34

6.62 10

Å=

E in

= 8.2

3 108

151 . 106

Å

10–3 Å which is choice (c).

52. E rect choice is (d). 53.

-

1 264.2 m0 2 = 132.1 m0. Energy of photon = mc2 = 132.1 m0 c2 = 132.1 ( m0 c2 m) =

1 2

1

or

2

1

/T

1 2

–

1

= T = 20 min.

6.25 1 1 = = . Hence 4 hours are 100 16 (2) 4 equal to 4 half lives. Therefore, the half life of the substance is 1 hour, which is choice (a). The deln(2) cay constant = = ln(2) per hour, which is half life choice (c).

/

1 decay constant

N 100 33 67 N0 100 100 If 1 is the time for 33% decay, then 1 2

1

= (i)

33 1 2/ = 100 2 Dividing (ii) by (i), we have 2

1

(ii)

60. Alpha particles are positively charged helium nuclei and beta rays are negatively charged electrons. Hence the correct choice is (b).

(iii)

61. Negative

/T

33 1 ~ . Hence Eq. (iii) reduces to 67 2

Hence choice (b) is incorrect. 1 After further 4 hours (i.e. after 8 hours), = 8 (2) 1 100 = % = 0.39% of the substance remains 256 256 undecayed, which is choice (d). Thus the only incorrect choice is (b).

N 100 67 33 N0 100 100 If 2 is the time for 67% decay, then

1 2

1 1 hour = hour ln(2) 0.693

/

For 67% decay,

33 = 67

2

Hence the correct choice is (b). 55. Beta particle is an electron which is emitted from a nucleus when a neutron decays into a proton and an electron within a nucleus. Hence the correct choice is (c). 56. For each species, the number of radioactive nuclei decreases exponentially with time. Hence, for both the species taken together, the total number of radioactive nuclei will decrease exponentially with time. This is best represented in plot (d). 1 1 , it follows that the time taken for 57. Since 16 24 1 the radioactivity to decay to th of its initial val16 ue = four times the half-life of the sample = 4 1/2 = 4 100 s = 400 s. Thus, the correct choice is (a).

For 33% decay,

67 = 100

T

59. 6.25% =

54. Given T = 20 min. We know that N = N0

2

58. During the emission of a gamma radiation, both the mass number and atomic number remain the same. Hence the correct choice is (c).

Hence the correct choice is (a).

Now

=

4

51. Energy released in annihilation = 0.51 + 0.51 =

=

1

decay is expressed by the equation: n p + e– + v

Hence the correct choice is (c).

29.16 Comprehensive Physics—JEE Advanced

62. If the original nucleus Th is at rest, i.e. if the momentum of the system before -decay is zero, the total momentum after the decay will also be zero. Thus Ra and He will have equal and opposite linear momenta. mHe vHe = – mRa vRa m2He v2He=

or Now

m2Ra v2Ra

(i)

2 1/ 2mHe vHe K.E.(He) = 2 K.E.(Ra) 1/ 2mRa vRa

mHe vHe mRa vRa

=

=

mRa mHe

2

226 4

mRa mHe 113 [use Eq. (i)] 2

Hence the correct choice is (c). 63. The a-particle is a helium nucleus A 92 X

4 2 He .

Hence

228 4 Z Y+ 2 He

ln or

=

69.3 s = 100 s 0.693

Hence the correct choice is (c). 65. Let the reduced number further reduce to 6.25% in n half lives. Then 6.25 100

n

1 2

or

1 16

1 2

N = 20% of N0 = 0.2 N0. N = N0 e– , which gives

Now

N0 =e N or

ln

N0 N

=

1.61 10

2

momentum 2 mass

2

.

-particle (m number of daughter nucleus (M) = 220 – 4 = 216 units. If P and p denote the momenta of the daughter nucleus and the -particle respectively, then P2 p2 Q= 2M 2m Since momentum is conserved, P = p. Hence

n

which gives n = 4. Therefore, the time taken would be = 4T1/2. Hence the correct choice is (d). 66. In each alpha decay, the mass number decreases by 4 and atomic number decreases by 2. In each beta decay, the mass number remains unchanged, but atomic number increases by 1. Hence the correct choice is (a). 0.693 = = 67. T1/2 = 69.3 days. Therefore T1 / 2 –2 10 per day.

10

2

Hence the correct choice is (c). 68. The correct choice is (a). 69. If mp is the mass of a proton and A, the atomic number of uranium nucleus, then the mass of a uranium nucleus is m = mp A and the volume of uranium nucleus is 4 3 4 4 3 r = (r0 A1/3)3 = r0 A V= 3 3 3 ( r = r0 A1/3) mp A 3 mp m = = = constant 4 3 V 4 r03 r0 A 3 Thus m V, which is choice (c).

Hence the correct choice is (a).

half life ln(2)

ln(5)

= 161 days

70. Kinetic energy =

64. In 69.3 s, the number of nuclei reduces to half (from 25% to 12.5%). Hence half life = 69.3 s

N0 N

Q= Now

or

p2 2

1 M

1 M

=

p2 m 2m M

1

p2 = KE of -particle = E . Thus, 2m m M Q =E M E =

QM m M

=

55

216 4

216

Hence the correct choice is (B). 71. In 12 hours, the activity of the sample decreases from 2000 dps to 1000 dps, i.e. it becomes half in 12 hours. Hence the half life of the sample is 12 hours. Now, 24 hours = 2 half lives. Hence the initial activity = 2000 dps (2) 2 = 8000 dps. Hence the correct choice is (d). 72. Four He nuclei fuse to produce one oxygen nucleus according to equation

Nuclear Physics 29.17

4( 42 He)

16 8O

(42 He)

where

m(168 O)

m = 4m – = 4 4.0026 – 15.9994 = 0.011 amu Energy released per oxygen nucleus is E = mc2 = 0.011

Therefore =

3.0

=

0.693 0.693 = = 0.1 per day T1/ 2 6.93

=

3.0 = 30 days. Thus the cor0.1

rect choice is (c). 75. From the binding energy per nucleon versus mass number curve, it follows that

Hence the correct choice is (c). 73. The probability at time that a radioactive nucleus

and

94 36 Kr

137 97 140 53 I , 39Y , 56 Ba

have higher binding energy than

236 92 U

.

Hence the nuclei of I, Y, Ba and Kr are more stable N0 N N0 where N0 = number of nuclei present initially at time = 0 N = number of nuclei left undecayed at time time . Now 8 days = 2 half lives. After 2 half lives

than the nucleus of U. Therefore, the rest mass energy of U is greater than the sum of the rest mass energies of I and Y or Ba and Kr. Hence the correct choice is (a). 76. Activity R = – N where = decay constant and N = number of radioactive nuclie present in the sample. Also

P=

N=

N0 ( 2) N0

2

=

N0 4

=

N0 4

0.693 T

where T = half life of the sample. Hence

3 P= = N0 4 Hence the correct choice is (d). N 74. = e– .Thus 1 = e or 20 = e or N0 20 = loge (20) = 3.0

T=

0.693

=–

0.693 N R

T1 N1 R1 2 10 = T2 N 2 R2 1 5 So the correct choice is (a).

20 5

II Multiple Choice Questions with One or More Choices Correct 1. For a certain radioactive substance, it is observed that after 4 hours, only 6 25% of the original sample is left undecayed. It follows that (a) the half life of the sample is 1 hour (b) the mean life of the sample 1/(ln 2) hour (c) the decay constant of the sample is (ln 2) hour 1 (d) after a further 4 hours, the amount of the substance left over would be only 0.39% of the original amount. IIT, 2003 2. The half-life of a radioactive substance does not depends upon (a) its temperature

(b) the external pressure on it (c) the mass of the substance (d) the strength of the nuclear force between the nucleons of its atoms. 3. In nuclear reactions, which of the following are conserved? (a) mass (b) energy (c) momentum (c) charge 4. When a nucleus emits a photon (a) its actual mass decreases (b) its actual mass remains the same (c) its atomic number decreases (d) its atomic number remains the same.

29.18 Comprehensive Physics—JEE Advanced

5. A tiny positively charged particle is moving headon towards a heavy nucleus. The distance of closest approach depends upon (a) the number of protons in the nucleus (b) the number of nucleons in the nucleus (c) the mass of the incident particle (d) the charge and velocity of the incident particle. 6. A uranium nucleus (atomic number 92, mass number 238) emits an -particle and the resultant nucleus emits a -particle. If Z and A are the atomic (a) Z = 90 (c) A = 234

(b) Z = 91 (d) A = 233

IIT, 1982 7. The decay rate of a radioactive element is found to be 1600 disintegrations per second. If the half life of the element in 1 second, then (a) after 1 second the decay rate will be 800 disintegrations per second. (b) after 2 seconds the decay rate will be 400 disintegrations per second. (c) after 3 seconds the decay rate will be 200 disintegrations per second. (d) after 4 seconds the decay rate will be 100 disintegrations per second. IIT, 1983 8. Pick the possible nuclear fusion reactions from the following: (a)

13 6C

1 1C

14 6C

(b)

12 6C

1 1C

13 7N

(c)

14 1 7 N + 1H

(d)

235 1 92 U + 0 n

(b) The rest mass of a stable nucleus is greater than the sum of the rest masses of its separated nucleons. (c) In nuclear fusion energy is released by fusing two nuclei of medium mass (approximately 100 amu). (d) In nuclear fusion, energy is released by fragmentation of a very heavy nucleus. IIT, 1994 11. In the nuclear reaction 2 1H

+ 21 H

4 2 He + Q

2 the mass of 1 H = 2.0141 u, mass of 42 He = 4.0024 u and Q is the energy released. (a) It is a fusion reaction

(c) Q

Q

IIT, 1996 12. Let mp be the mass of a proton mn the mass of a neutron, M1 the mass of 20 10 Ne nucleus and M2 the 40 mass of 20 Ca nucleus. Then (b) M2 > 2M1 (d) M1 < 10(mp + mn) IIT, 1998 13. Assume that the nuclear binding energy per nucleon (B/A Use this plot to choose the correct choice (s) given below. (a) M2 = 2M1 (c) M2 < 2M1

15 8O 140 94 1 54 Xe + 38Sr + 2 0 n

200

IIT, 1984 9. In the uranium radioactive series the initial nucleus 206 . If m is the is 238 92 U 82 Pb number of -particles and n the number of -particles emitted when the uranium nucleus decays to lead, then (a) m = 8 (b) m = 6 (c) n = 10 (d) n = 6 IIT, 1985 10. Which of the following statement (s) is/are correct? (a) The rest mass of a stable nucleus is less then the sum of the rest masses of its separated nucleons.

(a) Fusion of two nuclei with mass numbers lying in the range of 1 < A < 50 will release energy (b) Fusion of two nuclei with mass numbers lying in the range of 51< A < 100 will release energy (c) Fission of a nucleus lying in the mass number range of 100 < A < 200 will release energy when broken into equal fragments (d) Fission of a nucleus lying in the mass number range of 200 < A < 260 will release energy when broken into equal fragments IIT, 2008

Nuclear Physics 29.19

SOLUTIONS tions are (b) and (c). 9. In -decay, the mass number decreases by 4 and the atomic number decreases by 2. In -decay, the mass number remains unchanged but the atomic number increases by 1. In the decay of uranium to lead, the mass number decreases by (238 -206) = 32. Hence the number of -particles emitted is m = 32/4 = 8. But atomic number decreases by (92–82) = 10. In the emission of 8 -particles, the mass number decreases by 16. Hence the number of -particles emitted is n = 16 – 10 = 6. So the correct choices are (a) and (d). 10. The correct choices are (a) and (d). 2 11. It is a fusion reaction because two light nuclei 1 H 4 fuse to form a heavier nucleus 2 He

6.25 1 1 100 16 24 The given time of 4 hours thus equals 4 half lives so that the half life is 1 hour. 1n 2 and mean life = Since half life = decay constant 1 , statements (b) and (c) are easily decay constant seen to be correct. After a furthur 4 hours, the amount left over would 1 1 1 100 i.e., or or 0.39% of the origbe 4 4 256 256 2 2 inal amount. Hence all the four choices are correct. The correct choices are (a), (b) and (c). All the four choices are correct. A photon carries energy which is equivalent to mass. Hence the correct choices are (a) and (d). The distance of closest approach is given by 1 qZe r0 = 2 0 mv 2 where q, m and v respectively are the charge, mass and velocity of the incident particle and Z is the atomic number of the target nucleus. Hence the correct choices are (a), (c) and (d). The correct choices are (b) and (c).

1. We have 6.25% =

2. 3. 4. 5.

6.

7. Decay rate at time = =

decay rate at time 2

m = 2 × mass of 21 H – mass of 42 He = 2 × 2.0141 – 4.0024 = 0.0258 u

So the correct choices are (a) and (d). 12. The mass of a nucleus is less than the sum of the masses of its nucleons. In the nucleus 20 10 Ne , there are 10 protons and 10 neutrons. Hence M1 < 10 (mp + mn), which is choice (d). The heavier nucleus has more mass defect than the lighter nucleus, i.e.

0

n

40 20 Ca

> mass defect of 20 10 Ne 20 (mp + mn) – M2 > 10 (mp + mn) – M1 M2 < M1 + 10 (mp + mn) But M1 < 10 (mp + mn) Hence M2 < 2M1, which is choice (c) 13. Energy is released if the total binding energy of the products is greater than the total binding energy of the reactants. This is not possible in choices (a) and (b) The correct choices are (b) and (d)

1600

2n where n = number of half lives in time . Hence all the four choice are correct. 8. Reaction (a) is not possible because the atomic number is not conserved. Reactions (b), (c) and (d) are possible because both atomic and mass numbers are conserved, but reaction (d) is a

III Multiple Choice Questions Based on Passage Questions 1 to 3 are based on the following passage 1. The half life of the sample is Passage I (a) 10 s (b) At a given instant there are 25% undecayed radioactive (c) 20 s (d) nuclei in a sample. After 10 seconds the number of un 2. The mean life of the nuclei is decayed nuclei reduces to 12.5% (a) 6.93 s (b) IIT, 1996 (c) 9.36 s (d)

5s 30 s 7.21 s 14.43 s

29.20 Comprehensive Physics—JEE Advanced

3. The time in which the number of undecayed nuclei will further reduce to 6.25% of the reduced number is

(a) 10 s (c) 30 s

(b) 20 s (d) 40 s

SOLUTION 1. Since the number of undecayed nuclei reduces to half (from 25% to 12.5%) in 10 s, the half life is T = 10 s. 1

2.

T log e(2)

10 = 14.43 s 0.693

Questions 4 to 6 are based on the following passage Passage II In an ore containing uranium, the ratio of U-238 to Pb – 206 is 3. Assume that all the lead present in the uranium – 238 is 4.5 × 109 years. Take loge (2) = 0.693 and 4 = 0.288. loge 3 IIT, 1997 4. The decay constant is (a) 2.22 × 10–9 per year (b) 6.49 × 10–8 per year

3. Let the reduced number further reduce to 6.25% in n half lives, then 6.25 = 100

1 2

n

1 16

1 2

n

n=4

time taken n T = 4 × 10 = 40 s. (c) 3.11 × 10–9 per year (d) 1.54 × 10–10 per year 5. If N0 is the original number of radioactive nuclei and N is the number of nuclei left undecayed, then the ratio N/N0 is 1 2 (a) (b) 3 3 3 4 (c) (d) 4 5 6. The age of the ore is of the order of (b) 1010 years (a) 109 years 11 (d) 1012 years (c) 10 years

SOLUTION 4. Decay constant

=

log e (2) 0.693 = T 4.5 109 years

= 1.54

10

–10

6. If the age of the ore is years, then 3 = e– 4 4 =e 3

per year

5. If x nuclei of U-238 have decayed to Pb–206, then N = N0 – x N0 x N Given =3 x= 0 x 4 N 3N 0 N = N0 – x = N0 – 0 = 4 4 N 3 = N0 4 Questions 7 to 9 are based on the following passage Passage III The element seconds. Its fission and 8% and the

13 Curium 248 96 Cm has a mean life of 10 primary decay modes are spontaneous -decay, the former with a probability of latter with a probability of 92%. Each

1020Cm atoms. The masses involved in -decay are as follows: 244 248 Pu = 244.0641 u and 96 Cm = 248.072220 u, 94

= log e =

0.288

= 1.87

4 3

= 0.288 (given)

=

0.288 1.54 10

10

109 years

4 2 He

2

).

IIT, 1997 7. The energy released in each (a) 0.0514 (b) 0.514 (c) 5.14 (d) 51.4 8. The total energy released in one -decay and one (a) 20.72 (c) 205.86

(b) 207.2 (d) 210.14

Nuclear Physics 29.21

9. The power output from the sample of 1020 atoms is of the order of

(a) 105 W (c) 10–2 W

(b) 10 W (d) 10–5 W

SOLUTION 7. The equation governing the -decay is 218 69 Cm

244 94 Pu

+ 42 He m = m (Pu) + m (He) – m (Cm) = 244.064100 + 4.002603 –248.072220 = – 0.005517u

9. Since there are 1020 atoms of Cm, the total energy released in 1020 reaction is Etotal = 20.725

1020

= 20.725

1026

= 20.725

1026

Energy released per -decay is E = m c 2 = 0.005517

= 3.32 8.

108 J = 1013 s. Therefore, the

Given, mean life of Cm, power output is

E Since the probability of

-decay is 92% and that

transformation is E = 0.92 E + 0.08 E = 0.92 Questions 10 to 12 are based on the following passage Passage IV Nuclei of a radioactive element are being produced at a constant rate . The element has a decay constant . At time = 0, there are N0 nuclei of the element. IIT, 1998 10. If N is the number of nuclei at time , the net decay rate of the element is (a) (b) – N (c) ( – N) (d) ( + N) 11. The number N of nuclei at time is 1 (a) ( N 0 )e

P=

Etotal

10–19 J

1.6

3.32 108 J

=

1013 s

= 3.32 10–5 Js–1 (or W) So the correct choice is (d).

(b) (c) (d) 12. If

1

N 0 )e

(

1 1

(

N 0 )e

(

N 0 )e

= 2 N0 , the number N of nuclei as

N0 4 (c) 2 N0

N0 2 (d) 4 N0

(a)

(b)

SOLUTION 10. The rate of production of radioactive element is –

N

=

N

N0

The rate of decay of element is

1

or –

= N or

=– N

The net rate is =

or

1

– N

11. To calculate the number N of nuclei at time (given that the number at = 0 is N0 ), we integrate the above expression from N = N0 to N = N and from = 0 to = , We have

or log e or

= 0 N N ) N = | |0

log e (

0

N N0

log e N N0

=

=– N=

1

N = e– N0

or [

–(

–

N0)e– ]

is

29.22 Comprehensive Physics—JEE Advanced

12. If

= N0 (2 – e– )

= 2N0 , then we have N=

1

When

N 0)e– ]

[2N0 – (2N0 –

N = N0 (2 – e– ) = N0 (2 – 0) = 2 N0.

Questions 13 to 15 are based on the following passage Passage V A nucleus at rest undergoes a decay emitting an -particle of de Broglie wavelength = 5.76 × 10–15 m. The mass of the daughter nucleus D = 223.610 amu and the mass of an -particle = 4.002 amu. Take 1 amu = 1.656 × 10–27 kg and Planck’s constant = 6.63 × 10–34 Js. IIT, 2001 13. The momentum of the -particle (in kg ms–1) is (a) 1.15 × 10–19 (b) 1.6 × 10–19 –21 (c) 8.69 × 10 (d) 4.002 × 10–27

14. The total kinetic energy of the two particles is of the order of (a) 10–6 J (b) 10–8 J (c) 10–10 J (d) 10–12 J 15. The mass of the parent nucleus is (a) (223.610 + 4.002) amu (b) (223.610 – 0.0067) amu (c) (223.610 + 4.002 + 0.0067) amu (d) (223.610 + 4.002 – 0.0067) amu

SOLUTION = 6.637 10–27 kg mD = 223.610 1.656 = 370.3 10–27 kg

13. The given decay reaction may be represented as A A –4 P D=4 The given de-Brogile wavelength of -particle is = 5.76 10–15 m The momentum of -particle will be p =

=

= 1.151

6.63 10

34

5.76 10

15

Substituting the values in eq. (1) and solving we get E = 1.02 10–12 J 15. The mass equivalent of energy E is m =

10–19 kg m/s

14. Due to the conservation of linear momentum, we will also have pD = 1.151 10–19 kg m/s

c2

=

1.02 10 3 108

=

12 2

10–29 kg

1.113 10

29

1.656 10

27

= 0.0067 a mu

E= Now

E

= 1.113

The total kinetic energy of these two particles is p2 pD2 2m 2mD m = 4.002 a mu = 4.002 1.656

10–27

(1) m = mD + m + m = (223.610 + 4.002 + 0.0067) amu

10–27

Questions 16 to 18 are based on the following passage Passage VI A radioactive element decays by beta emission. A detector records n n beta particles in the nest 2 seconds. IIT, 2003 16. The decay constant of the element is 1 4 1 3 (b) = ln (a) = ln 2 3 2 4

1 3 (d) = ln(2) ln(2) 2 4 17. The mean life of the element is (given ln (2) = 0.693 and ln (3) = 1.1) (a) 5 s (b) 6 s (c) 7 s (d) 8 s 18. The half life of the element is (a) 4.8 s (b) 5.8 s (c) 6.8 s (d) 7.8 s (c)

=

Nuclear Physics 29.23

SOLUTION 16. N = N0 e –

17.

–4

n = N0 – N2 = N0 (1 – e–2 )

Given and

Therefore N2 = N0 e–2 and N4 = N0 e

(1)

0.75 n = N2 – N4 = N0 e–2 – N0 e–4

=

1 1 (ln 4 – ln 3) = (2 ln 2 – ln 3) 2 2 1 = [2 0.693 – 1.1] 2 = 0.143 s–1

0.75 n = N0 (e–2 – e–4 ) 0.75 n = N0 e–2 (1 – e–2 )

1

(2)

=

1 =7s 0.143

From Eqs. (1) and (2) we get 0.75 = e–2 =

e

2

=

4 3

18. Half life =

ln(2)

=

0.693 = 4.8 s 0.143

1 4 ln 2 3

Questions 19 to 21 are based on the following passage Passage VII Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, 21H, known as deuteron and denoted by D, can be thought of as a candidate for 3 fusion reactor. The D-D reaction is 21H + 21H 2He + n + energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of 21H nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place. Usually, the temperatures in the reactor core

plasma for a time 0 core. If n is the density (number/volume) of deuterons, the product 0 is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than 5 1014 s/cm3. It may be helpful to use the following: Boltzmann e2 = 1.44 10–9 constant k =8.6 10–5 4 0 IIT, 2009 19. In the core of nuclear fusion reactor, the gas becomes plasma because of (a) strong nuclear force acting between the deuterons (b) Coulomb force acting between the deuterons

(c) Coulomb force acting between deuteron-electron pairs (d) the high temperature maintained inside the reactor core 20. Assume that two deuteron nuclei in the core of fusion reactor at temperature T are moving towards each other, each with kinetic energy 1.5 kT, when the separation between them is large enough to neglect Coulomb potential energy. Also neglect any interaction from other particles in the core. The minimum temperature T required for them to reach a separation of 4 10–15 m is in the range (a) 1.0 109 K < T < 2.0 109 K (b) 2.0 109 K < T < 3.0 109 K (c) 3.0 109 K < T < 4.0 109 K (d) 4.0 109 K < T < 5.0 109 K 21. Results of calculations for four different designs of a fusion reactor using D-D reaction are given below. Which of these is most promising based on Lawson criterion? (a) deuteron density = 2.0 1012 cm–3 ment time = 5.0 10–3 s (b) deuteron density = 8.0 1014 cm–3 ment time = 9.0 10–1 s (c) deuteron density = 4.0 1023 cm–3 ment time = 1.0 10–11 s (d) deuteron density = 1.0 1024 cm–3 ment time = 4.0 10–12 s

SOLUTION 19. The correct choice is (d). 20. At a large separation, potential energy is zero and total energy = 1.5 kT + 1.5 kT = 3 kT.

From conservation of energy, 3 kT

=–

U=

e2 4

0r

29.24 Comprehensive Physics—JEE Advanced

T=

=

e2 4

0

21. According to Lawson, deuteron density (n) and 0 must satisfy the criterion

1 3kr 1.44 10

n

9

1014 cm–3 s

>5

0

3 (8.6 10 5 ) (4 10 5 )

= 1.4

which 0 = (8.0 1014) (9.0 10–1) = 7.2 cm–3 s. Hence the correct choice is (b).

109 K

So the correct choice is (a).

1014

IV Matching 1. released. Column I

Column II

(a) Alpha decay

(p)

235 92 U

(b) Beta decay

(q)

3 1H

1 0n 2 1H

4 2 He

230 9 0 Th

(d) Nuclear fusion

(s)

141 56 Ba

226 9 0 Ra

137 55 Cs

137 56 Ba

92 36 Kr

3 10 n

Q

Q 4 2 He

e

Q Q

ANSWER (a) (c)

(r) (p)

(b) (d)

(s) (q)

Column I

Column II

2.

(b) Nuclear fusion (c) -decay (d) Exothermic nuclear reaction

(q) involves conversion of matter into energy (r) atoms of higher atomic number are used (s) atoms of lower atomic number are used IIT, 2006

ANSWERS (a) (c)

(q), (r) (p)

(b) (d)

(q), (s) (q)

V Assertion-Reason Type Questions In the following questions, Statement-1 (Assertion) is followed by Statement-2 (Reason). Each question has the following four choices out of which only one choice is correct.

(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.

Nuclear Physics 29.25

(c) Statement-1 is True, Statement-2 is False. (d) Statement-1 is False, Statement-2 is True. 1. Statement-1 A nucleus at rest splits into two nuclear parts having radii in the ratio 1 : 2. Their velocities will be in the ratio 8 : 1. Statement-2 The radius of a nucleus is proportional to the cube root of its mass number. 2. Statement-1 The half life of radioactive sample is T. It will 1 of its initial value in a time 8T. decay to 16 Statement-2 The half life of a radioactive sample is the time in which half of the number of nuclei decay. 3. Statement-1 A freshly prepared radioactive sample has a half life of 3 hours and emits radiation of intensity which is 64 times the permissible safe value. The minimum time after which it would be safe to work with the sample is 18 hours. Statement-2 The intensity of the radiation falls by a factor of 2 every 3 hours. 4. Statement-1 Two radioactive sources A and B initially contain equal number of radioactive nuclei. Source A has a half life of 1 hour and source B has a half life of 2 hours. At the end of 2 hours, they will have the same rate of disintegration. Statement-2 ber of disintegrations taking place in the source per second. 5. Statement-1 The nucleus 22Ne absorbs energy and decays into two alpha particles and an unknown nucleus. The unknown nucleus must be carbon. Statement-2 In a nuclear reaction, the atomic number is conserved. 6. Statement-1 The number of and particles emitted when 238 92U Pb is 6 and 8 respectively. decays into 206 82

Statement-2 In a nuclear reaction, the mass number and the atomic number are both conserved. 7. Statement-1 The radioactive decay of nucleus X to nuclei Y and K is represented by the equation A Z

X

A Z+1

Y

A 4 z 1 K

The sequence of emitted radiations is ,

and .

Statement-2 In a nuclear reaction, the mass number and the atomic number are both conserved. 8. Statement-1 The distance of the closest approach of an alpha p is r0. The distance of the closest approach when the momentum 2p will be r0/2. Statement-2 The distance of closest approach from a given target element is inversely proportional to the kinetic energy of the incident particle. 9. Statement-1 The binding energy of deuteron (21 per nucleon and an alpha particle (24He) has a bindreaction 2 1H

+ 12 H

4 2 He

+Q

Statement-2 Total energy is conserved in a nuclear reaction. 10. Statement-1 A nucleus X, initially at rest, decays into a nucleus Y with the emission of an -particle and energy Q is released. If m is the mass of an alpha particle and M that of nucleus Y, the energy of the emitted -particle will be E =

QM ( M m)

Statement-2 process.

SOLUTIONS 1. The correct choice is (a). Let A1 and A2 be the mass numbers of the two nuclear parts. Their radii are given by R1 = R0 (A1)1/3 and R2 = R0 (A2)1/3

A 4 z 1K

Dividing, we get R1 = R2

A1 A2

1/ 3

or

29.26 Comprehensive Physics—JEE Advanced

A1 = A2

R1 R2

3

1 2

3

1 8

Hence the ratio of their masses is m1 1 = m2 8 From the principle of conservation of momentum, the magnitude of p1 = magnitude of p2 or m1 v1 = m2 v2, which gives v1 m = 2 v2 m1

8 1

unity; mass number remaining the same. Hence an electron ( -particle) is emitted. In transition A A– 4 Z + 1Y Z – 1 K, the mass number decreases by 4 and charge number decreases by 2. Hence an -particle is emitted. In the third transition, mass and charge numbers do not change. Hence a -ray is emitted. 8. The correct choice is (d). The distance of the closest approach is given by r0 = where K =

1 1 = 4 , it 16 2 follows that the time taken for the sample to decay to 1/16 of its initial value = four half lives = 4 T. 3. The correct choice is (a). Since the half life is 3 hours, the intensity of radiation falls by a factor of 2 every three hours. In 18 hours, it will fall by a factor of (2)6 = 64. 4. The correct choice is (a). The rate of disintegration is proportional to the number of radioactive nuclei present initially in the source. 5. The correct choice is (a). The given nuclear reaction is given by the equation 2. The correct choice is (d). Since

+ 24 He + ZX Since the atomic number is conserved, we have 10 = 2 + 2 + Z which gives Z = 6. The nucleus having Z = 6 is carbon. 6. The correct choice is (d). Let x and respectively be the number of and particles emitted. The equation of the decay is 22 10 Ne

238 92 U

4 2 He

206 82 Pb

+ x( 42He) + ( –10 e)

From conservation of charge, we have 92 = 82 + 2x – Conservation of mass number gives 238 = 206 + 4x

(i)

K=

4

0

1 m v 2. 2

1 1 m v2 = 2 2m r0 =

Thus r0

1 p2

mv

4

2

=

p2 . Therefore, 2m

4 m Z e2

1

p2

0

. When p is doubled, r0 becomes one-

fourth 9. The correct choice is (a). Binding energy of 21H = 1.15 number of nucelons = 1.15 Total binding energy of reactants = 2.3 + 2.3 = 4 num2 He = 7.1 ber of nucleons = 7.1 Q choice is (c). 10. The correct choice is (d). K = p2/2m. Therefore, P2 p2 (P = from conservation of energy, Q = 2 M 2m momentum of Y, p = momentum of -particle) Since momentum is conserved, P = p. Hence Q=

p2 1 2 M

1 m

=

p2 m 2m M

where E = energy of E =

QM ( M m)

1

m 1 M -particle. Hence =E

(ii)

which gives 4x = 32 or x = 8. Using this value of x in (i) we get = 6. 7. The correct choice is (b). In transition AZ X A Z + 1Y, the atomic (or charge) number increases by

2 Z e2 K

1

Nuclear Physics 29.27

VI Integer Answer Type Questions 1. To determine the half life of radioactive element, a ()

student plots a graph of ln

versus .

Here

()

is the rate of radioactive decay at time

. If the number of radioactive nuclei of this element decreases by a factor of p the value of p. IIT, 2010 2. The activity of a freshly prepared radioactive sample is 1010 disintegrations per second, whose mean life is 109 s. The mass of an atom of this radioisotope is 10–25 kg. The mass (in mg) of the radioactive sample is

SOLUTIONS 1. N = N0 e– = ln

=–

N0 e– . Therefore

N0 e– = ln ( N0) –

The slope of the graph is = 1 = per year 2

4 3 4 6

1 2

0.693 = 1.386 year 1/ 2 4.16 =3 Number of half lives = 1.386 3 Hence p = (2) = 8. N ; = mean life 2. Activity |R| = N = Half life T1/2 =

N=

|R| = 109

Total mass = 1019

1010 = 1019 atoms

10–25 = 10–6 kg = 1 mg.

MODEL TEST PAPER—I

The questions in the practice papers are based on questions asked in previous years’ Physics Question Papers of IIT-JEE. Answer key and complete solutions of questions are provided at the end of each practice paper. Each paper contains 30 questions to be answered in 45 minutes.

SECTION I (Single Correct Answer Type) This section contains 13 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONLY ONE is correct. 1. Students I, II and III perform an experiment for measuring the acceleration due to gravity (g) using a simple pendulum. They use different lengths of the pendulum and/or record time for different number of oscillations. The observations are shown in the table. Least count for length = 0.1 cm Least count for time = 0.1 s

I

64.0

8

128.0

16.0

II

64.0

4

64.0

16.0

III

20.0

6

36.0

6.0

If EI, EII and EIII are the percentage errors in g. i.e. Ê Dg ˆ ÁË g ¥ 100˜¯ for student I, II and III, respectively, (a) EI = 0 (c) EI = EII

(b) EI is minimum (d) EII is minimum

2. A block of mass m is held stationary against a wall by applying a horizontal force F on the block. Which of the following statements is false? (a) The frictional force acting on the block is f = mg (b) The normal reaction force acting on the block is N = F (c) No net torque acts on the block (d) N does not produce any torque. Wall Block

F

O

M has outer radius 4R and inner radius 3R. The work required to take a unit mass from point P P 4R 3R

Time Total time for (n) oscillations period (s) (s)

Number of oscillations (n)

Student Length of the pendulum (cm)

3.

4R

(a)

2GM (4 2 - 5) 7R

(b) -

(c)

GM 4R

(d)

2GM (4 2 - 5) 7R

2GM ( 2 - 1) 5R

4. A block A of mass m is placed on a frictionless horizontal surface. Another block B of the same mass is kept on A and connected to the wall with the help of a spring of force constant k, as shown blocks A and B is m. The blocks move together executing simple harmonic motion of amplitude a. The maximum value of frictional force between A and B is k B

m

A

(a) ka (b) ka/2 (c) zero (d) m mg 5. A bi-convex lens is formed with two thin planoindex n second lens is 1.2. Both the curved surface are of the same radius of curvature R = 14 cm. For this

MTPI.2 Comprehensive Physics—JEE Advanced

bi-convex lens, for an object distance of 40 cm, the image distance will be n = 1.5

n = 1.2

(c) 44 s (d) 88 s 10. A proton moving with a speed u along the positive x-axis enters at y = 0 a region of uniform magnetic B = B0 k which exists to the right of y-axis as

R = 14 cm

(a) – 280.0 cm

9. An RC circuit consists of a resistance R W. and a capacitance C = 1.0 mF connected in series with a battery. In how much time will the potential difference across the capacitor become 8 times that across the resistor? (Given loge (3) = 1.1)

(b) 40.0 cm

6. Three very large plates of same area are kept parallel and close to each other. They are considered as ideal black surfaces and have very high

after some time with a speed v at co-ordinate y. Then y B (out of page)

maintained at temperatures 2T and 3T respectively. The temperature of the middle (i.e. second) plate under steady state condition is 65 (a) ÊÁ ˆ˜ Ë 2¯

1/ 4

97 (c) Ê ˆ Ë 2¯

1/ 4

1/ 4

T

97 (b) ÊÁ ˆ˜ Ë 4¯

T

(d) (97)1/4 T

T

7. In the given circuit, a charge of + 80 mC is given to the upper plate of the 4 mF capacitor. Then in the steady state, the charge on the upper plate of the 3 mF capacitor is +80 mC 4 mF

2 mF

3 mF

(a) + 32 mC

(b) + 40 mC

(c) + 48 mC

(d) + 80 mC

of the water level in the column is (c) 16.4 cm

(d) 17.6 cm

x

O

(a) v > u, y < 0 (b) v = u, y > 0 (c) v > u, y > 0 (d) v = u, y < 0 11. A cylindrical conducting rod is kept with its axis along the x-axis. Also there exists a uniform x-axis. The current induced in the cylinder is (a) clockwise as seen from the + x axis (b) anticlockwise as seen from the + x axis (c) along the axis towards – x direction (d) zero 12. n0 n0 n , and 0 , 2 6 8 respectively. The angle of incidence q for which the beam just misses entering region IV is n 0,

Region I

8. A student is performing the experiment of resonance column. The diameter of the column tube is The air temperature is 38°C in which the speed of sound is 336 m/s. The zero of the meter scale co-

u

+q

q

Region II

Region III

n0 2

n0 6

Region IV

n0 8

n0 0

0.2 m

0.6 m

-1 Ê 3 ˆ (a) sin Á ˜ Ë 4¯

-1 Ê 1 ˆ (b) sin ÁË ˜¯ 8

-1 Ê 1 ˆ (c) sin Á ˜ Ë 4¯

-1 Ê 1 ˆ (d) sin ÁË ˜¯ 3

MTPI.3

13. Electrons with de-Broglie wavelength l fall on the target in an X-ray tube. The cut-off wavelength of the emitted X-ray is

(a) l0 =

2mc l 2 h

(b) l0 =

(c) l0 =

2m 2 c 2 l 2 h2

(d) l0 = l

2h mc

SECTION II (Multiple Correct Answer Type) This section contains 7 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONE or MORE than one choice/choices is/are correct. 14.

inclined plane PQ which makes an angle q with the horizontal. A horizontal force of 1 N acts of the block through its centre of mass as shown in

16. A cubical region of side a has its centre at the oriq at (0, – a/4, 0), + 3q at (0, 0, 0) and – q at (0, + a/4, 0). Choose the correct options(s). z

g = 10 m/s2). Q a

1N

-q

-q

y

3q

q P

O

x

(a) q (b) q towards P. (c) q towards Q. (d) q towards Q. 15.

is

2W

S 2W

2W 1W

4W

I1

4W

q . e0 z =

+a the plane x = + a/2.

17. A disc of mass M and radius R is rolling with angular speed w on a horizontal surface as shown in

1W

4W Q

y = + a/2

plane y = – a/2.

choose the correct option(s). P I2

x =

+a the plane x = – a/2

T

12 V

(a) The current through PQ is zero. (b) I1 = 3A (c) The potential at S is less than that at Q. (d) I2 = 2A

the disc about the origin O is (here v is the linear velocity of the disc) y w O

v R

O

x

MTPI.4 Comprehensive Physics—JEE Advanced

(a)

3 MR 2w 2

(c) MRv

(c) y¢min (d) y¢min

(b) MR2w (d)

3 MRv 2

off 1 voltage V0 and for three metals 1, 2 and 3, where l l is the wavelength of the incident radiation in nm.

20.

18. The displacement x of a particle varies with time t as x – A sin2 wt + B cos2 wt + C sin wt cos wt. For what values of A, B and C is the motion simple harmonic? (a) All values of A, B and C with C π 0. (b) A = B, C = 2B (c) A = – B, C = 2B (d) A = B, C = 0

Metal 1 Metal 2

Metal 3

V0

19. A beam of light consisting of two wavelengths

0.001

0.002

0.004

1 (nm-1) l

fringes in a Young’s double slit experiment. The separation between the slits is 1 mm and the distance between the plane of the slits and the screen is 100 cm. The least distance from the central maximum where the bright fringes due to both the wavelengths coincide is ymin and y¢min is the corresponding distance where the dark fringes due to both the wavelengths coincide. Then

If W1, W2 and W3 are the work functions of metals 1, 2 and 3 respectively, then (a) W1 : W2 : W3 = 1 : 2 : 4 (b) W1 : W2 : W3 = 4 : 2 : 1 (c) The graphs for metals 1, 2 and 3 are parallel to each other and the slope of each graph is hc/e, where h = Planck’s contant, c = speed of light and e = charge of an electron. (d) Ultraviolet light will eject photoelectrons from metals 1 and 2 and not from metal 3.

(a) ymin (b) ymin = 2.0 mm

SECTION III (Linked Comprehension Type) This section contains 4 questions based on a paragraph. Each question has four choices (a), (b), (c) and (d) out of which only ONE choice is correct.

Questions 21 and 22 are based on the following paragraph.

Questions 23 and 24 are based on the following passage.

A light rod of length L having a body of mass M attached to its end hangs vertically. It is turned through 90° so that it is horizontal and then released.

A point particle of mass M is attached to one end of a massless rigid non-conducting rod of length L. Another point particle of the same mass is attached to the other end of the rod. The two particles carry charges + q and – q. This

21. The centripetal acceleration when the rod makes an angle q with the vertical is (a) g cos q

(b) 2g cos q

(c) g sin q

(d) 2g sin q

E such that the rod makes a small angle q A +q

22. The tension in the rod when it makes an angle q with the vertical is (a) Mg cos q

(b) 2 Mg cos q

(c) 3 Mg cos q

(d) zero

q O -q B

E

MTPI.5

23. When the rod is released, it will rotate with an angular frequency w equal to qE ˆ 1 / 2 (a) ÊÁ Ë ML ˜¯

2qE ˆ (b) ÊÁ Ë ML ˜¯

qE ˆ 1 / 2 (c) ÊÁ Ë 2ML ˜¯

(d)

24. The minimum time taken by the rod to align itself given by

1/ 2

(a)

1 Ê qE ˆ 1 / 2 ˜ Á 2 Ë ML ¯

p Ê ML ˆ 2 ÁË 2qE ˜¯

1/ 2

Ê 2 ML ˆ (c) 2p Á Ë qE ˜¯

Ê ML ˆ (b) 2p Á Ë qE ˜¯

1/ 2

1/ 2

Ê ML ˆ (d) 2p Á Ë 2qE ˜¯

1/ 2

SECTION IV (Assertion-Reason Type) This section contains 2 questions. In each question, Statement-1 is followed by Statement-2. Each question has the following four choices (a), (b), (c) and (d) out of which only ONE choice is correct.

(a) Statement-1 is true, Statement-2 is true and Statement-2 is the correct explanation for Statement-1.

PV T

(b) Statement-1 is true, Statement-2 is true but Statement-2 is not the correct explanation for Statement-1.

B T1 T2

(c) Statement-1 is true, Statement-2 is false.

P

(d) Statement-1 is false, Statement-1 is true. 25. Statement-1 PV versus P graph for T a certain mass of oxygen gas at two temperatures T1 and T2. It follows from the graph that T1 > T2.

Statement-2 At higher temperature, real gas behaves more like an ideal gas. 26. Statement-1 A particle of mass M at rest decays into two particles of masses m1 and m2 which move with velocities v1 and v2 respectively. Their respective de Broglie wavelengths are l1 and l2. If m1 > m2, then l 1 > l 2. Statement-2 The de Broglie wavelength of a particle having momentum p is l = h/p.

SECTION V (Integer Answer Type) This section contains 3 questions. The answer to each question is single digit integer, ranging from 0 to 9 (both inclusive) 27. A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a uniform current density J P is given by

N m0 aJ , 12

a P

O

N.

2a

MTPI.6 Comprehensive Physics—JEE Advanced

28. A binary star consists of two stars A (mass 2.2 Ms) and B (mass 11 Ms), where Ms is the mass of the sun. They are separated by distance d and are rotating about their centre of mass, which is stationary. Find the ratio of the total angular momentum of the binary star to the angular momentum of star B about the centre of mass.

29. Two spherical bodies A (radius 6 cm) and B (radius 18 cm) are at temperature T1 and T2, respectively. The maximum intensity in the emission spectrum of A B Considering them to be black bodies, what will be the ratio of the rate of total energy radiated by A to that of B?

SECTION VI (Matrix Match Type) This section contains 1 question. Each question has four statements (a, b, c and d) given in Column I and (p, q, r, s and t) in Column II. Any given statement in Column I can have correct matching with ONE or MORE statement(s) given in Column II. For example, if for a given question, statement B matches with the statements given

30. One mole of a monatomic gas is taken through a cycle ABCDA as shown in the P-V diagram. Column II gives the characteristics involved in the

25. (a)

26. (d) Section-V

Column I. P 3P

B

28. (6)

27.

A

29. (9)

Section-VI

1P

C 0

1V

30. (a) Æ (p, r, t); (b) Æ (p, r), (c) Æ (q, s), (d) Æ (r, t).

D 3V

9V

Column I

Solutions

V

Column II

(a)

Process A Æ B

(p)

Internal energy decreases

(b)

Process B Æ C

(q)

Internal energy increases

(c)

Process C Æ D

(r)

(d)

Process D Æ A

(s) (t)

Work is done on the gas

Answers Section-I 1. 4. 7. 10. 13.

Section-IV

(b) (b) (c) (d) (a)

14. (a, c) 17. (a, d) 20. (a, d)

2. 5. 8. 11.

(d) (b) (b) (d)

3. 6. 9. 12.

(b) (c) (b) (b)

Section-II 15. (a, b, c, d) 16. (a, c, d) 18. (a, b, c) 19. (a, d)

Section-I 1. T = 2p

L g

t , when n = number of oscillan tion and t is the total time for n oscillation. In terms of measured quantities,

Time period T =

t L 4p 2 L n 2 . Therefore, = 2p fig= n g t2 Dg DL 2 Dt (∵ D n = 0; there is no error = + g L t in counting the number of oscillation) For student I,

EI =

Section-III 21. (b) 24. (a)

22. (c)

23. (b)

=

0.1 2 ¥ 0.1ˆ Dg ¥ 100 = ÊÁ + ˜ ¥ 100 Ë 64 128 ¯ g 5 % 16

MTPI.7

Putting r2 + 16 R2 = x2, we get 2r dr = 2x dx or rdr = x dx.

Ê 0.1 + 2 ¥ 0.1ˆ ¥ 100 EII = ÁË ˜ 64 64.0 ¯

For student II,

When r

15 = % 32

When r = 4 R, x =

0.1 2 ¥ 0.1ˆ + For student III, EIII = ÊÁ ˜ ¥ 100 Ë 20.0 36 ¯ =

VP = -

\

19 % 18

Thus the percentage error in the measurement of g is minimum for student I. 2. Since the block is held stationary, it is in transla-

mass from P Vp, where VP is the gravitational potential at P VP, we divide the disc into small elements, each of thickness dr. Consider one such element at a distance r from the centre of the disc, as shown

2MG 7R2

VP = -

16 R 2 + 16 R 2 = 4 2 R

4 2R

Ú

dx = -

5R

2 MG (4 2 - 5) R 7R2

2MG (4 2 - 5), which is choice (b). 7R

4. The blocks will move together as long as the frictional force of block B = mass of block B ¥

net force and no net torque acts on the block. No net force will act on the block if f = mg and N = F. No net torque will act on the block, if torque by frictional force f about centre O = counter torque by normal reaction N about centre O choice (d) is false.

f = mw2a where Thus

w=

k = ( m + m)

f= m¥

k 2m

k ¥a 2m

= ka/2 1 Ê 1.5 - 1ˆ =Á ˜ f1 Ë 1 ¯

5.

Ê 1 - 1 ˆ = 1 cm -1 ˜ ÁË 14 • ¯ 28

1 Ê 1.2 - 1ˆ Ê 1 1 ˆ 1 = =Á cm -1 ˜¯ Á Ë Ë • - 14 ˜¯ 70 f2 1

P

2

+1 6R

2

1 1 1 1 1 = + = + fi F = 20 cm F f1 f 2 28 70

r

3.

9 R 2 + 16 R 2 = 5 R

x=

dr

Now

r

fi M (2p rdr ) p (4 R)2 - p (3R) 2

dm = = 4R

VP = -

Ú

3R

2M rdr 7R2

G dm r 2 + 16 R 2

2 MG = 7R2

4R

Ú

3R

rdr 2 (r + 16 R 2 )1 / 2

1 1 1 - = v u F 1 1 1 = fi v = 40 cm v - 40 20

6. In the steady state, the rate at which the middle plate receives heat energy is equal to the rate at which heat energy is emitted by the other plates. Let A be the area of each plate and T0 be the steady state temperature of the middle plate. Since both sides of the middle plate receive heat energy, the total area of the middle plate receiving energy is 2A. 2T

To

3T

MTPI.8 Comprehensive Physics—JEE Advanced

From Stefan’s law 4

4

s (2A) (To) = sA (2T ) + sA (3T )

4

10. When the proton enters the region of the magnetic F given by F = q (u ¥ B)

2 To4 = 16 T 4 + 81 T 4 = 97 T 4

fi

97 To = ÊÁ ˆ˜ Ë 2¯

fi

1/ 4

where q is the charge of the proton. The force F is perpendicular to both u and B. Since the force is perpendicular to the velocity of the particle, it

T

7. Let q mC be the charge on the upper plate of 3 mF capacitor. Then the charge on the upper plate of 2 mF capacitor will be (80 – q) mC. Since potential difference across 2 mF capacitor = potential difference across 3 mF capacitors,

the velocity of the particle will remain unchanged;

q 80 - q = fi q = 48 mC. 3 2 8. End correction e = 0.3d = 0.3 ¥ 4 = 1.2 cm v 336 = Wavelength l = n 512 Now fi

L+e=

l 4

l 65.6 L= –e= –1.2 4 4

11.

v = u. Since u is perpendicular to B, the proton moves in a circular path. Since the charge of proton is positive, u is along positive x-axis and B is directed out of the page, the proton will move in a circle in the x-y plane in the clockwise direction. y coordinate will be negative, when it leaves the region. Thus the correct choice is (d). emf and current are zero. So the correct choice is (d).

12. The beam will not enter region IV if the angle refraction in region IV equals 90°. Apply Snell’s law at the interfaces, we have I

II

III

IV q3 = 90°

9. At instant of time t, the charge on the capacitor is given by

q2

q = q0 (1 – e–t/RC) and the potential drop across the capacitor is given by (∵ V = q/C)

q2

q1 q1

VC = V0 (1 – e–t/RC) where V0 is the voltage of the battery. The potential drop across the resistor is

q

VR = V0 – VC = V0 – V0 (1 – e–t/RC) = V0 e–t/RC \ Given

VC 1 - e-t / RC = et / RC - 1 = VR e-t / RC VC = 8. Therefore, VR

=

8 = et/RC – 1 or

et/RC = 9 = (3)2

or

t = 2 loge (3) RC t = RC ¥ 2 loge (3) 6

n0 sin q1 2 n0 n sin q 2 = 0 sin 90∞ 6 8

which gives sin q =

1 . 8

13. de-Broglie wavelength is l= –6

¥ 10 ) ¥ (1 ¥ 10 ) ¥ 2 ¥ 1.1 = 11 s

n0 sin q =

h 2mE

where E is the kinetic energy of the electrons. The out-off wavelength is

MTPI.9

l0 =

2a + c – 4b = 0

hc E

2(a – c) – c – 4 (b + c) – c = 0 h2 . 2 ml 2

From Eq. (1) E =

2mc l 2 l0 = h

a – 2b – 4c = 0

and

4b + 4 (b + c) + 4b – 12 = 0

fi

3b + c = 3

Also VS – VQ = – c – 4 (b + c) = – 4b = – 4 ¥ 1 = – 4V (∵ c S is less than that at Q.

q

16.

F q

the cubical region is

mg sin q

f=

mg q

O

P

Given F = 1 N, m = 0.1 kg and g = 10 ms–2. Let f be the frictional force between the block and the plane surface PQ. The block will be stationary if F cos q = mg sin q fi 1 ¥ cos q = 0.1 ¥ 10 ¥ sin q fi tan q = 1 fi q f=0 If q q > cos q. mg sin q > F cos q (∵ F = mg = 1 N). Therefore frictional force acts up the block towards Q. If q q < cos q mg sin q < F cos q. Therefore, in this case, frictional force f acts down in the plane towards P.

P 2W a

(a - c) 2 W

S

a 2W

c 1W

A

(3)

Thus the current through PQ is zero. Also I1 = a + b = 3 A and I2 = a – c = 2 A.

F cos q

15.

(2)

a = 2 A, b = 1 A and c = 0

Q

mg cos q

fi

Solving Eqs. (1), (2) and (3), we get

Section II 14.

(1)

1W

qnet 3q - q - q q = = e0 e0 e0

x = a/2 and the x = – a/2 is the same. Further, the positions of charges with respect to x = a/2 and z = a through the planes x = a/2 and z = a/2 is the same y = a/2 and y = – a/2 is the same. 17. The angular momentum about O is     LO = LCM + M ( R ¥ v)   Its magnitude is (∵ R ^ v) and L = Iw LO = Iw + MRv Ê1 2ˆ = ÁË MR ˜¯ w + MR ¥ Rw (∵v = Rw ) 2 =

3 MR 2w 2

=

v 3 3 MR 2 ¥ ÊÁ ˆ˜ = MRv Ë R¯ 2 2

B

b 4W Q (b + c) 4 W

I1

V

4W

c

12 V

b T (a + b)

U

Applying Kirchhoff’s loop rule to loops APQA, PSTQP and AQTBUVA, we get

18. The displacement equation can be rewritten as A B C x = (1 - cos 2w t ) + (1 + cos 2w t ) + sin 2w t 2 2 2 or 1 C 1 x = ( A + B) + ( B - A) cos 2w t + sin 2w t (1) 2 2 2 Choice (a): Equation (1) can be written as x = x0 + a cos 2wt + b sin 2wt

(2)

MTPI.10 Comprehensive Physics—JEE Advanced

where

C . 2 Equation (2) can be recast as x = x0 + A0 sin(2w + f) and

For dark fringes to coincide, the condition is

1 1 ( A + B), a = ( B - A) 2 2

x0 =

Ê y¢ = ÁË n -

b=

2

(3)

2 1/2

where A0 = (a + b ) and tan f = a/b. Equation (3) represents a simple harmonic motion of angular frequency 2w, amplitude = x0 + A0 and phase constant f. Choice (b): For A = B and C = 2B, Eq. (1) becomes x = B + B sin2 wt = B(1 + sin2 wt)

which represents a simple harmonic motion of amplitude 2 B, angular frequency 2w and phase constant p/4. Choice (d): For A = B and C = 0, Eq. (1) reduces to x= A which does not represent simple harmonic motion. 19. Let nth bright fringe of wavelength ln and the mth bright fringe of wavelength lm coincide at a distance y from the centre of the screen. Then y=

n ln D m lm D = d d

or

n ln = m lm

or

m ln = n lm

or

750 m = 450 n

1ˆ 750 2¯ fi 5n = 3m + 1 fi = 450 Ê n - 1ˆ Ë 2¯ The minimum integral values which satisfy this condition are n = 2 and m Ê y¢min = ÁË n -

1 ˆ ln D ˜ 2¯ d

Ê 2 - 1 ˆ ¥ 750 ¥ 10-9 ¥ 1 Ë 2¯ = 103 ¥ 10–3 The correct choices are (a) and (d). 20. Work function W = hn0 =

hc , where l0 is the l0

W1 : W2 : W3 =

hc hc hc : : (l0 )1 (l0 )2 (l0 )3

=

1 1 1 : : (l0 )1 (l0 ) 2 (l0 )3

= 0.001 : 0.002 : 0.004 =1:2:4 sion, the relation between V0 and l is given by eV0 = hn - W = or

V0 =

hc Ê 1 ˆ W Á ˜e Ë l¯ e

Therefore, the minimum value of y is n ln D 3 ¥ 750 ¥ 10-9 ¥ 1 = d 10-3 ¥ 10–3 m

1 l

hc which is the same for all metals. Therefore, e choice (c) is correct. The threshold wavelength for the three metals are 1 = 0.001 nm–1, therefore (l0)l = 1000 nm (l0 )1 = 10,000 Å 1 = 0.002 nm–1, therefore (l0)2 ( l0 ) 2 is

m 5 = . The minimum integral values of m and n 3 n that satisfy this equation are m n = 3.

-

hc -W l

V0 and

or

ymin =

1 ˆ lm D ˜ 2¯ d

Êm Ë

This equation represents a simple harmonic motion of amplitude 2B and angular frequency 2w. Choice (c): For A = – B and C = 2B, Eq. (1) becomes x = B cos2 wt + B sin2 wt

1 ˆ ln D Ê = Ám ˜ Ë 2¯ d

MTPI.11

1 = 0.004 nm–1, therefore (l0 ( l 0 )3 For photoelectric emission, the wavelength of the incident radiation must be less than the threshold wavelength. Since the wavelength of ultraviolet light is about 1200 Å, it will eject photoelecchoices are (a) and (d). Section III L O T

Mg cos q

q

L cos q C

B

q q

qE

-q

q

Since q is small, sin q  q, where q is expressed in radian. Thus t = qELq t = – qELq

experiences a torque which tends to align it with F = qE each acting at A and B constitute a couple whose torque is given by t = force ¥ perpendicular distance = F ¥ AC = F ¥ AB sin q = qEL sin q

(1)

If a is the angular acceleration of the rotatory motion, t = Ia where I is the moment of inertia of the two masses at A and B about an axis passing through the centre O and perpendicular to the rod. Since the rod is massless, I = M ¥ (AO)2 + M ¥ (BO)2

The loss of PE when the body falls from A to B = Mg ¥ OC = MgL cos q. If v is the velocity of the body at B, then

23. A non-conducting rigid rod having equal and opposite charges at the ends is an electric dipole.

E

B

Mg

1 M v 2 = MgL cos q or v2 = 2gL cos q (1) 2 v 2 2 gL cos q = centripetal acceleration = L L = 2g cos q, which is choice (b). 22. The centripetal force when the body is at B is M v2 Fc = L Thus, we have M v2 (2) T – Mg cos q = L Using (1) in (2), we get M ¥ 2 gL cos q = 2 Mg cos q T – Mg cos q = L or T = 3 Mg cos q Thus the correct choice is (c).

qE

A

O

\

21. A

So the correct choice is (a).

L 2 L 2 ML2 = M ¥ ÊÁ ˆ˜ + M ¥ ÊÁ ˆ˜ = Ë 2¯ Ë 2¯ 2 Thus

t=

ML2a 2

(2)

Using Eq. (2) in Eq. (1), we get Ê 2qE ˆ q = - w 2q a = - ÁË ˜ ML ¯ where

2qE ˆ w= Ê Ë ML ¯

1/ 2

, which is choice (b).

24. The time period of oscillation is T=

2p Ê ML ˆ = 2p Á Ë 2qE ˜¯ w

1/ 2

time taken by the rod to align itself parallel to one-fourth of angular oscillation, i.e. tmin =

T p Ê ML ˆ = 4 2 ÁË 2qE ˜¯

1/ 2

So the correct choice is (a). Section IV 25. The correct choice is (a). The line AB is parallel to the P-axis. This means that PV/T is a constant, AB corresponds to an ideal gas for which PV/T = constant.

MTPI.12 Comprehensive Physics—JEE Advanced

At higher temperatures, a real gas behave more T1 is greater than T2. 26. The correct choice is (d). The law of conservation of linear momentum gives m1v1 + m2v2 = 0 or

29. From Wien’s displacement law lm T = constant, we have T l lA TA = l B TB fi A = B TB l A From Stefan’s law, E = sA ◊ T 4 = s (4 pR2) T 4, we have

m2 v2 = 1.0 m1v1

l1 m2 v2 = 1.0 = l2 m1v1

m0 ( J p a 2 ) p 0 Ja = 2p a 2 P due to cavity is

B2 =

m0 ( J p a 2 / 4) m0 Ja = 3a ˆ 12 Ê 2p Ë 2¯ P is

\ B = B1 - B2 =

m0 Ja m0 Ja 5m0 Ja = 2 12 12

which gives N 28. Given M1 = 2.2 Ms and M2 = 11 Ms. Let R1 and R2 be their respective distances from the centre of mass. The total angular momentum about the centre of mass is Ltotal = (I1 + I2) w and the angular momentum of B is L2 = I2 w \

= 1+

4

=9 Section VI 30.(a) Process A Æ V μ T. Therefore TA > TB. DU = nCv DT = nCv(TB – TA). Since TB < TA, DU is negative, i.e. internal energy decreases. DQ = nCpDT DW = 3P(VB – VA) = – 6PV. Which is negative. \

(b) Process B Æ P μ T. Therefore TB > TC. DU = nCv DT = nCv (TC – TB) is negative, i.e. internal energy decreases. DW = PDV = 0 (∵ DV = 0) DQ = DU + DW), DQ = DU. Since DU is negative, heat is lost. \ (b) Æ (p, r)

DQ = nCp(TD – TC gained by the gas.

M1 R12

M 2 R22

\

M1 v12 M 2 v22

(a) Æ (p, r, t)

(c) Process C Æ D is isobaric, i.e. V μ T TD > TC. DU = nCv(TD – TC internal energy increases.

I1 + I 2 I Ltatal =1+ 2 = I2 I1 L2 = 1+

2

2 4 Ê 6 ˆ Ê 1500 ˆ = ÁË ˜¯ ¥ ÁË ˜¯ 18 50

P due to complete cylinder is B1 =

4

ÊR ˆ Êl ˆ = Á A˜ ¥Á B˜ Ë RB ¯ Ë l A ¯

Section V 27.

2

EA Ê R ˆ ÊT ˆ = Á A˜ ¥Á A˜ EB Ë RB ¯ Ë TB ¯

Since de Broglie wavelength l = h/mv, we will have

(∵ v = R w ) 2

M Ê M1 v1 ˆ ¥ 2 = 1+ Á ˜ Ë M 2 v2 ¯ M1 11 =6 = 1+1¥ 2.2 (∵ M1 v1 = M2 v2)

(c) Æ (q, s)

(d) In process D Æ A, the gas is returned to the iniDU = 0. Therefore DQ = DW. Since the gas is compressed, work is done on the gas, i.e. DW DQ is negative. \ (d) Æ (r, t) Answer: (a) Æ (p, r, t) (c) Æ (q, s)

(b) Æ (p, r) (d) Æ (r, t)

MODEL TEST PAPER—II

SECTION I (Single Correct Answer Type) This section contains 13 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONLY ONE is correct. 1. In the determination of Young’s modulus 4MLg ˆ Ê ÁË Y = ˜ by using Searle’s method, a wire of p d2 ¯ length L = 2 m and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension = 0.25 mm in the length of the wire is observed. Quantities d and are measured using a screw gauge and a micrometer, respectively. They have same pitch of 0.5 mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the Y measurement. (a) due to the errors in the measurement of d and are the same. (b) due to the error in the measurement of d is twice that due to the error in the measurement of . (c) due to the error in the measurement of is twice that due to the error in the measurement of d. (d) due to the error in the measurement of d is four times that due to the error in the measurement of . 2. A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The Ê v (helium) ˆ is ratio of the rms speeds Á rms Ë vrms (argon ) ˜¯ (a) 0.32 (b) 0.45 (c) 2.24 (d) 3.16 3. Two large vertical and parallel plates having a separation d are connected to a battery of voltage V. A particle of charge q is released at rest between the two plates. It is found to move at an angle q to the vertical just after release. Then V is given by (a)

mgd sinq q

(b)

mgd cosq q

(c)

mgd tanq q

(d)

mgd q

4. A ball of mass (m) 0.5 kg is attached to the end of string having (L) 0.5 m. The ball is rotated on a horizontal circular path about vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of ball (in radian/s) is

L

m

(a) 9 (b) 18 (c) 27 (d) 36 5. A ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, travelling with a velocity V m/s in a horizontal direction, hits the centre of the ball. After the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 m and the bullet at a distance of 100 m from the foot of the post. The velocity V of the bullet is V m/s

0

20

100

(a) 250 m/s (b) 250 m/s (c) 400 m/s (d) 500 m/s 6. A wooden block performs SHM on a frictionless surface with frequency, v0. The block carries a charge+Q on its surface. If now a uniform electric E the SHM of the block will be (a) of the same frequency and with shifted mean position.

MTPII.2 Comprehensive Physics—JEE Advanced

(b) of the same frequency and with the same mean position. (c) of changed frequency and with shifted mean position. (d) of changed frequency and with the same mean position. E +Q

7.

tor has capacitance C. The equivalent capacitance between A and B is A

B C

C

C C

(a) C (c) 8.

4C 3

(b)

2C 3

(d)

5C 3

with inner radius R/2 and other radius R carries a uniform current density along its length. The B as a function of the radial distance r from the axis is best represented by

(a)

(c)

r

|B|

0

R/2

R

r

9. A student is performing the experiment of resonance column. The diameter of the column tube is 4 cm. The frequency of the tuning fork is 512 Hz. The air temperature is 38°C in which the speed of sound is 336 m/s. The zero of the meter scale coincides with the top end of the Resonance the reading of the water level in the column is (a) 14.0 cm (b) 15.2 cm (c) 16.4 cm (d) 17.6 cm 10. Young’s double slit experiment is carried out by using green, red and blue light, one colour at a time. The fringe widths recorded are bG, bR and bB respectively. Then, (b) bB > bG > bR (a) bG > bB > bR (c) bR > bB > bG (d) bR > bG > bB 11. A loop carrying current I lies in the x-y plane as kˆ is coming out of the plane of the paper. The magnetic moment of the current loop is y

I

r

a

x

|B| a

0

(b)

(d)

R/2

R

r

r

|B|

0

R/2

R

0

R/2

R

r

r

|B|

r

(a) a2 Ikˆ

p (b) ÊÁ + 1ˆ˜ a 2 Ikˆ Ë2 ¯

p (c) - ÊÁ + 1ˆ˜ a 2 Ikˆ Ë2 ¯

(d) (2p + 1)a 2 Ikˆ

12. A small block is connected to one end of a massless spring of un-stretched length 4.9 m. The other tem lies on a horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at t = 0. It then executes simple harmonic motion with angular frequency w = p/3 rad/s. Simultaneously at t = 0, a small pebble is projected with speed v from point P at an angle of 45° as shown

Model Test Paper II MTPII.3

P is at a horizontal distance of 10 m from O. If the pebble hits the block at t = 1 s, the value of v is (take g = 10 m/s2)

calculated about O and P are denoted by LO and LP respectively, then z

z P

v 45° x

O 10 m

(a)

50 m / st

(b)

m

O

P

w

51 m / s (a) LO and LP do not very with time. (b) LO varies with time while LP remains constant. (c) LO remains constant while LP varies with time. (d) LO and LP both vary with time.

(c)

(d) 53 m / s 52 m / s 13. A small mass m is attached to a massless string P as shown in the the x-y plane with centre at O and constant angular speed w. If the angular momentum of the system,

SECTION II (Multiple Correct Answer Type) This section contains 7 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONE or MORE are correct. 14. Consider the motion of a positive point charge in a region where there are simultaneous uniform elecE = E0 ˆj and B = B0 ˆj . At time t = 0, this charge has velocity v in x-y plane, making an angle q with the x-axis. Which of the following option(s) is(are) correct for time t > 0? (a) If q = 0°, the charge moves in a circular path in the x-z plane. (b) If q = 0°, the charge undergoes helical motion with constant pitch along the y-axis. (c) If q = 10°, the charge undergoes helical motion with its pitch increasing with time, along the y-axis. (d) If q = 90°, the charge undergoes linear but accelerated motion along the y-axis. 15. A person blows into open-end of a long pipe. As a result, a high pressure pulse of air travels down the pipe. When this pulse reaches the other end of the pipe, (a) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is open. (b) a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is open. (c) a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed.

(d) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed. 16. In the given circuit, the AC source has w = 100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is (are) (a) The current through the circuit, I is 0.3 A. (b) The current through the circuit, I is 0.3 2A . (c) The voltage across 100 W resistor = 10 2V . (d) The voltage across 50 W resistor = 10 V. 100 mF

100 W

0.5 mF

50 W

I 20 V

17. Six point charges are kept at the vertices of a regular hexagon of side L and centre O, as shown 1 q K= , which of 4pe 0 L2 the following statement(s), is (are) correct?

MTPII.4 Comprehensive Physics—JEE Advanced L

F +q

E –q

P A + 2q

S

T O

19.

D – 2q

R B +q

C –q

O is 6K along OD. (b) The potential at O is zero. (c) The potential at all points on the line PR is same. (d) The potential at all points on the line ST is same. 18. Two solid cylinders P and Q of same mass and plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated near the axis. Which statement(s) is (are) correct? (a) Both cylinders P and Q reach the ground at the same time. (b) Cylinders P has larger acceleration than cylinder Q.

(c) Both cylinders reach the ground with same translational kinetic energy. (d) Cylinder Q reaches the ground with larger angular speed. along the diameter of a circular wire loop, without touching it, the correct statement(s) is (are) (a) The emf induced in the loop is zero if the current is constant. rent is constant. (c) The emf induced in the loop is zero if the current decreases at a steady rate.

current decreases at a steady rate. 20. Two spherical planets P and Q have the same uniform density r, masses MP and MQ and surface areas A and 4A respectively. A spherical planet R also has uniform density r and its mass is (MP + MQ). The escape velocities from the planets P, Q and R are VP, VQ and VR, respectively. Then (a) VQ > VR > VP

(b) VR > VQ > VP

(c) VR /VP = 3

(d) VP /VQ =

1 2

SECTION III (Paragraph Type) This section contains 4 multiple choice questions based on two paragraphs. Each question has four choices (a), (b), (c) and (d) out of which only ONE is correct.

Questions 21 and 22 are based on the following paragraph. The b-decay process, discovered around 1900, is basically the decay of a neutron (n). In the laboratory, a proton (p) and an electron (e–) are observed as the decay product of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has continuous spectrum. Considering a three-body decay process, i.e. n Æ p + e– + n e , around 1930, Pauli explained the observed electron energy spectrum. Assuming the antineutrino ( n e ) to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this

calculation, the maximum kinetic energy for the electron is 0.8 ¥ 106 eV, The kinetic energy carried by the proton is only the recoil energy. 21. What is the maximum energy of the anti-neutrino? (a) Zero (b) Much less than 0.8 ¥ 106 eV (c) Nearly 0.8 ¥ 106 eV (d) Much larger than 0.8 ¥ 106 eV 22. If the anti-neutrino had a mass of 3 eV/c2 (where c is the speed of light) instead of zero mass, what should be the range of the kinetic energy, K, of the electron? (a) 0 £ K £ 0.8 ¥ l06 eV (b) 3.0 eV £ K £ 0.8 ¥ 106 eV (c) 3.0 eV < K £ 0.8 ¥ 106 eV (d) 0 < K £ 0.8 ¥ 106 eV

Model Test Paper II MTPII.5

Questions 23 and 24 are based on the following paragraph. q1

Most materials have the refractive index, n > 1. So, when a light ray from air enters a naturally occurring sin q1 n2 = , it is undermaterial, then by Snell’s law, sin q 2 n1 stood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation, c n = ÊÁ ˆ˜ = ± e r m r where c is the speed of electromagË v¯ netic waves in vacuum, v its speed in the medium, er , and mr, are relative permittivity and permeability of the medium respectively. In normal materials, er and mr , are positive, implying positive n for the medium. When both er and mr , are negative, one must choose the negative root of n. Such negative refractive index materials can now be

(b)

Air Meta-material

q2

q1

(c)

Air Meta-material q2

q1

violating any physical laws. Since n is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials. 23. For light incident from air on a meta-material. the appropriate ray diagram is

q1

(a)

Air

(d)

Air Meta-material

q2

24. Choose the correct statement. (a) The speed of light in the meta-material is v = c|n| (b) The speed of light in the meta-material is c v= n (c) The speed of light in the meta-material is v = c. (d) The wavelength of the light in the meta-material (lm) is given by lm = lair |n|. where lair is wave1ength of the light in air.

Meta-material q2

SECTION IV (Assertion-Reason Type) This section contains 2 questions. Each question has statement-1 followed by statement-2. Only one of the following four choices (a), (b), (c) and (d) is correct. (a) Statement-1 is true, statement-2 is true and is the correct explanation of statement-1. (b) Statement-1 is true and statement-2 is true but is not the correct explanation of statement-1. (c) Statement-1 is true but statement-2 is false. (d) Statement-1 is false but statement-2 is true.

25. Statement-1 A sphere is rolling on a rough surface in the diforce of friction at the point of contact will be in the direction of the arrow.

MTPII.6 Comprehensive Physics—JEE Advanced

Statement-2 Friction opposes motion. 26. Statement-1 No induced emf is developed across the ends of a

w

Statement-2 No force acts on the free electrons of the conductor.

A

SECTION V (Integer Answer Type) This section contains 3 questions. The answer to each question is single digit integer ranging from 0 to 9 (both inclusive) 27.

nucleus with charge Q = 120e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is: (take the proton mass, mp = (5/3) ¥ 10–27 kg: h/e = 4.2 ¥ 1 = 9 ¥ 109 m/F: 1 fm = 10–15 m) 10–15 J.s/C; 4pe 0 28. A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass density and radius 2R The moment of inertia of this lamina about axes passing though O and P is IO and IP respectively. Both these axes are perpendicular to the plane of the lamina. The ratio IP /IO to the nearest integer is

29. A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a uniform current density J P is given by N m0 aJ , the value of N is 12

P

O

a

2a 2R

O

2R

P

SECTION VI (Matrix Match Type) 30. In Column I are listed some charged bodies and current carrying conductors. Match them with the effects they produce listed in column II Column I Column II (c) A coil carrying a current I = I0 sin w

(r) Magnetic moment

Model Test Paper II MTPII.7

Answers Section-I 1. 4. 7. 10. 13.

(a) (d) (d) (d) (c)

2. 5. 8. 11.

Also E =

(d) (d) (d) (b)

3. 6. 9. 12.

(a) (a) (b) (a)

Section-II

14. (c), (d) 16. (a), (c) 18. (d)

15. (b), (d) 17. (a), (b), (c) 19. (a), (c) 20. (b), (d) Section-III

21. (c) 24. (b)

22. (d)

23. (c)

Section-IV 25. (a)

(+)

V . Hence d

P qE mg q

mgd tanq q

Q

A

Section-V 28. (3)

R

4. Radius of the circular d path is BC = r = L sinq, where L = AB is the length of the string. The vertical component T cosq of tension T balances with the weight mg and the horizontal component T sinq provides the necessary centripetal force for circular motion. Hence

26. (a)

27. (7)

r E

q

V mg tanq = d q fiV =

(–)

29. (5)

q

Section-VI 30. (a) Æ (p), (b) Æ (p), (q), (r), (c) Æ (p), (r), (s), (d) Æ (q)

T T cos q

Solutions Section-I

0.5 mm = 0.005 mm. 100 Since M and L g is constant, D M = 0, D L = 0 and D g = 0. Hence

C

B

1. Least count of screw gauge =

mg

T sin q

T sin q = mrw2 = m(L sin q)w2

DY D 2Dd = + Y d

fi

T = mLw2

0.005 mm = 0.02 0.25 mm

\

Tmax = mL w 2max

fi

324 = 0.5 ¥ 0.5 ¥ w 2max

2Dd 2 ¥ 0.005 mm = 0.02 = d 0.5 mm

fi

D

2. vrms =

=

3RT M

vrms (helium) = vrms (argon ) 3. tanq = fiE=

5.

QR qE = PQ mg mg tanq q

wmax = 36 rad s–1 tf ) =

2h = g

2 ¥ 5 = 1s 10

Horizontal range (R) = horizontal velocity ¥ time M Ar = M He

40 = 10 = 3.16 4

\ Horizontal velocities of the bullet and of the ball after the collision respectively are (v)bullet = 100 = 100 ms–1 1 (v)ball = 20 = 20 ms–1 1

MTPII.8 Comprehensive Physics—JEE Advanced

From conservation of momentum,

Outside the cylinder. (i.e. for r > R), B=

(m)bullet ¥ V = (m)bullet ¥ (v)bullet + (m)ball ¥

fi

(v)ball 0.01 V = 0.01 ¥ 100 + 0.2 ¥ 20

fi

V = 500 ms–1

fi

6. The force exerted on charge +Q by the electric E is F = QE in the direction of E . Since F is constant, a constant force is added to the applied force. Hence only the mean position will change and the frequency of oscillation will remain the same. 7. The circuit can be redrawn as follows. 1

A

2

3

4

5

7

A

1

2

4

3

6

B

5

I is the current density and A is the A area of the shaded region. Now

where J =

Ê Rˆ A = pr2 - p Á ˜ Ë 2¯ \

8

B=

Wavelength l = 2C 3

C A

B

B

fi C

m0 J È r 2 - R 2 / 4 ˘ Í ˙ 2 Î r ˚

È R2 ˘ r Í ˙ 4r ˚ Î Hence the correct graph is (d). 9. End correction e = 0.3d = 0.3 ¥ 4 = 1.2 cm

B

A

2

È p R2 ˘ B ¥ 2p r = m0 J Íp r 2 ˙ 4 ˚ Î

=

fl

Ceq =

B ¥ 2p r = m0I

8

6

2C

R In the shaded region (i.e. for < r < R). From 2 Ampere’s circuital law,

fi

fl 7

m0 I 2p r

v ◊336 = 0.656 m = 65.6 cm = n 512

Now L + e = l 4 L = l – e = 65.6 – 1.2 4 4 = 16.4 – 1.2 = 15.2 cm

fi C

2C 5C +C = 3 3

m0 J 2

10. Since lR > lG > lB and b =

lD , b R > b G > b B. d

Æ

11. Magnetic Moment M = current ¥ area of the loop S r

8.

O

P

Q

R , OQ = R, OS = r, 2 Inside the cavity (i.e. for r lying between zero R and ); B = 0 2 OP =

Æ

= IA 2 È ˘ Ê aˆ = I ¥ Í a 2 + p Á ˜ ¥ 2˙ kˆ Ë 2¯ ÍÎ ˙˚

È p˘ = Ia 2 Í1 + ˙ kˆ Î 2˚ Æ The direction of area vector A is along kˆ . 12. When the pebble hits the block, the distance travelled by the pebble (Sp) = distance travelled by the block (Sb). Sp = 4.9 + 0.2 cos w t

Model Test Paper II MTPII.9

p = 4.9 + 0.2 cos ÊÁ ˆ˜ Ë 3¯

(

t – 1 s)

= 4.9 + 0.2 ¥ 1 = 5.0 m 2 Sb = 10 – horizontal range = 10 -

v 2 sin(2q ) g

= 10 -

v 2 sin(90 ) 10

= 10 -

v2 g

(

q = 45°)

Section-II Fe = qE . 14. Force on charge q Since E is along +y direction and q is positive, the charge will accelarate along the y-axis. Force on charge q = Fm = q(v ¥ B) which is perpendicular to both   v and B . If q = 90°, v will be parallel to B (which is along y-axis), Fm = 0 and the charge will accelerate along the y-axis due to the electric q lies between zero and 90°, the path is a helix with increasing pitch along the y-axis due E. 15. When a compression reaches the open end of a of the medium just outside the open end. There is no reversal of amplitude of the pressure wave on

Now Sb = Sp, Therefore, 10 v=

fi

a compression due to reversal of the amplitude of

v2 = 5.0 fi v2 = 50 g

50 ms

Hence the correct choices are (b) and (d). 16.

–1

C = 100 mF

R1 = 100 W

I1

13. T cosq = mg r LP

L = 0.5 H I

R2 = 50 W

I2 V

P q r LO

T q

O

T cos q T sin q

w mg

So T cos q balances with weight mg. T sin q produces no torque about O but produces a non-zero torque about P. Now Æ

Æ

Æ

Æ

XC =

XL = wL = 100 ¥ 0.5 = 50 W Since XC = Rl, current I1 leads V by 45°. Since XL = R2, current I2 lags behind V by 45°. So the phase difference between I1 and I2 is f = 90°. I1 =

Æ

dw L = I w and t = I a = I dt

and I2= Æ

Æ dw Since torque about O is zero, = 0, i.e. w = dt Æ constant. Hence angular momentum L about O is constant. But torque about P is non-zero. Hence

V X c2

+

R12

V X L2 + R22

=

\

I=

(100) + (100)

2

=

2 A 10

20 (50) 2 + (50) 2 2 A 5

I12 + I 22 + 2 I1 I 2 cos 90 2

=

20 2

= =

Æ

Æ Æ dw π 0, i.e. w changes with time. Hence L about dt P varies with time.

1 1 = = 100 W 100 ¥ 100 ¥ 10-6 wC

Ê 2ˆ Ê 2ˆ Á 10 ˜ + Á 5 ˜ Ë ¯ Ë ¯

2

=

0.1 = 0.316

MTPII.10 Comprehensive Physics—JEE Advanced

a= P.D. across R1 = I1R1 =

2 ¥ 100 = 10 2 V 10

P.D. across R2 = I2R2 =

2 ¥ 50 = 10 2 V 5

g sinq I 1+ MR 2

where I is the moment of intertia of the body about its centre of mass. Therefore, aP =

17.

and

aQ =

g sinq I 1+ P 2 MR g sinq IQ 1+ MR 2

Since cylinder P has most of its mass concentrated near its surface and Q has most of its mass concentrated near its axis, IP > IQ. Hence E1

E2

O due to –q at E directed from O to E O due to +q at B directed from O to E. 1 q = 4pe 0 L2 O due to + 2q at A directed from O to D O due to –q at D directed from O to D 1 2q = 4pe 0 L2

O is E = 2E1 cos 60° + 2E1 cos 60° + 2E2 = E1 + E1 + 2E2 = 2E1 + 2E2 = =

1 2q 1 4q 2 + 4pe 0 L 4pe 0 L2 6 2q along OD 4pe 0 L2

So choice (a) is correct. Net potential at O is V=

-q q q q + + + 4pe 0 L 4pe 0 L 4pe 0 L 4pe 0 L 2q 2q =0 4pe 0 L 4pe 0 L

18. The linear acceleration of a body of mass M and radius R rolling down an inclined plane of inclination q is given by

aP < aQ aP < aQ

\

19.

(

a = Ra)

Since w2 = 02 – 2a (q); here q = angular displace1 at 2, it follows that wP < wQ. So ment and q = 2 the only correct choice is (d). of the plane of the loop and below the wire into the plane of the loop. Hence the net magnetic induced in the loop is zero, irrespective of whether the current is changing or not changing. Hence the correct choices are (a) and (c). r B (out of page)

e e I

ƒ

e

Loop

e

ƒ ƒ

I

ƒ

r B (into page)

20. Let M be the mass of planet P and R its radius. 4p 3 Then Mass of P is M = r ¥ R 3 4p (2 R)3 = 8M ( surface area 3 of Q = 4 times that of P, therefore radius of Q = 2 (radius of P) 2R Mass of R = mass of P + mass of Q = M + 8M = 9M \ Radius of R = 91/3R Mass of Q is r ¥

Model Test Paper II MTPII.11

The escape velocities of P, Q and R are

\ Also

VP =

2GM R

VQ =

2G ¥ 8M = 2VP R

VR =

2G ¥ 9 M = 91/3VP 91 / 3 R

VP

=2

21. The mass of a proton is very large compared to electron and antineutrino. So all the energy is shared by the electron and anti-neutrino. When the kinetic energy of anti-neutrino is zero, the maximum kinetic energy of electron is 0.8 ¥ 106 eV and vice versa. Hence the total kinetic energy of electron + anti-neutrino is 0.8 ¥ 106 eV. 22. If the anti-neutrino has a mass m = 3 eV/c2, it will have kinetic energy = mc2 = 3 eV. Therefore, the maximum kinetic energy of the electron = (0.8 ¥ 106 –3) eV, which is only slightly less than 0.8 ¥ 106 eV. The minimum kinetic energy of the electron is still zero. Hence correct choice is (d) 23. The refractive index n for meta-materials is negasin q1 is negative. tive. Hence sin q 2 Thus if q1 is positive, q2 will be negative. So the current choice is (c). c c , which is choice (b) 24. N = fi v = v |n| v Since n remains unAlso frequency n = l changed,

= lair ¥ = lair

vm vair

Section-V 27.

cˆ Ê ÁË v = ˜¯ n

z = 120 e

Proton P

v

O

Nucleus

r0

The proton reaches a point P and is then repelled back by the nucleus. Loss of kinetic energy = gain in potential energy. Ze 120e2 1 2 = = mP v 4pe 0 r0 4pe 0 r0 2 If p is the linear momentum, then ( fi

K.E. =

p2 ) 2m

p2 120e2 = 2m p 4pe 0 r0

fi

Ê 240e2 m p ˆ p= Á ˜ Ë 4pe 0 r0 ¯

Now

l=

1/ 2

h h Ê 4pe 0 r0 ˆ = P e ÁË 240 m ˜¯

1/ 2

(1)

h 1 = 4.2 ¥ 10–15 Js/C, = 9 ¥ 109 m/F, e 4pe n 1 fm = 10–15 m, and r0 = 10 fm = 10 ¥ 10–15 m in Eq. (1), we get r0 = 7 ¥ 10–15 m = 7 fm

Putting

vm c ¥ c vair

nair nm

nm = |n| and nair = 1)

25. The correct choice is (a). The direction of the linear velocity at A, the point of contact is to the left (opposite to the direction of the arrow). Since friction opposes motion, the direction of the frictional force at A will be in the direction of the arrow, i.e. in the direction along which the sphere is rolling. 26. The correct choice is (a). Let v be the velocity B . Since the free electrons in the conductor are moving with   it, force F = e ( v = B ) is zero because v is parallel to B . Consequently, no induced emf is developed between the ends of the conductor.

vair v = m lair lm lm = lair ¥

(

Section-IV

Section-III

fi

lair |n|

So choice (d) is wrong.

VR > VQ > VP VQ

=

MTPII.12 Comprehensive Physics—JEE Advanced

28. Let M be the mass of the complete disc. The mass of the cut-out disc is

\

M M ¥ p R2 = 4 p (2 R)2

\

m=

Moment of inertia of the complete disc about O 1 = M ¥ (2 R)2 = 2 MR 2 2 Moment of inertia of the cut-out disc about O = \

Ip I0

=

2

3MR 13MR = 8 8

3 (to nearest integer)

B1 =

m0 ( J p a 2 ) p 0 Ja = 2p a 2 P due to cavity is

B2 =

2

Moment of inertia of the complete disc about 3 P = M (2 R)2 = 6 MR 2 2 Moment of inertia of the cut-out disc about P 2

37 13

11 MR 2 = 37/8 MR2 8

P due to complete cylinder is

29.

1 2MR 2 3MR 2 ¥ = 2 4 8

I0 = 2MR 2 -

=

Ip = 6MR2 –

11 MR M + ( R 2 + 4 R 2 ) = MR 2 8 4 8

m0 ( J p a 2 /4) p 0 Ja = 12 Ê 3a ˆ 2p Á ˜ Ë 2¯ P is

\ B = B1 - B2 =

m0 Ja m0 Ja 5m0 Ja = 2 12 12

Which gives N = 5. 30. (a) Æ (p), (b) Æ (p), (q), (r) (c) Æ (p), (r), (s), (d) Æ (q)

IIT-JEE 2012: PAPER-I

SECTION I (Single Correct Answer Type) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONLY ONE is correct. 1. In the determination of Young’s modulus Ê 4 MLg ˆ ÁË Y = p d 2 ˜¯ by using Searle’s method, a wire of length L = 2 m and diameter d = 0.5 mm is used. For a load M = 2.5 kg, an extension = 0.25 mm in the length of the wire is observed. Quantities d and are measured using a screw gauge and a micrometer, respectively. They have same pitch of 0.5 mm. The number of divisions on their circular scale is 100. The contributions to the maximum probable error of the Y measurement (a) due to the errors in the measurement of d and are the same. (b) due to the error in the measurement of d is twice that due to the error in the measurement of . (c) due to the error in the measurement of is twice that due to the error in the measurement of d. (d) due to the error in the measurement of d is four times that due to the error in the measurement of . 2. A small mass m is attached to a massless string P as shown in the the x-y plane with centre at O and constant angular speed w. If the angular momentum of the system,  calculated about O and P are denoted by LO and  L P respectively, then

  (a) LO and L P do not very with time.   (b) LO varies with time while L P remains constant.   (c) LO remains constant while L P varies with time.   (d) LO and L P both vary with time. 3. A bi-convex lens is formed with two thin planoindex n second lens is 1.2. Both the curved surface are of the same radius of curvature R = 14 cm. For this bi-convex lens, for an object distance of 40 cm, the image distance will be

(a) – 280.0 cm (b) 40.0 cm (c) 21.5 cm (d) 13.3 cm 4. A thin uniform rod, pivoted at O is rotating in the horizontal plane with constant angular speed w, as t = 0, a small insect starts from O and moves with constant speed v, with respect to the rod towards the other end. It reaches the end of the rod at t = T and stops. The angular speed of the system remains w throughout. The magnitude of the torque (|t |) about O, as a function of time is best represented by which plot?

IJI.2 Comprehensive Physics—JEE Advanced

7. Three very large plates of same area are kept parallel and close to each other. They are considered as ideal black surfaces and have very high maintained at temperatures 2T and 3T respectively. The temperature of the middle (i.e. second) plate under steady state condition is (a)

(b)

(c)

65 (a) ÊÁ ˆ˜ Ë 2¯

1/ 4

97 (c) ÊÁ ˆ˜ Ë 2¯

1/ 4

T

97 (b) ÊÁ ˆ˜ Ë 4¯

1/ 4

T

(d) (97)1/4 T

8. A small block is connected to one end of a massless spring of un-stretched length 4.9 m. The other tem lies on a horizontal frictionless surface. The block is stretched by 0.2 m and released from rest at t = 0. It then executes simple harmonic motion with angular frequency w = p /3 rad/s. Simultaneously at t = 0, a small pebble is projected with speed v from point P at an angle of 45° as shown P is at a horizontal distance of 10 m from O. If the pebble hits the block at t = 1 s, the value of v is (take g = 10 m/s2)

(d)

5. A mixture of 2 moles of helium gas (atomic mass = 4 amu) and 1 mole of argon gas (atomic mass = 40 amu) is kept at 300 K in a container. The Ê v (helium) ˆ is ratio of the rms speeds Á rms Ë vrms (arg on ) ˜¯ (a) 0.32 (b) 0.45 (c) 2.24 (d) 3.16 6. Two large vertical and parallel metal plates having a separation of 1 cm are connected to a DC voltage source of potential difference X. A proton is released at rest midway between the two plates. It is found to move at 45° to the vertical JUST after release. Then X is nearly (b) 1 ¥ 10–7 V (a) 1 ¥ 10–5 V –9 (d) 1 ¥ 10–10 V (c) 1 ¥ 10 V

(a)

50 m/s

(b)

51 m/s

(c)

52 m/s

(d)

53 m/s

9. Young’s double slit experiment is carried out by using green, red and blue light, one colour at a time. The fringe widths recorded are bG, b and bB respectively. Then, (a) bG > bB > b

(b) bB > bG > b

(c) b > bB > bG

(d) b > bG > bB

10. Consider a thin spherical shell of radius R with centre at the origin, carrying uniform positive surface charge density. The variation of the mag| E ( r ) | and the electric potential V(r) with the distance r from the centre, is best represented by which graph?

IIT-JEE 2012 Paper-I

(a)

(c)

(b)

(d)

IJI.3

SECTION II (Multiple Correct Answer(s) Type) This section contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONE of MORE are correct. 11. Consider the motion of a positive point charge in a region where there are simultaneous uniform elecŸ Ÿ   E = E0 j and B = B0 j . At time t = 0, this charge has velocity v in the x-y plane, making an angle q with the x-axis. Which of the following option(s) is(are) correct for time t > 0? (a) If q = 0° the charge moves in a circular path in the x-z plane. (b) If q = 0°, the charge undergoes helical motion with constant pitch along the y-axis. (c) If q = 10°, the charge undergoes helical motion with its pitch increasing with time, along the y-axis. (d) If q = 90°, the charge undergoes linear but accelerated motion along the y-axis. 12. A cubical region of side a has its centre at the q at (0,–a/4, 0), +3q at (0, 0, 0) and –q at (0, +a/4, 0). Choose the correct options(s).

+a the plane x = –a/2 +a ing the plane y = –a/2. is

q . e0

x = y = -

z = +a the plane x = +a/2. 13. A person blows into open-end of a long pipe. As a result, a high pressure pulse of air travels down the pipe. When this pulse reaches the other end of the pipe, (a) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is open. (b) a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is open. (c) a low-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed. (d) a high-pressure pulse starts travelling up the pipe, if the other end of the pipe is closed. 14. inclined plane PQ which makes an angle q with the horizontal. A horizontal force of 1 N acts of the block through its centre of mass as shown in g = 10 m/s2).

IJI.4 Comprehensive Physics—JEE Advanced

15.

(a) q = 45° (b) q > 45° and a frictional forces acts on the block towards P. (c) q > 45° and a frictional forces acts on the block towards Q. (d) q < 45° and a frictional forces acts on the block towards Q.

choose the correct option(s).

(a) (b) (c) (d)

The I1 = The I2 =

current through PQ is zero. 3A potential at S is less than that at Q. 2A

SECTION III (Integer Correct Answer Type) This section contains 5 questions. The answer to each question is single digit integer, ranging from 0 to 9 (both inclusive) P, which is at a distance 2R from the axis of the 23r R . The cylinder, is given by the expression 16k e 0 value of k is

16. A circular wire loop of radius R is placed in the x-y plane centered at the origin O.A square loop of side a (a<
18.

17.

R has a uniform volume charge density r. It has a spherical cavity of radius R/2 with its centre on

nucleus with charge Q = 120e, where e is the electronic charge. It makes a closest approach of 10 fm to the nucleus. The de Broglie wavelength (in units of fm) of the proton at its start is: (take the proton mass, mp = (5/3) ¥ 10–27 kg; 1 = 9 ¥ 109 m/F; h/e = 4.2 ¥ 10–15 J.s/C; 4pe 0 1 fm = 10–15 m) 19. A lamina is made by removing a small disc of diameter 2R from a bigger disc of uniform mass den-

IIT-JEE 2012 Paper-I

sity and radius 2R, as shown

D

inertia of this lamina about axes passing though O and P is IO and IP respectively. Both these axes are perpendicular to the plane of the lamina. The ratio IP /IO to the nearest integer is 20. A cylindrical cavity of diameter a exists inside a cylinder of diameter 2a

2Dd d

=

0.005 mm = 0.02 0.25 mm

=

2 ¥ 0.005 mm = 0.02 0.5 mm

IJI.5

2. (c) T cos q = mg

uniform current density J P is given by

N m0aJ, the value of N is 12

Answers Section-I 1. 4. 7. 10.

2. (c) 5. (d) 8. (a)

(a) (b) (c) (d)

3. (b) 6. (c) 9. (d)

1 Ê 1.2 - 1ˆ Ê 1 1 ˆ 1 =Á cm–1 ˜Á ˜= f 2 Ë 1 ¯ Ë • -14 ¯ 70

12. (a, c, d) 13. (b, d) 15. (a, b, c, d)

1 1 1 1 1 = + = + fi F = 20 cm F f1 f 2 28 70

Section-III 16. 7 19. 3

17. 6 20. 5

18. 7

Now

Hints and Solutions Section-I 1. (a) Least count = DY Y

=

D

1 Ê 1.5 - 1ˆ Ê 1 1 ˆ 1 =Á cm–1 ˜Á - ˜ = f1 Ë 1 ¯ Ë 14 • ¯ 28

3. (b)

Section-II 11. (c, d) 14. (a, c)

So T cos q balances with weight mg. T sin q produces no torque about O but produces a non-zero torque about P. Now      dw L = I w and t = I a = I dt   dw Since torque about O is zero, = 0, i.e. w = dt  constant. Hence angular momentum L about O is constant. But torque about P is non-zero. Hence    dw w changes with time. Hence L about dt P varies with time.

+

2Dd d

0.5 mm = 0.005 mm 100

fi

1 1 1 - = v u F

1 1 1 fi v = 40 cm = v - 40 20

4. (b) L = Iw x = vt I = mx2 = mv2t2

IJI.6 Comprehensive Physics—JEE Advanced

\

L = (mx2) w = mwv2t2

Êpˆ = 4.9 + 0.2 cos ÁË ˜¯ 3

dL \ t= = 2mwv2t dt Hence, as long as the insect is moving, the torque varies linearly with time. When the insect steps, v = 0, and torque becomes zero. 5. (d) vrms = \

6. (c) tan q =

M Ar = M He

40 = 10 = 3.16 4

QR qE = PQ mg

Given q = 45°. Therefore, qE fi qE = mg mg X Also E = . Hence d qX = mg d mgd fi X= q tan 45° =

(1.67 ¥ 10-27 ) ¥ 9.8 ¥ (1 ¥ 10-2 ) = 10–9 V 1.6 ¥ 10-19 7. (c) In the steady state, the rate at which the middle plate receives heat energy is equal to the rate at which heat energy is emitted by the other plates. Let A be the area of each plate and To be the steady state temperature of the middle plate. Since both sides of the middle plate receive heat energy, the total area of the middle plate receiving energy is 2A.

t = 1 s)

1 = 5.0 m 2 Sb = 10 – horizontal range = 4.9 + 0.2 ¥

3RT M

vrms (helium) = vrms (argon )

(

= 10 –

v 2 sin ( 2q ) g

= 10 –

v 2 sin (90∞) 10

(

q = 45°)

v2 10 Now Sb = Sp. Therefore, = 10 –

10 –

v2 = 5.0 fi v2 = 50 fi v = 10

9. (d) Since l > lG > lB and b =

50 ms–1

lD , d

b > b G > b B. 10. (d) For a spherical shell Ï 0  Ô | E (r) | = Ì Q Ô 4p e r 2 Ó 0

for r < R for r > R

and

Ï Q Ô 4 pe R for r < R Ô 0 V (r) = Ì Ô Q for r > R ÔÓ 4p e 0 r Hence the correct graph is (d). The graph shown with broken lines represents V(r) vs r graph.

Section-II

From Stefan’s law (2A) (To)4 = sA (2T)4 + sA (3T)4 fi

2 To4 = 16T 4 + 81T 4 = 97 T 4

fi

97 4 To = ÊÁ ˆ˜ T Ë 2¯

1

8. (a) When the pebble hits the block, the distance travelled by the pebble (Sp) = distance travelled by the block (Sb). Sp = 4.9 + 0.2 cos w t

11. (c, d) Force on charge q    F e = qE. Since E is along +y direction and q is positive, the charge will accelarate along the y-axis. Force on charge q    F m = q (v ¥ B) which is perpendicular to both   v and B . If q = 90°, v will be parallel to B (which is along y-axis), Fm = 0 and the charge will accelerate along the y-axis due to the electric q lies between zero and 90°, the path is a helix with increasing pitch along the y-axis due  E.

IIT-JEE 2012 Paper-I

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12. (a, c, d) According to Gauss’s law, the electric f=

qnet 3 q - q - q q = = e0 eo e0

x = a/2 and the x = –a/2 is the same. Further, the positions of charges with respect to x = a/2 and z = a through the planes x = a/2 and z = a/2 is the same y = a/2 and y = –a/2 is the same. 13. (b, d) When a compression reaches the open end openness of the medium just outside the open end. There is no reversal of amplitude of the pressure compression reaches the closed end of a pipe, it at the closed end. Hence the correct choices are (b) and (d). 14. (a, c)

Applying Kirchhoff’s loop rule to loops APQA, PSTQP and AQTBUVA, we get 2a + c – 4b = 0, (1) 2 (a – c) – c – 4 (b + c) – c = 0 fi a – 2b –4c = 0 (2) and 4b + 4 (b + c) + 4b – 12 = 0 fi 3b + c = 3 (3) Solving Eqs. (1), (2) and (3), we get a = 2 A, b = 1 A and c = 0 Thus the current through PQ is zero. Also I1 = a + b = 3 A and I2 = a – c = 2 A. Also VS – VQ = – c – 4 (b + c) = – 4b = – 4 ¥ 1 = – 4V ( c = 0). Hence potential at S is less than that at Q.

Section-III 16. (7) f = NBA cos q Given N = 2, A = a2, and q = 45°. 2p IR 2 m0 3 4p [ R 2 + ( 3R ) 2 ] 2 mI = 0 16 R

Also B = Given F = 1 N, m = 0.1 kg and g = 10 ms–2. Let f be the frictional force between the block and the plane surface PQ. The block will be stationary if F cos q = mg sin q fi 1 ¥ cos q = 0.1 ¥ 10 ¥ sin q fi tan q = 1 fi q = 45° and f = 0 If q > 45°, sin q > cos q. Hence mg sin q > F cos q ( F = mg = 1 N). Therefore frictional force acts up the block towards Q. If q < 45°, sin q < cosq. Hence mg sin q < F cos q. Therefore, in this case, frictional force f acts down in the plane towards P. 15. (a, b, c, d)

1 Ê m0 a 2 ˆ Êm Iˆ = 7 \ f = 2 ¥ Á 0 ˜ ¥ a2 ¥ I Ë 16 R ¯ 2 ÁË 2 2 R ˜¯ f m0 a 2 M= = 7 fip=7 I 2 2R 17. (6)

We take the Gaussian surface to be a cylinder of radius r and length l. From Gauss’s law,

IJI.8 Comprehensive Physics—JEE Advanced

   q Ú E.ds = e 0

h h Ê 4pe 0 r0 ˆ Now l = = p e ÁË 240 m ˜¯ q r ¥ p R 2l = e0 e0

fi E ¥ 2p rl = fiE=

(1) P due to solid cylinder

is (putting r = 2R) rR2 rR = Ec = along +y direction 2e 0 ( 2 R ) 4e 0

Es =

P due to solid sphere

=

Q 4pe 0 r 2

=

(1)

h 1 = 4.2 ¥ 10–15 Js/C, = 9 ¥ 109 e 4pe 0 m/F, 1 fm = 10–15 m, and r0 = 10 fm = 10 ¥ 10–15 m in Eq. (1), we get r0 = 7 ¥ 10–15 m = 7 fm

19. (3) Let M be the mass of the complete disc. The mass of the cut-out disc is M M ¥ p R2 = m= 2 4 p (2 R ) Moment of inertia of the complete disc about O 1 M ¥ (2R)2 = 2MR2 = 2 Moment of inertia of the cut-out disc about O 1 3MR 2 3MR 2 = ¥ = 2 4 8

4p ( R / 2)3 3 4pe 0 (2 R) 2

r¥

rR along–y direction. 96e 0 P is

Ep = Ec – Es =

2

Putting

rR2 2e 0 r

(since r > R) is

1

I0 = 2MR2 –

\

23r R rR rR = 4e 0 96e 0 96e 0

3MR 2 13MR 2 = 8 8

Moment of inertia of the complete disc about 3 M (2R)2 = 6 MR2 P= 2 Moment of inertia of the cut-out disc about P

which gives k = 6. 18. (7)

MR 2 M 2 11 (R + 4R2) = MR2 + 8 4 8 11 \ IP = 6MR2 – MR2 = 37/8 MR2 8 I P 37 \ 3 (to nearest integer) = I 0 13

=

The proton reaches a point P and is then repelled back by the nucleus. Loss of kinetic energy = gain in potential energy.

20. (5) P due to complete cylinder is m ( J p a 2 ) p 0 Ja B1 = 0 = 2p a 2 P due to cavity is

Ze 1 120e2 m p v2 = = 2 4pe 0 r0 4pe 0 r0 If p is the linear momentum, then ( fi

120e2 p2 = 2m p 4pe 0 r0

fi

Ê 240e2 m p ˆ p= Á ˜ Ë 4pe 0 r0 ¯

p2 K.E. = ) 2m

B2 =

m0 ( J p a 2/4 ) m0 Ja = 12 Ê 3a ˆ 2p Á ˜ Ë 2¯ P is

\ 1

2

B = B1 – B2 =

m0 Ja m0 Ja 5m0 Ja = 2 12 12

which gives N = 5.

IIT-JEE 2012: PAPER-II

SECTION I (Single Correct Answer Type) This section contains 8 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONLY ONE is correct. 1. Two identical discs of same radius R are rotating about their axes in opposite directions with the same constant angular speed w. The discs are in the same horizontal plane. At time t = 0, the point P and Q are facing each other as shown in the P and Q is vr . In one time period (T) of rotation of the discs, vr as a function of time is best represented by

(d)

2. A loop carrying current I lies in the x-y plane as k is coming out of the plane of the paper. The magnetic moment of the current loop is

(a)

(a) a 2 I k p (c) - ÊÁ + 1ˆ˜ a 2 I k Ë2 ¯

(b)

p (b) ÊÁ + 1ˆ˜ a 2 I k Ë2 ¯ (d) ( 2p + 1) a 2 I k

3. with inner radius R/2 and other radius R carries a uniform current density along its length. The B as a func(c)

tion of the radial distance r from the axis is best represented by (a)

IJII.2

6. Two moles of ideal helium gas are in a rubber balloon at 30°C. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to 35°C. The amount of heat required in raising the temperature is nearly (take R

(b)

(c) 7. Consider a disc rotating in the horizontal plane with a constant angular speed w about its centre O. The disc has a shaded region on one side of the diameter and an unshaded region on the other (d)

4. A thin uniform cylindrical shell, closed at both

the orientation as shown, two pebbles P and Q are simultaneously projected at an angle towards R y-z plane and is same for both pebbles with respect to the disc. Assume that (i) they land back on the disc 1 before the disc has completed rotation, (ii) their 8 range is less than half the disc radius, and (iii) w remains constant throughout. Then

rc with respect to water, then the correct statement is that the shell is rc is less than 0.5. rc is more than 1.0. rc is more than 0.5. rc is less than 0.5. 5. mC to the upper plate of the 4 mF capacitor. Then in the steady state, the charge on the upper plate of the 3 mF capacitor is

(a) P lands in the shaded region and Q in the unshaded region. (b) P lands in the unshaded region and Q in the shaded region. (c) Both P and Q land in the unshaded region. (d) Both P and Q land in the shaded region. 8. A student is performing the experiment of resonance column. The diameter of the column tube is 4 cm. The frequency of the tuning fork is 512 speed of sound is 336 m/s. The zero of the meter scale coincides with the top end of the Resonance

m m

mC mC

(a) 14.0 cm (c) 16.4 cm

(b) 15.2 cm (d) 17.6 cm

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SECTION II (Paragraph Type) This section contains 6 multiple choice questions relating to three paragraph with two questions on each paragraph. Each question has four choices (a), (b), (c) and (d) out of which ONLY ONE is correct.

Paragraph for Questions 9 and 10 x-z plane and lies in the plane of the disc for case (b). (c) It is horizontal of case (a); and is at 45° to the x-z plane and is normal to the plane of the disc for case (b).

The general motion of rigid body can be considered to be a combination of (i) a motion of its centre of mass about an axis, and (ii) its motion about an instantaneous axis passing through the centre of mass. These axes need not be stationary. Consider, for example, a thin uniform disc

is rotated about the origin on a horizontal frictionless plane with angular speed w, the motion at any instant can be taken as a combination of (i) a rotation of the centre of mass of the disc about the z-axis, and (ii) a rotation of

10.

x-z plane and is normal to the plane of the disc for case (b). angular speed about the instantaneous axis (passing through the centre of mass) is correct? 2w for both the cases. w for case (b). (b) It is w for case (a); and 2 (a) It is

through its centre of mass (as is seen from the changed orientation of points P and Q the same angular speed w in this case

(c) It is w for case (a); and 2w for case (b). (d) It is w for both the cases.

Paragraph for Questions 11 and 12 The b the decay of a neutron (n). In the laboratory, a proton (p) and an electron (e– of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should x-z plane; Case (b) the disc with its face making an angle of 45° with x-y plane and its horizontal diameter parallel to x-axis. In both the cases, disc is welded at point P, and the system are rotated with constant angular speed w about the z-axis.

the electron kinetic energy has continuous spectrum. Considerng a three-body decay process, i.e. nÆ p ne

e–

energy spectrum. Assuming the anti-neutrino ( n e ) to be massless and possessing negligible energy, and the principles are applied. From this calculation, the maximum ¥ 106 eV. The kinetic energy carried by the proton is only the recoil energy. 11. (a) Zero ¥ 106 eV 6 ¥ 10 eV ¥ 106 eV

9. instantaneous axis (passing through the centre of mass) is correct?

12. If the anti-neutrino had a mass of 3 eV/c2 (where c is the speed of light) instead of zero mass, what should be the range of the kinetic energy, K, of the electron?

IJII.4

(a) (b) (c) (d)

0
(b)

Paragraph for Questions 13 and 14 n > 1. So, when a light ray from air enters a naturally occurring material, sin q1 n2 = , it is understood that then by Snell’s law, sin q 2 n1 the refracted ray bends to emerges on the same side of the normal as the incident ray.

(c)

c n = ÊÁ ˆ˜ = ± e r m r , Ë v¯

the me where c v its speed in the medium, er and mr In normal materials, er and mr n

(d) er and mr n

laws. Since n similar to normal materials, the frequency of light remains 13. For light incident from air on a meta-material, the appropriate ray diagram is (a)

14. Choose the correct (a) The speed of v = c|n| (b) The speed of c v= n (c) The speed of v = c. material (lm lair

statement. light in the meta-material is light in the meta-material is

light in the meta-material is lm = lair |n|, where

SECTION III Multiple Correct Answer(s) Type This section contains 6 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONE or MORE are correct. 15.

w = 100 rad/s. Considering the inductor and capacitor to be ideal, the correct choice(s) is (are) (a) The current through the circuit, I is 0.3 A.

(b) The current through the circuit, I is 0.3 2 A W resistor = 10 2 V. W resistor = 10 V.

IJII.5

11 3 Rw i + Rw k 4 4 (c) the point P 13 3 Rw i Rw k 4 4 (d) the point P Ê 3ˆ 1 ÁË 3 - 4 ˜¯ Rw i + 4 Rw k

16. regular hexagon of side L and centre O, as shown 1 q , which of K= 4pe 0 L2 the following statement(s) is (are) correct?

O is 6K along OD. (b) The potential at O is zero. (c) The poteltial at all points on the line PR is same. (d) The potential at all points on the line ST is same. 17. Two spherical planets P and Q uniform density r, masses MP and MQ and surface areas A and 4A R also has uniform density r and its mass is (MP P, Q MQ and R are VP, VQ and VR

18.

(a) VQ > VR > VP

(b) VR > VQ > VP

(c) VR /VP = 3

(d) VP /VQ =

19. Two solid cylinders P and Q of same mass and plane from the same height at the same time. Cylinder P has most of its mass concentrated near its surface, while Q has most of its mass concentrated

1 2

of outer radius 3R rolling clockwise without slipping on a horizontal surface with angular speed w and (ii) an inner disc of radius 2R rotating anticlockwise with angular speed w/2. The ring and disc are separated by frictionless ball bearings. The point P on the inner disc is at a distance R from the origin, where OP makes an angle of 30° with the horizontal. Then with respect to the horizontal surface, Rwi (a) the point O (b) the point P

20.

(a) Both cylinders P and Q reach the ground at the same time. (b) Cylinders P has larger linear acceleration than cylinder Q. (c) Both cylinders reach the ground with same translational kinetic energy. (d) Cylinder Q reaches the ground with larger angular speed. along the diameter of a circular wire loop, without touching it, the correct statement(s) is (are) (a) The emf induced in the loop is zero if the current is constant. current is constant. (c) The emf induced in the loop is zero if the current decreases at a steady rate. current decreases at a steady rate.

IJII.6

Answers Section-I 1. (a) 4. (a) 7. (c)

2. (b) 5. (c) 8. (b)

3. (d) 6. (d)

 2. (b) Magnetic Moment M = current ¥ area of the  loop = I A È ˘ a 2 = I ¥ Ía 2 + p ÊÁ ˆ˜ ¥ 2˙ k ¯ Ë 2 Î ˚ p = Ia 2 ÊÁ1 + ˆ˜ k Ë 2¯

Section-II 9. (a) 12. (d)

10. (d) 13. (c)

11. (c) 14. (b)

Section-III 15. (a,c) 18. (a, b)

16. (a, b, c) 19. (d)

3. (d)

 A is along k .

17. (b,d)) 20. (a, c)

Hints and Solutions Section-I 1. (a)

OP =

v r = v P – vQ = v P

vQ )

R ); B = 0 2 Outside the cylinder, (i.e. for r > R), m I B= 0 2p r R In the shaded region, (i.e. for < r < R). From 2 Ampere’s circuital law, B ¥ 2p r = m0I

Thus v r is the resultant of v P and – vQ . vr =

v P2

+

vQ2

+ 2v P vQ cos (180∞ - 2w t )

=

v 2 + v 2 - 2v 2 cos 2wt

=

2v 2 (1 - cos 2wt )

= 4v 2 sin 2 wt fi vr = 2v sin wt = 2Rw sin wt

R , OQ = R, OS = r, 2 r lying between zero and

= m0 JA I is the current density and A is the A area of the shaded region. Now

where J =

R 2 A = p r2 – p ÊÁ ˆ˜ Ë 2¯ ( v = Rw) P and Q is

2pt ˆ \ |vr | = 2Rw |sin wt| = 2Rw sin ÊÁ Ë T ˜¯ Thus, |vr| becomes zero twice in one time period and becomes maximum twice in one time period. So the correct choice is (a).

\ fi

È 2 p R2 ˘ B ¥ 2p r = m0 J Íp r ˙ 4 ˚ Î B=

m0 J È r 2 - R 2 4 ˘ Í ˙ 2 Î r ˚

m0 J È R2 ˘ Ír ˙ 2 Î 4r ˚ Hence the correct graph is (d). =

IJII.7

4. (a) Let Vc = Volume of the shell dc = density of the material of the shell = rc dw where dw is the density of water. Va Vw placed = upthrust, i.e. Ê Vc + Va + Vw ˆ ÁË ˜¯ dwg = Vc dc g 2 fi

If

Now fi

= Vc rc dw g Vw dw g Va Vw = Vc (1 – 2rc 1 = 2Vc ÊÁ - rc ˆ˜ + Va ¯ Ë2 1 rc > , Vw < Va 2

7. (c)

= 25 R = 25 ¥

l 4

L=

l 65.6 –e= –1.2 4 4

Section II 9. (a) In case (a), when the disc-stick system has rotated by half a cycle (or p radian), point B comes to the front and point A goes towards the back as

mC. Hence the disc is rotating about the axis in the w.

6. (d) DQ = nCv DT =2¥

e=

= 16.4 – 1.2 = 15.2 cm

rc <

80 - q q = fiq 2 3

L

Vw dw g

1 , Vw > Va 2 Hence the correct choice is (a). 5. (c) Let q mC be the charge on the upper plate of 3 mF capacitor. Then the charge on the upper plate of 2 m q) mC. Since potential difference across 2 mF capacitor = potential difference across 3 mF capacitors, If

pR < R, it will also land in the 4 unshaded region. 8. (b) End correction e = 0.3d = 0.3 ¥ 4 = 1.2 cm v 336 = = 0.656 m = 65.6 cm l= n 512 For pebble Q,

5R ¥ (35 – 30) 2 ª

In case (b), when the disc-stick system has rotated by p/2, the points A and B p/2

T At t = 8 is x = vt =

has rotated by half a cycle (p radian), A comes to the front and B goes towards the back as shown in Fig. (b). Hence in both cases, the disc is rotating

P w RT 2p RT pR = = ¥ 8 T 8 4

pR where R is the radius of the disc. Since < 4 R , pebble P will land in the unshaded region. 2

the stick. 10. (d) 11. to electron and antineutrino. So all the energy is the kinetic energy of anti-neutrino is zero, the

IJII.8

¥ 106 ¥ 106 eV. 12. (d) If the anti-neutrino has a mass m = 3 eV/c2, mc2 = 3 eV. Therefore, the maximum kinetic energy of the ¥ 106 –3) eV, which is only slightly ¥ 106 eV. The minimum kinetic energy of the electron is still zero. Hence correct choice is (d) 13. n for meta-materials is sin q1 nega sin q 2 q2 Thus if q1 current choice is (c). 14. (b) N =

Since XC = R1, current I1 leads V by 45°. Since XL = R2, current I2 lags behind V by 45°. So the phase difference between I1 and I2 is f 20 V I1 = = (100)2 + (100)2 X c2 + R12 = I2 =

and

= \

I=

v Also frequency n = . Since n remains unl changed, vair v = m lair lm v fi lm = lair ¥ m vair = lair = lair

nair nm

=

lair n

Ê ÁË (

c v = ˆ˜ n¯

X L2 + R22

=

20 2

(50) + (50)2

2 A 5 I12 + I 22 + 2 I1 I 2 cos 90∞ 2

c c fiv= , which is choice (b) v n

v c ¥ m¥ c vair

2 A 10 V

=

Ê 2ˆ Ê 2ˆ ÁË 10 ˜¯ + ÁË 5 ˜¯

=

0.1 = 0.316

2

-

R 1 = I 1R 1 =

2 ¥ 100 = 10 2 V 10

R 2 = I 2R 2 =

2 ¥ 50 = 10 2 V 5

16. (a, b, c)

nm = |n| and nair = 1)

So choice (d) is wrong. Section III 15. (a, c) E1

O due to –q at E directed from O to E O

q at B directed

from O to E. =

1 q 4pe 0 L2

E2

O

q at A directed

from O to D 1 1 XC = = = 100 W 100 ¥ 100 ¥ 10-6 wC XL = wL = 100 ¥ 0.5 = 50 W

O due to –q at D directed from O to D 1 2q = 4pe 0 L2

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O is E1 E = 2E1 = E1 E1 E2 = 2E1 E2 =

E2

1 2q 1 4q + 2 4pe 0 L2 4pe 0 L

6 q along OD 4pe 0 L2 So choice (a) is correct. Net potential at O is q q q -q + + V= 4pe 0 L 4pe 0 L 4pe 0 L 4pe 0 L 2q 2q + =0 4pe 0 L 4pe 0 L So choice (b) is also correct. q at F, –q at E), q at B, –q at C q at A, –2q at D). Line PR is the perpendicular bisector of all the dipoles. Hence the potential at any point on line PR due to each dipole is zero. 17. (b, d) Let M be the mass of planet P and R its radius. Then Mass of P is M = r ¥ 4p R3 3 4 p 3 Mass of Q is r ¥ (2R) M ( surface area 3 of Q = 4 times that of P, therefore radius of Q = 2 (radius of P) = 2R Mass of R = mass of P Q =M M M 1/3 R \ Radius of R P, Q and R are =

VP =

2GM R

VQ =

2G ¥ 8M = 2VP R

VR =

2G ¥ 9M = 91/3 V P 91/3 R

\ Also

VR > VQ > VP VQ

=2 VP So the correct choices are b and d. 18. (a, b)

Fig. 1

O VO = 3 Rw along the x-axis.

 Hence V O = 3 Rwi (see Fig. 1)

Fig. 2

P is  Rw Rw V P = VO i + sin 30∞ i cos 30° k 2 2 = 3Rw i + =

Rw 3 Rw 1 ¥ k ¥ i 2 2 2 2

11 3 Rw i + Rw k 4 4

19. (d) The linear acceleration of a body of mass M and radius R rolling down an inclined plane of inclination q g sinq a= I 1+ MR 2 where I is the moment of intertia of the body about its centre of mass. Therefore, g sinq aP = I 1+ P 2 MR and

aQ =

g sinq IQ 1+ MR 2

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Since cylinder P has most of its mass concentrated near its surface and Q has most of its mass concentrated near its axis, IP > IQ. Hence aP < aQ \ aP < aQ ( a = Ra)

20.

Since w2 = 02 – 2a (q); here q = angular displace1 at2, it follows that wP < wQ. So ment and q = 2 the only correct choice is (d). of the plane of the loop and below the wire into the plane of the loop. Hence the net magnetic

the current is changing or not changing. Hence the correct choices are (a) and (c).

JEE ADVANCED 2013: PAPER-I Instructions Question Paper Format Section 1 contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONLY ONE is correct. Section 2 contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONE or MORE are correct. Section 3 contains 5 questions. The answer to each question is a single-digit integer, ranging from 0 to 9 (both inclusive). Marking Scheme For each question in Section 1, you will be awarded 2 marks if you darken the bubble corresponding to the correct answer and zero mark if no bubbles are darkened. No negative marks will be awarded for incorrect answers in this section. For each question in Section 2, you will be awarded 4 marks if you darken all the bubble(s) corresponding to only the correct answer(s) and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark will be awarded. For each question in Section 3, you will be awarded 4 marks if you darken the bubble corresponding to only the correct answer and zero mark if no bubbles are darkened. In all other cases, minus one (–1) mark will be awarded.

SECTION-1 (Only One Option Correct Type) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONLY ONE is correct. 1. The work done on a particle of mass m by a force, ˘ È y x i + j ˙ (K being a K Í 3/ 2 ˙ Í 2 2 3/ 2 x2 + y 2 ÍÎ x + y ˙˚ constant of appropriate dimensions), when the particle is taken from the point (a, 0) to the point (0, a) along a circular path of radius a about the origin in the x-y plane is:

(

)

(

The time to transport the same amount of heat in Configuration II

)

2K p Kp (b) (a) a a Kp (c) (d) 0 2a 2. Two rectangular blocks, having identical dimen-

the blocks has thermal conductivity K and other 2K. The temperature difference between the ends along takes 9s to transport a certain amount of heat from

Configuration I 2K K

2K

K x

(a) 2.0s (b) 3.0s (c) 4.5s (d) 6.0s 3. Two non-reactive monoatomic ideal gases have their atomic masses in the ratio 2 : 3. The ratio of their partial pressures, when enclosed in a vessel kept at a constant temperature, is 4 : 3. The ratio of their densities is: (a) 1 : 4 (b) 1 : 2 (c) 6 : 9 (d) 8 : 9 4. A particle of mass m is projected from the ground with an initial speed u0 at an angle a with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another

JAI.2 Comprehensive Physics—JEE Advanced

identical particle, which was thrown vertically upward from the ground with the same initial speed u0. The angle that the composite system makes with the horizontal immediately after the collision is: (a)

p 4

(b)

p +a 4

(c)

p -a 4

(d)

p 2

5. A pulse of light of duration 100 ns is absorbed completely by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is 3 ¥ –1 ) of the 108 ms–1 object is (b) 1.0 ¥ 10–17 (a) 0.3 ¥ 10–17 –17 (c) 3.0 ¥ 10 (d) 9.0 ¥ 10–17 6. In the Young’s double slit experiment using a monochromatic light of wavelength l, the path difference (in terms of an integer n) corresponding to any point having half the peak intensity is: l l (b) (2n + 1) (a) (2n + 1) 2 4 l l (c) (2n +1) (d) (2n + 1) 8 16 7. The image of an object, formed by a plano-convex lens at a distance of 8 m behind the lens, is real and is one-third the size of the object. The wavelength 2 times the wavelength in of light inside the lens is 3

free space. The radius of the curved surface of the lens is: (a) 1m (b) 2m (c) 3m (d) 6m 8. 2L and radius 2R is welded to an end of another horizontal thin copper wire of length L and radius R. When the arrangement is stretched by applying forces at two ends, the ratio of the elongation in the thin wire to that in the thick wire is: (a) 0.25 (b) 0.50 (c) 2.00 (d) 4.00 1 9. A ray of light travelling in the direction i + 3 j 2

(

travels along the direction

(

)

)

1  i - 3 j . The angle 2

of incidence is: (a) 30° (b) 45° (c) 60° (d) 75° 10. The diameter of a cylinder is measured using a Vernier callipers with no zero error. It is found that the zero of the Vernier scale lies between 5.10 cm and 5.15 cm of the main scale. The Vernier scale has 50 divisions equivalent to 2.45cm. The 24th division of the Vernier scale exactly coincides with one of the main scale divisions. The diameter of the cylinder is: (a) 5.112 cm (b) 5.124 cm (d) 5.136 cm (d) 5.148 cm

SECTION-2 (One or More Options Correct Type) This section contains 5 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONE or MORE are correct. 11.

lel plate capacitors C1 and C2, each of capacitance C The switch S1 pacitor C1 and then released. Switch S2 is then pressed so that the capaciton C2 is also charged. After some time S2 is released and switclh S3 is pressed. Then (a) the charge on the upper plate of C1 is 2 CV0 (b) the charge on the upper plate of C1 is CV0 (c) the charge on the upper plate of C2 is zero (d) the charge on the upper plate of C2 is – CV0 S1

S2

S3

12. A particle of mass M and positive charge Q. moving with a constant velocity u = 4i ms–1 enters a 1

region the x – y extends from x = 0 to x = L for all values of y, After passing through this rgion, the particle emerges on the other side after 10 milliseconds with a velocity u 2 = 2 3i + j ms -1 . The correct statement(s) is

(

(are) direction.

2V0

C1

C2

V0

direction.

)

JEE Advanced 2013 Paper-I

plete arrangement is placed in a liquid of density 2r and is allowed to reach equilibrium. The correct statement(s) is (are)

50p M 3Q

units.

(a) the net elongation of the spring is

100p M units. 3Q

JAI.3

4p R3 r g 3k

8p R3 r g 3k (c) the light sphere is partially submerged. (d) the light sphere is completely submerged. 15. Two non-conducting solid spheres of radii R and 2R, having uniform volume charge densities r1 and r2 respectively, touch each other. The net electric R from the centre of the smaller sphere, along the line joining the centres of the (b) the net elongation of the spring is

13. tion, y(x, t) = (0.01m) sin [(62.8 m–1)x] cos[628 s–1) t]. Assuming p = 3.14, the correct statements(s) is (are) (a) The number of nodes is 5. (b) The length of the string is 0.25 m. (c) The maximum displacement of the midpoint of the string, from its equilibrium position is 0.01m. (d) The fundamental frequency is 100 Hz 14. A solid sphere of radius R and density r is attached to one end of a mass-less spring of force constant k. The other end of the spring is connected to another solid sphere of radius R and density 3r. the com-

spheres, is zero, the ratio

r1 can be. r2

(a) – 4

(b) -

32 25

(d) 4

(c)

32 25

SECTION-3 (Integer Value Correct Type) This section contains 5 questions. The answer to each question is a single digit integer, ranging from 0 to 9 (both inclusive). 16. A bob of mass m, suspended by a string of length l1, is given a minimum velocity required to complete a full circle in the vertical plane. At the highest point, it collides elastically with another bob of mass m suspended by a string of length l2, which is initially at rest. Both the strings are mass-less and inextensible. If the second bob, after collision acquires the minimum speed required to complete a full circle l in the vertical plane, the ratio 1 is l2 17. A particle of mass 0.2 kg is moving in one dimension under a force that delivers a constant power 0.5 W to the particle. If the initial speed (in ms–1) of the particle is zero, the speed (in ms–1) after 5 s is 18. The work functions of Silver and Sodium are 4.6 and 2.3 eV, respectively. The ratio of the slope of the stopping potential versus frequency plot for Silver to that of Sodium is 19. A freshly prepared sample of a radioisotope of half-life 1386 s has activity 103 disintegrations per second. Given that ln 2 = 0.693, the fraction of the initial number of nuclei (expressed in nearest intepreparation of the sample is

20. A uniform circular disc of mass 50 kg and radius 0.4 m is rotating with an angular velocity of 10 rad s–1 about its own axis, which is vertical. Two uniform circular rings, each of mass 6.25 kg and radius 0.2 m, are gently placed symmetrically on the disc in such a manner that they are touching each other along the axis of the disc and are horizontal. Assume that the friction is large enough such that the rings are at rest relative to the disc and the system rotates about the original axis. The new angular velocity (in rad s–1) of the system is

Answers Section-1 1. 4. 7. 10.

(d) (a) (c) (b)

2. (a) 5. (b) 8. (c)

3. (d) 6. (b) 9. (a)

Section-2 11. (b, c) 14. (a, d)

12. (a, c) 15. (b, d)

13. (b, c)

JAI.4 Comprehensive Physics—JEE Advanced

4. At the highest point A, both the particles have the same potential energy (= mg h). Since the total energy (K.E. + P.E) is conserved, the two particles has equal kinetic energy and hence equal speed at A. Speed of particle 1 at A is ux = u cos a. Hence the speed of 2 at A will be v = u0 cos a ity of the composite mass (since the particles stick after perfectly inelasric collision) will be along the resultant along AC. Therefore,

Section-3 17. 5 20. 8

16. 5 19. 4

18. 1

Hints and Solutions    1. W = Ú F . dr

Ú Fx dx + Ú Fy dy

=

(0, a )

= K

Ú

xdx

(

2

( a , 0) x + y

= -

K 2 x + y2 2

(

)

v = u0 cos a C

(0, a )

)

2 3/ 2

+K

-1 / 2 (0, a )

( a , 0)

-

Ú

ydy

(

2

2 ( a , 0) x + y

K 2 x + y2 2

(

)

)

3/ 2

u0

-1 / 2 (0, a )

( a , 0)

A

q

h

u0

ux = u cos a D

= –0–0 O m1 = m

= 0 2.

Q DT = ; (R = thermal resistance) t R Q t = (i) DT R t = constant. Hence Since Q and DT are the same, R fi

tan q = fi

E Power ¥ Dt = c c

(30 ¥ 10 ) ¥ (100 ¥ 10 ) = -9

3 ¥ 108

l l 3l + = KA 2 KA 2 KA

1 2 KA KA 3KA = + = R2 l l l fi

R2 =

Using these in (i),

l 3KA

t2 9 = fi t2 = 2s l / 3KA 3l / 2 KA PM = n RT fi PM = n r RT r

Since R and T are constant, then for given n PM = constant r

fi

CD v u0 cos a = = =1 AD u x u0 cos a

-3

R1 =

or

B

q = 45°

5. Momentum =

t1 t = 2 R1 R2

3. PV = n RT fi

m2 = m

P1 M1 P M = 2 2 r1 r2 ÊPˆ ÊM ˆ 4 2 8 r1 = Á 1˜ ¥Á 1˜ = ¥ = r2 Ë P2 ¯ Ë M 2 ¯ 3 3 9

= 1.0 ¥ 10–17 kg ms–1 Ê Df ˆ 6. I = 4I0 cos2 Á ˜ ; Df = phase difference Ë 2 ¯ 1 Imax = 2I0 \Imax = 4I0 . For I = 2 Ê Df ˆ 2I0 = 4I0 cos2 Á ˜ Ë 2 ¯ 1 Ê Df ˆ fi cos Á ˜ = ± Ë 2 ¯ 2 Df p 3p 5p p , , .. = (2n + 1) = , 2 4 4 4 4 where n = 0, 1, 2, …

fi

or fi fi

Df = (2n +1)

p 2

2p p D x = (2n + 1) ; Dx = path difference l 2 D x = (2n +1)

l 4

JEE Advanced 2013 Paper-I

7. Give

 1  9. Given a = i + 3 j 2

(

v 1 = – fi u = – 3v = – 3 ¥ 8 = – 24m u 3

A

R1

JAI.5

)

Æ

a

R2 = •

Æ

b

i q

B¢ B A¢

u

v

 1  b = i - 3 j 2  1 1/ 2 Magnitude of a is a = (1 + 3) = 1 2  1 1/ 2 Magnitude of b is b = (1 + 3) = 1 2  If q is the angle between a and b then   a .b cos q = ab 1  1 i + 3 j . i - 3 j 2 2 = 1¥1

1 1 1 = v u f 1 1 1 – = fi f = 6m 8 -24 f If l is the wavelength of light in the lens and l0 in air, then v = nl

fi

c = nl0 l c l0 = fil= 0 m v l

\

m=

Given

2 l0 3 l= . Hence m = = 1.5 3m 2

1 0.5 = fi R1 = 3m 6 R1 FL FL = 8. Y = AD L p r 2 D L fi

FL fi DL = p r 2Y Since Y and F is the same for both wires,

( D L )1 ( D L )2

Ê L ˆ Êr ˆ = Á 1˜ ¥Á 2˜ Ë L2 ¯ Ë r1 ¯

2

Ê 2L ˆ Ê r ˆ = Á ˜ ¥Á ˜ Ë L ¯ Ë 2r ¯ =2 ¥ \

( D L )2 ( D L )1

=2

1 1 = 4 2

)

(

) (

)

1 2 q = 120° =–

Ê 1 1 1ˆ = (m – 1) Á - ˜ f Ë R1 R2 ¯ Ê 1 1ˆ 0.5 = (1.5 – 1) Á - ˜ = R1 Ë R1 • ¯

(

fi

180∞ - q 180∞ - 120∞ = = 30∞ 2 2 10. Least count = value of 1 MSD – value of 1 VSD

\

i=

= 0.05 cm –

2.45 cm 50

= 0.05 – 0.049 = 0.001 cm Diameter = 5.10 cm + 24 ¥ 0.001 cm = 5.10 cm + 0.024 cm = 5.124 cm 11. When only S1 is pressed, capacitor C1 is charged. Charge on the upper plate of C1 is + 2CV0 (see Fig. 1.)

2

2 CV0

C1

(+) (-)

2CV0

Fig. 1. When S1 is released and S2 is pressed, charge will C1 to C2 until they acquire equal charge

JAI.6 Comprehensive Physics—JEE Advanced

because they have the same capacitance (conservation of charge) as shown in Fig. 2.

(+) 2CV0

C1

(-)

C2

(+) (-)

we get 2p = 62.8 l

k = 62.8 fi fi

CV0 = q1

l=

2 ¥ 3.14 = 0.1m 62.8

and w = 628 fi 2pn = 628 fi n = Fig. 2. When S2 is relased and S3 is pressed (see Fig. 3),

C2

q=0

5l = L (L = length of string) 2

V0

the upper plate of C2 acquires a charge q2 = – CV0 so that the net charge on this plate is q = q1 + q2 = CV0 – CV0 = 0 12. From Fleming’s left-hand rule, the direction of the B will be along negative z-direction.

N

A

L=

N

5 ¥ 0.1 = 0.25 m 2

A

N

B

90° q

0

Æ

x=L

u1

x

x

 u1 = 4i ms–1  u 2 = 2 3 i + j ms -1

(

tan q =

2 2 3

1 3

fi q = 300 =

p rad 6

U1

fi10 ¥ 10

A

N

Êpˆ ÁË ˜¯ ¥ M 6 = QB

50 p m B= units 3Q

13. Comparing the given equation with the standing wave equation y(x, t) = 2a sin (kx) cos (wt)

A

A m1g

qM t= QB

–13

N

rVg

)

=

A

N

There is an antinode at the mid-point of the string where y is maximum. From the given equation, ymax = 0.01 m v nl 100 ¥ 0.1 = = Fundemental frequenty n0 = 2L 2L 2 ¥ 0.25 = 20 Hz 4p R3 14. Volume of each sphere is V = 3 Mass of sphere A is m1 = rV and of sphere B is m2 = 3rV (see Fig).

u2

q

x=0

A

L Æ

fi

fi

Æ

y

\

Hz

628 = 100 2 ¥ 3.14

U2

B

B m2g (a)

rVg (b)

Fig (a) The forces acting on m1 are its weight m1g and upthrust U1 = weight of the liquid displaced by it = 2rVg. The forces acting on m2 are its weight m2g and upthrust U2 = 2rVg (Fig. a). Net force acting on

JEE Advanced 2013 Paper-I

A = U1 – m1g = 2rVg – rVg = rVg acting upwords. Net force acting on B = m2g – U2 = 3rVg – 2rVg = rVg acting downwards (Fig. b). so the tension in the spring when equilibrium is attained = rVg. If x is the extension of the spring, then 4p R3 ¥g 3

kx = rVg = r ¥

4p R3 r g 3k In equilibrium, the total force acting on A is FA = kx + W1 = rVg + rVg = 2rVg which is equal to upthrust U1. Hence it is completely submerged. 15. Let P and Q be the two points at a distance 2R on the line joining the centres O1 and O2 of the spheres. fi

x=

Sphere 2 Sphere 1

Q

O2

17. Power P = 0.5 W. Work done in 5 s is (∵ P = constant) W=

Q1 = Q1

R

4pe 0 (O1 P)

r2 R 3e 0

=

4pe 0 (O1Q) 2

+

Q2

4pe 0 (O2Q)2

=0

4p 4p 3 R r1 ( 2 R )3 r 2 3 3 fi + =0 4p e 0 (2 R) 2 4pe 0 (5R) 2 r1 32 =r2 25 16. The minimum velocity which the bob must have is given by fi

m1 =

0

0

1 ¥ 0.2 ¥ v 2 = 2.5 fi v2 = 25 fi v = 5 ms–1 2

= eV0 + W0 fi

W  V0 = ÊÁ ˆ˜ n - 0 Ë e¯ e

 which e is constant independent of work function W0. 19. Half life T is related to disintegration constant l as 0.693 T= l Therfore, the slope of V0 versus n graph is

4p 3 R r1 r R r 3 fi 1 =4 = 2 2 e r 3 4pe 0 ¥ (2 R) 0 2 Q1

t

Ú P dt = P Ú dt = P ¥ t = 0.5 ¥ 5 = 2.5 J

18 From Einstein’s photoelectric equation n = Kmax + W0

4p 3 4p R r1 and Q2 = (2R)3 r2 3 3 P will be zero if 2

t

From work energy theorem, work done = change in kinetic energy. Hence

fi

The charges on spheres 1 and 2 are

fi

Since the bobs are identical and the collision is elastic, their merely exchange their velocities on collision. Hence the second bob will have a velocity u2 = v1 = gl1 . In order to complete the vertical circle, l1 gl1 = 5 gl2 fi =5 l2

P

2R

gl1

v1 =

1 2 mv = W 2

O1

2R

JAI.7

5 gl1

It is then at the lowest point of the circle. From energy conservation, its velocity at the highest point will be

fi

l=

0.693 0.693 = = 5 ¥ 10-4 s -1 T 1386

Now N = N0 e–lt The fraction of nuclei disintegrated in t = 80 s is f= fi

-4 N0 - N N =1= 1 - e- lt = 1 - e-5 ¥ 10 ¥ 80 N0 N0

f = 1 – e–0.04

Using e–a = 1 – a for a << 1, we have f = 1 – (1 – 0.04) = 0.04 Percentage disintegrated = 0.04 ¥ 100= 4% 20. Moment of inertia of the disc before the rings are placed (Fig. a) about O is

JAI.8 Comprehensive Physics—JEE Advanced

w = 10 rad/s

M

R = 0.4 m O

M

I1

I2

m r

m

w¢

() (

=

R 1 MR 2 + 4m 2 2

=

1 MR 2 + mR 2 2

O

2

Fig. (b)

1 MR 2 2 Using parallel-axes theorem, the moment of inertia of each ring about O is (Fig. b) I1 = I2 = mr2 + mr2 = 2mr2 I=

fi

1 MR 2w Iw 2 = w¢ = I¢ 1 ( M + 2m ) R 2 2

Moment of inertia of the system about O after the rings are placed is

=

Mw M + 2m

1 I¢ = I + I1 + I2 = MR 2 + 2mr 2 + 2mr 2 2

=

50 ¥ 10 50 + 2 ¥ 6.25

1 = MR 2 + 4mr 2 2

R 2

)

1 ( M + 2m ) R 2 2 From conservation of angular momentum, we have Iw = I¢w ¢ =

Fig. (a)

\r =

=8

JEE ADVANCED 2013: PAPER-II Instructions Question Paper Format Section 1 8 multiple choice questions ONE or MORE Section 2 4 paragraphs. MORE are correct. Section 3 4 multiple choice questions. ONE or MORE are correct. Marking Scheme Section 1 zero mark Section 2 mark

ONE or

2 marks No negative 4 marks

zero mark

Section 3 zero mark

all

minus one (–1)

4 marks

minus one (–1) mark

SECTION-1 (One or More Options Correct Type) 8 multiple choice questions. ONE or MORE 1. c

-

T -

2. 3h 2p

C

R

100

200

300 T(K)

400

500

9 32R

9 16R

9 5R

4 3R

h

JAII.2

3.

d sin q = l f >f.

d q q

l

q

q

f >f f

f

d d

f

f

d d

7.

4.

M L

R

R

m -

r

r

G m 4 r

-r

R1

GM L m

R2

2

GM L m 2GM L

5.

m

I R R

-

8.

m k

n

-

I P

t

r

u r

R

R

r

R

R

r

R

u

u r

r

t= p

R

6.

m k

u t= w f f V

t=

5p 3

m k

4p 3

m k

JAII.3

SECTION-2 (Paragraph Type) 4 paragraphs

Eight questions one correct answer

Paragraph for Questions 9-10

12.

-

Q g

y R 30°

) Paragraph for Questions 13-14 Q x-y

P

R

Q

Qw 2p

x

O

9.

R w

Q

z B

10 3 10. Q g. 13. Paragraph for Questions 11-12 BR 4 BR

BR 2 BR

14.

-

-g

g BQR g

BQR 2 2

BQR 2 2

g BQR

Paragraph for Questions 15-16 A Z

A Z

11.

Z

W M m

m m

m m

m

M

JAII.4

M¢

( 36 Li) m ( 12 H ) 206 m ( 82 Pb)

m

m ) > M¢. The

210 84

( 42 He) 210 m ( 84 Po)

m

m

70 30

82 34

-

16. 210 84

15.

-

6 3

SECTION-3 (Matching List Type ) 4 multiple choice questions. Each question has matching lists. ONLY ONE 17.

List-I EÆFÆGÆE

EÆFÆHÆE

P F

32P0

List-II

(P)

(1)

15 15 8 O Æ7

(Q) b

(2)

238 92 U

234 Æ90 Th + ....

(R)

(3)

185 83 Bi

Æ184 82 Pb + ....

(S)

(4)

239 94 Pu

Æ140 57 La + ....

R

S

N + ....

Code: P

Q

(a) (b) (c) (d) P0

E V0

19.

G

H VH

V

VG

List I

m

GÆE

(1)

(Q)

GÆH

(2)

P V

(R)

FÆH

(3)

P V

(S)

FÆG

(4)

P V e

Q

R

-

m m m ef eg eh

P V

Code: P

m

e

List

(P)

m

S

f

45°

g

m1

(a) (b) (c) (d)

ei

h

i

m2

18. -

m3

-

JAII.5

List I

List

DT dt

[P] e Æ f

(1) m1 > 2 m 2

[Q] e Æ g

(2) m > m

[R] e Æ h

(3) m = m

[S] e Æ i

(4) m 2 < m1 < 2 m 2 and m 2 > m3

T

m >m

C

Codes: P

Q

R

C

nh L= 2p 3h L= 2p

(a) (b) (c) (d)

List II T ] (1) T ] (2) (3) ] T ] (4)

[P] (Q) [R] [S]

n rn = a0 ¥ rn

Codes: P

Q

R

n2 Z2

a

32 fiZ z2 1 1 1 = 4 R ÊÁ 2 - 2 ˆ˜ Ë l 3 ¯ n a = a0 ¥

S

(a) (b) (c) (d)

\

n n

Answers

\

Section-1 1. 4. 7.

2. 5. 8.

3. 6. 3.

Section-2 9. 12. 15.

10. 13. 16.

11. 14.

17. 20.

18.

Hints and Solutions DQ

fi

fi l1 =

1 1 1 = 4 R ÊÁ - ˆ˜ Ë 22 32 ¯ l2

fi l2 =

9 32 R 9 5R

d=

l 2sin q q

Dd l l cot q = ( -cosec q cot q ) = sin q Dq 2

19. fi

Dd = -

l cot q Dq sin q

m DT

DQ = mCDT DQ DT = mC Dt dt

1 1 1 = 4 R ÊÁ 2 - 2 ˆ˜ l1 Ë1 3 ¯

q= l

d

Section-3

1.

n

Ê 1 1 1ˆ = RZ 2 Á 2 - 2 ˜ l Ë n2 n1 ¯

List I

T

2.

S

20.

T

Dd d

Dd d Dq

q) Dq

JAII.6

4.

m

R

P

r1 O1

1 2 mu 2

r2

u m) A P

O2

r

P

B

2

2GmM GM 2 GmM GmM GM =(2 L) 2L L L L m

4p 3 Q= R1 r 3

\  Er

A

4p 3 R2 r 3

P   r r1 Q r1 = = 4pe 0 R13 3e 0

1 Ep

L

G

r  r r2 = 3 e0

O

P -

P

u= 2

∵Q

GM L -

P

O. 8.

P

 1 rr p     E = E p + E p = 3e r 1 + r 2 = 3e 0 0    r = O1 O2 E

(

B

1 2 2GMm mu = 2 L

Q

Q

\

P

GMm L

R Q = -

A

B

u0

)

wall 0.5u0

A

O

-

5. O 6.

u

wind

w

t

Source

Observer

V + w + uˆ È (V + w) - (-u ) ˘ f =Í f1 = ÊÁ ˜ f1 fi f 2 > f1 ˙ Ë V + w - u¯ Î (V + w) - (+u ) ˚ È (V - w) - (-u ) ˘ ÊV - w + uˆ f =Í ˙ f1 = ÁË V - w - u ˜¯ f1 fi f 2 > f1 V w ( + u ) Î ˚ 7.

wt

x =a

u

V=

dx = Aw dt

wt) = u

m k fi Time period T = 2p k m

w= V

u t

u =u wt =

wt )

p 1 fi w t1 = 2 3

u A

B

P

M

m L

fi t1 =

p 3

m k -

O M

wt

L

t = 2t1 =

2p 3

m k

JAII.7

-

t = t2 + =

12.

Vp Vs

2p 3

m 2p + k 4

m 7p = k 6

m k

Np Ns

T 2

2p 3

\

m m 5p +p = k k 3

9.

O

m k

=

=

dB B-0 df = p R2 ¥ = p R2 B = p R2 1 dt dt

Q

E= 14.

BR 2

g¥

M = gL ;

g

DM = g D L

v m O N

DL fi DL = t D t Dt t = RF = RQE ∵ F = QE)

P

30°

R/2

40000 = 200 200    f = Ú B.ds = B ¥ p R 2

Vs

  df = p R2 B E ◊ dl = Ú dt E ¥ pR = pR B

40 1 = ¥ 1 ¥ v 2 + 150 2 2

fi

Vp

E

R 1 mg ¥ = mv 2 + 150 2 2 1 ¥ 10 ¥

fi V p = 40000 V

Vs

13.

t = t2 +

fi

Ns 4000 1 = Vp 10

O

fi

Np

T 4

-

=

=

t=

R R

Q

\

DL = RQE D t

mg cos 60° v

10.

DM = g R Q E D t

N

Q mv R

fi fi

E=

2

=N

mg

R P = VI I=

W

¥

Ê Q B R2 ˆ gÁ Ë 2 ˜¯

P¢ = I R

Ê Q B R2 ˆ DM = -g Á Ë 2 ˜¯

W

P 600 ¥ 103 = = 150A 4000 V

=

BR ¥1= 2

z

N

11.

Dt

DM = g R Q ¥

1 1 ¥ (10) 2 = N - 1 ¥ 10 ¥ 2 40

fi

BR 2

15. ¥

180 kW p¢ ¥ 100 = ¥ 100 = 30% 600 kW p

6 3 Li 6 3 Li

Æ 42 He + 12 H

4 2 He

+ 12 H

6 3 Li .

JAII.8

Ê Vo ˆ ÁË V ˜¯ H 70 30 Zn

=

1 32

5 3

82 34 Se

Ê Vo ˆ ÁË V ˜¯ H

16. 210 206 4 84 Po Æ 82 Pb + 2 He

fi

+Q

5/3

=

Vo

WGH

Po

Vo Vo

Vo

Po Vo

Q

¥

Q

1 1 = 5 32 2

VH

Dm \

g

For item (R) in List I FÆH f WFH = nR (TF - TH ) 2 f

Pa = PPo 2ma Ka =

2mPo K Po

ma Ka = m

3 nR (TF - TH ) 2 3 = ( PF VF - PH VH ) 2 3 = (32 PoVo - 8PoVo ) 2

K

=

mPo Q 206 ¥ 542.2 keV =  5319 keV mPo + ma 206 + 4

Ka = 17.

G Æ E

G Æ H

FÆE

FÆH FÆG FÆH

PoVo R

FÆG

For item (S) in List I

For item (P) in List I

FÆG

GÆE P0 VG VE

WGE

ÊV ˆ WFG = PV ln Á 2 ˜ Ë V1 ¯

P0 VGE VO

FÆG

ÊV ˆ = 32 PoVo ln Á G ˜ Ë VF ¯

PF VF = PG VG fi

Po Vo = Po VG

fi

VG

∵ PV = nRT)

Ê 32Vo ˆ = 32 PoVo ln Á Ë Vo ˜¯

Vo

= 32 PoVo ln (32) \

WGE

Po

V

V

= 160 PoVo ln ( 2)

Po Vo S

For Item (Q) in List I WGH

P VG VH

Po

Vo VH )

P) Q) R) S)

____

FÆH PV

g

PF Vog = Po VHg fi

g

Po Vo = Po VH

g

18. Z

Æ Æ Æ Æ A)

JAII.9

238 92 U

b

b

( 0+1 e)

m

r

234 Æ 90 Th + 42 He

For ray e A B i A 1 ˆ = sin -1 ÊÁ Ë 2 ˜¯

b

n

b

15 15 8 OÆ 7 N

239 94 Pu

Æ140 57

185 83 Bi

+

0 +1e

La +

eAEh

B

ic =

b

2 m

ic =

m2 m1

Êm ˆ sin - 1Á 2 ˜ < 45∞ fi m1 > 2 m 2 . Ë m1 ¯

+n

P) Æ

99 37 X

Q) Æ

R) Æ

-

S) Æ

20.

q 1 2E E = kq or k = 2 q

1 Æ 184 82 Pb + 1p

k Æ

Æ

Æ

p) For ray e A C f

A m >m

Æ

\

19.

m

C m

k= =

AC

dimensions of E dimension of q

[ML2 T -2 ] = [ML2 T -2 K -1 ] [K ]

Q

v

r h 45°

e

C A

D

F

phrv

g

E

m1 i

f

h

\

h=

B

= m3

m2

h=

F 6p r v

dimension of F dimension of r ¥ dimension of v [MLT -2 ] = [ML-1T -1 ] [L ¥ LT -1 ]

E = hn For ray e A D g A

m =m

For ray e A E h m

m

m

m

m 2

A E

m m1

[h] =

A r

= m 2 sin r fi m1 = 2 m 2 sin r

[ E ] [ML2 T -2 ] = = [ML2 T -1 ] n T -1 Q t A L q q

Q kA(q1 - q 2 ) ; = t L

q >q )

k

[ML2 T -2 ] [k ] ¥ [L2 ] ¥ [K ] = fi [k ] = [MLT -3 K -1 ] [T ] [ L]

Physics JEE Advanced 2014–Paper-I  PI.1

Physics Jee Advanced—2014 Paper-I—Model Solutions SECTION I (One or More than One Options Correct Type) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONE or MORE THAN ONE are correct. 1. A student is performing an experiment using a resonance column and a tuning fork of frequency 244 s–1. He is told that the air in the tube has been replaced by another gas (assume that the column remains filled with the gas). If the minimum height at which resonance occurs is (0.350 ± 0.005)m, the gas in the tube is (Useful information: 1/2

–1/2

1/2

–l/2

10 for each gas as M

10 7ˆ = ˜¯ 20 10



Ê (a) Neon Á M = 20, Ë



Ê (b) Nitrogen Á M = 28, Ë

10 3 ˆ = ˜ 28 5 ¯



Ê (c) Oxygen Á M = 32, Ë

10 9 ˆ = ˜ 32 16 ¯



Ê (d) Argon Á M = 36, Ë

10 17 ˆ = ˜ 36 32 ¯

2. At time t = 0, terminal A in the circuit shown in the figure is connected to B by a key and an alternating current I(t) = I0 cos (w t), with I0 = 1A and w = 500 rad/s starts flowing in it with the initial 7p direction shown in the figure. At t = , the key 6w is switched from B to D. Now onwards only A and D are connected. A total charge Q flows from the battery to charge the capacitor fully. If C = 20 mF, R = 10W and the battery is ideal with emf of 50 V, identify the correct statement(s).

D A C = 20 mF

167 RT

= 640 J mole ; 140 RT = 590 J mole . The molar masses M in grams are given in the options. Take the values of given there.)

B

50 V

R = 10 W



(a) Magnitude of the maximum charge on the 7p capacitor before t = is 1 ¥ 10–3 C. 6w (b) The current in the left part of the circuit just 7p before t = is clockwise. 6w (c) Immediately after A is connected to D, the current in R is 10  A. (d) Q = 2 ¥ 10–3 C. 3. A parallel plate capacitor has a dielectric slab of dielectric constant K between its plates that covers 1/3 of the area of its plates, as shown in the figure. Q1 E1 The total capacitance of the capacitor is C while that of Q2 E2 the portion with dielectric in between is C1. When the capacitor is charged, the plate area covered by the dielectric gets charge Q1, and the rest of the area gets charge Q2. The electric field in the dielectric is E1 and that in the other portion is E2. Choose the correct option/options, ignoring edge effects. E1 E1 1 (a) = 1 (b) = E2 K E2 Q1 3 C 2+ K (c) = (d) = Q2 K C1 K

PI.2  Comprehensive Physics—JEE Advanced

4. One end of a taut string of length 3m along the x axis is fixed at x = 0. The speed of the waves in the string is 100 ms–1. The other end of the string is vibrating in the y direction so that stationary waves are set up in the string. The possible waveform(s) of these stationary waves is (are) (a) y(t) = A sin

50p t px cos 6 3

(b) y(t) = A sin

100p t px cos 3 3

(c) y(t) = A sin

5p x 250p t cos 6 3

(d) y(t) = A sin

5p x cos 250p t 2

5. A transparent thin film of uniform thickness and refractive index n1 = 1.4 is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index. n2 = 1.5, as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance f1 from the film, while rays of light traversing from glass to air get focused at distance f2 from the film. Then n1

Air



(a) | f1 | = 3R (c) | f2 | = 2R

n2

(b) | f1 | = 2.8R (d) | f2 | = 1.4R

6. Heater of an electric kettle is made of a wire of length L and diameter d. It takes, 4 minutes to raise the temperature of 0.5 kg water by 40 K. This heater is replaced by a new heater having two wires of the same material, each of length L and diameter 2 d. The way these wires are connected is given in the options. How much time in minutes will it take to raise the temperature of the same amount of water by 40K ? (a) 4 if wires are in parallel (b) 2 if wires are in series (c) 1 if wires are in series (d) 0.5 if wires are in parallel. 7. Two ideal batteries of emf V1 and V2 and three resistances R1, R2 and R3 are connected as shown in the figure. The current in resistance R2 would be zero if

V1

R1 R2

V2 R3

(a) V1 = V2 and R1 = R2 = R3 (b) V1 = V2 and R1 = 2R2 = R3 (c) V1 = 2V2 and 2R1 = 2R2 = R3 (d) 2V1 = V2 and 2R1 = R2 = R3 8. Let E1(r), E2(r) and E3(r) be the respective electric fields at a distance r from a point charge Q, an infinitely long wire with constant linear charge density l, and an infinite plane with uniform surface charge density s. If E1(r0) = E2(r0) = E3(r0) at a given distance r0, then (a) Q = 4spr02 l (b) r0 = 2ps (c) E1(r0 /2) = 2E3(r0 /2) (d) E2(r0 /2) = 4E3(r0 /2) 9. A light source, which emits two wavelengths l1 = 400 nm and l2 = 600 nm, is used in a Young’s double slit experiment. If recorded fringe widths for l1 and l2 are b1 and b2 and the number of fringes for them within a distance y on one side of the central maximum are m1 and m2, respectively, then (a) b2 > b1 (a) m1 > m2 (c) From the central maximum, 3rd maximum of l2 overlaps with 5th minimum of l1. (d) The angular separation of fringes for l1, is greater than l2 10. In the figure a ladder of mass m is shown leaning against a wall. It is in static equilibrium making an angle q with the horizontal floor. The coefficient of friction between the wall and the ladder is m1 and that between the floor and the ladder is m2. The normal reaction of the wall on the ladder is N1 and that of the floor is N2. If the ladder is about to slip, then m1

q m2

Physics JEE Advanced 2014–Paper-I  PI.3

mg 2 mg (b) m1 π 0 m2 = 0 and N1 tan q = 2 (a) m1 = 0 m2 π 0 and N2 tan q =

(c) m1 π 0 m2 π 0 and N2 =

mg 1 + m1m 2

(d) m1 = 0 m2 π 0 and N1 tan q =

mg 2

SECTION II (One Integer Value Correct Type) This section contains 10 questions. Each question, when worked out will result in one integer from 0 to 9 (both inclusive). 11. During Searle’s experiment, zero of the Vernier scale lies between 3.20 ¥ 10–2 m and 3.25 ¥ 10–2 m of the main scale. The 20th division of the Vernier scale exactly coincides with one of the main scale divisions. When an additional load of 2 kg is applied to the wire, the zero of the Vernier scale still lies between 3.20 ¥ 10–2 m and 3.25 ¥ 10–2 m of the main scale but now the 45th division of Vernier scale coincides with one of the main scale divisions. The length of the thin metallic wire is 2 m and its cross-sectional area is 8 ¥ 10–7 m2. The least count of the Vernier scale is 1.0 ¥ 10–5 m. The maximum percentage error in the Young’s modulus of the wire is 12. Airplanes A and B are flying with constant velocity in the same vertical plane at. angles 30° and 60° with respect to the horizontal respectively as shown in the figure. The speed of A is 100 3 ms–1. At time t = 0 an observer in A finds B at a distance of 500 m. This observer sees B moving with a constant velocity perpendicular to the line of motion of A. If at t = t0, A just escapes being hit by B, t0 in seconds is A B

30°

60°

13. A thermodynamic system is taken from an initial state i with internal energy U1 = 100 J to the final state f along two different paths iaf and ibf as schematically shown in the figure. The work done by the system along the paths af, ib and bf are Waf = 200 J, Wib = 50 J and Wbf = 100 J respectively. The heat supplied to the system along the path

iaf, ib and bf are Qiaf, Qib and Qbf respectively. If the internal energy of the system in the state b is Ub = 200 J and Qiaf = 500 J, the ratio Qbf/ Qib is a

f

P b

i V

14. Two parallel wires in the plane of the paper are distance X0 apart. A point charge is moving with speed u between the wires in the same plane at a distance X1 from one of the wires. When the wires carry current of magnitude I in the same direction, the radius of curvature of the path of the point charge is R1. In contrast, if the currents I in the two wires have directions opposite to each other, the radius of curvature of the path is X R R2. If 0 = 3 the value of 1 is R2 X1 15. To find the distance d over which a signal can be seen clearly in foggy conditions, a railways engineer uses dimensional analysis and assumes that the distance depends on the mass density r of the fog, intensity (power/area) S of the light from the signal and its frequency f. The engineer finds that d is proportional to S1/n.The value of n is 16. A rocket is moving in a gravity free space with a constant acceleration of 2 ms–2 along +x direction (see figure). The length of a chamber inside the rocket is 4 m. A ball is thrown from the left end of the chamber in +x direction with a speed of 0.3 ms–1 relative to the rocket. At the same time, another ball is thrown in –x direction with

PI.4  Comprehensive Physics—JEE Advanced

a speed of 0.2 ms–1 from its right end relative to the rocket. The time in seconds when the two balls hit each other is

of mass 1 kg is pulled along the rail from P to Q with a force of 18 N, which is always parallel to line PQ (see the figure given). Assuming no frictional losses, the kinetic energy of the block when it reaches Q is (n ¥ 10) Joules. The value of n is (take acceleration due to gravity = 10 ms–2)

a = 2 ms-2 0.3 ms-1

0.2 ms-1

x

Q 4m

17. A galvanometer gives full scale deflection with 0.006 A current. By connecting it to a 4990 W resistance, it can be converted into a voltmeter of 2n range 0 – 30 V. If connected to a W resistance, 249 it becomes an ammeter of range 0 – 1.5 A. The Value of n is 18. A uniform circular disc of mass 1.5 kg and radius 0.5 m is initially at rest on a horizontal frictionless surface. Three forces of equal magnitude F = 0.5 N are applied simultaneously along the three sides of an equilateral triangle XYZ with its vertices on the perimeter of the disc (see figure). One second after applying the forces, the angular speed of the disc in rad s–1 is

90° O

X

O Z

3m

P

Answers Section-I

1. (d) 4. (a, c, d) 7. (a, b, d) 10. (c, d)

2. (c, d) 5. (a, c) 8. (c)

3. (a, d) 6. (b, d) 9. (a, b, c)

Section-II

F

Y

4m

11. 4 14. 3 17. 5 20. 5

12. 5 15. 3 18. 2

13. 2 16. 2 19. 4

F

Hints and Solutions

F

19. A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to rotate about its axis. Two massless spring toy-guns, each carrying a steel ball of mass 0.05 kg are attached to the platform at a distance 0.25 m from the centre on its either sides along its diameter (see figure). Each gun simultaneously fires the balls horizontally and perpendicular to the diameter in opposite directions. After leaving the platform the balls have horizontal speed of 9 ms–1 with respect to the ground. The rotational speed of the platform in rad s–1 after the balls leave the platform is

20. Consider an elliptically shaped rail PQ in the vertical plane with OP = 3 m and OQ = 4 m. A block

Section-I g RT . For a closed pipe, l M = 4L (fundamental mode). Also v = nl. Thus v = n ¥ 4L 1. Speed of sound v =



fi



fi

g RT = v ¥ 4L M L =

1 g RT (1) 4n M

For Neon: M = 20 g = 20 ¥ 10–3 kg = 2 ¥ 10–2 kg. Neon is a monatomic gas for which g = 1.67. Substituting the given values in (1), we have [since 10 1 7˘ 167RT = 640 J1/2 mol–1/2 and = = 20 2 10 ˙˚

L =

1 167 RT 4 ¥ 244 2 ¥ 10-2

Physics JEE Advanced 2014–Paper-I  PI.5

1 167 RT 4 ¥ 244 2 1 7 ¥ 640 ¥ = 4 ¥ 244 10 = 0.459 m =

For Nitrogen: M = 28 g = 2.8 ¥ 10–2 kg. Since nitrogen is a diatomic gas, g = 1.4. Given 140RT 10 3 = 590 J1/2 mol–1/2 and = . 28 5

\

L =

=

1 1.4 RT 4 ¥ 244 2.8 ¥ 10-2 1 140 RT ¥ 10 ¥ 4 ¥ 244 28

1 3 ¥ 590 ¥ 4 ¥ 244 5 = 0.363 m Similarly, for oxygen (g = 1.4), we get L = 0.340 m and for argon (g = 1.67), we get L = 0.348 m. It is given that L = (0.350 ± 0.005) m. Hence correct choice is (d) 2. Given I = I0 cos (w t) where I0 = 1A and w = 500 rad s–1. Since the voltage lags behind the current, the voltage across the capacitor varies with t as =



(

V = V0 cos w t -

p 2



fi

dq = I0 cos (w t) dt dq = I0 cos (w t)dt



fi

Ú dq = I0 Ú cos (w t)dt



fi

Now

q =

)

I0 w \ Magnitude of maximum charge on the capacitor 7p before t = (i.e., before A is connected to D) is 6w I0 1A Q = w = = 2 ¥ 10–3 C 500 s -1 So choice (a) is incorrect.

fi

q = Q sin w t, where Q =

Current in the left past of the circuit just before 7p t= is 6w 7p ˆ Ê I = I 0 cos Ëw ¥ 6w ¯

7p 6

p 3 =A 6 2 The negative sign shows that the current is anticlockwise. So choice (b) is also wrong. After A and D are connected, the battery of voltage V = 50V is in parallel with the capacitor, So the voltage VC across the capacitor is also 50 V. Hence the current in the circuit is V + V 50 + 50 I = c = = 10A R 10 So choice (c) is correct. Maximum charge Q = CVC = (20 ¥ 10–6) ¥ 50 = 2 ¥ 10–3C So choice (d) is also correct. 3. The capacitor is charged by connecting its plates to a battery of voltage, say, V. Since the battery is kept connected, the voltage between the plates is V equal to V. Now E = . Since d is not changed, d V E1 = E2 = . d = - cos

E1

E2

= V0 sin w t

I0  sin(w t) w

( ) ()

= 1 ¥ cos

K

A/3

Air

2A/3

d

E1 = 1 E2 Capacitance of dielectric filled part is A K Œ0 3 = K Œ0 A C1 = 3d d Capacitance of the air filled part is 2A Œ0 3 = 2 Œ0 A C2 = d 3 d Since the two capacitors are in parallel, the total capacitance is C = C1 + C2

\

Ê C ˆ = C1 Á1 + 2 ˜ Ë C1 ¯ 2 Aˆ Ê Œ0 d ˜ = C1 Á1 + 3 Á A˜ K Œ0 ÁË ˜ 3d ¯

PI.6  Comprehensive Physics—JEE Advanced

( )

= C1 1 + C K +2 = C1 K

fi



2 K



fi



fi

According to sign convention, f1 and R are both positive. Case (b) For refraction from glass to air

Now Q1 = C1V,  Q2 = C2V Q1 C K = 1 = Q2 C2 2 So the correct choices are (a) and (d). \



n1 = 1.4

A

n1 = 1.4

Second face n2 = 1.5

B

Air (n = 1)

v1

O u

I1

I¢

v¢

The ray OA from an object O, refracts along AB forming the image I ¢. For refraction at first face, we have n1 n n -n - = 1 (1) R v¢ u The ray AB suffers refraction at the second face, forming the final image I1. For refraction at second face, I ¢ serves at the virtual object. Thus, we have n2 n1 - = n2 - n1 (2) v1 v ¢ R Adding (1) and (2), we get



n2 n n - n n2 - n1 - = 1 + v1 u R R

By difinition v1 = f1 when u = •. Therefore

n2 n - n n2 - n1 = 1 + f1 R R

Second face

First face

4. Since the end at x = 0 of the string is fixed, it is a node, i.e. y (t) = 0 at x = 0. This condition is satisfied in all the four choices given in the question. The other end at x = 3m is free, it is an antinode, i.e. y (t) π 0 at x = 3 m. This condition is satisfied in choices (a), (c) and (d) but not in choice (b) for which y (t) = 0 at x = 3 m. Hence the correct choices are (a), (c) and (d). 5. The incident ray suffers two refractions — one at each face. Case (a) For refraction from air to glass First face

1.5 1.4 - 1 1.5 - 1.4 = + f1 R R f1 = 3R

n2 = 1.5 A Air (n = 1)

B

O u

I2

I¢

v2 v¢

For refraction second face, we have n n n -n 1 - 2 = 1 2 (3) R v¢ u For refraction at first face, we have n n1 - = n - n1 (4) v2 v ¢ R Adding (3) and (4), we get n n2 n -n n - n1 = 1 2 + v2 u R R Now v2 = –f2 when u = –•. Also R = –R. Therefore 1 1.4 - 1.5 1 - 1.4 = + - f2 -R -R fi f2 = –2R \ |  f2 | = 2R Hence the correct choices are (a) and (c). 6. Let R be the resistance of the wire of the kettle and let V be the voltage of the supply. The heat energy consumed in time t is V 2t H = (1) R If the heater is replaced by a new heater having two wires of the same material, the same length but twice the diameter, the resistance of each wire R becomes . 4 If the two wires are connected in series, the total R R R resistance of the heater is R1 = + = . If t1 4 4 2 is the time required to consume the same heat energy, then

Physics JEE Advanced 2014–Paper-I  PI.7

V 2t1 V 2t1 = (2) R1 R/2 Equating (1) and (2),

i.e. R2 can have any value. Choices (a), (b) and (d) satisfy condition (3) but choice (c) does not.

H =



V 2t1 V 2t = R/2 R 1 4 min fi t1 = t ¥ = = 2 min 2 2 So choice (b) is correct. If the two wires are connected in parallel, the total R resistance R2 = . If t2 is the required time, 8

V 2 t2 V 2t = R /8 R t 4 min = 0.5 min , which is fi t2 = = 8 8 choice (d). So the correct choices are (b) and (d). 7. The following figure shows the currents I1, I2 and I3 in the three branches of the circuit. I1

a

b

V1

R1

I1

I2

R2 c

F I3

V2 R3 e

I3 d

Using the junction rule, I2 + I3 = I1. If I2 = 0, then I 1 = I 3. Applying the loop rule to loop abcfa, we have I 1 R 1+ I 2 R 2 – V 1 = 0 Setting I2 = 0, we get V1 = I1R1 (1) Applying the loop rule to loop fcdef, we get –I2 R2 – V2 + I3 R3 = 0 Putting I2 = 0, we get V2 = I3R3 But I3 = I1. Therefore V2 = I1R3 (2) From (1) and (2), we get V1 R1 = (3) V2 R3 So no current will flow through R2 if condition (3) holds. This condition is does not contion R2,

8. E1 (r0) =

Q 4p Œ0 r02

E2 (r0) =

l 2p Œ0 r0

E3 (r0) =

s 2 Œ0

At r =

r0 , 2

Êr ˆ E1 Ë 0 ¯ = 2 Êr ˆ E2 Ë 0 ¯ = 2

Q 4p Œ0 l 2p Œ0

=

() r0 2

2

() r0 2

=

Q = 4E1 (r0) p Œ0 r02

l = 2E2 (r0) p Œ0 r0

s Êr ˆ = E3 (r0 ) E3 Ë 0 ¯ = 2 Œ0 2 Given E1 (r0) = E2 (r0) = E3 (r0). It follows from the above equations that the only correct choice is (c). lD 9. Fringe width b = . Since l2 > l1, b2 > b1. d Distance of mth bright fringe from the central maximum is ml D ym = d

\

ym1 = m1l1 D1 d

ym2 = m2 l2 D2 d Since ym1 = ym2 and l2 > l1, it follows that m1 > m2. and

Distance of mth dark fringe from the central maximum is 1 lD ym* = m 2 d

( )

Now y3 (for l2) =

3l2 D d

( )

1 l1 D 9l1 D = 2 d 2d Putting l1 = 400 nm and l2 = 600 nm, we find that and y5*  (for l1) = 5 -



y3 =

3 ¥ (600 nm ) ¥ D 1800 D = nm d d

PI.8  Comprehensive Physics—JEE Advanced

9 ¥ ( 400 nm ) ¥ D 1800 D and = nm 2d d Finally, angular separation q of conscentive fringes is given by l l tan q = fi q = (Q q is small) d d Since l2 > l1;  q2 > q1. Hence the correct choices are (a), (b) and (c). 10. The ladder AB of length l is in contact with the wall at A where the normal reaction is N1 and with the floor at B where the normal reaction is N2. Let f1 and f2 the frictional force at A and B. Then



y5* =



f1 = m1N1 and f2 = m2N2

The ladder is in translational as well as rotational equilibrium. f1

N1

A

Using (3) in (5) we get



Choice a: m1 = 0, m2 π 0.



From (2),

N2 = mg – m1N1



fi

N2 = mg (Q m1 = 0)

So

Choice (b): m1 π 0, m2 = 0. In this case, N2 = mg



Choice (c): m1 π 0, m2 π 0. Then from (4)



\

C

translational equilibruim f2 – N1 = 0 fi N1 = f2 = m2 N2(1) and f1 + N2 = mg fi N2 = mg – m1 N1  Solving (1) and (2) we get

(2)

m 2 mg (3) 1 + m1m 2



N1 =

and

mg N2 = (4) 1 + m1m 2

For

rotational equilibruim Total anticlockwise torque about B = total clockwise torque about B. i.e. N1 ¥ AC + f1 ¥ BC = mg ¥ BD l fi N1l sin q + m1N1 l cos q = mg ¥ cosq 2 mg fi N1 tan q + m1N1 = 2 fi N1 (m1 + tan q) =

N1 tan q =

m 2 mg mg = 2m2 2

Section-II

For



mg 1 + m1m 2

which agrees with (5) if m1 = 0. Hence the corrent choices are (c) and (d).

q f2



N2 =

Choice (d): m1 = 0, m2 π 0. Then from (3) and (6), we get 1 N1 = m2 mg and tan q = 2m2

mg



N2 tan q = mg tan q



N2

D

1 - m1m 2 (6) 2m2





C

B

tan q =

mg (5) 2

11. Let x metre be the main scale reading and VC be the vernier constant. First measurement of the change in length of the wire is l1 = x + 20 ¥ VC Second measurement of the change in length of the wire is l2 = x + 45 ¥ VC \ Change in the length is l = l2 – l1 = x + 45 ¥ VC – (x + 20 ¥ VC)

= 25 ¥ VC = 25 ¥ 1.0 ¥ 10–5m In the given experiment, only the change in the lengths is measured. Now

\

Y =

FL lA

DY Dl = Y l

The maximum error (Dl) in the measurement of change in length = one VC = 1.0 ¥ 10–5m. Hence, the maximum percentage error in Y is

Physics JEE Advanced 2014–Paper-I  PI.9



In process ibf, Wibf = Wib + Wbf = 50 J + 100 J = 150 J

DY Dl ¥ 100 = ¥ 100 Y l

1.0 ¥ 10-5 ¥ 100 = 4% = 25 ¥ 1.0 ¥ 10-5

(DU)ibf = Qibf – Wibf

= (Qib + Qbf) – (Wib + Wbf)

Æ Æ Æ 12. Since ÊÁ V B - V A ˆ˜ is perpendicular to V A ¯ Ë



= (Qib + Qbf) – (50 J + 100 J)

Æ

Æ

Æ

ˆ Ê ÁVB - VA ˜ ◊ VA = 0 ¯ Ë

fi

VB VA cos q – VA2 = 0



fi

VB cos q = VA

But (DU)ibf = (DU)iaf = 300 J Qib + Qbf = 300 J + 150 J = 450 J

\



Also Qbf – Qib= 150 J

Æ

Æ

where q is the angle between VA and VB . Given that q = 60° – 30° = 30°. Therefore VB cos 30° = VA VB =

Qbf



Qib

t0 =

relative distance between A and B at t0 relative velocity between A and B at t0

fi

t0 =

S AB S A - S B = v AB v A - v B

Wire 2

Æ

cosq

(100 3 ) 2 + (200) 2 = -2 ¥ 100 3 ¥ 200 ¥ cos 30∞ = 100 ms \

S AB 500 m = = 5 ms -1 t0 = v AB 100 ms -1

50 J, Wbf = 100 J, and Qiaf = 500 J. In Process iaf, Wiaf = Wia + Wa f = 0 + 200 J = 200 J

(DU)iaf = Qiaf – Wiaf = 500 J – 200 J = 300 J

\

Wire 1

Wire 2

r2

Uf = (DU)iaf + Ui = 300 J + 100 J = 400 J

I

I r1

X0 case 1

P

r2

X0 case 2

X0 2X0 . Therefore r2 = . 3 3

Case 1: Magnetic field at point P due to current I in wire 1 is

m0 I 3m0 I = directed into the page. 2p r1 2p X 0

Magnetic field at P due to current I in wire 2 is

–1

13. Given Ui = 100 J, Ub = 200 J, Wa f = 200 J, Wib =



P

Given r1 =

Æ

vAB = v A - v B

= v 2A + vB2 - 2v A vB

I r1

At time t = t0, A just escapes being hit by B, therefore, SAB = 500 m – 0 = 500 m.



300J =2 150J

VA 100 3 = = 200 ms -1 cos 30∞ 3 2





=

14.

I



(2)

From (1) and (2), Qbf = 300 J and Qib = 150 J. Therefore

Wire 1

fi

(1)



m0 I 3m0 I = directed out of the page 2p r2 4p X 0

\  Net magnetic field at P is

B1 =

3m0 I 3m0 I 3m0 I = (1) 2p X 0 4p X 0 4p X 0

Case 2: Magnetic field at P due to current I in wire 1 is

m0 I 3m0 I = directed into the page. 2p r1 2p X 0

PI.10  Comprehensive Physics—JEE Advanced

Magnetic field at P due to current I in wire 2 is m0 I 3m0 I = directed into the page. 2p r2 4p X 0



\  Net megnetic field at P is B2 =



3m0 I 3m0 I 9 m0 I + = (2) 2p X 0 4p X 0 4p X 0

As B1 and B2 depend on the values of r1 and r2 it is clear that B1 and B2 are not uniform in the region between the wires. Hence the trajectories of the charged particle are not circular. But the radius of curvature of the trajectory at point P is inversely proportional to magnetic field at that point. Thus R1 B2 = = 3 [use (1) and (2)] R2 B1



a

b

c

15. Let d = k r s f   where k is a dimensionless constant. [  r ] = [ ML–3 ] -3

2

power [ML T ] [  s ] = = = [MT -3 ] area L2 [  f  ] = [ T–1 ]



The second ball thrown from the right with a velocity u2 = 0.2 ms–1 will suffer a displacement s = –4 + 0.0225 = –3.997 m to meet the first ball. Since its acceleration is –2 ms–2 –3.997 = -0.02t -



fi  t2 + 0.2t – 3.997 = 0 _ 4 m, we have Since  3.997 m ~ t2 + 0.2 t – 4 = 0 The positive root of this equation is slightly greater than 1.9s. Thus t is approximately equal to 2s. 17. For voltmeter, V R+G



Ig =

fiG=

V 30 -R= - 4990 = 10 W Ig 0.006

For ammeter S =



IgG I - Ig



=

0.006 ¥ 10 1.5 - 0.006

= 0.04 W

Given S =

[  d ] = [ L ]

1 ¥ 2 ¥ t2 2

2n W . Thus 249

 

[ L ] = [ ML–3 ]a ¥ [MT–3]b ¥ [T –1]c



fi

[ L ] = [Ma+b ] ¥ [L–3a] ¥ [T –3b–c]

Equating powers of M, L and T, we have a + b = 0,   –3a = 1   and –3b – c = 0 Solving, We get

1 a= - , 3

\ d = k r

b=+ -1 3

Thus  s1 n = s1 3

s

1 3

+1 3

fi

f

0.4 =

F

and c = -1. R

-1

30°

r^ = R sin 30°

n = 3.

16. The ball thrown from the left with a velocity u1 = 0.3 ms–1 will have an acceleration a1 = –2 ms–2 towards the left, i.e. it will be retarded and will come to rest after travelling a distance x given by 2a1 x = 0 –u12 2 ¥ (–2) ¥ x = 0 – (0.3)

2n _5   fi  n = 4.98 ~ 249 1 18. Moment of inertia of disc is I = MR 2 . The total 2 torque on the disc by the three forees is (see figure)

x = 0.0225 m

F F

t = 3 F r^ = 3 F R sin 30° Also t = I a I a = 3 F R sin 30°



2



fi

a =

3 F R sin 30∞ (1) I

Physics JEE Advanced 2014–Paper-I  PI.11

Putting the values of M = 1.5 kg, F = 0.5 N, R = 1 0.5 m, I = ¥ 0.5 ¥ (0.5) 2 in Eq. (1) and solving 2 we get a = 2 rad s–2 Now w = w0 + a t = 0 + 2 ¥ 1



fi

w = 2 rad s–1

19. Given R = 0.5 m, M = 0.45 kg, m = 0.05 kg, r = 0.25 m and v = 9 ms–1. Since no external torque acts, the angular momentum of the system about the centre of rotation is conserved, i.e. Li = Lf If w is the angular speed, then i.e. Iw = 2 m r v 1 or MR 2 ¥ w = 2 m r v 2



fi



fi

1 ¥ 0.45 ¥ (0.5) 2 ¥ w = 2 ¥ 0.05 ¥ 0.25 ¥ 9 2 w = 4 rad s–1

20. Potential energy when the block is at Q is U = mgh Form work-energy principle, Gain in KE =  Work done in moving the block from P to Q – P.E. when the block reaches Q = F ¥ P Q – U

(



= F ¥ 5 – mgh Q PQ = 42 + 32 = 5 m



= 18 ¥ 5 – 1 ¥ 10 ¥ 4



= 90 – 40 = 50 J

Hence n = 5.

)

Physics JEE Advanced 2014–Paper-II  PII.1

Physics Jee Advanced—2014 Paper-II—Model Solutions SECTION I (Only One Option Correct Type) This section contains 10 multiple choice questions. Each question has four choices (a), (b), (c) and (d) out of which ONLY ONE option is correct. 1. Charges Q, 2Q and 4Q are uniformly distributed in three dielectric solid spheres 1, 2 and 3 of radii R/2, R and 2R respectively, as shown in figure. If magnitudes of the electric fields at point P at a distance R from the centre of spheres 1, 2 and 3 are E1, E2 and E3 respectively, then P P

R R R/2

Q

2Q

Sphere 1 Sphere 2

2S (b) cos (q + a ) br g 2S (c) cos (q - a / 2) br g 2S (d) cos (q + a / 2) br g 3. If l cu is the wavelength of Ka X-ray line of copper (atomic number 29) and l Mo is the wavelength of the Ka X-ray line of molybdenum (atomic number 42), then the ratio l cu/l Mo is close to (a) 1.99 (b) 2.14

P

(c) 0.50

(d) 0.48

R 4Q 2R

Sphere 3

(a) E1 > E2 > E3 (b) E3 > E1 > E2 (c) E2 > E1 > E3 (d) E3 > E2 > E1 2. A glass capillary tube is of the shape of truncated cone with an apex angle a so that its two ends have cross sections of different radii. When dipped in water vertically, water rises in it to a height h, where the radius of its h cross section is b. If the surface tension of water is S, its density is r, and its contact angle with glass is q the value of h will be (g is the acceleration due to gravity) 2S (a) cos (q - a ) br g

1 ¥ (radius of Earth) has 10 the same mass density as Earth. Scientists dig a R well of depth on it and lower a wire of the 5 same length and of linear mass density 10–3 kgm–1 into it. If the wire is not touching anywhere, the force applied at the top of the wire by a person holding it in place is (take the radius of Earth = 6 ¥ 106 m and the acceleration due to gravity of Earth is 10 ms–2) (a) 96 N (b) 108 N

4. A planet of radius R =



(c) 120 N

(d) 150 N

5. A tennis ball is dropped on a horizontal smooth surface. It bounces back to its original position after hitting the surface. The force on the ball during the collision is proportional to the length of compression of the ball. Which one of the following sketches describes the variation of its kinetic energy K with time t most appropriately? The figures are only illustrative and not to the scale.

PII.2  Comprehensive Physics—JEE Advanced

(a)

of 90 W, as shown in the figure. The least count of the scale used in the metre bridge is 1 mm. The unknown resistance is

(b)

K

K

t

(c)

t

R

(d)

90 W

K

K

t

40.0 cm

t

6. A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The maximum speeds of the photoelectrons corresponding to these wavelengths are u1 and u2, respectively. If the ratio u1 : u2 = 2 : 1 and hc = 1240 eV nm, the work function of the metal is nearly (a) 3.7 eV (b) 3.2 eV (c) 2.8 eV (d) 2.5 eV 7. A wire, which passes through the hole in a small bead, is bent in the form of quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is reA leased from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, 90° B the force it applies on the wire is (a) always radially outwards. (b) always radially inwards. (c) radially outwards initially and radially inwards later. (d) radially inwards initially and radially outwards later. 8. During an experiment with a metre bridge, the galvanometer shows a null point when the jockey is pressed at 40.0 cm using a standard resistance

(a) 60 ± 0.15 W (b) 135 ± 0.56 W (c) 60 ± 0.25 W (d) 135 ± 0.23 W 9. Parallel rays of light of intensity I = 912 Wm–2 are incident on a spherical black body kept in surroundings of temperature 300 K. Take StefanBoltzmann constant s = 5.7 ¥ 10–8 Wm–2 K–4 and assume that the energy exchange with the surroundings is only through radiation. The final steady state temperature of the black body is close to (a) 330 K (b) 660 K

(c) 990 K

(d) 1550 K

10. A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is Liquid

Block S



(a) 1.21 (c) 1.36

(b) 1.30 (d) 1.42

SECTION II Comprehension Type (Only One Option Correct) This section contains 3 paragraphs, each describing theory, experiments, data etc. Six questions relate to the three paragraphs with two questions on each paragraph. Each question has only one correct answer among the four given options (a), (b), (c) and (d).

Paragraph for Questions 11 and 12 The figure shows a circular loop of radius a with two long parallel wires (numbered 1 and 2) all in the plane of the paper. The distance of each wire from the centre of the

loop is d. The loop and the wires are carrying the same current I. The current in the loop is in the counterclockwise direction if seen from above.

Physics JEE Advanced 2014–Paper-II  PII.3

Q

Wire 1

S d

d

Wire 2

a

P

R

11. When d ª a but wires are not touching the loop, it is found that the net magnetic field on the axis of the loop is zero at a height h above the loop. In that case

(a) current in wire 1 and wire 2 is in the direction PQ and RS, respectively and h ª a



(b) current in wire 1 and wire 2 is in the direction PQ and SR, respectively and h ª a



(c) current in wire 1 and wire 2 is in the direction PQ and SR, respectively and h ª 1.2a



(d) current in wire 1 and wire 2 is in the direction PQ and RS, respectively and h ª 1.2a

12. Consider d >> a, and the loop is rotated about its diameter parallel to the wires by 30° from the position shown in the figure. If the currents in the wires are in the opposite directions, the torque on the loop at its new position will be (assume that the net field due to the wires is constant over the loop)

diatomic gas at 400 K. The heat capacities per mole of an 3 5 ideal monatomic gas are Cv = R, CP = R, and those 2 2 5 7 for an ideal diatomic gas are Cv = R, CP = R. 2 2 13. Consider the partition to be rigidly fixed so that it does not move. When equilibrium is achieved, the final temperature of the gases will be

(a) 550 K

(b) 525 K



(c) 513 K

(d) 490 K

14. Now consider the partition to be free to move without friction so that the pressure of gases in both compartments is the same. Then total work done by the gases till the time they achieve equilibrium will be (a) 250 R (b) 200 R

(c) 100 R

(d) – 100 R

Paragraph for Questions 15 and 16 A spray gun is shown in the figure where a piston pushes air out of a nozzle. A thin tube of uniform cross section is connected to the nozzle. The other end of the tube is in a small liquid container. As the piston pushes air through the nozzle, the liquid from the container rises into the nozzle and is sprayed out. For the spray gun shown, the radii of the piston and the nozzle are 20 mm and 1 mm respectively. The upper end of the container is open to the atmosphere.

m0 I 2 a 2 m0 I 2 a 2 (a) (b) d 2d 3 m I 2a2 3 m I 2a2 (c) 0 (d) 0 2d d

Paragraph for Questions 13 and 14 In the figure a container is shown to have a movable (without friction) piston on top. The container and the piston are all made of perfectly insulating material allowing no heat transfer between outside and inside the container. The container is divided into two compartments by a rigid partition made of a thermally conducting material that allows slow transfer of heat. The lower compartment of the container is filled with 2 moles of an ideal monatomic gas at 700 K and the upper compartment is filled with 2 moles of an ideal

15. If the piston is pushed at a speed of 5 mms–1, the air comes out of the nozzle with a speed of (a) 0.1 ms–1 (b) 1 ms–1 –1 (c) 2 ms (d) 8 ms–1 16. If the density of air is ra and that of the liquid rl, then for a given piston speed the rate (volume per unit time) at which the liquid is sprayed will be proportional to ra (a) (b) ra rl rl r (c) l (d) rl ra

PII.4  Comprehensive Physics—JEE Advanced

SECTION III Match List Type (Only One Option Correct) This section contains four questions, each having two matching lists. Choices for the correct combination of elements from List-I and List-II are given as option (a), (b), (c) and (d) out of which one is correct. 17. A person in a lift is holding a water jar, which has a small hole at the lower end of its side. When the lift is at rest, the water jet coming out of the hole hits the floor of the lift at a distance d of 1.2 m from the person. In the following, state of the lift’s motion is given in List I and the distance where the water jet hits the floor of the lift is given in List II. Match the statements from List I with those in List II and select the correct answer using the code given below the lists. List I P.

List II

Lift is accelerating vertically up.

1.

d = 1.2 m

Q. Lift is accelerating vertically down with an acceleration less than the gravitational acceleration.

2.

d > 1.2 m

R.

Lift is moving vertically up with constant speed.

3.

d < 1.2 m

S.

Lift is falling freely.

4.

No water leaks out of the jar

Code: (a) (b) (c) (d)

P-2, P-2, P-l, P-2,

Q-3, Q-3, Q-l, Q-3,

R-2, R-l, R-l, R-1,

S-4 S-4 S-4 S-1

q (+0, b)

Q2 (-a, 0)

Q3 (+a, 0)

Q4 (+2a, 0)

List I P.

Q1, Q4 positive; Q2, Q3 negative

3.

+ y

S.

Q1, Q3 positive; Q2, Q4 negative

4.

– y

Code: (a) (b) (c) (d)

P-3, P-4, P-3, P-4,

Q-1, Q-2, Q-1, Q-2,

R-4, R-3, R-2, R-1,

S-2 S-1 S-4 S-3

19. Four combinations of two thin lenses are given in List I. The radius of curvature of all curved surfaces is r and the refractive index of all the lenses is 1.5. Match lens combinations in List I with their focal length in List II and select the correct answer using the code given below the lists. List I

18. Four charges Q1, Q2, Q3 and Q4 of same magnitude are fixed along the x axis at x = –2a, – a, + a and + 2a, respectively. A positive charge q is placed on the positive y axis at a distance b > 0. Four options of the signs of these charges are given in List I. The direction of the forces on the charge q is given in List II. Match List I with List II and select the correct answer using the code given below the lists.

Q1 (-2a, 0)

R.

Q1, Q2, Q3 Q4 all positive

Q. Q1, Q2 positive; Q3, Q4 negative

List II 1.

+ x

2.

– x

List II

P.

1.

2r

Q.

2.

r/2

R.

3.

– r

S.

4.

r

Code: (a) (b) (c) (d)

P-1, P-2, P-4, P-2,

Q-2, Q-4, Q-1, Q-1,

R-3, R-3, R-2, R-3,

S-4 S-1 S-3 S-4

20. A block of mass m1 = 1 kg another mass m2 = 2 kg, are placed together (see figure) on an inclined plane with angle of inclination q. Various values of q are given in List I. The coefficient of friction between the block m1, and the plane is always zero. The coefficient of static and dynamic friction between the block m2 and the plane are equal to m = 0.3. In List II expressions for the friction on the block m2 are

Physics JEE Advanced 2014–Paper-II  PII.5

given. Match the correct expression of the friction in List II with the angles given in List I, and choose the correct option. The acceleration due to gravity is denoted by g. [Useful information: tan (5.5°) ª 0.1; tan (11.5°) ª 0.2; tan (16.5°) ª 0.3]

q e0

fi  E1 ¥ 4 p r 2 = fi

E1 =

q 4p e 0 r 2 P r

m1

O

R1

m2 q

Sphere 1 Gaussian surface

List I P.

List II 1. m2g sin q

q = 5°

Q. q = 10°

2. (m1 + m2)g sin q

R.

q = 15°

3. mm2 g cos q

S.

q = 20°

4. m(m1 + m2)g cos q

Code: (a) (b) (c) (d)

P-1, P-2, P-2, P-2,

Q-1, Q-2, Q-2, Q-2,

R-1, R-2, R-2, R-3,

S-3 S-3 S-4 S-3

1. (c) 4. (b) 7. (d) 10. (c)

2. (d) 5. (b) 8. (c)

11. (c) 14. (d)

12. (b) 15. (c)

3. (b) 6. (a) 9. (a)

13. (d) 16. (a)

Section III

17. (c) 20. (d)

18. (a)

19. (b)

Hints and Solutions 1. For Sphere 1 Let R1 be the radius of sphere 1 and r (> R1) be the distance of P from O. From Gauss’s law,

  q ∫ E1 ⋅ dS = ε 0

where

E1 = k =

E2 ¥ 4 p R 22 =

Section II



Q = k (1) 4p e 0 R 2 Q 4p e 0 R 2

For Sphere 2 Let R2 be the radius of sphere 2. Then from Gauss’s law, the electric field E2 at a point on its surface is given by

Answers Section I

Given q = Q and r = R. Hence

fi

E2 =

q e0 q 4p e 0 R22

Given q = 2Q and R2 = R. Therefore

E2 =

2Q = 2k (2) 4p e 0 R 2

For Sphere 3 Let R3 be the radius of the sphere and Q3 be the charge. Q3 Charge per unit volume = 4p 3 R 3 3 Charge in the Gaussian sphere is

q =

=

Q3 4p 3 r ¥ 4p 3 3 R3 3 Q3 r 3 R33

PII.6  Comprehensive Physics—JEE Advanced



fi

R3

fi

Gaussian surface

Given Q3 = 4Q, r = R and R3 = 2R.

E =

q Q = e 0 2e 0



Q Q k = = (3) 2 2 2 8p e0 r 8p e 0 R

From (1), (2) and (3) we find that E2 > E1 > E3. So the correct choice is (c). 2. It is clear from the figure that –OAB + –BAD = 90° because OA = R is the radius of the meniscus and AD is the tangent to the meniscus at point A. a But –ADB = q + . 2 Therefore –BAD = 90° – –ADB. Thus –OAB + 90° – –ADB = 90° or –OAB = –ADB = q +

a 2

O R B

b

q+

a 2S cos Ê q + ˆ Ë 2¯ b

a 2S cos Ê q + ˆ Ë 2¯ h = br g

3. The Ka X-ray line corresponds to the transition n = 2 to n = 1. For an element of atomic number Z, the wavelength l of the Ka line is given by

From Gauss’s law,

fi

2S h r g = = R

So the correct choice is (d).

4Q ¥ R3 Q q = = 2 ( 2 R )3

\  E3 ¥ 4 p r2 =

2S R

P

r O

\

p =

a 2

A q Tangent



3R 1 1 1 = RH ( Z - 1) 2 ÊÁ 2 - 2 ˆ˜ = H ( Z - 1)2 Ë1 4 l 2 ¯

\

lcu Ê Z - 1ˆ = Á Mo ˜ Ë Z cu - 1 ¯ lMo

2

2

Ê 42 - 1ˆ = 2.14 = Á Ë 29 - 1˜¯ 4. At a depth r below the surface of the planet, r gr = g s R where gs = acceleration due to gravity on the surface of the planet. GM G 4p 3 4p G r R gs = R r= = 2 ¥ 2 3 3 R R

gr =

4p G r r 3

If F is the force needed to keep the wire at rest, then F = weight of the wire =

D

R

4p G r r ˆ (l dr ) ÊÁ ˜¯ Ë 3 4R

Ú 5

a 2

4p G r l r 2 = 3 2

R 4R 5

C

a Now b = R cos ÊÁ q + ˆ˜ and excess pressure on Ë 2¯ the concave side of the meniscus is

fi

F =

4p G r l 9 R 2 ¥ (1) 3 50

On the surface of the earth,

Physics JEE Advanced 2014–Paper-II  PII.7



ge =

Also

r =

G Me Re2

fi G=

g e Re2 (2) Me

Me (3) 4p 3 Re 3

R Given R = e (4) 10 Using (2), (3) and (4) in (1), we get F =



=

\

u12 =

2 hc 2 W0 (1) m l1 m

and

u22 =

2 hc 2 W0 (2) m l2 m

Dividing (1) and (2) we get u12 u22



9 ¥ 10 ¥ (6 ¥ 106 ) ¥ 10-3 = 108 N 5 ¥ 10+3

5. The ball is dropped from rest at say t = 0. It hits the ground at t = t1 when its velocity v = gt. Its kinetic energy in time interval 0 to t1 is given by 1 1 mv 2 = mg 2 t 2 2 2 2 i.e. K µ t  . So the slope of the K-t graph increases with time and the graph is not linear as shown in the figure. K =



2 vmax =

9 ge Re l 5 ¥ 103

2 hc 2W0 ml m

\

hc - W0 l1 = (3) hc - W0 l2

Given l1 = 248 nm, l2 = 310 nm, hc = 1240 eV nm and

u1 = 2. Using these values in (3), we have u2



1240 eV nm - W0 248 nm 4 = 1240 eV nm - W0 310 nm

fi

4 =

5 eV - W0 4 eV - W0

K

W0 = 3.7 eV

fi

O

t1

t2

t

At t = t1, the velocity is reversed and the ball begins to rise upwards with initial velocity v which decreases with time and its K-t graph has a negative slope until the ball reaches the highest point at time t2 when it is momentarily at rest and its kinetic energy K = 0. After t = t2, although the velocity is reversed, the 1 speed is v and K = mv 2 which is positive. So 2 the ball reaches the same height from which it was dropped. Hence the correct graph is (b) 1 2 m vmax + W0 2 (W0 = work function)

6. hn = Kmax + W0 =

fi

1 hc 2 + W0 = m vmax 2 l

7. Initially the bead exerts an inward radial force (centripetal force) on the wire and the wire exerts a normal reaction N radially outwards. At a certain instant during the motion, the normal reaction N becomes zero. After that instant, the normal reaction N will act radially outwards. So the correct choice is (d). 40 ¥ 90 Ê l ˆ ¥ 90 = 8. R = Á = 60 W ˜ Ë 100 - l ¯ 60 Since 90 W is exact, the fractional error in R is

DR 0.1 cm 0.1 cm 0.1 0.1 = + = + R 40 cm 60 cm 40 60

Since R = 60 W, fi  DR =

0.1 0.1 ¥ 60 + ¥ 60 = 0.15 + 0.1 = 0.25 W 40 60

\  R ± DR = (60 ± 0.25) W The correct choice is (c).

PII.8  Comprehensive Physics—JEE Advanced

9. When the steady state is reached, the rate of energy lost by the sphere = rate at which the energy is incident on it, i.e. (here R = radius of sphere)

Loop

Q

I

    s ¥ 4p R 2 [T 4 - (300) 4 ] = 912 ¥ p R 2 fi   5.7 ¥ 10 ¥ 4 ÈÎT - (300) ˘˚ = 912 4

fi   T 4 - (300) 4 =

4

fi   T = (121 ¥ 108 )1 / 4 = 332 K  330 K

fi

fi

ic = 30° ml mb

ml = mb sin ic

= 2.72 ¥ sin 30° = 1.36 ml

Liquid r

O ic

ic h mb

Wire 2

I P

R

The distance of point C from the centre O of the loop is r = h 2 + a 2 The magnet field at the height h due to each wire m I is 0 . The direction of this field is along the 2p r tangent at the given point to the field. The total field due to both wires is 2 m0 I Bw = along the tangent at the given point. 2p r But the magnetic field due the loop is

r 5.77 = = 3 h 10

Also sin ic =

Wire 1

912 = 40 ¥ 108 4 ¥ 5.7 ¥ 10-8

fi   T 4 - 81 ¥ 108 = 40 ¥ 108

10. tan ic =

O d=a

d=a

A

-8

S

Bl =

out of the page. 2r 3 The net field at the given point will be zero if component of Bw along the axis of the loop = Bl, i.e. Bw sin q = Bl, where q = angle between AC and CO.

ic

S

2 m0 I a m I a2 ¥ = 0 3 2p r r 2r

fi

2 a = 3 2 r pr

fi

r =

pa 2

fi h 2 + a 2 =

pa 2

Block

So the correct choice is (c). 11. If the currents in wires 1 and 2 are in the same direction and since the distance of point C (not shown in figure) at a height h above the centre O of the loop is the same from the two wires, the magnitudes of magnetic field at C are equal for both wires, but their directions are opposite. Hence, the magnetic field due to current I in the wires will be zero at C but the magnetic field due current I is not zero at point C. Hence the net magnetic field at C due to the net magnetic field at C due to the wires and due to the loop cannot be zero. Therefore, options (a) and (d) are wrong.

m0 I a 2

fi

Êp2 ˆ - 1˜ h = a Á Ë 4 ¯

1/ 2

 1.2 a

So the correct choice is (c) 12. If d >> a; the torque on the loop is t = IBA sin q = I ¥ =

2 m0 I ¥ p a 2 ¥ sin 30∞ 2p d

m0 I 2 a 2 2d

Physics JEE Advanced 2014–Paper-II  PII.9

13. Let T be the equilibrium temperature. Since the partition between then two compartments is regid, the transfer of heat from the lower compartment (which contains a monoatomic gas at 700 K) to the upper compartment takes place at constant volume. \  Heat transferred from lower to upper compartment is 3R    Q1 = n Cv DT = 2 ¥ ¥ (700 - T ) (1) 2 Since the piston and the top of the upper compartment are frictionless, heat Q2 is gained by the diatomic gas in the upper compartment at constant pressure, i.e.    Q2 = n Cp DT = 2 ¥

7R ¥ (T - 400) (2) 2

Equating (1) and (2), 3R 7R ¥ (700 - T ) = 2 ¥ ¥ (T - 400) 2 2 fi  3 ¥ (700 – T ) = 7 ¥ (T – 400) fi  T = 490 K     2 ¥

14. If the partition between the two compartments is frictionless, the heat exchange between the gases takes place at constant pressure. If T ¢ is the equilibrium temperature now, then       2¥

5R 7R ¥ (700 - T ¢ ) = 2 ¥ ¥ (T ¢ - 400) 2 2

fi T ¢ = 525 K Since the temperature of the gas in the lower compartment falls, the work done by the gas is W1 = – nR DT = – 2 ¥ R ¥ (700 – 525) = – 350 R Since the temperature of the gas in the upper compartment increases, the work done by the gas is W2 = + nR DT = + 2 ¥ R ¥ (525 – 400) = + 250 R \ Net work done = W1 + W2 = – 350 R + 250 R = – 100 R So the correct choice is (d). 15. From the equation of continuity of flow, a1 v1 = a2 v2 fi (p r12) v1 = (p r22) v2 fi

Êr ˆ v2 = v1 ¥ Á 1 ˜ Ë r2 ¯

2

Ê 20 mm ˆ = 5 mm s -1 ¥ Á Ë 1 mm ˜¯

2

= 2000 mm s–1 = 2 ms–1 So the correct choice is (c). 16. From Bernoulli’s principle,     P0 +

1 1 ra va2 = P0 + rl vl2 2 2

fi   vl2 = va2 fi   vl = va

ra rl ra rl

For given va, vl µ

ra , which is choice (a). rl

17. Velocity of efflux is v = 2 g h where h is the depth of the hole below the surface of water in the jar. The time taken by water emerging from the hole to hit the floor of the lift is given by

t =

2( H - h) g

where H is the height of water in the jar. \ Horizontal range d = vt = 2 h( H - h). Thus d is independent of g. Hence in P, Q and R, d remains the same = 1.2 m. But in S, geff = 0. Hence v = 0. Thus in case S, no water emerges out of the hole. So, the correct answer is P Æ 1, Q Æ 1, R Æ 1 and S Æ 4, which is option (c). 18. Case P: In this case, the net force F exerted on q by Q1 and Q4 and also by Q2 and Q3 are in the + y direction. So P Æ 3. Case Q: In this case, the net force F exerted on q by Q1 and Q4 and also by Q2 and Q3 are in the + x direction. So Q Æ 1. Case R: In this case, the net force F1 exerted on q by Q1 and Q4 is in the + y direction and the net force F2 exerted on q by Q2 and Q3 is in the – y direction. Since Q2 and Q3 are closer to q than Q1 and Q4, F2 > F1. Hence the net force on q due to all the four charges is in the – y direction. So R Æ 4. Case S: In this case, the net force F1 exerted on q by Q1 and Q4 is in the + x direction and the net force F2 exerted on q by Q2 and Q3 is in the – x direction. Since Q2 and Q3 are closer to q than

PII.10  Comprehensive Physics—JEE Advanced

Q1 and Q4, F2 > F1. Hence the net force on q due to all the four charges is in the – x direction. So S Æ 2. Thus the correct choice is (a). 19.

1 1ˆ Ê 1 = ( m - 1) Á - ˜ Ë R1 R2 ¯ f R1 = •

C2







R2

C1

R1

C2

Fig. 1

R2

Fig. 2

For a combination of lens,

1 1 1 = + . F f1 f 2

Case P:

1 1 1 2 r = + = fi F = .Therefore P Æ 2. F r r r 2

Case Q:

1 1 1 1 = + = fi F = r.Therefore Q Æ 4 F 2r 2r r

1 1 1 1 Case R:  = - - = - fi F = - r. Therefore F r 2r 2r RÆ3 Case S:

1 1 1 1 = - = fi F = 2r. Therefore S Æ 1. F r 2r 2r

So the correct option is (b).

R1 = •

20. R2

N1 = m1 g cos q

C2

N2 = m2 g cos q m1 m2

Fig. 3

For an equi-convex lens (Fig. 1)    R1 = + r, R2 = – r. Given m = 1.5. 1 2 1 Ê1 1 ˆ = 0.5 ¥ =     = (1.5 - 1) Á ˜ Ë r -r¯ f r r fi   f = r For a plano-convex lens (Fig. 2)    R1 = •, R2 = – r \  

1 1 ˆ 1 Ê1 = = (1.5 - 1) Á Ë • - r ˜¯ 2r f

fi   f = 2r For a plano-convex lens (Fig. 3)    R1 = •, R2 = + r \  

1 1 1 1 = (1.5 - 1) ÊÁ - ˆ˜ = ¯ Ë f • r 2r

fi   f = – 2r

(m1 + m2) g sin q



q

The block m2 will not slide down the plane if the frictional force on it has a maximum value fmax given by    fmax ≥ (m1 + m2) g sin q fi  m N2 ≥ (m1 + m2) g sin q fi  m m2 g cos q ≥ (m1 + m2) g sin q fi  0.3 ¥ 2 ¥ g cos q ≥ (1 + 2) g sin q fi  0.2 ≥ tan q Thus q must be less than tan–1(0.2) = 11.5°. For cases P and Q, q is less than 11.5°. Hence the force of friction on m2 is f = (m1 + m2) g sin q. But for cases R and S, q is greater than 11.5°. Hence the force of friction on m2 must be f = fmax = m m2 g sin q So the correct answer is: P Æ 2, Q Æ 2, R Æ 3, S Æ 3, which is option (d).

JEE ADVANCED 2015: PAPER–I (MODEL SOLUTIONS) SECTION I (Single Digit Integer Type) EIGHT SINGLE DIGIT INTEGER 1.

A

y

y-z

z =

C B

3 a 2

v

v D

v

v

(g

ABCD lL ( ne 0

x-y

0=

n

5.

2.

A

B A

n

B

A B n

hc

A

B

A

B

3.

6.

v

T v

N

= v N

4. AB

CD

nT n

JAI.2

8.

7. S

S x

=pm

–d d M

m p

M2 M1

M

SECTION II (One or More than One Options Correct Type) TEN FOUR

ONE OR MORE THAN ONE

9.

Mμ c

Mμ G

Lμ h

Lμ G

11.

E

E p

x a n b

10.

h

c G

L

M

12.

a n R

E1w1 = E2w 2

w2 = n2 w1

w1w 2 = n 2

E1 E2 = w1 w 2 M

R

JAI.3

O M 8

d

O

P 8 w 9 3 R 5

S

S d O O

2R 3 1R 3

15. I

x-y

3R 5

B

F

(L =0 (L =0

R

4R 5 13.

q

B B B B

q

x

z F x F y F z F

R

16. T

RT 6/5

1/ 2

q

17.

q q

¥

q 14.

S

S

P

Q

¥

JAI.4

18. V0 V0

2475 64

1875 64

1875 49

2475 132

SECTION III TWO Column I Column I Column II

Column II

four Column I Column I

Column II Column II

19.

60 27 Co

Column I

Column II

235 92 U

20.

x a Column I

Column II 2 2

U1 ( x) =

U0 È Ê x ˆ ˘ Í1 = Á ˜ ˙ 2 Î Ë a¯ ˚

U 2 ( x) =

U0 Ê x ˆ 2 Á ˜ 2 Ë a¯

x=a

x

0

JAI.5

U 3 ( x) =

È Ê xˆ 2 ˘ U0 Ê x ˆ 2 ÁË ˜¯ exp Í- ÁË ˜¯ ˙ 2 a Î a ˚

U 4 ( x) =

U0 2

x = – a.

È x 1 Ê xˆ3˘ Í - ÁË ˜¯ ˙ Îa 3 a ˚

x

|x| < a U0 4

x=–a 2. n 1. 4. 7. 3

2. 5. 8.

9. 11. 13. 15. 16. 17.

E = hv –

3. 6. 3

13.6 eV n2

hc 13.6 eV l n2

E=

E =

10. 12. 14.

hc 1242 eVnm 13.6 eV – 90 nm n2 13.6 n2

18.

n

n 19. 20.

Æ

Æ

Æ

Æ

3.

PA g

1.

1 (g 4

A i.e.

GM 1 GM = 4 R2 r2

ABCD z =

P=

r= R

3 a 2

A= P 1 2 GmM GmM =0– mv – 2 R r 1 2 GM GM v =( r = 2 R) R 2 2R

q f= e0 q AB

ABCD L

v=

q= L

GM R

ABCD f q lL = = 6 6e 0 6e 0 n

ve =

2GM = 2v R

JAI.6

ve = N v.

EA = 104 EB

N

RA = 400 RB

4

ÊT ˆ 10 = Á A ˜ ¥ (400) 2 Ë TB ¯ 4

TA 1 = TB 2 4.

M

lA lB

R 6.

=

1 2 1 I w + Mv 2 2 2

P P P

P

1 1 v 2 1 = ÊÁ MR 2 ˆ˜ ¥ ÊÁ ˆ˜ + mv 2 ¯ Ë R¯ 2Ë2 2 =

P 8 7.

(

S1 P = S2 P = x 2 + d 2

= TA = B = TB TA TB B l A TB = l B TA

A

A EA

E A Ê TA ˆ ÊR ˆ = ¥Á A˜ Ë RB ¯ EB ÁË TB ˜¯

(

)

(

)

=

4 ¥ x2 + d 2 3

=

1 ¥ x2 + d 2 3 m x=m

(

)

(

)

1/2

1/2

=m

1 ¥ x 2 + d 2 = m2 l 2 9 x

m

–d

(

- 1 ¥ x2 + d 2

P

¥ x2 + d 2

AB = 4p RA2 2

1/2

B

EB = s TB4 AB2

4

1/2

S S x = μw S P – μa S P

AA2

AA = 4p RA2

)

P

5.

TA4

T

n

1 2 1 2 3 2 mv + mv = mv 4 2 4

3 2 3 mv2 - Mv12 = Mg (h2 - h1 ) 4 4 3 2 (v2 - 32 ) = 10 ¥ (30 - 27) = 30 4 4 v22 = 30 ¥ + 9 = 49 3 v A

P 12.5 P ×P= = 3 8 100 2

P=

)

1/2

JAI.7

x =p m p

–d

Ê 1.5 - 7 ˆ 1 Á 6 ˜ ¥ 1 fi f ' = 35 cm =Á 2 7 ˜ 5 2 f 2' ÁË ˜¯ 6

p

8.

u f 1 1 1 + = v1 u1 f1 1 1 1 + = v1 -15 -10

m =–

f2' =

u 1 1 1 = v2 ' u2 ' f2'

v

35 cm 2

in

1 1 2 + = fi v2 ' = 140 cm v2 ' 20 35

u1 -30 =– v1 -15

m2' =

140 = -7 -20

M M 2 14 = = -7 M1 -2

f

M2 =7 M1

u = 1 1 1 = = v2 u2 f 2 1 1 1 = v2 20 10

1 8

v

4 5

v2 +20 = = -1 m = u2 -20 = M =m ×m

4 1 1 ¥ = 5 8 10

1 1 1 - = 8 10 40

7 6 m f

0.05cm 100

Ê 7 - 1.5 ˆ 1 Á6 1ˆ ˜Ê 1 = ÁË ˜¯ Á - ˜ f 2' 1.5 Ë R1 R2 ¯ 1 1 1 = (1.5 - 1) ÊÁ - ˆ˜ Ë R1 R2 ¯ f2 1 1 1 1 1 = = = R1 R2 0.5 f 2 0.5 ¥ 10 5

1 mm = 100

JAI.8

h, c L L ha cb Gc L = k [ML T k a – c

a

G

a

× (LT

b

L 3T

× [M

C

a –b – c = 0. 1 3 a = b = – c = 2 2

h–

c

1 m R 2w 22 2

()

Êw ˆ ¥Á 1˜ Ë w2 ¯

= n2 ¥

1 w1 ¥ n2 w 2

2

2

Ê ÁË

w1 1ˆ = 2˜ w2 n ¯

w1 w2

E1 E2 = w1 w 2

G

M M = khacbGc h c

G a =

1 1 , b= , 2 2

E2 =

=

g

h c

1 m a 2w12 2

E1 a = E2 R

b 3c

1 2 L=k

E1 =

12. O x I = I’ ’

1 c=– . 2

2 2 È Ê M ˆ Ê3R ˆ Mx ˘ 8w MR 2w = Í MR 2 + Ë ¯ ¥ Ë ¯ + ˙¥ Î 8 5 8 ˚ 9

M = kh c G

Ê 9R2 x2 ˆ 8 + ˜¥ R 2 = ÁË R 2 + 200 8 ¯ 9

11. a x

v=a

Ê x2 ˆ ÁË1 - 2 ˜¯ a

9 R 2 = 8R 2 +

a

1/2

x2 =

9R2 + x2 25

16 R 2 4R fix= , 25 5

13.

m

r

2 1/2

p = mv = ma

Ê x ˆ ÁË1 - 2 ˜¯ a

E= x

p = p.

b = ma

l nˆ 2pe 0 r nˆ

a = mw1 n = m b x

p = R

R

R = mR = mR w2 = n2 w1

q q O

P

x

JAI.9

<< d

S1

P u

l E1 = 2pe 0 (d + x)

R=– 1 1.5 1 - 1.5 = v1 -50 -10

x q

F = qE

1 3 1 + = v1 100 20

x

l 2pe 0 (d - x)

x

E >E F

1.5 1 1.5 - 1 = • -(d - 50) 10

F q

1 1 = (d - 50) 20

F = – (F – F =

ql Ê 1 1 ˆ Ë 2pe 0 d - x d + x ¯

d

F = I (l ¥ B) = I ( PQ ¥ B)

Ê 2q l ˆ x F = –Á Ë 2pe 0 d 2 ˜¯

PQ L

PQ B

q

+x F = –qE

L + R + R + L

R

q

(–x

x L + R iˆ B = Bkˆ

z

F = I [2( L + R) iˆ ¥ B kˆ ]

l 2pe 0 (d + x)

F

x

> F

x

y

I (L B

q

x-

B = Bjˆ F = 2 I ( L + R) B(iˆ ¥ ˆj )

14.

= 2 I ( L + R) B kˆ B = Bkˆ F = 2 I ( L + R) B(iˆ ¥ kˆ) = 2 I ( L + R)(- ˆj)

R B.

B = Biˆ

F = 2 I ( L + R) B(iˆ ¥ iˆ) = 0

x O

( iˆ ¥ kˆ = - ˆj )

= 2 I ( L + R) B (- ˆj )

q

F = – qE l E2 = 2pe 0 (d - x) F

l B

x << d

E1 =

d

15.

Ê 2q l ˆ x F = -Á Ë 2pe 0 2 ˜¯

f

v

u = – (d R v = μ

x

E2 =

v =?

S2

q F = qE

μ

( iˆ ¥ iˆ = 0)

JAI.10

1 RT 2

16.

R

R1 R2 R1 + R2

=

5 ¥ 10-3 4 = 5 3 ¥ 10-5 + ¥ 10-3 4 3 ¥ 10-5 ¥

3 5 RT + RT 2 2 =

4 RT 2

RT

RT

150 ¥ 10-6 1875 = 5.12 64

= 18.

hv = eV0 +

g RT v= M

hc = eV0 + f l

nHe (C p ) He + mH 2 (CP ) H 2

hc Ê 1 ˆ f e Ël ¯ e

nHe (Cv ) He + nH 2 (Cv ) H 2 5 7 RT + 1 ¥ RT 3 2 2 = = 3 5 2 1 ¥ RT + 1 + RT 2 2

1 l

V0

1¥

V0 V0

1¥ 2 +1¥ 4 =3 2 v in mixture = v in helium

M He r of mixture ¥ r of helium M mixture

=

3 / 2 4 ¥1 ¥ 5/3 3

=

6 5

vrms of mixture = vrms of hydrogen

R =

Æ

Æ

Æ

2 1 = 4 2

17. R = rl A

R1 =

19.

20. (2.7 ¥ 10-8 ) ¥ (0.05) (0.007) 2 - (0.002)2 (1.0 ¥ 10-7 ) ¥ (0.05) 5 = ¥ 10-3 4 (0.002) 2

2 dU1 d ÈU 0 Ï x 2 ¸ ˘ Fx = = - Í Ì1 - 2 ˝ ˙ dx dx Î 2 Ó a ˛ ˚

=

2U 0

Fx = -

a

2

( x - a) x ( x + a)

dU 2 d ÈU x 2 ˘ =- Í 0 2˙ dx dx Î 2 a ˚

=-

U0 ¥ 2x 2a

2

=-

U0 x a2

Æ

JAI.11

Fx Fx = -

–x

Ê x2 ˆ ˘ dU 3 d ÈU x 2 = - Í 0 2 exp Á - 2 ˜ ˙ Ë a ¯˚ dx dx Î 2 a Ê x ˆ exp Á - 2 ˜ Ë a ¯ 2

= U0

a2

¥ ( x)( x - a)( x + a)

Fx = =-

dU 4 d ÈU Ê x 1 x 3 ˆ ˘ =- Í 0Á ˜˙ dx dx Î 2 Ë a 3 a3 ¯ ˚ U0 2a 3

[( x - a)( x + a)]

JEE ADVANCED 2015: PAPER–II (MODEL SOLUTIONS) SECTION I (Single Digit Integer Type) EIGHT SINGLE DIGIT INTEGER 5.

1. A large spherical mass M

A

m M

R R=

–dA dt

M ÊM ˆ m = kË . 288 ¯

P

Q

M k is

t= t = 2 are RP n = , RQ e

RQ

RP

–dN dt

N(t

t r=3

A=

n is: 6. n

n

(n 2.

t is E(t

A

2

at

=

a

A E(t 3.

t A

k

R

() r R

r B (r ) = k

()

B r as A(r

r 5 , R

k is

7. R (= 2 I is

IA

IB n = I A 10

IB

n is:

4. p 2p , 3 3

I0 nI0

dq = m. dn

3 m is

n is:

I

JAII.2

8.

l = pp a0

2+

3h . 2p

a0 is

p is

SECTION II (One or More than One Options Correct Type) FOUR

9.

P

ONE OR MORE THAN ONE

Q

2

L2 , respec 2 P

L 2

Q

L Q

VP

L2 P L

L2 VQ 12. P

Q

y

VP h = 1 VQ h2

VP h2 = VQ h1

VP .VQ > 0

VP VQ < 0

10.

V 0

P P P

I,

0

c = 0V 2 I = 0cV 0I

11.

2

= 0V 0cI = 0V

O R2 OP = a = R

Q Q Q

13.

R P(r

P

R2

r(r < R P r=

P(r r

R2

R2

r a

a E a

( (

) )

3R 5 = 16 2R 21 P r= 5 P r=

E

( ) ( ) ( ) ( )

3R 4 = 63 2R 80 P r= 3

r is

E(r , E E

Q

P

0I

R

x

R 2 = 20 R 27 P r= 3 P r=

JAII.3

14.

V2 = 2V

S d

T2 = 3T PV T2 = 4T

C V2 = 3V

(

7 PV 1 1 3

2

T2 = 4T 17 PV 1 1 6

V2 = 3V

C2 is C1

C2

6 5

5 3

7 5

7 3

16.

236 92 U

+x+y 236 92 U

15.

x

Æ 140 54 Xe +

y K , KSr, K

K 236 140 92 U, 54 Xe

T

P

V P2

T2 V2

x x x x

x

V2 = 2V

94 38 Sr

T2 = 3T 1 PV 1 1 4

= = = =

n, p, p, n,

y = n, KSr y = e, KSr y = n, KSr y = n, KSr

KXe KXe KXe KXe

SECTION III TWO TWO FOUR

ONE OR MORE THAN ONE

Questions 17 and 18 are based on the following paragraph.

B y z-

I I, w

x d

PQRS PQRS z

94 38 Sr

JAII.4

19.

S n2

S2

n =

8 5

n =

n2 =

45 4

7 5

4 3

17. w2

w

d K

d 2,

M

NA xy

V2

V K

M I

NA

B w w w w

= = = =

18.

w2 w2 2w2 2w2

d = 2d2 d = 2d2 d = d2 d = d2

d

V2 = V2 = V2 = V2 =

y

NA

20.

NA S2

B2 B2 2B2 2B2

n = 2n2 n = 2n2 n = n2 n = n2

4 15 S

NA M I

= = = =

S

NA

(NA2 K

B B B B

S2

S2

B2 V2

V

3 15 S 16 15

2V V 2V V

w n2 B

n

S S2 16

V2 = 2V V2 = V V2 V V2 = V

Questions 19 and 20 are based on the following paragraph.

NA1 + NA2 NA1 + NA2

NA + NA2

NA

NA2

Answers Section-I 1. 4. 3 7.

2. 4 5. 2 8. 2

3. 6. 2

Section-II

n n2 n

n2 as i

9. 12. 15.

10. 3. 16.

11. 14.

Section-III

im NA

n im

17. 18.

19.

20.

NA2

JAII.5

Hints and Solutions

3. dr

Section-I

r O

1.

Mass m

r = 3l

m2 M

m is F = Mass m2 M

Gm1M (3l )

Gm1m2

-

2

l2

r = 4l

r2 dr dm = × 4 r2dr

m m2 is

F2 =

Gm2 M (4l ) 2

Gm1m2 + l m

O is 2 dI = dmr 2 3

m2

F = F2 Gm1M (3l ) 2

-

Gm1m2 l2

=

Gm2 M (4l ) 2

+

R

Gm1m2

O is

l2 I =

m = m2 = m GmM Gm 2 GmM Gm 2 + 2 = 9l 2 l2 16l 2 l

=

M M -m = +m 9 16 2m =

M M 7M = 9 16 144

kM m= 288 2.

m=

IA =

7M 288

E = A2 e

A at

E 2 DA = ± ± a Dt E A E 2 DA Dt = ±a ¥t E A t DA A

a

8p 3

s

IB IA t

Ú rr

4

Ú rr

2

dr r 2

0

dr

0

R

Úk 0

()

r 8p k R 6 ¥ r 4 dr = ¥ R 3R 6

8p k 5 R 18 8p 3

R

Úk 0

() r R

=

8p k 5 R 30

=

18 6 = 30 10

5

8p k R10 ¥ r 4 dr = ¥ 3R5 10

n

4. Let a 2

E =± E

R

R

at

IB = E

8p 3

=

k

2 4p ¥ 3 R

Ú dI =

y

=a

t

I0 = ka2 kx

a

k is a , =( t

kx

JAII.6

pˆ Ê Ëw t – kx + 3 ¯

y2 = a

RP = –

A0 –2 e t

RQ = –

A0 –1 e 2t

RP = 2e RQ

=

pˆ Ê = a sin Ëq + ¯ 3 2p ˆ Ê Ëw t – kx + 3 ¯

y3 = a

2p ˆ Ê = a sin Ëq + 3 ¯ y4 = a

t

kx +

a

2 e

n=2

+ 6.

y = y + y2 + y3 + y4 2p ˆ pˆ È Ê Ê ˘ sin q + sin Ëq + ¯ + sin Ëq + ˙ ¯ = aÍ 3 3 Í ˙ + sin (q + p )˚ Î

r+r

A

r

n

r

r+n

(

y = A sin w t – kx +

p 2

)

A 3 = A = 2a ¥ 2

)

I = kA2 = 3ka2 = 3I0 I = n I0 n=3 Note :

–t

–t

5. AP = A0 e t

AQ = A0 e 2t

dr dn

R n

r n

r n r

dq dn

3a

r

sin r dr = – n cos r dn

p Êp ˆ = 2a sin q + cos Ë ¯ 2 6

(

r

n

[ sin (q + p ) = – sin q ]

p Êp ˆ y = 2a cos Ë ¯ sin w t – kx + 6 2

r Q,

pˆ 2p ˆ˘ È Ê Ê = a Ísin Ëq + ¯ + sin Ëq + Î 3 3 ¯˙˚

( )

A

n

r

dr dn

RQ =

A dAP = – 0 et t dt dAQ

t =2

dt

–t

=–

A0 2t e 2t

dq dn

dr 1 È sin (60∞ – r ) – n cos (60∞ + r ) ˘˙ cos q ÎÍ dn ˚

Èsin (60∞ – r ) + n cos (60∞ + r )˘ 1 Í dq = sin r ˙ ˙ cos q Í dn Î n cos r ˚ Èsin (60∞ – r ) + cos (60∞ + r )˘ 1 Í dq = sin r ˙ ˙ cos q Í dn Î cos r ˚

–t

RP =

cos q

n =

3

r =

sin 60∞

r r r

3 sin 30∞

3

=

1 2

JAII.7

dq dn

=

1 cos 60∞

sin 30∞ ˘ È ÍÎsin 30∞ + cos 30∞ ¥ cos 30∞ ˙˚

2sin 30∞ = = cos 60∞

2¥ 1 2

1 2 =2

m=2 7.

Fig.4

Fig.1

Fig.5 ABCD 8 A =3

+2

C +4

Fig.2

2 I =

6.5V 2W + 4.5W

L =

nh 3h = 2p 2p

8.

n=3 mvr = Fig.3

3h 2p

mv =

3h 2p r

h h 2p r 3h = = = mv 3 2p r

JAII.8

n 2

r = a0

a (3) n = 0 Z Z

2

Ê e Aˆ Q = CV = Ë 0 ¯ ¥ V l Ql e 0V = A Ql e 0 cV = A [ AT ] ¥ [ L] ¥ [ LT -1 ] eV 0 [ L2 ] = [A 0cV = I,

( n = 3)

Li, Z = 3, 2p a0 (3) 2 = 2 a0 ¥ 3 3 = p a0 p=2

=

Section-II 9.

r

v vg = +

(

2

+ 2

vg

vg = ( + + 2= =

2vg

=

2

E(r ) =

2 2

2

>

2

2 r 2 (s 2 - r1 ) g h2 9

2 r 2 (s1 - r2 ) g h1 9

VQ

r ra = (OP) = 3e 0 3e 0

2

< VP =

11.

2vg

12.

VP s 2 - r1 h1 ¥ VQ = s1 - r2 h2 P

VP h1 = VQ h2 V P × VQ

P VP

P

VQ

Q 13.

10. 1 2 1 Ê m0 N 2 A ˆ 2 LI = ÁË ˜I 2 2 l ¯ 2El μ0I 2 = 2 N A [ML2 T -2 ] ¥ [L] [μ0I 2 [L2 ]

E=

1 1 Ê e Aˆ E = CV 2 = Ë 0 ¯ V 2 2 2 l e 0V 2 = e 0V 2 =

2El A

[ML2 T -2 ]×[L] [L2 ]

Q. Q.

r < R. r is 4p 3 M= r r 3 r is GM I= 2 = r

G¥

4p 3 r r 4p G r r 3 = 2 3 r

JAII.9

dr A

r

is m=

× V = dAdr dP

Choice (d) : R P at r = 2 = 21 R 32 P at r = 3

dA is dF = dP × dA 14. C =

mass dF 4p G r A ¥ r dAdr 3

dF = Im =

dp =

e0 S d

4p G r 2 rdr 3 R

P=

4p G r 2 rdr 3 Úr

4p G r 2 2 (R - r 2 ) 2¥3 P = k (r2 – r2

C =

P=

k=

S C is S e1e 0 2 = 2e 0 S = 2C C = 1 d d 2

2p G r 2 3

Choice (a) : Choice (b) : r=

P = kR2

r=

Choice (c) :

r=

=

C 3B , 5

2R 21kR 2 , P4 = 5 25

P3 16 25 16 = ¥ = P4 25 21 21

S S 2e 2 = 0 2 =C 1 d d

e1e 0 C

Ê 9 R 2 ˆ 16kR 2 P3 = k ÁË R 2 ˜= 25 ¯ 25 r=

C =

Ê 2 4 R 2 ˆ 5kR 2 ÁË R ˜= 9 ¯ 9

P1 7 9 63 = ¥ = P2 16 5 80

S 2 = 4e 0 S = 4C 1 d d 2

e 2e 0

Ê 3R 9R2 ˆ 7kR 2 , P = k ÁË R 2 r= ˜¯ = 16 16 4

2R , P2 = k 3

e0 S d

C C2 =

CC ¢ + C ¢¢ C + C¢

=

2C1 ¥ 4C1 + C1 2C1 + 4C1

=

4C1 7C1 + C1 = 3 3

C2 7 = C1 3

JAII.10

P1v1 P2 v2 = T1 T2

15. v2 = 2v

= P1 A +

I2 = 3T

P1v1 P2 ¥ 2v1 = T1 T2

P2 =

= 2 P1v1 + 3P1 2

A

k

16.

2v

140

1 1 ææ Æ54 X e +94 38 Sr + 0 n + 0 n

Q

v = xA v2 = xA

v = xA

x=

v1 A Q = Kxe + Ksr + Kx + Ky

1 2 kx 2

Q=Q

=

1 kx ¥ x 2

=

1 P1 A v1 P1v1 ¥ ¥ = 2 2 A A

Kxe > Ksr + Kx + Ky

a Case (b) : U = nCv T 3R = n¥ ¥ (3T1 - T1 ) 2

= n¥

17.

2mK

q Fm = Fe = qE Where

3R 2 P1v1 = 3P V , ¥ nR 2

P2 =

2v1 A

W = P Ax + kx ¥ x

4 P1 PA , kx = 1 3 3

q B

E K E =

b Case (c) :

P=

Section-III

3R ¥ 2T1 = n¥ 2

x

7 P1v1 9 41 P1v1 + P1v1 = 3 2 6

P1 A 2 xA

v2

9 P1v1 2

x=y=n 236 92 U

3 P1 kx = P1 + 2 A kx =

U=

Q=W+ U= kx A

P1v1 7P v = 11 3 3

Case (d) :

kx A P2 = P1 +

2v1 1 P1 A 2v1 + ¥ ¥ A 2 A 3

V V = KM w

V w qV w V B = w V vw = B

q B =

M

K

M

JAII.11

I I = neA Where A = wd I = ne (wd v vw =

BC n

I ned

r 1/2 = n1 Ê1 - co2 s ˆ Ë m¯ Ê n2 ˆ = n1 Á1 - 22 ˜ Ë n1 ¯

V Choice (a) :

1/2

NA

B V1 ned1 B ned 2

NA =

V2 = 2V d = d2

V = V2

B1 n1ed

1 Ê 45 9 ˆ1/2 9 = 4 Ë16 4 ¯ 16 3

S2 16 3 15

,

Choice (b): S 16 n= , 15

B2 n2 ed

1

V1 B1 n2 = ¥ V2 B2 n1

15 Ê 45 9 ˆ2 3 15 NA = = 16 Ë16 4 ¯ 64

B2 n2 1 ¥ = B2 2n2 2

S2

V2 = 2V

n= 1

3 Ê 64 49 ˆ2 3 15 NA = Ë - ¯ = 4 25 25 20

19.

Choice (c):

n

1 Ê 45 3 NA = Ë - ¯ = 1 16 4 4 S2 4 n= , 15

n2 n1 C,

BC

ic BC

S 1 9 ˆ2

ic ic =

4ˆ Ê , Ën = 3 ¯

S

3 15 ÈÊ8 ˆ2 Ê 7 ˆ2 ˘ 2 9 NA = Í ˙ = 16 ÎË5 ¯ Ë 5 ¯ ˚ 16

Choice (a) : V1 =

=

n=

(n12 = n22 )1/2

1 n

1

B ned

V2 =

im =

Choice (a):

d = d2,

Choice (c) :

1/2

= (n12 = n22 )

w d=d

V2 -

im

B rm

n = sin im = n

B V = ned

18. V =

C.

AB

n is B

n rm =

ic

1

NA =

15 Ê 64 49 ˆ 2 3 = 4 ÁË 25 25 ˜¯ 4

4 3

JAII.12

S Choice (d): 3 NA = 4

n

1

3 NA = 1 ÊÁ 45 - 9 ˆ˜ 2 = 4 1 Ë 16 4 ¯ Ê n = 4ˆ ÁË ˜ 3¯

S2 NA =

S

3 15 20

NA2 =

S

n1 =

45 4

45 9

1

20. Structure S1 3 n2 = 2

n1 =

S2

Structure S2 7 n2 = 5 n1 =

8 5

S2

4 Ê 64 49 ˆ 2 4 = 0.46 ÁË - ˜¯ = 25 25 45 5 3

NA

JEE ADVANCED 2016: PAPER-I (MODEL SOLUTIONS) SECTION I (Single Correct Answer Type) This section contains FIVE questions. Each question has FOUR options (a), (b), (c) and (d). ONLY ONE of these four options is correct.

1. A parallel beam of light is incident from air at an angle α on the side PQ of a right angled triangular prism of refractive index n = 2 . Light undergoes total internal reflection in the prism at the face PR when α has a minimum value of 45o. the angle θ of the prism is P  

n= 2

(a) 15o

Q

(b) 22.5o

(c) 30o

R

(d) 45o

2. In a historical experiment to determine Planck’s constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength (λ) of incident light and the corresponding stopping potential (V0) are given below:

λ(µm) 0.3 0.4 0.5

V0(Volt) 2.0 1.0 0.4

Given that c = 3 × 108 ms–1 and e = 1.6 × 10–19C, Planck’s constant (in units of J s) found from such an experiment is (b) 6.4 × 10–34 (c) 6.6 × 10–34 (d) 6.8 × 10–34 (a) 6.0 × 10–34 3. A water cooler of storage capacity 120 litres can cool water at a constant rate of P watts. In a closed circulation system (as shown schematically in the figure), the water from the cooler is used to cool an external device that generates constantly 3 kW of heat (thermal load). The temperature of water fed into the device cannot exceed 30oC and the entire stored 120 litres of water is initially cooled to 10oC. The entire system is thermally insulated. The minimum value of P (in watts) for which the device can be operated for 3 hours is: (Specific heat of water is 4.2 kJ kg–1 K–1 and the density of water is

1000 kg m–3)

P-I.2 Comprehensive Physics—JEE Advanced

(a) 1600

(b) 2067

(c) 2533

(d) 3933

4. An infinite line charge of uniform electric charge density λ lies along the axis of an electrically conducting infinite cylindrical shell of radius R. At time t = 0, the space inside the cylinder is filled with a material of permittivity ε and electrical conductivity σ. The electrical conduction in the material follows Ohm’s law. Which one of the following graphs best describes the subsequent variation of the magnitude of current density j(t) at any point in the material?

(a)

(b)

(c)

(d)

5. A uniform wooden stick of mass 1.6 kg and length  rests in an inclined manner on a smooth, vertical wall of height h(<  ) such that a small portion of the stick extends beyond the wall. The reaction force of the wall on the stick is perpendicular to the stick. The stick makes an angle of 30o with wall and the bottom of the stick is on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio h/ and the frictional force f at the bottom of the stick are (g = 10 ms–2)

h 3 16 3 ,f N = = (a)  16 3

h 3 16 3 ,f N (b)= =  16 3

h 3 3 8 3 ,f N = = (c)  16 3

h 3 3 16 3 ,f N = = (d)  16 3

SECTION II (One or More than One Options Correct Type) This section contains EIGHT questions. Each question has FOUR options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four option(s) is (are) correct.   6. The position vector r of a particle of mass m is given by the following equation: r= (t ) α t 3iˆ + β t 2 ˆj where

α = 10 / 3 ms–3, β = 5 ms–2 and m = 0.1 kg. At t = 1s, which of the following statement(s) is(are) true about the particle?

  (a) The velocity υ is given by= υ (10iˆ + 10 ˆj ) ms −1   (b) The angular momentum L with respect to the origin is given by L = −(5 / 3)kˆ Nms   (c) The force F is given by F= (iˆ + 2 ˆj ) N   (d) The torque τ with respect to the origin is given by τ = −(20 / 3)kˆ Nm

P-I.3

Physics Paper I—2016

7. A transparent slab of thickness d has a refractive index n(z) that increases with z. Here z is the vertical distance inside the slab, measured from the top. The slab is placed between two media with uniform refractive indices n1 and n2 ( > n1), as shown in the figure. A ray of light is incident with angle θi from medium 1 and emerges in medium 2 with refraction angle θf with a lateral displacement l.

i

n1 = constant

1

n (z)

z

d n2 = constant

Which of the following statement(s) is(are) true ? (a)  is dependent on n2

(b) n1 sin θi = n2 sin θ f

(c)  is independent of n(z)

θi (d) n1 sin=



f

2

( n2 − n1 ) sin θ f

8. A plano-convex lens is made of a material of refractive index n. When a small object is placed 30 cm away in front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection from the convex surface of the lens, another faint image is observed at a distance of 10 cm away from the lens. Which of the following statement(s) is(are) true? (a) The refractive index of the lens is 2.5 (b) The radius of curvature of the convex surface is 45 cm (c) The faint image is erect and real (d) The focal length of the lens is 20 cm 9. Highly excited states for hydrogen-like atoms (also called Rybderg states) with nuclear charge Ze are defined by their principal quantum number n, where n >> 1. Which of the following statement(s) is(are) true? (a) Relative change in the radii of two consecutive orbitals does not depend on Z (b) Relative change in the radii of two consecutive orbitals varies as 1/n (c) Relative change in the energy of two consecutive orbitals varies as 1/n3 (d) Relative change in the angular momenta of two consecutive orbitals varies as 1/n 10.

A length-scale (  ) depends on the permittivity (ε) of a dielectric material, Boltzmann constant (kB), the absolute temperature (T), the number per unit volume (n) of certain charged particles, and the charge (q) carried by each of the particles. Which of the following expression(s) for  is (are) dimensionally correct?

 nq 2  (a)  =    ε k BT 

  q2 (c)  =  2/3   ε n k BT 

 εk T  (b)  =  B2   nq 

 q2  (d)  =  1/3   ε n k BT 

11. Two loudspeakers M and N are located 20 m apart and emit sound at frequencies 118 Hz and 121 Hz, respectively. A car is initially at a point P, 1800 m away from the midpoint Q of the line MN and moves towards Q constantly at 60 km/hr along the perpendicular bisector of MN. It crosses Q and eventually reaches a point R, 1800 m away from Q. Let v(t) represent the beat frequency measured by a person sitting in the car at time t. Let vP, vQ and vR be the beat frequencies measured at locations P, Q and R, respectively. The speed of sound in air is 330 ms−1. Which of the following statement(s) is(are) true regarding the sound heard by the person? (a) The plot below represents schematically the variation of beat frequency with time v(t) P VQ

Q R

t

P-I.4 Comprehensive Physics—JEE Advanced

(b) νP + νR = 2νQ (c) The plot below represents schematically the variation of beat frequency with time v(t) P VQ

Q R

t

(d) The rate of change in beat frequency is maximum when the car passes through Q 12. A conducting loop in the shape of a right angled isosceles triangle of height 10 cm is kept such that the 90° vertex is very close to an infinitely long conducting wire (see the figure). The wire is electrically insulated from the loop. The hypotenuse of the triangle is parallel to the wire.

10 cm

90º

The current in the triangular loop is in counterclockwise direction and increases at a constant rate of 10 As–1. Which of the following statement(s) is(are) true? (a) The induced current in the wire is in opposite direction to the current along the hypotenuse

 µ0  (b) The magnitude of induced emf in the wire is   volt π  (c) There is a repulsive force between the wire and the loop

 µ0  (d) If the loop is rotated at a constant angular speed about the wire, an additional emf of   volt is π  induced in the wire 13.

An incandescent bulb has a thin filament of tungsten that is heated to high temperature by passing an electric current. The hot filament emits black-body radiation. The filament is observed to break up at random locations after a sufficiently long time of operation due to non-uniform evaporation of tungsten from the filament. If the bulb is powered at constant voltage, which of the following statement(s) is(are) true? (a) The temperature distribution over the filament is uniform (b) The resistance over small sections of the filament decreases with time (c) The filament emits more light at higher band of frequencies before it breaks up (d) The filament consumes less electrical power towards the end of the life of the bulb

SECTION II (Single Digit Integer Type) This section contains FIVE questions. This answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive 14. Two inductors L1 (inductance 1 mH, internal resistance 3Ω) and L2 (inductance 2 mH, internal resistance 4Ω), and a resistor R (resistance 12Ω) are all connected in parallel across a 5V battery. The circuit is switched on at time t = 0. the ratio of the maximum to the minimum current (Imax / Imin) drawn from the battery is ________ .

P-I.5

Physics Paper I—2016

15. A metal is heated in a furnace where a sensor is kept above the metal surface to read the power radiated (P) by the metal. The sensor has a scale that displays log2 (P/P0), where P0 is a constant. When the metal surface is at a temperature of 487oC, the sensor shows a value 1. Assume that the emissivity of the metallic surface remains constant. What is the value displayed by the sensor when the temperature of the metal surface is raised to 2767oC? 16. A hydrogen atom in its ground state is irradiated by light of wavelength 970Å. Taking hc = 1.237 × 10−6 eV m and the ground state energy of hydrogen atom as −13.6 eV, The number of lines present in the emission spectrum is ________ . 17. Consider two solid spheres P and Q each of density 8 gm cm–3 and diameter 1 cm and 0.5 cm, respectively. Sphere P is dropped into a liquid of density 0.8 gm cm–3 and viscosity η = 3 poiseulles. Sphere Q is dropped into a liquid of density 1.6 gm cm–3 and viscosity η = 2 poiseulles. The ratio of the terminal velocities of P and Q is ________ . 18. The isotope

(

12 6

C*

)

12 5

B having a mass 12.014 u undergoes β-decay to

at 4.041 MeV above its ground state. If

12 5

12 6

B decays to

C. 12 6

12 6

C has an excited state of the nucleus

C* , the maximum kinetic energy of the

β-particle in units of MeV is ________ . (1u = 931.5 MeV/c2, where c is the speed of light in vacuum).

ANSWERS 1. (a)

2. (b)

3. (b)

4. (a)

6. (a), (b), (d)

7. (a), (b), (c)

8. (a), (d)

9. (a), (b), (d)

11. (a), (b), (d)

12. (b), (c)

13. (c), (d)

16. 6

17. 3

18. 9

5. (d) 10. (b), (d)

14. 8

15. 9

Solutions 1. The adjoining figure shows the ray diagram. Applying Snell’s law at B,

2 sin β

1.0 sin 45º = sin β =

⇒

A

1 2

⇒ β = 30º For total internal reflection at C, the critical angle (iC) is given by sin ic =

1 2

⇒ iC = 45o

In triangle PBC,

θ + 90o + β + γ = 180º ⇒ ⇒ ⇒ ⇒

P

θ + 90º + β + (90º – 45º) = 180º θ + β = 45º θ + 30º = 45º θ = 15º, which is option (a).

 

B

Q

β

γ 45º

C R

P-I.6 Comprehensive Physics—JEE Advanced

hν = eV0 + w0

2.

hc

⇒

λ

= eV0 + w0

V0 =

⇒ So the graph V0 vs

1

λ

hc w0 – eλ e

is a straight line of slope m =

...(1) hc . Using the first two sets of observations given in the e

question, we have

1 1  (V0)1 – (V0)2 = m  −   λ1 λ2  ⇒

1 1   −1 m − (2.0 – 1.0) volt = m  −6 −6  0.4 × 10   0.3 × 10 m = 1.2 × 10–6 Vm

⇒ But m =

hc . Hence e

h = ⇒ So the correct option is (b).

me (1.2 × 10−6 ) × (1.6 × 10−19 ) = = 6.4 × 10–34 VCs c 3 × 108

h = 6.4 × 10–34 Js

3. Volume of water in the cooler is V = 120 litre = 120 × 10–3 m3. Density of water is ρ = 103 kg m–3. Therefore, the mass of water is m = ρV = (120 × 10–3) × 103 = 120 kg The temperature of water decreases from 30ºC to 10ºC in time t = 3 hours = (3 × 60 × 60)s Out of 3000 W, water loses heat energy = 3000 – P in t seconds. Hence (3000 – P)t = 120 × 4200 × (30 – 10) ⇒ (3000 – P) × (3 × 60 × 60) = 120 × 4200 × 20 ⇒ P = 2067 W So the correct choice is (b). 4. Let L be the length of the wire and Q0 be the initial (i.e. at t = 0) charge distributed uniformly on it. The initial linear charge density is λ0 =

Q0 . Let Q be the wire at an instant t. The linear charge density of this instant is L

Q . The magnitude of the electric field at a distance r from the wire placed in a medium of permittivity ∈ t at time t is

λ=

Q λ = 2π ∈ r 2π ∈ rL From Ohm’s law, the relation between current density j and electric field E is

E =

σQ − j = −σ E = 2π ∈ rL

...(1)

Physics Paper I—2016

P-I.7

The negative sign indicates that the electrons drift in a direction opposite to that of E. Since the shell is hollow, the current flows only along the surface of the cylinder whose area is A = 2πRL. Since j =

I , A

I = jA = j × 2πRL where I is the current at time t. Since I =

dQ , we have dt

dQ σQ × 2π RL = j × 2πRL = – 2π ∈ rL dt

⇒

dQ σR = –kQ, where k = dt ∈r

⇒

dQ Q = –kdt Q

⇒

∫

Q0

t dQ − k ∫0 dt Q =

Q ln   = kt  Q0  Q = Q0 e–kt Hence the instataneous current density is ⇒

j =

Q0 − kt Q = e 2π RL 2π RL

Hence j falls exponentially with time. It is clear that at t = 0, j =

Q0 and as t → ∞, t → 0. So the correct graph is (a). 2π RL

5. Refer to the adjoining figure which shows the forces acting on the stick AB with A touching the wall and end B touching the floor. To prevent slipping of the stick, the friction force must act at A in the upward direction and at B towards the left. Let R be the reaction force (which is the resultant of the normal reaction and the frictional force) at A and N be the normal reaction at B. The entire mass of the stick may be assumed to be concentrated at its centre O. Since the stick is in translational equilibrium, the total upward force = total downward force, i.e. R sin 30º R sin 30º + N = Mg R ⇒

A

R + N = 1.6 × 10 = 16 2

30º

It is given that N = R. Hence

⇒

R =

32 newton 3

32 3 16 3 × = newton 3 2 3

N

Mg

C

Also the net horizontal force on the stick must be zero, i.e. f = R cos 30º =

O

h

R + R = 16 2

30º R cos 30º

D

60º f

B

P-I.8 Comprehensive Physics—JEE Advanced

Since the stick is also in rotational equilibrium, the total clockwise torque about B = total counterclockwise torque about B, i.e. Mg × BD = f × 0 + R × AB Mg × ⇒

 h cos 60º = R × 2 sin 60o

1.6 × 10 ×

32 2h  1 × × = 3 3 2 2 h 3 3 =  16

⇒

So the correct option is (d). 6. Substituting the given value of α and β, we have →

r =

10 3 ˆ t i + 5t 2 ˆj 3 →

(a)

dr velocity υ = dt →

=

d  10 3 ˆ 2 ˆ 2  t i + 5t j  = 10t i + 10tjˆ dt  3

→

Putting t = 1s (given), υ = 10(iˆ + ˆj ) ms–1. So option (a) is correct. (b) Angular momentum with respect to the origin is →

→

→

→

→

→

→

L = r × p = r × (mυ ) =m ( r × υ )

 10 3 3 ˆ ˆ ˆ = m  3 t i + 5t j  × 10i + 10 j   

(

 100 3  t − 50t 2  kˆ = m   3 → 5 Putting t = 1s and m = 0.1 kg, we get L = − kˆ kgm2 s–1. 3

So option (b) is also correct. (c)

→

dυ F = ma = m dt →

→

= m

d 10t 2 iˆ + 10tjˆ dt

(

(

= m 20tiˆ + 10 ˆj

)

)

= 0.1  (20 × 1)iˆ + 10 ˆj 

)

P-I.9

Physics Paper I—2016

= (2iˆ + ˆj ) ms−2 So option (c) is not correct. →

(

→

→

)

 10 3  t i + 5t 2 ˆj  × m ( 20tiˆ + 10 ˆj )  3 

τ = r× F = 

(d)

 100 3 ˆ  t k − 100t 2 k  = m   3  100  × (1)3 kˆ − 100 × (1)2 k  = 0.1  3 = −

20 ˆ k Nm, which is option (d). 3

So the correct options are (a), (b) and (d). 7. The transparent slab can be assumed to consist of a large number N of slabs labelled I, II, III, ..... etc. of refractive indices nI > nII > nIII ... etc. as shown in the figure which shows refractions at the interfaces of I, and II, II and III and so on. Applying Snell’s law at each interface, we have i n1 1 nI sin θi = nI sin θI nI sin θI = n2 sin θII nI I I n2 sin θII = n3 sin θIII nII

nN–1 sin θN–1 = n2 sin θf From the above equations it follows that

nIII nIV

II

II

III

III

IV

IV t

2

n1 sin θi = n2 sin θf So option (b) is correct. Since N is large, the path of the refracted ray in the slab is a curve whose curvature depends on how n varies with z. Hence the lateral displacement  depends on how n depends on z. So option (c) is also correct. Since the curvature of the curve does not depend on n2,  is independent of n2. So option (a) is correct. Thus the correct options are (a), (b) and (c). 8. Given u = – 30 cm and υ = + 60 cm. Using the lens formula 1

υ

−

1 1 = f , we have u

1 1 − = ⇒ f = 20 cm 60 −30

So option (d) is correct. For reflection at the convex surface of radius of curvature R, u = – 30 cm and υ = +10 cm. Using the spherical mirror formula 1

υ

+

2 1 = , we have R u

1 1 2 + = ⇒ R = 30 cm 10 −30 R

P-I.10 Comprehensive Physics—JEE Advanced

Using lens maker’s formula 1  1 1 = (µ – 1)  +  ⇒ µ = 2.5 30 ∞ 20

So option (a) is also correct. Further, a convex reflecting surface never forms a real image. So option (c) is incorrect. Hence the correct options are (a) and (d). 9. The radius of nth orbit of the hydrogen like atom is

kn2 , where k = 0.53 Å n = Z ∴ The relative change in the radii of two consecutive states is

k ( n + 1)2 kn2 − 2 2 rn +1 − rn Z Z = ( n + 1) − n = kn 2 n2 rn Z 2n + 1 2 1 2 = + 2  = 2 n n n n Thus

( n >> 1)

rn +1 − rn 1 ∝ , which does not depend on Z. So options (a) and (b) are both correct. The energy of nth rn n

state is En = – C

∴

En +1 − En = En

−

Z2 where C = 13.6 eV n2

CZ 2 CZ 2 + (n + 1) 2 n2 CZ 2 − 2 n

=

n2 −1 (n + 1) 2

=

n 2 − (n + 1) 2 (n + 1) 2

=

−2n − 1 n2

= −

2 1 − 2 2 n n

= −

2 n

So option (c) is incorrect. Angular momentum is L = ∴

nh 2π

Ln +1 − Ln (n + 1) − n 1 = = Ln n n

So option (d) is correct. Hence the correct options are (a), (b) and (d).

( n >> 1)

( n >> 1)

Physics Paper I—2016

P-I.11

10. The dimensional formulae of the given quantities are: [n] = [L–3] [q] = [AT] [∈] = [M–1 L–3 T4 A2] [T] = [K]

[ ] = [L] [kB] = [M1 L2 T–2 K–1]

11.

Substituting these in the expression of  given in options (a), (b), (c) and (d) we find that options (b) and (d) have dimension of length [L]. The dimension of the expression in option (a) is [L–1] and in option (c) is [L3/2]. Hence the only correct options are (b) and (d). Let ν1 and ν2 be the frequencies of the sound of loud speakers N and M respectively. Let the car be at point O at time t. The point O lies between P and Q as shown in the figure. If u is the speed of the car (observer) and υ is the speed of sound. As the observer (in the car) travels from O to Q, he approaches the sources of sound M and N with speed u cos θ. At O, the observer hears sounds of apparent frequencies given by

 υ + u cos θ  v 1' = v1   υ   and

P O  

M

Q

N

10 m 1800 m

R

 υ + u cos θ  v '2 = v2   υ  

The beat frequencies v is

 υ + u cosθ  v = v 1' – v '2 = (v1 – v2) 1   υ    u cos θ  v = (v1 – v2) 1 +  υ   The rate of change of beat frequency with time is given by d u dθ (v) = −(v1 − v2 ) sin θ dt υ dt

...(1)

When the car crosses the point Q, the observer recedes from M and N with speed u cosθ and he hears a beat frequency v ' given by

 u cos θ  v ' = (v1 − v2 ) 1 −  υ   ∴

dθ d u (v ') = (v1 − v2 )   sin θ υ dt dt  

...(2)

dv = 0 , i.e. the slope of v versus t graph is zero. dt At R, θ = 0º. Hence the slope of the graph is again zero. But at Q, θ = 90º. Hence the slope of the graph at Q is maximum which is option (d). So the current graph is (a). At P, θ = 0º.

When the car is at P, θ = 0º as P is far away from Q. Hence

P-I.12 Comprehensive Physics—JEE Advanced

 u vp = (v1 − v2 ) 1 +   υ At R, θ = 0º.

 u vR = (v1 − v2 ) 1 −   υ ∴ vP + vR = 2 (v1 – v2), which is option (b). So the correct options are (a), (b) and (d). 12. Refer to the adjoining figure. Divide the conducting loop ABC into small elements each of small width dr. Consider one such element EF at a distance r =AO. Now EO = AO tan 45º = AO = r and OF = r so that EF = 2r. The area of the element is dA = 2rdr The magnetic field due to current I at a distance r from it is B =

P

45º 45º

h = 10 cm

O

E r B

Q

A

F r

D

dr C

µ0 I 2π r

Magnetic flux through the element is dφ =BdA =

µ0 I µ0 Idr × 2rdr = π 2π r

Magnetic flux through the triangular loop is = φ = ∫ dφ

Induced emf is ε =– Given

µ0 I r = h µ0 Ih = dr ∫ π r=0 π

µ h dI dφ = − 0 π dt dt

dI = 10 As −1 and h = 10 cm = 0.1 m. Thus dt

|ε| =

µ0 µ × 0.1 × 10 =0 volt π π

So option (b) is correct. It is given that the induced current in loop is counterclockwise. This is possible only if the current in the wire PQ is towards right because then the magnetic filed will be out of the plane of the loop. The magnetic field due to the triagular loop at any point on the wire PQ is into the page. Hence the force on the wire PQ is directed away from the loop, i.e., the force is repulsive. So the correct options are (b) and (c). 13. It is given that the evaporation of tungsten from the filament is non-uniform, the temperature at different points on the filament must be different. Hence option (a) is incorrect. Due to evaporation, the cross-sectional area of the wire decreases. Hence the resistance of different sections of the filament increases. So option (b) is also incorrect. To check whether option (c) is correct or not, consider a small section of length  and cross-sectional area A of the filament. If I is the current in the filament, the power radiated by the element is

Physics Paper I—2016

P = I 2R =

P-I.13

I 2 ρL A

where ρ is the resistivity of tungsten. From Stefan’s law, 4 P = ∈σ T A where ∈ is the emissivity. Equating the two expressions for P, we get

I 2ρL A

∈σ T 4 A =

 I 2 ρL  1 T =   ∈σ  A2 4

or

1

T ∝

or

A As the evaporation continues, A becomes smaller and smaller. Hence the temperature of the element becomes higher and higher. Hence the element radiates electro-magnetic waves of a high energy and hence of a high frequency (because E = hν). So option (c) is correct. The power is also given by P =

V2 R

Since voltage V remains constant and R becomes very large (due to decrease in cross-sectional area caused by evaporation), the electrical power P consumed by the filament becomes small. So option (d) is also correct. Thus the correct options are (c) and (d). 14. Refer to the following figures. I=0

1 mH 3Ω

I=0

4Ω I I

12Ω

2 mH

⇒

12Ω

5V

5V

Fig. 1 3Ω 1.5Ω 4Ω

⇒

Imax

12Ω

5V

5V

Fig. 2

P-I.14 Comprehensive Physics—JEE Advanced

Where the circuit is switched on at time t = 0, the current in both the inductors is zero because the induced back emf prevents the flow of current. The current flows only through the 12Ω resistor and is minimum (Fig. 1) Imin =

5V 5 = A 12Ω 12

After some time, the steady current flows in each inductor. Hence there is no back emf induced in the inductor (because

dI = 0 ) and, therefore, they behave as pure resistors. The equivalent resistance R is given by (Fig. 2) dt 1 1 1 1 ⇒ R = 1.5 Ω = + + R 3 4 12

Since the resistance is minimum = 1.5 Ω, the current is maximum.

5V 10 A Imax = 1.5Ω = 3

10 I max 3 = 5 =8 I min 12

∴ The correct answer is 8. 15. Given

First reading of the sensor is

T1 = 273 + 487 = 760 K T2 = 273 + 2767 = 3040 K  ∈σ AT14  S1 = log 2   P0 

Second reading of the sensor is  ∈σ AT24  S2 = log 2   P0   ∈σ AT24   ∈σ AT14  S2 – S1 = log 2  log − 2  P0   P0   ∈σ AT24 P0  × = log 2   ∈σ AT14   P0

T  = log 2  2   T1 

4

T  = 4 log 2  2   T1   3040   = 4 log2(4) = 4 log2 (22) = 4 × 2 = 8 = 4 log 2  700  Given The correct answer is 9.

S1 = 1. Hence S2 = 8 + 1 = 9

Physics Paper I—2016

16. Energy of incident photon = hν = =

P-I.15

hc

λ

1.237 × 10−6 eVm 970 × 10−10 m

= 12.75 eV Energy of hydrogen atom in the ground state = – 13.6 eV When the hydrogen atom in the ground state absorbs the photon, the electron jumps to a state En given by En = –13.6 + 12.75 = – 0.85 eV En =

Now, ⇒

−13.6eV n2

– 0.85 eV = –

−13.6eV ⇒ n = 4. n2

The following transitions can take place. n = 4 → n = 3, n = 4 → n = 2, n = 4 → n = 1, n = 3 → n = 2, n = 3 → n = 1 and n = 2→ n = 1 So the number of spectral lines = 6.

2 r 2 (σ − ρ) g 17. Terminal velocity Vt = 9 η For sphere P: r1 = 0.5 cm, σ1 = 8 g cm–3, ρ1 = 0.8 g cm–3 and , η1 = 3 P 

For sphere Q: r2 = 0.25 cm, σ2 = 8 g cm–3, ρ2 = 1.6 g cm–3 and , η2 = 2 P  2

(Vt ) P  r1   σ1 − ρ1   η2  × =    (Vt )Q  r2   σ 2 − ρ2   η1 

∴

2

 0.5   8 − 0.8   2  =   ×  ×  0.25   8 − 1.6   3  =3 The correct answer is 3. 18. The nuclear reaction is represented by the equation 12 5

B  → 126 C+ 10 e +ν + Q

By definition, mass of

12 6

C = 12.000 u

The energy released is Q = (mB − mA ) c 2 = (12.014 – 12.000) × 931.5 = 13.041 MeV Out of this 4.041 MeV of energy is taken by 126C*. Therefore, the maximum kinetic energy of β particle ( −01 e ) = 13.041 – 4.041 = 9 MeV. Then the kinetic energy of antinutrino (ν ) is zero. So the correct answer is 9.



JEE ADVANCED 2016: PAPER-II (MODEL SOLUTIONS) SECTION I (Single Correct Answer Type) This section contains FIVE questions. Each question has FOUR options (a), (b), (c) and (d). ONLY ONE of these four options is correct. 1. There are two Vernier calipers both of which have 1 cm divided into 10 equal divisions on the main scale. The Vernier scale of one of the calipers (C1) has 10 equal divisions that correspond to 9 main scale divisions. The Vernier scale of the other caliper (C2 ) has 10 equal divisions that correspond to 11 main scale divisions. The readings of the two calipers are shown in the figure. The measured values (in cm) by calipers C1 and C2, respectively, are : 2

3

4

2

C1

3

4

C2 0

(a) 2.87 and 2.86

5

10

0

(b) 2.85 and 2.82

(c) 2.87 and 2.87

5

10

(d) 2.87 and 2.83

2. The electrostatic energy of Z protons uniformly distributed throughout a spherical nucleus of radius R is given by E=

3 Z ( Z − 1)e2 5 4πε 0 R

The measured masses of the neutron, 11 H,

15 7

N and 158 O are 1.008665 u, 1.007825 u, 15.000109 u and

15 15 15.003065 u, respectively. Given that the radii of both the 7 N and 8 O nuclei are same, 1u = 931.5 MeV/ c2 2 (c is the speed of light) and e /(4πε0 ) = 1.44 MeV fm. Assuming that the difference between the binding

energies of

15 7

10–15 m) (a) 2.85 fm

N and 158 O is purely due to the electrostatic energy, the radius of either of the nuclei is : (1 fm =

(b) 3.03 fm

(c) 3.42 fm

(d) 3.80 fm

3. The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a length of 1 m at 10ºC. Now the end P is maintained at 10ºC, while the end S is heated and maintained at 400ºC. The system is thermally insulated from its surroundings. If the thermal conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of PQ is 1.2 × 10–5K–1, the change in length of the wire PQ is : (a) 0.78 mm (b) 0.90 mm (c) 1.56 mm (d) 2.34 mm

P-II.2 Comprehensive Physics—JEE Advanced

4. A small object is placed 50 cm to the left of a thin convex lens of focal length 30 cm. A convex spherical mirror of radius of curvature 100 cm is placed to the right of the lens at a distance of 50 cm. The mirror is tilted such that the axis of the mirror is at an angle θ = 30º to the axis of the lens, as shown in the figure. f = 30 cm

 (–50, 0)

x

(0, 0)

R = 100 cm 50 cm 50 + 50 3, – 50

If the origin of the coordinate system is taken to be at the centre of the lens, the coordinates (in cm) of the point (x, y) at which the image is formed are : (a) (25, 25 3 )

(c) (125/3, 25/ 3 )

(b) (0, 0)

(d) (50 – 25 3 ,25)

5. A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure Pi = 105| Pa and volume Vi 10–3 m3 changes to a final state at Pf = (1/ 32) × 105 Pa and Vf = 8 × 10–3 m3 in an adiabatic quasi-static process, such that P3V5 = constant. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps: an isobaric expansion at Pi followed by an isochoric (isovolumetric) process at volume Vf . The amount of heat supplied to the system in the twostep process is approximately : (a) 112 J (b) 294 J (c) 588 J (d) 813 J 6. An accident in a nuclear laboratory resulted in deposition of a certain amount of radioactive material of halflife 18 days inside the laboratory. Tests revealed that the radiation was 64 times more than the permissible level required for safe operation of the laboratory. What is the minimum number of days after which the laboratory can be considered safe for use? (a) 64 (b) 90 (c) 108 (d) 120

SECTION II (One or More than One Options Correct Type) This section contains EIGHT questions. Each question has FOUR options (a), (b), (c) and (d). ONE OR MORE THAN ONE of these four option(s) is (are) correct. 7. Two thin circular discs of mass m and 4m, having radii of a and 2a, respectively, are rigidly fixed by a massless, rigid rod of length l = 24 a through their centres. This assembly is laid on a firm and flat surface, and set rolling without slipping on the surface so that the angular speed about the axis of the rod is ω. The →

angular momentum of the entire assembly about the point ‘O’ is L (see the figure). Which of the following statement(s) is(are) true? 4m m

z

ω 

 O

2a a

Physics Paper II—2016

P-II.3

(a) The magnitude of angular momentum of the assembly about its center of mass is 17ma2ω / 2 (b) The center of mass of the assembly rotates about the z-axis with an angular speed of ω / 5 →

(c) The magnitude of the z-component of L is 55ma2ω (d) The magnitude of angular momentum of center of mass of the assembly about the point O is 81ma2ω 8. A rigid wire loop of square shape having side of length L and resistance R is moving along the x-axis with a constant velocity v0 in the plane of the paper. At t = 0, the right edge of the loop enters a region of length 3L where there is a uniform magnetic field B0 into the plane of the paper, as shown in the figure. For sufficiently large v0, the loop eventually crosses the region. Let x be the location of the right edge of the loop. Let υ(x), I(x) and F(x) represent the velocity of the loop, current in the loop, and force on the loop, respectively, as a function of x. Counter-clockwise current is taken as positive. × × × × × × × × × ×

R L υ0

× × × × × × × × × ×

× × × × × × × × × ×

× × × × × × × × × ×

× × × × × × × × × ×

× × × × × × × × × ×

0

× × × × × × × × × ×

× × × × × × × × × ×

× × × × × × × × × ×

× × × × × × × × × ×

× × × × × × × × × ×

× × × × × × × × × ×

× × × × × × × × × ×

× × × × × × × × × ×

×× ×× ×× ×× ×× ×× ×× ×× ×× ××

2L

L

× × × × × × × × × ×

x

3L

4L

Which of the following schematic plot(s) is(are) correct? (Ignore gravity) I(x)

I(x)

F(x)

υ(x) υ0

0

L

2L

3L

4L

x

(a)

0

L

2L

(b)

3L

4L

x

0

L

2L

(c)

3L

4L

x

3L 0

L

2L

4L

x

(d)

9. In an experiment to determine the acceleration due to gravity g, the formula used for the time period of a periodic motion is T = 2π

7( R − r ) . The values of R and r are measured to be (60 ± 1) mm and (10 ± 1) mm, 5g

respectively. In five successive measurements, the time period is found to be 0.52 s, 0.56 s, 0.57 s, 0.54 s and 0.59 s. The least count of the watch used for the measurement of time period is 0.01 s. Which of the following statement(s) is(are) true? (a) The error in the measurement of r is 10% (b) The error in the measurement of T is 3.57% (c) The error in the measurement of T is 2% (d) The error in the determined value of g is 11%

P-II.4 Comprehensive Physics—JEE Advanced

10.

Light of wavelength λph falls on a cathode plate inside a vacuum tube as shown in the figure. The work function of the cathode surface is φ and the anode is a wire mesh of conducting material kept at a distance d from the cathode. A potential difference V is maintained between the electrodes. If the minimum de Broglie wavelength of the electrons passing through the anode is λe, which of the following statement(s) is(are) true? (a) For large potential difference (V >> φ / e), λe is approximately halved if V is made four times (b) λe decreases with increase in φ and λph (c) λe increases at the same rate as λph for λph < hc/φ (d) λe is approximately halved, if d is doubled

11. Consider two identical galvanometers and two identical resistors with resistance R. If the internal resistance of the galvanometers RC < R / 2, which of the following statement(s) about any one of the galvanometers is(are) true? (a) The maximum voltage range is obtained when all the components are connected in series (b) The maximum voltage range is obtained when the two resistors and one galvanometer are connected in series, and the second galvanometer is connected in parallel to the first galvanometer (c) The maximum current range is obtained when all the components are connected in parallel (d) The maximum current range is obtained when the two galvanometers are connected in series and the combination is connected in parallel with both the resistors

z

Screen

12. While conducting the Young’s double slit experiment a student replaced the two slits with a large opaque plate in the x-y plane containing two small holes that act as two coherent point sources (S1,S2) emitting light of wavelength 600 nm. The student mistakenly placed the screen parallel to the x-z plane (for z > 0) at a distance D = 3m from the mid-point of S1S2, as shown schematically in the figure. The distance between the sources d = 0.6003 mm. The origin O is at the intersection of the screen and the line joining S1 S2. Which of the following is(are) true of the intensity pattern on the screen?

O S1

d

y

S2 D

(a) (b) (c) (d)

x

Hyperbolic bright and dark bands with foci symmetrically placed about O in the x-direction Straight bright and dark bands parallel to the x-axis Semi circular bright and dark bands centered at point O The region very close to the point O will be dark

Physics Paper II—2016

P-II.5

13. A block with mass M is connected by a massless spring with stiffness constant k to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude A about an equilibrium position x0; Consider two cases: (i) when the block is at x0; and (ii) when the block is at x = x0 + A . In both the cases, a particle with mass m(<M) is softly placed on the block after which they stick to each other. Which of the following statement(s) is(are) true about the motion after the mass m is placed on the mass M?

M , whereas in the second case m+M

(a) The amplitude of oscillation in the first case changes by a factor of

it remains unchanged (b) The final time period of oscillation in both the cases is same (c) The total energy decreases in both the cases (d) The instantaneous speed at x0 of the combined masses decreases in both the cases 14. In the circuit shown below, the key is pressed at time t = 0. Which of the 40 µF 25Ω following statement(s) is (are) true? – (a) The voltmeter displays –5V as soon as the key is pressed, and displays V + +5V after a long time 50Ω A 20 µF (b) The voltmeter will display 0 V at time t = In 2 seconds + – (c) The current in the ammeter becomes 1/e of the initial value after 1 second Key (d) The current in the ammeter becomes zero after a long time 5V

SECTION II (Single Digit Integer Type) This section contains FIVE questions. This answer to each question is a SINGLE DIGIT INTEGER ranging from 0 to 9, both inclusive PARAGRAPH FOR QUESTIONS 15-16 A frame of reference that is accelerated with respect to an inertial frame of reference is called a non-inertial frame of reference. A coordinate system fixed on a circular disc rotating about a fixed axis with a constant angular velocity ω is an example of



→

a non-inertial frame of reference. The relationship between the force F rot experienced by a particle of mass m moving on the rotating disc and the force in

R

m

→

F in experienced by the particle in an inertial frame of reference is →

→

→

→

→

→

R/2

→

F rot = F in + 2 m (υ rot × ω ) + m (ω × r ) × ω →

→

where υ rot is the velocity of the particle in the rotating frame of reference and r is the position vector of the particle with respect to the centre of the disc. Now consider a smooth slot along a diameter of a disc of radius R rotating counter-clockwise with a constant angular speed ω about its vertical axis through its center. We assign a coordinate system with the origin at the center of the disc, → the x-axis along the slot, the y-axis perpendicular to the slot and the z-axis along the rotation axis (ω = ω kˆ) . A small →

block of mass m is gently placed in the slot at ( r = ( R / 2)iˆ at t = 0 and is constrained to move only along the slot. 15. The distance r of the block at time t is : (a)

R 2ω t (e + e −2ω t ) 4

(b)

R ωt (e + e − ω t ) 4

(c)

R cos 2ω t 2

(d)

R cos ω t 2

P-II.6 Comprehensive Physics—JEE Advanced

16. The net reaction of the disc on the block is: 1 mω 2 R ( e 2ωt − e −2ωt ) ˆj + mgkˆ 2 1 mω 2 R ( eωt − e −ω t ) ˆj + mgkˆ (d) 2

(a) − mω 2 R cos ω t ˆj − mgkˆ

(b)

(c) mω 2 R sin ω t ˆj − mgkˆ

PARAGRAPH 2 Consider an evacuated cylindrical chamber of height h having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius r << h . Now a high voltage source (HV) is connected across the conducting plates such that the bottom plate is at +V0 and the top plate at –V0. Due to their conducting surface the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible.(Ignore gravity)

A

– +

HV

17. Which one of the following statements is correct ? (a) The balls will bounce back to the bottom plate carrying the opposite charge they went up with (b) The balls will stick to the top plate and remain there (c) The balls will execute simple harmonic motion between the two plates (d) The balls will bounce back to the bottom plate carrying the same charge they went up with 18. The average current in the steady state registered by the ammeter in the circuit will be : (a) proportional to V01/2

(b) zero

(c) proportional to V02

(d) proportional to the potential V0

ANSWERS 1. (d)

2. (c)

3. (a)

4. (a)

6. (c)

7. (a), (b)

8. (b), (d)

9. (a), (b), (d)

11. (a), (c)

12. (c), (d)

13. (a), (b), (d)

16. (d)

17. (a)

18. (c)

14. (a), (b), (c), (d)

5. (c) 10. (a) 15. (b)

Solutions 1. If n vernier scale divisions coincide with (n – 1) main scale divisions (as is the case for vernier calipers C1), the length of a small object is measured by the usual method which gives. L1 = main scale reading + n × V.C. where n is the number of the vernier scale division which coincides with a main scale division and V.C. is the vernier constant. For C1, n = 7 and V.C. = 0.1 mm = 0.01 cm. Hence [see figure]

Physics Paper II—2016

L1 = (2.8 + 7 ×0.01) cm ⇒ L1 = 2.87 cm For vernier calipers C2, n vernier scale divisions concide 2 with (n + 1) main scale divisions. In this case, the length of a vernier scale division is greater than that of a main C2 scale division. So, for C2 the usual method is not applicable. The following method is to be used. Refer to the adjoining magnified diagram of C2. In ths case, the value of x is [see figure] x = 8 × 1 mm – 7 × 1.1 mm = 0.3 mm = 0.03 cm The length measured by C2 is L2 = main scale reading + x = 2.8 cm + 0.03 cm = 2.83 cm Hence the correct option is (d). 2. For nitrogen (Z = 7). Therefore,

P-II.7

8 × 1mm 3

4

123456 0 x 7 × 1.1 mm

7 × 7 − 1)e 2 126 e 2 3 × 4π ∈ R = 5 4π ∈ R EN = 5 0 0 For oxygen (Z = 8). Therefore,

3 8 × (8 − 1)e 2 168 e2 EO = 5 × 4π ∈ R = 5 4π ∈ R 0 0

42 e 2 ∴ ∆E = EO – EN = 5 4π ∈0 R

...(1)

Binding energy of 57 N is (B.E.)N = 7 × 1.007825 + 8 × 1.008665 – 15.000109 = 0.123986 u Binding energy of

15 8

O is

(B.E.)O = 8 × 1.007825 + 7 × 1.008665 – 15.003065 = 0.12019 u Difference in binding energy = 0.123986 – 0.12019 = 0.003796 u This must be equal to ∆E in Eq. (1). Hence

42 e2 = (0.003796) × 931.5 MeV 5 4π ∈0 R ⇒

42 1 × (1.44 MeV fm) × = 0.003796 × 931.5 MeV 5 R

⇒ R = 3.42 fm So the correct option is (c). 3. Let T0 be the temperature of the junction of wires PQ and RS (see figure) below: T1 = 10 ºC

T0 P

Q, R

S

T2 = 400 ºC

P-II.8 Comprehensive Physics—JEE Advanced

In the steady state, the rate of flow of heat in wire PQ = rate of flow of heat in wire RS. Hence 2k (T0 − 10) A 2(400 − T0 ) A =  

⇒ 2(T0 – 10) = 400 – T0 ⇒ T0 = 140ºC Here we have assumed that the two wires have the same cross-sectional area A. The average temperature of wire PQ is T =

1 (140 + 10) = 75o C . Hence the increase in the length of PQ is 2

∆ = α (T − T1 ) = (1.2 × 10–5) × (1.0) × (75 – 10) = 78 × 10–5 m = 0.78 mm So the correct choice is (a).

1 1 1 , 4. For the image formed by the lens, u = – 50 cm, f = + 30 cm. Using the lens formula υ − u = f we have 1

1 1 = ⇒ υ = 75 cm. This image is inverted and is formed on the principal axis of the lens. The (x, y) υ −50 +30 coordinates of the image are x = 75 cm and y = 0 (because the object is very small). This image, which is formed at as distance of (75 – 50) = 25 cm to the right of the convex mirror, acts as a virtual object for the mirror. If the mirror were not titled, then for the mirror u = + 25 cm, R = + 100 cm. Using the spherical mirror −

formula

1

υ

+

1 2 = , we have u R 1

+

1 2 = ⇒ υ = – 50 cm 25 100

υ If the mirror is rotated by an angle of 30º as shown in the figure, the image will rotate by 60º. The new x-coordinate of the image will be x' = 50 – 50 cos 60º = 50 – 25 = 25 cm The new of y-cordinate will be y' = 0 – υ sin 60º 3 50 3 = 0 – (–50) ×= = 25 3 cm 2 2 So the correction choice is (a) 5. For the first process (which is adiabatic), we have γ γ PV = Pf V f i i

Pi Pf

⇒

⇒ ⇒

105

1 × 105 32

 Vf  =    Vi 

γ

 8 × 10−3  =   10−3 

32 = 8γ

γ

Physics Paper II—2016

⇒

25 = 23γ

⇒

5 = 3γ

⇒γ=

P-II.9

5 3

The second process is a two-step process. Work done in isobaric expansion at pressure Pi is W = Pi (V f − Vi )= 105 (8 × 10−3 − 10−3 )= 105 × 7 × 10−3 ⇒ W = 7 × 102 J The change in internal energy is the isochoric process at volume Vf is

1 i i) ∆U = (γ − 1) ( Pf V f − PV 1 5    − 1 3

=

31 2 2  × 10 − 10  2 4

= ⇒

 1 5 −3 5 −3   3 × 10  × (8 × 10 ) − 10 × 10   

∆U = –

9 × 102 J 8

∴ Heat supplied is Q = W + ∆U = 7 × 102 – ⇒

Q =

9 × 102 8

47 × 102 = 588J 8

So the correct choice is (c).

 1 6. The radioactivity of a sample decreases to  n  in n half lives. Since the initial activity is 64 times the 2 permissible level, it must become

1 th of the initial value. Therefore, 64

1 1 = n ⇒n=6 64 2

Thus in 6 half lives the activity will reduce to the permissible level. Therefore, the time taken is t = 6 T1/2 = 6 × 18 days = 108 days So the correct option is (c). 7. The magnitude of angular momentum of the system about its centre of mass is Lcm = Icmω

 ma 2 4m ( 2a ) 2  + ω =  2  2  17ma 2ω = 2 So option (a) is correct.

P-II.10 Comprehensive Physics—JEE Advanced 

2a

For option (b), refer to the adjoining figure. The ratio of the angular speed of the discs is

2π (2a) ω2 1 = 2π (10a) = ω1 5 Now ω1 = ω, so ω2 =

ω 5

10 a

. Hence option (b) is also correct.

Angular momentum about OZ is 2 1 2 1 Lz =  m + 4m ( 2)  ω 2 2  2 = m ×

17 ω 2

= m × 24a2 × 17ω ×

1 2

= 204 ma2ω So choice (c) is incorrect. It can be shown that the magnitude of angular momentum about the centre of mass 81 2 81 1944 m ω = × m × 24a 2ω = ma 2ω . So option (d) is also incorrect. Hence the correct 5 5 5 choice are (a) and (b). 8. Induced emf = BLυ, where υ is the velocity of the loop at a general position x. Induced current is

of the discs =

i =

BLυ R

...(1)

Force on the loop is BLυ B 2 L2υ F = – BiL = – B × ×L=– R R dυ dυ dx dυ = m m= mυ F = ma = dt dx dt dx

Now Using (3) in (2), we have mυ

dυ = − v

∫ dυ

v0

⇒

...(3)

dυ B 2 L2υ =– dx mR

⇒ ⇒

...(2)

=–

B 2 L2 dx mR x

B 2 L2 dx mR ∫0

 B 2 L2  υ – υ0 = –  x mR 

...(4)

From Equation (4) it follows that the velocity of the loop decreases linearly with x. When the loop is completely in the region of magnetic field, the current i is zero (... there is no change in magnetic flux.) Hence F = 0, therefore, velocity υ is constant. The moment, the loop comes out of the field the current becomes clockwise, the force will be to the left (i.e. along negative x direction) and velocity υ decreases linearly with x. Hence the correct options are (b) and (d).

Physics Paper II—2016

P-II.11

9. Mean time period is

T =

0.52 + 0.56 + 0.57 + 0.54 + 0.59 5

= 0.556 s ~ 0.56 s (upto appropriate significant figures) The absolute error in the five reading is.

T − T1 = 0.56 – 0.52 = 0.04 s, T − T2 = 0.56 – 0.56 = 0.00, T − T3 = 0.56 – 0.57 = 0.01 s, T − T4 = 0.56 – 0.54 = 0.02 s, T − T5 = 0.56 – 0.59 = 0.03 s, Mean error ∆T =

0.04 + 0.00 + 0.01 + 0.02 + 0.03 5

= 0.02 s Percentage error in T =

∆T T

0.02 × 100 = 3.57% 0.56

Percentage error in r =

0.001 × 100 = 10% 0.010

Percentage error in R =

0.001 × 100 = 1.67% 0.060

Now

T = 2π

⇒

g =

∴

× 100 =

∆g g

7( R − r ) 5g

4π 2 × 7( R − r ) 5T 2

∆( R − r ) ∆T = (R − r) + 2 T =

0.002 0.02 +2× 0.05 0.56

= 0.04 + 0.07 = 0.11 ∴

∆g Percentage error in g = g × 100 = 0.11 × 100 = 11%

So the correct options are (a), (b) and (d). 10. Maximum initial kinetic energy of photoelectrons immediately after emission from the cathode plate is k i = hν – φ =

hc

λ ph

−φ

P-II.12 Comprehensive Physics—JEE Advanced

The energy gained by the electrons on reaching the anode (wire mesh) = eV. Hence the final kinetic energy of the electrons on reaching the anode is

 hc  − φ + eV Kf = Ki + eV =    λ ph

...(1)

For a very high V, equation (1) gives Kf ~ eV The de-Broglie wavelength of the electron is (using Eq. 2)

λe =

h = p

h = 2mK f

h 2me

...(2)

×

1

...(3)

V

Hence if V is made four times, λe becomes half. So option (a) is correct. It follows the Eq. (1) that if λ ph and φ both increase, Kf will decrease (as it becomes less then eV). Hence λe increases [See Eq. (3)]. So option (b) is incorrect.

d λe d λ Ph ≠ . So option (c) is also incorrect. We know that λe is independent of d as it dt dt depend only on V and Ki . So option (d) is also incorrect. Thus the only correct option is (a). Option (a) It is easy to see that

11.

RC

I

A

RC

R

R

B

Equivalent resistance between points A and B is Req = RC + RC + R + R = 2 (RC + R) Let I be the current. Since I is the same in each component of a series combination, the potential difference between A and B is Va = I Req = 2I (RC + R) ...(1) ~ 2IR (... RC << R) Since Req is maximum for a series combination, for a given current I, Va is the maximum. So option (a) is correct. Option (b) RC

R

RC+ 2R

R

A I

B

A I

B RC

RC

The equivalent resistance between points A and B is Req = = Since RC <



RC ( RC + 2 R ) RC + ( RC + 2 R ) RC ( RC + 2 R ) 2 ( RC + R )

R , RC is much less than 2R and R. So 2

Physics Paper II—2016

Req ~

RC × 2 R = RC 2R

Vb = IReq = IRC

and

It follows from (1) and (2) that

...(2)

Vb RC << 1. So option (b) is incorrect. = Va 2 R

Option (c) The equivalent resistance between points A and B is given by

A

RC

IC

R

2 2 = R +R C Req =

B

RC

1 1 1 1 1 = R +R +R+R Req C C

⇒

P-II.13

R

RC R R  C ( R >> R ) C 2( R + RC ) 2

For a given voltage V, the current IC is IC =

2V V = RC Req

...(3)

Since the equivalent resistance of a parallel combination is the minimum, for a given voltage V, the current IC is the maximum. So option (c) is correct. Option (d) A Id

RC

RC

R

A Id

B

2RC

B

⇒

R

2R

The equivalent resistance between points A and B is Req =

2 R × 2 RC 4 RRC  2 R + 2 RC 2R

( R >> RC )

= 2RC ∴ 12.

V V Id = R = 2 R eq C

...(4)

It follows from (3) and (4) that Id > Ic . So option (d) is incorrect. Thus the correct options are (a) and (c). Consider points on a circle of radius r on the screen. All points on this circle subtend the same angle on the centre of S1 and S2. So the interferance fringes are fringes of equal inclination. Hence the fringes are circular. At origin O, r = 0. Since there is no screen below O, fringes are semi-circular bright and dark bands centred at O. So option (c) is correct but options (a) and (b) are incorrect. At O the path difference = d = 0.6003 mm = 0.6003 ×10–3 m which is equal to 1000.5 λ (... λ = 6 × 10–7 m). So

1 1  ∆  n +  λ with n = 1000. This is the  λ which satisfies the conditon =  2 2 condition for dark fringe. Hence there will be a dark fringe at O. So option (d) is also correct. Hence options (c) and (d) are correct.  the path difference ∆ = 1000 + 

P-II.14 Comprehensive Physics—JEE Advanced

13. Let ω be the angular frequency of the simple harmonic motion when the mass m is not placed on block of mass M and ω1 when m is placed on M. Then k ( M + m)

k and ω1 = M

ω=

Case (i) When m is placed on M when M is at the mean position x0. Let υ be the velocity of block M when it passes through the mean position x0 and υ1 be the velocity of (M + m) when the composite block passes through x0. From conservation of momentum, Mυ = (M + m)υ1 It A and A1 are the amplitudes in the two situations, υ = ωA and υ1 = ω1A1. Thus Mω A = (M + m)ω1A1 ⇒

M ω A1 = ( M + m) × ω A 1 =

⇒

M × ( M + m)

A1 = A

M +m M = M ( M + m)

M ( M + m)

i.e. in this case the amplitude decreases by a factor of

M . ( M + m)

Case (ii) When m is placed on M when M is at the extreme position (x0 + A). At the extreme position, the velocity of M and of (M + m) is momentarily zero. Hence, in this case, the mean position x0 and the extreme position (x0 + A) remain unchanged. Hence the amplitude of oscillation remains unchanged equal to A. So option (a) is correct. Since the total mass in oscillation is (M + m) in cases (i) and (ii), the time period in each case is T1 =

2π

ω1

= 2π

(M + m) k

So option (b) is also correct. The energy of the oscillator in case (i) is E1 =

1 ( M + m) A12ω12 2

Without m placed on M, the energy is E =

1 MA2ω 2 2 2

∴

E1 ( M + m)  A1   ω1  ×  ×  =  A  ω  E M

2

=

( M + m) M M × × M ( M + m) ( M + m)

=

M M +m

Physics Paper II—2016

P-II.15

Hence E1 < E is case (i). E2 =

In case (ii),

E =

and

1 ( M + m) A12ω12 2 1 MA2ω 2 2

E2 ( M + m )  ω1  ×  = ω E M

∴

=

2

( A1 = A)

( M + m) M × = 1 ( M + m) M

Hence E2 = E in case (ii) so option (c) is incorrect. In case (i) the velocity at x0 is υ1 = ω1A1 when m is placed on M at x0 and υ = ωA when m is not placed on M. Therefore,

ω1 A1 υ1 × = = ω A υ

M M M × = ( M + m) ( M + m ) ( M + m)

Thus, v1 = v. In case (ii), the velocity = ω1A1, the same as in case (i). Hence velocity at x0 decreases in both cases so the correct options are (a), (b) and (d). 14. Refer to the following figure. B At t = 0, when the key is pressed, there is no charge on capacitors I1 C1 and C2 . Therefore, current I1 and I2 are C1 = 40 µF I1 and

5V = = 0.2 × 10–3 A = 0.2 mA 25000Ω

I2 =

5V = 0.1 × 10–3 A = 0.1 mA 50000Ω

VD + 50000 I2 = VB VD – VB = – 50000 × I2 = – 50000 × 0.1 × 10–3 = – 5V After a long time (i.e. when the capacitors are fully charged), I1 = I2 = 0. At this time charge on C1 is Q1 = 5V × 40µF = 200 µC The charge on C2 is Q2 = 5V × 20µF = 100 µC Therefore, ⇒

Therefore

⇒

VD –

Q1 = VB C1

VD – VB =

200µC Q1 = 40µF = +5V C1

So option (a) is correct. The charge on capacitor C1 varies with t is Q1' = Q1 (1 − e − t /τ1 )

I1 A I

R1 = 25 kΩ

+ Q1

–

V

+

I2 R2 = 50 kΩ

C

+ Q2

A D

5V

C2 = 20 µF

P-II.16 Comprehensive Physics—JEE Advanced

where τ1 = time constant = C1R1 = (40 × 10–6) × 25 × 103 = 1s Q1' = Q1(1 – e–t ) = 200 (1 – et )µC

∴

The current through C1 varies with t as I1' =

dQ ' = 200 e–t µA = 0.2 e–t mA dt

For capacitor C2, Q2' = 100 (1 − e− t /τ 2 )

τ2 = C2R2 = 20 × 10–6 × 50 × 103 = 1s. Hence

where

Q2' = 100(1 – e–t )µC I 2' = 0.1 e–t mA

and

VD' − VB' =

Therefore

=

Q2' − I1' × 25 × 103 C2

100(1 − et ) × 10−6 − 0.2 × e− t × 10−3 × 25 × 103 20 × 10−6

= 5(1 – 2e–t) At t =  n (2) VD' − VB' = 5(1 − 2e − n (2) )

= 5 (1 – 2 × 0.5) = 5 (1–1) = 0 So option (b) is also correct. At t = 1s, I' = I'1 + I'2 = 0.2 e–1 mA + 0.1 e–1 mA =

0.3 mA e

Intial current at t = 0 is I = (0.2 + 0.1) mA = 0.3 mA So option (c) is also correct. After a long time when steady state is reached, I1 = 0 and I2 = 0. So option (d) is also correct. Hence all the four options are correct. 15. It is given that the rotational force experienced by the particle is →

→

→

→

→

→

→

F rot = F in + 2m (υ rot × ω ) + m (ω × r ) × ω →

→

→

→

→

→

The term m (ω × r ) × ω is directed radially outwards. Hence it increases υ rot . The term 2m (υ rot . × ω ) is perpendicular to the edge of the slot and is hence cancelled by the normal reaction exerted by the edge of the slot on the particle. The radial force is the centripetal force which is F = mω2r ⇒

m

d 2r = mω2r dt 2

Physics Paper II—2016

⇒

d 2r = ω2r dt 2

P-II.17

...(1)

The solution of the differential equation (1) is r = aeωt + be–ωt where a and b are constants. This can be checked as follows. From Eq. (2)

...(2)

dr ωt −ω t = aω e − bω e dt

Differentiating again, we have

d 2r 2 ωt 2 −ω t = aω e + bω e 2 dt Using (2), we get

d 2r 2 ωt −ωt ω 2r = ω (ae + be ) = 2 dt which is eq. (1). It is given that at t = 0, r =

R . Putting t = 0 in Eq. (1), we get 2 R =a+b 2

Also

...(3)

dr dt

A aω eω t − bω e −ω t υrot= =

At t = 0, υrot = (a – b)ω Given that at t = 0, υrot = 0. This gives (a – b)ω = 0 ⇒ a = b. So at t = 0

r = 2a ⇒ a =

Hence

r =

r R = 2 4

R   r =  2

R ωt ( e + e−ωt ) 4

So option (b) is correct. →

→

16. As stated in the above solution, the term 2m (υ rot × ω ) gives the normal reaction N from the edge of the slot. We have shown above that r = ∴

→

R ωt (e + e −ω t ) 4 dr dt

υ rot= =

Rω ω t ( e − e−ωt ) 4

P-II.18 Comprehensive Physics—JEE Advanced

So the normal reaction is N = 2m

Rω ω t ( e − e −ωt ) × ω sin 90o 4

mRω 2 ωt ( e − e −ω t ) = 2 This force is along the y-axis. Hence →

N =

mRω 2 ω t ( e − e−ωt ) ˆj 2

The normal reaction exerted by the bottom of the slot = mg kˆ . Hence the net normal reaction from the slot is →

→

Nnet = N + mg kˆ mRω 2 ( ωt e − e−ωt ) ˆj + mg kˆ = 2 So option (d) is correct. 17. Since the lower plate is connected to the positive terminal of HV, the balls wil acquire a positive charge and hence they will be attracted by the upper plate which is connected to the negative terminal of HV. On reaching the upper plate the balls will acquire a negative charge and hence they will be attracted by the lower plate and so on. Thus the balls will oscillate between the two plates. At any distance r from a plate, the force experienced by a ball is proportional to 1/r2 . Thus the restoring force is not proportional to r. Hence the motion is not simple harmonic. So the correct option is (a). 18. If all the balls acquire the same charge q then

q V0 = 4π ∈ r 0 ⇒

q = 4π ∈0 rV0

...(1)

If E is the electric field between the plates, E = p.d. between plates h

= V0 – (–V0) = 2V0 ⇒ E = 2hV0 If m is the mass of each ball, the acceleration is a =

qE 2qhV0 = m m

Using (1) and (2) we get

8π ∈0 rhV02 a = m

...(2)

Physics Paper II—2016

P-II.19

Time taken by a ball to reach the other plate is

2h = t = a

m 1 = 2 V0 4π ∈0 rV0

m 4π ∈0 r

If there are n balls, the average current is Iav =

4π ∈0 r nq =n × 4π ∈0 rV02 × t m

Thus In ∝ V02, which is option (c). 

JEE Advanced 2017: Paper - I (Physics) SECTION – I Multiple Choice Questions will ONE or MORE THAN ONE correct options.

1. A block of mass M has a circular cut with a frictionless surface as shown. The block rests on the horizontal frictionless surface of a fixed table. Initially the right edge of the block is at x = 0, in a co-ordinate system fixed to the table. A point mass m is released from rest at the topmost point of the path as shown and it slides down. When the mass loses contact with the block, its position is x and the velocity is v. At that instant, which of the following options is/are correct? R m R

y

x



M x=0

(a) The velocity of the point mass m is: v=

2 gR m 1+ M



(b) The velocity of the block M is:



m 2 gR M (c) The position of the point mass is:

L = 1H C = 1F V0 sinwt

(a) At w ~ 0 the current flowing through the circuit becomes nearly zero (b) At w >> 106 rad. s–1, the circuit behaves like a capacitor (c) The frequency at which the current will be in phase with the voltage is independent of R (d) The current will be in phase with the voltage if w = 104 rad. s–1 3. A block M hangs vertically at the bottom end of a uniform rope of constant mass per unit length. The top end of the rope is attached to a fixed rigid support at O. A transverse wave pulse (Pulse I) of wavelength l0 is produced at point O on the rope. The pulse takes time TOA to reach point A. If the wave pulse of wavelength l0 is produced at point A (Pulse 2) without disturbing the position of M it takes time TAO to reach point O. Which of the following options is/are correct?

v= –

mR M +m (d) The x component of displacement of the center mR of mass of the block M is: – M +m

Pulse 1

O

x =– 2



2. In the circuit shown, L = 1mH, C = 1 mF and R = 1 kW. They are connected in series with an a.c. source V = V0 sin wt as shown. Which of the following options is/are correct?

R = 1 kW

Pulse 2 A



M

(a) The velocities of the two pulses (Pulse 1 and Pulse 2) are the same at the midpoint of rope (b) The velocity of any pulse along the rope is independent of its frequency and wavelength (c) The wavelength of Pulse 1 becomes longer when it reaches point A (d) The time TAO = TOA

P-I.2  Comprehensive Physics—JEE Advanced

4. A flat plate is moving normal to its plane through a gas under the action of a constant force F. The gas is kept at a very low pressure. The speed of the plate v is much less than the average speed u of the gas molecules. Which of the following options is/are true? (a) The pressure difference between the leading and trailing faces of the plate is proportional to uv (b) The plate will continue to move with constant non-zero acceleration, at all times (c) At a later time external force F balances the resistive force (d) The resistive force experienced by the plate is proportional to v 5. A human body has a surface area of approximately 1m2. The normal body temperature is 10 K above the surrounding room temperature T0. Take the room temperature to be T0 = 300K. For T0 = 300 K, the value of σT04 = 460 Wm–2 (where s is the Stefan-Boltzmann constant). Which of the following options is/are correct? (a) Reducing the exposed surface area of the body (e.g. by curling up) allows humans to maintain the same body temperature while reducing the energy lost by radiation (b) If the body temperature rises significantly then the peak in the spectrum of electromagnetic radiation emitted by the body would shift to longer wavelengths (c) The amount of energy radiated by the body in 1 second is close to 60 Joules (d) If the surrounding temperature reduces by a small amount DT0 << T0, then to maintain the same body temperature the same (living) human being needs to radiate DW = 4s T03DT0 more energy per unit time 6. For an isosceles prism of angle A and refractive index µ, it is found that the angle of minimum deviation dm = A. Which of the following options is/are correct? (a) At minimum deviation, the incident angle i1 and the refracting angle r1 at the first refracting surface are related by r1 = (i1/2)





(b) For this prism, the refractive index µ and the 1 –1 Ê m ˆ angle of prism A are related as A = cos ÁË ˜¯ 2 2 (c) For the angle of incidence i1 = A, the ray inside the prism is parallel to the base of the prism (d) For this prism, the emergent ray at the second surface will be tangential to the surface when the angle of incidence at the first surface is È ˘ A i1 = sin–1 Ísin A 4 cos 2 – 1 – cos A˙ 2 Î ˚

7. A circular insulated copper wire loop is twisted to form two loops of area A and 2A as shown in the figure. At the point of crossing the wires remain electrically insulated from each other. The entire loop lies in the plane (of the paper). A uniform  magnetic field B points into the plane of the paper. At t = 0, the loop starts rotating about the common diameter as axis with a constant angular velocity w in the magnetic field. Which of the following options is/are correct? × × × × ×

×

×

× × Area A × ×

×

× × Area 2A × ×

×

×

×

×

×

×

×

×

×

×

×

×

w



(a) The amplitude of the maximum net emf induced due to both the loops is equal to the amplitude of maximum emf induced in the smaller loop alone



(b) The rate of change of the flux is maximum when the plane of the loops is perpendicular to plane of the paper



(c) The net emf induced due to both the loops is proportional to cos w t



(d) The emf induced in the loop is proportional to the sum of the areas of the two loops

Physics Paper I—2017  P-I.3

SECTION 2 Single Digit Integer Answer Questions 8. An electron in a hydrogen atom undergoes a transition from an orbit with quantum number ni to another with quantum number nf. Vi and Vf are respectively the initial and final potential energies v of the electron. If i = 6.25. then the smallest vf possible nf is _____________. 9. A monochromatic light is travelling in a medium of refractive index n = 1.6. It enters a stack of glass layers from the bottom side at an angle q = 30°. The interfaces of the glass layers are parallel to each other. The refractive indices of different glass layers are monotonically decreasing as nm= n – mDn. where nm is the refractive index of the mth slab and Dn = 0.1 (see the figure). m

n – mn

m–1

n – (m – 1) n

3 2

n – 3n

1

n – n

n – 2n q

10.

131

l is an isotope of Iodine that b decays to an isotope of Xenon with a half-life of 8 days. A small amount of a serum labelled with 131I is injected into the blood of a person. The activity of the amount of 131I injected was 2.4 ¥ 105 Becquerel (Bq). It is known that the injected serum will get distributed uniformly in the blood stream in less than half an hour. After 11.5 hours, 2.5 ml of blood is drawn from the person’s body, and gives an activity of 115 Bq. The total volume of blood in the person’s body, in liters is approximately. (you may use ex ≈ 1 + x for | x | << 1 and ln 2 ≈ 0.7).

11. A stationary source emits sound of frequency f0 = 492 Hz. The sound is reflected by a large car approaching the source with a speed of 2 ms–1. The reflected signal is received by the source and superposed with the original. What will be the beat frequency of the resulting signal in Hz? (Given that the speed of sound in air is 330 ms–1 and the car reflects the sound at the frequency it has received). 12. A drop of liquid of radius R = 10–2m having 0.1 surface tension S = Nm–1 divides itself into 4π

n

The ray is refracted out parallel to the interface between the (m – l)th and mth slabs from the right side of the stack. What is the value of m?

K identical drops. In this process the total change in the surface energy DU = 10–3 J. If K = 10a then the value of a is _____________.

SECTION 3 Matching Type Questions Answer Q.13, Q.14 and Q.15 by appropriately matching the information given in the three columns of the following table.   A charged particle (electron or proton) is introduced at electric field E and magnetic field B are given in the origin (x = 0, y = 0, z = 0) with a given initial columns 1, 2 and 3, respectively. The quantities E , B 0 0   velocity v . A uniform electric field E and a uniform are positive in magnitude.   magnetic field B exist everywhere. The velocity v ,

P-I.4  Comprehensive Physics—JEE Advanced

(I)

Column 1 Column 2 Column 3     E Electron with v = 2 0 x (i) E = E0 z (P) B = –B0 x B0

 E0  y Electron with v = B0  (III) Proton with v = 0  E (IV) Proton with v = 2 0 x B0 (II)

 (ii) E = –E0 y

 (Q) B = B0 x

 (iii) E = –E0 x

 (R) B = B0 y  (S) B = B0 z

 (iv) E = E0 x

13. In which case will the particle move in a straight line with constant velocity?

(a) (IV) (i) (S)

(b) (III) (ii) (R)

(c) (III) (iii) (P) (d) (II) (iii) (S) 14. In which case would the particle move in a straight line along the negative direction of y-axis (i.e., move along – y )?

(a) (III) (ii) (P)

(b) (III) (ii) (R)

(c) (IV) (ii) (S) (d) (II) (iii) (Q) 15. In which case will the particle describe a helical path with axis along the positive z direction?



Column 1



(a) (III) (iii) (P)

(b) (II) (ii) (R)

(c) (IV) (ii) (R) (d) (IV) (i) (S) Answer Q.16, Q.17 and Q.18 by appropriately matching the information given in the three columns of the following table. An ideal gas is undergoing a cyclic thermodynamic process in different ways as shown in the corresponding P-V diagrams in column 3 of the table. Consider only the path from state 1 to state 2. W denotes the corresponding work done on the system. The equations and plots in the table have standard notations as used in thermodynamic processes. Here g is the ratio of heat capacities at constant pressure and constant volume. The number of moles in the gas in n. Column 2

Column 3 P 1

(I)

W1 → 2

1 = ( P 2V 2 – p 1V 1) g –1

(i) Isothermal

2

(P) V P 1

(II)

W1 → 2 = – PV2 + PV1

(ii)

Isochoric

(Q) 2 V P 1

(III) W1 → 2 = 0

(iii) Isobaric

2

(R) V P 1

ÊV ˆ (IV) W1 → 2 = –nRT ln Á 2 ˜ Ë V1 ¯

(iv) Adiabatic

(S) 2 V

Physics Paper I—2017  P-I.5

16. Which of the following options is the only correct representation of a process in which DU = DQ – PDV ?

(a) (II) (iv) (R)

(b) (III) (iii) (P)

(c) (II) (iii) (P) (d) (II) (iii) (S) 17. Which one of the following options correctly represents a thermodynamic process that is used as a correction in the determination of the speed of sound in an ideal gas?

(a) (I) (iv) (Q)

(b) (III) (iv) (R)

(c) (I) (ii) (Q) (d) (IV) (ii) (R) 18. Which one of the following options is the correct combination?

(a) (IV) (ii) (S)

(b) (III) (ii) (S)



(c) (II) (iv) (R)

(d) (II) (iv) (P)



4. (a), (c), (d)

5. (a), (c), (d)

6. (a), (c), (d)

7. (a), (b)

8. 5

9. 8

10. 5

11. 6

13. (d)

14. (b)



15. (d)

16. (c)



17. (a)

18. (b)

Hence, option (a) is correct. V=

m m v= M M

2 gR mˆ Ê ÁË1 + ˜¯ M

Let Dx be the displacement of the centre of mass of m and DX be the displacement of the centre of mass of the block of mass M. Since no external force acts on the system, its centre of mass has no net displacement, i.e., mDx + MDX = 0

Now,

R h=R v

R–x

Since no net external force acts on the system, the momentum is conserved, i.e., mv MV = mv ⇒ V = M From conservation of energy,

2 gR mˆ Ê ÁË1 + ˜¯ M

Hence, option (d) is correct.

1. Let V be the velocity of the block when mass m loses contact with the block (see figure).

M –x

⇒ v =

⇒ m (R – x) + M(–x) = 0 mR ⇒ x = m+M

SOLUTIONS

V

1 Ê mv ˆ 2 1 MÁ + mv2 ˜ 2 ËM¯ 2

m ⇒ 2gR = ÊÁ1 + ˜ˆ v2 Ë M¯



12. 6

m

⇒ mgR =

along negative x-direction. So option (b) is incorrect.

ANSWERS KEY 1. (a), (d) 2. (a), (c)

3. (a), (b), (d)



Loss in gravitational P.E. of m = gain in K.E. of M + gain in K.E. of m 1 1 ⇒ mgh = MV2 + mv2 2 2

Dx = R – x

= R –

mR MR = m+M M +m

\ Final position of m = 0 – x MR mR =0– =– M +m M +m Hence, option (c) is incorrect. 2. Capacitative reactance XC =

1 and inductive wC

reactance XL = w l. At w = 0, XC Æ ∞ and XL = 0. Hence the current flowing through the circuit is nearly zero. So option (a) is correct. When w >> 106 rad s–1, XC → 0 and XL >> 1. Thus the circuit does not behave like a capacitor and option (b) is incorrect.

P-I.6  Comprehensive Physics—JEE Advanced

The current will be in phase with voltage at resonant frequency i.e. when XL = XC or wL = 1 1 ⇒w= which is independent of R. So wc LC

4. Let m be the mass of each molecule. Let velocity directed to the right be taken as positive and velocity to the left be taken as negative. plate

option (c) is correct. The resonant frequency is 1 1 w= = = 106 rad s–1 –6 –6 LC (1 ¥ 10 ) ¥ (1 ¥ 10 )

Face 1 +u –u

So option (d) is incorrect. 3. Mass of rope is m = µL where µ is the mass per unit length of the rope and L is its length. The velocity of a pulse is given by v=



T m

where T is the tension. Since the rope has a finite mass, the tension will not be the same at points on the rope. At A the tension is TA = Mg and at O, the tension is TO = (M + m)g. At the mid-point, m  the tension is  M +  g which is the same for 2  both pulses. Hence pulses 1 and 2 have the same velocity at the mid-point, so option (a) is correct. Since m is constant and T has nothing to do with wavelength or frequency, the velocity of any pulse along the rope is independent of wavelength and frequency. So option (b) is correct. Since v = nl l=

1 T v = n m n

–(u – 2v)

l1 µ l2

T

Since the tension decreases as we go from O to A, the velocity of pulse 1 decreases as it travels from O to A. So the wavelength of pulse 1 becomes shorter when it reaches point A. Hence option (c) is incorrect. Since the average speed of each pulse is the same, the time taken for pulse 1 to go from O to A = time taken for pulse 2 to go from A to O. So option (d) is correct. Thus the correct options are (a), (b) and (d).

(u + 2v) v

The rate of change of momentum of a molecule due to collision with face 1 is

Dp1 = M  omentum after collision – momentum before collision = –m(u – 2v) – m(u + 2v) = – mu + 2mv – mu – 2mv = –2m (u + v) directed to the left The rate of change of monentum of a molecule due to collision with face 2 is Dp2 = m (u + 2v) – (–mu) = 2m (u – v) directed to the right Let t1 and t2 be the time of collision with face 1 and 2 respectively. Then the forces on face 1 and Dp1 Dp2 on face 2 will be proportional to and t1 t2 Let R1 and R2 be the number of collisions per unit 1 time with faces 1 and 2 respectively, the R1 ∝ t1 and R2 ∝

The frequency n of the pulse cannot change as it depends only on the frequency of the source producing the pulse and since µ is also constant,

Face 2

1 . If F1 and F2 are the magnitudes of t2

forces on faces 1 and 2, then F1 ∝ R |Dp1|



or F1 ∝ R1 D p1 and F2 ∝ R2 Dp2 Now

R1∝ (u + v) and R2 ∝ (u – v). Hence

F1 ∝ 2m (u + v)2 and F2 ∝ 2m (u – v)2

\ Net Force acting on the plate is

F ∝ F1 – F2

  ∝ 2m (u + v)2 – 2m (u – v)2 or F ∝ 8muv Thus F ∝ uv. So option (a) is correct. Now F ∝ v. Hence the plate will accelerate and will eventually acquire a terminal velocity because

Physics Paper I—2017  P-I.7

the gas will tend to reduce the velocity until its acceleration becomes zero. Hence choice (c) is correct.



È Ê 4DT0 ˆ ˘ = sA ÍT 4 - T04 Á1 – ˙ T0 ˜¯ ˙˚ Ë ÍÎ

Froms Stokes’ law, it follows that for small velocities the resistive force experienced by the plates is proportional to its velocity v. So option (d) is correct. Thus the correct options are (a), (c) and (d).

or Q ¢ = sA [T4 – T04 + 4T03 DT0]

5. Assuming that human body behaves as a block body, the heat energy radiated per second from its surface is

6. Refer to the following figure.



= Q + 4sA T03DT0

Thus to maintain the same temperature T, Q¢ must be greater than Q. So option (d) is correct. Hence the correct options are (a), (c) and (d).

A

Q = sA (T 4 – T04), s = Stefan’s constant

A = exposed surface area of the body

d i1

For small temperature difference DT = T – T0 ⇒ T = T0 + DT \

r2

È 4 Ê 4DT ˆ 4 ˘ 4 T ˙ = sA ÍT0 Á1 + 0 T0 ˜¯ ÍÎ Ë ˙˚ 3

Q = 4sA T0 DT

d = i1 + i2 – A and r1 + r2 = A For minimum deviation, i1 = i2 and r1 = r2. Hence dm = 2i1 – A and 2r1 = A Given dm = A. Hence A = 2i1 – A → i1 = A = 2r1 ⇒ r1 =

It follows that Q decreases with decrease in A. So option (a) is correct.

So, option (a) is correct.

From Wien’s displacement law, it follows that the peak of the graph of Q versus l shifts to shorter wavelength if the temperature T rises. So option (b) is incorrect.

sin i1 = µ sin r1

Q = 4sA T03 DT = 4sA T03 (T – T0) ÊT ˆ = (sT04) ¥ 4A ¥ Á – 1˜ T Ë 0 ¯ Ê 310 ˆ = 460 ¥ 4 ¥ 1 ¥ Á –1 Ë 300 ˜¯ = 61.3 Js–1 (or W) So option (c) is correct. If T0 decreases to T0 – DT0, the heat energy radiated per second is given by Q¢ = sA[T4 – (T0 – DT0)4] or

i2

Q = sA [(T0 + DT)4 – T04]

È Ê DT ˆ 4 ˘ = sA ÍT04 Á1 + - T04 ˙ ˜ T0 ¯ ÍÎ Ë ˙˚

or

r1

4 È Ê DT ˆ ˘ = sA ÍT 4 - T04 Á1 – 0 ˜ ˙ T0 ¯ ˙˚ Ë ÍÎ

i1 2

From Snell’s law,  A ⇒ sin A = µ sin   2  A  A  A ⇒ 2 sin   cos   = µ sin   2 2 2  A ⇒ 2 cos   = µ 2 µ  A ⇒ cos   = 2 2

m ⇒ A = 2 cos–1 ÊÁ ˆ˜ Ë 2¯ So option (b) is incorrect. At minimum deviation, the ray inside the prism is parallel to the base. So option (c) is correct. Applying Snell’s to the second refraction. m sin r2 = sin i2

P-I.8  Comprehensive Physics—JEE Advanced

For tangential emergence, i2 = 90º. Hence m sin r2 = 1 ⇒ r2 =

1–

1

sin r1 = sin (A – r2) = sin A cos r2 – cos A sin r2 = (sin A)

2

⇒

nf ni

= 2.5 =

µ2 –1 1 – (cos A) µ µ

From Snell’s law applied at the first refraction,   sin i1 = m sin r1

⇒ sin r1 = sin (A) (m2 – 1)1/2 – cos A È ˘ Ê Aˆ or i1 = sin–1 Ísin A 4 cos 2 Á ˜ – 1 - cos A˙ Ë 2¯ ÍÎ ˙˚ So option (d) is correct. Thus the correct options are (a), (c) and (d). 7. At any time t, the area vector of each loop makes  an angle q = wt with B. The instantaneous magnetic flux in loop 1 is f1 = BA cos wt and in loop 2 is f2 = B(2A) cos wt. The magnitudes of induced emfs in loops 1 and 2 are

9. The refractive index of layers decreases by a constant amount Dn as we go from lower to upper layers. For total reflection at the interface between the (m – 1)th and mth layers, we have (∵ n sin θ = constant)

n sin q = (n – m ∆ n) sin 90 °

⇒ 1.6 ¥ sin 30º = 1.6 – m ∆ n ⇒ 0.8 = 1.6 – m ¥ 0.1 ⇒ m =

=

ln(2) 0.7 per day = per day 8 8

t = 12 hours = –

0.7 1 ¥ 8 2

1 day. So 2 0.7 16

and

15.3 Ê 0.7 ˆ R = R0 Á1 – ˜¯ = R0 ¥ Ë 16 16

The net instantaneous flux is (because the two emfs oppose each other) enet = 2BAw sin wt – BAw sin wt

= BAw sin wt

The amplitude of the enet = BAw which is equal to that due to loop 1 alone. So option (a) is correct. p When q = , enet is maximum. So option (b) is 2 correct. enet is proportional to sinwt. So option (c) is incorrect. Option (d) is incorrect because the orientation of the loop is such that e1 and e2 oppose each other. Hence enet is proportional to the difference in areas. So the correct options are (a) and (b). 1 8. V µ 2 . So n Ê nf ˆ V i = Á ˜ V f Ë ni ¯

2

= R0 e

–

R = R0 e

d f2 = 2 BAw sin wt dt

0.8 =8 0.1

10. Activity R = Roe–lt where the decay constant l is given by ln(2) l = ; T = half life T

d f1 e1 = = BAw sin wt dt e2 =

5 2

So the smallest integral value of nf = 5

µ2

Now, r1 = A – r2

 

Ê nf ˆ or 6.25 = Á ˜ Ë ni ¯

Using ex = 1 + x, we get

=

2.4 ¥ 105 ¥ 15.3 16

= 2.3 ¥ 105 Bq Now, V litre is the volume of the blood, then 115 =

R ¥ 2.5 ¥ 10 –3 V

=

2.3 ¥ 105 ¥ 2.5 ¥ 10 –3 V

⇒ V = 5 litre 11. Apparent frequency is Ê 330 + 2 ˆ m¢ = Á ¥ 492 = 498 Hz Ë 330 – 2 ˜¯ \ Beat frequency = 498 – 492 = 6 Hz 12. Let r be the radius of each small drop. Since the total volume remains constant, 4p 4p 3 R3 = K ¥ r ⇒ R3 = Kr3 3 3

Physics Paper I—2017  P-I.9

14. If v = 0, the proton will move along the negative y  direction if E is along negative y direction

Initial P.E. is Ui = S ¥ 4 π R

2



Final P.E. is

because if v = 0, magnetic force is zero or parallel  or antiparallel to E . So the correct choices are (III), (ii) and (R) which is option (b).

Uf = KS ¥ 4p r2

DU = Uf – Ui ⇒ 10–3 = KS ¥ 4pr2 – 5 ¥ 4pR2 = 4p S(Kr2 – R2)

Ê Kr 2 ˆ = 4p S R2 Á – 1˜ Ë R2 ¯ = 4p S R2 (k1/3 – 1)

(

)

µ

(∵ R3 = Kr3)

or 10–3 = 4pSR2 10 3 – 1 (∵ K = 10a) Substituting the given values of S and R and solving we get a  6



⇒

q vB = qE E v = B

The electric and magnetic field must be perpendicular to each other.  E If the particle is an electron, then = v = 0 y if B0   E = –E0 x and B = B0 z    In this case FB = q ( v × B) ÊE ˆ = –e Á 0 y ¥ B0 z ˜ = –e E0 x Ë B0 ¯   and FE = q E = –e ¥ –E0 x = eE0 x

(

are (IV), (i) and (S) which is option (d). 16. Since W represents work done on the gas, it is negative. For an isobaric process = (P = constant), V2

13. The charged particle will move in a straight line with a constant velocity if no net force acts on it, i.e. if magnetic force = electrical force or

15. The proton will describe a helical path with    ÊE ˆ velocity v = 2 Á 0 ˜ x if E = E0 z and B = B0 z Ë B0 ¯  because then FE will be along the +z axis and  FB will be along –y axis. so the correct choices

)

So the correct choices are (II), (ii) and (S) which is option (d).

W = –

Ú PdV = – p VÚ dV = P (V1 – V2), 1

which is (II). For an isobaric process, the P-V graph is parallel to the V-axis which is option (P). No other combination is possible. So the correct choices are (II), (iii) and (P), which is option (c). 17. Laplace gave the correct formula for the speed of sound in a gas. He corrected Newton’s formula by assuming that the propagation of sound in a gas is an adiabatic process for which 1 W= (P V – P1V1) which is choice (I). Also γ –1 2 2 choice (iv) is correct. In option (Q), process 1 → 2 is adiabatic. So the correct option is (a). 18. Option (a) is incorrect because process 1 → 2 is isochoric for which W = 0. Choice (c) is incorrect because W = P (V1 – V2) is wrong for an adiabatic process. For the same reason choice (d) is also incorrect. Hence the only correct option is (b).

JEE Advanced 2017: Paper - II (Physics) SECTION – I Multiple Choice Questions having ONLY ONE correction option.

1. A photoelectric material having work function f0 is hc ˆ Ê illuminated with light of wavelength l Á l < ˜ f0 ¯ Ë The fastest photoelectron has a de Broglie wavelength ld. A change in wavelength of the incident light by Dl results in a change Dld in ld. Then the ratio Dld/Dl is proportional to:

(a) l3d / l2

(b) l2d / l2

(d) ld3 /l    2. Three vectors P , Q and R are shown in the  figure. Let S be any point on the vector R  . The distance between the points P and is b | R |.The  S  general relation among vectors P , Q and S is:

(c) ld /l

m0 I (a) 6[ 3 – 1] 4p a

h

Ê 2p ˆ (a) D = h sin Á ˜ Ë n¯

  – b) P + b2 Q   – b 2) P + b Q   – b) P + b Q   – 1) P + b2 Q

3. A symmetric star shaped conducting wire loop is carrying a steady state current I as shown in the figure. The distance between the diametrically opposite vertices of the star is 4a. The magnitude of the magnetic field at the centre of the loop is:

m0 I 3[ 3 – 1] 4p a

m0 I m0 I (c) 6[ 3 + 1] (d) 3[2 – 3] 4p a 4p a 4. Consider regular polygons with number of sides n = 3, 4, 5K as shown in the figure. The centre of mass of all the polygons is at height h from the ground. They roll on a horizontal surface about the leading vertex without slipping and sliding as depicted. The maximum increase in height of the locus of the centre of mass for each polygon is D. Then D depends on n and h as:

h

 (a) S = (1  (b) S = (1  (c) S = (1  (d) S = (b

(b)

h

Êpˆ (b) D = h sin2 Á ˜ Ë n¯

Êpˆ 1 ˘ (c) D = h tan2 Á ˜ (d) D = h ÈÍ – 1˙ Ë 2n ¯ p Ê ˆ Í cos Á ˜ ˙ Ë n¯ ˚ Î 5. Consider an expanding sphere of instantaneous radius R whose total mass remains constant. The expansion is such that the instantaneous density r remains uniform throughout the volume. The rate Ê 1 drˆ of fractional change in density Á is conË r dt ˜¯ stant. The velocity v of any point on the surface of the expanding sphere is proportional to: (a) R2/3

(b) R 1 (c) R (d) R 6. A person measures the depth of a well by measuring the time interval between dropping a stone and receiving the sound of impact with the bottom of the well. The error in his measurement of time is dT = 0.01 seconds and he measures the depth of the well to be L = 20 meters. Take the acceleration due to gravity g = 10 ms–2 and the velocity 3

P-II.2  Comprehensive Physics—JEE Advanced

of sound is 300 ms–1. Then the fractional error in the measurement dL/L, is closest to: (a) 0.2% (b) 5% (c) 1% (d) 3% 7. A rocket is launched normal to the surface of the Earth, away from the Sun, along the line joining the Sun and the Earth. The Sun is 3 ¥ 105 times heavier than the Earth and is at a distance 2.5 ¥ 104 times larger than the radius of the Earth. The

escape velocity from Earth’s gravitational field is ve = 11.2 kms–1. The minimum initial velocity (vs) required for the rocket to be able to leave the Sun-Earth system is closest to (Ignore the rotation and revolution of the Earth and the presence of any other planet) (a) vs = 62 kms–1 (b) vs = 42 kms–1 –1 (c) vs = 72 kms (d) vs = 22 kms–1

SECTION 2 (Multiple choice Questions having ONE or MORE THAN ONE Correct Options) 8. A source of constant voltage V is connected to a resistance R and two ideal inductors L1 and L2 through a switch S as shown. There is no mutual inductance between the two inductors. The switch S is initially open. At t = 0, the switch is closed and current begins to flow. Which of the following options is/are correct?



(a) The electric flux passing through the curved Q Ê 1 ˆ surface of the hemisphere is 1Á ˜ 2e 0 Ë 2¯



(b) The component of the electric field normal to the flat surface is constant over the surface (c) Total flux through the curved and the flat Q surfaces is 0 e0



S R + – V







L1

L2

(a) At t = 0, the current through the resistance V R is R (b) After a long time, the current through L2, will V L1 be R L1 + L2 (c) After a long time, the current through L1 will V L2 be R L1 + L2



(d) The circumference of the flat surface is an equipotential surface 10. A uniform magnetic field B exists in the region 3R between x = 0 and x = (region 2 in the figure) 2 pointing normally into the plane of the paper. A particle with charge + Q and momentum p directed along x-axis enters region 2 from region 1 at point P1(y = –R). Which of the following option(s) is/are correct? Region 1 × ×

(d) The ratio of the currents through L1, and L2 is fixed at all times (t > 0) 9. A point charge +Q is placed just outside an imaginary hemispherical surface of radius R as shown in the figure. Which of the following statements is/are correct?



Region 2 × × × B ×

Region 3

×

×

×

×

×

×

×

×

× P2

+Q P1 × × (y = –R) ×

×

×

×

×

×

×

×

×

×

O

x

3R/2



(a) For B =

8 p , the particle will enter region 13 QR

3 through the point P2 on x-axis

Physics Paper II—2017  P-II.3



2 p , the particle will re-enter 3 QR

(b) For B >

region I (c) For a fixed B, particles of same charge Q and same velocity v, the distance between the point P1 and the point of re-entry into region 1 is inversely proportional to the mass of the particle (d) When the particle re-enters region 1 through the longest possible path in region 2, the magnitude of the change in its linear momentum between point P1 and the farthest point from y-axis is p/ 2 11. Two coherent monochromatic point sources S1 and S2 of wavelength l = 600 nm are placed symmetrically on either side of the centre of the circle as shown. The sources are separated by a distance d = 1.8 mm. This arrangement produces interference fringes visible as alternate bright and dark spots on the circumference of the circle. The angular separation between two consecutive bright spots is Dq. Which of the following options is/are correct?

It is connected between points X and Y and then between Y and Z. The reading(s) of the voltmeter will be 1 rms (a) V YZ = V0 2 rms (b) V XY = V0

3 2

rms (c) V XY = V0



(d) independent of the choice of the two terminals 13. A rigid uniform bar AB of length L is slipping from its vertical position on a frictionless floor (as shown in the figure). At some instant of time, the angle made by the bar with the vertical is q. Which of the following statements about its motion is/are correct? A q L

P1

B O

q S1

S2

P2

d



(a) The total number of fringes produced between P1 and P2 in the first quadrant is close to 3000 (b) A dark spot will be formed at the point P2 (c) At P2 the order of the fringe will be maximum (d) The angular separation between two consecutive bright spots decreases as we move from P1 and P2 along the first quadrant 12. The instantaneous voltages at three terminals marked X, Y and Z are given by Vx = V0 sin ωt, 2p ˆ Ê Vy = V0 sin Á w t + ˜ and Ë 3¯

4p ˆ Ê Vz = V0 sin Á w t + ˜ Ë 3¯

An ideal voltmeter is configured to read rms value of the potential difference between its terminals.

(a) The trajectory of the point A is a parabola (b) Instantaneous torque about the point in contact with the floor is proportional to sin q. (c) The midpoint of the bar will fall vertically downward (d) When the bar makes an angle q with the vertical, the displacement of its midpoint from the initial position is proportional to (1 – cos q) 14. A wheel of radius R and mass M is placed at the bottom of a fixed step of height R as shown in the figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without slipping. Consider the torque t about an axis normal to the plane of the paper passing through the point Q. Which of the following options is/are correct?

P-II.4  Comprehensive Physics—JEE Advanced



(a) If the force is applied tangentially at point S then t ≠ 0 but the wheel never climbs the step (b) If the force is applied normal to the circumference at point P then t is zero



(c) If the force is applied normal to the circumference at point X then t is constant (d) If the force is applied at point P tangentially then t decreases continuously as the wheel climbs

SECTION 3 (Paragraph Based Questions having ONLY ONE Correct Option) Paragraph for Questions 15-16 Consider a simple RC circuit as shown in Figure 1. Process 1 : In the circuit the switch S is closed at t = 0 and the capacitor is fully charged to voltage V0, (i.e., charging continues for time T >> RC). In the process some dissipation (ED) occurs across the resistance R. The amount of energy finally stored in the fully charged capacitor is EC. Process 2: In a different process the voltage is first set V to 0 and maintained for a charging time T >> RC. 3 2V0 Then the voltage is raised to without discharging 3 the capacitor and again maintained for a time T >> RC. The process is repeated one more time by raising the voltage to V0 and the capacitor is charged to the same final voltage V0 as in Process 1. These two processes are depicted in Figure 2. S R V

+ –

C

1Ê 1 2ˆ ÁË CV0 ˜¯ 3 2 16. In Process 1, the energy stored in the capacitor EC and heat dissipated ED across resistance are related by: 1 (a) EC = E D (b) EC = 2ED 2 (c) EC = ED (d) EC = ED ln 2 Paragraph for Questions 17-18 One twirls a circular ring (of mass M and radius R) near the tip of one’s finger as shown in Figure 1. In the process the finger never loses contact with the inner rim of the ring. The finger traces out the surface of a cone, shown by the dotted line. The radius of the path traced out by the point where the ring and the finger is in contact is r. The finger rotates with an angular velocity w0. The rotating ring rolls without slipping on the outside of a smaller circle described by the point where the ring and the finger is in contact (Figure 2). The coefficient of friction between the ring and the finger is m and the acceleration due to gravity is g. (c) ED = 3CV02

R

Fig. 1

r

V V0

(d) ED =

R Process 1

Fig. 1 Process 2

T

T >> RC

2T

t

Fig. 2

15. In Process 2, total energy ED dissipated across the resistance is: 1 1 (a) ED = 3 ÊÁ CV0 2 ˆ˜ (b) ED = CV02 Ë2 ¯ 2

Fig. 2

17. The minimum value of w0 below which the ring will drop down is : 2g g (a) (b) m(R – r) m(R – r) 3g g (c) (d) 2m ( R – r ) 2m ( R – r )

Physics Paper II—2017  P-II.5

18. The total kinetic energy of the ring is: 3 (a) Mw02(R – r)2 2 1 (b) Mw02(R – r)2 2 (c) Mw20 (R – r)2



1. (a) 3. (a) 5. (b)

Answers Section I

3. Refer to the following figure. 2. (c) 4. (d) 6. (c)

P

Section II 7. (b) 8. (b), (c), (d) 9. (a), (d) 10. (a), (b) 11. (a), (c) 12. (b), (d) 13. (b), (c), (d) 14. (b), (d) Section III 15. (d) 16. (c) 17. (b) 18. None. Hints and Solutions

1. If p is the momentum of the fastest photoelectron, its K.E. is p2 Kmax = ; m = mass of electron 2m 1 Ê hˆ = 2m ÁË ld ˜¯ Kmax =

or

2



h (∵ ld = ) p

⇒

2mld2

2

h hc = + f0 l 2mld2

⇒

–

hc l2

Dl = –

2h 2 2mld3

Dld

3 Dld Ê c ˆ Êl dˆ = Á ˜ Á Ë 2hm ¯ Ë l 2 ˜¯ Dl

ld3 Dld ∝ 2 , which is option (a). l Dl  2. Applying triangle law to DOPS, S is the resultant   of vectors P and b | R |, i.e.

\

i

a R

Q 60° 30°

a

90° 30° a

The magnitude of the magnetic field due to one segment PQ of the star at the centre of the loop is

B = –

m0 i 4p a

30∞

Ú

sinq dq

60∞

m i 30∞ = 0 cos q 60∞ 4p a =

h2

From Einstein’s photo-electric equation, hc = Kmax + f0 l ⇒

 S   S–P

   It is given in the figure that R = Q – P    or P = Q – R   1   or P = Q – ( S – P ) b    ⇒ S = (1 – b) P + b Q , which is option (c).

(d) Mw20 R2

  P + b | R | =  ⇒ b | R | =

m0i (cos 30° – cos 60°) 4p a

m0i ( 3 – 1) directed into the page. 8p a The magnitude of the magnetic field at the centre of the loop = 12B as all segments of stars produce magnetic field directed into the page. Hence 12m0i Bloop = 12B = ( 3 – 1) 8p a =

Ê m iˆ = 6 Á 0 ˜ ( 3 – 1) Ë 4p a ¯ So the correct option is (a). 4. Refer to the following figure. For n-sided regular polygon, the angle subtended at the centre of mass 2p = n

P-II.6  Comprehensive Physics—JEE Advanced

Since errors do not cancel each other,

Centre of mass 2p n

h

p n

a

D = a – h =

a

h

h Êpˆ = cos Á ˜ ⇒ a = Ë n¯ a

h Êpˆ cos Á ˜ Ë n¯

h È 1 ˘ –h=h Í –1 Êpˆ Êpˆ ˙ Í cos Á ˜ ˙ cos Á ˜ Ë n¯ Ë n¯ ˚ Î

4p R3 r = constant 3 or R3r = constant Differentiating with respect to time, dr dρ 3R2r + R3 =0 dt dt

v=–

⇒ It is given that

R 1 dr 3 r dt

1 dρ = constant. Hence v µ R. ρ dt

So the correct option is (b). 6. Time taken for the stone to reach the bottom of the well is given by 1 – L = – gt2 2 t =

2L g

Time taken for sound to travel from the bottom of the well to the observer is L t ¢ = ; v = speed of sound v Total time taken is

T = t + t¢ =

L 2L + v g

Differentiating partially with respect to L

1 2 1 + ¥ 300 10 2 ¥ 20 1 1 16 = + = 20 300 300 0.01 16 ⇒ = DL 300 0.01 × 300 3 ⇒ ∆L = = 16 16 3 1 ∆L \ = ¥ 16 20 L DL 3 1 Percentage error = ¥ 100 = ¥ ¥100 L 16 20 15 =  1% 16 So the correct option is (c). 7. Refer to the following figure. SUN

EARTH Rocket

dr 3R2rv + R3 =0 dt

or

∆T = ∆L

1 2 1 + ¥ v g 2 L

=

So the correct option is (d). 5. The total mass of the sphere is constant. Thus

⇒

∆T = ∆L



p n

2 Ê 1 ˆ 1 ¥Á– + ˜ g Ë 2 L¯ v

x = 2.5 × 104RE

m

v

ME

M5 = 3 × 105 ME

The escape velocity for earth is

ve =

2GM E = 11.2 kms–1 (given) RE

The gravitational potential energy of the SUN– EARTH system is (m = mass of the rocket) GM E m GM S m U = – – RE x = –

G ¥ 3 ¥ 105 M E m GM E m – 2.5 ¥ 104 RE RE

= –

GM E m RE

or

U = –

(1 + 12)

13GM E m RE

The minimum initial velocity v required for to escape from the sun-earth system is given by 13GM E m 1 2 mv = RE 2

Physics Paper II—2017  P-II.7

v =

⇒

=

13 ¥

2GM E RE

13 ve

= 13 ¥ 11.2 = 40.4 kms–1 The closest option is (b). 8. At time t = 0 when the switch S is open, the source of voltage is not connected to the circuit. Hence no current flows in the circuit at time t = 0. So option (a) is incorrect. Immediately after the switch is closed, a current begins to flow. Since the inductors offer reactance, it takes some time to grow to its steady state value = i0. Since the inductors L1 and L2 are connected in parallel, the induced voltage VL is the same across each inductor. If i1 and i2 are the currents through L1 and L2 at any time t during the growth of current, then di di VL = L1 1 = L2 2 dt dt or L1di1 = L2 di2 ⇒

Ú

+Q q x = R2 + r2

R

dr

r O

Ú

L1 di1 = L2 di2

⇒ L1i1 = L2i2 ⇒

E

i1 L2 = , which is independent i2 L1

of time. So option (d) is correct. Now, i0 = i1 + i2 or i2 = i0 – i1   i2 i0 or = –1 i1 i1 or

So option (b) and (c) are both correct. Thus the correct options are (b), (c) and (d). 9. The hemisphere consists of a curved surface of radius R and a flat (bottom) surface which is a disc of radius R. Since the charge + Q lies outside the hemisphere, the net electric flux (fe) through the curved surface and the flux (ff) through the flat surface is zero, i.e.. fc + ff = 0 ⇒ fc = – ff So option (c) is incorrect. To calculate flux through the flat surface (disc), we divide the disc into very small elements of length dr. Consider one such element at a distance r from the centre O of the disc as shown in the figure.

L1 i0 = – 1 L2 i1

⇒

Area of element is dA = 2p r dr. The electric flux through the flat surface is R

Ê i1 L2 ˆ ÁË∵ i = L ˜¯ 2 1

i L i1 = 0 2 L1 + L2

Since, in the steady state, the inductors offer no V reactance, i0 = . Hence R

L2 V i1 = R ( L1 + L2 )

i2 = i0 – i1 L2 V V = – R R ( L1 + L2 ) Now,

=

L1 V R ( L1 + L2 )

Electric Field due to + Q at the element is Q Q E = = 2 4pe 0 x 4pe 0 ( R 2 + r 2 )



ff =

Ú

  E ◊ dA =

0

Ú

R

E dA cos q =

0

R

=

R

R

Ú E dA ¥ x 0

Q

Ú 4pe0 ( R2 + r 2 ) ¥ 2prdr ¥ 0

2p QR = 4pe 0

R

rdr

Ú ( R2 + r 2 )3/2 0

R

QR È 1 ( R 2 + r 2 ) –1/2 ˘ = Í ˙ 1 2e 0 Í 2 ˙ – Î ˚0 2 =

QR È 1 – 2e 0 ÍÎ R

=

Q 2e 0

1

˘ ˙ R +R ˚

1 ˆ Ê ÁË1 – ˜ 2¯

2

2

R 2

( R + r 2 )1/2

P-II.8  Comprehensive Physics—JEE Advanced

Now fc = – ff = –

Q 2e 0

1 ˆ Ê ÁË1 – ˜ 2¯

r < x i.e. if r <

So option (a) is correct. Electric potential at any point on the circumference of the flat surface is Q Q 1 V= = 4pe 0 R 2 + R 2 4pe 0 ( 2 R) which is constant. Hence the circumference of the flat surface is equipotential. So option (d) is correct. The component of the electric field normal to the flat surface is Q Q cos 2 q 1 = , which 2 4pe 0 R 2 4pe 0 ( R /cos q )



varies with q and is constant. So option (b) is incorrect. Thus the correct options are (a) and (d). 10. The particle will describe a circle of radius r mv p = given by r = in the anticlockwise sense. QB QB Option (a): If B = r=

Refer to the following figure. It follows from the figure that y-axis

B Region 2

Region 3

13R/8

5R 2 r

x

Q

P2

x-axis

R + Q  P1 3R 2 2

. So option (b) is also correct. Option (c): The distance between P1 and P (the 2mv point of re-entry in region 1 = 2r = which QB is directly proportional to m (for given v and B). So option (c) is incorrect. Option (d): The particle leaves region 2 and reenters region 1 moving horizontal along the negative x direction with momentum – p, it entered region 2 with momentum +p. So the change in momentum = – p – (+p) = – 2p. So option (d) is incorrect. Thus the correct options are (a) and (b). 11. Refer to the following figure. P1 P

S1

q O d

S2

P2

8 p , 13 QR

p 13QR 13 ¥ = R Q 8p 8

Region 1 P

3R p 3R 2p < , i.e. if or B > QB 2 3QR 2

13R ˆ Ê 5R ˆ x = ÁÊ –Á ˜ Ë 2¯ Ë 8 ˜¯

2

=

3R which is equal to OP2. 2

Hence the particle will enter region 3 at point P2. So option (a) is correct. Option (b) The particle will re-enter region 1 if it describes a semicircle in region 2 of r a d i u s

At point P1, the path difference is Dx = S1P1 – S2P1 = 0. Hence there is a bright fringe at P1. So option (b) is incorrect. At point P2, Dx = S1P2 – S2P2 = S1S2 = d. There will be nth bright spot at P2 if d = nλ –3 d or n= = 1.8 ¥ 10 = 3000 l 600 ¥ 10 –9 So there will be 3000th bright spot at P2. So option (c) is correct. Hence there will be n – 1 = 3000 – 1 = 2999 bright spots between P1 and P2. So option (a) is correct. Let P be the point on the circle for which angle between OP and OP1 is, say q. Assuming that the circle has a large radius, the path difference at point P is Dx = d sinq The rate of change of path difference with angle q is d (∆x) = d cosq, dθ which decreases will increase in q. So the angular separation between consecutive bright spots increases as we go from P1 to P2. Hence option

Physics Paper II—2017  P-II.9

(d) is incorrect. Thus the correct option are (a) and (c). 12. When the voltmeter is connected between X and Y VXY = VX – VY 2p ˆ Ê = V0 sin ωt – V0 sin Á w t + ˜ Ë 3¯ È 2p ˆ ˘ Ê = V0 Ísin w t – sin Á w t + ˜ ˙ Ë 3 ¯˚ Î È p p ˘ = 2V0 Ícos ÊÁ w t + ˆ˜ sin ÊÁ – ˆ˜ ˙ Ë ¯ Ë 3 3¯˚ Î =–

3¥2 pˆ Ê V0 cos Á w t + ˜ Ë 2 3¯

pˆ Ê = + 3 V0 sin Á w t – ˜ Ë 3¯

Maximum value of VXY =

3V0

3V0

\ rms value of VXY =

. So option (b) is correct 2 and option (c) is incorrect. When the voltmeter is connected between Y and Z. VYZ = VY – VZ 2p ˆ 4p ˆ Ê Ê = V0 sin Á w t + ˜ – V0 sin Á w t + ˜ Ë ¯ Ë 3 3¯ È 2p 4p ˘ = V0 Ísin ÊÁ w t + ˆ˜ – sin ÊÁ w t + ˆ˜ ˙ Ë ¯ Ë 3 3 ¯˚ Î =

Option (a): The rod rotates about an axis passing through B and perpendicular to the plane of the page. Hence point A describes a circular trajectory. So option (a) is incorrect. Option (b): At any instant of time, the magnitude of the torque about B = mg ¥ BD = mg B < sin q  mgL  =   sin q  2  This is so because the entire mass m of the rod can be assumed to be acting at its centre of AB L mass C. Since the rod is uniform, BC = = . 2 2 So option (b) is correct. Option (c): Since there is no external horizontal force and the initial velocity of the rod is zero, the centre of mass will move vertically downwards. So option (c) is correct. Option (d): Initial y-coordinate of the centre of L mass = . When the rod subtends an angle q 2 L with the vertical, the y-coordinate = cosq. Shift 2 L L L in the mid-point = – cos q = (1 – cosq). 2 2 2 So option (d) is correct. Thus the correct options are (b), (c) and (d). S

3 V0 cos wt

Maximum value of VYZ =

Q

3 V0

3V0

\ rms value of VYZ =

R Mg

, which is the same

2 as the rms value of Vxy. So option (a) is incorrect and option (d) is correct. Hence the correct options are (b) and (d). 13. Refer to the following figure. A

F1

R

Fig. 1

q Q

P

N2

F2

L

q

C (90° – q)

D

O mg

Mg

B N1

Fig. 2

P-II.10  Comprehensive Physics—JEE Advanced

Option (a): If the force F1 is applied at S as shown in Fig. 1, torque of F1 about Q = F1R (anticlockwise). Torque of Mg about Q = MgR (clockwise). The net torque depends on the magnitudes of F1 and Mg is given by t = F1R – MgR = (F1 – Mg) R If F1 > Mg, t π 0 and the wheel will climb the step. So option (a) is incorrect. Option (b): If the force F2 is applied normal to point P as shown in Fig. 2, then F2 balances with normal reaction N2 of the wall at Q and force Mg balances with the normal reaction N1 of the base of the step as shown. Hence no net force acts on the wheel and t = 0. So option (b) is correct. Option (c): If force F3 is applied at X as shown in Fig. 3, the torque due to F3 about Q = F3QT (clockwise). The torque due to Mg about Q = MgR (anticlockwise). The net torque on the wheel is t = F3QT – MgR

Torque due to F4 about Q = F4 ¥ 2 R cosq (clockwise) Torque due to Mg about Q = Mg ¥ R cosq (anticlockwise) Net clockwise torque = F4 ¥ 2R cosq – Mg ¥ R cosq (clockwise) = (2 F4R – MgR) cosq. Initially q = 0. As the wheel rises q increase from p zero to . Hence cosq decreases from 1 to zero. 2 So option (d) is correct. Thus the correct options are (b) and (d). 15. At time T > RC, the capacitor is fully charged to the voltage of the battery (here RC is the time constant of the RC circuit). V When the voltage is V1 = 0 , the charge supplied 3 CV0 by the battery is Q1 = CV1 = 3 2V0 , the 3 additional charge supplied by the battery is

When the voltage is increased to V2 =

T R

Q

2V Ê 2V ˆ Q2 = C Á 0 ˜ – C 0 Ë 3 ¯ 3

X F3

Fig. 3 If F3QT > MgR, a net clockwise torque acts on the wheel, as a result the wheel begins to climb up the step. As it climbs QT increases. Hence t does not remain constant. So option (c) is incorrect. Option (d): If the force F4 is applied tangentially at P, as shown in Fig. 4, the normal reaction at Q will be absent. Hence the frictional force is zero. As a result the wheel will start slipping as it rises up the step. It is given that there is no slipping, the wheel will rise up causing point Q to come down as shown in the figure. F4

P Mg

R cos q

q

R cos q

Fig. 4

Q

=

CV0 3

When the voltage is increased to V3 = V0, the additional charge supplied by the battery is 2CV0 CV0 Q3 = CV0 – = 3 3 Total energy stored in the capacitor is 1 1 1 E = Q1V1 + Q2 V2 + Q3V3 2 2 2 =

1 Ê CV0 ˆ V0 1 Ê CV ˆ + Á 0˜ ÁË ˜¯ ¥ 2 3 2Ë 3 ¯ 3

  ¥

2V0 1 Ê CV0 ˆ + Á ˜ ¥ V0 3 2Ë 3 ¯

1 1 1 CV0 2 + CV0 2 + CV0 2 18 9 6 1 = CV0 2 3 Since the final voltage is V0, final charge on the capacitor is Q = CV0 and the final energy stored in the capacitor is 1 E ¢ = CV0 2 2 =

Physics Paper II—2017  P-II.11

\ Energy dissipated through the resistor is ED = E ¢ – E 1 1 1 = CV0 2 – CV0 2 = CV0 2 2 3 6 So the correct option is (d). 16. In this case, the charge supplied by the battery is Q0 = CV0 \ Energy supplied by battery = Q0V0 = CV02 Energy stored in the capacitor is 1 1 EC = Q0V02 = CV02 2 2 Energy dissipated through the resistor is 1 1 ED = CV02 – CV0 2 = CV0 2 2 2 \  EC = ED. So in this case the correct option is (c). 17. Refer to the following figure.

Let the angular speeds of the smaller and larger rings be w1 and w2 respectively. Let v be the linear speed of the centre C of the larger ring. The centre of the larger ring moves in a circle of radius (R – r). So v = (R – r)w2 ...(1) Since there is no slipping at the point of contact, rw1 = Rw2 – v       ...(2) Using (1) in (2), we have rw1 = Rw2 – (R – r) w2 = rw2 ⇒ w1 = w2 = w0 Normal reaction at the point of contact is given by N = Mac = Mw02 (R – r) For translational equilibrium in the vertical direction, Mg = mN = mMw02 (R – r) ⇒ ω0 =

g , which is option (b). m(R – r)

Larger Ring Smaller Ring w1 N

r R–r v

C

18. The total kinetic energy = w2

1 1 M vcm 2 + I cm ω20 2 2

1 1 Mw02(R – r)2 + (MR2)w02 2 2 1 = Mw02 ( R – r ) 2 + R 2  2 So no option is correct. =