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NUMERICAL EXAMPLESIN

PHYSICS

BRIJ LAL Dr. N. SUBRAHMANYAM M.Sc. M.Sc., Ph.D.

Reader in Physics (Retired) Reader in Physics (Retired)Hindu College, University of Delhi Kirori Mal College

Delhi – 110 007 University of DelhiDelhi - 110 007

Revised byDr. M.K. GUPTA

M.Sc., Ph.D., B.Ed.Physics Department.

S.D. Public School (Senior Wing)Muzaffarnagar – 251 001(U.P.)

S. CHAND & COMPANY LTD.(An ISO 9001 : 2000 Company)

RAM NAGAR, NEW DELHI - 110 055

[For the students of 10+2 (CBSE, ISC and other State Boards) and also useful forEngineering and Medical Entrance Examinations]

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© 1965, Brij Lal, Dr. N. Subrahmanyam

All rights reserved. No part of this publication may be reproduced, stored in a retrievalsystem or transmitted, in any form or by any means, electronic, mechanical, photocopying,recording or otherwise, without the prior permission of the Publishers.t

First Edition 1965Subsequent Editions and Reprints 1966, 70, 72, 74, 76, 79, 80, 81, 82, 83, 84 (Twice), 85,86, 87, 88, 89, 90, 92, 94, 95, 96, 98, 99, 2000, 2003, 2004, 2006, 2008 (Twice), 2009Twenty Fourth Revised & Enlarged Edition 2010

ISBN : 81-219-1257-1 Code : 16A 142

By Rajendra Ravindra Printers Pvt. Ltd., 7361, Ram Nagar, New Delhi -110 055and published by S. Chand & Company Ltd., 7361, Ram Nagar, New Delhi -110 055.

PREFACE TO THE FIRST EDITIONThis book entitled “Numerical Examples in Physics” is written to suit the needs of

students appearing for Higher Secondary, Pre-University, Pre-Medical and I.I.T. EntranceExaminations of Indian Universities or Boards. The main feature of the book is theintroduction of M.K.S.A. (Metre-Kilogram-Second-Ampere) system of units. Many solvedand unsolved examples are given in M.K.S.A. system of units, besides F.P.S. and C.G.S.units. With the introduction of the Metric System of units in India, it was felt necessary thatthe students in Physics also should understand and use this system of units in studying thesubject.

The book contains 40 chapters distributed over seven parts (Mechanics, Heat, Light,Sound, Magnetism, Frictional Electricity and Current Electricity). At the beginning of eachchapter necessary theory, formulae and units in the three systems are given so as to enablethe students to understand the solved examples and to apply them to solve the examplesgiven in the exercises. The solved examples are mainly drawn from the University or BoardExamination question papers of recent years and while framing the examples for theexercises, the same representative character has been kept in mind. Illustrative diagrams aregiven at necessary places.

We hope that this book will be found useful by the students and teachers of the variousinstitutions in India. We will appreciate any suggestions from teachers and students for theimprovement of the book.

In the end, we sincerely thank Mr. Shyam Lal Gupta, Managing Director, S. Chand& Company Ltd., Ram Nagar, New Delhi, for publishing this book.

BRIJ LALDr. N. SUBRAHMANYAM

DelhiMay 1, 1965

PREFACE TO THE TWENTY FOURTH EDITIONIt gives me great pleasure in revising the famous book “Numerical Examples

in Physics” by Late Prof. Brij Lal and Dr. N. Subrahmanyam. This book was firstpublished in 1965. Since then, the book has been revised several times and was inthe market with 23 editions upto 2007. In the time span of 1965 to 2001, this bookwas one of the most popular Numerical Physics book among the students of ClassesXI, XII and among the Competitive aspirants of Engineering and Medical EntranceExaminations. However after 2001, due to sudden demise of one of the authors,Prof. Brij Lal and ill health of the other author, Prof. Subrahmanyam, the book couldnot be revised as per new syllabi and approach.

Now, S. Chand & Company, showed confidence in me and asked me to revisethis important Numerical Physics book. I have taken up the task to revise this bookas per the present needs of the + 2 and Competitive Exam. (Engg. & Medical) students.The goal is to create interest of students in the numericals without which study ofPhysics is worthless. The numerical examples solved in the book are helpful inclarifying the exact meaning of terms in the equation.

Each chapter has been divided topic-wise and plenty of solved numericalexamples are given on each topic. Large number of Unsolved examples are given atthe end of each chapter. I have tried my best to arrange the numericals in the sequencefrom easier to tough. I hope this will help the students for preparing for their BoardExaminations, as well as for competitions.

I thank the editorial board and production department of S. Chand & CompanyLtd., New Delhi. I am thankful to my fellow teachers and my wife Dr. Kavita Guptafor the encouragement and valuable guidelines.

I would appreciate receiving feedback from students as well as teachers.I dedicate this book to my teachers and all those who are serving the cause of

education.

Dr. M.K. GUPTA [email protected]

CONTENTS

1. Basic Mathematical Tools . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 1 – 62. Physical World and Measurement . . . . . . . . . . . . . . . . . . . . . . .. 7 – 473. Motion in One Dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48 – 854. Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 86 – 1165. Motion in two and three Dimensions (Projectile . . . . . . . . . . . . 117 – 141

Motion, Circular Motion)6. Laws of Motion and Friction . . . . . . . . . . . . . . . . . . . . . . . . . . .. 142 – 1937. Work / Power and Energy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 194 – 2338. Rotational Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234 – 2789. Gravitation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 279 – 320

10. Properties of Solids . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 321 – 33811. Mechanics of Liquid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 339 – 38012. Thermal Properties of Matter . . . . . . . . . . . . . . . . . . . . . . . . . . 381 – 41613. Thermodynamics and Kinetic Theory of Gases . . . . . . . . . . . . . 417 – 45814. Oscillations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 459 – 49415. Wave Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 495 – 549

1

ALGEBRIC IDENTITIES(i) (a + b)2 = a2 + b2 + 2ab

= (a – b)2 + 4ab

(ii) (a – b)2 = a2 + b2 – 2ab

= (a + b)2 – 4ab

(iii) (a + b)3 = a3 + 3a2b + 3ab2 + b3

= a3 + b3 + 3ab (a + b)

(iv) (a – b)3 = a3 – 3a2b + 3ab2 – b3

= a3 – b3 – 3ab (a – b)

(v) a2 – b2 = (a + b) (a – b)

(vi) a3 + b3 = (a + b ) (a2 – ab + b2)

= (a + b)3 – 3ab (a + b)

(vii) a3 – b3 = (a – b) (a2 + ab + b2)

= (a – b)3 + 3ab (a – b)

(viii) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

QUADRATIC EQUATIONA quadratic equation is an equation of the second order.

ax2 + bx + c = 0

the roots of such a quadratic equation are

x =2– – 4

2

b b ac

a

TRIGONOMETRY(i) Sexagesimal system:

1° (degree) = 60 (minute)

1’ = 60 (second)

(ii) Circular System:

In circular system, the unit of angle is taken radian. The angle subtended at the centre of acircle is

=circumference

radius

=2 r

r

= 2 radian.

�� 1UNIT

2 Numerical Examples in Physics

radian = 180°

1 radian = 57° 16 22

TRIGONOMETRICAL RATIOS AND THEIR RELATIONS

(a) sin =Perpendicular

hypotenuse

=p

h

cosine (cos ) =base

hypotenuse

=b

h

tangent (tan ) =perpendicular

base =

p

b

cotangent (cot ) =base

perpendicular =

b

p

secant (sec ) =hypotenuse

base =

h

b

cosecant (cosec ) =hypotenuse

perpendicular =

h

p

(b) cosec =1

sin

sec =1

cos

cot =1

tan (c) sin2 + cos2 = 1

1 + tan2 = sec2 1 + cot2 = cosec2

(d) Values of Trigonometrical ratios

0° 30° 45° 60° 90°

sin 0 1/2 12

32 1

cos 1 32

12

12 0

tan 0 13 1 3

A B

C

ph

b

angle

T.R.

Basic Mathematical Tools 3

(e) T-ratios of Allied Angles

sin (–) = – sin cosec (–) = – cosec cos (–) = cos sec (–) = sec tan (–) = – tan cot (–) = – cot

sin (90° – ) = cos cosec (90° – ) = – sec cos (90° – ) = sin sec (90° – ) = cosec tan (90° – ) = cot Cot (90° – ) = tan

sin (90° + ) = cos cosec (90° + ) = sec cos (90° + ) = –sin sec (90° + ) = –cosec tan (90° + ) = –cot cot (90° + ) = –tan

sin (180° – ) = sin cosec (180° – ) = cosec cos (180° – ) = –cos sec (180° – ) = – sec tan (180° – ) = –tan cot (180° – ) = – cot

sin (180° + ) = – sin cosec (180° + ) = – cosec cos (180° + ) = – cos sec (180° + ) = – sec tan (180° + ) = tan cot (180° + ) = cot

sin (270° – ) = – cos cosec (270° – ) = – sec cos (270° – ) = – sin sec (270° – ) = –cosec tan (270° – ) = cot cot (270° – ) = tan

sin (270° + ) = – cos cosec (270° + ) = – sec cos (270° + ) = sin sec (270° + ) = cosec tan (270° + ) = –cot cot (270° + ) = – tan

sin (360° – ) = – sin cosec (360° – ) = – cosec cos (360° – ) = cos sec (360° – ) = sec tan (360° – ) = – tan cot (360° – ) = – cot

(f) Important Trigonometrical Formulae, to be used in Physics:

sin (A + B) = sin A cos B + cos A sin B

sin (A – B) = sin A cos B – cos A sin B

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

tan (A + B) =tanA + tanB

1–tanA tanB

tan (A – B) =tanA – tanB

1+ tanA tanB

sin 2A = 2 sin A cos A = 2

2tanA

1– tan Acos 2A = cos2A – sin2 A

= 1 – 2 sin2A

= 2cos2A – 1


=2

2

1–tan A

1+ tan A

sin A + sin B = 2 sin A+B

2 cos

A–B

2

sin A – sin B = 2 cos A + B

2 sin

A – B

2

cos A + cos B = 2 cos A + B

2 cos

A – B

2

cos A – cos B = – 2 sin A + B

2 sin

A – B

2.

BINOMIAL EXPANSIONLet n is any integer (positive, negative or a fraction) and x is any real number then

(1 + x)n = 1 + nx + – 1

2

nn x2 +

– 1 – 2

3

nn n x3 + ...

Here 2 = 2 × 1 = 2

3 = 3 × 2 × 1 = 6

In case x <<1, then

(1 + x)n� 1 + nx.

Mensuration:Perimeter of a circle = 2r (r is radius of circle )

Area of a circle = r2

Surface area of a sphere of radius r = 4r2

Volume of the sphere = 34

3r

Volume of a cube of side a = a3

Surface area of cube = 6a2

Diagonal of cube = side 3 = a 3

Surface area of a cylinder of radius r and height l = 2r (l + r)

Curved Surface area of the cylinder = 2rl

Volume of cylinder = r2l

Curved Surface area of a cone of base radius r, and slant height l = rl

Volume of cone =2

3

r h

Area of a regular polygon A =2

na r (a is side of polygon and r is the distance

of any side of polygon from the centre)

Area of trapezium:

A =1

2 × Sum of parallel sides × distance between them.

Basic Mathematical Tools 5

DIFFERENTIAL CALCULUSSome important formulae:

(i) ndx

dx= nxn–1

(ii) If a is any constant thend

dx (a) = 0

(iii) If y is a function of x [y = f(x)]. then

d

dx (ay) = a

dy

dx

(iv) y = u v : u = f(x), v = f(x)

thendy

dx=

du

dx ±

dv

dx(v) Product of functions

Let y = u.v

thendy

dx=

dvu

dx +

duv

dx(First function differentiation of second function + Second function differentiation of firstfunction.)

(vi) Division of functions

y =u

v

Thendy

dx= 2–

du dv

v u vdx dx

(vii) loged

xdx

=1

x

(viii)d

dx (loga x) =

1

x loge a

(ix)d

dx ex = ex

(x)d

dx (ax) = loge a

(xi)d

dx (sin x) = cos x

(xii)d

dx (cos x) = – sin x

(xiii)d

dx (tan x) = sec2x

(xiv)d

dx (cot x) = – cosec2x

(xv)d

dx (sec x) = sec x tan x

(xvi)d

dx (cosec x) = – cosec x cot x


INTEGRAL CALCULUSSome important formulae

(i) nx dx =1

1

nx

n

+ c

(ii)dx

x = loge x + c

(iii) sin x dx = – cos x + c

(iv) cos x dx = sin x + c

(v) 2cosec x dx = – cot x + c

(vi) 2sec x dx = tan x + c

(vii) sec tanx x dx = sec x + c

(viii) cosec cotx x dx = – cosec x + c

(ix) nax b dx =

1

1

nax b

a n

+ c

(x)dx

ax b =1

a loge (ax + b) + c

(xi) xe dx = ex + c

(xii) xa dx =log

x

e

a

a = ax.loga e + c

In above all formulae c is integration constant.

DEFINITE INTEGRALAn integral defined between two definite limits is called definite integral.

b

a

f x dx = 'b

af x

= f (b) – f (a)

where f (x) is the integral of f (x)

a is called lower limit.

b is called upper limit.

7

UNITThe reference standard of measurement of a physical quantity is called its unit.

The force acting on a body is 5N. Here 5 is numerical value of physical quantity (force) andN (newton) is its unit.

The numerical value of a physical quantity is inversely proportional to the size of unit.

n 1

u

SYSTEM OF UNITSLength Mass Time

FPS System foot pound second

CGS System centimetre gram second

MKS System metre kilogram second

International System of Units (SI Units). It consists of seven basic units and twosupplementary units.

(i) Basic Units

Physical Quantity Unit Symbol

Mass kilogram kg.Length metre m

Time second s

Current ampere A

Temperature kelvin K

Luminous Intensity candela Cd.

Amount of Substance mole mol.

(ii) Supplementary Units

Angle radian rad.

Solid angle steradian Sr.

Fundamental and Derived UnitsThose units which are independent of each other and cannot be interconverted are called

fundamental units.For example unit of mass, length and time are fundamental units.Those units which are obtained from fundamental units, by certain mathematical operations

(multiplication or division) are called derived units.For example unit of speed is ms–1. (Division of two fundamental units.)

��

�� 2UNIT


Indirect methods for measuring long distances(i) Echo method or reflection method:By noting the time ‘t’ taken by the sound to travel to the obstacle and back to the initial

position, the distance of the obstacle can be calculated.

t =2x

u

or x =.

2

u t

u is speed of sound.

(ii) Triangular method:(a) Measurement of the height of a tower (accessible):

AB is tower and c is any point at distance x from foot of thetower. is angle of elevation of the top of the tower

h

x= tan

or h = x tan (b) Measurement of the height of a tower (inaccessible):

An object whose distance from the observation point cannot bedetermined is called inaccessible object.

h is height of tower AB. The angle of elevation of top oftower is from point P and on moving x distance away fromtower at point Q angle of elevation is .

AP

h= cot

AP x

h= cot

orx

h= cot – cot

or h =cot – cot

x

Parallax Method: Consider is the parallactic angle of an

astronomical object, at two locations on the earth seperated by distance AB,then the distance of the object is given by

Parallactic angle = (1 + 2)

Angle =Arc

Radius

Radius CM =AB

Size of an Astronomical Object:O is an observation point, on earth. be the angle subtended by two

end points of diameter of a planet, at point O. d be the average distanceof moon from the surface of earth then,

=AB

dor AB = d

C A

B

x

h

A B

C

2 1

2 1

M

O

earth

A B

d

Physical World and Measurement 9

Abbreviations in powers of 10

Power of 10 Prefix Symbol

1018 exa E

1015 Peta P

1012 tera T

109 giga G

106 mega M

103 kilo K

102 hecto h

101 deca da

10–1 deci d

10–2 centi c

10–3 milli m

10–6 micro 10–9 nano n

10–12 pico P

10–15 femto f

10–18 atto a

Some Important Practical Units(a) Largest Practical unit of mass is

Chandra Shekar Limit (C.S.L.) = 1.4 times the mass of Sun.

(b) Solar day: Time taken by earth to complete one rotation about its axis w.r.t. Sun.

(c) Sedrial day: Time taken by earth to complete one rotation about its axis w.r.t. a fixedstar.

(d) Solar year: Time taken by earth to complete one revolution around the sun in its orbit

1 Solar year = 365.25 Solar day

(e) Lunar month: The time taken by the moon to complete one revolution around the earthin its orbit is called lunar month. It equals to 27.3 days.

(f) One shake = 10–8 sec.

(g) One atmospheric pressure= 1 bar

= 105 Nm–2

= 105 Pascal (Pa)

1 torr = 1mm of Hg column

1 bar = 760 torr.

DIMENSIONSThe dimensions of a derived unit are the powers to which the fundamental units must be

raised to obtain that derived unit. For example,

speed =distance L

time T

= [LT–1][LT–1] is called dimensional formula of speed. Dimensions of speed in length is 1 and in

time is – 1.


Principle of Homogeneity of dimensionsAccording to this principle–A physical equation will be dimensionally correct if the

dimensions of all the terms on both the sides of the equation are same.

Uses of Dimensions(a) Conversion of one system of unit into other system. e.g. c.g.s. in M.K.S.

(b) To derive a relationship between different physical quantities.

(c) To check the correctness of a physical relation.

Dimensions of some physical quantities

S.No. Physical Basic Unit Dimensions RemarkQuantity formula

1. Length L metre (m) [L]

2. Mass M Kilogram (kg) [M]

3. Time T second (s) [T]

4. Current A ampere (A) [A]

5. Temperature K or Kelvin (K) [K] or []

Mechanics

6. Area length × breadth m2 [L2]

7. Volume l × b × h m3 [L3]

8. Density mass/volume kg m–3 [M L–3]

9. Relative density density of substance — [MoLoTo]density of water at 4°C

10. Speed distance/time ms–1 [LT–1] velocity also

has samedim.

11. Acceleration change in velocity/time ms–2 [LT–2]

12. Momentum mass × velocity kg ms–1 [MLT–1]

13. Force mass × acceleration kg ms–2 (N) [MLT–2] Kg ms–2 iscalled

Newton (N)

14. Impulse of force force × time interval N–s [MLT–1]

15. Tension N [MLT–2]

16. Linear mass mass/length kgm–1 [ML–1]

density (mass

per unit length)

17. Force Constant force/extension in spring Nm–1 [MLoT–2]

18. Coefficient of force of friction — [MoLoTo]friction normal reaction

19. Work force × distance N–m (J) [ML2T–2]

20. Kinetic energy1

2 mass × (velocity)2 kg m2 s–2 (J) [ML2T–2] Kg m2s–2 is

or any other form called Jouleof energy (J)


21. Power work/time Js–1 (w) [ML2 T–3] Js–1 is calledWatt (W)

22. Gravitational F = 1 22

Gm m

rKg–1 m3 s–2 [M–1 L3 T–2]

constant

23. Gravitational Gm/r2 ms–2 [MoLT–2]

field strength

24. Gravitational –Gm/r J/kg [MoL2T–2]

Potential

25. Gravitational – GMm/r J [ML2T–2]

Potential energy

26. Escape velocity Ve = 2GM

Rms–1 [MoLT–1]

27. Orbital Velocity Vo = GM/R ms–1 [MoLT–1]28. Pressure or stress force/area Nm–2 [ML–1 T–2]

29. Strain change in configuration — [MoLoTo]original configuration

30. Modulus of stress NM–2 [ML–1 T–2]elasticity. strain

(a) Youngs modulus

(b) Bulk modulus

(c) Shear modulus

31. Compressibility 1/Bulk modulus (NM–2)–1, N–1m2 [M–1LT2]

32. Surface tension force/length Nm–1 [MT–2]

33. Coefficient of F = A dv

dxkg m–1 s–1 [ML–1 T–1]

viscosity (poise)

34. Angle arc/radius radian [MoLoTo]

35. Angular velocity angle/time rad/sec [MoLoT–1]

36. Angular accele- angular velocity rad/sec2 [MoLoT–2]

ration time

37. Angular = 2/T rad/sec [MoLoT–1]

frequency

38. Moment of mass × (distance)2 kgm2 [ML2To]inertia

39. Radius of distance m [MoLTo]

gyration

40. Moment of force × distance N – m. [ML2 T–2]

force or couple

41. Angular Linear momentum kgm2s–1 [ML2T–1]momentum distance


42. Rotational1

2 × I × 2 Joule (J) [ML2T–2]

kinetic energy

43. Time period T sec. [MoLoT]

44. Frequency 1/time period Hertz (Hz) [MoLoT–1]

45. Wave velocity ms–1 [MoLT–1]

46. Wavelength = c/ = Velocity

Frequencym [MoLTo]

47. Planck’s Constant E = h kgm2s–1 [ML2 T–1] Angular

or h = E/ momentum

also hassame

dimensions.

Thermodynamics48. Heat Joule [ML2T–2]

49. Specific heat Q = mS JKg–1 k–1 [MoL2T–2K–1]

or S = Q/m50. Latent heat Q = mL Jkg–1 [MoL2T–2]

or L = Q/m

51. Molar gas PV = nRT Kg m2 s–2K–1 [ML2T–2K–1]constant

R = PV

nT52. Bollzmann K = R/N (Avogadro’s) kgm2s–2k–1 [ML2T–2K–1]

constant K. number)

53. Coefficient of Q = 1 2KA – t

d

kgms–3 K–1 [MLT–3 K–1]

thermalconductivity or K =

Q.

A –

d

t Electrostatics54. Charge Current × time Coulomb (C) [AT]

55. Electric field force/charge NC–1 [MLT–3 A–1]

56. Permittivity of F = 1 22

1.

4q q

rNm2C–2 [M–1L–3T4A2]

free spaceor o =

1

4 F 1 2

2

q q

r57. Electric flux Electric field strength Nm2C–1 [MLT–3A]

× Area

58. Electric potential work/charge volt (v) [ML2T–3A–1]

59. Relative permi- Permittivity of medium — [MoLoTo]ttivity of a Permittivity of free space

medium

60. Capacitance Charge/Potential Farad (F) [M–1L–2T4A2]


61. Electric dipole Magnitude of charge C – m [MoATL]

Moment × distance between them

62. Energy density Energy/volume Jm–3 [ML–1 T–2]

(capacitor)

63. Resistance Potential difference ohm () [ML2T–3A–2]current

64. Resistivity R = L

( )A

– m [ML3T–3A–2]

or = RA

L

65. MobilityDrift velocity

Electric fieldCs kg–1 [M–1LoT2A]

66. Conductance C = 1/R (ohm)–1� [M–1L–2T3A2]

67. Conductivity = 1

��m–1 [M–1L–3T3A2]

Magnetism68. Magnetic field F = BIL Tesla (T) [ML0 T–2A–1]

or B = F/IL

69. Magnetic B = oI

2r

Wb A–1m–1 [MLT–2A–2]

permeabilityor o =

B 2r

I

70. Magnetic flux Magnetic field area Weber (Wb) [ML2T–2A–1]71. Coefficient of N = LI Henery (H) [ML2T–2A–2]

self inductionor L =

N

I

Self inductance72. Coefficient of N = MI H [ML2T–2A–2]

Mutual induction

or or M = N

I

.

Mutual inductance

SIGNIFICANT FIGURESSignificant figures are those digits in a measurement that are known reliably plus one digit

about which we are uncertain.

Rules to calculate the Number of Significant figure in a Measurement:(i) All non zero digits in a measurement are significant. 127.4 cm has 4 significant figures.

(ii) Zeros coming between two non zero digits are significant. 1002 has 4 significantfigures.

(iii) Unless otherwise indicated all zeros to the left of an understood decimal point but to theright of non zero digit are not significant. 30700 has 3 significant figures.

(iv) All zeros to the left of an expressed decimal point and to the right of non zero digit aresignificant 30.100 has 5. significant figures.

(v) All zeros to the right of a decimal point are significant 764.0 cm has 4 significantfigures.

(vi) All zeros to the right of a decimal point but to the left of a non zero digit are nonsignificant. 0.017 have 2 significant figures.

or


(vii) The number of significant figures does not vary with the choice of different unit.12.1g; 0.0121 kg both have 3 significant figures.

ADDITION AND SUBTRACTION INTERMS OF SIGNIFICANT FIGURESIn addition or subtraction the number of decimal places which are significant in the result

is same as the smallest number of decimal places in any number used in addition or subtraction.

For example, 273.46+45.3318.76

= 318.8 (Interms of significant figures)

MULTIPLICATION AND DIVISION INTERMS OF SIGNIFICANT FIGURESIn multiplication or division the number of significant figures in the product or the quotient

is the same as the smallest number of significant figures in any one of the factor.

Ex, 321 × 15 = 4815

= 4.815 × 103

= 4.8 × 103 (Interms of significant figures)

ErrorsRandom Error: The errors which are due to nature or by chance are called random error.

To minimise random errors, number of readings are taken then arithmetic mean is taken.

Suppose five readings of the time period of a simple pendulum are taken then real value(true value or mean value) of time period is given by

t = 1 2 3 4 5

5t t t t t

or t =1

1 n

ii

tn

Absolute Error: The magnitude of the difference of true value and the measured value iscalled absolute error.

absolute error = 1 true value – measured value

t1 = | t – t1 |

t2 = | t – t2 |

t5 = | t – t5 |

The arithmetic mean of all the absolute errors is called mean absolute error i.e.,

t = 1 2 5...

5t t t

Relative Error: The ratio of the mean absolute error and true value is called relative error,i.e.,

relative error =t

t

Percentage Error:Relative error expressed in percentage is called percentage error i.e.

percentage error = 100 %t

t

○

○

○


Errors in Mathematical Operations(a) Addition or Subtraction:When two quantities are added or subtracted the absolute error in the result is given by

Z = A + B

(b) Multiplication or Division: When two quantities are multiplied or divided their relativeerror in the product or quotient is given by

z

z=

A

A

+

B

B

(c) Errors in Power:Suppose a physical quantity is given by

X =A B

C

p q

r

Then the relative error in X is given by

X

X

= p

A

A

+ q

B

B

+ r

C

C

or percentage error in X is

X

X

× 100 = p

A

A

× 100 + q

B100

B

+ r C

100C

SOLVED EXAMPLES

(a) Indirect Method for Long Distances and MeasurementExample 1: A laser light beamed at the moon takes measurement 2.56 sec to return

after reflection at the moon’s surface. How much is the radius of the lunar orbit around theearth?

(NCERT)Solution: Here t = 2.56 sec

c = 3 × 108 ms–1

Radius of lunar orbit around the earth

= Distance of moon from earth

=1

2c t

= 813 10 2.56

2

= 3.84 × 108mExample 2. When the planet Jupiter is at a distance of 824.7 million kilometres from

the earth, its angular diameter is measured to be 35.72 s of arc. Calculate the diameter ofJupiter. (NCERT)

Solution: Here distance between earth and Jupiter = 824.7 million km

S = 824.7 × 106 km

Angular diameter of Jupiter is = 35.72 s of arc

= 35.72

=o

35.72

60 60

= 35.72

3600 180

rad.


Diameter of Jupiter = S ×

= 824.7 × 106 × 35.72

3600 180

= 1.48217 × 105 km.Example 3. In a submarine equipped with a SONAR, the time delay between

generation of a probe wave and the reception of its echo after reflection from an enemysubmarine is found to be 77s. What is the distance of the enemy submarine? (Speed of soundin water = 1450 ms–1) (NCERT)

Solution: Here v = 1450 ms–1

t = 77 sec.

Distance of enemy submarine =1

2v t

=1

1450 772

= 55825

= 5.5825 × 104 mExample 4. The nearest star (Alpha Centauri) to our solar system is 4.29 LY away.

How much is the distance in terms of parsec? How much parallax would this star showwhen viewed from two locations of earth, six month apart in its orbit around the sun?

(NCERT)Solution: Here, Distance of Alpha centauri from earth

= 4.29LY

= 4.29 × 9.46 × 1015 m.

=15

16

4.29 9.46 10

3.08 10

Parsec

= 1.32 Parsec

Distance between two locations of the earth six month apart

= Diameter of earth

= 2 AU

Parallax of the star =Arc

Radius

=2AU

1.32 parsec

= 1.515 s of arc.Example 5. One mole of an ideal gas at S.T.P. occupies 22.4 lt. What is the ratio of

molar volume to the atomic volume of a mole of hydrogen? Why is this ratio so large (radiusof hydrogen molecule in 1A°). (NCERT)

Solution: Radius of hydrogen molecule = 1A°

r = (10–10 m)

Volume (of a mole of hydrogen) = 34N

3r

=4

3.143 (10–10)3 × 6.023 × 1023


= 25.2 × 10–7 m3

Molar volume = 22.4 × 10–3 m3 (1lt = 10–3 m3)

Molar Volume

Atomic Volume=

–3

–7

22.4×10

25.2×10= 104

The ratio has large value because the actual size of gas molecules is negligible incomparison of intermolecular separation.

Example 6. Express 1 parsec interms of metres.Solution: Parsec is the distance at which an arc of length 1 AU makes an angle of 1 second

of an arc i.e.

r =l

Here l = 1 AU = 1.496 × 1011m.

and = 1 sec of an arc.

=60×60×180

rad.

= 4.85 × 10–6 rad.

2 Parsec = 2r = 2 × 11

–6

1.496×10

4.85×10= 6.16 × 1016 m

Example 7. The radius of hydrogen atom is about 0.5A°. What is the total atomicvolume in m3 of a mole of hydrogen atom?

Solution: Here, r = 0.5A°

= 0.5 × 10–10 m.

Volume of 1 mole of hydrogen atom

= 34

3r

N

=4 22

3 7 × (0.5 × 10–10)3 × 6.023 × 1023

= 3.154 × 10–7 m3

Based on Measurement of Density Magnification and TimeExample 8. The photograph of a house occupies an area of 1.75 cm2 on a 35 mm slide.

The slide is projected on to a screen and the area of the house on the screen is 1.55 m2. Whatis the linear magnification of the projector screen arrangement?

Solution:Area of house on slide = 1.75 cm2

Area of house on screen (image) = 1.55 m2

= 1.55 × 104 cm2

Area magnification =size of images

size of object

=41.55×10

1.75


= 8857

Linear magnification = Areal magnification

= 8857= 94.1

Example 9. Estimate the average density of a sodium atom assuming its radius to beabout 2.5 A°. Compare it with the density of sodium in crystalline phase: 970 kgm–3. Are thetwo densities of same order of magnitude? (NCERT)

Solution: Here

r is radius of sodium atom = 2.5 A°

= 2.5 × 10–10 m.

Volume of sodium atom V = 34

3r

= 3–104 222.5 10

3 7

= 65.42 × 10–30m3

Average density of sodium atom

=mass of sodium atom

volume

=m

V =

M/N

V =

M

NVM = 23g (N is Avogadro’s No.)

Average density = 23 –30

23

6.023 10 65.42 10 = 0.58 × 103 kg m–3

Density of sodium in crystalline phase

= 970 kg m–3

= 0.970 × 103 kg m–3.Thus average density of sodium atom and density of sodium in crystalline phase both have

same order of magnitude i.e. (103).

Example 10. It is claimed that two cesium clocks allowed to run for 100 years free fromany disturbance may differ by only about 0.02 s. What does this imply for the accuracy ofthe standard cesium clock in measuring a time interval of 1.0sec.

Solution:Here t = 100 yrs = 100 × 365.25 days

= 100 × 365.25 × 86400 s

t = 0.02 s

Relative error =t

t

=0.02

100 365.25 86400 = 0.63 × 10–11.

Thus there is an accuracy of 10–11 in 1 sec. or 1 sec in 1011 sec


Example 11. Find the number of times the human heart beats in the life of 50 years ofa man. Given heart beats once in 0.8 sec.

Solution: Here

t = 50 yrs.

= 50 × 365 × 24 × 60 × 60 sec

Time taken by heart to beat once = 0.8 sec.

No. of times heart will beat in 50 yrs.

=50 365 24 60 60

.8

= 1.97×109

BASED ON DIMENSIONS AND DIMENSIONAL FORMULAEExample 12. Deduce the dimensional formulaes for (i) Specific heat (ii) Latent heat

(iii) Universal gas constant (iv) Mechanical equivalent of heat (v) Boltzmann’s constant.Solution: (i) Q = ms

s =Q

m =

2 –2��

M K

= [L2T–2 K–1]

(ii) Q = mL

L = 2 –2ML TQ

Mm

= [L2T–2]

(iii) Universal Gas Constant:

PV = RT

R =PV

T =

–1 –2 3ML T L

K

= [ML2T–2K–1]

(iv) Mechanical equivalent of heat:

W = JH

J =W

H

=2 –2

2 –2

ML T

ML T

= [MoLoTo]

(v) Boltzmann’s Constant:

K = [ML2T–2K–1].

Example 13. Deduce the dimensional formulae of the following physical quantities:(i) Charge (ii) Electric field (iii) Permittivity of free space (iv) Electric potential

(v) Capacitance.Solution: (i) Charge:

q = it

= [AT]

(ii) Electric field

E =F

q


= –2MLT

AT

= [MLT–3A–1]

(iii) F =1 2

20

1.

4

q q

r

or o =1 2

2

1.

4 F

q q

r =

2 2

–2

A T

[MLT ]

= [M–1L–3T4A2](iv) Electric potential

V =W

q =

2 –2ML T

AT

= [ML2T–3A–1]

(v) Capacitance

C =Q

V =

2 –3 –1

AT

ML T A

= [M–1 L–2 T4A2]Example 14: Name the physical quantities whose dimensional formulae are (i) [ML2T–1]

(ii) [ML2T–2] (iii) [MT–2] (iv) [ML2T–3] (v) [MoL2T–2]Solution: (i) [ML2T–1] = [MLT–1] [L]

= Linear momentum × radius (Length)= Angular momentum

(ii) [ML2T–2] = Work

(iii) [MT–2] = –2MLT

L

= Force

Length = Surface tension

(iv) [ML2T–3] = 2 –2ML T

T

=

Work

Time = Power

(v) [M0L2T–2] = 2 –2ML T

M

=

Heat

Mass = Latent heat

Example 15. Taking momentum, mass and length as fundamental quantities find thedimension of time.

Solution: Momentum = mass × velocity

= mass × distance

time

or time =mass × distance

momentum= [MLP–1]

Example 16. If the velocity of light c, gravitational constant G and Planck’s constant hbe chosen as fundamental units, find the value of a gram, a centimetre and a second in termsof new units of mass, length and time respectively, Given c = 3 × 1010 cms–1, G = 6.67 × 10–8

dyne cm2g–2 and h = 6.6 × 10–27 erg-sSolution: Dimensional formula for c = [LT–1] = 3×1010. ...(i)

Dimensional formula for G = [M–1L3T–2] = 6.67 × 10–8 ...(ii)Dimensional formula for h = [ML2T–1] = 6.6 × 10–27 ...(iii)


Multiplying equation (i) and (iii) and dividing by equation (ii)–1 2 –1

–1 3 –2

LT ML T

M L T

=10 –27

–8

3 10 6.6 10

6.67×10

or [M2] = 2.968 × 10–9

or M = 0.5448 × 10–4g

or 1 g = 1.8355 ×104 units of mass.Multiply eq. (ii) and (iii) and divide by cube of eq. (i)

–1 3 –2 2 –1

3 –3

M L T ML T

L T

=

–8 –27

310

6.67 10 6.6 10

3×10

or [L2] = 1.6304 × 10–65

[L] = 0.4038 × 10–32 cm.

or 1 cm = 2.4765 × 1032 unit of lengthNow from equation (i),

[T–1] = 103 10

L

or [T] =

10

L

3×10 = –32

10

0.4038×10

3×10= 0.1346 × 10–42s

or 1 s = 7.429 × 1042 units of timeExample 17: Show that angular momentum has same units as that of Planck’s

Constant.Solution: Angular momentum = moment of linear momentum

= mv × r

= [MLT–1] [L] = [ML2 T–1] ...(i)Now, E = h.

or h =E

v =

Energy

Frequency

dimensions of h =2 –2

–1

ML T

T

= [ML2 T–1] ...(ii)

From equation (i) and (ii) the dimensions of angular momentum and planck’s constant aresame so they have same units.

Example 18. If the unit of length is 1km, unit of force is k N and the unit of time is 10 s.Determine the unit of mass.

Solution: F = ma

m =F

a =

2FT

L[a = 2s

m]

Here, F = 1 kN, T = 10s, L = 1 km.

m =21000N×10×10s

1000m= 102 kg.


BASED ON CONVERSION OF UNITSExample 19: A calorie is a unit of heat energy and it equals about 4.2J where

1 J = 1kgm2s–2. Suppose, we employ a system of units in which the unit of mass equals kg, the unit of length equals m, the unit of time is sec. Show that a calorie has amagnitude 4.2 –1 –2 2 in terms of new unit. (NCERT)

Solution: Given 1 calorie = 4.2J = 4.2 kg m2 s–2.

n1u1 = n2u2

or n2 = n1 1

2

M

M

a

1

2

L

L

b

1

2

T

T

c

= 4.2

1Kg

Kg

2m

m

–2sec

sec

= 4.2 –1 –2 2.

Hence 4.2J = 4.2 –1 –2 2 new units of heat.Example 20: The speed of an automobile is 90 km h–1. By method of dimensions convert

it in ms–1.Solution: n1u1 = n2u2

or n2 = n1 1

2

M

M

a

1

2

L

L

b

1

2

T

T

c

Here n1 = 90 dim. of speed = [MoLT–1]

n2 = 90

1Km

m

–1

sec

hr

= 90 1000m

m

sec

60 60sec

=90 1000

60 60

= 25

90 kmh–1 = 25 ms–1.Example 21: Density of mercury is 13.6 gcm–3. Find its value in kg m–3.Solution: n1 = 13.6

Dimensions of density [ML–3]

n2 = n1 1

2

M

M

a

1

2

L

L

b

1

2

T

T

c

= 13.6

1gm

kg

–3cm

m

= 13.6 1000

g

g

–3

100

cm

cm


= 13.6 × 31100

1000

= 13.6 × 103

13.6 gm cm–3 = 13.6 × 103 kg m–3.Example 22. Convert a force of 1N in dyne.Solution: N (newton) is S.1 unit of force

dyne is mks unit of force

n2 = n1 1

2

M

M

a

1

2

L

L

b

1

2

T

T

c

n1 = 1 dimensions of force = [MLT–2]

n2 = 1

1Kg

g

1m

cm

–2sec

sec

= 11000g

g

100cm

cm

= 1 × 105

1N = 105dyne.Example 23. Find the value of 60 Joule mm–1 on a system which has 100 g, 100 cm, and

1min as fundamental units.Solution: Dimensions of power [ML2T–3]

60 J/min = 1 watt

Now n2 = n1 1

2

M

M

a

1

2

L

L

b

1

2

T

T

c

= 1

1K

100

g

g

2

100

m

cm

–3sec

60sec

= 1 × 1000

100

g

g

2100 C

100 C

m

m

(60)3

= 1 × 10 × 1 × 60 × 60 × 60

= 2.16 × 106

= 2.16 × 106 watt. (in new system of units).

Example 24. Find the units of mass, length and time. Given fundamental units arevelocity of light in air (3 × 108 ms–1) the acceleration due to gravity = 10ms–2. Density ofmercury = 13.6 × 103 kg m–3.

Solution:Velocity of light LT–1 = 3 × 108 ms–1 ... (i)

Acceleration due to gravity LT–2 = 10 ms–2 ...(ii)

Density of mercury ML–3 = 13.6 × 103kg m–3 ...(iii)


Divide equation (i) by (ii),

T = 3 × 107s ...(iv)

Multiplying equation (i) and (iv)

L = LT–1 × T

= 3 ×108 × 3 × 107

= 9 × 1015m (v)

Multiplying equation (iii) by cube of equation (v)

M = [ML–3] [L3]

= 13.6 × 103 × (9 × 1015)3

= 13.6 × 729 × 1048

= 9914.4 × 1048

= 9.9 × 1051 kg

BASED ON – TO CHECK THE CORRECTIONS OF PHYSICAL RELATIONSExample 25. Check the dimensional consistency of following equation.

v2 = u2 + 2aswhere symbols have their usual meaning.

Solution: v2 = u2 + 2as

Writing dimensions of both sides

[LT–1]2 = [LT–1]2 + [LT–2] [L]

or [L2 T–2] = [L2 T2] + [L2 T–2]Dimensions of R.H.S. = Dimensions of L.H.S.

So equation is consistent.

Example 26. By method of dimensions check the accuracy of the equation h = 2Tcos

rdg

where h = height to which liquid rises in a capillaryT = Surface tension of liquidr = radius of the capillary tubed = density of liquidg = acceleration due to gravity

Solution: h =2Tcos

rdg

Writing dimensions of both sides,

[L] =

–2

–3 –2

MT

L ML LT

cos has no dimensions

[L] = [L]

i.e., Dimensions of L.H.S. = Dimensions of R.H.S. so equation is dimensionally correct.Example 27. Check the consistency of following equation by method of dimensions

=2h

mv


where = wavelength associated with moving particleh = Planck’s constantm = mass of the particlev = velocity of the particle

Solution. =2h

mvWriting dimensions of both sides,

[L] =

22 –1

–1

ML T

M LT

= [L3 T–1]

Dimensions of L.H.S. dimensions of R.H.S.

So given equation is inconsistent.Example 28. What are the dimensions of k in the following Bernaulli’s equation.

2P 1

2gh v k

Solution:2P

2v

gh

= k

Dimensions of k,

Dim. of P

= dimensions of gh

= dimensions of v2

= [LT–1]2

= [L2 T–2]Example 29. The equation of S.H. Progressive wave is given by y = a sin (t – kx) where

t stands for time and x for distance. Find the dimensions of and k.Solution: The argument of sin has no dimensions

So dimensions of (t) = MoLoTo

or dimensions of =o o oM L T

T = [MoLoT–1]

Similarly, dimensions of k = [MoL–1To]Example 30. The distance covered by a particle in time t is given by

S = at + bt2

where a and b are two constants. Find dimensions of a and b.Solution. According to Principle of Homogeneity

dimensions of at = dimensions of S

or dimensions of a = L

T = [LT–1]

Similarly, dimensions of b = [LT–2]Example 31: A book with many printing errors contains four different formulae for the

displacement y of a particle undergoing a certain periodic motion.


(i) y = a sin 2 tT

(ii) y = a sin vt

(iii) y = Ta

sin ta

(iv) y = 2

a

2 t 2 tsin + cos

T T

a = maximum displacement of particlev = speed of particleT = time period of motion

Rule out the wrong formula on the dimensional grounds. (NCERT)Solution: Given equations will be correct if dimensions of R.H.S. is [L] and argument of

sin is dimensionless

(i) y = a sin 2

T

t

It is correct [dim. of a is [L] and argument of sine is dimensionless]

(ii) y = a sin vt

It is dimensionally incorrect[vt = LT–1.T = [L] Argument of sine is not dimensionless]

(iii) y =T

a sin

t

aIt is dimensionally incorrectDimensions of R.H.S. is not equal to [L]

Argument of sine is not dimensionless.

(iv) y =a

(sin

2 2cos

T T

t t )

It is dimensionally correct.Example 32. A famous equation in physics relates moving mass m to the rest mass m°

of a particle in terms of its speed v and speed of light c. A boy recalls the relation almostcorrectly but forgets where to put the constant c. He writes

m =

o1/221 –

m

v

Guess, where to put the missing c. (NCERT)Solution: According to principle of homogeneity, the dimensions of R.H.S. is [M] i.e

dimensions of L.H.S.

Therefore (1 – v2)1/2 should be dimensionless i.e. v2 should be divided by c2.

Hence correct relation is

m =m

v

o

221–

c


BASED ON DERIVING RELATION BETWEEN PHYSICAL QUANTITIESExample 33. Experiments reveal that the velocity v of water waves may depend on

their wavelength , density of water and acceleration due to gravity g. Establish a relationfor velocity of water waves.

Solution: Let v a b gc

or v = ka bgc (i)

where K is proportionality constant which is dimensionless. Substituting dimensions of variousquantities on both sides.

[LT–1] = [L]a [ML–3]b [LT–2]c

or [LT–1] = [L]a – 3b + c [M]b [T]–2c

Equating the powers of M and T on both sides

a – 3b + c = 1

b = 0–2c = –1

or on solving c =1

2, a = 1

2 , b = 0

Putting these values in equation (i).

v = k1/2 g 1/2

v = k gExample 34. The time of oscillation of a small drop of liquid under surface tension

depends upon the density , radius a and the surface tensions ‘S’. Using dimension analysisderive an expression for time period of oscillations.

Solution: Let t x ay Sz.

or t = kx ay Sz ...(i)

where k is dimensions constant.

Writing dimensions of various quantities on both sides

[T] = [ML–3]x [L]y [MT–2]z

= [M]x + z [L]–3x + y [T]–2z

Equating the powers of M,L and T on both sides

x + z = 0

–3x + y = 0

–2z = 1

On solving, x =1

2, y =

3

2, z = –1

2Putting these values in equation (i)

t = k 1/2 a3/2 S–1/2

or t 3

Sa

Example 35: The time period T of a simple pendulum may depend upon the mass of thebob ‘m’ length of the string ‘l’ and acceleration due to gravity ‘g’ By dimension methodderive an expression for the time period of simple pendulum.

Solution: Let t ma lb gc

or t = k ma lb gc ...(i)


k is dimensionless constant.

Writing dimensions of both sides,

[T] = [M]a [L]b [LT–2]

= [M]a [L]b+c [T]–2c

Equating dimensions of M, L and T on both sides,

a = 0

b + c = 0–2c = 1 or c = –1/2, b = +1/2

Putting these values in equation (i),

t = k mo l1/2 g–1/2

or t lg

Example 36. A gas bubble from an explosion under water oscillates with a period Tproportional to pa db Ec where P is static pressure d is density and E is total energy ofexplosion. Find out values of a, b and c.

Solution: Let T PadbEc

or T = kPa db Ec

k is a dimensionless constant.

Writing dimensions of all the physical quantities,

[T] = [ML–1 T–2]a [ML–3]b [ML2 T–2]c

= [M]a+b+c [T]–2a–2c [L]–a–3b+2c

Equating powers of M, L and T on both sides,

a + b + c = 0

–2a –2c = 1

–a –3b + 2c = 0

Solving these equations, we get

a =–56

, b = 12

, c = 13

Example 37. The centripetal force F, acting on a particle moving in circular pathdepends on the mass ‘m’ of the particle, radius of circular path r and angular velocity . Bymethod of dimensions derive the relation for centripetal force. Take proportionalityconstant to be 1.

Solution: Let F ra mb c

or F = k ra mb c ...(i)


Writing dimensions of all the physical quantities on both the sides,

[MLT–2] = [L]a [M]b [T–1]c

equating powers of M, L and T,

a = 1, b = 1, c = 2

Equation (i) becomes F = mr2 (k = 1 given)

Example 38. The wavelength associated with a moving particle is given by = mp vq

hr where m is mass, v is velocity and h is Planck’s constant. Calculate the values of p, q andr.


Solution: Let mp vq hr

or = kmp vq hr. ...i)

where k is dimensionless constant.

Writing dimensions of the physical quantities on both the sides.

[L] = [M]p [LT–1]q [ML2T–1]r

= [M]p+r [L]q+2r [T]–q–r

equating the powers of M, L and T,

p + r = 0 ...(ii)

q + 2r = 1 ...(iii)

–q – r = 0 ...(iiv)

Solving equation (ii), (iii) and (iv)

q = –1, p = –1, r = 1

Equation (1) becomes,

= k m–1 v–1 h1

or = k h

mv

orh

mv

.

Example 39. Assuming that the mass m of the largest stone that can be moved by aflowing river depends only on v the velocity, the density of water and g the accelerationdue to gravity. Show that m varies with the sixth power of the velocity of flow.

Solution: Let m va b gc

or m = k va b gc


Writing dimensions of both the sides

[M] = [LT–1]a [ML–3]b [LT–2]c

= [M]b [L]+a–3b+c [T]–a–2c

Equating the powers of M L and T on both sides

b = 1

a –3b + c = 0

–a – 2c = 0

On solving these equations,

a = 6, b = 1, c = –3

m v6 g–3

or m v6

i.e. mass m varies with the sixth power of the velocity of flow.

Example 40. Reynold number NR (dimensionless quantity) determines the conditions oflaminar flow of a viscous liquid through a pipe. NR is a function of the density of the liquid, its average speed v, and the coefficient of viscosity of the liquid . Given that NR is alsodirectly proportional to d (the diameter of the pipe.) show that

NR vd


Solution: Let NR a vb c

also NR d

NR a vb c d

or NR = ka vb c d ...(i)


Writing dimensions of all the physical quantities

[MoLoTo] = [ML–3]a [LT–1]b[ML–1T–1]c [L]

= [M]a+c [L]–3a+b–c+1 [T]–b–c

Equating the powers of M L and T on both sides

a + c = 0

–3a + b – c + 1 = 0

– b – c = 0

Solving these equations, we geta = 1, b = 1, c = – 1.

Equation (i) becomes,NR = k 1 v1 –1 d

or NR = k vd

or NR vd

BASED ON SIGNIFICANT FIGURESExample 41. State the number of significant figures in the following.(i) 0.04 m2, (ii) 2.64 × 1024 kg, (iii) 0.2370 cm–3, (iv) 6.320 J, (v) 6.032 Nm–2,

(vi) 0.0006032, (vii) 2.000 m, (viii) 5100 kg, (ix) 0.050 cm. (NCERT)Solution: (i) one –4 (ii) Three – 2, 6, 4 (iii) Four – 2, 3, 7, 0

(iv) Four – 6, 3, 2, 0 (v) Four – 6, 0, 3, 2 (vi) Four – 6, 0, 3, 2

(vii) Four – 2, 0, 0, 0 (viii) Four – 5, 1, 0, 0 (ix) Two – 5, 0

Example 42. State the number of significant figures in the following.(i) 200.05, (ii) .0047, (iii) 43200, (iv) 4.00 m, (v) 122.00, (vi) 2.72 × 10–2

Solution: (i) Five, 2, 0, 0, 0, 5

(ii) Two, 4, 7

(iii) Three, 4, 3, 2(iv) Three, 4, 0, 0

(v) Five 1, 2, 2, 0, 0

(vi) Three, 2, 7, 2

Example 43. Solve the following to correct significant figures:(i) 9.65 + 15.237

(ii) 7.24 + 11.141 + 0.0025(iii) 7.987 – 5.45(iv) 29.412 – 29.4(v) Subtract 2.6 × 104 from 3.8 × 105


Solution: (i) 9.65

+15.237

24.887

= 24.89 (Rounded off upto second decimal place.)

(ii) 7.24

11.141

0.0025

18.3835

= 18.38 (Rounded off upto second decimal place)

(iii) 7.987

– 5.45

2.537

= 2.54 (Rounded off upto second decimal place)

(iv) 29.412

– 29.4

0.012

= 0.0 (Rounded off upto one decimal place)

(v) 3.8 × 105 – 2.6 × 104 = 3.8 × 105 – 0.26 × 105

= 3.54 × 105

= 3.5 × 105 (Rounded off upto first decimal place)

Example 44. Solve the following to correct significant figures.(i) 4.372 × 11 (ii) 6.32 × 7.4695Solution. (i) 4.372 × 11 = 48.092

= 48 (The product must contain two significant figures)

(ii) 6.32 × 7.4695 = 47.20724

= 47.2(The product must have least number of significantfigures, in any one of the factors which is three)

Example 45. Solve the following to correct significant figures.

(i) 210 5.5 (ii)–3 65.02×10 ×1.81×10.8926

Solution: (i) 210 5.5 =210

5.5 = 38.18

= 38 (Rounded off to two significant figures)

(ii)–3 65.02 10 1.81 10

.8926

= 10.179 × 103

= 10.2 × 103 (Rounded off to three significant figures)

Example 46. The length , breadth and thickness of a rectangular sheet of metal are4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correctsignificant figures. (NCERT)

Solution: Here l = 4.234 m

b = 1.005 m


h = 2.01 cm

= 0.0201 m.

Total area S = 2 (lb + bh + lh)

= 2(4.234 × 1.005 + 1.005 × 0.0201 + 0.0201 × 4.234)

= 8.7209478

= 8.72 m2. (Rounded off upto three significant figures)

Volume V = l. b. h.

= 4.234 × 1.005 × 0.0201

= 0.08552m2

= 0.0855 m2 (Rounded off upto three significant figures)

Example 47. The mass of a box measured by a grocer’s balance is 2.3 kg. Two goldpieces of masses 20.15 g and 20.17 g are added to the box. What is (a) total mass of the boxand (b) the difference in the masses of the pieces to correct significant figures.

Solution: Here m1 = 2.3 kg.

m2 = 20.15g = 0.02015 kg.

m3 = 20.17g = 0.02017 kg.

Total mass of the box = m1 + m2 + m3

= 2.3 + 0.02015 + 0.02017

= 2.34032 kg

= 2.3 kg(b) Difference in masses of two pieces

= m3 – m2

= 20.17 – 20.15= 0.02 g

Example 48. Calculate the area of a circle in appropriate number of significant figuresif the radius of the circle is 0.45 m.

Solution: Here r = 0.45 m

Area of circle = r2

= 3.14 × (.45)2

= 0.63585

= 0.64 m2

Example 49. The mass of a body is 242.7 g and its volume is 24.2 cm3. Calculate thedensity of the substance upto appropriate significant figures.

Solution: Here m = 242.7 g

v = 24.2 cm3

Density =mass

volume =

V

m

=242.7

24.2= 10.0289 g cm–3

= 10.0 g cm–3


Example 50. The radius of a sphere is 12.4 cm. Calculate its volume, with knowledgeof significant digits.

Solution: Here

r = 12.4 cm

Volume =4

3r3

=4

3.143 (12.4)3

= 7892.399 cm3

= 7.892399 × 103 cm3.

= 7.89 × 103 cm3.Example 51. The radius and length of a thin wire are 0.25 mm and 2.25 m, respectively,

Find the volume of the wire upto appropriate number of significant figures.Solution: Here r = 0.25 mm = 0.025 cm

l = 2.25 m = 225 cm.

Volume of wire = r2l

= 3.14 (0.025)2 × 225

= 0.4415625 cm3

= 0.442 cm3

Example 52. The side of a cube is 7.2 m. Calculate its surface area, by takingsignificant figures in to consideration.

Solution: Here l = 7.2 m

Surface area of a cube = 6 l2

= 6 × (7.2)2

= 311.04 m3

= 3.1104 × 102 m3

= 3.1 × 102 m3.

BASED ON ERRORS IN MEASUREMENTSExample 53: In an experiment to measure the time period of a simple pendulum, the

time period was found to be 2.54s, 2.67s, 2.72s, 2.63s, and 2.60s respectively. Determine.(a) True value of the time period(b) Mean absolute error(c) Relative error(d) Percentage error.Solution: Here t1 = 2.54s, t2 = 2.67 s, t3 = 2.72s, t4 = 2.63s, t5 = 2.60s(a) True value of the time period

= 1 2 3 4 5

5t t t t t

= 2.54 2.67 2.72 2.63 2.60

5

=13.16

5 = 2.63s


(b) Absolute errors in different measurements

T1 = | 2.62 – 2.54 | = 0.08s

T2 = | 2.62 – 2.67 | = 0.05s

T3 = | 2.62 – 2.72 | = 0.10s

T4 = | 2.62 – 2.63 | = 0.01s

T5 = | 2.62 – 2.60 | = 0.02s

Mean absolute error = T = 0.08 0.05 0.10 0.01 0.02

5

=0.26

5= 0.05 s

(c) Relative error =T

T

=0.05

2.63= 0.02

(d) Percentage error = 0.02 × 100 = 2%Example 54: The refractive index of water as measured by an observer was found to

have the values 1.29, 1.35, 1.32, 1.33, 1.36, 1.30, 1.33, 1.34 respectively. Find the true valueof refractive index, mean value of the absolute error, the relative error and the percentageerror?

Solution: Here 1 = 1.29, 2 = 1.35, 3 = 1.32, 4 = 1.33, 5 = 1.36, 6 = 1.30, 7 = 1.33,8 = 1.34

Mean value of refractive index,

=8

=8

=10.62

8= 1.3275

= 1.33(b) Absolute errors in different measurements

1 = |– 1| = |1.33 – 1.29|= 0.04

2 = |1.33 – 1.35| = 0.02

3 = |1.33 – 1.32| = 0.01

4 = |1.33 – 1.33| = 0.005 = |1.33 – 1.36| = 0.03

6 = |1.33 – 1.30| = 0.03

7 = |1.33 – 1.33| = 0.00

8 = |1.33 – 1.34| = 0.01

=..............

8


=0.04 0.02 0.01 0.00 0.03 0.03 0.0 0.01

8

=0.14

8= 0.02

(c) Relative error =Mean absolute error

True value

Relative error =0.02

1.33= 0.015

(d) Percentage error = 0.015 × 100

= 1.5%Example 55: Two resistances

R1 = 100 3R2 = 200 4

are connected in series. What is their equivalent resistance? (NCERT)Solution: Here R1 = 100 3

R2 = 200 4In series equivalent resistance is

R = R1 + R2

= (100 3) + (200 4)

= (300 ± 7)

Example 56: The resistance of a conductor is given by R = VI

, where V = (200 ± 5)V

and I = (20 0.4)A. Calculate total error in R.

Solution: Given R =V

I

orR

R

=

V

V

+

I

I

Percentage error in R is R

100R

= V

100V

+ I

100I

R100

R

=5

100200

+

0.4100

20

= (2.5 + 2)% = 4.5%.Example 57. Two different masses are given by

m1 = (24.6 ± 0.5) gm2 = (16.7 ± 0.3)g

Calculate (i) m1 + m2, (ii) m1 – m2.Solution: (a) m1 + m2 = (24.6 ± 0.5) + (16.7 ± 0.3)

= (24.6 + 16.7) ± (0.5 + 0.3)

= 41.3 ± 0.8g(b) m1 – m2 = (24.6 ± 0.5) – (16.7 ± 0.3)


= (24.6 – 16.7) ± (0.5 + 0.3)

= 7.9 ± 0.8 gExample 58. A Capacitor of the capacitance C = (2.0 ± 0.1)μF is charged to a voltage

V = (20 ± 0.2) V. What is the charge on the capacitor.Solution: Here C = (2.0 ± 0.1) F

V = (20 ± 0.2)V

Q = CV = (2.0 ×10–6) × 20

= 40.0 ×10–6 C.

NowQ

Q

=

C V

C V

or Q =0.1 0.2

2.0 20

× 40 × 10–6

= (0.05 + 0.01) × 40 × 10–6

= 0.06 × 40 × 10–6

= 2.4 × 10–6 C

Charge on the Capacitor Q = (40.0 ± 2.4) 10–6 CExample 59. The side of a cube is measured to be (5.5 ± 0.1) cm. Find the volume of

the cube.Solution: Here side of a cube = (5.5 ± 0.1) cm.

Volume = a3

= (5.5)3

= 166.375

= 166.4 cm3

V = a3

V

V

= 3

a

a

or V = 3 Va

a

= 3 × 0.1

166.45.5

= 9.076 = 9.1 cm3.

Volume of cube is V = (166.4 ± 9.1) cm3

Example 60. The mass of a body is (250.0 ± 0.5) g and its volume is (25.0 ± 0.2) cm3.Calculate the density of the body.

Solution: Here m = (250.0 ± 0.5) g

v = (25.0 ± 0.2) cm3

density d =V

m

=250.0

25.0= 10.0 g cm–3.


Now d =V

m

d

d

=

V

V

m

m

or10

d=

0.5 0.2

250 25

or d =5 2

250 25 = 0.02 + 0.08

= 0.10

Density = (100 ± 0.1) g cm–3.Example 61. The error in the measurement of radius of a sphere is 2%. What would

be the error in the calculation of surface area of the sphere?

Solution: Here 100r

r

= 2%

Surface area of the sphere s = 4r2

Percentage error in the surface area

S100

S

= 2 100

r

r

= 2 × 2%

= 4%.Example 62. The distance covered by a body in time (5.0 ± 0.6)s is (40.0 ± 0.4)m.

Calculate the speed of the body. Also determine the percentage error in the speed.Solution: Here S = (40.0 ± 0.4) m

t = (5.0 ± 0.6)s.

Speed is given by v =s

t =

40.0

5.0= 8.0 ms–1.

v =s

t

v

v

=

s

s

+

t

t

or8.0

v=

0.4 0.6

40 5.0

or v =0.4 0.6

8.040 5.0

= 1.04

Speed v = (8.0 ± 1.04) ms–1.

% error = 100v

v

=1.04

1008

= 13%


Example 63. The length and breadth of a field are measured asl = (100 ± 2)mb = (80 ± 1)m

What is the area of the field.Solution: Here l = (100 ± 2)m

b = (80 ± 1)m

Area of the field A = l.b

A = 100 × 80

= 8000 m2.

A = l.b

A

A

=

l

l

+

b

b

orA

8000

=

2

100 +

1

80

or A =2

8000100

+ 1

80 × 8000

= 160 + 100

= 260m2

Area of the field = A ± A

= (8000 + 260) m2

= (8.0 + 0.26) × 103 m2

Example 64. A physical quantity P is related to four variables a, b c and d as follows

P = a3 b2/ c .d

The percentage error in the measurement in a, b, c and d are 1%, 3%, 4% and 2%respectively. What is the percentage error in the quantity P? (NCERT)

Solution: Given P =3 2

.

a b

c d

P

P

= 3

a

a

+ 2

b

b

+

1

2

c

c

+

d

d

orP

100P

= 3 100

a

a

+ 2 100

b

b

+

1

2 100

c

c

+

100d

d

= 3(1%) + 2 (3%) + 1

2 (4%) + (2%)

= 13%.

PROBLEMS

(a) Indirect method for long distance and small distance Measurements1. How many astronomical units are there in 10 m. [Ans. 6.68 × 10–11 AU]

2. The Parallax of a heavenly body measured from two points diameterically opposite onearth’s equator is 60s. If the radius of the earth is 6.4 × 106m. Determine the distance of theheavenly body from the centre of the earth. Give answer in astronomical units.


[Ans. 0.293 AU]

3. A radar signal is beamed towards a planet from earth and its echo is received sevenminutes later. The distance between the planet and earth is 6.3 ×1010m. Calculate the speedof the signal. [Ans. 3 × 108 ms–1]

4. The angle of elevation of the top of a hill at a certain place was 60°. On moving 30 m awayfrom the hill, the angle of elevation becomes 30°. Find the height of the hill. [Ans. 26 m]

5. A star is located 12 light year away from us. Calculate this distance in Parsec. What is theParallax shown by this star when viewed from two locations 3 × 1010m apart?

Given 1 Parsec = 3.084 × 1016m

1 light year = 9.46 ×1015m. [Ans. 3.681 Parsec]

6. Calculate the distance of a quasar from which light takes 2 billion years to reach the earth.

[Ans. 1.89 × 1022 km.]

7. Calculate the distance of sun from earth if light takes 8 min to reach earth from the surfaceof sun. [Ans. 1.44 × 1011m].

8. A rock under water is 1595m deep. Find the time in which an ultrasonic signal returnsafter reflection from the rock. Given speed of ultrasonic wave in water = 1450 ms–1.

[Ans. 2.2 sec]

BASED ON MEASUREMENT OF DENSITY MAGNIFICATION AND TIME9. Two atomic clocks allowed to run for average life of an Indian (say 70 years) differ by 0.2s

only. Calculate the accuracy of standard atomic clock in measuring a time interval of 1 sec.

[Ans. accuracy of atomic clock is 10–10s in 1 sec.]

10. The radius of hydrogen atom is about 0.5A°. What is the total atomic volume in m3 of agram molecule of hydrogen atom. [Ans. 3.2 × 10–7 m3]

11. The nucleus is not a perfect sphere. It is slightly deformed. The radius of the nucleus istherefore given by

r = r0 A1/3

where A is mass number and r0 is constant having value r0 = 1.2 fm. Considering mass ofeach nucleon to be 1.67 ×10–27 kg, Calculate the density of nucleus.

[Ans. 2.3 × 1017 kg m–3]

12. The thickness of hair is measured using a microscope of magnification 100. The averagewidth of the hair is found to be 3.5 mm. Find the thickness of the hair. [Ans. 3.5 ×10–

2mm]

13. A neutron star has a density equal to that of nuclear matter = 2.8 × 1017 kg m–3. Assumethe star to be spherical find the radius of the neutron star whose mass is 4.0 × 1010 kg.

[Ans. 15 km]

14. Find the number of times, a human heart will beat in the life of 60 years of a man. Givenheart beats at the rate of 5 beats in 4 sec. [Ans. 2.37 × 109]

15. Age of universe is about 1010 years where as the mankind has existed for 106 years for howmany days would the man have existed if age of universe is 1 month (30 days)

[Ans. 3 × 10–3 days]

BASED ON DIMENSIONS AND DIMENSIONAL FORMULA16. The energy is expressed as E = Kv, where v is velocity. What is the dimensional formula

for k.[Ans. MLT–1]


17. If force F, acceleration A and time T are taken as fundamental physical quantities, then findout the dimensions of length. [Ans. MoLoAT2]

18. Deduce dimensional formulae for (i) viscosity (ii) Heat (iii) Impulse (iv) Angularmomentum (v) Torque.[Ans. (i) ML–1T–1, (ii) ML2T–2, (iii) ML1T–1, (iv) ML2T–1, (v) ML2T–

2]

19. L represents the self inductance of a coil and R is its resistance. Find out the dimensions of(L/R) [Ans. MoLoT]

20. C and R represents respectively the capacitance and resistance, then determine thedimensions of CR [Ans. MoLoTo]

21. What is the ratio of the dimensions of angular momentum and linear momentum.

[Ans. MoLTo]

22. If the velocity of light C, acceleration due to gravity g and atmospheric pressure P are thefundamental units. Find the dimensions of length. [Ans. C2g–1]

23. The number of particles crossing a unit area perpendicular to x axis in unit time is given by

n = –D 2 1

2 1

–

–

n n

x x

where n1 and n2 are number of particles per unit volume for the values of x, meant to x1 andx2. Find the dimensions of diffusion constant D. [Ans. MoL2T–1]

24. Show that dimensions of 0

1

is same as that of speed. o is called permiability of free

space and o is called permittivity of free space.

25. Which of the following groups have different dimensions?

(a) Potential difference, emf, voltage(b) Pressure, stress, young’s modulus.

(c) Heat, energy, torque

(d) Dipole moment, electric flux and electric field. [Ans. d]

26. Show that dimensions of resistance are same as that of h/e2.

27. Calculate the dimensions of 2

o

1.e

hc

symbols have their usual meaning. [Ans. MoAoLoTo]

28. The speed of light c, acceleration due to gravity ‘g’, and pressure P are taken asfundamental units. Find out the dimensions of gravitational constant G. [Ans. Cog2p–1]

29. A quantity X is defined by the equations

X = 3C B2

Where C is capacitance in farad and B represents magnetic field in tesla. Find out thedimensions of X. [Ans. ML–2]

BASED ON DIMENSIONAL CONSISTENCY OF PHYSICAL RELATION30. Check the dimensional correctness of the relation

= Iwhere = torque and I = moment of inertia, = angular acceleration [Ans. Correct]

31. Test by using dimensional analysis the accuracy of the physical relation

t =3

Tk

r


= density of liquid

r = radius of drop.

T = surface tension of liquid

t = time period of oscillation of drop. [Ans. Correct]

32. Check the dimensional correctness of the following physical relation.

v2 – u2 = 2a2s

where symbols have their usual meaning. [Ans. Wrong]

33. Using the principle of homogeneity of dimensions check the accuracy of followingequation.

W = mgh + 1

2 mv2

Symbols have their usual meaning. [Ans. Correct]

34. Determine the dimensions of a

b in the equation

P =2–a t

bxwhere P is pressure, x is distance and t is time.[Ans. MT–2]

35. In the relation, P =

e–z/k

P is pressure, z is distance, k is Boltzmann constant and is the temperature. Determinethe dimensional formula for . [Ans. MoL2To]

36. The velocity of a particle at time t is given by

v = at + +

b

t cFind the dimensions of constant a, b and c.[Ans. LT–2, L, T]

37. A force F is given by F = at + bt2 where t is time, what are the dimensions of a and b.

[Ans. MLT–3, MLT–4]

38. Applying the principle of homogeneity of dimensions, check the dimensional accuracy ofthe relation.

E =1

2 I2

where E = Rotational energy

I = moment of inertia

= angular velocity. [Ans. Correct]

39. The velocity of transverse wave on a stretched string is given by

v =2

2

4MgD

M = Mass producing tension in the string

g = acceleration due to gravity

D = diameter of string and


= density of the material of the string.

Is the physical relation dimensionally correct. [Ans. No]

40. By method of dimension show that the speed of longitudinal wave in a gaseous medium,given by

v =P

d

= sp. heat ratioP = Pressure of gasd = density of gas is correct

Check the consistency of given equation [Ans. Correct]

41. The volume of liquid flowing per second, through a capillary tube is given by

Q =4P

8r

l

P = (P1 – P2) = Pressure difference between the two ends of capillary

r = radius of capillary tube

= viscosity of liquid.

l = length of capillary tube

Check the consistency of given physical relation by the method of dimensions.

[Ans. Correct/Consistent]

42. De Broglie wavelength associated with an electron, having kinetic energy k is given by

=22m k

h

h = Planck’s constant

m = mass of electron.

Using method of dimensions, check whether the physical relation is correct or not.

[Ans. No, it is incorrect]

43. The viscous force acting in between two layers of liquid is given by

F = ± A v

x

= Coefficient of viscosityA = Area of two layers in contact

v

x

= velocity gradient.

Using dimensional analysis check the consistency of above physical relation.

[Ans. Consistent]

44. Using dimensional analysis check the consistency of following physical relation.

h =2T Cos

rdg

h = height of liquid column in the capillaryr = radius of capillaryd = density of liquidg = acceleration due to gravity

= angle of contact. [Ans. Consistent]


45. The centripetal force acting on a body of mass ‘m’ moving in a circular path of radius ‘r’with constant angular velocity ‘’ is given by

F = mr2Using method of dimension check whether the equation is correct or not.

[Ans. inconsistent]

46. The escape velocity of a body from the surface of earth is given by

Ve =2GM

RG = Universal gravitational constant

M = Mass of earth

R = Radius of earth.

Is the physical relation dimensionally correct. [Ans. Consistent]

47. The time period of revolution of a satellite around the earth is given by

T = 3

22 R+

GM

h

R = Radius of earth

h = Height of satellite above earth

G = Universal Gravitational Constant

M = Mass of earth.

Check the consistency of given physical relation by method of dimension.

[Ans. Correct]

48. The rate of flow of heat, through a metallic rod is given by

Q = 1 2KA –

d

K = Coefficient of thermal conductivity

A = Area of cross section of the rod.

(1 – 2) = Temperative difference between the two ends of rod.

d = length of rod.

Use method of dimensions to check the consistency of given physical relation.

[Ans. Correct]

BASED ON DERIVING RELATION BETWEEN PHYSICAL QUANTITIES49. The velocity v, of transverse waves on a string depends on the tension in the string ‘T’ and

its mass per unit length ‘m’. By method of dimension derive an expression for speed of

transverse wave. [Ans. V × Tm ]

50. Show by method of dimensions that the distance S covered by a body, in time t

S =1

2 a2t

is wrong. Derive the correct physical relation. [Ans. S = 1/2 at2]

51. A body of mass ‘m’ is revolving in a circular path of radius r with constant angular velocity‘’. Show that centripetal force acting on it is given by

F mr2


52. The kinetic energy k of a moving body depends on the mass m of body and its velocity v. Bymethod of dimensions derive an expression for K.E. Given proportionality constant is 1/2.

[Ans. K = 1

2 mv2]

53. The wavelength associated with a moving electron depends upon its mass m, its velocity

v and Planck’s constant h. Prove dimensionally that h

mv.

54. The velocity of sound v in a gas depends on its elasticity ‘E’ and density ‘d’. By method of

dimensions derive a relation for velocity of sound. [Ans. v Ed ]

55. Assuming that the escape velocity for a planet depends on gravitational constant G, radiusR of planet and its density . Derive formula for escape velocity from dimensional

knowledge. [Ans. ve R G ]

56. Assuming that the critical velocity vc of a viscous liquid flowing through a capillary tubedepends upon the radius a of the tube, density and coefficient of viscosity of the liquid.Obtain a relation for critical velocity. [Ans. vc /a]

57. The volume of the liquid flowing per second Q through a tube depends upon (i) coefficientof viscosity (ii) radius of the tube r (iii) the pressure gradient p/l. Deduce by method of

dimensions the formula for the volume of liquid flowing per second. [Ans. Q 4Pr

l]

58. The rotational K.E. ‘E’. of a body rotating about an axis depends on its moment of inertiaI and angular velocity . show that E I2.

59. The period of vibration t of a tunning fork depends upon the length l of the prongs, density‘’ and young’s modulus Y of the material. By method of dimension derive a relation for

time period. [Ans. t l Y

]

60. By method of dimension prove that acceleration due to gravity ‘g’ inside the surface ofearth is given by

g =4

3 GR

where symbols have their usual meaning.61. The frequency of vibration of a mass m suspended from a spring of spring constant k is

given by the relation = cmx ky

where c is a dimensionless constant. Calculate the value of x and y. [Ans. 12

1

2]

62. The orbital velocity Vo of a satellite revolving around earth depends on the mass of earth‘M’, radius of earth R, and gravitational constant ‘G’. By method of dimensions, derive a

relation for orbital velocity. [Ans. Vo GM

R]

BASED ON CONVERSION OF UNITS63. How many dyne are there in one newton? Use the method of dimension analysis. [Ans.

105]

64. Find the value of 90 km/h in a system based on kilogram, metre and second. [Ans. 25ms–1]


65. Convert 5 joule in erg with the help of dimensional analysis. [Ans. 5 × 107eg]

66. The density of mercury is 13.6 g cm–3. Convert its value into kgm–3. Using dimensionalanalysis. [Ans. 13.6 × 103kgm–3]

67. Using dimensional method calculate the value of G in S1 units. Given the value of G inCGS system is 6.67 × 10–8 dyne cm2 g–2. [Ans. 6.67 ×10–11 Nm2kg–2]

68. A new unit of length is chosen such that the speed of light in vacuum is unity. What is thedistance between sun and earth in terms of new unit if light takes 8 min and 20s to coverthis distance. [Ans. 500 new units]

69. Momentum P area A and time T are assumed to be fundamental quantities. What will bethe dimensional formula for energy? [Ans. P1T–1 A1/2]

70. The surface tensions of water is 72 dyne cm–1. Express it in SI units. [Ans. 0.072 Nm–1]

71. A body has uniform acceleration of 10 kmh–2. Using dimensional method calculate itsvalue in CGS units. [Ans. 0.0752 cms–2]

72. Find the value of force of 200 dyne on a system based on metre, kilogram, and minute asfundamental units. [Ans. 7.2 units]

73. Using dimensional knowledge convert 5.0 × 105 Nm–2 in dyne cm–2.

[Ans. 5.0 × 106 dyne-cm–2]

74. Using dimensional analysis calculate the value of Planck’s constant in C.G.S. unit. Givenh = 6.63 × 10–34 JS. [Ans. 6.63 × 10–27 erg sec]

MORE PROBLEMS ON DIMENSIONS75. The velocity of a particle depends upon the time t according to the equation

v = a + bt + c

d +tWrite the dimensions of a, b, c and d. [Ans. LT–1, LT–2, L and T]

76. Given that C = KL + MI

where L = angular momentum

I = moment of Inertia

= angular speed

Calculate the dimensional formula for K × M [Ans. T–4]

77. Find the dimensions of a

b in the following equation.

P =2a – t

bxwhere P is pressure, x is distance and t is time.[Ans. MT–2]

78. Bernoulli’s equation is given by

P + 2

v2 + gh = K.

What are the dimensions of K. [Ans. ML–1 T–2]

79. Given that P =RT

–v b ×

– V

RTe

Find out the dimensions of .


80. Progressive wave is represented by

M = A sin (t – kx)

where x is distance and t is time. Find the dimensions of and k. [Ans. M°L° T–1, M°L–

1T°]

81. The position of a particle travelling along x axis depends on time t according to relation

x = at – bt2

Find the units and dimensions of constant a and b [Ans. ms–1, ms–2, LT–1, LT–2]

82. When white light travels through the glass the refractive index ‘’ is found to vary with thewavelength according to relation:

= A + 2

B

where A and B are constants. Using the principle of homogeneity determine the dimensionsand SI units of A and B. [Ans. A = dimension [MoLoTo] Unit less, B: [MoL2To] Unit-m2]

BASED ON SIGNIFICANT FIGURES83. State the number of significant figures in the following measurements:

(i) 0.009 cm2, (ii) 0.00125, (iii) 86400s, (iv) 846.00, (v) 0.2450 cm–3, (vi) 20.10

[Ans. (i) 1 (ii) 3 (iii) 5 (iv) 5 (v) 4 (vi) 4]

84. Write the order of magnitude of:

(i) 765.4 g, (ii) 0.005 km., (iii) 327.6 m. [Ans. (i) 103 (ii) 10–2 (iii) 102]

85. Add 2.54 × 10–4 kg and 34.6 × 10–3 kg. [Ans. 34.8 × 10–3 kg]

Subtract 25.76 from 78.3. [Ans. 52.5]

86. The length and breadth of a card board are 8.7 m and 2.32 m. Find the area of card board.

[Ans. 20m2]

87. Each side of a cube is 7.203 m. Calculate the volume of the cube to appropriate significantfigures. [Ans. 373.7 m3]

88. Solve –2 4236 10 68 10

189.7

upto three significant figures. [Ans. 5.83 × 102]

89. Subtract 4.2 × 10–6 from 5.2 × 10–4 keeping in mind significant figures. [Ans. 5.2 × 10–4]

90. The diameter of a sphere is 3.14 m. Calculate its surface area with due regards tosignificant figures. [Ans. 31.0 m2]

91. A thin wire has a length of 22.4 cm and radius 0.50 mm. Calculate the volume of the wireto correct significant figures. [Ans. 0.18 m3]

92. A block of mass 5.74 g occupies a volume 1.4 cm3. What is the density to appropriatesignificant figures. [Ans. 4.1 g cm–3]

93. The diameter of a circle is 2.42 m Calculate its area in proper significant figures.

[Ans. 4.60 m2]

94. Calculate the surface area of a cube of side 2.42 m, with due regards to significant figures.

[Ans. 35.1 m2]

BASED ON ERRORS95. The time period of oscillation of a simple pendulum in an experiment is recorded as 2.63s,

2.56s, 2.58s, 2.45s, and 2.50s respectively. Find the time period, mean absolute error,relative error and percentage error. [Ans. 2.54 s, 0.06s, 0.02, 2.4%]


96. The refractive index of glass is found to have the values 1.35, 1.45, 1.50, 1.47, 1.45, 1.521.44 and 1.51. Calculate the true value, absolute error, the relative error and the percentageerror. [Ans. 0.04, 0.03 and 2.7%]

97. The length of two rods are (2.02 0.3)m and (1.07 0.1)m. What is the total length?

[Ans. (3.09 0.4) m ]

98. Two different masses are m1 = (27.5 0.2)g, m2 = (15.3 0.3)g. Calculate the differencein two masses. [Ans. (12.2 0.5) g]

99. The initial and final temperature of a body are (15 0.2)°C and (42 0.4)°C What is therise in the temperature of the body. [Ans. (27 0.6)°C]

100. Two resistances R1 = (42 1) and R2 = (200 ± 4) are connected in series. Calculatetheir equivalent resistance. [Ans. (242 5) ]

101. A potential difference V = (80 2)V is applied across a conductor. The current in theconductor is (10 0.5)A. Calculate the error in the measurement of resistance ofconductor.

[Ans. 6%]

102. The capacitance of a capacitor is C = (5.0 0.1) F which is charged to a potentialV = (10 1) V. Calculate the charge on the plates of capacitor. [Ans. (50 6) × 10–6 C]

103. The length and breadth of a field are measured as l = (120 2)m, b = (100 5)m. Whatis the area of the field? [Ans. (1.2 0.08) × 104 m2]

104. The mass of a body is m = (275.32 0.01)g and it occupies a volume V = (36.41 0.01)cm3. Find the percentage error in the measurement of density. [Ans. 0.03%]

105. The side of a cube is (10.0 0.1)m. Calculate its volume. [Ans. (1000 30)m3]

106. The side of a cube is measured as (10.5 0.1) cm. Find the volume of the cube.

[Ans. (11.58 0.33) × 102 m3]

107. A quantity is measured by X = MaLbTc. The percentage error in the measurement of M, Land T are %, % and %, respectively calculate the percentage error in X.

[Ans. (a + b + c%)]

108. The radius of the sphere is (5.3 0.1)cm. Calculate the percentage error in the volume ofthe sphere. [Ans. 5.66%]

109. The density of the sphere is measured by measuring the mass and diameter. If it is knownthat the maximum percentage error in the measurement of radius and mass are 3% and 2%then what is the maximum percentage error in the measurement of density. [Ans. 11%]

110. A physical quantity P is given by P = 2 3a b

c d.

The error in the measurement of a, b, c and d are 2%, 4%, 2% and 1% respectively.Calculate the percentage error in the value of P. [Ans. 18.5%]

111. The time period of oscillation of a simple pendulum is T = 2l

g, l is 10 cm and is known

to 1 mm accuracy. The time period of oscillation is about 0.5 sec. The time of 100oscillation is measured with an accuracy of 1s. Calculate the accuracy in the determinationof g. [Ans. 5%]

48

Types of Motion: Depending on the number of coordinates required to describe the motion,motion is classified as

(i) One dimensional motion: motion in a straight line.

(ii) Two dimensional motion: motion in a plane.

(iii) Three dimensional motion: motion in space.

DISTANCE AND DISPLACEMENTDistance is the actual length of the path covered by an object, between initial and final

positions.

Displacement is the shortest distance between initial and final positions.

Speed: It is defined as the rate of change of distance w.r.t. time.

speed v =s

t

It is scalar quantity. S.I. unit is ms–1.

Instantaneous speed: The speed of the particle at a particular instant of time is calledinstantaneous speed.

Instantaneous speed v =0

Limitt

s

t

= ds

dt.

Instantaneous speed is equal to differential derivative of distance w.r.t time.

Average speed: When the body is in nonuniform motion, the average speed is defined as theratio of the total path covered by the object to the total time taken.

Average speed =Total path covered

Total time taken

Velocity: It is defined as the rate of change of displacement w.r.t. time.

Velocity V

=s

t

It is vector quantity. S.I. unit ms–1 Dimension [LT–1].Average velocity: The average velocity of a particle over a interval of time is defined as the

ratio of the displacement to the time taken.

avV

=Total displacement

Total time taken

Important Average speed Average velocity

�� 3UNIT

Motion in One Dimension 49

Relative velocity: The relative velocity of an object with respect to other object, when bothare in motion, is the rate of change of position of first object w.r.t. second object. eg.

Relative velocity of object A w.r.t. B

ABV

= A BV – V

.

Acceleration: It is defined as the rate of change of velocity w.r.t. time.

a =

V

t

.

Vector quantity S.I. Unit ms–2 Dimension [LT–2]

Instantaneous Acceleration: It is defined as the acceleration of an object at a particularinstant of time.

Instantaneous acceleration

a

=0

VLim

t t

=V��

d

dt.

Average Acceleration: It is defined as the ratio of the change in velocity to the total timeinterval in which that change in velocity has taken place

aav = 2 1

2 1

V – V

–t t.

Accelerated Motion: That motion in which velocity of particle changes with time, is calledaccelerated motion. It is of two types.

(a) Uniformly accelerated motion: In this type of motion acceleration remains constant.Also the average acceleration equals to the instantaneous acceleration.

(b) Non-Uniformly accelerated motion: The velocity changes by unequal amounts in equalinterval of times, so acceleration is not constant.

Equation of Motion for Constant Acceleration: Let u be the initial velocity of the particle,a is uniform acceleration and in time t, after travelling a distance s, the velocity of the bodybecomes v. Then

v = u + at

s = ut + 21

2at

v2 = u2 + 2as and

distance travelled by body in nth second is

Sn = u + 2

a (2n – 1).

Motion Under Gravity: Motion under gravity means an object is in motion under the force ofgravity and it is called free fall. In case of free fall, the acceleration equals to acceleration due togravity i.e. g = 9.8ms–2. Motion under gravity is a uniformly accelerated motion, and equation ofmotions are


v = u + gt

h = ut + 21

2gt

v2 = u2 + 2gh.

Important: (i) When we throw a body vertically upward its velocity decreases and g is taken–ve.

(ii) For motion in vertical direction, time of ascent equals to time of descent.Graphs:In this unit there are mainly three types of graph.

(i) Distance – time graph

(ii) Velocity – time graph

(iii) Acceleration-time graph.

(a) Slope of position (distance, displacement) time graph gives us velocity.

(b) Slope of velocity-time graph gives us acceleration.

(c) Area bounded by velocity-time graph and time axis equals to displacement of the body.

(d) Area under acceleration-time graph gives change in velocity.

SOLVED EXAMPLES

BASED ON DISTANCE-DISPLACEMENT AVERAGE SPEED AND VELOCITYExample 1. An object moves on a semicircular path, calculate the ratio of distance to

displacement.Solution: Let radius of circular path is r

distance

displacement=

2

r

r

=

2

=3.14

2 = 1.57.

Example 2. Two towns A and B are 100 km apart. A bus travels from A to to B at40 kmh–1 and returns from B to A at 50 kmh–1. Calculate the average speed and averagevelocity of the bus.

Solution.

Time taken to go from A to B = 100

40 =

5hr.

2

Time taken to come from B to A = 100

50 = 2hr.

Average speed =100 100

5 2 2

= 200

9 2 =

400

9 = 44.4 kmh–1.

Average velocity =Total displacement

Total time taken

= 0


Example 3. An object covers one third of its journey with speed u1 next, one third with speedu2 and last one third with speed u3. Calculate the average speed of the body during the entirejourney.

Solution. Let total distance = 3x

Then Total time taken =1 2 3

x x x

u u u

Average speed =Total distance covered

Total time taken

=

1 2 3

3x

x x x

u u u

=2 3 1 3 1 2

1 2 3

3x

u u u u u ux

u u u

.

=

1 2 3

1 2 2 3 1 3

3u u u

u u u u u u

Example 4. A man walks on a straight road from his home to a market 2.5 km awaywith a speed of 5 kmh–1. Finding the market closed, he instantly turns and walks back homewith a speed of 7.5 kmh–1. What is the

(a) Magnitude of average velocity and(b) Average speed of the man over the interval of time.(i) 0 to 30 min (ii) 0 to 50 min (iii) 0 to 40 min. (NCERT)Solution. (i) 0 to 30 min.

Distance covered in 30 min.= 5 kmh–1 × 1

h2

= 2.5 km.

Displacement = 2.5 km

Average speed =Distance

Time =

2.5

1 2 = 5 kmh–1

Average velocity =Displacement

Time =

2.5

1 2 = 5 kmh–1.

(ii) 0 – 50 min.

Time taken to go to market = 2.5

5 =

1h

2 = 30 min.

Time taken to come back = 2.5 1

h7.5 3

= 20 min.

In 50 min.distance covered = 2.5 + 2.5


= 5 km.

displacement of body = 0 km

Average speed =distance

total time taken

=5

5 6 = 6 kmh–1.

Average velocity = 0(iii) 0 – 40 min:

Distance covered in 30 min. = 2.5 km

Distance covered in 10 min. = 1.25 km

Total path covered = 3.75km

Displacement of body = 2.5 – 1.25

= 1.25 km.

Average speed =Total distance

Time taken =

3.75

4 6

= 5.625 kmh–1.

Average velocity =Net displacement

Time taken

=1.25

4 6 = 1.875 kmh–1.

Example 5. A particle goes along a quadrant AB of a circle of radius 5 cm with a constantspeed 2.5 cms–1 Calculate the average velocity over the interval.

Solution.

Time taken =distance

speed

=2 1

4

r

v

=3.14 5 1

2 2.5

= 3.14 s

Displacement AB = 2 25 5 = 5 2 cm.

Average velocity =displacement

time

=5 2

3.14= 2.25 cms–1.

A O

B

5 cm

V B

V A


BASED ON RELATIVE VELOCITYExample 6. Rain is falling vertically with a speed of 35 ms–1. A woman rides a bicycle with

a speed of 12 ms–1 in east to west direction. What is the direction in which she should hold herumbrella?

Solution.Velocity of rain w.r.t woman is given by

RWv

= R W–v v

.

RWv

makes an angle with vertical direction. Which is given by

Tan =W

R

v

v =

12

35.

= 0.343.

= 19°.

She should hold her umbrella at an angle 19° with vertical.

Example 7. A car travelling at 58 kmh–1 overtakes another car travelling at 40 kmh–1.Assuming each car to be 5 m long calculate the time taken for overtaking.

Solution. Relative velocity of first car w.r.t. other car

= 58 – 40

= 18 km h–1

= 5 ms–1

Total length of the two cars = 5 + 5 = 10 m.

Time taken =10

5 = 2s.

Example 8. A policeman moving on a highway with a speed of 30 kmh–1 fires a bullet atthief’s car speeding away in the same direction with a speed of 192 kmh–1. If the muzzle speedof the bullet is 150 ms–1 with what speed does the bullet hit the thief’s car? (NCERT)

Solution. Relative velocity of thief’s car w.r.t. police van

= 192 – 30

= 162 kmh–1

= 45 ms–1.Muzzle speed of the bullet = 150 ms–1.

Relative velocity of bullet w.r.t. thief’s car

= 150 – 45

= 105 ms–1

Example 9. A jet airplane travelling at the speed of 500 kmh–1 ejects its products ofcombustion at the speed of 1500 kmh–1 relative to the jet plane. What is the speed of thelatter w.r.t an observer on the ground. (NCERT)

Solution. Let

vJ = Velocity of jet plane vJ = –500 kmh–1.

vPJ = Velocity of products of combustion w.r.t. jet = +1500 kmh–1

Velocity of products of combustion w.r.t. ground is vP.

vPJ = vP – vJ

5152

1845 /

km

hm s

v R W

– Wv

vR

v W


1500 = vP – (–500)

or vP = 1000 kmh–1.Example 10. On a two lane road, car A is travelling with a speed of 36 kmh–1. Two cars B and

C approach car A in opposite directions, with a speed of 54 kmh–1. At a certain instant, when thedistance AB is equal to AC, both being 1km, B decides to overtake A, before C does. Whatminimum acceleration of car B is required to avoid an accident? (NCERT)

Solution.

36 km /h = 10 m s–1

A

B

54 km /h = 15 m s–1 54 km /h = 15 m s–1

C

Relative velocity of car C w.r.t. car A = 15 + 10 = 25 ms–1.

Relative velocity of car B w.r.t. car A = 15 – 10 = 5 ms–1.

Time taken by car C to reach to car A

=AC

v =

1000

25= 40 sec.

For Car B, t = 40s , s = 1000 m,

s = ut + 21

2at

1000 = 5 × 40 + 1

2a × (40)2

or a = 1 ms–2.

Example 11. Two towns A and B are connected by a regular bus service with a busleaving in either direction every T min. A man cycling with a speed of 20 kmh–1 in thedirection A to B notices that a bus goes past him every 18 min, in the direction of his motion,and every 6 min in the opposite direction. What is the period T of the bus service and withwhat speed do the buses ply on the road? (NCERT)

Solution. Let speed of each bus is v.

For bus going from A to B.

Relative velocity of the bus w.r.t. man = (v – 20)

The bus plying in the direction A to B go past the cyclist after every 18 min.

Distance covered in 18 min = (v – 20)18

60 km.

Bus leaves the town every T min

(v – 20)18

60=

T

60 v

or v.T. = (v – 20) 18. ...(i)

For bus going from B to A.

Relative velocity of bus w.r.t. man = (v + 20)

vT = (v + 20)6 ...(ii)


diving equation (i) by equation (ii)

( 20)18

( + 20)6

vv = 1

or v = 40 kmh–1.

Using value of v in equation (i) or (ii)

T = 9 min.Example 12. Two trains 125 m and 75 m in length are running in opposite directions with

velocities 60 km–1 and 30 kmh–1. In what time they will completely cross each other.Solution. Relative velocity of first train w.r.t. second train

= 60 – (– 30)

= 90 kmh–1

= 25 ms–1.

Total length of the two trains be 125 + 75 = 200 m.

Time taken =200

25 = 8 sec.

BASED ON UNIFORMLY ACCELERATED MOTIONExample 13. A car moving along a straight high way with speed of 126 kmh–1 is brought to

stop with in a distance of 200 m. What is the retardation of the car and how long does it take forthe car to stop? (NCERT)

Solution. Here,

u = 126 kmh–1 = 126 × 5

18 = 35 ms–1.

s = 200 m

v = 0

Using third equation of motion:

v2 = u2 + 2as

0 = (35)2 + 2(a) (200)

or s = – 35 35

400

=

49–

16

= –3.06 ms–2

i.e. retardation is 3.06 ms–2,

Now time taken t =– u

a

v

=0 (35)

4916

= 11.43 s.Example 14. A truck is moving with constant velocity of 5 ms–1. Suddenly at t = 50 sec. it

starts accelerating uniformly at 1ms–2. Calculate the velocity of the truck at t = 80 sec.


Solution. Here,

u = 5 ms–1

t = 80 – 50 = 30 sec

a = 1 ms–2

v = u + at

= 5 + 1× 30

= 35 ms–1.Example 15: A truck’s motion is recorded for a period of 10 sec, i.e. starting from

t = 10 s to t = 20 sec. The truck is moving with constant velocity of 15 ms–1, startedacceleration at the rate of 2.5 ms–2 at t = 10 sec. Calculate the displacement of truck.

Solution: Here,

u = 15 ms–1

a = 2.5 ms–2

t = 10 sec.

s = ut + 2

2

at

= 15 × 10 + 1

2 (2.5) (10)2

= 275 m.Example 16. A body starting from rest, and moving with constant acceleration, covers

10 m in the first second. Find the distance travelled by it in the next second.Solution. Here, u = 0

n = 1, s = 10 m

Using equation s = u + (2 –1)2

an

10 = 0 + (2 –1)2

a

10 =2

a

or a = 20 ms–2.

Now distance travelled in second second is

S =20

0 (2 2 –1)2

= 30m.Example 17. The displacement of a particle is given by

s = 3t3 + 7t2 + 14t + 8 mCalculate its acceleration at t = 1 secSolution. Here,

s = 3t3 + 7t2 + 14t + 8

t = 1 s


Instantaneous speed

u =ds

dt =

d

dt [3t3 + 7t2 + 14t + 8]

= 9t2 + 14t + 14 ms–1

Now instantaneous acceleration

a =du

dt

= 2(9 14 14)d

t tdt

= 18t + 14

acceleration at t = 1 sec.

a = 18(1) + 14

= 32 ms–2

Example 18. A scooter moving at a speed of 10 ms–1 is stopped by applying brakes whichproduces a uniform acceleration of – 0.5 ms–2. How much distance will be covered by thescooter before coming to rest?

Solution: Here,

u = 10 ms–1

a = – 0.5 ms–2

v = 0

Using relation v2 = u2 + 2as

(0)2 = (10)2 + 2(–.5) (s)

or 100 = s

s = 100 m.Example 19. A bullet is fired at a target. Its speed is halved after penetrating 30 cm into it.

How much further will it travel before coming to rest?Solution. Here, u = u

v =2

u, s = 30 cm

Using relation

v2 = u2 + 2as

or2

2

u

= u2 + 2(a) (30)

2

4

u= u2 + 60a

23–

4u = 60a

or a =2

–80

u


For further motion, v = 0, u = 2

u

v2 = u2 + 2as

0 =2 2–

22 80

u us

2

–4

u=

2

–40

us

or s = 10 m

Example 20. A body starting from rest moves in a straight line with a constantacceleration a. It travels a distance s1 in time t second and s2 in rest t second. What is therelation between s1 and s2?

Solution. Here, u = 0, a = a,

from second equation of motion.

s = ut + 21

2at

s1 = 0 + 21

2at

s2 = 0 + 21(2 )

2a t

1

2

s

s=

2

24

t

t

or s1 : s2 = 1 : 4.Example 21. A body covers a distance of 25 m in 5th sec and 40 m in 8th sec of its

motion. Calculate the acceleration and initial velocity of the body.Solution. Here, S5 = 25 m n = 5

S8 = 40 m n = 8

Using equation

Sn = u + (2 –1)2

an

25 = u + (10 –1)2

a = u + 4.5a ...(i)

40 = u + (16 –1)2

a = u + 7.5a ...(ii)

Subtracting equation (i) from (ii)

15 = 3a

or a = 5 ms–2

Substituting for a in equation (i) or (ii)

u = 2.5 ms–1.


Example 22. A drives takes 0.3 sec to apply the brakes after he sees a need for it. He isdriving car at 72 kmh–1 and the brakes cause a retardation of 4.0 ms–2, find the distancetravelled by the car after he sees the need to put the brakes.

Solution. Here, u = 72 × 5

18 = 20 ms–1.

v = 0

a = – 4.0 ms–2

Using equation v2 = u2 + 2as

0 = (20)2 + 2 (– 4) s

s =400

8 = 50 m

distance travelled in 0.3 s = 0.3 × 20 = 6 m

Total distance covered = 50 + 6 = 56 m.

Example 23. A body covers 12 m in 2nd second and 20 m in 4th second. How much distancewill it cover in 4 seconds after the 5th second?

Solution. Here, s2 = 12 m

s4 = 20 m

From equation sn = u + (2 –1)2

an

s2 = u + (4 –1)2

a = 12 ...(i)

s4 = u + (8 –1)2

a = 20 ...(ii)

Solving equation (i) and (ii) a = 4 ms–2

u = 6 ms–1

Distance covered in 4 seconds after 5th second.

= s9 – s5

Here st = ut + 21

2at

= 216 9 4 9

2

– 216 5 4 5

2

= 216 – 80

= 136 m.Example 24. A body starting from rest accelerates at the rate of 10 cms–2, and retards

uniformly at the rate of 20 cms–2. Find the least time in which it can complete the journey of5 km, if the maximum velocity attained by the body is 72 kmh–1.

Solution. Here, u = 0, v = 572

18 = 20 ms–1,

a = 10 cms–2 = 0.1 ms–2


Using v = u + at

20 = 0 + 0.1 t1

t1 = 200 s

also v2 = u2 + 2as

(20)2 = 02 + 2 (0.1) s

or s = 2000 m

= 2 km.For motion with uniform retardation u = 20 ms–1, v = 0.

a = – 20 cms–2

= – 0.2 ms–2 t = t2 s = ?

v = u + at

0 = 20 – 0.2 t2

or t2 = 100 s

Also v2 = u2 + 2as

0 = (20)2 + 2 (–0.2) S

s = 1000 m

= 1 Km.

Remaining part of journey = 5 – (2 + 1)

= 2 km = 2000 m.

This journey occurs at the uniform maximum velocity of 20 ms–1.

For motion with uniform velocityu = 20 ms–1 s = 2000 m

t3 =s

u =

2000

20 = 100 s

t = t1 + t2 + t3

= 200 + 100 + 100= 400 s.

Example 25. A particle starting from rest travels a distance x in first 2 sec and a distanceY in next two second then what is the relation between X and Y.

Solution. Here, u = 0 t = 2s s = X

t = 4s s = (X + Y)

s = ut + 1

2at

X = 0 + 21(2)

2a = 2a ...(i)

(X + Y) = 0 + 21(4)

2a

X + Y = 21 X(4)

2 2

[Using equation (i)]


or Y = 3XExample 26. The placement x of a particle moving in one dimension is related to time by

equation.

t = 3x

Where x is in m and t is in second calculate the displacement when the velocity is zero.Solution. Given

t = 3x

or x = t – 3

Square both sides x = t2 – 6t + 9

dx

dt= 2t – 6

i.e. v = 2t – 6.

But velocity is zero

2t – 6 = 0

or t = 3 sec

Displacement of body at t = 3S

x = (3)2 – 6(3) + 9

= 9 – 18 + 9

= 0.Example 27. A hundred metre sprinter increases her speed from rest uniformly at the rate

of 1 ms–2 up to three quarters of the total run and covers the last quarter with uniform speed. Howmuch time does she take to cover the first and second half of her path.

Solution.

A50 m

C D

75 m 100 mB

For the motion from A to C.

s = ut + 21

2at

50 = 0 + 21

11

2t

or t1 = 10 sec.For the motion from A to D.

s = ut + 21

2at

75 = 0 + 22

1(1)

2t

or t2 = 150

t2 = 150 = 12.2 s.


Let at point D the velocity of sprinter becomes u

v = u + at

= 0 + (1) (12.2)

= 12.2 ms–1.

Time taken to go from D to B

=25

12.2 = 2 sec.

Time taken to go from C to D=12.2 – 10

= 2.2 sec.

Time taken to cover second half = 2 + 2.2 = 4.2 s.

BASED ON MOTION UNDER GRAVITYExample 28. A body falling from rest has acquired a velocity v after it falls through a distance

h. Calculate the distance it has to fall further, for its velocity to become tripple.Solution. Here, u = 0, v = vUsing relation v2 = u2 + 2gh

v2 = 0 + 2gh ...(i)

When the final velocity becomes tripple

(3v)2 = 0 + 2gh9v2 = 2gh ...(ii)

Dividing (ii) by (i)

9 =h

h

or h = 9h.

Additional height to fall = 9h – h = 8h.Example 29. A body is projected vertically upward with a velocity of 20 ms–1 and reaches a

maximum height h. What is its velocity when it reaches 34h

. (g = 10 ms–2).

Solution. Here, u = 20 ms–1

v = 0

v2 = u2 + 2gh

or 0 = (20)2 + 2(–10)h

or h = 20 m.

Let velocity at 3

4

h is v.

v2 = u2 + 2gh

v2 = (20)2 + 2(–10) × 320 4

= 400 – 300

= 100

or v = 10 ms–1


Example 30. A ball is thrown upward with some initial speed. It goes up to height of 19.6 mand then returns. Find (a) initial speed (b) the time taken to reach the highest point.

Solution. Here, h = 19.6 m. g = – 9.8 ms–2

Let initial speed is u.

v = 0

Using equation

v2 = u2 + 2gh

0 = (u)2 + 2(–9.8) (19.6)

or u = 19.6 ms–1.(ii) Also v = u + gt

0 = 19.6 – 9.8 t

t = 2 s.Example 31. A boy on a cliff 49 m high drops a stone. After one second he throws a second stone.

They both hit the ground at the same time. With what speed did he throw the second stone?Solution. Here, h = 49 m.

For first stone

h = ut + 21

2gt

49 = 0 + 21

2gt

or 49 = 21

2gt

or t = 10 s

= 3.16 sFor second stone

t = (t – 1) = 2.16 s, v = u

h = ut + 21

2gt

49 = u (t – 1) + 21( –1)

2g t .

or 49 = u (2.16) + 219.8(2.16)

2 .

On solving u = 12.1 ms–1.

Example 32. A stone is dropped from a certain height. It moves down to a distance of 24.5 min the last second before hitting the ground. Calculate the height from where the stone wasdropped.

Solution. Let stone takes n second to reach the ground.

sn = u + (2 –1)2

gn

Here,

sn = 24.5 m. u = 0


24.5 =.8

(2 –1)2

n

or (2n – 1) = 5

or n = 3s

Height h = ut + 21

2gt

= 0 + 219.8 3

2

= 44.1 m.Example 33. A ballon is moving vertically with a velocity of 4 ms–1. When is at a height

h a body is released from it, which reaches the ground in 4 seconds. Calculate h.Solution. Here, u = – 4 ms–1, g = 9.8 ms–2

t = 4 s.

From 2nd equation of motion

h = ut + 21

2gt

h = – 4 (4) + 1

2 × 9.8 × 4 × 4

= – 16 + 78.4

= 62.4 mExample 34. A particle is released from rest from a building of height 3h. t1, t2, and t3

be the times to cover successive heights h. Calculate the ratio t1 : t2 : t3.Solution. Here t1, t2 and t3 are the times taken by the particle to fall successive heights h.

h

h

h t1

(t + t )1 2

(t + t + t )1 2 3

h = ut + 21

2gt

or h = 21

10

2gt ...(i)

2h = 21 2

10 ( )

2g t t ...(ii)

3h = 21 2 3

10 ( )

2g t t t ...(iii)


Divide equation (i) by (ii)

1

2=

21

21 2( )

t

t t

or t1 + t2 = 12 t

or t2 = 12 – 1 t ...(iv)

Divide equation (i) by (iii)

1

3=

21

21 2 3( )

t

t t t

or (t1 + t2 + t3)2 = 3t1

2.

or 1 1 32 –1t t t = 13 t

or t3 = 13 – 2 t ...(v)

t1 : t2 : t3 = 1 : 2 – 1 : 3 – 2 .

Example 35. A ball is thrown upward from the top of a tower 39.2 m high with avelocity of a 9.8 ms–1. Calculate the speed with which the body will hit the ground and thetime when it strikes the ground. (g = 10 ms–2)

Solution. Here, u = 9.8 ms–2

h = – 39.2 m, g = –9.8 ms–2

Let ‘t’ be the time taken by ball to reach the ground.

h = ut + 1

2gt2

– 39.2 = 9.8t + 21(–9.8)

2t

or t2 – 2t – 8 = 0

(t – 4) (t + 2) = 0

t = 4s, – 2s

t = 4s.Now v = u + gt

= 9.8 + (– 9.8)4

= – 29.4 ms–1

– ve sign shows velocity is in the downward direction.

Example 36. A ball is dropped from the top of a tower of 100 m height. Simultaneouslyanother ball is thrown upward from the bottom of tower with a speed of 50 ms–1. When andwhere the two balls will cross each other?

Solution: Let the two balls meet after time ‘t’ distance covered by 1st ball

h1 = 0 + 21

2gt ...(i)

distance covered by second ball, thrown in upward direction.


h2 = 50t – 21

2gt ...(ii)

Buth1 + h2 =100 m

2 21 150 –

2 2t gt gt

= 100

or t = 2 sFor position substitute for ‘t’ in equation (i)

h1 =1

9.8 2 22

= 19.6 mTwo balls will cross each other at a distance of 19.6 m from the top of tower.

Example 37. A parachutist bails out from an aeroplane and after dropping through a distanceof 40 m, he opens the parachute and decelerates at 2ms–2. If he reaches the ground with a speedof 2ms–1, how long is he in the air? Also calculate the height, at which he bails out from theplane?

Solution. Here, u = 0, g = 9.8 ms–2

s = 40 m,

Now s = ut + 21.

2gt

40 = 210 9.8

2t

or t2 =80

9.8

220

7

or t = 2.86 sAlso velocity of the parachutist after time t becomes

v = u + gt

=20

0 9.87

= 28 ms–1.Now falling through 40 m he decelerates at 2 ms–2 till its final velocity becomes 2 ms–1.

t =– u

a

v =

2 – 28

–2 = 13 sec.

s = ut + 21

2at

= 28 × 13 + 21(–2)(13)

2= 364 – 169= 195 m

Total time taken = 13 + 2.86 = 15.86 sTotal height = 195 + 40 = 235 m

u = 0

100 m

u = 50 ms–1


Example 38. A body is dropped from a certain height. After 2 sec., from the same heightanother body is projected vertically downward with a speed of 39.2 ms–1. When and where the twobodies will meet?

Solution. For the first body u = 0.

h1 = 210

2gt .

Second body travels for time (t – 2) sec.

h2 = u (t – 2) + 21( )( – 2)

2g t

h2 = 39.2 (t – 2) + 219.8( – 2)

2t .

= 39.2t – 78.4 + 4.9t2 – 19.6t + 19.6

= 4.9t2 + 19.6t – 58.8.

According to problem

h1 = h2

219.8

2t = 4.9t2 + 19.6t – 58.8

or t = 3 sec.Height at which they overlap each other

h1 = 4.9 t2

= 4.9 (3)2

= 44.1 mSecond body will overtake the first body aftert 1 sec. of its projection and a distance of 44.1m

from the top.

Example 39. A ship is sailing westward at 8 ms–1 while trying to fix a bolt at the top ofthe mast, a sailor drops the bolt. If the mast of the ship is 19.6 m where does the bolt hit thedeck?

Solution. Here, h = 19.6 m

g = 9.8 ms–2, u = 0

h = ut + 21

2gt

19.6 = 210 9.8

2t

or t2 = 4

or t = 2 sec.

i.e. the bolt hits the deck after 2 sec. The distance travelled by ship in 2 s = 8 × 2 = 16 m.

Example 40. A ball is thrown upward from the ground with an initial speed u. The ball is ata height of 80 m at two times, the time interval being 6 s. Find the initial velocity of the body.(Given g = 10 ms–2).

Solution. Here,

u = u, g = –10 ms–2, h = 80 m.


h = ut + 21

2gt

80 = ut + 21(–10)

2t

80 = ut – 5t2.

or 5t2 – ut + 80 = 0.

Solving this quadratic equation.

t =2 –1600

10

u u and

2– –1600

10

u u

Given the time interval is 6 sec.

2 –1600

10

u u

– 2– –1600

10

u u

= 6

or22 –1600

10

u= 6

or u2 – 1600 = 900

u2 = 2500

or u = ± 50 ms–1

Ignore – ve sign so u = + 50 ms–1.

BASED ON VARIOUS GRAPHS (S-t GRAPH, V-t GRAPH, etc.)Example 41. The velocity-time graph of an object is shown in figure. Calculate the

displacement of body and maximum acceleration during the journey.Solution. The displacement of the body equals to area enclosed by v-t graph and time axis

200 40 60 80t (sec)

v (m /s)

20

40

60

80

D

C

BA

1 2 3 4

s = Ar 1 + Ar 2 + Ar 3 + Ar 4

=1 1 1

20 20 (10 20) (20 80) 10 40 802 2 2

= 200 + 200 + 500 + 1600

= 2500 m.


Maximum acceleration in the body is from 30 to 40 s

Max acclr = slope of BC graph

=(80 – 20)

40 – 30

=60

10 = 6 ms–2.

Example 42. The speed time graph of a particle moving along a fixed direction is shownbelow. Obtain the distance travelled by the particle between (i) t = 0 to t = 10s (ii) t = 2 to 6 S.What is the average speed of the particle in intervals in (i) and (ii).

Solution.Distance travelled in 0 to 10 s

= Area OAB

=1

OB × AP2

=1

10 122

= 60 m.

Average speed =60

10 = 6 ms–1.

(ii) During OA, acceleration is given by

v = u + at

12 = 0 + 5a or a = 12

5 = 2.4 ms–2.

Similarly for journey A to B a = – 2.4 ms–2.

Velocity of the particle after 2 sec from start.

v = u + at

= 0 + 2.4 × 2 = 4.8 ms–1.

Distance travelled by the particle between t = 2 to 5s

s1 = ut + 21

2at

= 4.8 × 3 + 212.4 (3)

2

= 25.2 m.

Distance travelled by the particle in t = 5 to 6 sec.

s2 = ut + 21

2at

0 5 10t (sec)

v (m s )–1

12

B

A

P


= 12 × 1 + 21(–2.4) (1)

2

= 10.8 m

Total distance travelled in t = 2 to 6 sec.

s = s1 + s2

= 25.2 + 10.8

= 36 m.

Average speed from 2 to 6 sec.

=36

4 = 9 ms–1.

Example 43. A car starts from rest and moves with uniform accelerations and attains avelocity of 90 kms–1. in 15 s. Now it move with constant speed for 15 s and finally the car isbrought to rest in 15 sec. Find the distance travelled using velocity time graph.

Solution.

150 30 45t (sec)

v (m /s)

25

C

BA

v =km

90hr

= 5

9018

= 25 ms–1.

Total distance covered

= Area OABC.

=1

(45 15) 252

= 30 × 25

= 750 m.

Example 44. A three wheeler starts from rest, accelerates uniformly with 1 ms–2 on astraight road for 10s, and then moves with uniform velocity. Plot distance covered by thevehicle during the nth sec. (n = 1, 2....) versus n. What do you expect this plot to be during,accelerated motion, a straight line or parabola? (NCERT)

Solution. Here, u = 0, a = 1 ms–2

Distance travelled in nth sec

s = u + (2 –1)2

an

or s =1

(2 – 1)2

n


nth sec. s (m)

1 0.5

2 1.5

3 2.5

4 3.5

5 4.5

6 5.5

7. 6.5

8. 7.5

9. 8.5

10. 9.5

10 2 3 4 5 6 7 8 9 10n (s)

s (m )

0 .5

1 .5

2 .5

3 .5

4 .5

5 .5

6 .5

7 .5

8 .5

Uniform

Acce

leratio

n

After 10th second the three wheeler moves with uniform speed of 10 ms–1.

Example 45. Two stones are thrown up simultaneously from the edge of a cliff 200 m highwith initial speed of 15 ms–1 and 30 ms–1. Verify that the following graph correctly representsthe time variation of the relative position of the second stone w.r.t. first. Neglect air resistanceand assume that the stones do not rebound after hitting the ground (g = 10ms–2). Give the equations for the linear and curved parts of the plot. (NCERT)

Solution. Using equation

s(t) = s(0) + ut + 21

2gt

Taking the edge of cliff as origin,

s1 = 200 + 15t – 2110

2t

s2 = 200 + 30t – 2110

2t


20 4 6 8 B Ct (s)

20

40

60

80

100

120

(s –

s)

m2

1

When first stone strikes the ground

s1 = 0

200 + 15t – 5t2 = 0

t2 – 3t – 40 = 0

On solving above eqn. t = 8 s or – 5 s

Time can not be negative = t = 8 s.

Relative position of 2nd stone w.r.t. first stone

s2 – s1 = 15t.

(s2 – s1) t

It shows graph is a straight line (for t = 8 s). After 8 s, only the second stone is in motion.So the graph is parabolic.

s2 = 200 + 30t – 5t2.The second stone will hit the ground when s2 = 0.

200 + 30t – 5t2 = 0

On solving t = 10 s.

After 10 sec, the separation between the balls is zero. Which is denoted by part BC of thegraph.

Example 46. The velocity-time graph for an automobile is shown below. Drawacceleration time graph.

1 2 3 40 5 6 7t (s)

v (m s )–1

20

30

10

C

BA

Solution. From O to A, acceleration of the vehicle

a = slope of OA

=20 – 0

2.5 – 0

= 8 ms–2.

From A to B velocity time graph is parallel to time axis so acceleration = 0.


For graph BC

a = slope of BC

=0 – 20

7 – 5 = – 10 ms–2.

1 2 3 4

5

6 7t (s)

a (m s )–2

0

–5

–10

5

10a = 8 m s–2

a = 0

a = –10 m s–2

Example 47. A ball is thrown upward with an initial velocity of 50 ms–1. After how muchtime will it return? Draw velocity time graph for the ball. Use this graph to calculate (i) themaximum height attained by the ball and (ii) height of the ball after 7.5 s.

Solution. Here, u = 50 ms–1, g = –10 ms–2.

at highest point v = 0.

v = u + gt

0 = 50 – 10t or t = 5 sec.

The ball will return to ground in 10 sec.

(i) Maximum height attained by the ball = Area OAB

=1

5 502

= 125 m

2.5 5

A C

7.5 10t (s)

v (m s )–1

0

–25

–50

50

D

B

(ii) Height after 7.5 sec

= Area OAB + Area ACD

= 125 + 1

(7.5 – 5) (–25)2

= 125 + (2.5) (–25)

2


= 125 – 31.25

= 93.75 mExample 48. The velocity time graph of a body moving in a straight line is shown below.

Calculate the distance and displacement travelled by the body in 6 sec.Solution. Distance travelled in 6 sec.

= Ar 1 + Ar 2 + Ar 3

= (2 × 4) + (2 × 2) + (2 × 2)

= 8 + 4 + 4

= 16 m.

1 2 3 4 5 6t (s)

v (m /s )–1

0

2

4

4

2 1

2

3

Displacement of body in 6 sec = Ar 1 + Ar 3 – Ar 2

= 8 + 4 – 4

= 8 m.Example 49. A lift is going up. The speed time graph for the lift is shown below.

Calculate the height to which the lift takes the passangers.Solution. Height to which the lift takes the passenger

v(m /s)

4 .5

0 2

A B

10 12c

t (sec)

= Area OABC

= (12 + 8) × 4.5 × 1

2= 45 m.

Example 50. Figure shows the displacement-time graph of a body moving in a straight line.With the help of it draw velocity time graph. Also calculate velocity for

(i) first five second(ii) time interval t = 5 s to t = 15 s(iii) last five second.Solution. (a) Velocity for 5 second


12

8

4

0

A B

M N C

5 10 15 20t (sec)

s (m )

= slope of graph OA

=12 – 0

5 – 0 =

12

5

= 2.4 ms–1.

(b) Velocity for t = 5 to t = 15 sec.

= Zero (body is rest).

(c) Velocity for last 5 sec.

= slope of graph BC.

=0 –12

20 –15

=–12

5 = –2.4ms–1.

0

0.6

1 .2

1 .8

2 .4

10

15 20

5t (sec)

Example 51. A car accelerates from rest at a constant rate for some time, after whichit declerates at constant rate to come to rest. If the total time elapsed is t sec. Calculate(i) maximum velocity reached (ii) the total distance travelled.

Solution. Let the car accelerates for time t1 and declerates for time t2

0 t1 t2

B

A

t

vm

v


Then

=1

m

t

v

or t1 = m

v

...(i)

Similarly =2

m

t

v

or t2 = mv

...(ii)

As total time taken is t

t = t1 + t2

= m m v v

or vm =t

Total distance = area under v – t graph

=1

2 mt v .

=1

2

tt

=2

2( )t

.

Example 52. A particle is moving eastward with a velocity of 5 ms–1. If in 10 s the velocitychanges to 5 ms–1 northwards what is the average acceleration in this time? (I.I.T.)

Solution. v = 2 1

v v

= 2 1( )

v v

= 2 25 5 = –15 2 ms due N – W.

aav =t

v =

5 2

10

= –21ms

2 (N – W).


Example 53. A body falling freely from a given height H hits an inclined plane in its path ata height h. As a result of this impact the direction of the velocity of the body becomes horizontal.For what value of h/H the body will take maximum time to reach the ground.

(IIT) (MNR).Solution. Time taken by the body to fall from A to B i.e. t1 is given by

h = ut + 21

2gt

(H – h) = 0 + 21

1

2gt

or t1 =2(H – )h

g.

h

B

A

H

After striking the inclined surface, the body becomes horizontal. Let t2 is the time taken bybody to reach the ground.

t2 =2h

g

Total time taken t = t1 + t2

t =2(H – ) 2h h

g g

t =2

(H – h hg .

For t to be maximum.

dt

dh= 0.

or 2H –

dh h

dh g

= 0.

or1 12 2

– –1 1(H – ) (–1)

2 2h h = 0.

or h = H – h


or h =H

2

Example 54. The acceleration experience by a moving boat after its engine is cut off is givenby

dudt

= – Ku3

where K is constant. If u0 is the magnitude of the velocity at cut off. Find the magnitude of thevelocity at cut off. Find the magnitude of the velocity at time t after the cut off.

Solution. Here,du

dt= – Ku3

or 3

du

u= – Kdt

Integrating both sides

0

3

u

u

du

u =0

Kt

dt

or0

2

1

2

u

uu

= – K(t)t

0

or 2 20

1 1–

u u= 2 Kt

or u = 0

201 2K

u

u t

Example 55. What is the retardation of a particle if the relation between time and positionis

t = x2 + x where and are constants.(MNREC – 82)

Solution. Given t = x2 + x

dt

dx= 2x +

Velocity u =dx

dt = (2x + )–1

Now acceleration a =du

dt =

du dx

dx dt

=du

udx


= 12

du x

dx

= 1 –2(2 ) 1 – (2 ) 2x x

= 3–2 (2 )x .

Retardation = – a

= 2 (2 x + )–3.

EXERCISE

BASED ON DISTANCE, DISPLACEMENT, AVERAGE SPEED AND VELOCITY1. A man walks at a speed of 7ms–1 for 56m and 8ms–1 for next 56m. What is the average

speed for the entire journey? [Ans. 7.47 ms–1]

2. A car covers the first half of the distance between two places at a speed of 40 km/hr andthe second half at 60 km/hr. What is the average speed of the car? [Ans. 48 kmh–1]

3. A person moves on a semi circular path of radius 20 m. If he starts at one end A of thepath and reaches other end B diamatrically opposite to A find the distance covered anddisplacement during this motion. [Ans. 62.8m, 40m]

4. A car moves the first half of the distance between two places at a speed of 40 kmh–1 andsecond half of 60 kmh–1. What is the average speed of the car? [Ans. 48 kmh–1]

5. A car moves on a straight road for first half time with constant speed of 20 ms–1 and nexthalf time with constant speed of 60 ms–1. Calculate the average speed. [Ans. 40 ms–1]

6. A car travels the first half of the total distance with velocity u1 and second half with

velocity u2, calculate the average velocity of the car. [Ans. 1 2

1 2

2u u

u u]

7. A body travelling along a straight line covers one third of the distance with a velocity V0.The remaining part of the distance was covered with velocity V1 for half the time andwith velocity V2 for the other half of the time. Calculate the mean velocity of the body

over the whole journey. [Ans. 0 1 2

1 2 0

3V (V V )

V V 4V

].

8. A table clock has its minute hand 4.0 cm long. Find the average velocity of the tip of theminute hand between 6.00 a.m to 6.30 a.m. [Ans. 4.4 × 10–3 cms–1]

BASED ON RELATIVE VELOCITY9. A bus moves with a uniform velocity 80 kmh–1. Another bus B moves in the same

direction with 100 kmh–1 calculate.

(a) relative velocity of bus A w.r.t. bus B. (b) relative velocity of bus B w.r.t. bus A.

[Ans. – 20 kmh–1, 20 kmh–1]

10. A man is moving due east with a speed of 1 kmh–1 and rain is falling vertically with a

speed –13 kmh . At what angle from vertical the man has to hold his unbrella to keep

the rain away. Also find the speed of rain drops w.r.t. man.

[Ans. 30° with vertical, 2 kmh–1]


11. Two buses A and B are moving with speed 54 kmh–1 and 90 kmh–1 respectively in thesame direction. Calculate relative velocity of bus B w.r.t. A. Also find out how far bus Bwill be from bus A after 2 sec. [Ans. 36 kmh–1, 72 km]

12. Two trains A and B each of length 100 m are running on parallel tracks. One overtakesthe other in 20s and one crosses the other in 10s. Calculate the velocities of each train.

[Ans. 15ms–1, 5ms–1]

13. A boat covers certain distance between the spots on a river taking 4 hours going downstream and 6 hrs going up stream. What time will be taken by the boat to cover samedistance in still water? [Ans. 4.8h]

BASED ON EQUATIONS OF MOTION (UNIFORMLY ACCELERATED MOTION)AND INSTANTANEOUS VELOCITY, ACCELERATION14. On applying brakes a car slows down from V1 to V2 and travels a distance s. Futher the

car slows down from V2 to V3 and travels a distance s. What is the relation between V1,

V2 and V3. [Ans. V2 = 2 2

1 3V V

2

]

15. A particle is moving with uniform acceleration along a straight line ABC. Its velocity atA and C are 5ms–1 and 25ms–1 respectively. If BC = 2 AB find the ratio of time to travelfrom A to B, B to C. [Ans. 1 : 1]

16. A particle moving with a constant acceleration describes in the last second of its motion

in 9

th25

of the whole distance. If it starts from rest how long is the particle in motion

and through what distance does it move if it describes 6cm in the first second?

[Ans. 5 sec, 150 cm]

17. A riffle bullet loses 1

th20

of its velocity in passing through a plank. Calculate the least

number of such planks required to stop the bullet. [Ans. 11]

18. A particle starting from rest from O accelerates uniformly along a straight line andcrosses two points A and B which are 5m apart in 2s. If velocity at B is 2ms–1 more thanthe velocity at A. Find the distance OA. [Ans. 9/8 m]

19. A train starting from rest accelerates uniformly for 100 s runs at a constant speed for 5minutes, and then comes to stop with uniform retardation in next 150 second. During thismotion it covers a distance of 4.25 km. Find its (a) constant speed (b) acceleration

(c) retardation. [Ans. 10 ms–1, –2 –21 –1ms , ms

10 15]

20. A body starts from rest and travels with a uniform acceleration of 3 ms–2 and thendecelerates at a uniform rate of 2ms–2 again to come to rest. Total time of travel is 10s.Find the maximum velocity during the journey and the total displacement.

[Ans. 12 ms–1, 60 m]

21. The driver of a train travelling at 115 kmh–1 sees on the same track 100 m in front of hima slow train travelling in the same direction at 25 kmh–1. What is the least retardation thatmust be applied to the faster train to avoid a collision? [Ans. –3.125 ms–2]


22. A car starts from rest and attains a speed of 8 ms–1 in 2 sec. It travels with uniform speedfor 3 seconds. Calculate the total displacement of car in 5 seconds. [Ans. 32 m]

23. The position ‘s’ of a particle varies with time t

as s = at2 – bt3

Calculate the time, when the acceleration of the particle will be zero. [Ans. t = 3

a

b]

24. A body covers 10m in 2nd second and 25m in 5th second, of its motion. If the motion isuniformly accelerated how far will it go in 7th second? [Ans. 35 m]

25. An electron is emitted with a velocity of 5 × 106 ms–1. It is accelerated by an electricfield in the direction of initial velocity at 3 × 1014 ms–2. If its final velocity is 7 × 106 ms–1;calculate the distance covered by the electron. [Ans. 0.04 m]

26. A bullet travelling with a speed of 350ms–1 enters a target and penetrates through adistance 5cm. Find the retardation produced in the bullet. [Ans. 1.22 × 106 ms–2]

27. An object is moving with uniform aceeleration. Its velocity after 5 second is 25 ms–1 andafter 8 sec, it is 34 ms–1. Find the distance travelled by the object in 12 th second.

[Ans. 44.5 m–1]

28. A particle starts with a velocity of 200 cms–1 and moves in a straight line with aretardation of 10 ms–2. Find the time it takes to describe 1500 cm. [Ans. 10 s or 30 s]

29. A man loses 20% of his velocity after running through 108 m. Prove that he cannot runmore than 192 m, further if his retardation is uniform.

30. A tired man has moved through a distance s in t minutes. If s = 80t – 5t2 find hisretardation. What is the time elapsed and distance covered before he comes to rest.

[Ans. 8s, 320 m]

31. A bullet loses 1

th20

of its velocity in penetrating through a plank of wood. How many

such planks would be required to bring the bullet to rest. [Ans. 11]

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

t = 3x where x is in m and t is in second calculate the displacement when velocityis zero. [Ans. 0 m]

33. The displacement of a particle varies with time

as x = t + 7.

Calculate the instantaneous velocity at t = 1 sec. [Ans. 16 ms–1]

34. A particle moves along x axis in such a way that its X coordinate varies with time t as

x = 2 – 5t + 6t2.

Find the initial velocity of the particle. [Ans. – 5 units]

35. The position of a particle is given by theequation x(t) = 3t3.

Find the instantaneous velocity at instant t = 2s. [Ans. 36 ms–1]

36. A particle is moving along x axis, its position varying with time as X (t) = 2t3 – 3t2 + 1.

(a) At what time is its velocity zero.

(b) What is the velocity when it passes through origin?

[Ans. (a) 0, 1 sec (b) 0, 4.5 ms–1]


37. A particle moves along a straight line such that its displacement at any time t is givenby

s = t3 – 3t2 + 2 m

What will be the displacement of the particle when the acceleration is zero. [Ans. 0 m]

38. The displacement s of a particle along x axis is given by

s = 6 + 8t + 7t2 m

Obtain its velocity and acceleration at t = 2s. [Ans. 36 ms–1, 14 ms–2]

39. A body travels 2 m in the first 2 s and 2.2 m in the next 4 s. What will be the velocity atthe end of the 7th second from the start? [Ans. 0.1 ms–1]

40. The coordinates of a moving particle at any time t are given by x = ct2, y = bt2. Calculate

the speed of the particle at time t. [Ans. 2 22t b c ]

BASED ON MOTION UNDER GRAVITY:41. A stone is thrown vertically upward with a velocity 24.5 ms–1.

(i) Calculate the height to which it rises.

(ii) Calculate also time it takes to reach the highest point. [Ans. 30.625 m, 2.5 s]

42. A body is thrown vertically upwards and rises to a height of 10 m. Calculate (i) thevelocity with which the body was thrown upward and (ii) the time taken by the body toreach the highest point. [Ans. 14 ms–1, 1.43 s]

43. A stone falls from the top of a building and travels 122.5 m in the last second before itreaches the ground. Find the height of the building. [Ans. 828.1 m]

44. A stone is allowed to fall from the top of a tower 100 m high and at the same timeanother stone is projected vertically upward from the ground with a velocity of25 ms–1. Find when and where the two stones will meet. [Ans. 4 sec, 21.6 m from ground]

45. A ball thrown vertically up returns to the thrower after 6 s. Find the velocity with whichit was thrown and its position after 4 s. [Ans. 29.4 ms–1, 4.9 m from top]

46. A rocket is fired vertically from the ground with a resultant vertical acceleration of 10ms–2. The fuel is finished in 1 min. and it continues to move up. What is the maximumheight reache? [Ans. 36.4 km]

47. A balloon is ascending at the rate of 14 ms–1 at a height of 98 m above the ground whenthe food packet in released from the balloon. After how much time and with whatvelocity does it reach the ground? g = 9.8 ms–2. [Ans. 6.12 s, 45.98 ms–1]

48. A stone is dropped from a balloon at a height of 300 m above the ground and it reachesthe ground in 10 seconds. What was the velocity of the balloon at the moment the stonewas dropped? [Ans. 19 ms–1 upward]

49. A ball is allowed to fall from a tower 80 m high. As soon as the ball reaches half the waythe acceleration due to gravity suddenly disappears. Discuss the condition of the ballafter this moment. [Ans. The ball will continue to move with velocity of 28 ms–1]

50. A stone is dropped from the top of a tower and is found to reach the ground in twoseconds. Calculate the height of the tower and the velocity with which it strikes theground? [Ans. 19.6 m, 19.6 ms–1]

51. A ball projected vertically upwards returns to the thrower after 10 seconds. Calculate(1) the maximum height to which it rises (2) the velocity with which it was projected(3) its position after 7 sec. [Ans. 122.5 m, 49 ms–1, 19.6 m from the top]


52. A stone is projected vertically upwards with a velocity of 50 ms–1. Calculate the distancemoved in the 3rd second. (g = 9.8 ms–2) [Ans. 25.5 m]

53. A stone is dropped from the top of a tower and one second later another stone isdropped from a balcony 20 m below the top. If both the stones reach the ground at thesame instant determine the height of the tower. (g = 10 ms–2) [Ans. 31.25 m]

54. A body falling freely from rest from the top of a tower describes 60.1 m in the last secondof its fall. Find the height of the tower. [Ans. 216 m]

55. A parachutist after bailing out falls 50 m, without friction. When the parachute opens hedecelerates downwards 2.0 ms–2. He reaches the ground with a speed of 3.0 ms–1.

(a) How long is the parachutist in the air?

(b) At what height did he bail out? [Ans. 17.46 m, 297.39 m]

PROBLEMS BASED ON VARIOUS GRAPHS56. A car accelerates from rest at the rate of 1 ms–2 for 5 sec and then retards at the same

rate till it comes to rest. Draw the v – t and a – t graph.

57. A train travels in 3 minutes a distance of 3.15 km from rest at one station to rest atanother station. It is uniformly accelerated for the 1st 30 seconds and uniformly retardedfor the last 15 seconds, the speed being constant for the remaining time. Find themaximum velocity; acceleration and retardation using graph.

[Ans. 20 ms–1, 0.667 ms–2, – 1.334 ms–2]

58. A rocket is launched from the earth surface whose velocity time graph is shown in figure.Calculate the maximum height attained by the rocket. [Ans. 60 km]

20 40 60 80 100 120 140t (s)

v (m /s)

1000

59. The velocity verses time graph of a body moving along a straight line is shown in figure.Calculate the ratio of displacement and distance covered by body in 5 sec.

[Ans. 3 : 5]

1 2 3 4 5t

v (m /s)

0

–1

–2

1

2

60. Figure shows the acceleration time graph for a particle in rectilinear motion. Calculate theaverage acceleration in first 20 s. [Ans. 30 ms–2]


10 20 30t (sec)

a (m s )–2

20

40

61. Figure shows the position time graph find

(i) Velocity from O to C

(ii) Displacement from C to D(iii) Velocity from D to E.

1 2 3 4 5 6t (sec)

s (m )

5

10

15

20

25

30

C D

A B

F

E

O

62. The speed time graph of an object moving along a fixed direction is shown in figure.Calculate

10 2 3 4 5 6 7t (sec)

v (m /s)

2

4

6

8

10

12

(i) The distance covered by the object in 7 sec.

(ii) The acceleration of object in last 5 sec. [Ans. 35 m, – 2 ms–2]


63. Figure represents the velocity time graph of a body moving in a straight line.

40 8 12 16 20t (sec)

v (m /s)

5

10

15

20 A B

E D C

(a) How much distance does it travel during its motion?

(b) How much distance does it travel during the last 10 second?

(c) Calculate the acceleration in the body during first four second.

[Ans. 260 m, 100 m, 5ms–2]

86

Scalar Quantities: A physical quantity which can be explained by its magnitude only andneed no direction is known as scalar quantity. e.g. distance, mass, time, speed.

Vector Quantities: A physical quantity which requires magnitude and a particular direction,when it is expressed e.g. displacement, velocity and acceleration.

(a) Equal vectors: Vectors which have equal magnitude and same direction are called equalvectors.

(b) Negative vectors: The negative of a vector is defined as a vector which is of samemagnitude but opposite in direction.

(c) Coplanar vector: Those vectors which are located in the same plane are called coplanarvectors.

(d) Collinear vectors: Vectors acting along same straight line are called collinear vectors.

(e) Null vector: A vector having zero magnitude is called null vector.

eq. A – A = O

.

(f) Unit vector: A vector having unit magnitude is called unit vector. It is represented by cap

(^) unit vector in the direction of A

is

a =A

A

Triangle law of vector Addition: According to triangle law of addition of vectors if twovectors are represented by the two sides of a triangle in same order then their sum or resultantvector is given by the third side of the triangle taken in opposite order.

R

= P Q

R = P + Q

P

Q

Parallelogram Law of Vector Addition: If two vectors acting at a point are represented bytwo adjacent sides of a parallelogram (both in magnitude and direction) then their resultant isrepresented completely both in magnitude and direction by the diagonal of the parallelogram,passing away through the common point.

��4UNIT

Vectors 87

D

AP

B

Q

C

R

AB BC

= AC

or P Q = R

The magnitude of the resultant vector is given by

R = 2 2P Q 2PQ cos

and tan =Q sin

P Q cos

where is the angle which R

makes with P

and is the angle between P

and Q

.

Subtraction of vectors: Subtraction of vectors B

from vector A

may be considered as

addition of vector – B

to A

i.e.

A B = A ( B)

.

A– B

= 2 2A B – 2ABcos .

and tan =B sin

A–Bcos

.

Resolution of vectors: Reverse process of addition of vectors is called resolution of vectors.When a vector in splitted into two components which are at right angle to each other, then thecomponents are called rectangular components.

Consider A

makes an angle with x axis. Ax and Ay are the rectangular components of A

along x and y axis respectively. OM is called projection of A

on x axis and it is represented byAx.

O

AyA

N

Ax MX

Y


ON is the projection of A

along y axis and it is represented by Ay.

According to vector addition.

A

= A Ax y

.

or A

= ˆ Â Ax yi j .Also Ax = A cos and Ay = A sin .

A = 2 2A Ax y

and tan =A

Ay

x.

A = ˆˆ Â A Ax y zi j k

In three dimensions A = 2 2 2A A Ax y z

Directional Cosine: Consider a vector A

which makes angle , and with x, y and z axisrespectively. The directional cosines of the given vector are given by

cos =A

Ax

cos =A

Ay

and cos = A

Az .

Scalar Product (Dot Product): The scalar or dot-product of two vectors is defined as theproduct of their magnitudes with cosine of the angle between them i.e.

A B = A B

cos .

Example: Work is dot product of force and displacement

i.e. W = F .d

Important: 1. Dot product is commutative i.e.

A B = B A

.

2. Dot product of two mutually perpendicular vectors is zero i.e.

A B = 0 if A B

.

3. For two parallel vectors = 0.

A B = AB.

4. A A = A2.

5. ˆ ˆˆ ˆ ˆ î i j j k k = 1.

6. ˆ ˆˆ ˆ ˆ î j j k k i = 0.

Vectors 89

7. A

= ˆˆ Â A Ax y zi j k , B

= ˆˆ ˆB B B .x y zi j k

A B = ˆˆ Â A Ax y zi j k ˆˆ ˆB B Bx y zi j k

= AxBx + AyBy + AzBz.

8. Let be the angle between two vectors A

and B

then

cos =A B

A B

Vector or Cross Product: The vector product of two vectors is also a vector.

Consider two vectors A

and B

then

C

= A B = Â B sin .n

where is the angle between A

and B

. n is a unit vector perpendicular to plane of A

and

B

.

Direction of n is given by right hand thumb rule.Example: 1. Moment of force is called torque given by

= Fr

.

2. Linear velocity and angular velocity are correlated as

V

= r .

Important: 1. Vector product of two parallel or anti-parallel vectors is zero.

2. A B = – B A .

3. ˆ ˆˆ ˆ ˆ î i j j k k = 0

4. ˆˆ ˆ ,i j k ˆˆ ˆ,j k i ˆ ˆ ˆ.k i j

5. Unit vector perpendicular to A

and B

is

n =A B

A B

6. A B =

ˆˆ Â A AB B B

x y z

x y z

i j k.

= ˆˆ ˆ(A B – A B ) – (A B – A B ) (A B – A B )y z z y x z z x x y y xi j k

7. The area of a parallelogram is given by the vector product of two vectors representing itsadjacent sides.


Area of parallelogram PQRS

= A B

P Q

RS

B

A

SOLVED EXAMPLES

BASED ON VECTOR ADDITIONExample 1. A girl travels 3 m due north and then 4m due west. Find her displacement.

Solution. The magnitude of the displacement is

BN

3 m5 m

W O

4 m

A

OB = 2 2OA AB

= 2 23 4 = 5 m.

Also tan =4

1.333 .

= tan–1 (1.33) = 53°.Displacement 5 m along N53° W.

Example 2. Figure shows a parallelogram ABCD. Prove that AC— —AC BD =

—2 BC

.

Solution.

A B

CD

Vectors 91

—AC

=— —AB BC ...(i)

—BD

=— —BC CD ...(ii)

Adding equation (i) and (ii)

— —AC BD =

— — — —AB BC BC CD

=— — —AB 2 BC CD

But—CD

=—

– AB

— —AC BD =

— — —AB 2 BC – AB =

—2 BC

.

Example 3. A particle has a displacement 5m towards north, 12m towards east and then 13mvertically upward. Calculate the magnitude of resultant displacement.

Solution. Here 1S��

= 5m towards North

2S��

= 12m towards East

3S��

= 13m vertically upward.

All the three displacements are mutually perpendicular to each other so magnitude ofresultant displacement is

S = 2 2 21 2 3S S S .

= 2 2 212 5 13

= 13 2 m .

Example 4. A man goes 10 m towards north then 20 m towards east. Calculate the magnitudeof displacement

Solution. Here S1 = 10 m due northS2 = 20 m due east.

S1 and S2 are perpendicular to each other so resultant displacement is

S = 2 21 2S S

= 2 210 20 = 22.5 m

Example 5. Two equal velocities have their resultant equal to either. What is the inclinationbetween them.

Solution. Here V1 = V2 = VAlso R = V.

From formula R = 2 2A B 2AB cos

V = 2 2V V 2 V V cos

or cos =1

–2

= cos 120°

or = 120°.


Example 6. Two forces whose magnitudes are in the ratio of 3:5 give a resultant of 35N. Ifthe angle of inclination be 60°. Calculate the magnitude of each force.

Solution. Let the two forces be 3F and 5F = 60°.R = 35 N

R = 2 2A B 2AB cos

35 = 2 2(3F) (5F) 2 (3F) (5F) cos 60°

= 2 2 19F 25F 2(3F) (5F) 2

=249 F

35 = 7F

or F = 5.

F1 = 3F = 15NF2 = 5F = 25N.

Example 7. Calculate the angle between 3N and 4N force so that their resultant is 5N.Solution. Here F1 = 3N, F2 = 4N, R = 5N

R = 2 2A B 2 AB cos

5 = 2 2(3) (4) 2(3 4) cos

25 = 25 + 24 cos or cos = 0 = cos 90° = 90°

Example 8. Prove that — — — — —AB BC CD DE EA = 0.

Solution. — — — — —BB BC CD DE EA

= — — — — —AB BC CD DE EA

= — — — —AC CD DE EA

= — — — —AC CD DE EA

= — — —AD DE EA

= — —

– EA EA

= 0 = R.H.S.Example 9. Establish following vector inequalities.

(i) a b | | | |a b

(ii) a b –a b

when does the equality sign apply?

A B

C

D

E

Vectors 93

Solution. (i) If be the angle between a

and b

then

a b =

2 2

2 cos .a b a b

for a b = to be maximum cos = 1 or = 0°.

max

a b =

2 2

2 cos0 .a b a b

=2 2

2a b a b

= a b

.

a b a b

.

Inequality sign (<) is applicable when > 0° i.e. when a

and b

are in the same dir.

(ii) Again a b =

2 2

2 cos .a b a b

The value of a b will be minimum when cos = – 1

or = 180°

min

a b =

2 2

– 2a b a b

=2

–a b

= –a b

Equality sign is applicable when = 180°.Example 10. Establish following vector inequalities:

(i) –a b

a b

(ii) –a b

a b

.

When does the equality sign apply?

Solution. Let be the angle between a

and b

. Then angle between a

and – b

is ( – )

–a b

= a b

.

=2 2

2 cos( )a b a b

=2 2

2 cosa b a b

cos ( – ) = cos (180 – ) = – cos .


(a) –a b

will be max. when

cos = – 1 i.e. = 180°.

–a b

= a b

Hencemax

–a b

a b

.

The equal sign is used when = 180°.

(ii) –a b

will be minimum when cos = 1 or = 0.

–a b

=2 2

– 2 cos0a b a b

.

= –a b

Hence –a b

–a b

.

The equality sign is used when = 0°.

Example 11. The resultant vector of P

and Q

, is R

. On reversing the direction of Q

, the

resultant vector becomes S

Show that R2 + S2 = 2(P2 + Q2).

Solution. Let be the angle between P

and Q

.

R = 2 2P Q 2PQ cos .

or R2 = P2 + Q2 + 2PQ cos ...(1)

When direction of Q is reversed let the resultant is S. S2 = P2 + Q2 + 2PQ cos (180 – )

S2 = P2 + Q2 – 2PQ cos . ...(2)

Adding Eq. (1) and (2)

R2 + S2 = 2(P2 + Q2).

Example 12. At what angle should the two forces 2P and P 2 act so that the resultant force

is equal to P 10 ?

Solution. Here F1 = 2P, F2 = P 2 , R = P 10 .

The resultant is given by

R = 2 2A B 2ABcos

P 10 = 2 2(2P) (P 2) 2 P 2 2P cos

10P2 = 4P2 + 2P2 + 24 2 P cos

Vectors 95

4P2 = 24 2 P cos .

cos =1

2or = 45°.

Example 13. Two equal forces act at a point. The square of their resultant in 3 times of theirproduct of two vectors. What is the angle between them?

Solution. Here A = F, B = F and R2 = 3F2.

from formula

R = 2 2A B 2ABcos

R2 = A2 + B2 + 2AB cos 3F2 = F2 + F2 + 2F2 cos

F2 = 2F2 cos

or cos =1

2 = cos 60°.

= 60°.

Example 14. Two vectors A

and B

are represented as

A

= ˆˆ ˆ10 –12 5i j k . B

= ˆˆ ˆ7 8 –12 .i j k Find the resultant vector, calculate the magnitudeand direction cosines.

Solution. A

= ˆˆ ˆ10 –12 5i j k

B

= ˆˆ ˆ7 8 –12i j k

R = A B = ˆˆ ˆ17 – 4 – 7i j k .

Also A B = 2 2 2(17) (4) (7) = 18.82 units

Direction cosines are given by

cos =R

Rx

= 17

18.82

cos =R

Ry

= – 4

18.82

cos =R

Rz

= – 7

18.82

Example 15. The x and y components of vector A

are 4 and 6 m respectively. The x and y

components of vector A B

are 10 and 9 m respectively. Calculate for the vector B

(a) its x and y component

(b) its length (c) the angle it makes with x axis.


Solution. Here Ax = 4 mAy = 6 m

(Ax + Bx) = 10 m(Ay + By) = 9 m

(a) Now Bx = (Ax + Bx) – Ax = 10 – 4 = 6 mBy = (Ay + By) – Ay = 9 – 6 = 3 m

(b) B = 2 2B Bx y = 2 2(6) (3) = 45 m

(c) = –1 Btan

By

x

= –1 3tan

6

= tan–1 (0.5) = 26.6°.

Example 16. The magnitude of two mutually perpendicular vectors A

and B

are 6 and 8

units respectively. Find the magnitude and direction of A – B

.

Solution. Here A

= 6 unit

B

= 8 unit

R = 2 2A B 2AB cos ( – )

= 2 2A B .

= 2 26 8= 10 units.

Tan =B

A =

8

6 =

4

3.

= –1 4tan 3 = 53°.

R

A

–B

BASED ON RELATIVE VELOCITYExample 17. Rain is falling vertically with a speed of 30 ms–1. A woman rides a bicycle with

a speed of 10 ms–1 in the north to south direction. What is the relative velocity of rain w.r.t. thewoman? What is the direction in which she should hold her umbrella to protect herself from therain?

B

A

Vectors 97

Solution.

— —– V OAw

= 10 ms–1 velocity of woman

—

RV OR

= 30 ms–1 velocity of rain

Vertica l

10 m s –1

30 m s –1

V wS

A

V R

R Q

PO – V w

Fig. 3.10

Velocity of rain w.r.t. woman

RV w

= RV – Vw

= RV + – Vw

VRw = 2 2(30) (10) = 10 10 = 31.6 ms–1.

It is represented by —OQ

. Which makes an angle with the vertical. Then

tan =QR

OR =

10

30 = 0.333

or = tan–1 (0.333) = 18° 26.The woman should hold her umbrella inclined at 18°26 with vertical toward South.

Example 18. In a harbour, wind is blowing at the speed of 72 km h–1 and the flag on the mastof a boat anchored in the harbour flutters along the N.E. direction. If the boat starts moving at aspeed of 51 km h–1 to the north, what is the direction of the flag on the mast of the boat?

Solution. The flag on the mast of boat anchored in the harbour flutters along N-E direction,therefore wind is blowing in N–E direction.

B A

E

D

S

C

OW

N

V B

VW

VR

–VB

45°


Vw

= 72 km h–1 (due N–E)

Velocity of boat BV

= 51 km h–1

Relative velocity of wind w.r.t. boat is given by

B BV V – Vw w

= BV – Vw

=—OD

(Vector Addition)

As shown in figure angle between —OA

and —OC

= 135°.

Vw

makes an angle with

—OD

.

Then

tan = B

B

V sin

V V cosw

=51sin 135

72 51 (cos 135)

=51sin 45

72 51 (– cos 45 ) =

15121

72 – 512

= 1.0037.

= tan–1 (1.0037) = 45.01°.

Angle w.r.t. east direction is 45.01 – 45 = 0.01°.

Hence the flag will flutter almost in the east direction.

Example 19. A ship is streaming towards east with a speed of 4 ms–1. A child runs across thedeck at a speed of 3 ms–1 in the direction at right angles to the direction of motion of ship i.e.towards North. What is the velocity of the child relative to the sea?

Solution. As the child is in ship, he has two velocities simultaneously (i) velocity equal tovelocity of ship towards east and (ii) velocity towards north direction. Velocity of child relative tosea

Q

North

3 m /s

R

OEast

P40 m /s

= Resultant velocity of child = OQ

Vectors 99

= 2 2OP PQ = 2 24 3 = 5 ms–1.

Let resultant velocity makes angle with the direction of ship.

tan = 34 = 0.75

or = tan–1 (0.75). (North of east).Example 20. A motarboat is racing towards north at 25 km h–1 and water current in that

regions is 10 km h–1 in the direction 60° east of south. Find the resultant velocity of the boat.

Solution. As shown in the figure the angle between Vb and

Vw = 120°.

120°

O

W E

60°

V b

N

V

V w

Resultant velocity is

VR = 2 2V V 2V V cosb w b w

= 2 2(25) (10) 2(25 10) cos 60

= 21.8 km h–1.

Let V

makes an angle with the dir. of Vb

then.

tan =V sin

V V cosw

b w

=10 sin 120

25 10 cos120

= 310 2

125 10 – 2

tan = 0.433

or = 23.4°.Example 21. A boat man can row with a speed of 10 km/h in still water. In which direction

should the boat man row in order to reach a point on the other bank directly opposite to the pointfrom where he started. The width of the river is 2 km.

Solution. Let Vb

is velocity of boat and Vr

is velocity of the river respectively.


B C

A

Vr

VVb

As shown in figure, to reach a point C on the other bank directly opposite to point A, the

boatman should travel along AB. From fig. sin = V

Vr

b

= 5

10 =

1

2 or = 30°.

Angle made by river current = 90 + = 90 + 30 = 120°.

BASED ON VECTORS IN TERMS OF BASE VECTORS

Example 22. Find a unit vector parallel to A

given by

A

= ˆˆ ˆ2 4i j k

Solution. Here A

= ˆˆ ˆ2 4i j k

A = 2 2 22 4 1 = 21.Using relation

A =A

A

= ˆ ˆ ˆ2 4

21

i j k

Example 23. Two forces 1F

= ˆˆ ˆ2 4i j k and 2F = ˆ ˆ2i j are acting simultaneously at apoint. What is the magnitude of the resultant force?

Solution. Here 1F

= ˆˆ ˆ2 4i j k

2F

= ˆ ˆ2i j

Resultant force F

= 1 2F F

= ˆˆ ˆ3 3 4i j k .

and F

= 2 2 23 3 4

= 9 9 16 = 34 N .

Example 24. If A

= ˆ ˆ3 4i j and B

= ˆ ˆ7 24i j find a vector having the same magnitude B

and parallel to A

.

Vectors 101

Solution. A

= ˆ ˆ3 4i j

A

= 9 16 = 5

Unit vector in the direction of A

.

A =A

A

= ˆ ˆ3 4

5

i j

Now B

= 2 27 24 = 25.

Required vector is ˆB.A.

=ˆ ˆ(3 4 )

255

i j

= ˆ ˆ15 20i j .

Example 25. A unit vector is represented by ˆˆ ˆ0.5 0.8i j ck . Calculate the value of c.

Solution. A

= ˆˆ ˆ0.5 0.8i j ck

A

is unit vector, the magnitude of A is should be 1.

1 = 2 2 2(0.5) (0.8) ( )c

or C = .0 11 .

Example 26. The coordinates of two points A and B respectively are (–1, 5, 7) and

(3, 2, – 5). Represent the vector AB��

.

Solution. Coordinate of point A and B are (–1, 5, 7) and (3, 2, – 5) respectively.

—AB

= 2 1 2 1 2 1ˆˆ ˆ( – ) ( – ) ( – )x x i y y j z z k

= ˆˆ ˆ(3) – (–1) (2) – (5) (–5) – (7)i j k

= ˆ ˆ ˆ4 – 3 – 12i j k

Example 27. Find a unit vector in the direction of sum of vectors

a

= ˆˆ ˆ– i j k

b

= ˆˆ ˆ2i j k .

Solution. a b = ˆ ˆˆ ˆ ˆ ˆ(– ) (2 )i j k i j k

= ˆˆ ˆ2 2i j k .


Unit vector in the direction of a b is

=ˆˆ ˆ2 2

1 4 4

i j k

= 1 ˆ ˆ ˆ( 2 2 )3

i j k .

BASED ON RECTANGULAR COMPONENTS OF VECTORSExample 28. One of the rectangular component of a velocity of 100 ms–1 is 50 ms–1.

Find other component.Solution. Here v = 100 ms–1

vx = 50 ms–1

vx = v cos

cos = xv

v =

50

100 =

1

2

or = 60°

vy = v sin

= 100 sin 60° = 100 3

2

= –150 3 ms .

Alternate MethodV2 = Vx

2 + Vy2

(100)2 = (50)2 + Vy2.

Vy = 2 2100 – 50 = 150 50

= –150 3 ms .

Example 29. An aeroplane takes off at angle of 60° to the horizontal. If muzzle speed of theplane is 400 km h–1. Calculate the horizontal and vertical components.

Solution. Here u = 400 km/h, = 60°

Here component is ux = u cos = 400 cos 60°

= 200 km/h.Vertical component is uy = u sin

= 400 sin 60°

=3

4002

= 200 3 km/h .

Example 30. i and j are unit vectors along x and y axis respectively. What is the magnitude

and direction of the vector ˆ ˆ( )i j and ˆ ˆ( – )i j .

Solution. A

= ˆ ˆi j

Vectors 103

B

= ˆ ˆ–i j

A

= 2 21 1 = 2 .

If 1 is the angle made by A

with x-axis then

tan 1 =A

Ay

x

= 1

1 = 1

or = 45°.

i.e. the magnitude of ˆ ˆi j is 2 units and its direction is inclined at 45° with x-axis.

Magnitude of B

is 2 21 (–1) = 2 .

If B

makes an angle 2 with x axis then

tan 2 =B

By

x

= –1

1 = 1 or 2 = 135°

i.e. magnitude of B

is 2 and it makes an angle of 135° with x-axis.

BASED ON DOT PRODUCT OF VECTORSExample 31. The magnitude of two vectors be 4 and 5 and the value of their scalar product

is 10. Find the angle between vectors.

Solution. Here A

= 4

B

= 5

A B = AB cos .

cos =A B

AB

= 10

4 5

=1

2 = cos 60°

or = 60°.

Example 32. A force F

= ˆˆ ˆ3 7 10 Ni j k acts on a body and the displacement of body is10 m along y axis. Calculate the work done.

Solution. Here F

= ˆˆ ˆ3 7 10i j k

S

= ˆ10 j

W = F S

= ˆˆ ˆ ˆ(3 7 10 ) 10i j k j = 70 J.


Example 33. Find the value of p so that vectors

A

= ˆˆ ˆ6 3 4i j k

and B

= ˆˆ ˆ– 2i pj k are mutually perpendicular.

Solution. Here A

= ˆˆ ˆ6 3 4i j k

B

= ˆˆ ˆ– 2i pj k

A

is perpendicular to B

A B = 0

ˆ ˆˆ ˆ ˆ ˆ(6 3 4 ) (– 2 )i j k i pj k = 0

– 12 + 3p + 4 = 0

or p = 83 .

Example 34. Find the angle between vectors

ˆˆ ˆA i j k , ˆˆ ˆB – 2 – 2 – 2i j k

.

Solution. Let be the angle between two vectors then

A B = AB cos

or cos =A A

AB

cos =2 2 2

ˆ ˆˆ ˆ ˆ ˆ( ) (– 2 – 2 – 2 )

3 2 2 2

i j k i j k

=– 2 – 2 – 2

3 12 = – 1

cos = – 1 = cos 180°

or = 180°.Example 35. The sum and difference of two vectors are equal in magnitude. Calculate the

angle between them.

Solution. We consider two vectors A

and B

.

According to problem A B = A – B

.

Squaring both sides2

A B =

2

A – B

A A B B 2 A B = A A B B– 2 A B

Vectors 105

2 2A B 2 A B

= 2 2A B – 2 A B

or 4 A B = 0

4 0

A B = 0

A

B

.

i.e. angle between A

and B is 90°.

Example 36. What are the components of vector A

= ˆ ˆ2 3i j along direction ˆ ˆ( )i j ?

Solution. Component A

along the direction of B

is given by

A cos B .

B is unit vector along B

and is the angle between A

and B

cos =A B

A B

.

A cos B =A B B

AA B B

.

= 2

A B B

B

.

Here A

= ˆ ˆ2 3 ,i j B

= ˆ ˆi j .

A cos B =

2ˆ ˆ ˆ ˆ(2 3 ) ( ) ˆ ˆ( )

1 1

i j i ji j

=2 3 ˆ ˆ( )

2i j

= ˆ ˆ5( )

2i j .

Example 37. Three vectors A, B

and C

are such that C

= A B and their magnitudes are

3, 4 and 5 units find the angle between A

and C

.

Solution. A

= 3, B

= 4, C

= 5

Now A B = C

B

= C – A

.


Squaring both sides,

B2 = C2 + A2 – 2 A C

(4)2 = (5)2 + (3)2 – 2AC cos 16 = 25 + 9 – 2(3) (5) cos

– 18 = – 30 cos

cos =18 3

30 5 = 0.6

or = cos–1 (0.6)

= 53°.Example 38. A particle moves from (5, – 3, – 4) m to the point (–1, 3, 4) m under the

influence of a force ˆˆ ˆF – 3 – 2i j k N. Calculate the work done by the force.

Solution. 1r

= ˆˆ ˆ5 – 3 – 4i j k

2r

= ˆˆ ˆ– 3 4i j k

Displacement of particle = 2 1–r r

= ˆ ˆˆ ˆ ˆ ˆ(– 3 4 ) – (5 – 3 – 4 )i j k i j k

= ˆˆ ˆ– 6 6 8i j k .

W = F S

= ˆ ˆˆ ˆ ˆ ˆ(–3 – 2 ) (– 6 6 8 )i j k i j k

= 18 – 6 + 16

= 28 J.

Example 39. If unit vectors a and b are inclined at angle then prove that

–a b

= 2 sin2

.

Solution.2ˆˆ –a b = ˆ ˆˆ ˆ( – ) ( – )a b a b

= ˆ ˆ ˆ ˆ ˆˆ ˆ ˆ– –a b a b b a b b

= ˆˆ1 – 2 1a b = 2 – 2 × 1 × 1 × cos = 2 (1 – cos )

= 22 2 sin2

= 24 sin2

Vectors 107

or ˆˆ –a b = 2 sin2

.

Example 40. Prove that A

= ˆ ˆ– 2i j and B

= ˆˆ ˆ2 8i j k are perpendicular to each other.

Solution. Here A

= ˆ ˆ– 2i j

B

= ˆˆ ˆ2 8i j k

A B = ˆˆ ˆ ˆ( – 2 ) (2 8 )i j i j k

= 2 – 2 + 0

= 0

A B = 0 A B

.

Example 41. A force of ˆ ˆ3 4i j N makes a body move on a rough plane with a velocity

ˆˆ ˆ6 5 2i j k ms–1. Calculate the power in watt.

Solution. Here, F

= ˆ ˆ3 4i j N

V

= –1ˆˆ ˆ6 5 2 msi j k

Power = F V

= ˆˆ ˆ ˆ ˆ(3 4 ) (6 5 2 )i j i j k = 18 + 20

= 38 Watt.

Example 42. If A B C and 2 2 2A + B = C then prove that A

and B

are perpendicular

to each other.

Solution. A B = C

A B A B

= C

C

2 2A 2 A B B

= C2

But A2 + B2 = C2

2 A B = 0

or A B = 0

or A

B

.

BASED ON CROSS PRODUCT OF VECTORS

Example 43. Prove that the vectors A

= ˆˆ ˆ2 2 2i j k and B

= ˆˆ ˆ– 2 – 2 – 2i j k are parallel.

Solution. Here A

= ˆˆ ˆ2 2 2i j k


B

= ˆˆ ˆ– 2 – 2 – 2i j k

A B =

ˆˆ ˆ2 2 2

–2 –2 –2

i j k

= ˆˆ ˆ(– 4 4) (– 4 4) (– 4 4)i j k

= 0

Here A B = 0

A ||

B

Example 44. If A B = 3 A B

calculate the value of A B

.

Solution. Given A B = 3 A B

AB sin = 3 AB cos

or tan = 3 = tan 60°.

= 60°.

A B = 2 2A B 2 AB cos

= 2 2A B 2 AB cos 60

= 2 2A + B + AB .

Example 45. If ˆˆ ˆ2 – 3 4i j k and ˆˆ ˆ3 μi j k be collinear vectors then find the value of and μ.

Solution: a

= ˆˆ ˆ2 – 3 4i j k

b

= ˆˆ ˆ3 μi j k

Since a

and b

are collinear vectors therefore,

a b = 0

ˆˆ ˆ2 – 3 43 μ

i j k

= 0

or ˆˆ ˆ(–3μ – 4 ) (12 – 2μ) (2 9)i j k = 0.

Equating coefficient of ˆ ˆi j and k on both sides

– 3μ – 4 = 0

Vectors 109

12 – 2μ = 0

2 + 9 = 0

i.e. =9

–2

and μ = 6.

Example 46. Find a unit vector perpendicular to both the vectors

A

= ˆˆ ˆ3 2 ,i j k , B

= ˆˆ ˆ2 – 2 4i j kSolution.

A B =

ˆˆ ˆ3 1 22 – 2 4

i j k

= ˆˆ ˆ(4 4) (4 –12) (– 6 – 2)i j k

= ˆˆ ˆ8 – 8 – 8 .i j k

A B = 2 2 28 8 8 = 8 3.

Unit vector to both A

and B

is

n =A B

A B

=

ˆˆ ˆ8 – 8 – 8

8 3

i j k

= ˆ ˆ ˆ( )– –3

i j k .

Example 47. Calculate the area of the parallelogram whose two adjacent sides are formed bythe vectors.

A

= ˆˆ3 4i k B

= ˆˆ2 .j k

Solution. Area of a prallelogram = A B

A B =

ˆˆ ˆ3 0 40 2 1

i j k

= ˆˆ ˆ(0 – 8) (0 – 3) (6 – 0)i j k

= ˆˆ ˆ–8 – 3 6 .i j k

A B = 2 2 28 3 6 = 64 9 36

= 109 units .


Example 48. Find a vector whose length is 14 and which is perpendicular to each of twovectors.

A

= ˆˆ ˆ2 – 3 6i j k , B

= ˆˆ ˆ – .i j kSolution.

A B =

ˆˆ ˆ2 –3 61 1 –1

i j k

= ˆˆ ˆ(3 – 6) (6 2) (2 3)i j k

= ˆˆ ˆ– 3 8 5 .i j k

A B = 2 2 2(3) (8) (5) = 98 = 7 2.

Unit vector to the plane containing A B

=ˆˆ ˆA B – 2 8 5.

7 2A B

i j k

Vector in the direction perpendicular to A

and B

having magnitude 14.

=ˆˆ ˆ– 2 8 5

147 2

i j k

= ˆ ˆ ˆ( )2 – 2 8 5i j k .

Example 49. In any triangle ABC prove that

sin A sin B sin C

a b c .

Solution. ,a b

and c

are cyclic vectors. Therefore

A

B Ca

c b

a b c = 0

a b = – c

( )a b c = – c c

0c c

Vectors 111

or ( ) ( )a c b c = 0

– ( )c a b c = 0

b c = c a

...(i)

Similarly, a b = b c

...(ii)

A

B C

c b

a

From equation (i) and (ii)

a b = b c

= c a

.

or a b = b c

= c a

ab sin (180 – C) = bc sin (180 – A) = ac sin (180 – B)

ab sin C = bc sin A = ac sin B.

orsin C

c =

sin A

a =

sin B

b.

orsin A sin B sin C

a b c .

EXERCISE

BASED ON VECTOR ADDITION OR SUBTRACTION1. A particle has a displacement of 5 m towards east, and 12 m towards south. Finally the

displacement is 6 m vertically upward. Find the magnitude of resultant displacement.

[Ans. 14.32 m.]

2. ABCDEF is a regular hexagon.

Prove. — — — — —AB AC AD AE AF =

—6 AO

A B

C

D

OF

E


3. At what angle do the two forces P Q

and P Q

act so that the resultant is

2 23P Q . [Ans. 60°]

4. A man walks from a point 2 km east, then he walks 4 km along a distance east north andfinally he walks 3 km towards west. Find the displacement from the starting point.

[Ans. 3.4 km, 55° North of east]

5. Two forces of 3 N and 9 N act on a particle at an angle 60°. Find the magnitude anddirection of resultant force. [Ans: 10.8 N at 46° with the direction of 3 N]

6. While travelling from one station to another, a car travels 75 km north, 60 km north east and20 km east. Calculate the magnitude of the displacement. [Ans. 133 km]

7. Two forces whose magnitudes are in the ratio of 3:5 give a resultant of 35 N. If the angle ofinclination be 60°, calculate the magnitude of each force. [Ans. 15 N, 25 N]

8. A body is acted upon by two forces 30 N due east and 40 N due north. Calculate themagnitude and direction of resultant force.

[Ans. 50 N, 53° 8 with the direction of 30 N force]

9. The maximum and the minimum resultant force of two forces acting at a point are 29 N and5 N respectively. If each force is increased by 3N find the resultant of two new forces actingat right angle to each other.

10. A vector A

and B

make angle of 20° and 110° respectively with the x-axis. The magnitudesof these vectors are 5 m and 12 m respectively. Find the resultant vector.

[Ans. 13 m, the resultant vector makes angle –1 12tan

5

with A

]

11. The resultant of two forces A

and B

is a force of 12 N. The direction of resultant force

makes an angle of 60° with the direction of A

. If the magnitude of A

is 8 N. Find the

magnitude of B

. [Ans. 10.6 N]

12. The resultant of two equal vectors acting at right angle to each other is 1414 dyne. Find themagnitude of either force. [Ans. 1000 dyne]

13. Determine the vector which added to the resultant of ˆˆ ˆA 4 – 3i j k and ˆˆ ˆB 2 4 ,i j k

gives unit vector k . [Ans. ˆ ˆ– 2 – 6i j ]

14. Find the resultant of velocities, one is 6 ms–1 due east and the other is 8 ms–1 due north.

[Ans. 10 ms–1, 53.8 with 6 ms–1 velocity]

15. The resultant of two forces P

and Q

is of magnitude R. Show that if P is doubled, Q

remaining unaltered then the new resultant will be of magnitude

2 2 22P 2R – Q .

16. Two forces 3 N and 4 N act at a point at an angle of 90° with each other. Determine themagnitude and direction of the resultant force by drawing a vector diagram.

[Ans. 5 N, 53° from 3 N force]

Vectors 113

BASED ON RELATIVE VELOCITY17. A river 800 m wide flows at the rate of 5 km h–1. A swimmer who can swim at 10 km h–1 in

still water, wishes to cross the river straight. (i) Along what direction must he strike?(ii) What should be his resultant velocity (iii) How much time he would take?

[Ans. 30°, 2.4 ms–1, 333.3 s]

18. On a certain day, rain was falling vertically with a speed of 35 ms–1. Wind starts blowingafter some time with a speed of 12 ms–1 in east to west direction. In which direction shoulda boy waiting at a bus stop hold his umbrella? [Ans. 37 ms–1, 19°]

19. To a driver going east in a car with a velocity of 40 km h–1, a bus appears to move towards

north with a velocity of 40 3 km h–1. What is the actual velocity and direction of motion of

bus? [Ans. 30° east of north]

20. A plane is travelling eastward at a speed of 500 km h–1. But wind is blowing at 90 km h–1

southward. What is the direction and speed of the plane relative to ground.

[Ans. 10.2° south of east, 508 km h–1]

21. A train is moving with a velocity of 54 km h–1 due north and a bus is moving with a velocityof 36 km h–1 due west. What is the relative velocity of bus w.r.t. train. [Ans. 18 ms–1]

22. A man is walking due east at the rate of 3 km h–1. Rain appears to fall down vertically at therate of 3 km h–1. Calculate the actual velocity and direction of rainfall.

[Ans. 3 2 km h–1, 45° east of vertically]

23. A bird is flying due east with a velocity of 8 ms–1. The wind starts to blow with a velocity of6 ms–1 due north. What is the magnitude of relative velocity of bird w.r.t. wind? Also give itsdirection. [Ans. 5 ms–1, 37°]

BASED ON VECTORS INTERMS OF BASE VECTORS

24. Find a unit vector parallel to A

given by

A

= ˆˆ ˆ2 2 3i j k . [Ans. ˆˆ ˆ2 2 3

17

i j k ]

25. A

= ˆˆ ˆ0.3 0.4i j ck calculate the value of c if A

is a unit vector. [Ans. 0.87]

26. Given A

= ˆˆ ˆ3 – 4 – 5i j k , ˆˆ ˆB 2i j k . Find A B

and A – B

.

[Ans. ˆ ˆˆ ˆ ˆ ˆ4 – 2 – 4 , 2 – 6 – 6i j k i j k ]

27. The x and y component of A

are 4 m and 6 m respectively. The x and y component of

A B

are 10 and 9 m respectively. Calculate x and y component of B

. Also calculate its

length. [Ans. Bx = 6 m, By = 3 m, B = 45 m ]

28. If A

= ˆ ˆ3 4 ,i j B

= ˆ ˆ7 24i j find a vector having same magnitude as B

and parallel to

A

. [Ans. ˆ ˆ15 20i j ]


29. A vector P

when added to the resultant of A

= ˆˆ ˆ2 – 5 6i j k and B

= ˆˆ ˆ3 4 – 2i j k gives a

unit vector along x axis. Find the value of P

. [Ans. ˆˆ ˆ– 4 – 4i j k .]

30. Find the vector —AB

and its magnitude if it has initial point A (1, 1, – 1) and final point

B (3, 2, 2). [Ans. ˆˆ ˆ2 3 , 14i j k ]

31. Find a unit vector parallel to the resultant of A

= ˆˆ2i k , B

= ˆˆ2i j k .

[Ans. ˆˆ ˆ(4 2 )21

i j k ]

BASED ON RECTANGULAR COMPONENTS OF VECTOR32. One of the rectangular components of a vector of 100 unit is 50 unit. Find the other

component. [Ans. 50 3 units ]

33. A velocity of 40 ms–1 has Y component –120 2 ms . Calculate its X component.

[Ans. –120 2 ms ]

34. A force is inclined at 50° to the horizontal. If its rectangular components in the horizontaldirection be 50 N, find the magnitude of the force and its vertical component (Given cos 50°= 0.6428, sin 50° = 0.7660). [Ans. 77.78, 59.58 N]

35. A velocity of 10 ms–1 has y component –15 2 ms . Calculate its x component.

[Ans. –15 2 ms ]

BASED ON SCALAR OR DOT PRODUCT OF TWO VECTORS

36. Prove that A 2 B

2 A – 3B

= 2A2 + AB cos – 6B2.

37. Show that ˆˆ Â 2 2i j k and ˆˆ ˆB – – 4i j k

are perpendicular to each other.

38. Prove that vectors ˆˆ Â 2 4 6i j k and ˆ ˆB 4 – 2i j

are perpendicular to each other.

39. Calculate the value of P

so that ˆˆ Â 2i j pk and ˆˆ ˆB 2 3 –i j k

are perpendiculat to

each other. [Ans. p = 7]

40. If unit vectors a and b are inclined at angle then prove that ˆˆ –a b = 2 sin .2

41. Find the angle between the vectors ˆˆ Â 2 –i j k and ˆˆ ˆB – – 2 .i j k

[Ans. 60°.]

42. Two vectors of magnitude 4 and 5 units are inclined at an angle of 60° calculate themagnitude of their dot product. [Ans. 10 units]

43. If A B = A – B

find the angle between A

and B

. [Ans. 90°]

Vectors 115

44. A body constrain to move along x axis of coordinate system is subjected to a constrant force

F

= ˆˆ ˆ4 3 Ni j k .

What is the work done by this force in moving the body over a distance of 10 m along x axis.[Ans. 40 J.]

45. A force F

= ˆˆ ˆ4 3i j k Newton acts on a body of mass 10 kg and displaces it from the

position ˆˆ ˆ6 5 – 3i j k to ˆˆ ˆ10 3 4 .i j k Calculate the work done. [Ans. 17 J]

46. A vector P, Q and R

have magnitude 6, 8 and 10 units P Q = R . Find the angle between

Q

and R

. [Ans. = cos–1 4/5]

47. Show that the vectors A

= ˆˆ ˆ3 – 2i j k , B

= ˆˆ ˆ– 3 5i j k and C

= ˆˆ ˆ2 – 4i j k form a rightangle triangle.

BASED ON CROSS PRODUCT

48. Find a unit vector perpendicular to both ˆˆ ˆA = 2i j k

and ˆˆ ˆB = – 2i j k

.

[Ans. ˆˆ ˆ(5 – – 3 )35

i j k ]

49. Given A B = 3 A B

. Find angle between A

and B

. [Ans. 60°.]

50. Show that vectors A

= ˆˆ ˆ2 – 3 –i j k

B

= ˆˆ ˆ– 4 6 2i j k are parallel.

51. Find the area of a parallelogram whose adjacent sides are ˆˆ ˆ2 3i j k and ˆˆ ˆ– 2 .i j k

[Ans. 84 units ]

52. If A

= ˆ ˆ3 4i j

B

= ˆ ˆ7 24i j . Find a vector having same magnitude as B

and parallel to A

.

[Ans. ˆ ˆ15 20i j .]

53. Find A B if A

= 10

B

= 2 and A B = 16. [Ans. 12]

54. Prove that A B A – B

= 2 B A

.

55. Find the area of a parallelogram whose two adjacent sides are represented by two vectors.

ˆˆ ˆA i j k ˆˆ ˆB – 3 .i j k

[Ans. 4 2 square units]


56. Find a vector whose length is 5 and which is perpendicular to each of the vectors.

A

= ˆˆ ˆ– 3 6i j k

B

= ˆˆ ˆ – .i j k

[Ans. �

5 ˆˆ( 3 7 4 )74

i j k ]

57. The diagonals of a vector are given by the vectors ˆˆ ˆ2 3i j k and ˆˆ ˆ– 2 2 .i j k Find the area

of the parallelogram. [Ans. 5.52 sq. units]

58. Determine a unit vector perpendicular to both ˆˆ ˆA 2i j k and ˆˆ ˆB – 2i j k

.

[Ans. 1 ˆˆ ˆ( – – )3

i j k ]

59. Prove that the vectors ˆˆ ˆA 4 3i j k and ˆˆ ˆB 12 9 3i j k

are parallel to each other.

60. For two vectors can A B = 0 and A B

= 0 Simultaneously.

61. A force ˆˆ ˆF – 3 5i j k acts at a point having position vector ˆˆ ˆ7 3 .r i j k

What is the

torque? [Ans. ˆˆ ˆ–14 38 –16i j k ]

62. Given A B = 3 A B

then calculate the value of A B

. [Ans. (A2 + B2 + AB)½]

117

GROUND TO GROUND PROJECTION: Consider a projectile projected into space withconstant speed u, at an angle to horizontal. The projectile moves under the effect of gravityalone.

The path followed by a projectile is called its trajectory which is always a parabola.

Component of initial velocity ux = u cos , uy = u sin Component of acceleration at any instant ax = 0, ay = g.

O

u sin

Y

u A

u cos BX

(a) Equation of trajectory of projectile.

y = 22 2

tan – .2 cos

gx x

u

= tan 1 – .Rxx

(b) Coordinate of projectile after time ‘t’.

x = uxt

y = 21– .

2yu t gt

net displacement in time t

s = 2 2 .x y

(c) Time of flight T =2 sinu

g

(d) Maximum Height H =2 2sin

2u

g

��

��

�� 5

UNIT


(e) Range R =2

sin 2u

g

for range to be max sin 2 = 1 = sin 90°i.e. = 45°

Rmax =2

.u

g

(f) Velocity of projectile at time ‘t’

ux = u cos uy = u sin – gt.

V = 2 2x yu u = 2 2 2 – 2 sinu g t u gt

Let velocity V makes an angle with ground then

Tan =V

Vy

x

= sin –

cosu gt

u

.

PROJECTILE THROWN PARALLEL TO HORIZONTALIn horizontal direction

ux = u ax = 0

In vertical direction

uy = 0 ay = g

O

H

Y

u

X

V x

V y

(a) Equation of Trajectory

y = 22

.2

gx

u

(b) Vx = ux = u

Vy = uy = gt

Resultant Velocity is u = 2 2V Vx y

= 2 2 2 .u g t

Motion in Two and Three Dimensions 119

Also tan =V

Vy

x =

gtu

.

(c) Displacement

s = 2 2x y

when x = ut

and y = 21

2gt

also�

s = 21ˆ ˆ( ) .2

ut i gt j

(d) Time of flight

t =2h

g

(e) Range =2h

ug

OBLIQUE PROJECTION FROM HEIGHT HHere ux = u cos , ax = 0

uy = u sin , ay = g

x = u cos t

h = 21

2ut gt

H = 21– sin .

2u gt

gt2 – 2 u sin – 2H = 0.

H

u sin u

u cos

or t =22 sin (2 sin ) 8Hg

2

u u

g


=2

sin sin 2Hu u

g g g

.

UNIFORM CIRCULAR MOTION: When a body moves on a circular path with constantspeed its motion is called uniform circular motion.

Angular Velocity: Angular displacement of the body per unit time is called angular velocityi.e.

= .d

dt

If is the angle swept out by radius vector in time t then

= .t

Also = 2

T

where T is time period of one revolution.

S.I. unit radian per second.

Relation Between v and v = r.

or in vector formv =

� �

r .

Angular Acceleration: Rate of change of angular velocity is called angular acceleration i.e.

=0

Lim .t

d

t dt

S.I. unit radian/sec2.

Relations Between Angular and Linear Accelerationa = r

Linear acceleration = radius of circular path × angular acceleration.

CENTRIPETAL ACCELERATION: A particle moving with constant speed on a circularpath, is acted upon by an acceleration which is directed towards the centre of circular path. Thisacceleration is called centripetal acceleration. It is given by

ac =2

r

v = r2.

Important. Centripetal acceleration is always perpendicular to the velocity or displacementat each point, so that work done by centripetal force is always zero.

BASED ON MOTION IN A PLANEExample 1. A cyclist travels from centre O of a circular park of radius 1 km and reaches

point P. After cycling along ¼ of the circumference along PQ, he returns to the centre of the parkalong QO. If the total time taken is 10 minutes calculate (i) net displacement (ii) average velocityand (iii) average speed of the cyclist.

Solution. (i) Net displacement is zero as initial and final position are same.


(ii) Average velocity = Total displacement

Total time taken

But displacement is zero.

avV��

= O

(iii) Total path covered

= (OP + PQ + QO)

=2

4

rr r

=2 22 1

1 17 4

=25

km7

Time taken = 10 min.

=1

hr6

.

Vav =Total path covered

time taken = 25

71

6 = 21.4 km/h

Example 2. An air craft is flying at a height of 3400 m above the ground. If the anglesubstended at a ground observation point by the aircraft position 10 s apart is 30. What is the speedof the aircraft?

Solution. A is initial position of air craft and B is final position.

AB = ?

OC = 3400 km

Let AC = x

3400

x= tan 15°

x = 3400 × 0.2679

= 910.86 m

AB = 2x = (2 × 910.86) m

speed =AB

t =

2 910.86

10

= 182.2 ms–1

Example 3. On an open ground, a motorist follows a track that turns to his left by an angleof 60° after every 500 m. Starting from a given turn, specify the displacement of the motorist atthe third, sixth and eight turns. Compare the magnitude of the displacement with the total pathlength covered by the motorist in each case.

15°

30°

0

A C B

O P

Q


Solution. Let he starts from point A

In figure

AC = 2 2A B 2AB cos

= 2 2(500) (500) 2(500) (500) cos 60

= 500 3 m.

A B

C

DE

F

30°

30°

60°

60°

By geometry ACD = 90°

AD = 2 2AC CD

= 2 2(500 3) (500) = 500 × 2

= 1000 m.

At third turn

distance = 500 × 2 = 1000 m.

displacement = AC = 500 3 m .At sixth turn

distance = 500 × 6 = 3000 m.displacement = (zero).

At eight turndistance = 500 × 8 = 4000 m.

displacement = AD = 1000 m.Example 4. The position of a particle is given by

r

= 2 ˆˆ ˆ3.0 – 2.0 4.0 mt i t j k Where t is in second and the coefficients have proper units (b) what is the magnitude and

direction of velocity of the particle at t = 2s.

Solution. Here r

= 2 ˆˆ ˆ3.0 – 2.0 4.0t i t j k

v (t) =

d r

dt

�

= �2 ˆˆ[3.0 – 2.0 4.0 ]

dt i t j k

dt

= ˆ ˆ3.0 – 4.0 m/si t j


Also ( )a t�

= d v

dt

�

= ˆ ˆ(3.0 – 4.0 )d

i t jdt

= –2ˆ– 4.0 msj

(b) At t = 2s

v�

= ˆ ˆ3.0 – 8.0i j

The magnitude of velocity is

v = 2 2x yv v = 2 2(3) – (–8) = –173 8.54 ms

The direction of velocity is given by

= –1 –1 –8tan tan 3y

x

v

v

= – tan–1 (2.6667)

= – 70° with x axis.

GROUND TO GROUND PROJECTIONExample 5. A cricketer can throw a ball to a maximum horizontal distance of 100 m. How

high above the ground can the cricketer throw the same ball?

Solution. Max horizontal distance Rmax = 2u

g.

i.e. Rmax =2

100u

g (1)

To throw the body upward

V2 = u2 + 2gh

0 = u2 – 2gh

or h =2 21 1

1002 2 2

u u

g g

= 50 m

Example 6. The ceiling of a long hall is 25 m high. What is the maximum horizontal distancethat a ball thrown with a speed of 40 ms–1 can go without hitting the ceiling of the hall?

Solution. Here H = 25 m, u = 40 ms–1.

Max Height H =2 2sin

2

u

g

or sin2 = 2 2

2H 2 25 9.80.306

(40)

g

u

or sin = .306 .554


Range is given by

R =2

(2sin cos )u

g

=2(40)

2 (0.554) 1– (0.306)9.8

2cos 1– sin

= 150.7 mExample 7. A projectile is projected at 45° with horizontal. Show that the horizontal range

is four times the maximum height.

Solution. Here = 45°

Rmax =2u

g.

Maximum Height =2 2sin

2

u

g

H =2

2(sin 45 )2

u

g

=2 21 1

2 2 4

u u

g g

Rmax =2u

g

H = maxR

4

or Rmax = 4H.Example 8. Show that the range of a projectile remains same when it is projected at an angle

or (90 – ) with horizontal.

Solution: Range of projectile, when projected at an angle with horizontal is

R =2

sin 2 .u

g

Now angle of projection is (90 – )

R =2

sin 2(90 – )u

g

=2

sin 2(180 – 2 )u

g


=2

sin 2 Ru

g

R = R.Example 9. The range of projectile when launched at an angle of 15° with horizontal is 1.5

km. What is the range of the projectile when launched at an angle 45° to horizontal.

Solution. Here 1 = 15° R1 = 1.5 km

2 = 45° R2 = ?

Range R1 =2

1sin 2u

g

1.5 =2

sin 2(15)u

g or

2

3.0u

g

Now R2 =2

2sin 2u

g

=2 2

sin 2(45 ) 3.0 km.u u

g g

Example 10. Two projectiles are thrown with equal speeds at angle and (90 – ) thencalculate the ratio of their (i) time of flights, (ii) maximum heights of projectile.

Solution. Time of flight when angle of projection is

T =2 sin

.u

g

Time of flight when angle of projection is (90 – )

T =2 sin (90 – ) 2 cosu u

g g

T

T=

sin

cos

= tan .

(ii) Max. height of projectile at projection angle

H =2 2sin

2

u

g

Max. height of projectile at projection angle (90 – )

H =2 2sin (90 – )

2

u

g

=2

2cos .2

u

g

H

H= tan2 .


Example 11. A ball is thrown at angle and another ball is thrown at angle (90 – ) with thehorizontal direction from the same point with velocity 40 ms–1. The second ball reaches50 m higher than the first ball. Find their individual heights. g = 10 ms–2.

Solution. For the first ball angle of projection = Velocity of projection u = 40 ms–1

Max height =2 2sin

2

u

g

i.e. h =2 2(40) sin

2 10

...(i)

For second ball, angle of projection = (90 – )

Velocity of projection u = 40 ms–1

Max. height reached = (h + 50) m.

(h + 50) =2 2sin (90 – )

2

u

g

=

22 cos

(40)2 10

...(ii)


2h + 50 =2(40)

2 10 = 80

or h =80 50

215 m.

Example 12. Find the angle of projection for which horizontal range and the maximumheight are equal.

Solution. Horizontal range =2

sin 2 .u

g

Maximum height =2 2sin

2

u

g

Horizontal range = Maximum Height.

2

sin 2u

g =

2 2sin

2

u

g

2 sin cos =2sin

2

or tan = 4

= 75° 58.

Example 13. A ball is projected with a velocity of –140 2 ms at an angle of 45°. Find the

position and velocity of ball after 2 sec.


Solution. Here u = –140 2 ms , = 45°

ux = u cos 40 2 cos 45° = 40 ms–1

uy = u sin = 40 2 sin 45° = 40 ms–1.

After 2 sec

ux = 40 ms–1.

for vertical component.

v = u + gt

vy = 40 – 9.8 × 2

= 20.4 ms–1.

2 2x yv v v = 2 2(40) (20.4)

= 44.90 ms–1.

Also tan =y

x

v

v =

20.4

40 = 0.51

= tan 27.01°

= 27.01°In 2 sec

Horizontal displacement isx = 40 × 2 = 80 m.

Vertical displacement is

y = 40 × 2 – 1

9.8 2 22

= 60.4 m.

BASED ON PROJECTILE FIRED HORIZONTALLYExample 14. From the top of a building 49 m high a projectile is projected horizontally with

a velocity of 9.8 ms–1. Calculate (1) the time taken by projectile to reach the ground (2) horizontalrange.

Solution. Here u = 9.8 ms–1 h = 49 m.

(i) Time of flight =2h

g

=2 49

9.8

= 3.16 sec.

(ii) Horizontal range

R = u × T

= 9.8 10 = 31m.


Example 15. A body is projected horizontally from the top of a tower and strikes the groundafter three seconds at an angle of 45°, with the horizontal. Find the height of the tower and thespeed with which the body was projected (g = 9.8 m/s2).

Solution. Here g = 9.8 ms–2, u = 0, t = 3 s.

h = 21

2ut gt

=1

0 9.8 3 3.2

= 44.1 m.(b) Vertical component of velocity is

vy = 0 + 9.8 × 3= 29.4 ms–1.

Horizontal component vx = u

tan =y

x

v

v

But = 45°

tan 45° =29.4

u

or u = 29.4 ms–1.Example 16. A projectile is projected horizontally with a velocity 98 ms–1 from the top of a

tower 490 m high. Find (i) the time taken to reach the ground, (ii) the distance of the point fromthe tower where projectile will hit the ground, (iii) the velocity with which the projectile hits theground.

Solution. Here, u = 98 ms–1.

h = 490 m.

(i) Initial velocity in

vertical direction = 0

h =21

2ut gt

490 = 210 9.8 .

2t

t2 = 100or t = 10 sec.(ii) OA = Horizontal velocity × time

= 98 × 10= 980 ms–1.

(iii) vx = 980 m/svy = u + gt

= 0 + 9.8 × 10= 98 ms–1.

v = 2 2 2 2(98) (98)x yv v = –198 2 ms .

O A

u

h = 490 m

45°

h


Example 17. Two tall towers face each other and are at a distance of 180 m from each other.With what velocity must a ball be thrown horizontally from a window 55 m above the ground inone tower, so that it enters a window 10.9 m above the ground in second tower?

Solution.Vertical displacement of ball = 55 – 10.9

= 44.1 m.Initial vertical velocity of ball uy = 0.

As y = 21

2yu t gt

44.1 = 210 9.8

2t

t2 =44.1 2

9.8

= 9

t = 3 sec.

Required horizontal velocity =Horizontal distance

time

=180

3 –160 ms .

Example 18. A bomb is dropped from an aeroplane when it is directly above a target at aheight 2000 m. The aeroplane is moving horizontally with a speed of 360 km h–1. By how muchdistance will the bomb miss the target.

Solution. Here, u = –1560 100 ms .

18

Time taken by bomb to reach the ground

h = 21.

2ut gt

2000 = 210 10 .

2t

t = 20 s.

Distance by which bullet will miss the target

s = 100 × 20

= 2000 m.Example 19. A stone is thrown from the top of a tower at an angle of 30° up with the

horizontal with a velocity of 16 ms–1. After 4 s of flight it strikes the ground. Calculate the heightof the tower from the ground and the horizontal range of the stone. (g = 9.8 ms–2)

Solution. t = 4 sec.

Vertical component of velocity is 16 sin 30° = 8ms–1.

h = 21

2ut gt .

180 m

u

55 m

10.9 m .


h =1

–8 4 10 4 42

= – 32 + 80

= 48 m.Horizontal range = Horizontal velocity × time

= 16 cos 30° × 4 = 3

16 42

= 32 3 m.

Example 20. A small ball A suspended from a string, OA is set in oscillation. When the ballpasses its lowest position with velocity 0.8 ms–1 at a height of 5 m from the ground, the string iscut. Find the horizontal distance travelled by the ball just before it hits the ground. (g = 10m/s2)

Solution. Time taken to fall 5 m in vertical downward direction

h = 21

2ut gt

or 5 = 210 10

2t

10 = 10 t2

or t = 1 sec.

x = BC

= Horizontal velocity × time of flight

= 0.8 × 1

= 0.8 mExample 21. A bomb is dropped from an aircraft when it is directly above a target at a height

2000 m. The air craft is moving with a speed of 540 km h–1. Will the bomb hit the target? If notby how much distance will it miss the target?

Solution. No. The bomb will not hit the target.

Time taken by bomb to fall through 2000 m.

h = 21

2ut gt

2000 = 210 10

2t

or t = 20 sec.

Distance by which the bomb will miss the target

= u × t

=5

540 2018

= 3000 m = 3 km.

Example 22. A body is projected horizontally from the top of a cliff with a velocity of9.8 ms–1 what time elapses before horizontal and vertical velocities becomes equal? (g = 9.8 ms–2).

30°

16m /s

A

O

0.8 m/s

5 m

B Cx


Solution. Horizontal velocity is 9.8 ms–1.

i.e. ux = 9.8 ms–1.

Vertical velocity after time t is

uy = u + gt = 0 + 9.8 t.

If horizontal and vertical velocities are equal then

ux = uy

9.8 t = 9.8 or t = 1 sec.Example 23. A ball rolls off from top of a stair way with a horizontal velocity u. If each step

has h height and b width, the ball will just hit the edge of nth step. Find the value of n.

Solution. The ball hits the nth step.

Horizontal distance travelled = nb

Vertical distance travelled = nh.

Let the time taken = t.

Velocity along horizontal direction is V

nb = Vt. ...(i)

for vertical displacement.

h = 21

2ut gt

nh = 210 .

2gt

Using eq (i) t = V

nb

nh =2 2

2

1.

2 V

n bg

or n =2

2

2 V.

h

gb

Example 24. Two paper screens are seperated by a distance of 100 ms. A bullet pierces A andB. The hole in B in 10 cm below the hole in A. If the bullet is travelling horizontally at the time ofhitting the screen A calculate the velocity of the bullet when it hits the screen A.

Solution. Let the bullet hit the screen A with velocity u and pierces the screen B after time t.

P

u Q

R

A100 m

0.1m

B


Horizontal distance

PQ = ut = 100

or t =100

.u

Vertical distance QR = 2

21 1 1009.8 0.1

2 2gt

u

or4

2

1 109.8

2 u = 0.1

or u2 = 49 × 104

or u = 700 ms–1.

Example 25. A projectile follows the trajectory y = 21– .

2x x Determine its range.

Solution. The equation of trajectory is

y = tan 1R

xx

If = 45°

y = 1– .Rxx

Given y =2

– 1– .2 2

x xx x

Comparing the two equations.

R = 2 units.Example 26. A ball is thrown from the top of a tower with an initial velocity of 10 ms–1 at an

angle of 30° above the horizontal. It hits the ground at a distance of 17.3 m from the base of tower.Calculate the height of the tower. (g = 10 ms–2).

Solution. The angle of projection of the ball is = 30° and the velocity of projection isu = 10 ms–1

ux = u cos = 10 cos 30° = 8.65 m/s.

Vertical component.

uy = u sin = 10 sin 30° = 5 m/s.

Let ball hits the ground after time t then

ux × t = R

t =R

xu =

17.3

8.65 = 2 sec.

h

u y

u x

30°

h = 17 .3 m


If h is height of tower then h = 21.

2ut gt

h = (– 5) (2) + 2110 (2)

2

= 10 m.Height of tower = 10 m.Example 27. A fighter plane flying horizontally at an altitude of 1.5 km with a speed

720 km h–1 passes directly overhead an anticraft gun. At what angle from the vertical should thegun be fired for the shell muzzle speed 600 ms–1 to hit the plane. At what maximum altitude shouldthe pilot fly the plane to avoid being hit? (g = 10 ms–2).

Solution.

600 m s–1

600 s in

600cos

Speed of plane = 720 km/h

=5

72018

= 200 ms–1.

Let bullet hits the plane after time t.

Horizontal distance travelled by the plane = Horizontal distance travelled by shell.

200 × t = 600 sin × t

or sin =1

3 = 0.333.

or = 19.30°.

Maximum height attained by shell

=2 2cos

2

u

g

=2(600 cos 19 30 )

2 10

= 16000 m = 16 km.The pilot should fly the fighter plane above 16 km to avoid being hit.


BASED ON CIRCULAR MOTIONExample 28. Calculate the angular speed of a wheel, making 480 revolution per minute.

Solution. Here,

= 480 rpm

=480

60 = 8 rps.

Angular speed

= 2= 2 × 8 = 16

=22

167

= 50.3 rad/s

Example 29. A stone ties to the end of a string 80 cm long is whirled in a horizontal circlewith a constant speed. If the stone makes 14 revolution in 25 seconds what is magnitude anddirection of acceleration of the stone?

Solution. Here, r = 80 = 0.8 m.

=14

rps.25

Acceleration of stone a = 2u

r = r2 = r (2)2

a =2

22 140.8 2

7 25

.

� 9.9 ms–2.Example 30. Calculate the angular velocity of seconds hand of a clock.

Solution. Here, T = 60 sec. (seconds hand of a clock completes one

rotation in 60 s)

Angular velocity,

=2

T

=2

60

= rad/ .

30s

Example 31. An aircraft executes a horizontal loop of a radius 1 km, with a steady speed of900 km h–1. Compare its centripetal acceleration with acceleration due to gravity.

Solution. Here, r = 1 km = 1000 m.

u = 900 km h–1

= 900 × 5

18 = 250 ms–1

Centripetal acceleration =2u

r =

2(250)

1000

= 62.5 ms–2.


Centripetal acceleration

Acceleration due to gravity=

62.5

9.8

= 6.38.Example 32. An insect trapped in a circular groove of radius 12 cm moves along the groove

steadily and completes 7 revolution in 100 s (i) What is the angular speed and linear speed of themotion (ii) Is the acceleration vector, a constant vector? What is its magnitude?

Solution. Here, r = 12 cm. v = 7

rps.100

(a) angular speed = 2v

=22 7

2 × 0.44 rad/sec.7 100

Linear speed v = r= 12 × (4.4)

= 5.28 cm s–1.(b) The direction of acceleration in circular motion changes continuously so acceleration

vector is not constant vector.

Magnitude of acceleration

a = r2

= 12(0.44)2

= 2.32 cm s–2.Example 33. A particle is revolving in a circular path completes first one third of

circumference in 2 s, while next one third in 1 sec. Calculate the average angular velocity ofparticle.

Solution. Total angular displacement is 2 2

3 3

=4

3

Total time taken t = 2 + 1 = 3 s.

av =4 / 3 4 rad

.3 9 sec

Example 34. The angular displacement of a particle is given by = 6t2 + 4t + 7 rad. Find theangular velocity of particle at t = 1 sec.

Solution. Here, = 6t2 + 4t + 7

angular velocity =d

dt

=2(6 4 7)

dt t

dt

= 12t + 4

when t = 1 sec

= (12 × 1) + 4 = 16 rad/sec.


Example 35. The angular velocity of a particle is given by = 1.5t – 3t2 + 2. Find the timewhen its angular acceleration is zero.

Solution. Here, = 1.5t – 3t2 + 2.

Angular acceleration =d

dt

.

2(1.5 – 3 2)d

t tdt

= 0

1.5 – 6t = 0

or t =1.5 1

6 4

= 0.25 sec.Example 36. A stone of mass 0.1 kg tied to one end of a string 1 m long is revolved in a

horizontal circle at the rate of 10

rev/s. Calculate the tension in the string.

Solution. Here, m = 0.1 kg, l = 1m,

= 2 = 10

2

= 20 rad/s.

Tension in the string T = 2mv

r = mr2

= 0.1 × 1 × (20)2

= 40 N.Example 37. A body of mass 1kg revolves in a circle of radius 0.20 m completes 100

revolution in one minute. Calculate its linear velocity and centripetal acceleration.

Solution: Here, m = 1 kg. r = 0.20 m, v = 100

Hz.60

Linear velocity v = r

=100

0.2 260

= 2

/3

m s

Centripetal acceleration

a =2v

r =

22

3(0.2

=2

–220�ms .

9

Example 38. A string can with stand a tension of 50 N what is the greatest speed at which abody of mass 1 kg can be whirled in a circle using 0.7 m length of the string.


Solution. Here, T = 50 N, m = 1 kg, r = 0.7 m

T =2mv

r

or v =. .r

m

=

50 0.7

1

= 5.9 ms–1

Example 39. The moon completes one revolution in a circular orbit around earth in 27.3days. Calculate the acceleration of the moon towards the earth. The radius of the circular orbit canbe taken as 3.85 × 105 km.

Solution. Here, T = 27.3 days

= 27.3 × 24 × 60 × 60 sec.R = 3.85 × 105 km

= 3.85 × 108 m.

ac =22

2V 2

R Tr r

=2

8 23.85 10

27.3 24 60 60

= 2.73 × 10–3 ms–2.

EXERCISE

BASED ON MOTION IN A PLANE1. A particle is moving on a circular path of radius r with constant speed V. It describes angle

of 120° at the centre calculate its

(a) Magnitude of displacement

(b) Magnitude of change in velocity

(c) Magnitude of average acceleration [Ans. 23 3

3 , 3,2

vr v

r]

2. Two towns A and B are connected by a regular bus service, leaving in either direction everyT minutes. A man cycling with a speed of 20 km h–1 in the direction A to B notices that a busgoes past him every 18 minute in the direction of his motion, and every 6 minute in theopposite direction. What is the period T of the bus service and with what speed do the busesply on the road? [Ans. 9 min, 40 km h–1]

3. The position of a particle in given by

r

= 2 ˆˆ ˆ3.0 2.0 5.0t i t j k .Where t is seconds and the coefficients have the proper units for r to be in metre (a) Findv(t) and a(t) of the particle. (b) Find the magnitude and direction of v(t) at t = 3.0s.

[Ans: (a) v (t) = ˆ ˆ3.0 4.0i t j

a

(t) = ˆ4.0 j(b) 12.4 ms–1, 76° with x axis]


4. The position of a particle is given by

r

= ˆˆ ˆ(4 cos 2 ) (4 sin 2 ) 6t i t j t k .

Calculate the acceleration at t = .4 [Ans. – –1ˆ16 msj ]

GROUND TO GROUND PROJECTION5. A cricketer can throw a ball to a maximum horizontal distance 150 m. How high above the

ground can the cricketer throw the same ball? [Ans. 75 m]

6. A cricket ball is thrown at a speed of 28 ms–1 in a direction 30° above the horizontal.Calculate (a) the maximum height (b) the time taken by the ball to return to the same level(c) the distance from the thrower to the point where the ball returns to the same level.

[Ans. 10.0 m, 2.9 s, 69. 3m]

7. Find the angle of projection for which the horizontal range and maximum height are equal.[Ans. 75° 58]

8. Show that for two angles of projection for which range is same, the sum of maximum heightsfor these two angles is independent of angle of projection.

9. A particle is projected over a triangle from one end of a horizontal base and grazing thevertex falls on the other end of the base. If and be the base angle and the angle ofprojection prove that.

tan = tan + tan 10. Two projectiles are projected from the same point with same speed at angles and (90 – )

with the horizontal. What will be the ratio of (i) maximum heights (ii) horizontal range.

[Ans. tan2 , 1]

11. A boy projects a ball with velocity u1 from point A as shown. At the same time another ball

is projected vertically upwards with velocity u2 from point B. What should be value of 1

2

u

u

for both the balls to collide. [Ans. 1

2

23

u

u ]

60°

u 1

A B C

u2

12. A ball is thrown with a velocity of 15 ms–1 at an angle of 30° with horizontal. Calculate

(a) time of flight.

(b) maximum height attained by the ball (g = 10 ms–2)

[Ans. 1.5 s, 2.8 m]

13. A ball is projected with a velocity of 320 ms–1 at an angle of 30° with the horizontal. Find

(i) Horizontal range

(ii) Maximum range. (g = 10 ms–2) [Ans. 8.87 km, 10.24 km]

14. A bullet fired from a riffle attains a maximum height of 25 m and a range 200 m. Find theangle of projection. [Ans. 25.56°]


15. The speed with which a bullet can be fired is 150 ms–1. Calculate the greatest distance towhich it can be projected and also the maximum height to which it would rise.

[Ans. 2295.14 m, 573.91 m]

16. For a projectile show that

(a) gT2 = 2R tan

(b) Hmax = R

tan4

(c) gT2 = 8Hmax

17. If R is the horizontal range and h, the greatest height of a projectile prove that its initialspeed is

2 25 (16 R )

4

h

h

(g = 10 ms–2)

18. A riffle shoots a bullet with a muzzle speed of 500 m/s at a small target 50 m away. Howabove the target must the riffle be aimed so that the bullet will hit the target? Takeg = 10 ms–2. [Ans. 5 cm]

19. The height of a hall is 20 m. A boy can throw a ball with a velocity of 30 ms–1. Calculate themaximum horizontal distance, the ball can go without hitting the ceiling of the hall.

[Ans. 89.4 cm]

20. A bullet is fired with a velocity 100 ms–1 in a direction making an angle of 60° with thevertical. Calculate the time of flight. [Ans. 10.2 s]

21. Show that trajectory of projectile is also given by Y = x tan (1 – x/R) symbols have theirusual meaning.

22. A body is projected with a velocity of 30 ms–1 in a direction making an angle of 60° withhorizontal. Calculate

(i) Position after 0.5 s

(ii) Velocity after 0.5 s

[Ans. x = 7.5 m, 11.76 m, 25.88 ms–1, = 54.6°]

23. A body is projected with a velocity of 20 ms–1 in a direction making an angle of 30° with the

horizontal. Calculate velocity after 1 sec. [Ans. –110 3 ms ]

24. The maximum vertical height of a projectile is 15 m. If the magnitude of initial velocity is20 ms–1 what is the direction of initial velocity. (g = 10 ms–2) [Ans. = 60°]

25. A bullet fired at an angle of 60° with the vertical hits the ground at a distance of 2 km. Byadjusting the angle can the bullet hit the target situated 3 m away. [Ans. No.]

26. The horizontal range of a projectile is four times the maximum height attained by aprojectile. Calculate angle of projection. [Ans. 45°]

27. The range of projectile when launched at an angle of 15° with the horizontal is 1.5 km. Whatis the range of the projectile when launched at 45° to the horizontal. [Ans. 3 km]

28. A cricket ball is projected with a kinetic energy E making an angle of 45° with horizontal.Calculate its K.E. at highest point. [Ans. K/2]

29. A particle of mass 0.01 kg is projected with velocity u = –1ˆ2 msi from point (x = 0, y = 0).

Find out its position coordinate after 2 sec. [Ans. 4 m and 20 m]


30. A bullet is fired from a canon with a velocity 500 ms–1. If the angle of projection is 15° andg = 10 m–2, then calculate the range. [Ans. 12.5 × 103 m]

31. A projectile A of mass m is thrown with velocity V at an angle of 30° to the horizontal andanother projectile B of the same mass is thrown with velocity V at an angle of 60° to thehorizontal. Find the ratio of horizontal range and maximum height of A and B.

[Ans. 1:1, 1:3]

32. A ball of mass m is thrown vertically up. Another ball of mass 2 m is thrown at an angle with the vertical. Both of them stay in air for same period of time. What is the ratio of heightattained by the two balls. [Ans. 1]

33. Two projectiles are projected at angle and –2

to the horizontal respectively with

same speed 20 m/s one of them rises 10 m higher than the other. Find the angle of projection(g = 10 ms–2). [Ans. 30° and 60°]

34. A ball is thrown at an angle 45° to the horizontal which just touches the top edge of abuilding that is 20 m away and 5 m high. Find the projected velocity? [Ans. 16.33 ms–2]

BASED ON PROJECTILE FIRED HORIZONTALLY35. An aeroplane is flying in a horizontal direction with a velocity 600 km/h at a height of

1960 m. When it is vertically below point A of the ground a bomb is dropped from it. Thebody strikes the ground at point B. Calculate distance AB. [Ans. 3.33 km]

36. A stone is thrown from the top of a tower at an angle 30° up with the horizontal with avelocity of 16 ms–1. After 4 sec of flight it strikes the ground. Calculate the height of thetower above ground and horizontal range of stone (g = 9.8 ms–2) [Ans. 46.4 m, 55.4 m]

37. A ball is projected horizontally from a tower with a velocity of 5 ms–1. Find the velocity of

the ball after 0.5 s. (given g = 10 ms–2) [Ans. –15 2 ms ]

38. A projectile is fired horizontally with a velocity of 98 ms–1 from the top of a hill 490 m high.Find.

(i) the time taken to reach the ground

(ii) the distance of the target from the hill.

(iii) the velocity with which the projectile hits the ground. [Ans. 10 s, 980 m, 138.59 ms–1]

39. A body is thrown horizontally with a velocity 2gh from the top of a tower of height h. It

strikes the level ground through the foot of the tower at a distance x from the tower.Calculate x. [Ans. 2h]

40. A ball is projected upwards from the top of a tower with a velocity of 50 ms–1 making anangle of 30° with the horizontal. The height of the tower is 70 m. After how much time fromthe instant of throwing will the ball reach the ground? [Ans. 7 s]

41. A particle is projected horizontally with speed 20 ms–1 from the top of a tower. After whattime velocity of particle will be at 45° angle from initial direction of projection (g = 10 ms–2).

[Ans. 2 s]

42. A bomb is dropped from an aeroplane when it is directly above a target at a height of1960 m. The aeroplane is moving horizontally with a speed of 20 ms–1. By how muchdistance will the bomb miss the target. [Ans. 400 m]


43. A body is thrown horizontally from the top of a tower and strikes the ground after threeseconds at an angle of 45° with the horizontal. Find the height of the tower and speed withwhich the body was projected. [Ans. 44.1 m, 29.4 ms–1]

44. A ball is projected horizontally from a tower with a velocity of 4 ms–1. Find the magnitudeof velocity of ball after 0.7 s. [Ans. 8.06 ms–1]

BASED ON CIRCULAR MOTION45. A string can with stand a tension of 200 N what is the greatest speed at which a body of mass

4 kg can be whirled in a circle of radius 0.7 m. [Ans. 5.9 ms–1]

46. What is the acceleration of a sprinter running at 10 ms–1 when rounding a bend with a turnof radius 25 m. [Ans. ]

47. The angular speed of a wheel is 88 rad/sec. Calculate the number of revolution made by it inone second. [Ans. 14]

48. Calculate the angular velocity in rad s–1 of a particle which makes 300 rev min–1. What is thelinear velocity if the radius is 4 cm? [Ans. 10 rad s–1, 40 cm s–1]

49. A string can with stand a maximum tension of 70 N. What is the greatest speed at which abody of mass 1 kg can be whirled in a circle using 0.7 m length of the string?

[Ans. 7 ms–1]

50. Calculate the angular speed of the earth about its own axis. [Ans. /43200 rad s–1]

51. A particle moves in a circle of radius 20 cm. Its linear speed is given by V = 3t2 + 5t wheret is in seconds and V in ms–1. Find the resultant acceleration at t = 1 s. [Ans. 320.2 ms–2]

142

FORCE: is a push or pull which tries to change or changes the state of rest or motion of abody. Forces are of two types basically.

(i) Balanced force (ii) Unbalanced force

Intertia: The property of a body by virture of which it opposes any change in its state ofmotion or rest is called inertia.

Mass is the measure of inertia: Greater the mass of body, more is its inertia.

NEWTON’S FIRST LAW OF MOTION: (Law of inertia): According to this law, an objectat rest remains at rest and an object in motion will continue to move in a straight line with constantvelocity unless it experiences an external force.

NEWTON’S SECOND LAW OF MOTION: The acceleration of an object is directlyproportional to net force acting on it and inversely proportional to its mass.

a F

a 1

m

or a F

mF ma

The rate of change of linear momentum of a body is directly proportional to the external forceapplied on the body. i.e.,

F p

t

S.I. Unit of force:S.I. unit of force is Newton (N) and CGS unit is dyne. Dimension [MLT–2]

1 N = 105 dyne.Unit of force is also given in kg wt or kg fi.e. 1 kg wt = 9.8 N

1 kg f = 9.8 N.MOMENTUM: Momentum of a body is measured as the product of mass and its velocity.

The momentum of a body of mass m moving with velocity V is given by

P = mV.

Vector form P��

= V��

m .

It’s S.I. unit is kg ms–1. Dimension [MLT–1].Change in momentum P = m V.

FRAME OF REFERENCE: System of coordinates in terms of which we study a particularevent is called frame of reference.

�� 6UNIT

Laws of Motion and Friction 143

Frame of reference are of two types:(i) Inertial frame of reference—are those frame of reference in which Newton’s laws of

motion are applicable. A frame of reference moving with constant speed is an example of inertialframe of reference.

(ii) Non-inertial frame of reference—are those in which Newton’s laws of motion are notapplicable. An accelerated frame of reference is a non-inertial frame of reference. Rotating frameof reference is a non inertial frame of reference.

PSEUDO FORCE: The force on a body due to acceleration of non-inertial frame ofreference is called pseudo force.

NEWTON’S THIRD LAW OF MOTION: For every action there is an equal but oppositereaction.

Action = – reactionContact forces: Contact forces arise when one body is in contact with another body. These

forces are the result of electromagnetic interaction between the surface atoms of the two bodies.Impulse: When a large force acts on a body for a short interval of time, then total effect of

force is called Impulse. Numerically impulse is equal to the product of force applied to theduration for which it is applied, i.e.,

Impulse J = F tAlso F t = P J = F t = PIt is a vector quantity.S.I. unit N–S.Dimension [MLT–1]LAW OF CONSERVATION OF MOMENTUM: In the absence of an external force, total

momentum of a body or system of bodies remains constant.

i.e. F =P

t

When no external force is appliedP

t

= 0

P = constantSpecial case (i) When two bodies collide then

momentum before collision = momentum after collision.m1u1 + m2u2 = m1V1 + m2V2

(ii) When a bullet of mass m is fired with velocitiy v from a gun of mass M, then recoilvelocity of gun is given by

Mv = – mv

v =–

.M

mv

APPARENT WEIGHT OF A BODY IN LIFT:

a

m g

a

m g

a = 0

m g

R R R


Case (i) Lift is going upward with uniform acceleration ‘a’. Then apparent weight is

R – mg = ma

R = m (g + a).

Case (ii) When lift is going downward with uniform acceleration ‘a’ then apparent weight is

mg – R = ma

or R = m (g – a)

Case (iii) In case lift is moving with constant velocity a = 0

R = mg.

Case (iv) In case of free fall (a = g) the apparent weight

W = m (g – g) = 0

It is called weightlessness.

CONCURRENT FORCES: Two or more than two forces acting at a point on passingthrough the same point are called concurrent forces.

Concurrent forces will be in equilibrium if their resultant is zero, i.e.,

In case of equilibrium of concurrent forces.

1 2F F Fn

= 0

or1

Fn

n

i

= 0.

Let 1 2F , F� �

and 3F�

be three concurrent forces which are in equilibrium then

1 2F F

sin sin

= 3F

sin When , and are shown in figure.

F2 F 3

F 1

This is Lamis theorem.FREE BODY DIAGRAM: A free body diagram is a diagram showing body by itself free of

its surroundings, with vectors drawn to show the magnitude and direction of all the forces applied tothe body.

Free Body Diagram of Two Bodies in Contact


Free Body Diagram of two Bodies attached by a String Passing over a Pulley:For body m1 m1g – T = m1a

For body m2 T – m2g = m2a

FRICTION: It is a kind of force which opposes relative motion between twosurfaces in contact.

Friction is of two types (i) Rolling friction (ii) Sliding friction.

Limiting Friction: The maximum value of static friction that comes into playwhen a body just begins to slide over another surface. It is denoted by (fs)max.

Coefficient of Limiting (Max. Static Friction): The maximum value of static friction isgiven by

(fs)max = sN.

or μ s = max( ).

Nsf

where N = mg.

Coefficient of Kinetic friction: It is the ratio of kinetic friction fk to the normal reaction

μk = .N

kf

where N = mg.

Angle of Friction: tan =N

sf

or tan =N

Ns

tan = s. ...(1)

Ffs A O

BC

W

N

m g cos

m g

m g sinfs

N

Angle of Repose: fs = mg sin N = mg cos

tan = sf

N

tan = s. ...(2)


From eqn. (1) and (2)

μ s = tan = tan .

Motion of a Body along a rough horizontal surface:

F

f

Force of friction is given by f = μN = μmg

Retardation a = μf mg

m m = μg

a = μg

Workdone against friction

W = f s= (μmg) s

Motion of a Body along a Rough inclined Plane: On an inclined plane component ofweight mg which acts along the plane is mg sin .

m g cosm g

m g sin

N

(i) Body moves down an inclined plane with uniform velocity then

F = mg sin – f

= mg sin – μN

= mg sin – μmg cos F = mg (sin – μcos ).

(ii) Body moves up an inclined plane with uniform velocity: Here

F = mg sin + μN

= mg (sin + μcos )

m g cos

m g

m g sin

NF

SOLVED EXAMPLES

BASED ON NEWTON’S SECOND LAW OF MOTIONExample 1. A pebble of mass 0.05 kg is thrown vertically upwards. Give the direction and

magnitude of the net force on the pebble

(i) during its upward motion.

(ii) during its downward motion.(iii) at the highest point where it is momentarily at rest.


Do your answer change if the pebble were thrown at an angle say 45° with the horizontaldirection.

Solution. Here m = 0.05 kg, g = 9.8 ms–2.

(i) Net force on pebble = mg

= 0.05 × 9.8 = 0.49 N (downward)(ii) Net force on pebble = mg

= 0.05 × 9.8 = 0.49 N (downward)(iii) Net force on pebble = mg

= 0.05 × 9.8 = 0.49 N (downward)No, because the horizontal component of velocity remains constant.

Example 2. A constant force acting on a body of mass 5 kg changes its velocity from 3 ms–1 to6 ms–1 in 15s. The direction of the force remains unchanged. What is the magnitude and thedirection of force.

Solution. Here m = 5 kg

u = 3 ms–1, v = 6 ms–1, t = 15 sec.

from formula

F = ma = ( – )m v u

t

=5(6 – 3)

15

= 1.0 NExample 3. A constant retarding force of 50N is applied on a body of mass 20 kg moving

with a speed of 15 ms–1. How long does the body take to stop?

Solution. Here, F = – 50 N, m = 20 kg.

v = 0, u = 15 ms–1.

a =F

m =

– 50

20 = – 2.5 ms–2.

Using 1st equation of motion

v = u + at

0 = 15 + (– 2.5)t

or t = 6 sec.Example 4. A bullet of mass 0.04 kg moving with a speed of 90 ms–1 enters a heavy wooden

block and is stopped after a distance of 60 cm. What is the average resistive force exerted by theblock on the bullet?

Solution. Here, m = 0.04 kg

u = 90 m/s

v = 0

s = 60 cm = 0.6 m

Using 3rd equation of motion

v2 = u2 + 2as

02 – 902 = 2 × a × 0.6


a =– 90 90

2 0.6

= – 6750 ms–2. Retarding force = mass × retardation.

= 0.04 (6750)

= 270 NExample 5. Fuel is consumed at the rate of 100 kg per sec in a rocket. The exhaust gases are

ejected at the rate of 5 × 104 ms–1. What is the thrust experienced by the rocket.

Solution. Rate of consumption of fuel = 100 kg/sec.

v = 5 × 104 m/sec.

Thrust on the rocket = mu

i.e. F = (100) (5 × 104)

= 5 × 106 N.Example 6. A body of mass 5 kg is acted upon by two perpendicular forces of 8N and 6N.

Give the magnitude and direction of the acceleration of the body.

Solution. Here, F1 = 8N, F2 = 6N, m = 5kg.

F1 and F2 are perpendicular to each other so resultant force is

F = 2 21 2F F

= 2 26 8= 10N

The magnitude of acceleration produced is

a =F 10

5m

a = 2 ms–2

Let F makes an angle with F1 then

cos = 1F 80.8

F 10

= cos–1 (0.8)

= 36.87°.Example 7. The driver of a three wheeler moving with a speed of 36 kmh–1 sees a child

standing in the middle of the road and brings his vehicle to rest in 4s just in time to save the child.What is the average retarding force on the vehicle. The mass of the three wheeler is 400 kg andmass of the driver is 65 kg.

Solution. Here, u =5

3618

= 10 ms–1, v = 0, t = 4s

m = 400 + 65 = 465 kg

As v = u + at

0 = 10 + a × 4

a = – 2.5 ms–2.

F2

F

F1


Magnitude of force F = ma

= 465 × 2.5

= 1162.5 N.Example 8. A force acts on a body of mass 5 kg for 4 sec. The force then stops acting and

the body covers 40 m in the next 4 sec. Calculate the force.

Solution. Here, m = 5 kg, t = 4 sec, s = 40 m.

After 4 sec. the force stops acting on the body and body covers 40 m in 4 sec.

u =distance

time

=40

4 = 10 ms–1.

Now v = u + at

10 = 0 + a × 4

or a =10

4 = 2.5 ms–2.

F = ma = 5 × 2.5 = 12.5 N.Example 9. Give the magnitude and direction of the net force acting on

(a) a drop of rain falling down with constant speed.

(b) a cork of mass 10g floating on water.

(c) a kite skilfully held stationary in the sky.

(d) a car moving with a constant speed of 30 km/h on a rough road.

(e) a high speed electron in space far from all material objects and free of electric andmagnetic fields.

Solution. (a) Rain drop is falling down with constant speed so force acting on it is zero.

(b) Cork is floating on water. Its apparent weight is zero. Hence net force is zero.

(c) As kite is stationary the force acting on it is zero.

(d) Car is moving with constant velocity so force acting on it is zero.

(e) As there is no field close to electron, net force experienced by it is zero.

Example 10. A particle of mass 10g is acted by a force of 150 dyne. Assuming that theparticle starts from rest. Calculate the distance travelled in 10s. What is the velocity at the end ofthis period.

Solution. Here F = 150 dyne m = 10 g

F = ma

a =F 150

10m = 15 cm s–2.

From second equation of motion

s = ut + 1

2 at2

=1

0 15 10 102

= 750 cm


v = u + at

= 0 + 15 × 10

= 150 cm s–1.Example 11. A body of mass 5 kg traverses in successive seconds distances of 8, 9 and 10

metres respectively. What force is acting on it.

Solution. The ditances travelled in successive seconds are 8 m, 9 m and 10 m.

It means that the velocity in the first second = 8 ms–1

It means that the velocity in the second second = 9 ms–1

It means that the velocity in the third second = 10 ms–1

The change of velocity in one second = 9 – 8 = 1 ms–1

a =Change of Vel. 1

time 1 = 1 ms–2

F = ma = 5 × 1 = 5N.

Example 12. A motor car running at the rate of 7 ms–1 can be stopped by its brakes in 10metres. Prove that the total resistance to the car’s motion, when the brakes are on is one quarter ofthe weight of the car.

Solution. Here, u = 7ms–1 v = 0, s = 10m, a = ?

Applying the formula

v2 – u2 = 2as

02 – 72 = 2 × a × 10

20a = – 49

a = – 2.45 ms–2.

– ve sign indicates retardation .

Let the mass of the car be m.

Retarding force F = ma = m × 2.45 N. (i)

Weight of car W = mg = m × 9.8 N. (ii)

Divide (i) by (ii),

F

W=

2.45

9.8 =

1.

4

F =W

.4

Example 13. A force of 20N acts on a block inclined at an angle of 60° with the horizontal.If the mass of the block is 2 kg. Find the horizontal acceleration produced?

Solution. Here, F = 20 N, = 60°

Horizontal force = F cos

Horizontal acceleration =F cos

m

=cos60

202

= 5 ms–2.


Example 14. A car starts from rest and accelerates uniformly with 2ms–2. At t = 10s a stoneis dropped out of the window 1 m high of the car. What are the (i) velocity (ii) acceleration of stoneat 10.1 s ? (g = 9.8 ms–2)

Solution. Here, u = 0, a = 2 ms–2, g = 9.8 ms–2, t = 10 s.

(i) During first 10s, the horizontal component of velocity is

vx = 0 + 2 × 10 = 20 ms–1.

From 10.0s to 10.1s the vertical component of velocity is

vy = u + gt = 0 + 9.8 (0.1) = 0.98 ms–1.

Vx and Vy are perpendicular to each other so resultant velocity is

v = 2 2x yv v

= 2 2(20) (0.98)

= 20.02 ms–1

The direction is given by

tan =y

x

v

v =

0.98

20 = 0.049

or = tan–1 (0.049).(ii) The acceleration of the stone = 9.8 ms–2.

BASED ON MOMENTUM LAW OF CONSERVATION OF MOMENTUMAND IMPULSE

Example 15. A cricket ball of mass 100 gm moving with a velocity of 20 ms–1 is brought torest by a player in 0.05 sec. Find the impulse of the ball and the average force applied by theplayer?

Solution. m = 100 g = 0.1 kg

u = 20 m/s, v = 0, t = 0.05s

Impulse J = m v

= 0.1 × 20 = 2N–s

Also J = F T

F =J 2

0.05t

= 40 N.

Example 16. A ball of mass 150 g moving with velocity 12 ms–1 strikes with the bat andrebounds back with velocity 20 ms–1. If duration of contact between bat and ball is 0.1 sec. Thenfind (i) change in momentum of ball (ii) average force on ball.

Solution. Here, m = 150 g = 0.15 kg

u = 12 ms–1, v = – 20 ms–1

t = 0.1 sec.

Change is momentum

P = m(v – u) = 0.15(– 20 – 12) = – 4.8 kg ms–1.

(ii) P = F t

or F =P 4.8

0.1t

= 48 N.


Example 17. A ball of mass 1 kg is dropped from height 9.8 m strikes with ground andrebounds to height of 4.9 m. If the time of contact between ball and ground is 0.1 s then findimpulse and average force acting on ball.

Solution. Here, m = 1 kg h1 = 9.8 m

h2 = 4.9 m

t = 0.1 sec.

Velocity with which ball hits the ground

v2 = u2 + 2gh

v12 = 0 + 2 × 9.8 × 9.8

v1 = 9.8 2 m/s.

Velocity with which ball goes up after striking the ground

v2 = u2 + 2gh

0 = v22 – 2 × 9.8 × 4.9

or v2 = 9.8 m/s

Impulse of force = Change in momentum.

= m v

= 1 (9.8 2 9.8)

= 9.8( 2 1)

= 9.8 × 2.414

= 23.65 N – S

Also F =J

t =

23.65

0.1

= 236.5 N.Example 18. A body of mass 0.25 kg moving with velocity of 12 ms–1 stopped by applying

a force of 0.6 N. Calculate the time taken to stop the body. Also calculate the impulse of this force.

Solution. Here, m = 0.25 kg, u = 12 ms–1.

F = 0.6 N.

as F = ma

Retardation a =F 0.6

.25m

v = u + at

0 =0.6

12 –0.25

t

t =12 0.25

0.6

= 5 sec.

Impulse of force J = F t

= 0.6 × 5 = 3 NS.


Example 19: A ball of mass 0.1 kg is thrown against a wall. It strikes the wall normally witha velocity of 30 ms–1 and rebounds with a velocity of 20 ms–1. Calculate the impulse of the forceexerted by wall.

Solution. Here, m = 0.1 kg

u = 30 ms–1

v = – 20 ms–1

Impulse = m (v – u)

= 0.1 (– 20 – 30) = – 5 NS.Example 20. Two billiard balls each of mass 0.05 kg moving in opposite directions with

speed of 6 ms–1 collide and rebound with the same speed. What is the impulse imparted to eachball by the other?

Solution.A

6 m s–1

B

– 6 ms–1

– 6 ms–1 + 6 ms–1

For ball A

Pi = Momentum before collision

= 0.05 × 6 = 0.3 kg ms–1

Pf = Momentum after collision

= 0.05 (– 6) = – 0.3 kg ms–1

Impulse imparted to ball A due to ball B

= Pf – Pi = (– 0.3) – (0.3) = – 0.6 kg ms–1.For ball B

Pi = 0.05 (– 6) = – 0.3 kg ms–1

Pf = 0.05 (6) = 0.3 kg ms–1

Impulse imparted to ball B

= Pf – Pi = (0.3) – (– 0.3)

= 0.6 kg ms–1

Example 21. A ball moving with a velocity of 10 ms–1 strikes against a wall at an angle of45°. It is reflected at the same angle. If mass of ball is 0.5 kg. Calculate the change in momentum.

Solution. Here, m = 0.5 kg, u = 10 m/s, = 45°

45°

45°O

m v cos 45°

C

A

B

m v cos 45°

m u

m v m u sin 45°m u sin 45°


Various components of momentum before striking the wall and after striking are shown inchange in momentum along y-axis.

= mu sin 45° – mu sin 45°

= 0.

Change in momentum after collision

= (– mu cos 45°) – (mu cos 45°)

= – 2mu cos 45°

=1

– 2 0.5 102

= – 7.07 kg ms–1.Example 22. Figure shows the position time graph of

a particle of mass 4 kg. What is the (i) force acting on theparticle for t < 0, t > 4s, 0 < t < 4s, (ii) Impulse at t = 0 andt = 4s. Assume that motion is one dimensional.

Solution. (i) For t < 0 body is at rest as displacementis zero. So no force is acting on the particle.

0 < t < 4s. The velocity of body is constant, i.e.,acceleration is zero.

F = ma = zero.

For t > 4s the displacement of body is not changing i.e. body is at rest. So force acting onbody is zero.

(ii) Impulse at t = 0:

For t < 0 u = 0

t > 0 u =3

4 = 0.75 ms–1

Impulse J = m (v – u)

=3

4 – 04

= 3 N-S.Impulse at t = 4s

For t < 4s u = –13 ms4

and for t > 4s v = 0

Impulse J = m (v – u)

= 34 0 – 4

= – 3.0 NS.Example 23. A body of mass 2 kg collides with a wall with speed 100 ms–1 and rebounds

with the same speed. If time of contact was 1

s50

calculate the force exerted on the wall.

Solution. Here, m = 2 kg, u = 100 ms–1, v = 100 ms–1

4O t(s)

x(m )

3


t =1

sec.50

Force =J P m v

t t t

=2 200

150

= 2 × 104 N.Example 24. If a force of 250 N acts on a body the momentum acquired is 125 kg ms–1. What

is the period for which force acts on the body.

Solution. Here, F = 250 N P = 125 kg ms–1.

t = ?

By definition of impulse

J = F t = P

t =P 125

F 250

= 0.5 sec.Example 25. A stream of water flowing horizontally with a speed of 15 ms–1, gushes out of

tube of cross sectional area 10–2 m2 and hits at a vertical wall nearby. What is the force exerted onthe wall by the impact of water, assuming it does not rebound. (Density of water = 103 kg m–3)

Solution. Volume of water flowing per sec = AV

Mass of water flowing per sec = AVForce applied by water on wall = Change in momentum per second.

F =P

t

= Vm

t

=AV

Vt

= (10–2) (15) (103) (15 – 0)

= 2250 N.Example 26. Compute the momentum of a 100 kg weight which has fallen from rest for two

second.

Solution. Here, m = 100 kg, t = 2 sec, u = 0

Velocity gained by body in 2 sec.

v = u + at

= 0 + 9.8 × 2

= 19.6 ms–1

Momentum P = mv= 100 × 19.6

= 1960 kg ms–1


Example 27. A force acts for 10s in the direction of motion. The force-time graph is shownin figure. Calculate the impulse of force.

F(N )

0

2

4

2 4 6 8 10t(s)

1 2 3

Solution. Impulse of force = Area between the force time graph and time axis

= Ar (1) + Ar (2) + Ar (3)

=1 1

(4 – 0) (4 – 0) (4 2) (6 – 4) (10 – 6) 22 2

= 8 + 6 + 8

= 22 N-S

BASED ON LAW OF CONSERVATION OF LINEAR MOMENTUMExample 28. A shell of mass 0.02 kg is fired by a gun of mass 100 kg. If the muzzle speed

of the shell is 80 ms–1. What is the recoil speed of gun?

Solution. Mass of shell = m = 0.02 kg

Mass of gun = M = 100 kg

Speed of shell v = 80 ms–1

Let recoil speed of gun = v.

mv + Mv = 0

or v =V

–M

m

v =0.02 80

–100

= – 0.016ms–1.– ve sign indicates recoil velocity which is in a direction opposite to the direction of velocity

of shell.

Example 29. A body of mass 1 kg initially at rest explodes and breaks into three fragmentsof masses in the ratio 1 : 1 : 3. The two pieces of equal mass fly off perpendicular to each otherwith a speed of 30 ms–1 each. What is the velocity of the heavier fragment.

Solution. Here, m1 : m2 : m3 = 1 : 1 : 3.

m1 + m2 + m3 = 1 kg

m1 = m2 = 0.2 kg

m3 = 0.6 kgv1 = v2 = 30 ms–1

v3 = ?


As shown in the figure, applying law of conservation of linear momentum.

m3 v3 = m1 v1 cos 45° + m2 v2 cos 45°

0.6 v3 = 2 × 0.2 × 30 × 0.707

or v3 =2 0.2 30 .707

0.6

= 14.14 ms–1

Example 30. A nucleus is at rest in the laboratory frame of references. Show that if it dis-integrates into two smaller nuclei the products must be emitted in opposite directions.

Solution. No external force is applied on the system and the system is at rest so

Initial momentum of the system is zero. If 1P��

and 2P��

are momentum of two parts of nucleusafter disintegration then from conservation of momentum

1 2P P��

= 0

or 1P��

= 2– P��

Let m1 and m2 be the masses of the two fragments and 1v�

and 2v�

be their velocities then

1P��

= 11m v�

and 2P��

= 22m v�

11m v�

= 22– m v�

or 2v�

=1

1

2

– .m

vm

�

– ve sign shows that two fragments must be emitted in opposite directions.

Example 31. A machine gun of 25 kg fires 30 g bullet at the rate of 10 bullet per second, witha speed of 400 ms–1. What force in Newtons is required to hold the gun in position?

Solution. Here, mass of bullet m = 30 g

u = 400 ms–1, v = 0.Change in momentum per second per bullet

= mv

=30

4001000

= 12 kg ms–1.

Change in momentum of 10 bullet per second

= 10 × 12

= 120 kg ms–1.

By definition change in momentum per second is called force

F = 120 N.


Example 32. A neutron has a mass 1.67 × 10–27 kg and moving at 108 ms–1 collides with adeutron at rest and sticks to it. If the mass of the deutron is 3.34 × 10–27 kg, find the speed of thecombination.

Solution. Here, mn = 1.67 × 10–27 kg vn = 108 ms–1.

md = 3.34 × 10–27 kg vd = 0

Let speed of combination is u then

mnvn + mdvd = (mn + md) u

u =–27 8

–27 –27

(1.67 10 ) (10 ) 0

(1.67 10 3.34 10 )

= 3.33 × 107ms–1.

BASED ON NEWTON’S THIRD LAW AND MOTION IN A LIFTExample 33. An elevator weights 4000 kg. When the upward tension in the supporting cable

is 48000 N. What is upward acceleration? Starting from rest how far does it rise in 3 sec?

Solution. Mass of elevator M = 4000 kg.

T = 48000 N

g = 9.8 ms–2.

Net force on cable in upward direction = T – mg

= 48000 – (4000) (9.8)

= 8800 N.

Upward acceleration a =8800

4000

= 2.2 ms–2

For upward motion u = 0, a = 2.2 ms–2

t = 3 s

s = 21.

2ut at

=1

0 3 3 2.22

= 9.9 m.Example 34. A man weighs 70 kg. He stands on a weighing machine in a lift which is moving

(i) upwards with a uniform speed of 10 ms–1

(ii) downwards with a uniform acceleration of 5 ms–2

(iii) upwards with a uniform acceleration of 5 ms–2. What should be the readings on the scalein each case?

What would be the reading if the lift mechanism failed and it came down freely under gravity?Solution. (i) When the lift moves upward with uniform velocity, reaction of the lift is equal

to the weight of man, i.e.,

W = mg

= 70 × 9.8

= 686 N or 70 kg wt.


(ii) When the lift is going downward with a uniform acceleration of 5 ms–2, the apparentweight is

W = m (g – a)

= 70 (9.8 – 5)

= 70 × 4.9 = 336 N

= 34.29 kgwt.(iii) When the lift moves upwards with uniform acceleration a = 5 ms–2, the apparent weight is

W = m (g + a)

= 70 (9.8 + 5)

= 70 × 14.8

= 1036N = 105.7 kgwt.In case lift mechanism is failed, the lift falls freely under gravity a = g

Apparent weight W = m (g – a)

= m (g – g)

= 0.It is called weightlessness.

Example 35. A monkey of mass 40 kg climbs on a rope which can with stand a maximumtension of 600 N. In which of the following cases will the rope break, the monkey

(i) climbs up with an acceleration of 6 ms–2.

(ii) climbs down with an acceleration of 4 ms–2.

(iii) climbs up with a uniform speed of 5 ms–1.

(iv) falls down the rope nearly free under gravity? take g = 10 ms–2.

Solution. (i) When the monkey climbs up with an acceleration a then tension in the rope isT = m (g + a)

= 40 (10 + 6) = 640 N

(ii) When the monkey climbs down with an acceleration of 4 ms–2, then tension in the rope is

T = m (g – a)

= 40 (10 – 4)

= 240 N.

(iii) In case of uniform speed

T = mg

= 40 × 10 = 400 N.

(iv) When monkey falls down freely

T = m (g – g)

= 0 N.

Only in first case T > 600 N (maximum tension, the rope can with stand)

So, rope will break in first case.Example 36. Ten one rupees coins are put on top of each other on a table. Each coin has a

mass m kg. Give the magnitude and direction of

(i) the force on the 7th coin (counted from bottom) due to all the coins on its top.

(ii) the force on the 7th coin by the eight coins

(iii) the reaction of the 6th coin on the 7th coin


Solution. (i) Force on 7th coin = force due to all the coins above it

= 3 mg.

(ii) Force on the 7th coin by the 8th coin

= 3 mg

(iii) Reaction of the 6th coin on the 7th coin

= force on 6th coin due to other coins above it

= 4 mg.Example 37. A rocket burns 0.25 kg of fuel per second and ejects it as a gas with a velocity

of 15 km s–1. Calculate the force exerted by the injected gas on the rocket.

Solution. Here, m = 0.25 kg/sec.

Velocity of ejection = 15 × 103 ms–1

Change in momentum per sec.

= mu

= 0.25 × 15 × 103

= 3750 NChange in momentum per second is called force, so F = 3700 N.Example 38. While launching a rocket of mass 2 × 104 kg, a force of 5 × 105 N is applied

for 10s. Calculate the velocity attained by rocket at the end of 10s.

Solution. Here, m = 2 × 104 kg

F = 5 × 105 N t = 10s

Impulse of force = F × t = m (v – u)

5 × 105 × 10 = 2 × 104 (v – 0)

or v =5

4

5 10 10

2 10

= 2.50 × 102 ms–1.

BASED ON MOTION OF CONNECTED BODIESExample 39. Two masses M and m are connected at the two ends of an extensible string. The

string passes over a smooth frictionless pulley. Calculate the acceleration in the system and tensionin the string. (M > m).

Solution. Let acceleration in the two masses be a and tension in the stringsbe T.

Using force body diagram for mass M and m we have

Mg – T = Ma ...(1)

and T – mg = ma ...(2)

Add equation (1) and (2)

(M – m)g = (M + m)a

or a =M –

M

mg

m

...(3)

Using Equation (2) and (3)

T = mg + ma


=M –

M

mmg m g

m

=2M

M

mg

m.

Example 40. Two masses 8 kg and 12 kg are connected at the two ends of a light inextensiblestring that goes over a frictionless pulley. Find the acceleration of the masses and the tension in thestring, when the masses are released.

Solution. Here M = 12 kg, m = 8 kg

Acceleration in the system is

a =M –

M

mg

m

=12 – 8

1012 8

=8

1020

= 4 ms–2

Also tension in the string in

T =2 M

M

mg

m

=2 12 8

1012 8

= 96 N.Example 41. Two bodies of masses m1 = 40 kg and m2 = 60 kg are tied by a light string and

are placed on a frictionless horizontal surface. When m1 is pulled by a force F an acceleration of5 ms–2 is produced in both the bodies. Calculate the value of F and tension in the string.

Solution. Here, m1 = 40 kg

m2 = 60 kg

a = 5 ms–2

Using free body diagram

For mass m2

T = m2 a

= 60 × 5

= 300 N.

For mass m1 F – T = m1 a

F – 300 = 40 × 5

F = 500 N.Example 42. Three blocks of masses m1, m2 and m3 are placed in contact on a smooth

horizontal surface. Force F is applied on mass m1. Determine contact force between m1, m2 andm2, m3.

a

m 1 m 2FT

m 1F Ta

m 2T


Solution. Acceleration in the system is

a =1 2 3

F F

M m m m

Drawing free body diagram of m1, m2 and m3

m 3R 2 m 2

R 1 R 2 m 1F R 1

R2 = m3a

= 31 2 3

Fm

m m m

= 3

1 2 3

Fm

m m m

F – R1 = m1 a

R1 = F – m1 a

= 11 2 3

FF – m

m m m

R1 = 2 3

1 2 3

( )F.

m m

m m m

Example 43. Two masses 10 kg and 20 kg respectively are connected by a massless springas shown. Force of 300 N acts on the 20 kg mass. At the instant shown in figure the 10 kg masshas 15 ms–2. What is the acceleration of 20 kg mass.

Solution. Force on mass m1

= m1 a

= 10 × 15 = 150 N.

Using free body concept for mass m2

300 – F = 20 × a2

300 – 150 = 20 a2

or a2 =150

20

= 7.5 ms–2.Example 44. A uniform rope of 2 m is pulled by a force of 4N. What is the tension in the

rope at a distance of 0.5 m from the end where force is applied?

Solution. Let T be the tension in the rope at point O

A B2 m

0.5 m4 N

Now a =F 4

Mm

m 1

10 K g

m 2

20 K gF F

300 N


Writing equation of motion of part OB

(4 – T) = m (0.5) a

Where m is mass per unit length = M

2

4 – T =M 4

(0.5)2 M

4 – T = 1

T = 3 N.Example 45. Three blocks are connected by strings as shown, and are pulled by a force of

T3 = 150 N. If m1 = 5 kg, m2 = 10 kg and m3 = 15 kg. Calculate the acceleration of the system T1and T2.

m 1 m 2 m 3T1 T2 T 3

Solution. Acceleration of the system a = 1 2 3

F

m m m

a =150

5 10 15 = 5 ms–2.

(ii) T2 = (m1 + m2) a= (5 + 10) × 5 = 75 N.

(iii) T1 = m1a

= 5 × 5 = 25 N.Example 46. A block of mass m is placed on a horizontal table. It is connected to another

mass M (M > m) with the help of inextensible string passing over a pulley as shown. Calculate theacceleration in the system and tension in the string.

Solution. Using concept of free body diagram.

Mg – T = Ma

and T = ma

Mg – ma = Ma

Mg = a (M + m)

a =M

M

g

m .

Also tension in the string T = M

M

gm

m

T =M

.M

mg

m

Example 47. Three masses m1, m2 and m3 of the three bodies are 5, 2 and 3 kg respectively.Calculate the values of the tension T1, T2 and T3 when the system is going upward with anacceleration of 2 ms–2. (ii) System is at rest (given g = 10 ms–2).

m

M

T

a

M g

a

T


Solution. The force with which the system is pulled upward,

T1 = (m1 + m2 + m3) (g + a)

(as system is going upward with acceleration a):

T1 = (5 + 2 + 3) (10 + 2)

= 120 N.Similarly,

T2 = (m2 + m3) (g + a)

= (2 + 3) (10 + 2)

= 60 Nand T3 = (m3) (g + a)

= 3 (10 + 2) = 36 N(ii) When the system is at rest a = 0. T1 = (m1 + m2 + m3)g = (5 + 2 + 3) × 10

= 100 N.T2 = (m2 + m3) g

= (3 + 2) 10

= 50 NT3 = m3 g = 3 × 10 = 30 N.

Example 48. In the figure masses A, B and C are 1 kg,3 kg and 2 kg respectively.

Find (a) the acceleration of the system (b) tension in thestring (g = 10 ms–2).

Solution. Resolving weights of masses A, B and C in thecomponents along the inclined planes and perpendicular to them

The net pulling force

= mAg sin 60° + mBg sin 60° – mCg sin 30°

= (mA + mB)g sin 60° – mCg sin 30°

= (1 + 3) × 10 × 3 1

– 2 102 2

= 20 3 –10= 24.64 N.

Acceleration of system =24.64

6

= 4.1 ms–2.

(b) For tension in the string between A and B

mAg sin 60° – T1 = mAa

T1 = mAg sin 60° – mAa

= mA (g sin 60° – a)

=3

1 10 – 4.12

= 4.56 N

60° 30°

A

BC

T2T2

T1

T1

m 1

m 2

m 3

T2

T3


For tension in the string between B and C

T2 = mCg sin 30° + mCa

T2 = mC(a + g sin 30°)

=1

2 4.1 10 .2

= 18.2 N.

Example 49. In the figure shown (a) What is the acceleration ofthe masses.

(b) What is the tension in each string.

Solution. The acceleration in every part of the system will besame say it is ‘a’.

Applying Newton’s law

mg – T = ma ...(i)

and 2T – mg = ma ...(ii)

multiply equation (i) by 2 and add to equation (ii)

mg = 3 ma

or a = .3g

(b) Tension in the string T = m (g – a)

= m –3

gg

=2

.3

mg

Example 50. Figure shows a uniform rope of mass 3 kg and length 30 cm. It is pulled byconstant forces 20 N and 32 N. Find out its acceleration and tension at point P.

Solution. Mass per unit length of the rope is 3

30

=1

kg/cm.10

mA =1

1010

= 1 kg.

mB = 2 kg.

T T

m

m ga

T T

m

m ga

ma


acceleration of the rope =32 – 20

3 = 4 ms–2

Let tension at point P is T.

T – 20 = 1 × 4

or T = 24 N.Example 51. In terms of masses m1, m2 and g find the

acceleration of both the blocks shown in figure. Neglect all frictionand masses of the pulley.

Solution. As mass m2 moves downward m1 moves towardsright.

Mass m1 moves towards right through distance x, then mass

m2 moves through 2

x . In otherwords of acceleration of m1 is a

then that of m2 will be 2

a . Applying Newton’s second law to masses m1 and m2.

T1 = m1a

mg – T2 = 2 2

am

Also T2 = 2T1 = 2 m1a

m2g – 2m1a = 2 2

am

or 2m2g = (4m1 + m2)a

or a = 2

1 2

2

4

mg

m m

Acceleration of mass m2 = 2

a

acceleration of mass m2 = 2

1 2

21

2 4

mg

m m

=

2

1 24m

gm m

Example 51. Two blocks connected by a cord passing over a small frictionless pulley rest ona smooth plane as shown.

(i) Which way will the system move

(ii) What is the acceleration of blocks

(iii) What is the tension in the cord.

30° 53°

A B 50 kg

100 kg

T1

T1T1

T2

m 2

m 1


Solution. (i) A moves down the plane and B up the plane

(ii) Resolving components of weight

mg sin 1 – T = m1a

100 g × sin 30° – T = 100 a ...(i)

Also T – 50 g sin 53° = 50 a ...(ii)


100 g sin 30° – 50 g sin 53° = 150 a

1100 – (50 ) (0.7986)

2g g

= 150 a

In solving

a = + 0.65 ms–2.Example 52. What is the acceleration of the block and trolly system shown in figure, if the

coefficient of kinetic friction between the trolley and the surface is 0.04? What is the tension in thestring? (g = 10 ms–2)

Solution. Let a be the acceleration in the system.

Applying second law of motion (in case of block)

f20 K g

30 N

3 Kg

30 – T = 3a ...(i)

(in case of trolley)

T – fk = 20 a ...(ii)

Here, fk = kN = 0.04 × 20 × 10 = 8 N

T – 8 = 20a ...(iii)

Solving equation (i) and (iii)

a =22

23 = 0.956 ms–2.

and T = 27.13 N.Example 53. Two masses P and Q equal to m1 and m2 respectively are connected to one

another via a massless string passing over a pulley. Mass Q rests on a fixed horizontal surface. If is the coefficient of friction between any two surfaces. Show that the minimum force F neededto keep P and Q moving with uniform speed is μ(3m1 + m2) g.

P

Q

F

Solution. Various forces acting on the two bodies and the surface are shown below.

P

Q

FFT

f2

f1f 1

f 2

R 1

R 2

m g1

m g2


For mass m1

R1 = m1g

f1 = μm1g = f1 ...(i)

T + f1 = F

or F = T + μm1g ...(ii)

For mass m2

R2 = (m1 + m2)g

f2 = μR2 = μ(m1 + m2)g = f2 ...(iii)

Total force of friction

f = f1 + f2= μ(m1 + m2)g + μm1g

= μ(2m1 + m2)g

As P and Q moving with uniform speed

T = f = μ(2m1 + m2)g

Putting this value in equation (ii)

F = μ(2m1 + m2)g + μm1g

= μ(3m1 + m2)g.Example 54. Two bodies of masses 11 kg and 11.5 kg are connected by a long light string

passing over a smooth pulley. Find

(i) The velocity of each body at the end of 4s.

(ii) The height ascended or descended by each as the case may be during that time.

(iii) If at the end of 4s, the string be cut. Find the position of each body in next 2 seconds.Solution: For the upward motion of body A.

T – 11 g = 11 a ...(i)

Similarly for the downward motion of B

11.5g – T = 11.5 a ...(ii)

Solving equation (i) and (ii)

a = 0.218 ms–2.(ii) At the end of 4s velocity of body A and B is

v = u + at

= 0 + 0.218 × 4 = 0.87 ms–1

Distance covered by each body is

h = 21

2ut at

= 210 (0.218) (4)

2

= 1.74 m.(iii) When the string is cut, both the bodies will fall freely under gravity with u = 0.87 ms–1

s = 21

2ut gt

A

11g

T

a

B

11.5 g

Ta


=1

0.87 (2) 9.8(2 2)2

= 21.34 m.Example 55. A box of mass 4 kg rests upon an inclined plane.

The inclination of the plane to the horizontal is gradually increased. Itis found that when the slope of the plane is 1 in 3, the box slides downthe plane, (g = 9.8 ms–2)

(i) Find the coefficient of friction between the box and plane

(ii) What force applied to the box parallel to the plane will justmake it move up the plane?

Solution. Here, m = 4 kg

sin = 13

and g = 9.8 ms–2

(i) On an inclined plane

= tan

=1

8

= 0.35(ii) Minimum force required to move the box up the plane.

= mg sin + f= mg sin + mg sin = 2 mg sin

=1

2 4 9.83

= 26.13 N.Example 56. A body m1 of mass 9 kg and another body m2 of mass 6 kg are connected by a

light string. Consider a smooth inclined plane 30° over which one of them can be placed while theother one hangs vertically and freely. Show that m1 will drag m2 up the whole length of the planein half the time that m2 hanging vertically would take to draw m1 up the plane.

Solution.

30°

m 2m 1

m g1

T9 KgT m 1 6 Kg

m 2

Case (i) Let a1 be the acceleration in the system when 9 kg mass hangs freely and T be thetension in the string.

m1g – T = m1a1

and T – m2g sin 30° = m2a1

g[m1 – m2 sin 30°] = a1(m1 + m2)

Here, m1 = 9 kg, m2 = 6 kg

m g cos W =m g

mg sin

fN

31


On solving a1 = –22 ms5g .

Case (ii) When 6 kg mass hangs freely then acceleration of the system is a2 and tension instring is T.

m2g – T = m2a2

T – m1g sin 30° = m1a2

or g(m2 – m1 sin 30°) = a2(m1 + m2)

On solving a2 =10

g

Let length of plane be s

In case (i) s = 21 1

10

2a t .

or t1 =1

2s

a

Similarly in case (ii) t2 =2

2s

a

1

2

t

t= 2

1

a

a =

1.

2

Example 57. A mass of 15 kg and another mass of 6 kg are attachedto a pulley system as shown. A is a pulley while B is a movable one bothare considered light and frictionless. Find the acceleration of 6 kg mass.

Solution. In the string tension every where will remain same.

M1 will move downward and M2 upwards.

Now M1g – 2T = M1a ...(i)

T – M2g = M2 2a ...(ii)


a = 1 2

1 2

– 2

4

M Mg

M M

=15 –12

15 24g

=

3

39g

or a = .13

g

Acceleration of 6 kg mass = –22g13

ms .

EQUILIBRIUM OF CONCURRENT FORCES (LAMI’S THEOREM)Example 58. Calculate the tension T1, T2 and T3 in the massless strings shown in figure.

(g = 10 ms–2).

Solution. Here, T3 = 5 × 10 = 50 N.

T

1.5 kg

T

m 26 kg

m 1

B

TA


Applying Lami’s theorem

1T

sin (90 60) = 2 3T T

sin (90 30) sin90

or1T

sin150 =2T

sin120 = 50

sin 90 = 50

T1 = sin 150° × 50

= cos 60° × 50 = 1

502 = 25 N.

T2 = 50 sin 120°

= 50 × cos 30° = 3

502

25 3 N.

Alternate Method: As point O is in equilibrium we have

T2 sin 60° + T1 sin 30° = 50 ...(i)

T1 cos 30° = T2 cos 60° ...(ii)

On solving equation (i) and (ii) we have

T1 = 25 N

T2 = 25 3 N.

Example 59. If tension T2 in the string is 100 N. Find out tension T1 and weight of the block W.

Solution.

30° 60°

T 2T 1

W

Point O is in equilibrium so

T1 cos 30° = T2 cos 60°

T1 =1 2

1002 3

= 1003 N

Also W = (T1 sin 30° + T2 sin 60°)

W =100 1 3

1002 23

= 115.47 N.

30° 60°

30° 60°

T Cos 60°2

T2T1

5kg

OT Cos 30°1

T Sin 60°+

2

T Sin 30°1


Example 60. Determine tension T1 and T2 in the figure shown.

Solution.

T2

T1

60°

4 kg w t

As shown in figure

T1 sin 60° = 4 kg wt

or T1 =4 kg wt

sin 60 =

4

32

=8

kg wt3

Also T2 = T1 cos 60°

=8 1

23 =

4kg wt.

3

Example 61. A mass of 6 kg is suspended by a rope of length 2 m from a rigid surface. Aforce of 50 N in the horizontal direction is applied at the mid point of the rope as shown. What isthe angle the rope makes with the vertical direction in equilibrium. (g = 10 ms–2)

Solution. Suppose the rope makes an angle with vertical in equilibrium. Three forces actingat the point P are T1, T2 and T3 respectively.

Resolving T1 into its horizontal and vertical components

T1 sin = T3 = 50 N ...(i)

T1 cos = T2 = 60 N ...(ii)

T2

T1

T2

T Cos 1

T S in 1 T = 50 N3

60 N

Dividing equation (i) by (ii)

Tan = 5 .6

or = 39.8°.


Example 62. A body of weight 100 N is suspended with the help of strings as shown. Findtension T1 and T2.

Solution. Using Lami’s theorem in the given figure

1T

sin135 = 2T 100.

sin120 sin105

T1 =sin135

100sin105

= 146.6 N.

T2 =sin120

100sin105

= 179.3 N.

Example 63. A mass m is hung with a light inextensible string as shown. Find the tension inhorizontal string.

Solution. Let T1 and T2 be the tension in string PA and PB respectively.

As point P is in equilibrium

T2 sin 30° = mg

T2 = 2 mg

In horizontal direction,

T1 = T2 cos 30°

=3

22

mg

= 3 mg

BASED ON FRICTION, ANGLE OF FRICTION AND MOTION ALONG ROUGHINCLINED PLANES

Example 64. A block of weight 100 N lying on a horizontal surface just begins to move whena horizontal force of 25 N acts on it. Determine the coefficient of static friction.

Solution. We know fs = μ5N.

Here, fs = 25 N N = 100 N.

μ s =N

sf = 25

100

= 0.25

Example 65. A block lying on an inclined plane has a weight of 50 N. It just begins to slidedown when inclination of plane with the horizontal is 30°. Find μs.

Solution. Here Weight = 50 N.

= 30° (Angle of repose)

μ s = tan

= tan = 1

3.


Example 66. A block of mass 2kg in placed on the floor. The coefficient of static friction is0.4. A force of 2.5 N is applied on the block as shown. Calculate the force of friction between theblock and the floor.

Solution. Here, m = 2 kg F = 2.5 N

μ s = 0.4 g = 9.8 ms–2

Now fs = μ sN

= μ smg

= 0.4 × 2 × 9.8 = 7.84 N.

F, the applied force is less than the limiting friction, so the block does not move.

In this case force of friction = applied force = 2.5 N.

Example 67. Length of a chain is L and coefficient of static friction is μ. Calculate the lengthof chain which can be hang from the table without sliding.

Solution. Let M is mass of L length of chain.

Length of the chain hanging = y

Length of the chain on table = (L – y)

Weight of the part of chain on table = M

(L – )L

y g

and weight of hanging part =M

Ly g

For the chains in equilibrium μN = W

Mμ ( – )

LL y g

=M

Ly g

or y =μ

1 μL

Example 68. How much work will be done in dragging a body of mass 200 kg through a

distance of 80 m on a level road, if the coefficient is 0.25.

Solution. Here, μ = 0.25 M = 200 kg

F = ? s = 80 M

μ =F

N

or F = μN = μMg

= 0.25 × 200 × 9.8

= 490 N Work done W = F × s

= 490 × 80

= 39,200 JExample 69. A body weighing 20 kg just slides down a rough plane that rises 5 cm in every

12 cm. Find, the coefficient of friction.

Solution. sin =5

12(because the plane rises 5 cm in every 12 cm)

F

f

L – Y

Y


cos = 2 25 1191 sin 1–

144 12

μ = tan = 5

119 = 0.4583

Example 70. A 10 kg block slides without acceleration down a rough inclined plane makingan angle of 20° with the horizontal. Find the acceleration when the inclination of the plane isincreased to 30°.

Solution. In the first case, force of friction is equal and opposite to the component of theweight of the block along the inclined plane. Component of the weight of the block along theinclined plane = mg sin

Let the force of friction be F

F = mg sin Here = 20°

F = mg sin 20° ...(i)

In the second case, the component of the weight of the block along the inclined plane = mgsin 30° and the force of friction F = mg sin 20°, acts in the opposite direction.

Net force along the inclined plane

= mg sin 30° – mg sin 20°

But Force = ma

ma = mg (sin 30° – sin 20°)

a = g (sin 30° – sin 20°)a = 9.8 (0.5 – 0.3420) = 1.55 m/s2

Example 71. A cubical block rests on an inclined plane with the four edges horizontal. If

μ = 1

,3

determine the angle when the block just slides.

Solution. μ =1

3

Formula μ = tan , 1

3 = tan ,

But tan =1

3 = 30°.

Example 72. The weight of a train is 2500 quintals and the resistance due to friction andmotion through air is 0.5 newton per quintal of weight. Find the power of the locomotive whichcan maintain a speed of 72 kilometers per hour on a level track.

Solution. Here, mass of the train = 2500 quintals

Force of friction = 0.5 newton/quintal

Total force of friction = 0.5 × 2500 = 1250 newton

Speed = 7.2 kilometers/hour = 72000

3600= 20 m/s


Work done/second = F × S = 1250 × 20 Joules/s

= 25000 watts

Power = 25000 watts = 25 kilowatts.Example 73. Find the distance travelled by a body before coming to rest if it is moving on

the ground with a velocity of 36 km/hour and the coefficient of friction between the body and theground is 0.2.

Solution. Velocity = 36 km/hour = 10 m/s

Coefficient of friction,

μ =F

N

where F is the force of friction and N is the normal reaction

F = μN

But, F = ma

N = mg

ma = μmg

a = μg = 0.2 × 9.8 = 1.96 m/s2

Here a is retardation

a = – 1.96 m/s2

Now u = 10 m/s, a = – 1.96 m/s2

v = 0, s = ?

Applying the formula

v2 – u2 = 2as

0 – (10)2 = 2(– 1.96) × s

s =10 ×10

2×1.96 = 25.51 m

Example 74. A body is sliding down a rough inclined plane which makes an angle of 30°with the horizontal. Calculate the acceleration. (Coefficient of friction = 0.25)

Solution. Force along the inclined plane

= F = mg sin Force of friction = f = μN = μmg cos Net downward force along the plane

F = F – f

F = mg sin – μ mg cos But F = ma

ma = mg sin – μ mg cos

a =3

sin 30° – 0.25×2

g

= 9.8 [0.5 – 0.2165] = 2.778 m/s2.Example 75. A body of mass 1 kg is placed on an incline of 1 in 10. If the coefficient of

friction is 0.3. What force must be applied (i) to just set the body in motion up the plane and(ii) to prevent the body from running down the plane.


Solution. Here, mass = 1 kg, sin = 1 .10

g = 9.8 ms–2.

Weight of the body down the inclined plane due to gravity

F1 = mg sin

=1

1 9.810

= 0.98 N

cos = 21– sin = 2

11–

10

= 99

1100

� (App.)

Normal reaction N = mg cos = 1 × 9.8 × 1

= 9.8

μ = 0.3

Force of friction F2 = μN

= 0.3 × 9.8 = 2.94 N.

To move the body up the plane, the force applied

= F1 + F2

= 0.98 + 2.94 = 3.92 N(ii) As the force of friction F2 up the plane is more then the force of gravity F1 down the

plane, the body does not slide. It will be at rest. Hence force required is zero.Example 76. A block of weight 100 N lying on a horizontal surface is pushed by a force F

acting at an angle 30° with horizontal. For what value of F will the block begin to move if μs =0.25.

Solution. Free body diagram of the block is shown below.

N = mg + F sin 30°

F cos 30° = μ s N

or F cos 30° = μ s (mg + F sin 30°)

or F =μ ( )

cos 30 –μ sin30s

s

mg

=0.25 100 2

3 – (0.25) 1

= 33.74 N.

30°F


Example 77. A block of mass 4 kg rests on a plane inclined at an angle of 37º with thehorizontal. The coefficient of friction between the block and the surface is 0.7 (i) What will be thefrictional force acting on the block? (ii) What is the force applied by inclined plane on block.

Solution. Normal reaction of the surface

N = mg cos 37°

Limiting force of friction fs = μ sN

= μ s mg cos 37°

=4

0.7 4 105

= 22.4 N

mg sin 37° = 3

4 105

= 24 N

Force of friction = 11.2N.Example 78. A horizontal force of 12 N is applied to a 1.5 kg block which rests on a

horizontal surface. If the coefficient of friction is 0.3, calculate the acceleration produced in theblock.

Solution. Here, m = 1.5 kg, F = 12 N, μ = 0.3.

Force of friction

f = μmg

= 0.3 × 1.5 × 10 = 4.5 N.

Net force on body = 12 – 4.5 = 7.5 N.

a =Fnet

m

=7.5

1.5 = 5m/s2.

Example 79. Two bodies A and B of masses 5 kg and 10 kg in contact with each other reston a table against a rigid partition. The coefficient of friction between the bodies and the table is0.15. A force of 200 N is applied horizontally at A. What are (i) the reaction of the partition(ii) the action-reaction forces between A and B? What happens when the partition is removed?Does answers to (ii) change, when the bodies are in motion? Ignore difference between μs and μk.

Solution: Here mA = 5 kg, mB = 10 kg.

μ = 0.15, F = 200 N

(i) Force of limiting friction f = μN

= μ(mA + mB)g

= .15 (5 + 10) × 9.8

= 22.05 N.

200 N

A B


When force F is applied, the net force applied on the partition

F – f = 200 – 22.05

= 177.95 N.Reaction of partition = 177.95 N (towards left)

(ii) Force of limiting friction on body A

f1 = μmAg

= .15 × 5 × 9.8

= 7.35 N

Force exerted by body A on body B is

= 200 – 7.35

= 192.65 N (towards right).

Reaction of body B on A

= 192.65 N (towards left).

When partition is removed, the system moves under the action of net force i.e. Fnet = 177.95 N

Acceleration produced in the system

a =177.95

10 5 = 11.86 ms–2.

Force on body A = mA a

= 5 × 11.86= 59.3 N.

Net force exerted by A on B after the removal of partition = fnet = mAa

= 192.65 – 59.3

= 133.35 N. (towards right)

Reaction of the body B on A = 133.35 N (towards left)

Example 80. Find the force required to move a train of 2000 quintal up an incline of 1 in 50with an acceleration of 2 ms–2, the force of friction being 0.5 N/per quintal.

Solution. Here, M = 2000 quintal = 2 × 105 kg.

sin =1

,50

a = 2 ms–2.

f = 0.5 × 2000 = 1000 N.Force required in moving the train up the inclined plane

= mg sin

= 5 12 10 9.8

50

= 39,200 N.

Force required to produce an acceleration of 2 ms–2

= ma

= 2000 × 102 × 2

= 4 × 105 N.

Total force required

= 1000 + 39,200 + 4 × 105.

= 4.402 × 105 N.


Example 81. A block of mass 2 kg slides down an inclined plane which makes an angle of

30° with the horizontal. The coefficient of friction between the block and surface is 3

.2

(i) What force must be applied to the block so that the block moves down the plane withoutacceleration?

(ii) What force should be applied to the block so that it can move up without anyacceleration?

Solution. When the block moves down the plane withoutacceleration then.

F + mg sin – f = 0

or F = f – mg sin = μmg cos – mg sin = mg (μ cos – sin )

=3

2 9.8 cos30 – sin 302

=3 3 1

2 9.8 –2 2 2

= 3 12 9.8 –4 2

= 4.9 N.(ii) To move the block upwards

F = mg sin + μmg cos = mg(sin + μ cos )

=3 1

2 9.84 2

=3 1

19.64 2

= 24.5 N.

Example 82. A body of mass 100 g is sliding from an inclined plane of inclination 30°. Whatis the frictional force experienced if μ = 1.7?


= 30°, = 1.7

Friction force f = μN

= μmg cos = 1.7 × 0.1 × 10 × cos 30°

=3

1.72

= 1.47 N.

Example 83. A ball rolls on ice with a velocity of 5.6 ms–1 and comes to rest after travelling10 m. Determine the coefficient of friction.

Solution. Here, u = 5.6 ms–1

s = 10 m

m g cosm g

m g sin

NF

f = N


Using third equation of motion

v2 = u2 + 2as

0 = (5.6)2 + 2(a) (10)

a = 1.568 ms–2.

Now a = μg

or μ =a

g =

1.568

9.8

= 0.16.Example 84. When an automobile moving with a speed of 36 km h–1 reaches the foot of an

upward inclined road of angle 30°, its engine is switched off. The coefficient of friction involvedis 0.1. How much distance will the automobile move before coming to rest. (g = 10 ms–2)

Solution. Here, u = 36 km h–1 = 5

3618

= 10 ms–1

v = 0

= 30°, μ = 0.1

g = 10 ms–2.

When automobile moves up an inclined plane, the downward force acting on it

F = mg (sin + μ cos )

a = F

m= 10 (sin 30° + 0.1 cos 30°)

= 5.9 ms–2

Using third equation of motionv2 – u2 = 2as

0 – (+ 10)2 = 2 × 5.9 s

or s � 8.5 m.Example 85. Calculate the time required by a block to come to rest from a speed of 10 ms–1

moving on a horizontal surface where k = 0.2. What is the distance covered before coming to rest?(g = 10 ms–2)

Solution. u = 10 ms–1, μk = 0.2

Retardation in the block due to friction

a = μg

= 0.2 × 10

= 2 ms–2

Now v = u + at

0 = 10 – 2 t

t =10.0

2 = 5 sec.

Let displacement is s

s =2 2–

2

v u

a =

2 20 – (10)

2 (–2)

= 25 m.


Example 86. Two blocks of mass 2kg and 5 kg are connected by an ideal string passing overa pulley. The block of mass 2 kg is free to slide on a surface inclined at an angle of 30° withhorizontal. Where as 5 kg block hangs freely. Find the acceleration of the system and the tensionin the string (μ = 0.30)

Solution. Let acceleration in the system is a, as shown in figure.

30°

m 1

T

5 Kg

Ta

5 Kg2gf + 2g S in

2g C os

N

Using concept of free body diagram the equation of motion may be written as

5g – T = 5a ...(i)

T – 2g sin – f = 2a ...(ii)

Add equation (i) and (ii)

5g – 2g sin – f = 7a.

But f = μN = μmg cos .

5g – 2g sin – μmg cos = 7a.

or 10[5 – 2 × sin 30° – .30 × 2 × cos 30°] = 7a

or 10[5 – 1 – 0.6 × 0.866] = 7a

or 10[5 – 1 – .5196] = 7a

or 10 × 3.48 = 7a

or a =34.8

7

� 5 ms–2.Also 5g – T = 5a

5 × 10 – T = 5 × 5

or T = 25 N.Example 87. A block of mass 100 g where placed over an inclined plane, inclined at 30°

slides down without acceleration. If the inclination is increased by 30° what would be theacceleration in the block.


= 30°

= 30 + 30 = 60°.

Now μ = tan = tan 30°

=1

3 = 0.577

Let acceleration produced in the block is a

f = mg sin – f = mg sin – μmg cos

m g cos

m g

mg sin

Nf

60°


or a =F sin – μ cosmg mg

m m

= g sin – μg cos = 10 × sin 60° – 0.577 × 10 cos 60°

=3 1

10 – 0.577 102 2

= 5 3 – 2.885

= 5.78 ms–2.Example 88. In the figure the body is moving with constant speed, determine the coefficient

of friction.

Solution. Free body diagram is shown below.

30°

f

F S in 30°

F = 125 N

F Cos 30°

m g

N

30°

F = 125 N

125 k

N + F sin = mg ...(i)

f = F cos ...(ii)

N = mg – F sin

μ =N

f =

F cos

– Fsinmg

=o125 cos 30

25 9.8 125 sin 30 = 0.59.

Example 89. Find the power of an engine which can maintain a speed of 50 ms–1 for a trainof mass 4 × 105 kg on a rough plane. The coefficient of friction is 0.04. (g = 10 ms–2).

Solution. Here, u = 50 ms–1,

m = 4 × 105 kg, μ = 0.04

Force of friction is

f = μN = μmg

= 0.04 × 4 × 105 × 10= 1.6 × 105 N

Power of engine

P = FV= 1.6 × 105 × 50

= 8 × 106 W

= 8000 KW


Example 90. A block of mass 2 kg rests on an inclined plane which makes an angle of 30°

with the horizontal. The coefficient of friction between the block and the surface is 3

2.

(i) What force should be applied to the block so that it moves down without any acceleration.

(ii) What force should be applied on the block so that it moves up without any acceleration.

(iii) Calculate the ratio of the powers in the two cases if the block moves with uniform speedin both the cases.

Solution. (i) Let F1 is the force applied down the plane then to move the block withoutacceleration

F1 + mg sin = f

or F1 = f – mg sin = μmg cos – mg sin .

= mg (μcos – sin )

Here m = 2 kg, μ = 3

2, = 30°.

F1 =3

2 9.8 cos30 – sin302

=3 3 1

2 9.8 –2 2 2

=3

9.8 –12

= 11.0 N.

(ii) Let F2 is the force required to move the block upward without acceleration

F2 = f + mg sin .

m g cosm g

m g sin

NF 2

f

or F2 = mg (μ cos + sin )

=3

2 9.8 cos30 sin 302

=3 3 1

2 9.82 2 2

= 30.6 N.

(iii) Speed is same in both the cases

1

2

P

P= 1

2

F

F

u

u

= 1

2

F 11

F 30.6

= 0.36.

m g cosm g

mg sin

NF


EXERCISE

BASED ON NEWTON’S SECOND LAW OF MOTION1. An automobile of mass 2.5 × 103 kg is moving with a velocity of 36 km h–1. By the

application of brakes it is brought to rest in a distance of 25 m. Find the average force ofresistance in Newton. [Ans. 5 × 103 N]

2. Two balls A and B of masses 100 g and 250 g respectively are connected by a stretchedstring of negligible mass and placed on a smooth table. When the balls are releasedsimultaneously, the initial acceleration of ball B is 10 cm s–2 westward. What is themagnitude and direction of the initial acceleration of the ball A. [Ans. 25 cm s–2 Eastward]

3. In the given figure three bodies are shown connected to each other with strings. The mass ofthe bodies are m, 3m and 5m respectively and they are being pulled with a force F on africtionless horizontal surface. The tension P in the first string is 16N. Calculate (i) accelera-tion of the bodies (ii) tension Q in second string (iii) the force F.

[Ans. 2m (ii) 10 N (iii) 18 N]

4. A force of 10 N acts on a body for 10 s and the body acquires a velocity of 50 ms–1. Find themass of the body. [Ans. 2 kg]

5. A particle of 50 gm is acted by a force of 100 dynes. Assuming that the particles starts fromrest. Calculate the distance travelled and velocity after 20 seconds. [Ans. 40 m, 4 ms–1]

6. A constant force acts for 5 seconds on a body of mass 5 quintals and then ceases to act.During the next 5 seconds, the body describes 100 meters. Find the magnitude of the forcein newtons. [Ans. 2000 N]

7. A body of mass 20 kg is moving with a velocity of 100 ms–1. Calculate the magnitude of theforce required to stop it in 25 seconds. Calculate also the distance it moves before coming torest. [Ans. 80 N, 1250 m]

8. A body of mass 10 kg traverses 10, 11 and 12 m in successive seconds. Calculate theacceleration and the force acting on the body. [Ans. 1 ms–2, 10 N]

9. A force acts for 10s on a body of mass 10 kg after which the force ceases and the bodydescribes 100 m in next 5 sec. Find the magnitude of the force. [Ans. 20 N]

10. A force of 10 N produces an acceleration of 8 ms–2 in a body of mass m1 and an accelerationof 24 ms–2 in a body of mass m2. What will be the acceleration if same force is applied onboth the masses tied together. [Ans. 6 ms–2]

11. A body is acted upon by two mutually perpendicular forces of 3N and 4N respectively. Ifmass of body is 1 kg. Calculate the acceleration produced in it.

[Ans. a = 5 ms–2 making an angle = tan–1 43

from the direction of 3 N force]

12. A ship of mass 3 × 107 kg and initially at rest can be pulled through a distance of 3m bymeans of a force of 5 × 104 N. The water resistance is negligible. Find the speed attained bythe ship. [Ans. 0.1 ms–1]

13. The velocity of a particle of mass 5 kg is given by V��

= 2ˆ ˆ2 65t i j ms–1. Calculate the force

experienced by the particle, at t = 10 s. [Ans. ˆ200 Ni ]

14. A force of 72 dyne is inclined to the horizontal at an angle of 60°, find the acceleration in amass of 9 g which move in the effect of this force in a horizontal direction. [Ans. 4 cms–2]

m 3 m 5 mF P Q


15. A machine gun fires a bullet of mass 50 g with velocity 1000 ms–1. If average force acting ongun is 200 N then find out maximum number of bullets fired per minute. [Ans. 240]

16. In an X-ray machine an electron is subjected to a force of 10–23 N. Calculate the time takenby the electron to cover a distance of 10 cm. (me = 10–30 kg). [Ans. 1.4 × 10–4 s]

BASED ON MOMENTUM LAW OF CONSERVATION OF MOMENTUM AND IMPULSE17. A projectile weighing 200 kg is fired from a gun weighing 8000 kg with a velocity of 5 ms–1.

Find the recoil velocity of gun. [Ans. 12.5 ms–1]

18. Two balls of different masses have same kinetic energy. Which ball (lighter or heavier) willpossess greater momentum? [Ans. Heavier ball]

19. A 20g bullet pierces through a plate of mass M1 = 1 kg and then comes to rest in secondplate of mass M2 = 2.98 kg as shown. It is found that the two plates, initially at rest, nowmove with equal velocities. Find the percentage loss in the initial velocity of the bullet whenit is between M1 and M2.[Ans. 25%]

A B

M 1 M 2

20. A body of mass 10 kg is allowed to fall freely under gravity. Calculate the momentum of thebody 3 seconds after it starts falling. (g = 9.8 ms–2) [Ans. 294 kg ms–1]

21. A cricket ball of mass 100 g moving with a velocity of 500 cm s–1 is brought to rest by aplayer in 0.1 second. Find the impulse and the average force applied by the player.

[Ans. 5 × 104 g cm/s, 5 × 105 (dyne)]

22. A rifle bullet of mass 125 g leaves the rifle with a velocity of 500 ms–1. The velocity of recoilof the rifle is 5 ms–1. Calculate the mass of the rifle. [Ans. 12.5 kg]

23. A 40 g ball hits a wall, at an angle of 45°. With a velocity of 15 ms–1 and it rebounds at 90°to the direction of incidence. Calculate the impulse received by the ball.

[Ans. 0.84 kg ms–1]

24. A hammer weighing 3 kg moving with a velocity of 10 ms–1 strikes against the head of a nailand drives it into a block of wood. If the hammer comes to rest in 0.025 s find (i) the impulse(ii) the average force on the nail. [Ans. (i) – 30 Ns (ii) – 3000 N]

25. A body of mass 100 kg falls from a height of 40 m. Find the momentum before striking theground. [Ans. 2800 kg ms–1]

26. A bullet of mass 0.05 kg and moving with a velocity of 100 ms–1 is stopped with in 0.1 m ofthe target. Find the average resistance offered by the target in Newton.

[Ans. 2.5 × 103 N]

27. A machine gun of mass 10 kg fires 20 g bullets at the rate of 10 bullets per second with aspeed of 500 ms–1. What force is required to hold the gun in position. [Ans. 100 N]

28. A gun weighing 10 kg fires a bullet of 30 g with a velocity of 330 ms–1. With what velocitydoes the gun recoil ? With is the resultant momentum of the gun and the bullet before firingand after firing. [Ans. 0.99 ms–1, momentum = zero]


29. A rifleman together with his riffle weighs 75 kg. He stands on roller skates and fires 5 shotshorizontally in 5 sec. Each bullet weighs 0.05 kg. The muzzle velocity of the bullet is750 ms–1. Calculate (a) the velocity of the riffleman at the end of 5 shots. (b) average forceexerted on him. [Ans. 2.5 ms–1, 37.5 N]

30. A ball of mass 200 g is hit by a hockey stick giving it an impulse equal to 2500 dyneseconds. Calculate the velocity with which the ball moves, if it was initially at rest.

[Ans. 12.5 cms–1]

31. A bullet of mass 10 g starts from rest from one end of the barrel of the gun 1 m long andweighing 5 kg. The bullet leaves at the other end with a velocity of 400 ms–1. Calculate(i) time taken by the bullet to leave the gun and (ii) the recoil velocity of gun.

[Ans. 0.005 sec, 0.8 ms–1]

32. A body of mass 100 g and velocity 100 cms–1 collides with another body of mass 50 g andvelocity 40 cms–1 coming from the opposite direction. After collision the two bodies movetogether. Find the velocity after collision. [Ans. 53.33 cms–1]

33. A cricket ball of mass 0.5 kg moving with a speed of 10 ms–1 is hit by a bat. The ball turnsback with the same speed. The ball is in contact with bat for 10–2 s. What is the force exertedby the bat on the ball? [Ans. 103 N]

34. A bomb at rest rexplodes into three fragments of equal masses. Two fragments fly off at rightangles to each other with velocities 9 ms–1 and 12 ms–1 respectively. Calculate the speed ofthe third fragment. [Ans. 15 ms–1]

35. A force of 10N acts on a body for 3 μs. If mass of the body is 5g. Calculate the impulse andchange is velocity. [Ans. 3 × 10–5 NS, 6 × 10–3 ms–1]

36. A cricket ball of mass 150 g is moving with a velocity of 12 ms–1 and is hit by a bat so thatthe ball is turned back with a velocity of 20 ms–1. If the duration of contact between the balland bat is 0.01s find the impulse of force. [Ans. 4.8 NS]

37. Figure shows the position-time graph of a particle of mass 0.04 kg. Suggest a suitablephysical context of this motion. What is the time between two consecutive impulses receivedby the particle? What is the magnitude of each impulse?

2 4 6 8 10 12t (s)

x

[Ans. 8 × 10–4 N-S]

38. A bats man hits back a ball straight in the direction of the bowler without changing its initialspeed of 12 ms–1. If the mass of the ball is 0.15 kg, determine the impulse imparted to the ball.

[Ans. – 3.6 N-S]

39. A certain force exerted for 1.2 s raises the speed of an object from 1.8 ms–1 to 4.2 ms–1. Laterthe same force is applied for 2 sec. How much does the speed change in 2s?

[Ans. 4 ms–1]

40. A body of mass 1 kg initially at rest explodes into three fragments of masses in the ratio1 : 1 : 3. The two pieces of equal mass fly off perpendicular to each other with a speed of30 ms–1 each. What is the velocity of the heavier fragment? [Ans. 14.1 ms–1]


41. A batsman hits a ball of mass 0.25 kg that was thrown towards him with a speed of 10 ms–1

back towards the bowler with a speed of 20 ms–1. If the process of hitting the ball takes 0.05 s.Compute the impulse of the force and the average force exerted on the ball by the bat.

[Ans. 7.9 kg ms–1, 150 N]

BASED ON NEWTON’S THIRD LAW AND MOTION IN A LIFT42. An elevator’s weight is 4000 kg. When the upward tension supporting cables is 48000 N.

What is the upward acceleration? Starting from rest how far does it rise in 2 sec?

[Ans. 2.2 ms–2, 4.4 m]

43. A boy of mass 60 kg stands on a weighing machine inside a lift. The lift starts to ascend withan acceleration of 2.45 ms–2.

(i) What is the reading of the machine.

(ii) Also calculate the reading when lift moves with

(a) uniform velocity.

(b) a retardation of 4.9 ms–2 [Ans. 75 kg, 60 kg, 30 kg]

44. A man weighing 50 kg is standing on a lift. Find his weight as recorded by the weighingmachine when

(a) lift moves upward with a uniform velocity of 5 ms–1

(b) the lift moves upward with a uniform acceleration of 2.1 ms–2.

(c) lift moves downward with a uniform accleration of 2.1 ms–2.

[Ans. 50 kg wt, 60.7 kg wt, 39.3 kg wt]

45. A woman weighing 50 kg f stands on a weighing machine placed in a lift. What will be thereading of the machine. When the lift is (i) moving upward with a uniform velocity 5 ms–1

(ii) moving downward with a uniform acceleration of 1 ms–2. (g = 10 ms–2)[Ans. (i) 500 N (ii) 450 N]

46. By what minimum acceleration can a man of 48 kg descend by means of a parachute ifstrings of parachute can with stand a maximum tension of 72 kg wt. [Ans. 4.9 ms–2]

47. A rocket with a lift off mass 20,000 kg is blasted upward with an initial acceleration of 5.0 ms–2.Find the initial thrust of the blast (g = 10 ms–2) [Ans. 3.00 × 105 N]

48. A massless string pulls a mass of 50 kg upward against gravity. The string would break ifsubjected to a tension greater than 600N. What is the maximum acceleration with which themass can be moved upward.

[Ans. 2.2 ms–2]

49. Satun V rocket develops an initial thrust of 3.3 × 107N and has a lift off mass of 2.8 × 106 kg.Find the initial acceleration of the rocket at lift off (g = 10 ms–2] [Ans. 1.8 ms–2]

EQUILIBRIUM OF COCURRENT FORCES50. If tension T2 in the string is 100 N. Find out T1 and weight of block W.

30° 60°

T1 T2

W

[Ans. T1 = 57.73 N

W = 115.47 N]


51. A mass of 6 kg is suspended by a rope of length 2 m from a ceiling. A force of 50 N in thehorizontal direction is applied at the mid point of the rope. What is the angle the rope makeswith the vertical in equilibrium. (g = 10 ms–2) [Ans. 39.8°]

T 1

T2

T 3

T S in 1

T Cos 1

1 m

1 m

T2

60 N

52. A weight of 1000 N is supported by two chains as shown in figure. Determine the tension ineach chain. [Ans. 500 N, 866 N]

30° 60°

T 1 T2

1000 N

53. A weight of 1 kg is suspended as shown in figure. Calculate the value of tension T1.

[Ans. 1.0 kg wt]

54. A 20 kg block is hung by two chains as shown in figure. Find the tension T1 and T2 in thestrings.

[Ans. T1 = 14.66 kg wt.

T2 = 17.92 kg wt.]

30° 45°

T1 T2

20 kg

PROBLEMS ON CONNECTED BODIES55. Two masses 7 kg and 12 kg are connected at the two ends of a light inextensible string that

goes over a frictionless pulley. Find the acceleration of the masses and the tension in thestring when the masses are released. [Ans. 2.58 ms–2, 86.66 N]

56. Two blocks of masses 2.9 kg and 1.9 kg are suspended from a rigid support S by two wireseach of length 1 m. The upper wire has negligible mass and the lower wire has a uniformmass of 0.2 kg m–1. The whole system of blocks, wire and support have an upwardacceleration of 0.2 ms–2. (g = 9.8 ms–2).


(i) Find the tension at the mid point of the lower wire.

(ii) Find the tension at the mid point of the upper wire. [Ans. 50 N, 20 N]

57. Two bodies of masses 11 kg and 11.5 kg are connected by a long light string passing over asmooth pulley find

(i) the velocity of each body at the end of 4s.

(ii) the height ascended or descended by each, as the case may be during that time.

(iii) at the end of 4s the string is cut, find the position of each body in next 2s.

[Ans. 0.872 ms–1, 1.744 m, 21.344 m]

58. A 4 kg block A is placed at the top of a 8 kg block B which rests on a smooth table. A justships on B when a force of 12 N is applied on A. What is the horizontal force F required tomake both A and B move together. [Ans. 36 N]

59. A horizontal force of 600 N pulls two masses 10 kg and 20 kg (lying on a frictionless table)connected by a light string. What is the tension in the string? Does the answer depend onwhich mass, the pull is applied? [Ans. 200 N, 400 N, Yes]

60. Three blocks of masses m1 = 10 kg, m2 = 20 kg and m3 = 30 kg are connected by a string ona smooth horizontal table and pulled to the right; with a force T3 = 60 N. Find theacceleration of system and tension T1 and T2.

[Ans. 1 ms–2, T1 = 10 N, T2 = 30 N]

m 360 N m 2 m 1T 2 T1

30kg 20kg 10kg

61. In the figure the ends P and Q of an unstretchable string move downwards with uniform

speed v. Calculate the speed with which mass M moves upwards. [Ans. v/cos ]

P

M

Q

v v

62. In the adjoining figure calculate the acceleration of 1 kg mass. [Ans. 2

g upwards]

63. Two masses of 10 kg and 20 kg respectively are connected by a massless spring as shown inthe figure. A force of 200 N acts on the 20 kg mass. At the instant shown the 10 kg mass hasacceleration 12 ms–2. What is the acceleration of 20 kg mass. [Ans. 4 ms–2]

10 kg 20 kg 200 N


64. In the given diagram the surface are frictionless. Find the ratio (T1 : T2). [Ans. 5 : 1]

T 2 T 13kg 12kg 15kg

30°

65. Find the acceleration of the blocks A and B in the given figure.

[Ans. 7.57 ms–2, 3.7 ms–2]

2kg

A

5kg B

66. A horizontal force of 300 N pulls two masses of 10 kg and 20 kg connected by a light string.What is the tension in the string if force is applied on 10 kg mass. Consider the surfaces arefrictionless. [Ans. 100 N]

BASED ON FRICTION, COEFFICIENT OF FRICTION AND INCLINED PLANE68. A force of 490 N is required to pull a body of mass 600 kg on a glass surface. Calculate the

coefficient of friction between the surfaces in contact. [Ans. 0.08]

69. A force of 15 kg wt is required to just slide a sledge weighing 500 kg over ice. What iscoefficient of friction. [Ans. (0.03)]

70. A body is sliding down a rough inclined plane which makes an angle of 45° with horizontal.Calculate the acceleration (μ = .1414) [Ans. 5.951 ms–2]

71. A body of mass 10 kg is lying on a plane inclined at an angle of 30° to the horizontal.Coefficient of friction = 0.5. Find the least force required to pull the body up the plane.

[Ans. 91.43 N]

72. What is the acceleration of block and trolley system shown in figure. If the coefficient ofkinetic friction between the trolley and the surface is 0.04, what is the tension in the string(g = 10 ms–2) [Ans. 0.96 ms–2, 27.1 N]

73. A block rests on a horizontal surface and weighs 20 N. The horizontal force P can be increasedto 8 N before the block starts to slide along the horizontal surface. A force of 7 N keeps theblock in uniform motion, once it has started moving. Calculate (a) the coefficient of static andkinetic friction, (b) force of friction when a horizontal force of 5 N is applied on it.

[Ans. 0.40, 0.35]

f P

M g

N

20 K g

30 N

3 Kg

T

T


74. A heavy uniform chain lies on a horizontal table top. If the coefficient of friction between thechain and the table surface is 0.25. What is the maximum fraction of the length of chain thatcan hung over one edge. [Ans. 20%]

75. A block sliding initially with a speed of 10 ms–1 on a rough horizontal surface, comes to restin a distance of 7.0 m. What is the coefficient of friction between the block and the surface.

[Ans. 0.72]

76. A block of mass 2 kg rests on a rough inclined plane making an angle 30° with thehorizontal. The coefficient of static friction between the block and the plane is 0.7. Calculatethe frictional force on the block. [Ans. 11.8 N]

77. The coefficient of static friction μs between the block A of mass 2 kg and table is 0.2. Whatwould be the maximum value of mass of block B so that the two blocks do not move. Thestring and the pulley are assumed to be smooth, and massless. (g = 10 ms–2) [Ans. 0.4 kg]

78. A block rests on a rough inclined plane making an angle of 30° with the horizontal. Thecoefficient of static friction between the block and the plane is 0.8. If the frictional force onthe block is 10 N. Calculate the mass of the block. (g = 10 ms–2). [Ans. 2 kg]

79. A block of mass 1 kg lies on a horizontal surface in a truck. The coefficient of static frictionbetween the block and the surface is 0.6. If the acceleration of the truck is 5 ms–2. Calculatethe frictional force acting on the block. [Ans. 5 N]

80. Calculate the power of an engine which can maintain a speed of 50 ms–1 for a train of mass3 × 105 kg on a rough track. The coefficient of friction is 0.05. Take g = 10 ms–2.

[Ans. 7500 KW]

81. A block slides down an incline of 30° with an acceleration equal to 4

g. Find the coefficient

of Kinetic friction. [Ans. 1

2 3]

82. A block of mass 1 kg sits on an incline as shown in figure.

m a

(a) What must be the frictional force between block and incline if the block is not to slidealong the incline when the incline is accelerating to the right at 3 ms–2.

(b) What is the least value μs can have for this happen? [Ans. 3.48, 0.36]

83. A force of 49 N just able to move a block of wood weighing 10 kg on a rough horizontalsurface. Calculate the coefficient of friction and angle of friction. [Ans. 0.5, tan–1 (0.5)]

84. A block of mass 1 kg lies on a horizontal surface in a truck, the coefficient of static frictionbetween the block and the surface is 0.6. What is the force of friction on the block. If theacceleration of the truck is 5 ms–2. [Ans. 5.88 N]

85. A body is moving along a rough horizontal surface with a velocity 6 ms–1. If the body comesto rest after travelling 9m, then calculate the coefficient of sliding friction (g = 10 ms–2)

[Ans. 0.2]

86. A block B is pushed momentarily along a horizontal surface with an initial velocity V. If isthe coefficient of sliding friction between B and the surface, then calculate the time taken byblock B to come to rest. [Ans. V/ug]

VB


87. The coefficient of static friction between block A and table is 0.2. The mass of block A is2 kg. What should be the maximum mass value of block B so that the two blocks do notmove (g = 10 ms–2) [Ans. 0.4 kg]

2 KgA

B

88. A block of mass 10 kg is moving uniformly on a ramp with inclination of = 30° withhorizontal. Find the force of friction acting on the block. (g = 10 ms–2) [Ans. 50 N]

89. A block start moving on rough horizontal surface with a velocity of 8 ms–1 and comes to restafter travelling 40 m. Find the coefficient of friction between block and surface (g = 10 ms–2).

[Ans. 0.08]

90. A horizontal force of 1.2 kg f is applied to a 1.5 kg block which rests on a horizontal surface.If the coefficient of friction is 0.3 find the acceleration produced (g = 10 ms–2).

[Ans. 5 ms–2]91. A block of 100 kg is placed on a rough inclined plane of angle 30°. The coefficient of

limiting friction is 1

3. Calculate the greatest and least forces acting along the plane to keep

the block in equilibrium. [Ans. 0 N, 1000 N]

194

Work: Work is said to be done, if the force applied produces a displacement in the body, inany direction, except perpendicular to the direction of force.

If force F�

is applied on a body and s�

is the displacement of body, at an angle , to thedirection of force.

W = (Fcos )s

= FS cos = F S� �

Also F�

= ˆˆ ˆF F Fx y zi j k

s�

= ˆˆ ˆs s sx y zi j k .

W = F s� �

= Fx sx + Fy sy + Fz sz

Nature of Work:(a) Positive Work: W = Fs cos .

When 0 90°

The work done is said to be positive. e.q. when a body falls freely under gravity the workdone by gravity is positive.

(b) Negative Work: > 90°

In this case work done is negative. e.q. work done by frictional force is negative.

(c) Zero Work: When = 90°

W = Fs cos 90° = 0.

e.q. In circular motion work done by centripetal force is zero.

Work Done by Variable Force: In general, the force is not constant. The magnitude and

direction of the force changes with position. The work done by variable force F

in changing the

position of a body from 1s

to 2s

is

W =2

1

S

S

F ds��

= Area enclosed by (F – S) graph and displacement axis.

Unit of Work:S.I. Unit = Joule

C.G.S. Unit = erg.1J = 107 erg.

Dimension = [ML2T–2]

�� 7UNIT

s

F

Work Power and Energy 195

Energy: The capacity of doing work of a body is called its energy.

Various forms of energy are – Mechanical energy, Sound energy, Light energy, Electricalenergy, Solar energy, Chemical energy etc.

Energy can be transformed from one form to another.

Einstein’s Mass Energy Relation: According to this relation, if m mass of a substancedisappears then energy produced is given by

E = mc2

c = velocity of light = 3 × 108 ms–1

Energy is a scalar quantity.

Dimension [ML2T–2]

S.I. Unit Joule

Other Units 1erg = 10–7 J

1 ev = 1.6 × 10–19 J

1 kWh = 36 × 105 J

1 calorie = 4.2 J.

Kinetic Energy: The energy possessed by a body by virtue of its motion is called Kineticenergy.

The kinetic energy of a body of mass m moving with speed v is given by

K.E. = 21.

2mv

Work Energy Theorem: The work done by net force on a body is equal to change in itsKinetic energy.

W = 2 21 1– .

2 2m muv

Potential Energy: It is defined as the energy which a body has by virtue of its position orconfiguration in a conservative field. The increase in potential energy of a body of mass m, raisedthrough a height h, above the ground is

U = mghPotential energy may be + ve or – ve depending upon nature of field.

For attraction = – ve

For repulsion = + ve.

Potential Energy of a Spring: According to Hooke’s law, when a spring is compressed orstretched through distance x then restoring force produced in the spring is

F = – kx

The work done is compressing or stretching the spring is stored in it in the form of potentialenergy of spring. It is given by

U = 21.

2kx

Law of Conservation of Energy: It states that energy can neither be created nor bedestroyed but can only be transformed from one form to another i.e. total energy of the universe isconserved.

Power: Rate of doing work is called power


i.e. Power =work

.time

or P =w

t.

S.I. Unit: Watt 1W = 1 Joule/1sec.

Other Units of Power are

1 Kilowatt = 1000 watt = 103 W.

1 MW = 106 W.1 HP = 746 W.

Efficiency: Efficiency of a machine is defined as the ratio of output work to input energy

i.e. =work done

100%.input energy

CONSERVATIVE AND NON CONSERVATIVE FORCEConservative Force: If work done by the force on a particle moving between two points

does not depend on the path followed by the particle, then force is called conservative force. eq.gravitational force.

Nonconservative force: A force is said to be nonconservative if work done by the force ona particle in between two points depends on the path followed.

eq. Force of friction.

Collision of Bodies: A collision between two bodies is said to occur when they physicallystrike against each other or the path of one body is influenced due to presence of other body.

Collision is of following two types:

(a) Elastic Collision: Those collisions in which momentum as well as kinetic energy of thesystem is conserved are called elastic collision.

(b) Inelastic Collision: In inelastic collision, momentum of the system is conserved butkinetic energy is not conserved. If two bodies after collision stick together then it is calledperfectly inelastic collision.

Head on Elastic Collision: That collision in which the colliding bodies move in the samedirection in which they were coming is called elastic collision.

Oblique Collision: If two bodies do not move along the same straight line in which they werecoming before collision, then it is called oblique collision.

Velocities in One Dimensional Elastic Collision: Suppose two bodies of masses m1 and m2moving with velocities u1 and u2 (u1 > u2) in the same direction and collide. After collision, theymove with velocities v1 and v2.

m 1 m 2u 1m 1u 2

m 2v 1 v 2

Using law of conservation of momentum

m1u1 + m2u2 = m1v1 + m2v2 ...(i)

Using law of conservation of Kinetic energy

2 21 1 2 2

1 1

2 2m u m u = 2 2

1 1 2 21 1

2 2m mv v ...(ii)


Solving Equation (i) and (ii)

v1 = 1 2 21 2

1 2 1 2

– 2.

m m m

m m m m

u v

and v2 = 1 2 11 2

1 2 1 2

2 –.

m m m

m m m m

u v

SOLVED EXAMPLES

BASED ON WORK DONEExample 1. A body of mass 30 kg is at rest. A force of 5 N is applied on it. Calculate the

work done in first second.Solution: Here, m = 30 kg, F = 5N, t = 1 sec.

Displacement in 1 sec.

= 2 21 1.

2 2

Fat t

m

=1 5 1

1 1 .2 30 12

m

Work done = F × s

=1

5 .12

= 5

J.12

Example 2. Calculate the amount of work done in raising a block of mass 5 kg througha height of 40 cm. (g = 10 ms–2)

Solution. Work done = force × distance

= (mg) h

= (5 × 10) × 0.4

= 20 J.Example 3. A person pushes a lawn roller through a distance of 10 m. If he applies a

force of 25 N in a direction inclined at 60° to the ground, find the work done by him.Solution. Here, s = 10 m,

F = 25 N, = 60°

Work done

W = Fs cos = 25 × 10 × cos 60°

=1

25 102

= 125 J.

Example 4. A force F��

= ˆ ˆ ˆ5 7i j k acts on a particle and displaces it through S�

=

ˆ ˆ6 9i k . Calculate the work done if the force is in Newton and displacement in metre.


Solution. Here, F�

= ˆˆ ˆ5 7i j k

s�

= ˆˆ6 9 .i k Work done

W = F s� �

= ˆ ˆˆ ˆ ˆ( 5 7 ) (6 9 )i j k i k

= 6 + 63

= 69 J.

Example 5. A force F��

= ˆ ˆ ˆ2i j k N is applied on a body. The body moves from point

A(2, 3 – 4) to B(4, 5, 0)m. Calculate the work done.Solution. Here displacement

s�

= ˆ ˆˆ ˆ ˆ ˆ(4 5 0 ) – (2 3 – 4 )i j k i j k

= ˆˆ ˆ2 2 4i j k m.

F�

= ˆˆ ˆ2 .i j k

Work done W = F s� �

= ˆ ˆˆ ˆ ˆ ˆ( 2 ) (2 2 4 )i j k i j k

= 2 + 4 + 4

= 10 J.Example 6. A uniform chain of length L and mass M is lying on a smooth table and one

third of its length is hanging vertically down over the edge of the table. If g is accelerationdue to gravity. Calculate work required to pull the hanging part on the table.

Solution. Weight of the chain = Mg.

Weight of the hanging part of chain = Mg

.3

The centre of gravity of the hanging part will be at L6 so work done to pull the chain up

on the table

= (force) (distance)

=3 6

Mg L

=L

18

Mg

.

Example 7. The force displacement graph of a body moving under the influence of avariable force is shown below. Calculate the work done displacing the body from A to B.

Solution. Work done = Area under F – s graph


= Area (1) + Area (2) + Area (3)

= (4 – 2) (4 – 0) + 1

(10 4) (8 – 4)2

+ 1

(10 2) (12 – 8)2

10

8

6

4

20

2 4 6 8 10 12 S(m )

1 2 3B

F(N )

=1 1

8 14 4 12 42 2

= 8 + 28 + 24 = 60 J.Example 8. A body constrained to move along z axis of coordinate system is subjected

to a constant force.��

F = ˆ ˆ ˆ– + 2 + 3i j k N where ˆ ˆ ˆandi j k are unit vectors along

x, y and z axis respectively. What is the work done by this force in moving the body througha distance of 4 m along z axis.

Solution. Here, F��

= ˆˆ ˆ– 2 3i j k N and S�

= ˆ4k .

W = F s��

= ˆ ˆˆ ˆ(– 2 3 ) (4 )i j k k

= 12 J.Example 9. A 5 kg object slides 10 m along a horizontal surface. How much work is

done on the object by the frictional force. Given coefficient of friction is 0.2. (g = 10 ms–2)Solution. Here, m = 5 kg, s = 10 m, μ = 0.2.

Force of friction = (μmg)

Work done by the frictional force

= – (μmg) s

= – 0.2 × 5 × 10 × 10

= – 100 J.Example 10. A force f = 4x3 + 2x + 1 N acts on a body in x direction. Calculate the work

done if the particle moves from x = 0 to x = 2m.Solution. Here, F = 4x3 + 2x + 1.

Work done to move the body from x = 0 to x = 2 m.

W = F dx

=2

3

0

(4 2 1)x x dx


=

24 2

0

4 24 2

x xx

= 16 + 4 + 2 = 22 J.Example 11. A rain drop of radius 2 mm falls from a height of 500 m above the ground.

It falls with decreasing acceleration (due to viscous resistance of the air) until at half itsoriginal height, it attains its maximum speed and moves with uniform speed there after.What is the work done by the gravitational force on the drop in the first and second half ofits journey.

Solution. Here, r = 2 mm = 2 × 10–3 m.

distance moved in each half of journey = 250 m.

density of water d = 103 kg m–3.

Mass of rain drop = 34

3r

d

= –3 3 34 22(2 10 ) 10 .

3 7

= –64 228 10

3 7 kg.

Work done by gravitational force in each part of journey is

= –64 228 10 250

3 7

= 0.0083 J.Example 12. A body of mass 2kg initially at rest moves under the action of an applied

horizontal force of 7 N on a table with coefficient of kinetic friction = 0.1 Calculate the(i) Work done by the applied force in 10 s(ii) Work done by friction in 10 s(iii) Work done by the net force on the body in 10 s.Solution. Here, m = 2 kg, u = 0, F = 7N, μ = 0.1, t = 10 s.

Force of friction Fk = μmg

= 0.1 × 2 × 9.8

= 1.96 N.

Net force Fnet = F – Fk

= 7 – 1.96

= 5.04 N

Acceleration a = netF

m

=5.04

2

= 2.52 ms–2


Distance travelled in 10 s

S = 21

2ut at

=1

0 2.52 10 102

= 126 m(i) Work done by applied force W = F × S

= 7 × 126

= 88 2J

(ii) Work done by friction

= Fk × S

= – 1.96 × 126

= – 246.96 J(iii) Work done by net force

= Fnet × S

= 5.04 × 126

= 635.04 J.Example 13. A particle moves along x axis from x = x1 to x = x2 under the influence of

a force given by F = 2x calculate the work done, in the process.Solution. Here, F = 2x,

Work done W =2

1

x

x

F dx

=2

1

2x

x

x dx

=2

1

2

22

x

x

x

= 2 22 1( – ).x x

Example 14. A box is dragged across a floor by a rope which makes an angle of 45° withthe horizontal. The tension in the rope is 100 N. Calculate the work done if box is draggedby 10 m.

Solution. Force acting in the direction of displacement

= (F cos )

W = (F cos ) s= 100 cos 45° × 10

= 707.1 J


BASED ON KINETIC ENERGY POTENTIAL ENERGY AND TOTAL ENERGYExample 15. A body of mass 5 kg initially at rest is subjected to a force of 20 N. What is

the kinetic energy acquired by the body at the end of 10 S.Solution: Here, m = 5 kg, F = 20 N, t = 10 sec.

Now a =20

5

F

m = 4ms–2.

va = u + at = 0 + 4 × 10 = 40 ms–1.

Kinetic energy acquired by the body

= 21

2mv

=1

5 40 402

= 4 KJ.Example 16. The kinetic energy of a body is increased by 200%. Calculate the

percentage change in the momentum of the body.Solution. Kinetic energy of a body is given by

K 2

.2

p

m

1

2

K

K=

2122

.p

p

or1

3=

2122

p

p

or p2 = 13 .p

% Increase in momentum

= 2 1

1

–100

p p

p

= 1 1

1

3 –100

p p

p

= 73.2%.Example 17. The momentum of a body of mass 4 kg is 400 kg ms–1. Calculate its K.E.Solution. Here, m = 4 kg, p = 400 kgms–1.

From formula P = mv

or v = .p

m

=400

4 = 100 ms–1


K.E. = 21

2mu

=1

2 100 1002 = 104 J.

Example 18. A bullet of mass 0.05 kg is moving with a velocity of 100 ms–1. It justpenetrates a wall 10 cm thick. Calculate the resistive force of the wall on the bullet.

Solution: Here,

m = 0.05 kg, u = 100 ms–1.

x = 0.10 m, F = ?

21

2mu = F x

F =2 0.05 100 100

2 2 0.1

mu

x

= 2500 J.Example 19. A toy rocket of mass 0.1 kg has a small fuel of mass 0.02 kg which it burns

out in 3 sec. Starting from rest on a horizontal smooth track it gets a speed of 20 ms–1 afterthe fuel is burnt out. What is the approximate thrust of the rocket? What is the energycontent per unit mass of the fuel?

Solution. Here, m = 0.1 kg, u = 0, v = 20 ms–1, t = 3 s

Thrust of the rocket = ma

=–v u

mt

=20 – 0 2

0.1 N.3 3

K.E. gained by the rocket

K = 21

2mu

= 210.1 (20)

2 = 20 J.

Energy content per unit mass of the fuel

=Total energy

Mass of fuel

=20

0.02 = 1000 Jkg–1.

Example 20. A bullet weighing 10 g is fired with a velocity of 800 ms–1. After passingthrough a mud wall 1 m thick its velocity decreases to 100 ms–1. Find the average resistanceoffered by the mud wall.

Solution. Here, m = 10 g = 10 × 10–3 kg


u = 800 ms–1,

v = 100 ms–1, s = 1 m.

Work done by force offered by mud wall = change in K.E.

F.s. = 2 21( – )

2m uv

or F =2 2( – )

2

m u

s

v

=2 20.01 (800 –100 )

2 1= 3150 N.

Example 21. An electron and a proton are detected in a cosmic ray experiment, the firstwith kinetic energy of 10 kev and the second with 100 kev. Which is faster, the electron orproton? Obtain the ratio of their speeds.

me = 9.11 × 10–31 kg, mp = 1.67 × 10–27 kg.1eV = 1.6 × 10–19 J.Solution. Kinetic energy of electron

10 kev =1

2 meve

2. ...(i)

Similarly for proton

100 kev = 21

2 p pm v ...(ii)

dividing eq. (i) by (ii)

1

10=

2

2.e e

p p

m

m

v

v

or1

10=

–31 2

–27 2

9.11 10

1.67 10e

p

v

v

e

p

v

v= 13.53.

Electron moves faster than proton.

Example 22. If the linear momentum of a body increases by 20% what will be thepercentage increase in the kinetic energy of the body?

Solution. We know,

K =2

.2

p

m

Momentum is increased by 20%.

P = 20

P P100

=6P

.5


Final kinetic energy = K = 2P

.2m

=2(6P / 5)

2m =

2 26.

5 2

p

m

=2

6K

5

% Increase in kinetic energy

=K –

100k

k

=36 K –25 100

k

k

= 44%.Example 23. A ball of mass 0.3 kg falls from the ceiling of an elevator moving down with

a uniform speed of 7 ms–1. It hits the floor of the elevator (length of the elevator 3 m) anddoes not rebound. What is the heat produced by the impact. Would your answer be different,if the elevator were stationary.

Solution. The elevator is moving down with a uniform speed, therefore g = g.

P.E. of the ball = mghHere m = 0.3 kg

h = 3 m

g = 9.8 m/s2

P.E. = 0.3 × 9.8 × 3

= 8.82 J.

The ball does not rebound, the energy is converted into heat.

Heat produced = 8.82 J.The answer will not change if elevator is stationary.

Example 24. The bob of a pendulum is released from a horizontal position A as shown.If the length of the pendulum is 1.5 m. What is the speed with which the bob arrives at thelower most point B, given that it dissipates 5% of its initial energy.

Solution. Here, h = 1.5 m.

P.E. = mgh

= (m × 9.8 × 1.5)

5% of P.E. is lost so K.E. of bob at B

K.E. = 21

2mv =

95( 9.8 1.5)

100m

or v =2 9.8 1.5 95

100

= 5.3 ms–1.

B

AO

1.5m

1.5m


Example 25. The displacement x of a particle moving in one dimension, under the action

of constant force is related to time t by equation t = 3.x Where x is in metre and t in sec.Calculate:

(a) The displacement of the particle when its velocity is zero.(b) The work done by the force in first 6 sec.

Solution. t = 3.x or x = (t – 3)2

Velocity v =dx

dt = 2(t – 3) ...(i)

(a) When velocity is zero 2 (t – 3) = 0 or t = 3 sec.

X = (3 – 3)2

= 0.

i.e. displacement is zero.(b) From equation (1) v0 = 2(0 – 3) = – 6m/s

According to work-energy v6 = 2(6 – 3) = 6 m/s.

theorem W = 2 21( – )

2m v u = 2 21

[(6) – (– 6) ]2

m = 0 J.

Example 26. A uniform chain of length L and mass M is held on a smooth table with onesixth of its length hanging over the edge. Calculate the work done to pull the hanging partof the chain.

Solution. Weight of hanging part of chain = Mg

6

This weight acts at the centre of gravity of the hanging part which is at L12

below the edge

of table.

Work done W =Mg L

6 12

=MgL

J.72

Example 27. A body falling on the ground from a height of 10 m, rebounds to a height2.5 m. Calculate

(i) the percentage loss in K.E.(ii) Ratio of the velocities of the body just before and after the collision.Solution. Consider Vi and Vf be the velocities of the body just before and after the collision.

Ki = 21

1

2 im mghv ...(i)

Kf = 22

1

2 fm mghv ...(ii)


Dividing (i) by (ii)

2

2

V

Vi

f

= 1

2

h

h =

10

2.5 = 4.

orV

Vi

f= 2.

Now percentage loss in K.E.

= 1 2

1

( – )100

mg h h

mgh

=10 – 2.5

10010

= 75%Example 28. A bullet of mass 0.012 kg and horizontal speed 70 ms–1 strikes a block of

wood, of mass 0.4 kg and instantly comes to rest; with respect to the block. The block issuspended from the ceiling by means of thin wires. Calculate the height to which the blockrises. Also estimate the amount of heat produced in the block.

Solution. Here,

mass of bullet m = 0.012 kg.Mass of block M = 0.4 kg

Speed of the bullet = 70 ms–1.

Let velocity v of the combination after collision

i.e. mu = (M + m)v.

or v =M

mu

m

=0.012 70 0.84

0.4 0.012 0.412

= 2.04 ms–1.Let the block rises to a height h.

Applying law of conservation of energy P.E. of the combination = K.E. of the combination.

(M + m)gh = 21( )

2M m v

or h =2

2g

v =

2(2.04)

2 9.8

= 0.212 m.Heat produced = Loss in K.E.

= 2 21 1– (M )V .

2 2mu m


= 2 21 1(0.012) (70) – 0.412 (2.04)

2 2

= 28.54 J.Example 29. A body of mass 2kg initially at rest moves under the action of an applied

horizontal force of 7 N on a table with coefficient of kinetic friction 0.1. Compute the(i) Work done by the applied force in 10 s(ii) Work done by the friction in 10 s(iii) Work done by the net force on the body in 10 s(iv) Change in K.E. of the body in 10 S.Solution. Here, m = 2 kg, u = 0

uk = 0.1 t = 10 s, and f = 7N.

Force of friction on the body

= μmg

= 0.1 × 2 × 9.8

F = 1.96 N.Net force acting on the body = 7 – 1.96

F = 5.04 N.

acceleration a = F

m

=

5.04

2 = 2.52 ms–2

Distance travelled in 10 sec

s = 21

2ut at

= 210 (2.52) (10)

2

= 126 m

(i) Work done by applied force = F × s

= 7 × 126 = 882J.(ii) Work done by the friction force = – Fs

= – 1.96 × 126 = – 246.9J(iii) Work done by the net force = F × s

= 5.04 × 126 = 635 J(iv) Change in K.E. = Work done by net force

= 635 J.Example 30. A bullet, travelling at a speed of 300 ms–1 is just able to pierce a block

9 cm thick. What velocity is required to pierce a block 16 cm thick? Resistive force of theblock in the two cases remain same.

Solution. The K.E. of bullet equals to the work done against resistive force of block, i.e.

21

1

2mu = F × s1


and 22

1

2mu = F × s2.

or2122

u

u= 1

2

s.

s

or u2 = 21

1

s.

su

=16

3009

= 400 ms–1.Example 31. A body of mass 2 kg initially at rest is subjected to a force of 10 N. What is

the work done by the force in 20 sec? Show this is equal to change in kinetic energy of thebody.

Solution. Here, m = 2 kg, F = 10 N, t = 20 s.

a =F

m =

10

2 = 5 ms–2.

Distance travelled S = 21

2ut at

=1

0 5 20 202

= 1000 mWork done W = F × s

= 10 × 1000= 104 J.

Velocity of body after 20 Sv = u + at

= 0 + 5 × 20 = 100 ms–1.

Change in K.E. = 2 21( – )

2m uv

= 2 212(100 – 0 )

2

= 104 J. = Work done.Example 32. A block of mass 20 kg is pulled up a slope by applying a force parallel to

the slope. The slope makes an angle of 30° with the horizontal.(i) Calculate the work done in pulling the block up a distance of 2m.(ii) Calculate the increase in the potential energy of the block.Solution. Here, m = 20 kg

= 30°

s = 2m.


(i) Force required to pull the block = mg sin .

W = f × s

= mg sin s

= 20 × 9.8 sin 30° × 2

= 196 J.

(ii) Increase in P.E. = Work done = 196 J.Example 33. A body of mass 0.3 kg is taken up an inclined plane of length 10 m and

height 5 m and then allowed to slide down to the bottom again. The coefficient of frictionbetween the body and plane is 0.15. What is the

(i) Work done by the gravitational force over the round trip.(ii) Work done by the applied force over the upward journey.(iii) Work done by frictional force over the round trip.(iv) Kinetic energy of the body at the end of the trip.How is the answer to (iv) related to the first three answers?Solution. Here, m = 0.3 kg

l = 10 m

h = 5 m.

and uk = 0.15

As sin =Perp. 5 1

Hypt 10 2

or = 30°.

(i) Work done by gravitational force in moving the body up the inclined plane

W = F s= – mg sin × l

= – 0.3 × 9.8 × 1

102

= – 14.7 JWork done in moving the body down the plane

= + 14.7 J.Work done over the round trip = (–14.7) + (14.7)

W1 = 0 J.(ii) Force needed to move the body up the inclined plane

F = mg sin + Fk = (mg sin + uk mg cos )Work done by force F in moving the body up the inclined plane

W2 = mg (sin + uk cos ) × l

= 0.3 × 9.8 (sin 30° + 0.15 cos 30°) 10

=1 3

0.3 9.8 0.15 102 2

= 1.85 J.(iii) Work done by force of friction over the round trip.

W3 = – Fk (2l)

10m

F

5m

mg cos mg

N

m g cos

m g30°

mg sin


= –μmg cos . 2l

= – 2 × 0.15 × 0.3 × 9.8 × cos 30° × 10

= – 7.6J.(iv) K.E. of the body at the end of round trip

= work done by the net force in moving the body down theinclined plane

= (mg sin – μk mg cos ) l

= mg (sin – μk cos )l

= 0.3 × 9.8 (sin 30° – 0.15 cos 30°)10

= 0.3 × 9.8 3

0.5 0.15 102

= 10.9 J.Example 34. A uniform rod of length 1m and mass 200 g is pivoted at one end is hanging

vertically. It is displaced through 60° from the vertical position. Find the increase in itspotential energy.

Solution. The mass of the body can be supposed to be concentrated at its centre.

Increase in P.E.= mgh (h is height by which C.M. rises)

=1 – cos60

0.2 102

h

60°CG

= 0.5 J.Example 35. A stone of mass 1 kg falls from the top of a cliff 50 m high and buries itself

1 m deep in sand. Find the average resistance offered by sand and time of penetration.Solution. When the stone is at a height of 50 metres it possesses potential energy.

= mgh = 1 × 9.8 × 50 = 490 joules

When it falls, its potential energy is converted into kinetic energy at the bottom.

Kinetic energy = 490 joules

This energy is spent in passing through 1 metre of sand.

If F is the resistance offered by sand and F the net upward force, then

F = (F – mg)

(F – mg) × s = 490 joulesHere, s = 1 metre

(F – mg) = 490 newtons

F = 490 + mg newtons


= 490 + 1 × 9.8 = 499.8 newtons

But F = ma

490 = 1 × a

a = 490 m/s2

As a is retardation it is negative.

When it falls through a height of 50 metres its velocity

v = 2gh

= 2 9.8 50 = 145 m/s

As soon as the stone enters the sand,

u = 145 m/s, v = 0, a = – 490 m/s2

v = u + at

t =– 0 14 5 14 5

–490

u

a

v

= 0.06389 s

= 0.064 sExample 36. A bullet of mass 10 grams is fired from a spring field rifle with a velocity

of 800 m/s. After passing a mud wall 100 cm thick, its velocity drops to 100 metres/s.Calculate the average resistance of the wall. Neglect friction due to air.

Solution. Here,m = 10 grams = 0.01 kg

S = 100 cm = 1 metre

Initial velocity v1 = 800 m/s

Initial K.E. = 2 21

1 10.01 (800)

2 2mv

= 3200 joules

Final velocity v2 = 100 m/s.

Final K.E. = 22

1

2mv

= 210.01 (100)

2 = 50 joules

Hence change in K.E.

= 3200 – 50 = 3150 joules

But the change in K.E. is equal to the work done in overcoming the resistance of the wall.

W = F × s = 3150

F × 1 = 3150

F = 3150 newtons.


Example 37. The mass of a pendulum bob is 100 grams and the string is one metre long.The bob is held so that the string is horizontal and it is then allowed to fall. Find its kineticenergy when the string makes an angle of (i) 0° and (ii) 30° with the vertical.

(I.I.T. Entrance Exam. 1963)Solution. Mass = m = 100 grams = 0.1 kg

(i) When the string makes an angle of 0° with the vertical, the pendulum bob has fallenthrough a vertical height of 1 metre. The loss in potential energy will be equal to the gain in kineticenergy.

Loss in potential energy = mgh

= 0.1 × 9.8 × 1 = 0.98 joules

KE. = 0.98 joule(ii) When the string makes an angle of 30° with the vertical the pendulum bob has fallen

through a vertical height

= 1 cos 30° = 3

metres2

Loss in P.E. = mgh metres

=3

0.1 9.80

= 0.49 × 3

K.E. = 0.49 × 3 = 0.849 joule.Example 38. ABC is a wire 8 m long with A at a height 0.5 m above that of B and C at

a height 0.3 m above B. A 3.0 g particle is released from rest at A. It slides along the wire tocome to rest at C. What is the average friction force opposing its motion?

Solution. Here, h1 = 0.5 m, h2 = 0.3 m.

s = 8 m, m = 3.0 g

Work done by the frictional force

= Change in K.E.

FS = mg (h1 – h2)

F8 = 3 × 10–3 × 9.8 (0.5 – 0.3)

F =–33.0 10 9.8 0.2

8

= 0.74 × 10–3

= 7.4 × 10–4 N.Example 39. An automobile moving horizontally at a speed of 54 kmh–1 reaches the foot

of an inclined smooth plane and the engine is switched off. How much distance does theautomobile go up before coming to rest. The inclination of the plane to horizontal is 30°.(g = 10 ms–2)

Solution. Here, u = 554 18 = 15 ms–1.

= 30° g = 10ms–2.The K.E. of the automobile is used up in doing work against component of gravitational force

i.e. mg sin .

0.3 m

B

0.5 m

A


21

2mu = F S

or S =2

2 sin

mu

mg

=2(15)

2 10 sin 30 = 22.5 m.

Example 40. Two springs have their force constants K1 and K2 (K2 > K1). On whichspring is more work done (i) When their length are increased by same amount (ii) When theyare stretched by same force?

Solution. For first case :

f1 = k1xf2 = k2x

1

2

f

f= 1

2

k

k f1 > f2.

Work done is more in the first spring.

(ii) Here, f1 = f2

k1x1 = k2x2.

k1 > k2

x1 < x2

Work done in second spring will be more.

Example 41. A ball of mass m is dropped from a height h on a pan fixed on a spring. Thepain is depressed by distance x. Calculate spring constant.

Solution. Loss in the P.E. of ball = mg (h + x).

x

h

x

It is stored in spring so,

21

2kx = mg (h + x).

k = 2

2 ( ).

mg h x

x

Example 42. A body of mass 50 g moving with a speed of 20 ms–1 compresses a springthrough a distance 4 cm before coming to rest calculate the spring constant, of spring.

Solution. Here, m = 50 × 10–3 kg

x = 4 cm = 4 × 10–2 m, u = 20 ms–1.


K.E. of body = Energy stored in the spring

21

2mu = 21

.2

kx

K =–3 2

–2 2

50 10 (20)

(4 10 )

= 1.25 × 104 Nm–1.Example 43. The potential energy of a spring when stretched through a distance x is

10 J. What is the amount of work done on the same spring to stretch it through an additionaldistance x?

Solution. P.E. of spring when stretched through a distance x.

u = 21

2kx = 10

Now the spring is further stretched through distance x then potential energy

u = 21(2 )

2k x

= 21(4 )

2k x

= 214

2kx

= 4 × 10

= 40 J.Work done = 40 – 10 = 30 J.Example 44. The force constant of a spring is 100 Nm–1. The spring is stretched by

2 cm. What is the change of elastic potential energy.Solution. Here, k = 100 Nm–1, x = 2 cm = 2 × 10–2 m.

Elastic potential energy

u = 21

2kx

= –2 21100 (2 10 )

2

= 0.02J.Example 45. A spring stretches by 1.0 cm when it is loaded with a weight of 5.0 kg.Calculate (i) force constant of the spring (ii) work done in stretching the spring.Solution. Here, x = 1.0 cm = 10–2 m.

M = 5 kg.

(i) We knowF = kx


k = –2

5 10

10

F

x

= 5000 Nm–1.(ii) Work done in stretching the spring.

= 21

2kx

= –2 215000 (10 )

2

= 0.25 J.Example 46. The spring constant of a coiled spring is 100 N cm–1. What should be the

height of a block of mass m = 7.5 kg above the spring through which it should be allowed tofall to compress it by 2.0 cm. (g = 10 ms–2).

Solution. Let block falls from a height h cm and x is the compression in the spring.

Loss in P.E. of block = mg(h + x).

mg(h + x) = 21

2kx

7.5 × 10 (h + 2) × 10–2 = 2 –2 21100 10 (2 10 )

2

0.75 (h + 2) = 2

h = 0.667 cm.

BASED ON POWERExample 47. A car of mass 2000 kg is lifted up a distance of 30 m by a crane in 1 min.

A second crane does the same job in 2 min. Do the cranes consume same or different amountof fuel? What is the power supplied by each crane?

Solution. Here, m = 2000 kg, S = 30 m, t1 = 1 min = 60 S t2 = 2 min = 120 S.

Work done by cranes = mgh

= 2000 × 9.8 × 30

= 5.88 × 105 J.

As the work done is same, the fuel consumed by two cranes will also be same.

Power supplied by first crane P1 = 1

W

t =

55.88 10

60

= 9800 W.

Power supplied by second crane P2 = 2

W

t =

55.88 10

120

= 4900 W.

Example 48. A pump on the ground floor of a building can pump up water to fill a tankof volume 30 m3 in 15 min. If the tank is 40 m above the ground, and the efficiency of thepump is 30%, how much electric power is consumed by the pump.

Solution. Here, V = 30 m3 d = 1000 kg m–3.

t = 15 min = 15 × 60 = 900 S.


h = 40 m.

Mass of water = V.d. = 30 × 1000 = 3 × 104 kg.

Output power =mgh

t =

43 10 9.8 40

900

=39200

W.3

Efficiency =Output power

×100Input power

Input power =39200

1003 30

= 43.6 × 103 W = 43.6 kW.Example 49. A large familyused 8 kW of power (i) Direct solar energy is incident on the

horizontal surface at an average of 200 W per square metre. If 20% of this energy can beconverted to useful electrical energy how large an area is needed to supply 8 kW? Comparethis area to that of the roof of a house constructed on a plot of size 20 m × 15 m with apermission to cover up to 70%.

Solution. Let area required = A m2

Energy striking per unit area = 200 W.

Energy striking area A = (200 × A)W.

Solar energy converted into electrical energy

=20

200 A100

20

200 A100

= 8000

or A = 200 m2.(b) Area of the roof of given house = 20 m × 15 m = 300 m2.

A =70

300100

= 210 m2.

A

A=

200

210 20

.21

Example 50. An electric motor of power 300 W is used to drive the stirrer in a waterbath. 60% of the energy supplied to the motor is expended in stirring the water. Calculatethe work done on the water in one min.

Solution. Power of motor = 300 W.

Energy used to stir water = 60%.

Useful power =60

300100

= 180 W.


Work done in minute = 180 × 60

= 10800 J.Example 51. A man cycles up a hill whose slope is 1 in 25 at the rate of 10 km/hr. The

weight of the cycle and the man is 100 kg. Find the power of man.

Solution. m = 100 kg, sin = 1

.25

Speed v = 10 km/hr = 10 × 1000

60 × 60 m/s.

Power =work done

time = force × speed

= mg sin × speed

=100 × 9.8 × 10000

25 × 3600 = 108.89 W.

Example 52. An engine of 98 kilowatts power is used to pump water from a mine 50metres deep. Calculate the quantity of water in kilolitres which it can pump out in one hour.

Solution. Power = 9.8 kilowatts

= 9.8 × 1000 = 9800 watts

Work done in one second

= 9800 joules

Work done in one hour= 9800 × 3600 joules ...(i)

Let the mass of water pumped out be m kg

Total work done = mgh

= m × 9.8 × 50 joules ...(ii)

Equating (i) and (ii)

m × 9.8 × 50 = 9800 × 3600

m =9800 × 3600

9.8 × 50

= 72000 kg

Volume of water pumped out in one hour

= 72000 litres

= 72 kilolitres.Example 53. A well 20 metres deep and 7 metres in diameter contains water to a depth

of 10 metres. A pumping engine empties it in one hour. Find the power of the engine.Solution. Height of water in the well = 10 metres

Diameter of the well, d = 7 metres

Volume of water in the well

=2

height4

d


=22 7 7 10

7 4

= 385 cubic metres

But 1 cubic metre of water has a mass of 1000 kgMass of water in the well

m = 385 × 1000 = 385000 kg.

The total depth of water is 20 metres and water is up to a depth of 10 metres. Therefore, the

water can be considered at an average depth of 20 10

2

= 15 metres from the top

m = 385000 kg

h = 15 metres

t = 3600 seconds

Total work done = mgh = 385000 × 9.8 × 15 joules

Work done/second =385000 9.8 15

watts3600

Power = 15720.83 watts

= 15.72 kilowatts.Example 54. A pump can hoist 10,000 kg of coal per hour from a coal mine 120 m deep.

Calculate the power of the engine in watts assuming that its efficiency is 80%.Solution. Here, m = 10,000 kg, g = 9.8 ms–2, h = 120 m.

= 80%, t = 3600 S.

Total work done = mgh = 10,000 × 9.8 × 120 J.

=10,000 9.8 120

W.3600

=useful (output) power

Input power

Input power =10,000 × 9.8 × 120

1003600 80

= 4083.33 W.Example 55. Calculate the power of a man who can chew ice at the rate of 30 g per

minute. J = 4.2 J cal–1, L = 80 cal/g.

Solution. Here, m = 30 g/min = 30

60

g

s

=1

.2

g

s

Power =work L 1

80 4.2time 2

m

t

= 168 W.


Example 56. The human heart forces 70 cc of blood per second through the arteriesunder a pressure of 12.5 cm of Hg. Calculate the power of the heart in watts.

Solution. Here, Pressure = 12.5 cm of Hg

= 12.5 × 13.6 × 980 dynes per sq. cm.

Volume = 70 cc

Work done per second = P dv P V = 12.5 × 13.6 × 980 × 70 ergs/second

= 7

12.5 13.6 980 70joules/second

10

= 1.166 watts.Example 57. A train weighing 106 kg is moving upon inclined plane rising 1 in 200, with

a uniform speed of 72 kmh–1. If the frictional resistance amounts to 0.4 kg per metric ton,find the power of engine.

Solution. Here, m = 106 kg = 103 metric ton.

sin = 1 .200

v = 72 km h–1 = 72 × 5

18 = 20 ms–1.

Fk = μmg

= 0.4 × 103 × 9.8

= 3920 N.

Force needed to pull the train upon inclined plane

= Fk + mg sin

= 6 13920 10 9.8

200

= 52920 N.

Power = F.v= 52920 × 20 = 1058.4 × 103 W = 1058.4 kW.

BASED ON COLLISIONExample 58. Two ball bearings of mass m each moving in opposite directions with equal

speeds v collide head on with each other. Predict the outcome of the collision; assuming it tobe perfectly elastic.

Solution. Here, m1 = m2 = m (let).

u1 = u, u2 = –u.

For elastic collision.

u1 = 1 2 21 2

1 2 1 2

– 2m m m

m m m m

u u

=– 2

(– )m m m

um m m m

u

= – u.


v 2 = 1 2 11 2

1 2 1 2

2 –.

m m mu

m m m m

u

=2 –

(– )m m m

um m m m

u

= u.

The two balls bounce back with equal speeds.

Example 59. A nucleus of radium (88Ra226) decays to 86Rn222 by the emission of particle (2He4) of energy 4.8 MeV. If mass of 86Rn222 = 222.0 u and mass of 2He4 = 4.003 u,then calculate the recoil energy of the daughter nucleus 86Rn222.

Solution.

K = ? 4 .8 mev

2He4Rn 222Ra 226

A t rest

We know K =2

.2p

m

or p = 2 .mK

As momentum is conserved

mK = constant.

mRKR = mK.

KR =4.003 4.8

222

= 0.0866 Mev.

Example 60. The nucleus Fe57 emits a ray of energy 14.4 kev. If the mass of the nucleusis 56.935 u, calculate the recoil energy of the nucleus. (1 u = 1.66 × 10–27 kg).

Solution. Fe57 Fe57 + h.

Momentum of a photon of energy E is

p =E

C = 8

14.4

3×10

kev

=3 –19

8

14.4 10 1.6 10

3 10

= 7.68 × 10–24 kgms–1.Using law of conservation of momentum.

p = momentum of ray photon

= 7.68 × 10–24 kg ms–1.

The recoil energy of the nucleus will be

K =2 –24 2

–27

(7.68 10 )

2 2 56.935 1.66 10

p

m

= 0.312 × 10–21 J.


Example 61. Two blocks of masses 0.3 kg and 0.2 kg are moving towards each otheralong a horizontal frictionless surface with velocities of 0.5 m/s and 1 m/s respectively.

(i) Find the velocity if the blocks collide and stick together.(ii) Find the loss in kinetic energy during the collision.(iii) Find the final velocity of each block, if the collision is completely elastic.

[Indian School Certificate (12), 1977]Solution. Here, m1 = 0.3 kg

m2 = 0.2 kg

v1 = 0.5 m/s

v2 = – 1m/s

(1) Let the final velocity be v (m1 + m2) v = m1 v 1 + m2 v 2.

v = 1 1 2 2

1 2

m m

m m

v v

v =0.3 0.5 0.2(–1)

0.3 0.2

V v = – 0.1 m/sIt means that the combined block will move in the direction of the second block.

(ii) Loss in kinetic energy

= 2 2 21 1 2 2 1 2

1 1 1– ( )

2 2 2m m m m v v v

= 2 2 210.3 (0.5) (0.2)(–1) – 0.5(–0.1)

2

= 0.135 J(iii) When the collision is completely elastic, let v 1 and v 2 be the velocities of the two

blocks after collision.

v 1 = 1 2 21 2

1 2 1 2

– 2m m m

m m m m

v v

v 1 =0.3 – 2 2 0.2

0.5 (–1)0.5 0.5

v 1 = – 0.7 m/s

v 1 = 1 2 11 2

1 2 1 2

2 –m m m

m m m m

v v

=0.6 – 0.1

0.5 (–1)0.5 0.5

= 0.8 m/s


Example 62. A ball of mass 0.1 kg makes an elastic collision with a ball of unknownmass, that is initially at rest. If the 0.1 kg ball rebounds at one third of its original speed,calculate the mass of second ball.

Solution. Here, m1 = 0.1 kg, u2 = 0, u1 = u, u1 = – .3u

We know

u1 = 1 2 2 21

1 2 1 2

– 2 UU .

m m m

m m m m

–3

u= 2

2

0.1–.

0.1

mu

m

1–

3=

2

2

0.1–

0.1

m

m

or on solving m2 = 0.2 kg.Example 63. A ball of mass 8 kg and moving with velocity 4 ms–1 collides with another

ball of mass 10 kg moving with a velocity of 2 ms–1 in the same direction. Find the velocitiesof the balls after the impact, if coefficient of restitution is 1/2.

Solution. Let V1 and V2 be the velocities of two balls after collision.

Here, m1 = 8 kg, v1 = 4 ms–1, m2 = 10 kg, v2 = 2 ms–1

Applying law of conservation of momentum,

(8 × 4) + (10 × 2) = 8 v 1 + 10 v 2

4 v 1 + 5 v 2 = 26 ... (i)

Also, e =velocity of separation

velocity of approach

1

2= 2 1 2 1

1 2 4 2

v v v vv v

or v2 – v1 = 1 ... (ii)

On solving equation (i) and (ii),

v1 = 2.33 ms–1

v2 = 3.33 ms–1

Example 64. A 40 kg skatter travelling at 4 ms–1 overtakes a 60 kg skatter travelling at2 ms–1 in the same direction and collides with him. If the two skatters remain in contact, howmuch KE is lost.

Solution. Here, m1 = 40 kg, m2 = 60 kg,v1 = 4 ms–1 v2 = 2 ms–1.

Let find velocity after collision is V. Then

m1u1 + m2u2 = (m1 + m2)v.

40 × 4 + 60 × 2 = (40 + 60)v.

or v = 2 8 ms–1.


Initial K.E. of the system

= 2 21 1 2 2

1 1

2 2m u m u

= 2 21 140(4) 60(2)

2 2 = 440 J.

Final K.E. of the system

= 21(40 60) (2.8)

2

= 392 J.

Loss in K.E. = 440 – 392 = 48 J.Example 65. A spherical ball of mass 2g is initially at rest. It is struck head on by

another steel ball of mass 16 g moving with a velocity of 81 cms–1. Calculate the velocities ofthe two balls after collision, consider the collision to be elastic.

Solution. m1 = 2g m2 = 16 g

u1 = 0 u2 = 81 cms–1.

Let velocities of the balls after collision be v1 and v2 then

m1u1 + m2u2 = m1v1 + m2v2

(2 × 0) + (16 × 81) = 2v1 + 16v2.

or v1 + 8v2 = 648

Also v1 – v2 = 81

On solving v1 = 144 cms–1

v2 = 63 cms–1.

Example 66. A moving ball of mass m undergoes a head on collision with another

stationary ball of mas 2m. Show that the colliding balls losses 89 th of its energy after

collision.Solution. Here, m1 = m, m2 = 2m, u2 = 0.

u1 = 1 2 21 2

1 2 1 2

– 2.

m m m

m m m m

u u

= 1– 2

02

m m

m m

u

= 1–3

m

m

u = 1– .3

u

Loss in kinetic energy = Ki – Kf

=2

2 11

1 1– – .

2 2 3mu m

u

= 21

1 11 – 92mu


=218

9 2

m

v

=8

.9 inK

= 89 (initial K.E. of balls).

Example 67. A ball is dropped vertically from a height of 20 m. It rebounds from theground to a height of 15 m. Find the coefficient of restitution of the material of ball.

Solution. Here, h1 = 20 m, h2 = 15 m.

Velocity, with which ball strikes the ground

u = 12gh = 2 10 20 = 20 m/s.

Velocity with which ball rebounds back

v = 22 2 10 15 300gh = 17.3m/s

e =17.3

20uv

= 0.866.

Example 68. A bullet of mass 0.01 kg and travelling at a speed of 500 ms–1 strikes ablock of mass 2kg suspended by a string of length 5 m. The centre of mass of the block risesthrough 0.1 m. What is the speed of the bullet after it emerges from the block?

Solution. Here, initial velocity of bullet u1 = 500 ms–1.

Let v1 and v2 be the speeds of bullet and block after collision then

22

1

2mv = mgh

v2 = 2gh

= 2 9.8 0.1 = 1.4 ms–1.

Applying law of conservation of linear momentum

m1u1 + 0 = m1v1 + m2v2

0.01 × 500 + 0 = 0.01v1 + 2 × 1.4

0.01v1 = 5 – 2.8

or v1 =2.2

.01

= 220 ms–1.Example 69. There is a head on elastic collision between a 10 kg mass moving with a

velocity of 5 ms–1 and a 10 kg mass at rest. Calculate the velocities of the masses after thecollision.

Solution. Here, m1 = m2 = 10 kg, u1 = 5 ms–1, u2 = 0

v 2

h

u 1m 1

v 1


Let after collision velocities of the two bodies be v1 and v2 then applying law of conservationof momentum and kinetic energy.

(5 × 10) + 0 = 10v1 + 10v2

or 50 = 10v1 + 10v2

or v1 + v2 = 5 ...(i)

2110(5) 0

2 = 2 2

1 21 1

10 10 .2 2 v v

125 + 0 = 2 21 25 5v v

or 2 21 2v v = 25 ...(ii)

On solving equation (i) and (ii)

v1 = 0, v2 = 5 ms–1.

Example 70. A pendulum bob of mass 10–2 kg is raised to a height 5 × 10–2 m and thenreleased. At the bottom of its swing, it picks up a mass 10–3 kg. To what height will thecombined mass rise. (g = 10ms–2).

Solution. Here, m = 10–2 kg, h = 5 × 10–2 m, m = 10–3 kg

Velocity of bob at the bottom

u = –22 2 10 5 10gh

= 1.0 ms–1

When it picks up mass m, let its velocity be v then applying law of conservation ofmomentum

Mv = (M + m) v

v =M

M mv

=0.01 1

(0.01 0.001)

= 0.91 ms–1.

Let combined mass rises to height h then.

21(M )

2m u = (M + m)gh

or h =2 0.91 0.91

2 2 10

u

g

= 4.14 × 10–2 m.

Example 71. A ball of mass 4 kg moving with a velocity of 12 ms–1 impinges directly onanother ball of mass 8 kg moving with a velocity of 4ms–1 in the same direction. Find theirvelocities after impact, if e = 0.5.

Solution. Here, m1 = 4 kg, m2 = 8 kg

u1 = 12 ms–1 u2 = 4 ms–1

Let after collision the velocities of the two bodies be v1 and v2 respectively.


Applying law of conservation of momentum,

m1u1 + m2u2 = m1v1 + m2v2

(4 × 12) + (8 × 4) = 4v1 + 8v2

48 + 32 = 4v1 + 8v2

v1 + 2v2 = 20 ...(i)

Also

e = 2 1

1 2

–

–u u

v v

0.5 (12 – 4) = v2 – v1

or v2 – v1 = 4 ...(ii)


v 1 = 4 ms–1

v 2 = 8 ms–1.

Example 72. A railway carriage of mass 10,000 kg moving with a speed of 54 kmh–1

strikes a stationary carriage of same mass. After collision the two carriage get coupled andmove together. What is their common speed after collision.

Solution. Here, m1 = m2 = 10,000 kg

u1 = 54 km/h = 5

5418

= 15 ms–1.

u2 = 0.

Let after collision they move with speed v m1u1 + m2u2 = (m1 + m2)v

10000 × 15 + 0 = (10,000 + 10,000)vor V = 7.5 ms–1.

EXERCISE

I. Based On Work Done:1. A body constrained to move along the z-axis of a coordinate system is subjected to a

constant force F�

given by ˆˆ ˆF – 2 3 .i j k �

N when ˆ ˆ,i j and k are unit vectors along,

x, y and z axis. What is the work done by this force in moving the body a distance of a4m along z axis? [Ans. 12 J]

2. A body of mass 2 kg initially at rest moves under the action of an applied horizontalforce of 7 N on a table with coefficient of kinetic friction = 0.1 compute the

(a) Work done by applied force in 10 S

(b) Work done by friction in 10 S [Ans. 882 J, 247 J]

3. A particle is displaced from a position ˆˆ ˆ(2 – )i j k m to another position ˆ ˆ(3 2 )i j m

under the action of a force ˆˆ ˆ(2 – ) N.i j k Calculate the work done by the force.

[Ans. 6 J]

4. A force ˆˆ ˆF 3 2i cj k �

acting on a particle causes a displacement ˆˆ ˆS –4 2 3i j k �

in

it. If the work done by the force is 6 J then find the value of C. [Ans. 6]


5. A force ˆF 10 i �

is acting on a particle of mass 2kg. Calculate work done by force forthe displacement of particle from X = 0 to X = 5 m. [Ans. 125 J]

6. A person carries a load of 80 kg on his head through a distance of 15 m in

(i) the horizontal direction

(ii) the vertical direction. Calculate work done in each case.[Ans. 0 J, 11.76 kJ]

7. A 10.0 kg object slides 50 cm along a horizontal surface. How much work is done on theobject by the horizontal surface, if coefficient of friction is 0.5? [Ans. 24.50 J]

8. A particle of mass 0.5 kg travels in a straight line with velocity U = 3

2ax where

a = 1– –125m s . What is the work done by the net force during its displacement from

x = 0 to x = 2m. [Ans. 3.50 J]9. A man of mass 50 kg supports a body of 20 kg on his head. Calculate the work done

when he moves a distance of 15 m up an incline of 1 in 10 m. (g = 10 ms–2).

[Ans. 1.05 kJ]

10. A bullet of mass 0.02 kg moving with a speed of 100 ms–1 penetrates 2 cm into awooden block. Calculate the average force exerted by the block on the bullet.

[Ans. 5 kJ]

11. Calculate the work done by the force of 40 N, which acting on an object displaces it by10 m at an angle of 60°. [Ans. 200 J]

12. A block of mass 10.0 kg is pulled up at a constant speed from the bottom to the top of asmooth incline 5.0 m long and inclination 30° from the horizontal. Calculate the workdone by the force applied parallel to the incline. (g = 9.8 ms–2) [Ans. 245 J]

13. Calculate the work done by a force F = 2

a

x where (a = 9Nm–2) in displacing a particle

from x = 1 to x = 3m. [Ans. 6 J]

14. An automobile of mass 750 kg moving at a speed of 30ms–1 is brought to rest with adistance of 50 m. Calculate

(i) The force exerted on the automobile by brakes.

(ii) The work done by brakes. [Ans. (i) 6.75 KN (ii) – 337.5 kJ]

15. A coolie carries a load of 60 kg on his head and moves a distance of 40 m in

(a) the horizontal direction

(b) the vertical direction.

Calculate the work done in each case. [Ans. 0, 24 kJ]

16. An automobile of mass 1000 kg moving at a speed of 20 ms–1 is brought to rest with ina distance of 50 m. Calculate

(i) the force exerted on the automobile by brakes

(ii) the work done by the brakes. [Ans. 4000 N, – 2 × 105 J]

17. A man weighing 60 kg carries a 40 kg box to the top of a building 20 m high. Calculatethe work done by the man. [Ans. 19.6 kJ]

18. A body is acted upon by a force along x axis. Force-displacement graph is shown infigure. Find the work done by the force in moving the body x = 0 to x = 14 m.

[Ans. 40 J]


2 4 6 8 10 12 X(m)140

1

2

3

4

19. A 10–kg weight is taken to a height of 2.0 m with an acceleration of 3 ms–2. Computethe work done if g = 10 ms–2. [Ans. 460 J]

20. A particle is acted upon by constant forces 1ˆˆ ˆF 2 – 3 4i j k

��

and 2ˆˆ ˆF – 2 – 3i j k

��

is

displaced from the point A(2, 1, 0) to the point B (–3, – 4, 2) find the total work doneby these forces. [Ans. 2 Units]

II. Based on Kinetic Energy Potential Energy and Total Energy:21. A body of mass 2 kg initially at rest is moved by a horizontal force of 0.5 N on a smooth

frictionless table. Obtain the work done by the force in 8 S. Also show that it equals tochange in K.E. [Ans. 4 J]

22. A light body and a heavy body have same linear momentum. Which one has greaterK.E.? [Ans. Light body]

23. An object of 10 kg initially at rest is subjected to a force of 40 N. What is the kineticenergy acquired by the body at the end of 10 S. [Ans. 8000 J]

24. A pendulum bob of mass 10–2 kg is raised to a height 5 × 10–2 m and then released. Atthe bottom of its swing it picks up a mass 10–3 kg. To what height will the combinedmass rise? (given g = 10ms–2) [Ans. 4.14 × 10–2 m]

25. Calculate the gravitational potential energy of a molecule of gas at a height of 1 m abovethe surface of earth. Given mass of gas molecule is 5.2 × 10–26 kg.

[Ans. 5.0 × 10–25 J]

26. A particle of mass 1 mg moves with constant speed and is found to pass two points2.0 m apart in a time interval of 1 ms. Find the kinetic energy of the particle.

[Ans. 2.0 J]

27. A body of mass 5 kg is moving on a horizontal smooth surface at a velocity of 10 ms–1. Aconstant force of 20 N is applied on the body for 10 s in the direction of the velocity .Find (a) the work done by the force (b) the increase in kinetic energy of body and(c) the power of the agent that exerts that force. [Ans. 6000 J, 6000 J, 600 W]

28. The mass of a pendulum bob is 100 g and the string is 1.0 m long. The string is heldhorizontal and the bob is allowed to fall. Find the kinetic energy when the string makesan angle of 30° with vertical. [Ans. 0.847 J]

29. A solid of mass 2 kg moving with velocity 10 ms–1 strikes an ideal weighless spring andproduces a compression of 25 cm on it. Calculate the force constant of the storing.

[Ans. 3200 Nms–1]

30. A body of mass 10 kg moving with a speed of 4.0 ms–1 on a smooth surface, strikes aspring and comes to rest. If the force constant of the spring be 4 × 105 Nm–1 then howmuch is the spring compressed. [Ans. 2 cm]

31. The displacement x of a particle moving in one dimension under the action of a constantforce is related to the time t by the equation

t = 3x where x is in metres and time in seconds. Find


(i) the displacement of the particle when its velocity is zero and

(ii) work done by the force in first six second. [Ans. zero, 18 J, Joule]

32. The force acting on a particle is given by the equation

f = 1 + 2x + 3x2 + 4x3 + 5x4.

Calculate the work done when x changes from 0 to 5. Force is in newton and x is inmetres. [Ans. 3905 J]

33. Calculate the kinetic energy of a body of mass 0.5 kg and momentum 150 kg ms–1.

[Ans. 22.5 KJ]

34. A body of mass 2.50 kg falls freely under the action of gravity from a height of 100metres. Find the velocity with which it strikes the ground and its KE before striking theground. [Ans. 44.28 ms–1, 2.45 KJ]

35. A force of 2000 dyne acts on a body of mass 500 g for 8 second. Find the velocity,momentum, and kinetic energy of the body after 8 sec.

[Ans. (a) 0.32 ms–1 (b) 0.16 kg ms–1 (c) 0.0256 J]

36. An electron strikes the screen of a cathode ray tube with a velocity of 106 m/s. Calculatethe kinetic energy of the electron. (me = 9.1 × 10–31 kg] [Ans. 4.55 × 10–5 J]

37. Calculate the mass of the electron that strikes the cathode ray screen with a velocity of104 km/s. The kinetic energy of the electron at the instant of impact = 4.55 × 10–17 J.

[Ans. 9.1 × 10–31 kg]

38. Two bodies A and B of masses 1 kg and 3 kg respectively have equal momentum. Findthe ratio of their kinetic energy. [Ans. 3 : 1 cm]

39. A bullet of mass 10 g travels with a speed of 500 ms–1. It penetrates a target which offersa constant resistive force of 1500 N to the bullet. Calculate.

(i) initial K.E. of the bullet

(ii) the distance through which the bullet penetrates the target. [Ans. 1250 J, 83.3 cm]

40. A body of mass 5 kg is dropped from the top of a tower of height 100 m. Calculate theK.E. of the body at the end of 4 sec. (g = 10 ms–2) [Ans. 4000 J]

41. A girl of mass 30kg sits in a swing formed by a rope 6m long. A person pulls the swingside so that the rope makes an angle of 60° with the vertical. What is the gain in potentialenergy of the girl. [Ans. 882 J]

42. A body of mass 5 kg falls from a height of 2m on a spring of force constant4 × 104 Nm–1. Calculate compression of the spring if the energy of the falling body istotally transmitted to spring. [Ans. 7 cm]

43. A spring requires 10 J of work to stretch it through 20 cm. Calculate the value of springconstant. [Ans. 500 Nm–1]

44. The potential energy of a spring when stretched through a distance X is 15 J. What isthe amount of work done on the same spring to stretch it through an additional distance2X ? [Ans. 120 J.]

45. A block of mass 2 kg , is dropped from a height of 40 cm on a spring whose springconstant is 1960 Nm–1. What will be the maximum distance X through which the springis compressed. [Ans. 10 cm]

46. A ball is dropped from rest from a height 10 m on striking the ground it loses 20% of itskinetic energy. What is the height to which the ball will rise? [Ans. 8 m]


III. Based On Power:47. Calculate the power of a person weighing 80 kg climbing a vertical stairway at the rate

of 2 metres/s (g = 9.80 metres/s2). [Ans. 1568 W]

48. An engine of power 8 kilowatts is used to pump water from a mine 50 metres deep.Calculate the mass of water which it can pump out in one hours. [Ans. 58780]

49. An elevator is designed to lift a load 1000 kg through four floors of a building averaging5 m per floor in ten seconds. Calculate the power of the elevator.

[Ans. 19.6 kW]

50. A train of mass 2000 quintals is moving up an incline of 1 in 200 at the rate of 72kilometres per hour. Calculate the power of the engine required to pull the train,neglecting friction.

51. Calculate the work done by a person weighing 70 kg in going up a tower 40 metres high.If he takes one minute to go up, what is the power? [Ans. 23520 J, 392 W]

52. Calculate the power of the engine required to lift 1000 quintals of coal per hour from amine 200 metres deep. The efficiency of the engine is 7% (1 quintal = 100 kg, g = 9.80m/s2). [Ans. 77.778 Kw]

53. A man weighing 90 kg lifts a mass of 30 kg to the top of a building 10 metres high inone minute. Calculate the power. [Ans. 196 W]

54. A person weighing 60 kg carries a baby weighing 10 kg to the top of a building 7.5 mhigh in 15 seconds. Find the power. [Ans. 343 W]

55. Find the useful power expended in pumping 2400 kg of water per minute from a well 45metres deep assuming that 25% of the power of the engine is wasted. What is the totalpower in kilo-watts developed by the engine? [Ans. 17.64 KW 23.52 KW]

56. A pump is required to lift 600 kg of water per minute from a well 25 metres deep and toeject it with a speed of 50 metres per second. Calculate

(a) the work done per minute in lifting the water,

(b) the work done in imparting kinetic energy to water, and

(c) the power of the engine in kilowatts. [Ans. 1.47 × 104 J, 7.5 × 105 J, 14.95 kW]

57. A pump can hoist 9 × 103 kg of coal per hour from a mine 120 m deep. Calculate thepower of the pump in watts assuming that its efficiency is 75%.

[Ans. 3.92 × 103 W]

58. A cyclist climbing up an inclined road maintains a constant speed of 18 kmh–1. The totalmass of the cyclist and cycle is 120 kg. If the angle of inclination is 15°. What is theminimum power expended by the cyclist. [Ans. 1521.75 W]

59. The human heart discharges 75 ml of blood at each beat against a pressure of 0.1 m ofHg. Calculate the power of heart assuming that the pulse frequency in 80 beats perminute. [Ans. 1.33 W]

60. A train weighing 106 kg is moving upon inclined plane rising 1 in 200, with a uniformspeed of 72 kmh–1. If the frictional resistance amounts to 0.4 kg per metric ton. Find thepower of engine. [Ans. 1058.4 KW]

IV. Based On Collision:61. A spherical metal ball A of mass 0.5 kg moving with a speed of 0.5 ms–1 on a smooth

surface collides head on with another identical ball B at rest. Assuming the collision tobe perfectly elastic what are the speeds of A and B after the collision.

[Ans. 0, 0.5 ms–1]


62. A body of mass 2 kg makes an elastic collision with another body at rest and continuesto move in the original direction with a speed equal to one third of its original speed.Find the mass of the second body. [Ans. 1 kg]

63. A moving ball of mass m undergoes a head on collision with another stationary ball of

mass 2m. Show that the colliding ball loses 89 of its energy after collision.

64. A ball is dropped vertically from a height of 3.6 m. It rebounds from a surface to a heightof 1.6 m. Find the coefficient of restitution of the material of the ball.

65. A 40 kg body travelling at 4 ms–1 overtakes a 60 kg body travelling at 2ms–1 in the samedirection and collide with him. If the two bodies remain in contact, how much kineticenergy is lost. [Ans. 48 J]

66. A neutron moving with a speed of 106 ms–1 suffers a head on collision with a nucleus ofmass number 80. What is the fraction of energy retained by nucleus. [Ans. 79/81]

67. A body of mass M at rest is struck by a moving body of mass m. Prove that fraction of

the initial K.E. of the mass m transferred to the struck body is 24 M

(M )m

m in an

elastic collision.

68. A ball of mass 2.4 kg suffers an elastic head on collision with another ball B at rest. After

collision, the ball A continues moving in the same direction with a speed 1

5 of its

original speed while the ball B starts moving forward. Find the mass of the ball B.

[Ans. 1.6 kg]

69. What percentage of kinetic energy of a moving particle is transferred to a stationary

particle it strikes when the stationary particle has a mass (a) 19 times (b) 1

19 times the

mass of the moving particle? [Ans. 19%, 100%, 19%]70. A moving ball of mass 0.1 kg undergoes an elastic head on collision with another ball at

rest. After collision, the first ball rebounds at one third of its original speed while thesecond ball starts moving forward. Find the mass of the second ball. [Ans. 0.2 kg]

V. Extra Numericals:71. Estimate the amount of energy released in the nuclear fusion reaction.

1H2 + 1H2 2He3 + 0n1

Given

M(1H2) = 2.0141 u

M(2He3) = 3.0160 u

M(0n1) = 1.0087 u

1U = 1.661 × 10–27 kg.

Express your answer in units of MeV. [Ans. – 3.27 MeV]

72. A nucleus of radium 88Ra226 decays to radon (86Rn222) by the emission of an particleof energy 4.8 MeV. Calculate the recoil energy of the daughter nucleus.

M(2He4) = 4.003 U

M(86Rn222) = 222.0 U [Ans. 0.085 Mev]


73. An alpha particle strikes a beryllium target to produce 6C12. 5.7 Mev energy is released.

Write down the nuclear reaction and find the mass of the alpha particle. M(4Be9)

= 9.012218 U. [Ans. 4Be9 + 2He4 6C12 + 0n1 + 5.7 MeV; M(2He4) = 4.002569]

74. Calculate the energy produced in KWh when 1 kg of 3Li6 is converted into 2He4 byproton bombardment. [Ans. 6.616 × 107 KWh]

75. How much mass is converted into energy per day at a nuclear power plant operated at108 kW? [Ans. 96 g]

234

CENTRE OF MASSThe centre of mass of a system of particles is that point at which its total mass is supposed to

act or concentrated. It is a point that represents the entire body and moves in the same way as apoint mass having mass equal to that of the object when subjected to an external force. For a two

particle system, let 1r�

and 2r�

be the position vectors of particles of mass m1 and m2 respectively.

Then position of centre of mass is

cmR�

= 1 1 2 2

1 2

m r m r

m m

� �

For n particles system.

cmR�

= 1 1 2 2

1 2

...

...n n

n

m r m r m r

m m m

� � �

=1

1

M

n

i ii

m r �

where M = m1 + m2 ... mn.

COORDINATES OF CENTRE OF MASSThe coordinates of centre of mass are given by

Xcm =1

M i im x

Ycm =1

M i im y

Zcm =1

M i im z

VELOCITY OF CENTRE OF MASSThe velocity of centre of mass is given by

cmv�

=1 1 2 2

1 2

m v m v

m m

� �

.

For n particles system

cmv�

= 1 1 2 2

1 2

...

...n n

n

m v m v m v

m m m

� � �

.

��8UNIT

Z

X

Ym 2

m 1

r1

r2

Rotational Motion 235

MOMENT OF INERTIAThe property by virtue of which a body revolving about an axis opposes the change in

rotational motion is known as rotational inertia. Which is measuredin terms of moment of inertia.

The moment of inertia of a body free to rotate about an axis isthe sum of the product of masses of constitute particles and squareof their distances from axis of rotation.

i.e. I = m1r12 + m2r2

2 ... mnrn2

= mr2.

S.I. Unit Kg m2

Dimension [ML2].Imp. The moment of inertia of a body depends on the choice of axis of rotation and the

distribution of mass about axis of rotation.

RADIUS OF GYRATIONIt is the distance from axis of rotation, the square of which when multiplied by the mass of

the body, gives the moment of inertia of the body.

Let K is radius of gyration and M is mass of body then

I = MK2.

I = mr2.

MK2 = mr2.

K2 =2 2 2

1 1 2 2 ...

Mn nm r m r m r

K =2 2 2

1 1 2 2

1 2

...

...n n

n

m r m r m r

m m m

. Also K =

I.

M

Radius of gyration is a scalar quantity.S.I. Unit : m

THEOREMS OF MOMENT OF INERTIA(i) Theorem of Parallel Axis : According to this theorem, the moment of inertia of a body

about an axis is equal to the moment of inertia about a parallel axis passing through centre of massplus product of mass of the body and the square of the distance between the two axis. i.e.

I = Icm + mr2.

Where Icm is moment of inertia about the axis passing through centre of mass.

(ii) Theorem of Perpendicular Axis : The moment of inertia of a plane lamina about an axisperpendicular to its plane is equal to the sum of the moment of inertia of the lamina about any twomutually perpendicular axis in its own plane intersecting each other at the point through whichthird axis passes i.e.

IZ = IX + IY.

When

IX is moment of inertia about X axis

IY is moment of inertia about Y axis

IZ is moment of inertia about Z axis

m 2

m 1r1

r2

m rr3

rn

m n


TORQUEMoment of couple or turning effect of force about the axis of rotation is called torque.

Torque = Force × lever arm

= F r

Also in vector form �

= Fr �

�

.

or = r F sin

where is the angle between F and .r�

�

S.I. Unit N-mDim. [ML2 T–2].

WORK DONE BY A TORQUE AND POWER OF TORQUEIf a torque is applied on a body which rotates it through an angle then

W = .

Now power associated with torque is given by

P =W

t t

= = Torque × angular velocity.

ANGULAR MOMENTUMMoment of linear momentum of a particle about axis of rotation is called angular momentum

Angular momentum is given by

L�

= r p� �

or L = rp sin .

Where is the angle between and .r p� �

Also angular momentum may be given by

L = mvr

S.I. Unit Kg m2s–1.Dim. [ML2T–1].

ROTATIONAL KINETIC ENERGYThe rotational kinetic energy of a body of mass m and moment of inertia I, about axis of

rotation is given by

Rotational K.E. =21

I2

is angular velocity.

KINETIC ENERGY OF A ROLLING BODYThe kinetic energy of a rolling body is the sum of its translational K.E. and rotational K.E.

i.e

Total K.E = Translational K.E + Rolling K.E.


=2 21 1

I .2 2

mv

ANGULAR MOMENTUM IN TERMS OF MOMENT OF INERTIAThe angular momentum of a body of moment of inertia I about axis of rotation, rotating with

uniform angular velocity is given by

L = IRate of change of angular momentum equals to the torque acting on the body

i.e.dL

dt= (I )

d

dt

= Id

dt

= where d

dt

Also =

dL

dt= I = .

LAW OF CONSERVATION OF ANGULAR MOMENTUMAccording to this law, if no external torque is applied on a system, total angular momentum

of the system remains constant. i.e.

In the absence of external torque

= constant.

EQUATIONS OF ROTATIONAL MOTION:f = i + t

= i t + 21

2t

f2 = i

2 + 2 Where i = Initial angular velocity

f = Final angular velocity

= angular displacement

= angular acceleration.

MOMENT OF INERTIA OF SOME REGULAR BODIES

1. Moment of Inertia of a Ring(a) About an axis passing through its centre and perpendicular to its plane (OZ)

I = MR2

(b) About diameter of ring : (YY)

I =2MR

2


(c) About an axis tangential to ring and parallel to any diameter of the ring (AB)

I = 23MR

2

R

Z

O

M

N

Z

O

Y

Y

A

B

diam eter

(d) About an axis tangential to the ring and parallel to the axis passing through centreand pendicular to plane of ring (MN).

I = 2 MR2.

2. Moment of Inertia of the Disc(a) About an axis passing through the centre and perpendicular to its plane.

=2MR

.2

(b) About a diameter I = 2MR

.4

(c) About an axis tangential to the rim, in its plane (AB)

I = 25MR .

4

(d) About an axis tangential to the rim but perpendicular to theplane of disc. (MN).

I = 23MR .

2

3. Moment of Inertia of Solid Sphere about

(a) Its diameter as axis I = 22MR .

5

(b) A tangent to the sphere I = 27MR .

5

4. Moment of Inertia of hollow sphere

(a) About its diameter as axis I = 22MR .

3

(b) About a tangent to the surface of sphere

I = 25MR .

3

RBA

R

XM

N

O

R

X


5. Moment of Inertia of hollow Cylinder(a) About its geometrical axis which is parallel to

its length

I = MR2

(b) About an axis which is perpendicular to itslength and passes through its centre of mass.

I =2 2R L

M2 12

.

(c) About an axis perpendicular to its length andpassing through one end of the cylinder.

I =2 2R L

M2 3

.

6. Moment of Inertia of Solid Cylinder:(a) About its geometrical axis which is parallel to its length.

I =2MR

2(b) About an axis passing through the centre of mass and perpendicular to geometrical axis

I =2 2L R

M12 4

.

7. Moment of Inertia of Thin rod:(a) About an axis passing through one end and perpendicular to its length.

I =2ML

.3

L

(b) About an axis passing through centre of mass and perpendicular to its length

I =2M

.12

L

R

L

R

L


8. Moment of Inertia of Rectangular Plate(a) About an axis passing through centre of mass and

perpendicular to side b in its plane.

I =2M

.12

b

(b) If axis is perpendicular to side a and parallel to sideb.

I =2M

12

a

(c) About an axis passing through centre of mass andperpendicular to its plane

I = 2 2M( )

12a b .

VEHICLE TAKING CIRCULAR TURN ON A LEVEL ROADThe maximum safe velocity with which the vehicle can cross a circular turn of radius r

v = rg

= coefficient of friction between tyres and road.

BANKING OF ROADIt is the process of providing an inclination sloping to the roads at curved path.

The maximum safe velocity with which a vehicle can cross a circular turn (with out takingfriction into account) of radius r banked at an angle is given by

v = tanrg or tan = v2/rg.

If friction is taken into account then

v =tan

1 tanrg

.

MOTION IN VERTICAL CIRCLEConsider a body of mass m is rotating in a vertical circle with the help of a string of length r

(a) Tension in the string at lowest point Q

TQ = 2( )m

u grr

.

(b) Tension in the string at highest point.

TH = 2( 5 )m

u grr

.

(c) Tension in the string at any point P

TP = 2( 3gh gr)m

ur

.

a

b

a

b

O

PN

r – h

h

Q m g

H


(d) Velocity of body at point P, (at height h from lowest point Q).

v = 2 2ghu .

(e) Minimum velocity at the lowest point Q for looping the loop

v = 5gr

(f) Critical velocity at the highest point for looping the loop

vC = vH = gr .

Motion of a Point Mass attached to String Wound Over a Horizontal Cylinder.

T

m ga

T

m g

R

Consider a light string is wound on a cylinder of mass M and radius R free to rotate abouthorizontal axis. The free end of the string is attached to a mass m. Mass m is released from restand it falls downward with acceleration a.

(a) Linear acceleration a = 2

g

1 / RI m.

(b) Tension in string T = 2R1

I

mg

m

ROLLING WITH OUT SLIPPINGConsider a solid cylinder of mass M is rolling down an inclined plane, inclined at an angle Q

to horizontal. R is radius of the cylinder.

(a) Linear acceleration of cylinder

a = 2

g sin

1 / MRI

(For solid cylinder I = 2MR

2)

a =2

g sin3

.

(b) Force of friction between inclined plane and cylinder

f =Mg sin

3

(c) Condition for rolling of cylinder without slipping

s tan

3

.

N

f

m g cos

m g

m g sin


SOLVED EXAMPLES

BASED ON CENTRE OF MASSExample 1. In the HCl molecule the seperation between the nuclei of two atoms in about

1.27 Aº. Find the approximate location of the centre mass of the molecule, given that the chlorineatom is about 35.5 times as massive as hydrogen atom and nearly all the mass of an atom isconcentrated in all its nucleus.

Solution. The position of cm is

X = 1 1 2 2

1 2

m x m x

m m

=(1 0) (35.5) (1.27)

1 35.5

= 1.235A°.Example 2. Three bodies of equal masses are placed at the three vertices of an equilateral

triangle of side a. Find out the coordinates of centre of mass.

Solution. Coordinates of three vertices P, Q and R are (0, 0) (a, 0) 3

,2 2

a a

Coordinates of cm are

X = 1 1 2 2 3 3

1 2 3

m x m x m x

m m m

=(0) ( ) ( / 2)

3

m m a m a

m

=2a

.

Y =

3(0) (0)

2

3

am m m

m

=3

6a

.

Example 3. Two particles of masses 100 g and 200 g has position vectors ˆˆ ˆ2i j k andˆˆ ˆ2 4i j k m and velocities are ˆˆ ˆ2 2 3i j k and ˆˆ ˆ3 4 2i j k ms–1 respectively. Calculate

the instantaneous position and velocity of centre of mass.

Solution. Here m1 = 100 g = 0.1 kg 1 1ˆ ˆˆ ˆ ˆ ˆ2 , V 2 2 3 m/s.r i j k m, i j k

�

�

m2 = 200 g = 0.2 kg 2 2ˆ ˆˆ ˆ ˆ ˆ2 4 , V 3 4 2 m/s.r i j k m, i j k

�

�

cmR�

= 1 1 2 2

1 2

m r m r

m m

� �

R a2

a 32

,

Q(a , 0 )

P(0 , 0 )

a


=ˆ ˆˆ ˆ ˆ ˆ0.1 ( 2 ) .2 ( 2 4 )

0.1 0.2

i j k i j k

cmR�

=ˆ ˆˆ ˆ ˆ ˆ( 2 4 8 2 )

3

i j k i j k

=ˆ ˆ ˆ3 10 3

3i j k

m.

Velocity of centre of mass

cmV�

= 1 1 2 2

1 2

m v m v

m m

� �

=ˆ ˆˆ ˆ ˆ ˆ0.1 (2 2 3 ) 0.2 (3 4 2 )

0.1 0.2

i j k i j k

=ˆ ˆˆ ˆ ˆ ˆ2 2 3 6 8 4

3

i j k i j k .

= 1ˆ ˆ ˆ8 6 7

ms .3

i j k

Example 4. Two bodies of masses 1 kg and 2 kg are located at (1, 2) and (– 1, 3) respectively.Calculate the coordinates of centre of mass.

Solution. Here m1 = 1 kg x1 = 1, y1 = 2 units

m2 = 2 kg x2 = – 1, y2 = 3 units

Coordinates of centre of mass

X = 1 1 2 2

1 2

m x m x

m m

=(1 1) (2 –1)

1 2

=1 2

3

=

13

Y = 1 1 2 2

1 2

m y m y

m m

=

(1 2) (2 3)

1 2

=

83

.

Example 5. From a uniform disc of radius R a circular hole of radius R/2 is cutout. Thecentre of the disc is at R/2 from the centre of original disc. Locate the centre of gravity of theresulting flat body.

Solution. Let O be the centre of mass of disc and O is the centre of mass of the scooped outportion.


Consider is mass per unit area.

Mass of disc M = R2Mass of scooped out portion m = (R/2)2 .

X coordinate of cm

= 1 1 2 2

1 2

m x m x

m m

=2 2

22

( ) (0) ( / 4 ) / 2

R4

R R R

R

=3

2

4

8 3

R

R

= 6R

.

Example 6. Find the centre of mass of metallic letter F ofuniform density whose dimensions are shown in figure. The widthof the letter is 4 cm every where.

Solution. As shown here the letter F has been divided inthree strips with centre of masses p1, p2 and p3 respectively.

Coordinates of p1, p2 and p3 are p1 (6, 10) cm, p2 (2,10) cm p3 (6, 18) cm

Masses of various parts are

12 cm

4 cm

4 cm

4 cm

m 3 = 12 × 4 × units= 48 un its

m 1 = 4 × 4 × = 16 un its

4 cm

16 cm

m 2 = 4 × 16 × = 64 un its

X =1 1 2 2 3 3

1 2 3( )

m x m x m x

m m m

=(16 6) (64 2) (48 6)

48 16 64

=96 128 288 512

48 16 64 128

= 4 cm.

Y = 1 1 2 2 3 3

1 2 3

m y m y m y

m m m

R/2

O P

O

12 cm

P 3

P 1

P 2

20 cm

X

8 cm


=(16 10) (64 10) (48 18)

16 64 48

=160 640 864

128

= 13 cm.

Example 7. Find the centre of mass of a uniform semi-circular wire of radius R and mass M.

Solution. Consider a small element of circular wire of length dx.

Origin is taken at the centre of semicircular wire i.e.

Mass of element AB

=M

Rdx

=M

RR

d

=M

d

.

Now X coordinate of CM can be calculated as

X =1

Mx dm

=1

M

Md x

but x = R cos

X =1

cosM

MR d

= 00

cos (sin )R R

d

= 0.

Y =1 1 M

M My dm d y

but y = R sin

=1 M

R sinM

d

=0

Rsin d

= 0R

( cos )

=2 R

.

Position of centre of mass is 2

0, .R

B

A

XO

R cos d

R

R sin

dx


Example 8. Two particles of masses 2 kg and 4 kg are approaching towards each other withacceleration 1 ms–2 and 2 ms–2 respectively. Find the acceleration of the centre of mass.

Solution. Here m1 = 2 kg, m2 = 4 kg,

a1 = 1 ms–2 a2 = – 2 ms–2

Acceleration of centre of mass is given by

acm = 1 1 2 2

1 2

m a m a

m m

� �

=(2 1) (4) ( 2)

2 4

=2 8

6

= – 1 ms–2.

The direction of acceleration of centre of mass will be in the direction of a2.

Example 9. Show that centre of mass of a uniform rod of mass M and length L lies at themiddle point of the rod.

Solution. Consider a rod of length L is placed along x axis with its one end at origin. Wefurther consider a small element of the rod of thickness dx at distance x from origin.

Mass of element of width dx = M

Ldx

.

Centre of mass is given by

X =1

Mx dm

=1 M

M Ldx x

=L

0

1x dx

L

=

L2

0

1

L 2

x

=21 L

0L 2

=L2

.

BASED ON BANKING OF ROADSExample 10. A circular race track of radius 300 m is banked at an angle of 15°. If the

coefficient of friction between the wheels of a race car and the road is 0.2 what is the

(a) optimum speed of the race car to avoid wear and tear on its tyres

(b) maximum permissible speed to avoid slipping ?

Solution. Here r = 300 m, = 15° s = 0.2

dx

4 cmO


(a) The optimum speed of the car

V0 = tanrg

= o300 9.8 tan 15

= 300 9.8 0.2679

= 28.1 ms–1.(ii) Maximum permissible speed of race car will be

Vmax =tan

1 tans

s

rg

=0.2 0.2679

300 9.81 0.2 0.2679

= 38.1 ms–1

Example 11. An aircraft executes a horizontal loop at a speed of 720 km h–1 with its wingsbanked at 15°. What is the radius of the loop ?

Solution. Here u = 720 km h–1 = 720 × 5

18

= 200 ms–1.

= 15°, g = 9.8 ms–2.

We know tan =2v

rg

r =2

o

200 200

tan 9.8 tan 15

v

g

=200 200

9.8 0.2679

= 15.24 × 103 mExample 12. A cyclist goes round a circular track of 220 m length in 20 second. Find the

angle that the cycle makes with vertical to avoid slip.

Solution. Let r is radius of circular track

2r = 220

r =220 7

2 22

= 35 m.

Also v =2 2 22 35

T 7 20

r = 11 ms–1.

If is the angle which the cycle makes with vertical then

tan =2v

rg.


=11 11

35 9.8

= 0.3528

or = 19.43°.Example 13. Calculate the coefficient of friction between the tyres and the road, so that a car

travelling at 90 km h–1 goes round a turn of 40 m on level ground.

Solution. Here v = 90 km h–1 = 5

9018

= 25 ms–1

r = 40 m.

We know v = rg

=2 25 25

40 9.8

v

rg

= 1.59Example 14. A train has to negotiate a curve of 400 m. By how much should the outer rail

be raised with respect to the inner rail for a speed of 48 km h–1. The distance between the rails is1 m.

Solution. Here v = 48 km h–1

= 15 4048 ms

18 3

r = 400 m.Let the outer rail be raised through height ‘h’.

tan =2v

rg

or1

h=

2v

rg

or h =2

40 1

3 400 9.8

= 0.0454 m.

Example 15. A long playing record revolves with a speed of 1

333

rev. min–1, and has a radius

of 15 cm. Two coins are placed at 4 cm and 14 cm away from the centre of the record. If thecoefficent of friction between the coins and the record is 0.15 Which of the two coin will revolvewith record ? (g = 9.8 ms–2)

Solution. Here =1 100

33 rpm rps.3 3 60

= 2 = 22 100 1

27 3 60

h

l = 1 m


=220

rad / sec.63

r = 15 cm, = 0.15

The coin will revolve with the record if

mr2 mg

r 2

g

.

2

g

=2

0.5 9.8

22063

= 0.12 m = 12 cm.

The coin placed at 4 cm will revolve with the record.Example 16. A train rounds an unbanked circular track of radius 30 m at a speed of 54 km

h–1. The mass of the train is 106 kg. What provides the centripetal force required for this purpose?The engine or the rails? Which rail will wear out faster the outer or the inner rail? What is the angleof banking required to prevent wearing out of the rails?

Solution. Here r = 30 m v = 54 × 5

18 = 15 ms–1

The centripetal force is provided by the lateral thrust by the outer rail. The outer rail will wearout faster.

Angle of banking is given by

tan =2v

rg

=15 15

30 9.8

= 0.7653

or = 37.4°.Example 17. A bend in a level road has a radius of 100 metres. Find the maximum speed

which a car turning this bend may have without skidding, if the coefficient of friction between thetyres and road is 0.8.

Solution. Here r = 100 m, = 0.8

The car will not skid if force of friction provides the necessary centripetal force.

f = mg = 2mv

r

v = rg

= 0.8 100 9.8

= 28 ms–1.Example 18. A cyclist speeding at 18 km h–1 on a level road takes a sharp circular turn of

radius 3 m without reducing the speed. The coefficient of static friction between the tyres and theroad is 0.1. Will the cyclist slip while taking the turn?


Solution. Here u = 18 km h–1 R = 3 m = 0.1

For the cyclist not to slip speed is given by

v2 s Rg

0.1 × 3 × 9.8

2.94 (ms–1)2.

or v 1.71 ms–1

But here v = 18 Km/hr = 5

1818

= 5 ms–1

Hence the cyclist will slip.

Example 19. A road is banked at an angle of 50°. What is the optimum safe speed at whicha car can negotiate a curve of 200 m radius on this road?

Solution. Here = 50°r = 200 m

The optimum safe speed is given by

v = tanrg

= o200 9.8 tan 50

= 200 9.8 1.1917

= 48.3 ms–1.

BASED ON EQUATION OF ROTATIONAL MOTIONExample 20. The angular speed of a motor wheel is increased from 1200 rpm to 3120 rpm

in 16 seconds

(i) What is its angular acceleration assuming the acceleration to be uniform

(ii) How many revolution does the wheel make during this time?

Solution. Here i =1200

60 = 20 rps

i = 2 i = 2 × 20 = 40 rad s–1

Similarly f = 2f = 2 × 52 = 104 rad s–1

(i) Angular acceleration =f i

f

=104 40

16

= 4 rad s–2.(ii) Angular displacement in time t.

= it + 21

2t

= 40 × 16 + 1

2 × 4 × (16)2


= (640 + 512) rad.

= 1152 rad.

Number of revolution completed in 16 s

=1152

2 2

= 576.

Example 21. A flywheel rotating at a speed of 600 rpm about its axis is brought to rest byapplying a constant torque for 10 seconds. Find the angular retardation and angular velocity after4 seconds of application of torque.

Solution. Here i = 600 rpm = 600

60 = 10 rps.

f = 0 t = 10 sec.

=( )

2f i f i

t t

=2 (0 10)

10

= – 2 rad/s2

Also f = i + t

= 2 1 + t

= 2 × 10 – (2) 3

= 14 rad/s.Example 22. A constant torque is acting on a wheel. Wheel starts from rest. It makes 40

revolutions in 2 seconds. Calculate angular acceleration.

Solution. Here i = 0

= 2 × 40 = 80 radians

t = 2 sec.

= it + 1

2 t2

80 = 0 + 1

2 22

or = 40 rad/s2.Example 23. A grind stone start from rest at a constant acceleration of 5 rad s–2. Calculate

its angular speed after 3 sec.

Solution. Here = 5 rad s–1, i = 0

t = 3 sec.

f = i + t = 0 + 5 × 3

= 15 rad s–1.Example 23. A flywheel revolving at 15 rps slows down to 5 rps. Meanwhile it completes 50

revolutions. Calculate

(a) angular acceleration (b) time taken.

Solution. Here i = 21 = 2 15 = 30 rad/s–1

f = 22 = 2 5 = 10 rad/s–1.

n = 50 revolution. = 2n = 2× 50 rad.


From equation

f2 = i

2 + 2(10 )2 = (30)2 + 2 × (2 × 50)

or 1002 = 9002 + 200

or =2 2100 900

200

= – 4 rad/s2.

Time taken t =f i

=10 30

4

= 5 sec.

Example 24. A car is moving at a speed 72 km h–1. The diameter of its wheel is 0.50 m. Ifthe wheels complete 20 rotation before coming to rest, calculate the angular retardation.

Solution. Here V = 72 km h–1 = 72 × 5

18 = 20 ms–1

r = 0.25 m

=20

.25

V

r = 80 rad/s–1

Also angular displacement = (2)n

= 2 × 20 = 40 rad.

Using equation f2 = i

2 + 2 .0 = (80)2 – 2 (40)

or =2(80) 80

80

=80�

rad s–2.

Example 25. The earth completes one revolution around the sun in one year. The radius oforbit is 1.5 × 1011 m. Calculate the angular velocity and linear velocity of earth.

Solution. Here T = 1 year = 365 × 24 × 60 × 60 sec.

R = 1.5 × 1011 m.

Angular velocity

=2 2

T 365 24 60 60

= 1.99 × 10–7 rad s–1

Also V = r= 1.5 × 1011 × 1.99 × 10–7

= 2.985 × 104 ms–1.Example 26. The speed of a motor increases from 1200 rev/min to 1800 rev/min in 20 sec.

What is the angular acceleration? Also calculate the number of revolutions completed in this time.


Solution. Here i = 2i = 1200

260

= 40 rad/sec.

f = 22 = 1800

260

= 60 rad/sec.

t = 20 sec.

Using relation

=60 40

20f i

t

= rad/s2.

Now = i t + 21

2t

=1

(40 20) 20 202

= 800 + 200 = 1000.

Number of revolution =1000

2

= 500 revolution.Example 27. The radius of a wheel of a car is 0.5 m. The car accelerates for 30 s with angular

acceleration of 2.0 rad/s2. In this time interval calculate

(i) angular displacement of wheel(ii) linear distances covered by wheel

(iii) linear velocity of car.

Solution. Here R = 0.5 m, = 2 rad/s2

(i) Angular displacement of wheel in 30 sec

= it + 21

2t

= 210 2 (30)

2

= 900 rad.(ii) Linear displacement s = R

= 0.5 × 900

= 450 m.(iii) Angular velocity after 30 sec

f = i + t

= 0 + 2 × 30 = 60 rad/sec

Linear velocity v = R = 0.5 × 60

= 30 ms–1.


BASED ON MOMENT OF INERTIAExample 28. Calculate the moment of inertia of a body of mass 100 g revolving about an axis

in a circular path. The centre of mass of body is at 25 cm from axis of rotation.

Solution. Here m = 100 g = 0.1 kg

r = 25 cm = 0.25 m.

Moment of inertia I = mr2

= 0.1 (.25 × .25)

= 625 × 10–5

= 6.25 × 10–3 kg m2.Example 29. Calculate the moment of inertia of earth about its diameter. Consider the earth

to be a sphere of radius 6400 km and mass 1025 kg.

Solution. Here M = 1025 kg

R = 6400 km.

I =2

5 MR2

= 25 2210 (6400)

5

= 1.64 × 1038 kg m2.Example 30. Three point masses each of 1 kg are located at the vertices of an equilateral

triangle of side 1m. Find the moment of inertia of the system about an axis.

(i) Through one vertex and perpendicular to plane of triangle.

(ii) Through mid point of one side and perpendicular to the plane of triangle.

Solution. (i) Let the axis passes through point B then

I = m1 (AB)2 + m3 (BC)2 + m2 (0)2

= 1 (1)2 + 1 (1)2 + 0

= 2 kg m2.

(ii) Let the axis passes through D and is perpendicular to theplane of triangle. The moment of inertia about this axis

I2 = m1 (AD)2 + m2 (BD)2 + m3 (DC)2

I2 =

2 2 23 1 1

1 1 12 2 2

=3 1 1

4 4 4

=54

kg m2.

Example 31. An equilateral triangle is formed by three identical uniform rods each of length50 cm and mass 2 kg. Find the moment of inertia of this system about an axis passing through onecorner.

Solution. We consider axis passes through A.

M.I. of the system I = M.I. of AB about an axis through A + M.I. of AC about an axispassing through A + M.I. of BC about an axis passing through A.

A

CB

m 1

m 3m 2

D

60°


I =2 2 2

2M M MM AD

3 3 12

l l l

=

22 22 M M 3M

3 12 2

l ll

=2 2 22 M M 3M

3 12 4

l l l = 23M

2l

= 232 (.5)

2

= 0.75 kg m2.Example 32. Calculate the moment of inertia of a wheel of mass 10 kg and radius of gyration

20 cm.

Solution. Here M = 10 kg K = 20 cm = 0.2 m

Moment of inertia I = MK2

= 10 (.2)2

= 0.4 kg m2.Example 33. Calculate the moment of inertia of a solid cylinder of mass 5 kg and radius 40

cm. about its own axis.

Solution. Here M = 5 kg r = 40 cm = 0.4 m.

M.l. of cylinder I =2MR

2

=1

5 .4 .42

= 0.4 kg m2.Example 34. Four particles of masses 4 kg, 2 kg, 3 kg and 6 kg are situated at the four

corners A, B, C and D of a square of each side 1 m. Calculate the moment of inertia of the systemabout side (i) AB (ii) Diagonal AC.

Solution. M.I. of the system about side AB

= 3 (BC)2 + 6 (AD)2

= 3 (1)2 + 6 (1)2

= 3 + 6

= 9 kg m2

(ii) M.I. of the system about diagonal AC

= 6 (OD)2 + 2 (OB)2

= 62 2

1 12

2 2

= 3 + 1 = 4 kg m2.Example 35. Calculate moment of inertia of a circular disc of radius 20 cm. thickness 2 cm,

about its transverse axis through its centre. Given density of the disc is 8 g cm–3.

A

CBD

50 cm50 cm

50 cm

A B

D C

1 m4 kg

2 kg

6 kg

3 kg

1 m1 m

1 m

O


Solution. Here R = 20 cm = 0.2 m

Thickness, t = 2 cm = 0.02 m

Density, d = 8 g/cc.

Mass of the disc M = ( R2td)

= 222(20) (2) 8

7

=104800

7 gm.

M.I. of the disc about a transverse axis through its centre

I =2MR

2

=1 104800

(20 20)2 7

= 1.20 × 107 gm cm2.Example 36. Calculate the moment of inertia of a rectangular bar about an axis passing

through its centre and parallel to its thickness. Mass of magnet is 125 g, length 14 cm, breadth 4cm and thickness is 2 cm.

Solution. Here M = 125 g ; l = 14 cmb = 4 cm and t = 2 cm.

The M.I. of the bar magnet about an axis passing through its centre and parallel to itsthickness is

I =2 2

M12

l b

=2 2(14) (4)

12512

=196 16

12512

= 2208 g cm2

= 2.2 × 103 g cm2.Example 37. Calculate the radius of gyration of a rod of mass 200g and length 150 cm, about

an axis passing through its centre and perpendicular to its length.

Solution. Here M = 200 g, l = 150 cm.

M.I. of the rod about an axis through its centre and perpendicular to its length is

I =2M

12

l

If K is radius of gyration then

I = MK2

MK2 =2M

12

l

K =150 150

3.46412 12

l

= 43.3 cm.


Example 38. A uniform wire of length l and mass M bented in the shape of semicircle ofradius r as shown in figure. Calculate the M.I. about its diameter.

Solution. Length of the wire l = r

or r =l

M.I about axis XX =2M

2

r =

2

2

M

2

l

.

Example 39. Four spheres each of diameter 2a and mass M are placed with their centres onthe four corners of a square of side b. Calculate the M.I. of the system about one side of the squaretaken as axis.

Solution. Four spheres A, B, C and D each of radius ‘a’ are placed at the four corners of asquare of side ‘b’.

We have to calculate the moment of inertia of the system about axis BC.

M.I. of the system about axis BC

= M.I. of sphere A about BC + M.I. of D about axis BC

+ M.I. of sphere C about axis BC

+ MI of sphere D about axis BC.

=2 2 2 2 2 22 2 2 2

5 5 5 5Ma Mb Ma Mb Ma Ma

= 2 282

5Ma Mb

=25

(4a2 + 5b2) M.

Example 40. Three identical discs are arranged as shown in figure. Mass of each disc is Mand radius is R. Calculate the moment of inertia about axis XX.

Solution. M.I. of the system is

I = IA + IB + IC

=2 2 25 MR 5 MR MR

4 4 4

=11

4 MR2.

Imp. : M.I. of disc A or B about axis XXIA = IB = IG + MR2

=2 2

2MR 5 MRMR

4 4

BASED ON ROTATIONAL KINETIC ENERGYExample 41. A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad/

s. The radius of the cylinder is 0.25 m. Calculate its rotational K.E.

C

BA

X

X

B

X O

r

X

A Bb

bb

bD C

a

aa

a


Solution. Here M = 20 kg, = 100 rad/s.

R = 0.25 m.

Rotational K.E. = 21I

2

=2

21 MR

2 2

=1

4 × 20 × .25 × .25 × 100 × 100

= 3125 J.Example 42. A wheel of mass 10 kg and radius 0.40 m is rolling without sliding at an angular

speed 20 rad s–1. The moment of inertia of the wheel about the axis of rotation is 0.65 kg m2.Calculate the ratio of rotational K.E. to translational K.E.

Solution. Here M = 10 kg, R = 0.40 m, = 20 rad s–1

I = 0.65 kg m2.

Translational K.E = 21

2mv

=2 21

R2

m (v = R)

=1

2 × 10 × (.40)2 (20)2

= 320 J

Rotational K.E = 21I

2

=1

2 × 0.65 × (20)2

= 130 J

Rotational K.E.

Translation K.E=

130

320 = 0.41.

Example 43. A solid sphere of mass 1 kg and radius .25 m rolls without sliding with auniform velocity of 10.0 ms–1 along a straight line on a smooth horizontal table. Calculate its totalenergy.

Solution. Here M = 1 kg, R = 0.25 m, v = 10.0 ms–1.

Translational K.E. = 21

2Mv

=1

2 × 1 × 10 × 10

= 50 J


2


=2 21 2

MR2 5

=2

22

1 2MR

2 5

v

R

(V = R)

=21

M5

v = 1

1 10 105

= 20 J

Total energy = 50 + 20

= 70 J.Example 44. A metallic bar 1 m long, weighing 3 kg rotates at rate of one revolution per sec.

What is its K.E. if the axis of rotation passes through

(i) C.G.

(ii) One extreme end.

Solution. Here M = 3 kg, R = 1 m, = 1 Hz.

(i) M.I. of bar about an axis passing through C.G.

I =2M

12

l

=21 1

312 4

kg m2.


2

= 21 12 1

2 4

=2

J.2

(ii) M.I. of the bar about an axis through one end

I =2M

3

l

=3 1

3

= 1 kg m2.

K.E. of rotation = 21I

2

= 211 (2 1)

2

= 22 J.Example 45. A solid cylinder rolls down an inclined plane. Its mass is 2 kg and radius 0.1

m. If the height of the inclined plane is 4 m what is its rotational K.E. when it reaches the foot ofthe plane?


Solution. Here m = 2 kg, R = 0.1 m, h = 4 m.

When the cylinder reaches to bottom its P.E. is converted in to Rotational K.E andTranslational K.E is

mgh =2 21 1

M I2 2

v

or2

2 21 1 MRM (R )

2 2 2

= Mgh2MR

I , and V R2

or 2 23MR

4 = Mgh

or 2 =2

4 gh

3 R


2

=2

2

1 MR 4 gh

2 2 3 R

=Mgh

3

=2 9.8 4

3

= 26.13 J.

Example 46. A flywheel is rotating at a rate of 60 rev/min. To increase the rate of rotation to180 rev/min. 197.4 J of work is required. Find the moment of inertia of the fly wheel, about its axisof rotation.

Solution. Here 1 = 60 rpm = 1 rps

2 = 180 rpm = 3 rps

W = 197.4 J

The increase in rotational K.E. equals the work done on the flywheel.

1

2 I (f

2 – i2) = W

or I = 2 2 2 2

2 W 2 197.4

(2 3) (2 1)f i

= 2 2

2 197.4

36 4

=2 197.4

32 9.8

(2 = 9.8)

= 1.250 kg m2.Example 47. A solid sphere rolls without sliding, from rest down an inclined plane. Find its

velocity after it has suffered a vertical drop of 2 m.


Solution. Here h = 2 m.

Here loss in potential energy equals to the sum of translational K.E. and rotational K.E. andenergy equation is

mgh = 2 21 1V I

2 2m

or mgh =2

2 21 1 2 VV

2 2 5m mr

r

or gh =2 2

2V V 7V

2 5 10

or V =gh 10 9.8 2

107 7

= 5.29 ms–1.

Example 48. Considering the earth to be a sphere of radius 6400 Km and mass 6 × 1024 kg.calculate its rotational K.E.

Solution. Here M = 6 × 1024 Kg

R = 6400 Km = 6400 × 103 m.

Rotational K.E. of the earth = 21I

2

=2

21 2 2MR

2 5 T

22For earth I MR

5

Rotational K.E. =24 3 2 2

2

1 2 6 10 (6400 10 ) 4

2 5 (24 60 60)

= 2.6 × 1029 J.Example 49. Calculate the rotational K.E. of a circular disc of mass 2 kg. and radius 0.4 m

rotating about an axis passing through its centre and perpendicular to its plane. The disc is rotating

at 1

2 rps.


R = 0.4 m, = 1

rps2

.

M.I. of disc is I =2MR

2


2

= 2

21 MR2

2 2


=221 2 (0.4) 1

22 2 2

= 0.08 J.

BASED ON TORQUE AND POWER OF TORQUEExample 50. An automobile moves on a road with a speed of 54 Km h–1. The radius of its

wheels is 0.35 m. What is the average negative torque transmitted by its brakes to a wheel if thevehicle is brought to rest in 15 s? The moment of inertia of the wheel about the axis of rotation is3 kg m2.

Solution. Here u = 54 Km/h = 5

5418

= 15 ms–1

R = 0.35 m, t = 15 sec. I = 3 kg m2

Now i =15

R 0.35

u rad/s

f = 0

=

150

.3515

f i

t

=1

0.35 rad/s2

Torque transmitted by the brakes

= 1

= 3 × 1

0.35

= – 8.57 Kg m2 s–2.Example 51. A spherical body of moment of inertia 3 kg m2 is at rest. It is rotated for 20

seconds with a moment of force 6 Nm. Find the angular displacement of the body. Also calculatethe work done.

Solution. Here I = 3 kg m2, t = 20 s

= 6 Nm.

We know = I

=6

I 3

= 2 rad/s2

Angular displacement in 20 sec

= it + 1

2 t2

= 0 + 1

2 × 2 × 20 × 20

= 400 rad.


Work done W = = 6 × 400

= 2400 J.Example 52. To maintain a rotor at uniform angular speed of 200 rad/sec., an engine needs

to transmit a torque of 180 Nm. What is the power of the engine required?

Solution. Here = 200 rad/s.

= 180 Nm.

Power of the engine = = 180 × 200

= 36 × 103 J

= 36 kJ.Example 53. A rope is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What

is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N? What is thelinear acceleration of the rope?

Solution. Here M = 3 kg, R = 40 cm = 0.4 m.

F = 30 N.

Now = F R= 30 × 0.4

= 12 Nm

M.I. of hollow cylinder I = MR2

= 3 × (.4)2

= .48 kg m2.

Angular acceleration =12

I 0.48

= 25 rad/sec2.

Also linear acceleration a = R= 0.40 × 25

= 10 ms–2.Example 54. A flywheel of mass 25 kg has a radius of 0.2 m. It is making 240 rpm. What is

the torque necessary to bring it to rest in 20 sec? If the torque is due to a force applied tangentiallyon the rim of the fly wheel, what is the magnitude of the force?

Solution. Here M = 25 kg, R = 0.2 m,

i = 240 rpm = 4 rps

i = 2i = 2 × 4 = 8 rad/sec.

f = 0, t = 20 sec.

f = i + t

or =20 8 2

rad/s20 5

f i

t

M.I. of the wheel I =2MR

2 Torque acting on the flywheel

= I = 2MR

2


=25 .2 .2 2

2 5

= �

5 Nm. (–ve sign is due to retarding nature of torque).

Also = F × R

F =( / 5)

0.2R

= N.

Example 55. The moment of inertia of a ring is 0.50 kg m2. It is rotating at a rate of 2100rpm. It is brought to rest in 2.0 seconds. Calculate the torque required?

Solution. Here I = 0.50 kg m2.

i = 2100 rpm = 2100

60 = 35 rps

i = 2i = 22

2 357

= 220 rad/s.

f = 0

Angular acceleration =f i

t

=0 220

2

= – 110 rad/s2.

Retarding torque required = I= 0.5 × (110)

= 55 N-m.Example 56. A cord is wound around the circumfrance of a wheel of diameter 0.3 m. The

axis of the wheel is horizontal. A mass of 0.5 kg is attached at the end of the cord and it is allowedto fall from rest. If the weight falls 1.5 m in 4s. What is the angular acceleration of the wheel? Alsofind out the moment of inertia of the wheel.

Solution. Here R =D 0.3

2 2 = 0.15 m

M = 0.5 kg

h = 1.5 m t = 4 sec.

Let a is acceleration, with which the mass falls downward, so

h = ut + 1

2 at2

1.5 = 210 (4)

2a

or a =1.5 3

8 16 ms–2

Angular acceleration =3

R 16 .15

a

= 1.25 rad/s2


Torque = F R= mg R = 0.5 × 9.8 × .15

I =0.5 9.8 0.15

1.25

= 0.588 Kg m2.Example 57. A flywheel of mass 25 kg has a radius of 0.2 m. What force should be applied

tangentially to the rim of the wheel so that it acquires an angular acceleration of 2 rad/sec2 ?

Solution. Here m = 25 kg

R = 0.2 m

= 2 rad/sec2

M.I. of the flywheel is I = 21MR

2

=1

25 .2 .22

= 0.50 Kg m2

We know = F × R = I

F =I 0.50 2

R 0.2

= 5N.Example 58. The mass of a flywheel is 20 Kg. The radius of gyration of the flywheel is 40

cm. A torque of 10 Nm is applied to it. Calculate the resulting angular acceleration.

Solution. Here M = 20 Kg

K = 0.40 m = 10 Nm

= ?

We know = I = MK2

=2MK

= 2

10

20 (0.4)

= 3.125 rad/s2.

BASED ON ANGULAR MOMENTUM AND LAW OF CONSERVATION OF ANGULARMOMENTUM

Example 59. A solid cylinder of mass 20 kg rotates about its axis with angular speed 100 rad/sec. The radius of the cylinder is 0.25 m. What is the Kinetic energy associated with the rotationof the cylinder. What is the magnitude of the angular momentum of the cylinder about its axis.

Solution. Here M = 20 Kg

= 100 rad/sec.

R = 0.25 m


M.I. of the cylinder I =2MR

2

=20 (.25 .25)

2

= 0.625 Kg m2.


2

= 21(0.625) (100)

2

= 3125 J.

Angular momentum L = I = 0.625 × 100

= 62.5 Kg m2/s.Example 60. A disc of mass 0.10 kg and radius one decimetre has a little projection of

negligible mass on its circumference. A 10 g bit of putty moving with a velocity of 500 cm s–1

strikes this projection and stick fast. Find the angular velocity of disc.

Solution. Here M = 0.10 Kg

R = 1 dm = 0.1 m,

m = 10 g = 0.01 Kg

V = 500 cm s–1 = 5 ms–1

M.I. of disc = 21MR

2

= 210.1 (.1)

2 = 5 × 10–4 Kg m2

M.I. of putty = mR2

= 0.01 × (0.1)2 = 1 × 10–4 Kg m2.

Total moment of inertia = (5 × 10–4) + (1 × 10–4)

= 6 × 10–4 Kg m2

Let is the angular velocity of the disc then angular momentum of the system

= I= 6 × 10–4 × Kg m2 s–1

But L = MvR

= 0.01 × 5 × 0.1

= 5 × 10–3 Kg m2 s–1.

6 × 10–4 = 5 × 10–3

=3

4

5 10

6 10

= 8.33 rad/s.Example 61. A rope is wound round a hollow cylinder of mass 3 kg and radius 40 cm. What

is the angular acceleration of the cylinder if the rope is pulled with a force of 30 N. What is thelinear acceleration of the rope?


Solution. Here M = 3 Kg,

R = 40 cm = 0.4 m

F = 30 N.

M.I. of the cylinder I = MR2

= 3 (0.4)2

= .48 Kg m2

Now = 1 = F R

=F R 30 0.4

I 0.48

= 25 rad/s–2.Also a = R

= 0.4 × 25 = 10 ms–2

Example 62. A ring of diameter 0.4 m and of mass 10 kg is rotating about its axis at the rateof 2100 rpm. Find (i) moment of inertia (ii) angular momentum (iii) rotational K.E. of ring.

Solution. Here R =0.4

2 = 0.2 m.

M = 10 Kg,

= 2100 rpm = 2100

60 = 35 rps.

= 2 = 22

2 357

= 220 rad/s–1

(i) M.I. of the ring I = MR2

= 10 (0.2)2 = 10 × 0.04

= 0.4 Kg m2

(ii) Angular momentum L = I= 0.4 × 220 = 88 kg m2s–1

(iii) Rotational K.E = 21I

2

=1

2 × 0.4 (220)2

= 9680 J.Example 63. An electron of mass 9 × 10–31 kg revolves in a circle of radius 0.53Å around

the nucleus of hydrogen with a velocity of 2.2 × 106 ms–1. Show that its angular momentum isequal to h/2. Where h is Planck’s constant of value 6.6 × 10–34 Js.

Solution. Here m = 9 × 10–31 kg

r = 0.53 Å = 0.53 × 10–10 m

h = 6.6 × 10–34 Js

v = 2.2 × 106 ms–1

Now L = mvr

= 9 × 10–31 × 2.2 × 106 × 0.53 × 10–10

= 1.0494 × 10–34 kg m2s–1


Also2

h

=

346.6 10

2 3.14

= 1.05 × 10–34 kg m2s–1

L =2�h

.

Example 64. An electron revolves around the nucleus of an atom in a circular orbit of radius4.0Å with a speed of 6.0 × 106 ms–1. Calculate the linear kinetic energy and its angular momentum.

Solution. Here m = 9 × 10–31kg

r = 4.0 Å = 4.0 × 10–10 m

v = 6.0 × 106 ms–1.

Kinetic energy of electron = 21

2mv

= 31 6 219 10 (6 10 )

2

� 1.6 × 10–17 J.Angular momentum of the electron

L = I

=2( )

vmr

r

= mvr

= 9 × 10–31 × 6 × 106 × 4.0 × 10–10

= 2.16 × 10–33 kg m2s–1.Example 65. If earth suddenly contracts to half its present radius, what would be the length

of the day ?

Solution. Here I1 =21

2MR

5

1 =1

2

T

I2 =22 2

2

2 2MR ,

5 T

.

Applying law of conservation of angular momentumI11 = I22

2 21 2

2

2 2 2 2MR MR

5 T 5 T

T2 =22

2 11 2

11

R R / 2T (24 hr)

RR

= 6 hr.


Example 66. (i) A child stands at the centre of turn table with his two arms outstretched. Theturntable is set rotating with an angular speed of 40 rpm. How much is the angular speed of the

child if he folds his hands back and thereby reduces his moment of inertia to 2

5 times the initial

value? Assume that the turntable rotates with friction.

(ii) Show that the child’s new kinetic energy of rotation is more than the initial kinetic energyof rotation. How do you account for this increase in kinetic energy.

Solution. Here 1 = 40 rpm

I2 = 12

I5

Using law of conservation of angular momentum

I11 = I22

I1 × 40 = 1 22

I5

or 2 = 100 rpm.(ii) Now initial K.E. of rotation

Ei =2

1 11

I2

= 21

1I (40)

2= 800 I1

Final K.E.

Ef =2

2 21

I2

=2

11 2

I (100)2 5

= 2000 I1

E

Ei

f=

1

800 I1.

2000 Ii

Ei < Ef.Example 67. A flywheel of mass 500 kg and diameter 1m revolves about its axis. Its

frequency of revolution is increased by 18 in 5 sec. Calculate the torque applied.


R =1

2m = 18 rps

t = 5 sec

Now

Angular impulse = Change in angular momentum.

×t = I Angular impulseTorque time.


or =2I 1

MR2t t

=2

1 1 2 18500

2 2 5

= 1.4 × 103 Nm.Example 68. Two discs of moment of inertia I1 and I2 about their respective axis (Normal to

disc and passing through centre) and rotating with angular speeds 1 and 2 are brought intocontract face to face with their axes of rotation coincident

(a) What is the angular speed of the two disc system?

(b) Show that the kinetic energy of the system is less than the sum of the initial energies ofthe two discs. How do you account for this loss in energy? (Take 1 2).

Solution. Initial angular momentum of the two discs

= I11 + I22

When the discs come in contact then moment of inertia of the combinedsystem is I (let) when I = (I1 + I2) find angular velocity of the system is .

Applying law of conservation of angular momentum.

(I11 + I12) = (I1 + I2)

or = 1 1 2 2

1 2

I II I

.

(b) Initial K.E. of the sysem is

Ei =2 2

1 1 2 21 1

I I2 2

Final K.E. of the combined system is

Ef = 21 2

1(I I )

2

=

21 1 2 2

1 21 2

I I1(I I )

2 I I

=2

1 1 2 2

1 2

1 (I I )

2 (I I )

Loss in K.E. = Ei – Ej.

=2

2 2 1 1 2 21 1 2 2

1 2

(I I )1 1 1I I

2 2 2 (I I )

=2 2 2 2 2 2 2 2 2 21 1 1 2 1 1 2 2 2 2 1 1 2 2 1 2 1 2

1 2

1 I I I I I I I I 2 I I

2 (I I )

E = 21 21 2

1 2

I I( )

2 (I I )

.

I1

I1

(I + 1 I )2


Example 69. A circular metal disc of mass 4 kg and diameter 0.4 m makes 10 revolution persecond about an axis passing through its centre and perpendicular to its plane

(a) What is the angular momentum about the given axis?

(b) Calculate the magnitude of the torque which will increase the angular momentum by 20%is 10 sec.

Solution. Here M = 4 Kg R = 0.4

2 = 0.2 m

= 10 rps

= 2 = 2 × 10 = 20 rad/sec

(a) M.I. of the disc about given axis

I =2MR

2

=1

2 × 4 × .2 × .2

= 8 × 10–2 Kg m2

Angular momentum of the disc is

L = I= 8 × 10–2 × 20= 5.02 Kg m2/sec.

(b) =dL

dt

dL = 20% = 20

5.02100

� 1 Kg m2/sec

dt = 10 sec

=1

10= 0.1 Nm.

Example 70. Show that the angular momentum of a satellite of mass Ms revolving around theearth having mass Me in an orbit of radius r is

L = 2eGM M s r .

Solution. Satellite revolves around earth in circular orbit so centripetal force is provided bygravitational force between earth and satellite.

2Ms v

r= 2

GM Me s

r

or v2 =GMe

r

or v =GMe

r


Now angular momentum of satellite is

L = Msvr

=GM

M es r

r

= 2GM Me s r .

EXERCISE

BASED ON CENTRE OF MASS1. Two particles m, m, 2m and 2m are placed at the four corners of a square of side a. Find the

centre of mass of the system.

2,

2 3

a a Ans.

2. Two bodies A and B of 3 kg each are moving along Y axis. At a particular instant body A is1 m from the origin and has a velocity of 3 ms–1 and body B is 2m from the origin and has avelocity of 1ms–1. Find the position and velocity of the centre of mass and also find the totalmomentum. [Ans. 1.5 m, 1 ms–1, 6 Kg ms–1]

3. Three bodies of masses 1kg each are placed at (0, 0) (a, 0) and 3

,2 2

a a

. Find out the

coordinates of centre of mass.

3

6

a Ans.

4. Determine the position of centre of mass of a systems consisting of two particles of massesm1 and m2 seperated by a distance L apart from m1.

2

1 2

Lm

m m

Ans.

5. A circular hole of radius R

2 is scooped out from a circular disc of radius R, with the hole

tangent to the rim. Find the distance of centre of mass from the centre of the disc.

Ans. R

, 06

6. Four particles of masses m, 2m, 3m and 4m are kept at conersof a square of side a. Find coordinate of centre of massconsidering mass m to be at origin.

7,

2 10

a a Ans.

7. Locate the centre of mass of a system of three particles ofmasses 1.0 kg, 2.0 kg and 3.0 kg placed at the corners of an equilateral triangle of 1 m side.

Ans. 7 3

,12 4

m


8. Two bodies of masses 10 kg and 2 kg are moving with velocities ˆˆ ˆ4 7 3i j k andˆˆ ˆ20 35 6i j k ms–1 respectively. Find the velocity of the centre of mass.

[Ans. ˆ1.5 k ms–1]

9. Two bodies of mass 1 kg and 2 kg are lying in the X-Y plane at points (– 1, 2) and (3, 4)

respectively. Locate the centre of mass of the system.5 10

,3 3

Ans.

10. Two blocks of masses 10 kg and 4 kg are connected by a spring of negligible mass and areplaced, on a frictionless horizontal surface. An impulsive force gives a speed of 14 ms–1 tothe heavier block in the direction of the lighter block. Determine the velocity of the centre ofmass. [Ans. 10 ms–1]

BASED ON BANKING OF ROADS11. A string can with stand a maximum tension of 400 N. A mass of 1.5 kg is tied to one end of

a 1.5 m long string and rotated in a horizontal plane. Calculate maximum linear velocity withwhich mass can be rotated. [Ans. 20 ms–1]

12. Find the maximum speed at which a car can take turn round a curve of 30 m, radius on alevel road if the coefficient of friction between the tyres and road is 0.4. (g = 10 ms–2)

[Ans. 11 ms–1]

13. An aircraft executes a horizontal loop at a speed of 360 Km h–1 with its wings banked at 15°.Calculate the radius of the loop. [Ans. 3.81 Km].

14. A cyclist speeding at 18 Km h–1. on a level road takes a turn of radius 3m. The coefficient ofstatic friction between the tyres and the road is 0.1. Will he be able to take the turn.

[Ans. No, The cyclist will slip]

15. A circular race track of radius 300 m is banked at an angle of 15°. If the coefficient offriction between the wheels of a race car and road is 0.2. What is the

(a) optimum speed of the car to avoid wear and tear of its tyres.

(b) maximum permissible speed to avoid slipping. [Ans. 28.1 ms–1, 38.1 ms–1]

16. A sphere of mass 200 g is attached to an inextensible string of length 130 cm, whose upperend is fixed to the ceiling. The sphere is made to describe a horizontal circle of radius 50 cm.Calculate the periodic time of this conical pendulum and the tension in the string.

[Ans. 2.19 s, 2.12 N]

17. A motor cyclist goes round a circular track of diameter 320 m at 144 ms–1. How far from thevertical must he lean inwards to balance himself. (g = 10 ms–2). [Ans. 45°]

18. For traffic moving at 60 Km h–1, if the radius of the curve is 0.1 km. What is the correctangle of banking of the road? (g = 10 ms–2). [Ans. 15.5°]

19. An aeroplane travelling at a speed of 500 Km h–1 tilts through an angle of 30° as it makes aturn. What is the radius of the curve? [Ans. 3.409 Km]

20. A train has to negotiate a curve of radius 400 m. By how much the outer rail be raised ascompared with the inner rail for a speed of 48 km h–1. The distance between the rail is 1m.

[Ans. 4.53 cm].

21. Find the maximum speed at which a car can turn round a curve of 30 m radius on a levelroad if the coefficient of friction is 0.4. (Take g = 10 ms–2) [Ans. 11 ms–1].


22. The radius of the turn of a road is 500 m. The width of the road is 10 m. Its outer edge is0.15 m higher than the inner edge. For what speed of the vehicle running over the road thebanking is ideal? [Ans. 8.66 ms–1]

23. What should be the coefficient of friction between the tyres and the road when a cartravelling at 60 Km h–1 makes a level turn of radius 40 m? [Ans. 0.71]

24. An aeroplane executes a horizontal loop of radius 15.24 Km. The wings are banked at 15°.What is the speed of aeroplane? [Ans. 200 ms–1]

BASED ON EQUATIONS OF ROTATIONAL MOTION25. A constant torque is applied on a wheel, which turns through 200 radian in 10 sec. starting

from rest. Calculate

(i) Angular acceleration

(ii) What will be the angular velocity of the wheel after 20 sec from rest.

[Ans. 4 rad s–2, 80 rad s–1]

26. The radius of a wheel of a car is 0.4 m. The car is accelerated from rest by an angularacceleration of 3 rad/s–2, for 20 sec. How much distance the wheel covers in this time. Alsocalculate its linear velocity. [Ans. 240m, 24 ms–1]

27. A car is moving at a speed of 72 Km h–1. The diameter of its wheel is 0.50 m. If the wheelsare stopped in 20 rotations by applying brakes, calculate the angular retardation produced bybrakes. [Ans. – 25.5 rad/s2]

28. The wheel of a car is completing 1200 rotations per minute. On pressing the accelerator ofthe car, the wheel makes 4500 rpm after 10 sec. Compute its angular acceleration and theangular displacement in 10 sec. [Ans. 34.5 rad/s2, 475 rad].

29. The wheel of a train is rotating at a rate of 6 rps. On applying brakes it stops in 12 sec. Whatis the angular retardation produced by the brakes. [Ans. 3.14 rad/s–2]

30. A car initially at rest is accelerated by an angular acceleration of 4.0 rad s–2. In how muchtime will it gain an angular velocity of 800 rpm? What will be the angular displacement?

[Ans. 21s, 882 rad]

31. A car is moving at a speed of 144 Km/h. The diameter of its wheel is 1 m. If the wheels arestopped in 20 rotations by applying brakes, calculate the angular retardation produced by thebrakes.

2320160 rad / s, rad / s

Ans.

32. A constant torque is acting on a wheel. Wheel starts from rest. It makes 80 revolution in 2seconds. Calculate the angular acceleration. [Ans. 80 rad/s2]

BASED ON MOMENT OF INERTIA33. Three mass points m1 m2 and m3 are located at the vertices of

an equilateral triangle of side a. What is the moment ofinertia of the system about an axis, along the altitude of thetriangle passing through m1?

2

2 3( )4

am m

Ans.

a a

a/2 a /2

m 1

m 2 m 3


34 An equilateral triangle is formed by three identical uniform roads each of length 80 cm andmass 4 kg. Find the moment of inertia of this system about an axis passing through onecorner. [Ans. 3.84 Kg m2]

35. Calculate the radius of gyration of a rod of mass 100 g and length 100 cm about an axispassing through its centre of gravity and perpendicular to its length. [Ans. 28.8 cm]

36. Four bodies of masses 5 kg, 2 kg, 3 kg and 4 kg are respectively placed at position (0, 0, 0),(2, 0, 0) (0, 3, 0) and (– 2, – 2, 0). Calculate the moment of inertia about x axis, y axis and zaxis. [Ans. 43 unit, 24 unit, 67 unit]

37. Four rods are placed in the form of a square calculate the moment of inertia about an axispassing through the centre and perpendicular to the planes. Mass of each rod is M and lengthis L.

2ML

2

Ans.

38. Two spherical shells each of mass M and diameter R arejoined by a light and massless rod. Calculate the moment ofinertia about axis XX.

216MR

3 Ans.

39. A thin uniform disc is of mass 9M and of radius R. A small disc

of radius R

3 is cut as shown in figure. Find the moment of

inertia of the remaining disc about an axis passing through O andperpendicular to plane of disc. [Ans. 4 MR2].

40. What is the moment of inertia of a uniform circular disc ofradius R and mass M about an axis

(i) passing through the centre and normal to the disc.

(ii) passing through a point on the edge and normal to the disc. The moment of inertia of the

disc about any of its diameters is given to be 2MR

4.

2 2MR 3MR,

2 2

Ans.

41. Calculate the moment of inertia of a wheel of mass 10 kg and radius of gyration 30 cm.

[Ans. 0, 9 kg m2]

42. Calculate the moment of inertia of a solid cylinder of mass 5kg and radius 20 cm, about itsown axis. [Ans. 0.2 Kg m2]

43. Calculate the radius of gyration of a cylinderical rod of mass M and length L about an axis

of rotation perpendicular to its length and passing through its centre.L

12

Ans.

44. Four particles each of mass 2m are kept at the four corners of a square of side a. Find themoment of inertia of the system about an axis perpendicular to its plane and passing throughcentre of the square. [Ans. 4 ma2]

45. Calculate the moment of inertia of givensystem with respect to axis XX.[Ans. 1 kg-m2]


46. Calculate the moment of inertia of a circular disc of radius 10 cm, thickness 5 mm anduniform density 8 g/cc about a transverse axis through the centre of the disc.

[Ans. 6.28 × 104 g cm2]

47. The mass of a disc is 400 g and radius is 7 cm. Using theorem of parallel axis determinemoment of inertia about an axis passing through its edge and perpendicular to its plane.

[Ans. 2.94 × 104 g cm2]

48. The moment of inertia of a uniform circular disc about its diameter is 100 g cm2. What is itsmoment of inertia about a tangent in the plane of disc. [Ans. 500 g cm2]

49. The moment of inertia of a uniform circular disc of radius 10 cm about an axis, tangent tothe disc and parallel to the diameter is 12500 g cm2. Calculate the mass of the disc.

[Ans. 500 g]

50. Calculate the moment of inertia of a uniform disc of mass 200 g and radius 5 cm about anaxis passing through its edge and perpendicular to the plane of the disc.

[Ans. 7.5 × 10–4 Kg m2]

BASED ON ROTATIONAL KINETIC ENERGY51. An energy of 968 J is spent in increasing the speed of flywheel from 60 rpm to 360 rpm.

Calculate the moment of inertia of flywheel. [Ans. 1.4 kgm2]

52. A metre stick is held vertically with one end on the floor and is then allowed to fall. Find thevelocity of the other end when it hits the floor, assuming that the end on the floor does notslip. [Ans. 5.4 ms–1]

53. A solid spherical ball rolls on a table. What fraction of its total K.E. is rotational?

[Ans. 2/7].

54. Calculate the kinetic energy of a uniform circular disc of mass 100 g and diameter 10 cmmaking 100 rev min–1 about its axis. [Ans. 6.85 × 10–3 J.]

55. The rotational kinetic energy of two bodies of moment of inertia 9 kg m2 and 1 kg m2

respectively are same. Find the ratio of their angular momentum. [Ans. 3 : 1]

56. A solid sphere of mass 1 kg and radius 3 cm is rotating about its own axis with an angularvelocity of 50 radian per second. Find its kinetic energy of rotation. [Ans. 0.45 J]

57. A body of mass 4 kg is revolving in a horizontal circle of radius 2m at the rate of 1revolution per second. Calculate

(i) moment of inertia of the body

(ii) rotational K.E. of the body. [Ans. 16 kg m2, 16 2 J]

58. A uniform circular disc of mass m is set rolling on a smooth horizontal table with a uniform

linear velocity v. Find its total K.E.23

4mv

Ans.

59. A body of mass 1 kg is revolving in a horizontal circle of radius 1m. The frequency ofrevolution is 20 rps. Calculate the rotational K.E. of the body about axis of rotation.

[Ans. 800 2 J]

BASED ON TORQUE AND POWER OF TORQUE60. The torque associated with an automobile engine at 6000 rpm is 237.5 Nm. Calculate its

power. [Ans. 1.49 × 105 W]61. A wheel has moment of inertia 5 × 10–3 Kg m2 and is making 20 rev/sec. Calculate the torque

required to stop it in 10 sec. [Ans. 2 × 10–2 N-m]


62. The grinding stone a flour mill is rotating at 600 rad/sec. For this power of 1.2 Kw is used.Calculate the effective power. [Ans. 2 N-m]

63. A solid flywheel has a moment of inertia of 0.1 Kg-m2 about its axis of rotation. A tangentialforce of 2 kg is applied tangentially to it. If the radius of the wheel is 10 cm. Calculateangular acceleration produced. [Ans. 19.6 rad/s2]

64. A flywheel of mass 25 Kg has a radius of 0.2 m. A force of 5 N is applied tangentially to it.Calculate the acceleration produced in the flywheel. [Ans. 4 rad s–2]

65. A body is subjected to a torque of 12 Nm for 20 sec. The moment of inertia of the body is 6kg m2. Calculate the angular displacement of the body and work done in this duration.

[Ans. 400 rad, 2400 J]

66. A grindstone has a moment of inertia of 6 kg m2. A constant torque is applied and the grindstone is found to have a speed of 150 rpm, 10 sec after starting from rest. Calculate thetorque. [Ans. 3 Ns]

67. How much tangential force will be needed to stop the earth in one year, if it was rotatingwith angular velocity of 7.3 × 10–5 rad/sec. Given the moment of inertia of earth is 9.3 × 1037

kg m2 and radius of earth is 6.4 × 106 m. [Ans. 3.36 × 1019 N]

68. To maintain a rotor at a uniform angular speed of 400 rad s–1 an engine needs to transmit atorque of 18 Nm. What is the power of the engine. [Ans. 72 kW]

BASED ON ANGULAR MOMENTUM69. The frequency of revolution of a body is 10 rps. The moment of inertia of the body about the

axis of rotation is 1 kg m–2. Calculate angular moment of the body. [Ans. 20 J-s]

70. Calculate the angular momentum of the earth rotating about its own axis. Mass of earth is 6× 1027 kg and mean radius of the earth is 6.37 × 106 m. Consider earth to be a sphere ofuniform mass density. [Ans. 1.53 × 1034 kg m2 s–1]

71. The ratio of earth’s angular momentum (about the sun) to its mass is 4.4 × 1015 m2 s–1. Findthe area enclosed by earth’s orbit. [Ans. 6.93 × 1022 m2]

72. A cylinder of mass 5 kg radius 30 cm is free to rotate about its axis. It receives an angularimpulse of 3 kg m2 s–1 initially followed by a similar impulse after every 4 sec. What is theangular speed of the cylinder 30 s after the initial impulse? The cylinder is at rest initially.

[Ans. 106.7 rad/sec]

73. A particle of mass 0.01 kg having position vector ˆ10 6r i �

metre and moving with a

velocity ˆ5i ms–1. Calculate its angular momentum about origin. [Ans. ˆ0.3k kg m2/s]

74. A particle performs uniform circular motion with an angular momentum L. If the frequencyof particle motion is doubled and its kinetic energy is halved then calculate its new angularmomentum. [Ans. L/4]

75. Keeping the mass of earth constant, radius is halved then by how many years will theduration of day will decrease. [Ans. 18 hrs]

76. The maximum and minimum distance of a comet from the sun are 14 × 109 m and 7 × 107 mrespectively. If the maximum velocity of the comet is 6 × 102 km/sec. Calculate its minimumvelocity. [Ans. 3 km/sec]

77. A ring of mass 10 kg and diameter 0.4 m is rotating about its geometrical axis at 1200 rps.Calculate its angular momentum about the same axis. [Ans. 50.28 J-sec]

78. A fan of moment of inertia 0.6 kg m2 s–1 is to run upto a speed of 0.5 rps. Indicate the correctvalue of angular momentum, of the fan. [Ans. 0.6 kg m2 s–1]


79. A horizontal disc rotating about a vertical axis passing through its centre makes 180 rpm. Asmall piece of wax of mass 10 g falls vertically on the disc and stick to it at a distance of 8cm from its axis. If the frequency is thus reduced to 150 rpm, calculate the moment of inertiaof the disc. [Ans. 3.2 × 10–8 kg m2]

80. A flywheel of 100 kg is rotating at a speed of 20 rps. If the whole weight of the flywheel maybe considered to be concentrated at a distance of 3 m from the axis of rotation. Calculate its

K.E. and the average torque a brake will exert to stop it in 1

2 minute. [Ans. 117.7 N]

SOME ADDITIONAL PROBLEMS81. A ring, a disc and a sphere all of the same radius and mass roll down an inclined plane from

the same height h. Which of the three reaches the bottom

(i) earliest (ii) latest. [Ans. (i) sphere (ii) ring]

82. A solid cylinder of radius 5 cm. and mass 500 g rolls down an inclined plane (1 in 20)calculate the acceleration and total energy of the cylinder after 5 sec.

[Ans. 0.327 ms–2, 0.99 J]

83. An inextensible and flexible carpet of length 8 m iswound together and is placed on an inclined ramp asshown in figure, with one of its end fixed on it. When it isallowed to roll down the ramp, determine the time takenby the carpet to unwound completely

(sin 37° = 0.6 cos 37° = 0.8)

84. A steel cylinder of radius 3 cm and mass 2000 g is supported horizontally in frictionlessbearings on a light shaft passing through its axis. A light cord with one end fixed to thecylinder is wrapped round it and a weight of 100 g fastened to its free end. Find

(i) the linear acceleration of the weight

(ii) the tension in the cord

(iii) the angular acceleration of the cylinder. 1 2100ms , N, ms

11 11 33

g gg

Ans.

37°

279

GRAVITATION AND GRAVITYThe force of attraction between any two bodies is called gravitation force. If one of the two

bodies is earth then gravitation force is called gravity.

NEWTON’S LAW OF GRAVITATIONThis law states that

Every body in this universe attracts every other body with a force which is directlyproportional to the product of their masses and inversely proportional to the square of the distancebetween them.

Mathematically F = 1 22

G m m

r.

G = Universal gravitational constant

GRAVITATIONAL CONSTANTIt is a scalar quantityUnit S.I. G = 6.67 × 10–11 Nm2/kg2

CGS G = 6.67 × 10–8 dyne-cm2/g2.

Dimension [M–1 L3 T–2].

Gravitational Constant does not depend on the nature and size of the bodies, or the nature ofthe medium between them. It is universal constant.

It was first of all calculated by Cavendish using Torsion Balance.

ACCELERATION DUE TO GRAVITYThe acceleration produced in a freely falling body under the effect of gravity is called

acceleration due to gravity.

It is denoted by g

g = 9.8 ms–2

The acceleration due to gravity in terms of mass of earth is

g = 2

GM.

RR is radius of earth (R = 6400 km)

MASS OF EARTH

g = 2

GM

R

or M =2R

G

g.

��9UNIT


=

3 2

11

(9.8) (6400 10 )

6.67 10

� 5.98 × 1024 kg

DENSITY OF EARTH

M =2R

G

g.

Consider earth to be a sphere of radius R

M =34

R .3

34

R3 =

2R

G

g

or = 3

4 GRg

VARIATION IN THE VALUE OF ‘g’Various factors which affect the value of acceleration due to gravity are

(a) Shape of Earth

The radius of earth is nearly 24 km at equator than at poles. We know 2

1

Rg

Therefore value of g is maximum at poles and minimum at equator.

gp – geq = 0.02 m/s2

(b) Due to Rotation of EarthLet is the angular velocity of earth about its axis of rotation, then at latitude , the

aceleration due to gravity is given by

g = g – R2cos2.

(i) If latitude angle = 0, i.e. body is at equator

geq = g – R2.

(ii) If latitude angle = 90°, i.e. body is at pole

gp = g

gp – geq = R2 = 0.03 ms–2

(c) Due to Altitude (height)At a height h, the acceleration due to gravity is given by

g = g (1 + h/R)–2

= g (1 – 2h/R).

This formula is valid if h is upto 5% of earth radius. In case h is greater than 5% of earth then

g = g 2

2( )

R

R h

Gravitation 281

(d) Due to DepthAt a depth ‘d’ below the surface of earth, the acceleration due to gravity is given by

g = g(1 – d/R)

INERTIAL MASSInertial mass of a body is the measure of the difficulty to accelerate the body.

Inertial mass mi =F

a.

It is measured in kilogram

GRAVITATIONAL MASSIt is the force with which a body is attracted towards the centre of earth. It is also called

weight of the body i.e.

W = mg.

It is measured in Newton (N).

GRAVITATIONAL FIELDThe space surrounding a body in which its gravitational pull can be experienced by other

bodies is called its gravitational field.

The gravitational field intensity at a point in the gravitational field of a body is defined asthe force experienced by unit mass placed at that point.

Gravitational field intensity

I = 2

GMF

m r .

It is a vector quantity.

S.I. Unit N-kg–1

Dimension [LT–2].

The gravitational field intensity at a point is equal to the acceleration due to gravity at thatpoint.

GRAVITATIONAL POTENTIALThe gravitational potential at a point in the gravitational field of a body is defined as the work

to be done in bringing a body of unit mass from infinity to that point. It is scalar quantity.

v =GM

r .

S.I. Unit J kg–1

Dimension [L2T–2]

GRAVITATIONAL POTENTIAL ENERGYThe gravitational potential energy of a body at a point in the gravitational field of other body

is defined as the work done to bring the body from infinity to the point under consideration.

Gravitational P.E. =GMm

r .

S.I. Unit Joule.

Dimensions [ML2T–2]


Gravitational field intensity and gravitational potential at a point are related as

I =dV

dr

ORBITAL VELOCITYThe velocity required to put a satellite in its orbit around a planet is called orbital velocity.

The orbital velocity of a satellite revolving at a height h above the surface of earth is givenby

vo =GM

.R h

=2R

R ( )

g gR

h R h

.

Near the surface of earth

h << R

vo =GM

R .R

g

TIME PERIOD OF SATELLITETime taken by satellite to complete one revolution around the planet is called its time period,

i.e.,

T =2 (R )

Vo

h

=3/ 22 (R ) (R )

2 .GM GM

(R + )

h h

h

If satellite is revolving near the surface of earth then h << R.

T =3/ 22 (R)

GM

=

3/2

2

2 (R)

Rg

.

=R 3

2Gg

.

ESCAPE VELOCITYIt is the minimum velocity with which a body must be projected vertically upwards so that it

may escape from earth’s gravitational field. It is denoted by ve.

On the surface of earth escape velocity is given by

ve =2GM

2 R .R

g

Gravitation 283

ve = 11.2 km s–1 for the surface of earth.

Imp : vo = Rg

ve = 2 Rg

ve = 2 Vo

i.e. If orbital velocity of a satellite in its orbit increases by a factor 2 then it will escapeform the orbit.

HEIGHT OF SATELLITE ABOVE THE SURFACE OF EARTH

We know T =3

2

(R )2 .

R

h

g

On solving h =

1/ 32 2

2

T RR .

4

g

KEPLER’S LAWS

(i) Law of OrbitAll planets revolve around the sun in elliptical orbits with sun at one of its foci.

(ii) Law of areaThe radius vector joining the sun and planet sweeps out equal area in equal intervals of time,

i.e. area velocity remains constant.

(iii) Law of periodsSquare of time period (T) of a planet around the sun is proportional to the cube of semi major

axis (distance between sun and planet) i.e.

T2 r3

or T2 = Kr3

ENERGY OF A SATELLITEK.E. of satellite in its orbit

K =21 GM

2 2om

mvr

P.E. of satellite in its orbit

u =GM

.m

r

Total energy = u + K

=GM GM

2

m m

r r

= .GM

2m

r


BINDING ENERGY OF SATELLITEB.E. = – (T.E.)

= – GM

2

m

r

=GM

.2

m

r

GEOSTATIONARY SATELLITEIt is the satellite which revolves around the earth and have time period of revolution equal to

that of earth.

SPECIAL POINTS ABOUT GEOSTATIONARY SATELLITE(a) It rotates in equitorial plane.

(b) Its angular velocity and time period should be same as that of earth.

(c) Its height from earth surface is 36000 km.

(d) Its direction of rotation is from west to east.

(e) Its Orbit is called parking Orbit.

(f) The Orbital velocity of geostationary satellite is 3.1 km s–1.

SOLVED EXAMPLES

BASED ON NEWTON’S LAW OF GRAVITATIONExample 1. The mass of planet Jupiter is 1.9 × 1027 kg and that of the Sun is 1.99 × 1030 kg.

The mean distance of the Jupiter from Sun is 7.8 × 1011 m. Calculate the gravitational force whichthe Sun exerts on Jupiter.

Solution. Here M = 1.99 × 1030 kg

m = 1.9 × 1027 kgr = 7.8 × 1011 m.

From formula F = 2

GMm

r

F =11 30 27

11 2

6.67 10 1.99 10 1.9 10

(7.8 10 )

= 4.15 × 1023 NExample 2. The period of moon around the earth is 27.3 days and radius of the orbit is 3.9

× 105 km. G = 6.67 × 10–11 Nm–2 kg–2. Find the mass of the earth.

Solution. Here T = 27.3 days

= 27.3 × 24 × 3600 sec

r = 3.9 × 105 km

= 3.9 × 108 m

Moon revolves round earth in circular orbit. The centripital force required is provided byforce of attraction between earth and moon.

Gravitation 285

i.e.2mv

r= 2

GMm

r

or2

2

T

m r

r

= 2

GMm

r.

or M =2 3

2

4.

GT

r

M =8 3

11 2

4 9.8 (3.9 10 )

6.67 10 (27.3 24 3600)

= 6.31 × 1024 kgExample 3. Two particles of masses 2 kg and 4 kg are placed 50 cm apart. Assuming that the

only force acting on the particles are their mutual gravitation find the initial acceleration of heavierparticle.

Solution. Here m1 = 2 kg

m2 = 4 kg r = 50 cm = 0.50 m

From formula F =1 22

G m m

r

=11

2

6.67 10 2 4

(.5)

= 21.2 × 10–10 NAcceleration of heavier particle

a =2

F

m

=1021.2 10

4

= 5.3 × 10–10 ms–2

Example 4. Two stationary particle of masses M1 and M2 are d distance apart. A third particleof mass m is lying on the line joining the particles, experiences no resultant gravitational forces.What is the distance of this particle from M1.

Solution. The force on m towards M1 is

F1 =1

2

GM m

rThe force on m towards M2 is

F2 = 22

GM

( )

m

d rNet force on m is zero

F1 = F2

12

GM m

r=

22

GM

( )

m

d r

dM 1 M 2

r m


or2

2

( )d r

r

= 2

1

M

M

or 1d

r =

2

1

M

M.

ord

r=

2 1

1

M M

M

.

or r =1

2 1

M.

M M

d

Example 5. A rocket is fired from the earth towards the sun. At what point on its path is thegravitational force on the rocket is zero? Mass of sun = 2 × 1030 kg, mass of earth = 6 × 1024 kg.Neglect the effect of other planets etc. Orbital radius = 1.5 × 1011 m.

Solution. Here Ms = 2 × 1030 kg

Me = 6 × 1024 kg

r = 1.5 × 1011 m

Let m is mass of rocket. Let the force on the rocket is zero at distance x from earth.

Force on rocket due to earth = Force on rocket due to sun

2

GMem

x= 2

GM

( )sm

r x

orr x

x

=

M

Ms

e =

30

24

2 10

6 10

or 1r

x =

310

3.

r

x=

3101

3

On solving x = 2.6 × 108 mExample 6. Two solid spheres of same size of a metal are placed in contact by touching each

other. Prove that the gravitational force acting between them is directly proportional to the fourthpower of their radius.

Solution. We know

F = 2

G MM

(2R)

=

3 3

2

4 4G

3 3

4R

R R

=2 2 44

G9

n R

F R4

R R

M M

Gravitation 287

Example 7. Three spheres each having mass m and radius R are kept as shown in figure.Calculate the magnitude of the gravitational force on any one sphere due to other two.

Solution. The force on sphere C due to sphere A

= 2

G

(2R)

mm

F1 =2

2

G

4R

m

Similarly force on sphere C due to sphere B

= 2

G

(2R)

mm

F2 =2

2

G

4R

m

Magnitude of resultant force on sphere C

F = 2 21 2 1 2F F 2F F cos

= 2 21 2 1 2F F 2F F cos60

= 13 F as (F1 = F2)

F =2

2

3 G4 4R

m

Example 8. Find the gravitational force of attraction between a uniform sphere of mass Mand a uniform rod of length L and mass m. The distance of the near point of rod from centre ofmass of sphere is R.

Solution. Consider a small element of length dx of mass dm

dm = L

mdx

Force of gravitation between the element and the sphere is

dF = 2

G Mdm

x

dF = 2

GM

L

mdx

x

=2GM

L

mx dx

Total force of gravitation between the sphere and rod is

F = Fd = (R L) 2

R

GM

L

mx dx

=(R L) 2

R

GM

L

mx dx

r dx

L

x

R R

A

B C

60°

F1

F2C


=R L

R

GM 1

L

m

x

=GM 1 1 GM R L R

L R L R L R(R L)

m m

=GM

R(R L)

m .

BASED ON ACCELERATION DUE TO GRAVITYExample 9. The acceleration due to gravity at the moon’s surface is 1.67 ms–2. If the radius

of the moon is 1.74 × 106 m. Calculate the mass of the moon. Use the known value of G.

Solution. Here G = 6.67 × 10–11 Nm2/kg2

g = 1.67 ms–2 R = 1.74 × 106 m.

From formula

g = 2

GM

R

or M =2R

G

g

M =

6 2

11

1.67 (1.74 10 )

6.67 10

= 7.58 × 1022 kg

Example 10. Calculate the value of acceleration due to gravity on the surface of moon. Givenmass of moon is 7.4 × 1022 kg and its radius 1740 km.

Solution. Here m = 7.4 × 1022 kg

R = 1740 km = 1740 × 103 m

From formula g = 2

GM

R

g =

11 22

6 2

6.67 10 7.4 10

(1.7 10 )

= 1.64 ms–2.

Example 11. Calculate the mass of the earth from the following data.

Radius of earth R = 6×108 cm = 6 × 106 m.

Acceleration due to gravity = 9.8 ms–2

Gravitational constant G = 6.67 × 10–11 Nm2/kg2.

Solution. We know

g = 2

GM

R

Gravitation 289

M =2R

G

g

M =

6 2

11

9.8 (6 10 )

6.67 10

� 5.4 × 1024 kgExample 12. The weight of a body on earth is 1800 N. Calculate its weight on a body which

has a mass twice that of earth and radius thrice that of earth.

Solution. Here M = 2 Me

R = 3 Re

g = 2

GM

R

= 2

G (2M )

(3R )e

e = 2

GM2

9 Re

e

=2

9g

mg =2 2

( ) 18009 9

mg

= 400 NExample 13. The radius of earth decreases by 3% keeping its mass constant. Calculate the

percentage change in the acceleration due to gravity.

Solution. HereR

R

d=

3

100

From formula g = 2

GM

R

or 100g

g

=R

2 100R

=3

3 100100

= 9%

The value of acceleration due to gravity will decrease by 9%.

Example 14. Given that mass of earth is 81.5 times the mass of the moon and the diameterof the moon is 0.27 times that of the earth. Find the value of acceleration due to gravity at thesurface of moon in terms of ‘g’.

Solution. HereM

Me

m= 81.5

R0.27

Rm

e


From formula g = 2

GM

R

ge = 2

GM

Re

e

, gm = 2

GM

Rm

m

m

e

g

g= 2

GM

Rm

m ×

2R

GMe

e

=

2 2R M 1 1

R M 0.27 81.5e m

m e

gm = 0.1683 ge

or gm = 0.1683 g

Example 15. An astronaut on the moon measures the acceleration due to gravity to be 1.7ms–2. He knows that the radius of the moon is about 0.27 times that of the earth. What is hisestimate on the ratio of the mass of the earth to that of the moon. The acceleration due to gravityon the earth’s surface is 9.8 ms–2.

Solution. Here gm = 1.7 ms–2 ge = 9.8 ms–2

Rm = 0.27 Re

orR

Re

m=

1

0.27

From formula g = 2

GM

R, M =

2R

G

g

M

Me

m=

2R

Re e

m m

g

g

M

Me

m=

29.8 1

1.7 0.27

= 79.

Example 16. A man can jump 2.0 m high on the earth. Calculate the approximate height hemight be able to jump on a planet whose density is one quarter that of earth and radius is one thirdof the earth’s radius.

Solution. In terms of density acceleration due to gravity on the surface of earth is

g =4

GR3

Similarly on the planet g = 4

GR3

Here R = R

3and =

4

Gravitation 291

g =4 R

G3 3 4

=1 4

GR12 3

=

12

g

Let the man can jump to height h on the planet

mgh = mgh

12

gm h = m × g × 2.0

h = 24 mExample 17. If earth were made of lead of relative density 11.3, what then would be the

value of acceleration due to gravity on the surface of earth?

Re = 6.4 × 106 m and G = 6.67 × 10–11 Nm2/kg2.

Solution. Here = 11.3 × 103 kgm–3

Re = 6.4 × 106 mG = 6.67 × 10–11 Nm2/kg2

From formula

g = 2

GM

R =

32

G 4R

3R

.

=11 6 36.67 10 4 3.14 (6.4 10 ) (11.3 10 )

3

= 20.21 ms–2

Example 18. A body weighs 180 N on the surface of earth. How much will it weigh on thesurface of Mars whose mass is one nineth and radius one half that of earth.

Solution. Here Mm =M

9, Rm =

R

2

From relation g = 2

GM

R

gm = 2

M 4G

9 R

= 2

4 GM

9 R

= 4

9 eg

Weight of body on Mass

= mgm = 4

9em g

=4 4

( ) 1809 9emg

= 80 N


BASED ON VARIATION IN THE VALUE OF g WITH ALTITUDEExample 19. A body weighs 63 N on the surface of earth. What is the gravitational force on

it due to the earth at a height equal to half the radius of earth.

Solution. The value of acceleration due to gravity at a height h above the surface of earth is

g = 2(1 / R)

g

h

Here h =R

2.

Now mg = 2(1 / R)

mg

h

= 2

63

R / 21

R

=4

639

= 28 N

Example 20. Calculate the value of acceleration due to gravity at Mount Everest. Givenheight of Mount Everest is 8848 m and acceleration due to gravity on the surface of earth is 9.8ms–2.

Solution. Here h = 8848 m

R = 6.4 × 106 m

g = 9.8 ms–2

here h << R, therefore

g =2

1R

hg

= 6

2 88489.8 1

6.4 10

= 9.8 (1 – 0.002765)

= 9.8 × 0.997235

= 9.773 ms–2

Example 21. Calculate the height above the surface of earth where the value of g will bereduced by 19% of its value on the surface of earth.

Solution. From formula

g = 2(1 / R)

g

h

g =81

100g (Reduced by 19% of g)

Gravitation 293

81

100g = 2

1R

g

h

or9

10=

R

R h

or 9R + 9h = 10R

or R = 9h

or h =R

9 km

=6400

9 = 711.1 km

Example 22. A body weighs 98 N at North pole of earth. What will be its true weight in ageostationary satellite distance 7R (R radius of earth) from centre of earth.

Solution. True weight in a satellite at height h is given by

mg =

2R

Rmg

h

=

2R

R 6Rmg

=98

49 49

mg = 2 N

Example 23. Calculate the height where the value of g will remain one fourth of its value onthe surface of earth.

Solution. Here g =4

g

From formula

g =

2

2

R

(R )g

h

4

g=

2R

Rg

h

or1

2=

R

R h

R + h = 2R

h = R = 6400 km


Example 24. Calculate the percentage decrease in the weight of the body when taken to aheight of 64 km above the surface of earth. Re = 6400 km.

Solution. Here h = 64 km R = 6400 km

From formula g =2

1R

hg

or2

1R

h =g

g

org g

g

=

2

R

h

or 100g g

g

=2

100R

h

=2 64

1006400

= 2%Example 25. A body of mass m is raised to a height h from the surface of earth where the

acceleration due to gravity is g. Prove that the loss in weight due to variation in g is approximately2 mgh/R.

Solution. At a height h (h << R) acceleration due to gravity is given by

g =2

1R

hg

or (g – g) =2

R

hg

Loss in weight is given by

= mg – mg= m(g – g)

=2

R

hm g

=2

R

mgh.

Example 26. Two equal masses m and m are hung from a balance whose scale pans differ invertical height h. Determine the error in the weighing in terms of density of earth .

Solution. We know at a height h (h << R)

g =2

1R

hg

Now W2 – W1 = mg2 – g1

Gravitation 295

= 1 22R R

h hmg

=2

R

mg (h1 – h2)

= 2

2 GM

R R

mh

Where h = h1 – h2

Error in weight

= W2 – W1 = 3

3

2 G 4R

3R

m h

=8

G3

m h .

BASED ON VARIATION IN THE VALUE OF g WITH DEPTHExample 27. Assuming the earth to be a sphere of uniform mass density how much would a

body weight half way down to the centre of earth if it weighs 250 N on the surface.

Solution. Here mg = 250 N, d = R/2

At a depth d

g = 1R

dg

or mg = 1R

dmg

=R / 2

250 1R

= 125 N

Example 28. Calculate the depth below the surface of earth. Where the acceleration due togravity becomes 2% of its value on the surface of earth. (Re = 6400 km)

Solution. Here g =2

100g

, R = 6400 km

From formula g = 1R

dg

2

100g = 1

R

dg

1

50= 1

R

d

orR

d=

49

50


or d =49

640050

= 6272 km

Example 29. Calculate the depth below the surface of earth where acceleration due to gravitywill remain half of its value on the surface of earth. (Re = 6400 km).

Solution. Here

g =2

g

From formula g = 1R

dg

2

g= 1

R

dg

or1

2= 1

R

d

orR

d=

1

2

or d =R

2 = 3200 km

Example 30. Calculate the value of acceleration due to gravity at a depth of 1600 km belowthe surface of earth. Given radius of earth is 6400 km and on the surface of earth g = 9.8 m/s2.

Solution. Here d = 1600 km

From formula g = 1R

dg

=1600

16400

g

=3

4g

=3

9.84

= 7.35 ms–2

Example 31. Calculate the percentage change in the value of acceleration due to gravity at adepth of 32 km below the surface of earth.

Solution. Here R = 6400 kmd = 32 km

From formula g = 1R

dg

Gravitation 297

g = 32 11 1

6400 200g g

or g – g =200

g.

% change in the value of g is

100g g

g

=1

100200

= 0.5%The value of acceleration due to gravity will decrease by 0.5%.

Example 32. Determine the percentage decrease in the weight of one quintal sugar whentaken to the bottom of mine 1 km deep. Radius of earth 6400 km.

Solution. From Formula

g = 1R

dg

g

g

= 1

R

d

orR

d= 1

g

g

orR

d=

g g

g

Consider m is mass of the sugar Percentage change in the weight of sugar

( )100

m g g

mg

= 100R

d

=1

1006400

= 0.0156%

BASED ON VARIATION IN THE VALUE OF g WITH ROTATION OF THE EARTHExample 33. Determine the speed with which the earth would have to rotate on its axis so

that a person on the equator would weigh 3/5th as much as at present. Take the equatorial radiusas 6400 km.

Solution. From relation

g = g – R2

or mg = mg – mR2

Here g =3

5

g


3

5mg = mg – mR2

or =3

2 2 9.8

5R 5 6400 10

g

= 7.8 × 10–4 rad/secExample 34. Calculate the imaginary angular velocity of the earth for which the effective

acceleration due to gravity at the equator becomes zero. Also calculate the length of the day in thiscondition. (Re = 6400 km, g = 10 ms–2)

Solution. From relation

g = g – R2

The value of acceleration due to gravity at equator is zero is g = 0

o = g – R2

or =R

g = 3

10

6400 10

= 1.25 × 10–3 rad/sec

Also time period

T = 3

2 2 3.14

1.25 10

= 5024 sec

= 1.4 hExample 35. Find the difference in the weight of a gram of a substance due to rotation of

earth as measured at the equator and at the poles. (Radius of earth = 6.378 × 108 cm).

Solution. We know

g =2 2R cos

1gg

is called latitude

At equator = 0 cos = 1

ge =2

2R1 Rg g

g

(1)

At poles gp =2R

1 cos90gg

= g (2)

Using equation (1) and (2)

ge = gp – R2

Hence the difference in the weight of a grass as measured at the equator and at pole is

mge = mgp – mR2

Gravitation 299

or (mgp – mge) = mR2

= R2 as m = 1 g

=2

6 26.378 10

86164

= 0.034 N

Example 36. Determine the speed with which the earth would have to rotate on its axis so

that a body placed at latitude 60° would weigh 3

4

th of its weight without rotation. (Re = 6400

km).

Solution. We know

g = g – R2 cos2or mg = mg – mR2 cos2

But mg =3

4mg and = 60°

3

4mg = mg – mR2 cos2

4

mg= mR2 cos2 = mR2 (cos 60)2

4

mg=

2

4

mR

or w = 3

9.8

R 6400 10

g .

� 1.24 × 10–3 rad/sec

BASED ON ORBITAL VELOCITYExample 37. Calculate the time period of a satellite revolving near the surface of earth.

Given radius of earth 6400 km.

Solution. Here R = 6400 km = 6.4 × 106 m

g = 9.8 m/s

Time period T =2 R 2

Ro

R

g

v

=66.4 10

2 3.149.8

= 5079 sec

� 1.41 hr

Example 38. An artificial satellite makes a rotation around earth at a height of 270 km, abovethe surface of earth. Determine its time period of satellite.


Solution. Here R = 6400 km = 6.4 × 106 m

g = 9.8 ms–2

h = 270 km = 0.270 × 106 m

From formula

T =3

2

(R )2

R

h

g

or T =6 6 3

6 2

(6.4 10 0.27 10 )2 3.14

9.8 (6.4 10 )

or T =6 3

2 12

(6.67 10 )2 3.14

9.8 (6.4) 10

� 5400 sec

= 1.5 hrExample 39. An artificial satellite revolves round the earth at a height of 1000 km. The

radius of the earth is 6.38 × 103 km. Mass of the earth is 6 × 1024 kg. G = 6.67 × 10–11 Nm2/kg2.Calculate the orbital velocity of satellite and its time period.

Solution. Here h = 1000 km

R = 6.38 × 103 km

M = 6 × 1024 kg

Orbital velocity

vo =GM

R h

=

11 24

6 6

6.67 10 6 10

(6.38 10 1.0 10 )

=

13

6

6.67 6 10

7.38 10

= 7364 ms–1

Now time period

T =

62 (R ) 2 (7.38 10 )

7364o

h v

= 6297 sec

= 1.75 hrExample 40. A satellite is revolving in a circular orbit around the earth. The time period of

the satellite is 8474 sec. Calculate the height of the satellite above the surface of earth.

Solution. Here T = 8474 sec

Gravitation 301

From formula T =2 (R )

Vo

h

Where vo =GM

(R )h

T =3/ 22 (R )

GM

h

or T2 =2 3 2 3

2

4 (R ) 4 (R )

GM R

h h

g

or (R + h) =3/ 22 2

2

T R

4

g

or h =3/22 2

2

T RR

4

g

=

3/ 22 6 239.8(8474) (6380 10 )

6380 104 9.8

� 2620 × 103

� 2620 kmExample 41. Consider earth to a sphere of uniform mass density . T is time period of

revolution of an artificial satellite revolving around earth close to its surface. Show that T2 isuniversal constant having value 1.41 × 1011 kgs2m–3

Solution. We know

T =3(R )

2GM

h

or M =2 3

2

4 (R )

GT

h

as h << R (near the surface of earth)

M =2 3

2

4 R

GT

or 34R

3

=2 3

2

4 R

GT

or T2 =3

G

(Which is Universal constant)


or T2 = 11

3 3.314

6.67 10

� 1.41 × 1011 kgs2m–3

BASED ON ESCAPE VELOCITYExample 42. Show that escape velocity of a body from the surface of earth is nearly

11.2 km/s.

Solution. Here R = 6400 km

= 6400 × 103 m

g = 9.8 ms–2

From formula

ve = 32 R 2 9.8 6400 10g

= 11.2 × 103 ms–1

= 11.2 km/sExample 43. Jupiter has a mass 318 times that of the earth and its radius is 11.2 times the

earth’s radius. Estimate the escape velocity of a body from Jupiter surface given that the escapevelocity from the earth’s surface is 11.2 kms–1.

Solution. MJ = 318 Me

RJ = 11.2 Re

Escape velocity from the surface of earth is given by

ve =2GM

Re

e

Similarly for Jupiter

vJ = J

J

2GM

R

J

e

vv

= J

J

M R

R Me

e

=318M R

11.2R Me e

e e

vJ =318

11.211.2

= 59.7 kms–1.

Example 44. Determine the escape velocity of a body from the moon. Take the moon to bea uniform sphere of radius 1.76 × 106 m and mass 7.36 × 1022 kg. (G = 6.67 × 10–11 Nm2/kg2).

Solution. Here G = 6.67 × 10–11 Nm2/kg2

M = 7.36 × 1022 kg

Gravitation 303

ve =

11 22

6

2GM 2 6.67 10 7.36 10

R 1.76 10

= 2375 ms–1

= 2.375 × 103 ms–1

Example 45. The escape velocity of a projectile on the earth’s surface is 11.2 kms–1. A bodyis projected out with thrice this speed. What is the speed of the body far away from the earth?Ignore the presence of the sun and other planets.

Solution. Here v e = 11.2 kms–1 v = 3ve

Let far away from the surface of earth the speed of projectile is v

Total energy on the surface of earth = Total energy far away from earth

2V GM

2 R

m m =

2

02

mv (Far away from earth P.E. is zero)

or v 2 = 2 2GM

Rv

or v 2 = 2 2ev v

2GM

Re

v

or v 2 = 2 2(3 )e ev v

or v 2 = 28 ev

or v = 2 2 e v

= 2 2 11.2

= 31.68 kms–1

Example 46. If the earth has a mass nine times and radius twice that of the planet Mars.Calculate the escape velocity from the surface of Mars if its value on the earth surface is 11.2 kms–1.

Solution. For Surface of earth

ge = 2

G M

Re

e

and for planet mass

gm = 2

G M

Rm

m

m

e

g

g=

2 2

2 2

M R M (2 R )

M 9 MR Rm e m m

e mm m

=4

9


Now ve = 2 Re eg

and vm = 2 Rm mg

m

e

vv =

R

Rm m

e e

g

g =

4 1

9 2

vm = 2 111.2

3 2

= 5.28 × 103 ms–1

Example 47. Find the escape velocity from the sun, if its mass is 1.89 × 1030 kg and itsdistance from earth is 1.59 × 108 km. (G = 6.67 × 10–11 Nm2/kg2)

Solution. Here M = 1.89 × 1030 kg R = 1.59 × 108 km

From formula

ve =2GM

R =

11 30

8

2 6.67 10 1.89 10

1.59 10

= 3.98 × 104 ms–1

BASED ON KEPLER’S LAW OF PLANETARY MOTIONExample 48. Compare the period of rotation of planet Mars about the Sun with that of the

earth about it. The mean distance of the Mars from the sun is 1.52 A.U.

Solution. According to Kepler’s third law

T2 R3

2

1

2

T

T

=

331

2

R(1.52)

R

1

2

T

T

= (1.52)3/2

= 1.8741Example 49. The mean orbital radius of the earth around sun is 1.5 × 108 km. Calculate the

mass of the sun if G = 6.67 × 10–11 Nm2/kg2.

Solution. Here r = 1.5 × 108 km = 1.5 × 1011 m

T = 365 days = 365 × 24 × 3600 s

From Kepler’s third law

T2 = Kr3

or T2 =2

34

MGr

where K =

24

MG

or M =

2 3

2

4

G T

r

Gravitation 305

M =

11 3

11 2

4 9.8 (1.5 10 )

6.67 10 (365 24 3600)

= 2.01 × 1030 kgExample 50. The period of moon around the earth is 27.3 days and radius of the orbit is 3.9

× 105 km. G = 6.67 × 10–11 Nm2/kg2. Find the mass of the earth.

Solution. Here r = 3.9 × 105 km = 3.9 × 108 m

T = 27.3 × 24 × 3600 s

Using formula

M =2 3

2

4

GT

r

=

8 3

11 2

4 9.8 (3.9 10 )

6.67 10 (27.3 24 3600)

= 6.3 × 1024 kgExample 51. Calculate the period of revolution of Neptune around the sun, given that

diameter of its orbit is 30 times the diameter of earth’s orbit around the sun both orbits beingassumed to be circular

Solution. Using Kepler’s third law

RN = 30 Re

Te = 1 yr.

2NT 3

NR

2eT 3

eR

2N2

T

Te

=3N3e

R

R

or 2NT =

3230R

. (1)R

e

e

or TN = 27000

= 164.3 yrs.

Example 52. The ratio of time periods of two planets round Sun is 10 10 . Calculate theratio of the distances of the two planets from Sun.

Solution. Here

1

2

T

T= 10 10

From Kepler’s third law

2122

T

T=

3132

R

R


2(10 10) =3132

R

R

or

3

1

2

R

R

= 1000

or 1

2

R

R= 10.

Example 53. The distances of two planets from Sun has ratio (100 : 1). Find the ratio of thespeeds of the two planets.

Solution. Here 1

2

R

R= 100

We know v1 = 1

1

2 R

T

and v2 = 2

2

2 R

T

1

2

vv

=

3/ 2

1 2 1 2

1 1 2 1

R T R R

T R R R

(T2 R3)

= 2

1

R 1

R 100 1

10

BASED ON GRAVITATIONAL POTENTIAL AND POTENTIAL ENERGYExample 54. A geostationary satellite orbits the earth at a height of nearly 36,000 km from

the surface of earth. What is the potential due to earth’s gravity at the site of the satellite? Me =6.0 × 1024 kg. Re = 6400 km.

Solution. Here r = R + h

= 6400 + 36000

= 42400 km

= 4.24 × 107 m

M = 6.0 × 1024 kg

Gravitational Potential

v =G M

r

v =11 24

7

6.67 10 6 10

4.24 10

= – 9.43 × 106 J/kg

Gravitation 307

Example 55. A satellite orbits the earth at a height of 500 km from its surface. Compute its(i) Kinetic Energy (ii) Potential Energy (iii) Total Energy. Mass of satellite = 300 kg, Mass of earth6 × 1024 kg, radius of the earth = 6.4 × 106 m

Solution. Here h = 500 km m = 300 kg

Me = 6 × 1024 kg

Re = 6.4 × 106 m

(i) Kinetic Energy

K.E. = eGM

2(R )

m

h

=11 24

6 3

1 6.67 10 6 10 300

2 (6.4 10 500 10 )

= 8.7 × 109 J(ii) Potential Energy

P.E. = eG M

(R )

m

h

=11 24

6 3

6.67 10 6 10 300

(6.4 10 500 10 )

= – 17.4 × 109 J(iii) Total Energy

= K.E. + P.E.

= (8.7 × 109) + (–17.4 × 109)

= – 8.7 × 109 JExample 56. The radius of the earth is 6.37 × 108 cm, its mean density 5.5 gcm–3 and the

gravitational constant is 6.66 × 10–11 Nm2/kg2. Calculate the earth’s surface potential.Solution. We know

v =GM

R

=3G 4

RR 3

= 24G R

3

or v = 11 6 2 343.14 6.67 10 (6.37 10 ) 5.5 10

3

= – 6.22 × 107 J/kgExample 57. A geostationary satellite orbits the earth at a height of nearly 36000 km from

the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite?Me = 6.0 × 1024 kg, Re = 6400 km


Solution. Here h = 36000 km = 3.6 × 107 m

Me = 6.0 × 1024 kg

Re = 6400 = 6.4 × 106 m

Potential due to earth at a height h is

v =GM

R h

v =11 24

6 7

6.67 10 6.0 10

6.4 10 3.6 10

=13

6

6.67 6.0 10

(6.4 36) 10

= – 9.44 × 106 J/kgExample 58. What is the gravitational potential energy of a body of 0.2 kg at a height 1600

km above the surface of the earth.

G = 6.67 × 10–11 Nm2/kg2

Me = 6 × 1024 kg

Solution. Gravitational Potential Energy

=eGM

(R )h

=

11 24

6 6

6.67 10 6 10 0.2

(1.6 10 6.4 10 )

= – 107 JExample 59. A satellite orbits the earth at a height of 400 km above the surface. How much

energy must be expended to a rocket to send it out of the earth’s gravitational influence. Mass ofsatellite = 200 kg, Mass of earth = 6 × 1024 kg, radius of earth = 6.4 × 106 m, G = 6.67 × 10–11

Nm2/kg2.

Solution. Total energy of the satellite in its orbit is

E =GM

2(R )

m

h

=11 24

6 6

6.67 10 6 10 200

2[6.4 10 0.4 10 ]

=

15

6

6.67 6 2 10

2 6.8 10

= – 5.9 × 109 J

Energy to be expended to rocket the satellite out of earth’s gravitation

= 0 – [– 5.9 × 109]

= 5.9 × 109 J

Gravitation 309

Example 60. A space ship is stationed on Mars. How much energy must be expended on thespace ship to rocket it out of the solar system?

Mass of the space ship = 1000 kg

Mass of the sun = 2 × 1030 kg

Mass of Mars = 6.4 × 1023 kg

Radius of Mars = 3395 km

Radius of the orbit of Mars = 2.28 × 108 km

Solution. Energy of the space ship in its orbit

= K.E. + P.E.

=21 GM

2 (R )m

mh

v

=GM GM

2(R ) (R )

m m

h h

vo =

GM

R h

=GM

2(R )

m

h

Energy expanded on the space ship to rocket it out of the solar system

=GM

2(R )

m

h

=11 30

8 3 3

6.67 10 1000 2 10

2(2.28 10 10 3395 10 )

= 2.9 × 1011 JExample 61. A rocket is fired vertically from the surface of Mars with a speed of 2 km s–1.

If 20% of its initial energy is lost due to mantain atmospheric resistance, how far will the rocketgo from the surface of Mars before returning to it?

Mass of Mars = 6.4 × 1023 kg, radius of Mars = 3395 km, G = 6.67 × 10–11 Nm2 kg–2.

Solution. Total energy of the rocket

E = 21 GM

2 Rm

m v

Since 20% of energy in lost.

Energy retained with in rocket =80 4

(E) E100 5

At maximum height ‘h’ total energy will become potential energy

or4

E5

=GM

(R )

m

h

or 24 1 GM

5 2 R

mm

v =

GM

(R )

m

h


or22 R 2GM

5 R

v=

GM

(R )h

or (R + h) = 2

5 RGM

2 ( R 2GM)

v

=3 11 23

3 2 3 11 23

5 3395 10 6.67 10 6.4 10

2 [(2 10 ) 3395 10 2 6.67 10 6.4 10 ]

=77.25 10

2 7.18

= + 5.05 × 106 m

R + h = 5.05 × 106 m

h = 5.05 × 106 – 3395 × 103

= 1.655 × 106 m

� 1655 kmExample 62. Two heavy spheres each of mass 100 kg and radius 0.1 m are placed 1.0 m

apart on a horizontal table. What is the gravitational field and potential at the mid point on the linejoining the centres of the spheres.

Solution.

A BP

0.5 m 0.5 m

Consider A and B are two spheres and P is the mid point of AB.

Gravitational field at P due to sphere A.

E1 = 2

GM

r(along PA)

Gravitational field at P due to sphere B

E2 = 2

GM

r(along PB)

E1 and E2 are equal in magnitude but opposite in direction so net Gravitational field is zero.

Gravitational potential is a scalar quantity so potential at point P is

v = v A + v B

=GM GM 2GM

r r r

=112 6.67 10 100

0.5

= – 2.67 × 10–8 J/kg

Gravitation 311

Example 63. Gravitational field instensity at a point 104 km from the surface of the earth is4.8 Nkg–1. Calculate the gravitational potential at that point.

Solution. Here I = 4.8 N/kg

h = 104 km = 107 m

Now

I = 2

GM

r = 2

GM

(R )h

v =GM

(R )h

v = – (I) (R + h)

= – 4.8 (6.4 × 106 + 10 × 106)

= – 4.8 × 16.4 × 106

= – 7.87 × 107 J/kgExample 64. Two bodies of mass 10 kg and 1000 kg are at a distance of 1 m. At what point

on the line joining them will the gravitational field intensity be zero.

Solution. Here m1 = 10 kg

m2 = 1000 kg

r = 1 m

m 1 m 2

x

Let gravitational field intensity will be zero at distance x from m1.

12

Gm

x= 2

2

G

( )

m

r x

Here m1 = 10 kg m2 = 1000 kg

r = 1 m

2

10

x= 2

1000

(1 )x

or1

x=

10

1 x

1 – x = 10x

or x =1

m11

Example 65. The mass of earth is 6.0 × 1024 kg and its radius is 6.4 × 106 m. How muchwork will be done in taking a 100 kg body from surface of earth to infinity?

Solution. Work done in taking a body of mass m from surface of earth to infinity.

W =GM

R

m


=

11 24

6

(6.67 10 ) (6 10 ) 100

6.4 10

= 6.25 × 109 JExample 66. Calculate the gravitational field strength and gravitational potential at the

surface of moon.

Given Mm = 7.34 × 1022 kg

Rm = 1.74 × 106 m

Solution. Here

Mass of moon, Mm = 7.34 × 1022 kg

Radius of moon, Rm = 1.74 × 106 mGravitational field intensity

I = 2

GM

R

I =

11 22

6 2

6.67 10 7.34 10

(1.74 10 )

= 1.62 N/kgGravitational potential

v =GM

R

v =11 22

6

6.67 10 7.34 10

1.74 10

= – 2.8 × 106 J/kgExample 67. Three mass points each of mass m are placed at the vertices of an equilateral

triangle of side l. What is the gravitational field and potential due to three masses at the centroidof the triangle.

Solution. Consider three mass points each of mass m aresituated on the three vertices of an equilateral triangle of side l. Ois the centroid of triangle ABC.

OA = OB = OC

In triangle ODB ODB = 90°

cos 30° =BD

OB

OB =BD / 2

cos30 3 / 2 3

l l

Similarly OC = OA = 3

l.

Gravitational field strength at O due to masses at point A, B and C are

120°

30°

O

B D C

A

F EE 1

E 3E 2

Gravitation 313

I1 = 2 2 2

GM GM 3GM

OA ( / 3)l l (

—along OA

)

Similarly I2 = 2

3GM

l(

—along OB

)

and I3 = 2

3GM

l(

—along OC

)

angle between 1I��

and 2I��

is 120°

Therefore resultant is I = 2 21 2 1 2I I 2I I cos

= 2

3 GM1 1 2( 1/ 2)

l cos 120° = – 1/2

= 2

3 GM

lalong OF

��

As shown is figure I is opposite to I3.

Therefore gravitational field at O is zero.

Gravitational potential is a scalar quantity, so the total gravitational potential at O is

v = v1 + v2 + v3

v =GM GM GM

( / 3) / 3 / 3l l l

=GM

3/ 3l

=GM

- 3 3l

Example 68. A rocket is launched vertically from the surface of the earth with an initialvelocity u. Find the height up to which the rocket can go from the surface of earth before fallingback to earth.

Solution. Total energy of the rocket at the surface of earth

= K.E. + P.E.

=21 GM

2 Rm

m v (GM = gR2)

=21

R2

m mgv

Consider h be the height up to which the rocket can go from the surface of earth.

Total energy of rocket at height h

=2GM R

0(R ) (R )

m mg

h h


Using law of conservation of energy

21 GM

2 R

mmu =

2R

(R )

mg

h

or21

R2

mu mg =2R

(R )

mg

h

21

2mu =

2RR

(R )

mgmg

h

=2 2R R R

(R )

mg mgh mg

h

2

2

u=

R

(R )

gh

hor 2

R 2 Rh g

h

v

R

h= 2

2 R1

g

u

or h =

2

R2 R

1g

u

= 2

12

Rg

u

Example 69. A rocket is fired vertically with a speed of 5 kms–1 from the earth’s surface.How far from the earth does the rocket go before returning to the earth? Mass of the earth = 6.0 ×1024 kg, mean radius of the earth = 6.4 × 106 m, G = 6.67 × 10–11 Nm2/kg2.

Solution. Here v = 5 km/s = 5000 m/s

M = 6.0 × 1024 kg

R = 6.4 × 106 m

G = 6.67 × 10–11 Nm2/kg2.

Let h is the height attained by rocket before returning to earth.

Total energy on the = Total energy at height h

surface of the earth

21 GM

2 Rm

m v =GM

0R

m

h

2 2GM

Rv =

2GM

(R )h

or2GM

R h = 22GM

R v

22GM

R e

v

Gravitation 315

2GM

(R )h= 2 2

e v v

(R + h) = 2 2

2GM

e v v

=11 24

3 3 3 3

2 6.67 10 6 10

(11.2 10 5 10 ) (11.2 10 5 10 )

=13

6

12 6.67 10

6.2 16.2 10

= 8 × 106 m

h = 8.0 × 106 – 6.4 × 106

= 1.6 × 106 mExample 70. Two stars each of 1 solar mass (= 2 × 1030 kg) are approaching each other for

a head on collision. When they are at a distance 109 km, their speeds are negligible. What is thespeed with which they collide? The radius of each star is 104 km. Assume the stars to remainundisorted until they collide. G = 6.67 × 10–11 Nm2/kg2.

Solution. Here M = 2 × 1030 kg

Radius of each star = 107 m

Initial distance between two stars

R = 1012 m

Initial total energy of the system

= K.E. + P.E.

= 0 + 2

12

GMM GM

R 10

When they are just to strike, total energy is

=21 GMM

M2 2r

v

= Mv2 – 2

7

GM

2 10

According to law of conservation of energy

2

12

GM

10

=

22

7

GMM

2 10

v

or Mv2 =2 2 2

7 12 7

GM GM GM

2 10 10 2 10

or v2 = 7

GM

2 10

2 2

12 7

GM GMto neglect it

10 2 10

10 km9 2 × 10 kg

30

2 r


v = 7

GM

2 10

=11 30

7

6.67 10 2 10

2 10

= 2.6 × 106 m/s.

EXERCISE

BASED ON NEWTON’S LAW OF GRAVITATION1. The force of gravitation between the earth and a body of mass M placed on its surface is 18

× 107 N. Calculated the mass M. [Ans. 18.4 × 106 kg]

2. Calcualte the gravitational force between two bodies of mass 1000 kg each seperated by 1 mon earth. What will be the change in gravitational force if bodies are taken on moon.

[Ans. 6.67 × 10–5 N, No change]

3. The earth completes one revolution around sun in one year. Calculate the value ofgravitational constant. Given Mass of Sun = 2 × 1030 kg, distance of earth from sun = 1.5 ×1011 m. [Ans. 6.69 × 10–11 Nm2 kg–2]

4. A sphere of mass 40 kg is attracted by a second sphere of mass 15 kg when their centres are20 cm apart, with a force of 9.8 × 10–7 N. Calculate the value of gravitational constant.

[Ans. 6.53 × 10–11 Nm2 kg–2]

5. The gravitational force of attraction between two bodies is 20 N. If each mass is doubled anddistance between them is also doubled then what is the force of attraction between them.

[Ans. 20 N]

6. A sphere of mass 40 kg is attracted by a second sphere of mass 80 kg, with a force equal to

weight of 1

4 of mg when their centres are 30 cm apart. Calculate the constant of gravitation.

[Ans. 6.88 × 10–11 Nm2 kg–2]

7. Compare the electrostatic force of attraction between an electron and a proton with thegravitational force of attraction between them when held at same distance.

[Ans. 2.29 × 1039]

8. Two solid spheres of same size of a metal are placed in contact by touching each other.Prove that the gravitational force acting between them is directly proportional to the fourthpower of their radius.

9. Three particles each of mass m are placed at the three corners of an equilateral triangle ofside a. Calculate the gravitational force due to these three particles on a particle of mass mplaced at centroid. [Ans. 0 N]

BASED ON ACCELERATION DUE TO GRAVITY10. Calculate the value of acceleration due to gravity on the surface of a planet, that has a radius

half that of earth and average density as that of earth. [Ans. 4.9 ms–2]

11. If earth ever made of lead of relative density 11.3 what then would be the value ofacceleration due to gravity on the surface of earth?

Re = 6.4 × 106 m, G = 6.67 × 10–11 Nm2 kg–2. [Ans. 20.21 ms–2]

Gravitation 317

12. The mass of moon is 7.6 × 1022 kg. The acceleration due to gravity at the moon’s surface is1.67 ms–2. Using known value of G calculate the radius of the moon. [Ans. 1.74 × 106 m]

13. A body weighs 180 kgf on the surface of the earth. How much will it weigh on the surface of

the moon whose mass is 1

9th and radius is half of that for the earth? [Ans. 80 kgf]

14. A man can jump 1.2 m on the earth. Calculate the approximate height up to which height hecan jump on a planet whose density is one quarter that of the earth and radius is one third ofthe earth. [Ans. 14.4 m]

15. The mass of the Jupiter is 314 times that of earth and the diameter is 11.35 times that ofearth. Calcualte the value of acceleration due to gravity on Jupiter if its value on earth is 9.8ms–2. [Ans. 23.90 ms–2]

16. Considering the earth to be a sphere of uniform mass density 5.5 gcm–3. Calcualte the valueof g on its surface. Given Re = 6400 km, G = 6.67 × 10–11 Nm2/kg2 [Ans. 9.82 ms–2]

17. Calculate the percentage change in the value of g if keeping the mass of earth fix, its radiusis decreased by 2%. [Ans. g will increase by 4 %]

18. The radii of two planets are in the ratio R1 : R2 their mean densities are in the ratio 1 : 2.Prove that the g1 : g2 = R11 : R22

Where g1 and g2 are acceleration due to gravity on the surface of two planets.

19. The radius of the earth is made four times. By what factor would its density have to bechanged to keep ‘g’ same. [Ans. Density should decrease by 1/4]

20. An astronaut on moon measures acceleration due to gravity to be 1.7 ms–2. He knows that theradius of moon is about 0.27 times that of earth. What is his estimate of the ratio of the massof earth to that of moon. Given ge = 9.8 ms–2. [Ans. Me : Mm = 79]

BASED ON VARIATION OF g WITH ALTITUDE21. A body weighs 81 N on the surface of earth. What will be the gravitational force on it due to

earth at a height equal to half the radius of the earth. [Ans. 36 N]

22. At what height above the earth’s surface the acceleration due to gravity will be 1

9th of its

value at the earth’s surface? Re = 6400 km. Ans. 1.28 × 104 km]

23. What will be your weight at 800 km above the surface of earth if on the surface of earth its

value is W N. [Ans. 8

9 W N]

24. Consider the earth to be a sphere of radius 6.37 × 106 m. It is rotating about its axis with aperiod of 8.64 × 104 s. Calculate the variation in the ‘g’ if one moves from pole to equator.

[Ans. 3.37 × 10–2 ms–2]

25. A body weighs 90 kg on the surface of earth. How much will it weigh on the surface of Mars

whose mass is 1

9 and radius

1

2 that of earth. [Ans. 40 kgwt]

26. A body is raised to a height of 500 km. What is the acceleration due to gravity at this point?

[Ans. 8.45 ms–2]

27. The mass of the planet is three times that of earth and the diameter of planet is also threetimes that of the earth. What is the acceleration due to gravity on the surface of planet?

[Ans. g = 3.267 ms–2]

28. Find the percentage decrease in the weight of a body when taken to a height of 16 km abovethe surface of earth. Radius of the earth is 6400 km. [Ans. 0.5%]


29. A body when suspended from a spring, stretches it by 1 cm on the earth’s surface. How muchwill the same body stretch spring at a place 1600 km above the earth’s surface? Re = 6400km. [Ans. 0.64 cm]

30. Calcualte the height above the surface of earth where the value of acceleration due to gravitywill remain one fourth of its value on the surface of earth. [Ans. 6400 km]

BASED ON VARIATION IN THE VALUE OF g WITH DEPTH31. Calculate the depth below the surface of earth where acceleration due to gravity will remain

half of its value on the surface of earth. [Ans. 3200 km]

32. Calcualte the depth below the surface of earth where acceleration due to gravity becomes 70% of its value on the surface of earth (Re = 6400 km) [Ans. 1920 km]

33. Assuming the earth to be a sphere of uniform mass density, how much would a body weighhalf way down to the centre of the earth, if it weighs 200 N on its surface. [Ans. 100 N]

34. A what depth below the earth’s surface value of g is same at a height of 50 km above thesurface of earth. [Ans. 100 km]

35. At what height from the surface of earth the acceleration due to gravity is the same as it is ata depth 160 km below the surface of earth. Re = 6400 km. [Ans. 80 km]

36. Calculate the percentage decrease in the weight of 100 kg sugar when taken to the bottom ofa mine, 1 km deep. Re = 6400 km. [Ans. 0.0156%]

37. Compare the weight of a 5 kg body 10 km above and 10 km below the surface of earth.

[Ans. 0.998]

38. Assuming the earth to be a sphere of uniform mass density, calculate the value ofacceleration due to gravity in a mine 100 km deep. Radius of the earth is 6380 km, and g =9.8 ms–2. [Ans. 9.66 ms–2]

VARIATION IN THE VALUE OF ‘g’ DUE TO SHAPE OF EARTH AND ROTATION OFEARTH

39. Determine the speed with which the earth would have to rotate on its axis so that a person on

the equator would weigh 3

5th as much as at present. Take the equatorial radius as 6400 km.

[Ans. 7.8 × 10–4 rad/s]

40. Calcualte the effect of rotation of earth on the weight of a body at a place at latitude 45°.

[Ans. wt. of body of .001 kg mass at pole will be greater than at latitude 45° by 0.0168 N]

41. Calculate the change in the value of acceleration due to gravity as one moves from equatorto pole. The radius of earth at equator is 21 km more than at poles. [Ans. 0.02 ms–2]

42. Calculate the change in the acceleration due to gravity if earth stops to rotate.

[Ans. 0.03 ms–2]

43. If the earth were a perfect sphere of radius 6.37 × 106 m rotating about its axis with a periodof 1 day, how much would the acceleration due to gravity differ from poles to the equator?

[Ans. 3.37 × 10–2 ms–2]

44. Calculate the value of acceleration due to gravity at a place of latitude 45°. (Given Re = 6.38× 106 m) [Ans. 9.783 ms–2]

BASED ON ESCAPE VELOCITY45. A space ship is launched into a circular orbit close to the earth’s surface. What additional

velocity has now to be imported to the space ship in the orbit to overcome the gravitationalpull? (Re = 6400 km, g = 9.8 ms–2) [Ans. 3.28 kms–1]

Gravitation 319

46. Calcualte the escape velocity for a particle 1600 km above the earth’s surface, given that theradius of the earth is 6400 km and acceleration due to gravity on the surface of earth is 9.8ms–2. [Ans. 10.02 kms–1]

47. The escape velocity from earth’s surface is 11.2 km/s. A certain planet has a radius twice thatof earth but mean density is same as that of earth. Find the value of escape velocity from thesurface of that planet. [Ans. 22.2 kms–1]

48. The escape velocity of the projectile on the surface of earth is 11.2 kms–1. If a body isprojected out with twice of this speed. Find the speed of the body far away from the earth.Ignore the presence of other planets and sun. [Ans. 19.4 kms–1]

49. If the earth has a mass sixteen times and radius twice that of the planet Mars. Calcualte themaximum velocity required by a rocket to pull it out of the gravitational force of Mars.Given escape velocity on the surface of Mars. (Ve = 11.2 kms–1) [Ans. 3.95 kms–1]

50. A body of mass 100 kg falls on earth from infinity, what will be its velocity and energy onreaching the earth? (Re = 6400 km) [Ans. 11.2 kms–1, 6.27 × 109 J]

BASED ON ORBITAL VELOCITY51. A remote sensing satellite of the earth revolves in a circular orbit at a height of 250 km

above the earth’s surface. What is (a) orbital velocity (b) period of revolution of satellite. (Re= 6.38 × 106 m [Ans. 7756.6 ms–1, 5373.16 sec]

52. Calcualte the orbital velocity of a satellite orbiting at a height of 500 km above the surfaceof earth. [Ans. 7.6 kms–1]

53. Calculate the ratio of escape velocity to the orbital velocity for a satellite revolving near the

surface of earth. [Ans. 2 ]

54. Calculate the height of a geostationary satellite rotating along the direction of rotation ofearth. Also calculate its speed. [Ans. 35850 km, 3.1 kms–1]

55. A small satellite revolves round a planet in an orbit just above the planet surface. Meandensity of planet is 8.0 gm cm–3. Calculate the time period of the satellite. [Ans. 4205 sec]

BASED ON KEPLER’S LAW56. A geostationary satellite is orbiting the earth at a height 6R above the surface of earth, where

R is the radius of the earth. Find the time period of an other satellite at a height of 2.5 R from

the surface of earth. [Ans. 6 2 h]

57. If the earth contracts slightly so that its radius decrease by R

n, where R is original radius of

earth. Show that the length of the day shortens by 48

n

hr.

58. If the distance between the earth and sun were half its present value then what would havebeen the number of days in a year [Ans. 129.1]

59. The distance of planet Jupiter from the sun is 5.2 times that of earth. Find the period ofJupiter’s revolution around the Sun. [Ans. 11.86 years]

60. Two satellite of a planet have periods 32 days and 256 days. If the radius of the orbit of firstplanet is 6000 km. Calculate the radius of the orbit of second planet. [Ans. 24000 km]


BASED ON GRAVITATIONAL POTENTIAL AND POTENTIAL ENERGY61. A rocket is fired vertically upwards with a velocity of 10 kms–1 from the surface of earth.

Calculate the maximum height attained by the rocket. [Ans. 2.56 × 104 km]

62. A 400 kg satellite is revolving in a circular orbit of radius 2R about the earth. Calcualte theenegy required to move it to a circular orbit of radius 4R. (Re = 6.37 × 106 m)

[Ans. 3.13 × 109 J]

63. A satellite of mass 300 kg, orbits the earth at a height of 500 km from its surface. Calcualte(i) Kinetic energy (ii) potential energy (iii) total energy of the satellite (Re = 6400 km)

[Ans. 8.7 × 109 J, – 17.4 × 109 J, – 8.7 × 109 J]

64. Calculate the gravitational potential and potential energy of a body of mass 0.1 kg. (i) at theearth’s surface (ii) at a height above the surface of earth equal to its mean radius, (R = 6.37× 106 m) [Ans. – 6.26 × 107 J kg–1, – 6.26 × 106 J; – 3.13 × 107 J kg–1, – 3.13 × 106 J]

65. An artificial satillite is moving in a circular orbit around the earth with a speed equal to halfthe magnitude of escape velocity from the earth.

(i) Determine the height of the satellite above the earth’s surface.

(ii) If the satellite is stopped suddenly in its orbit and allowed to fall freely on to the earth,find the speed with which it hits the surface of earth. [Ans. 6400 km, 7.92 kms–1]

66. If a body is projected vertically upwards from the earth’s surface to reach a height equal toten times the radius of the earth, with what velocity the body should be projected. (Re = 6 ×106 m) [Ans. 10.34 kms–1]

67. A satellite of mass 200 kg is to be orbited round the earth at a height of 1000 km from thesurface of earth in a circular orbit. Calculate the work done. [Ans. 7.12 × 109 J]

68. Two bodies of masses 1 kg and 100 kg are placed 1 m apart. At which point on the linejoining them with the resultant gravitational field instensity be flow? [Ans. 1/11 m]

69. Two bodies of masses 100 kg and 1000 kg are at a distance 1.00 m apart. Calculate thegravitational potential at the middle point of the line joining them.

[Ans. – 1.47 × 10–7 J/kg]

70. The radius of the earth is R and acceleration due to gravity at its surface is g. Show that workrequired to raise a body of mass m to a height h from the surface of earth is given by

1 / R

mgh

h

.

321

1. ELASTICITY AND PLASTICITYUnder the action of deforming force a body gets deformed. On removing the deforming force

the body may or may not regain its original configuration. The property of a body by virtue ofwhich a body regains its original configuration on removing the deforming force is calledelasticity.

The property by virtue of which a body does not regain its original configuration on removingthe deforming force is called plasticity.

STRESSWhen deforming force is applied on a body some internal force is developed in the deformed

body called restoring force. The restoring force per unit area of cross section of the deformedbody is called stress i.e.

stress =Restoring force

area

=F

A

S.I. Unit Nm–2

Dimension [ML–1 T–2]Stress is of three types

(1) Tensile stress

(2) Compressive stress

(3) Tangential stress

STRAINWhen deforming force in applied on a body then the ratio of change in configuration to

original configuration is called strain.

Strain =Change in Configuration

Original ConfigurationIt has no unit and hence no dimension.

Strain is also of following three types

(a) Longitudinal strain

=Change in length

Original length

=L

L

.

�� 10UNIT


(b) Volume strain

=Change in volume

Original volume.

=V

V

.

(c) Shear Strain

= = Relative displacement between two parallal planes

Distance between parallal planes .

ELASTIC LIMITThat limit with in which when deforming force is applied the body regains its original

configuration on removing the deforming force is called elastic limit.

HOOKE’S LAWWith in elastic limit, stress is directly proportional to strain.

Stress strain

orStress

Strain= Constant (E)

E is called modulus of elasticity.

YOUNG’S MODULUSIt is defined as the ratio of tensile stress to longitudinal strain.

It is denoted by Y.

Y =Tensile stress F/A

Longitudinal StrainL

l

=F L

A l

.

BULK MODULUSIt is defined as the ratio of compressive stress to volume strain i.e.

B =Compressive stress

Volume strain

=F/A

VV

P VV

The reciprocal of bulk modulus is called compressibility i.e.

Properties of Solids 323

K =1

B.

MODULUS OF RIGIDITYIt is defined as the ratio of tangential stress to shear strain i.e.

=Tangential stress

Shear strain

=F/ A F/ A

X/ L

=F L

A X

.

All the three types of modulus of elasticities have same unit i.e. N/m2 and same dimensions[ML–1T–2].

POISSON’S RATIOWhen tensile stress is applied on a body the length of the wire increases.

As a result the diameter of the wire decreases.

The ratio of change in diameter to original diameter is called lateral strain.With in elastic limit the ratio of lateral strain to longitudinal strain is called poisson’s ratio

i.e.

=Lateral strain

Longitudinal strain

=D / D

L/ L

It is a pure number. It has no unit or dimension.

ELASTIC POTENTIAL ENERGY OF A STRETCHED WIREWhen a wire is stretched, work is done against the internal restoring forces which is stored

in the form of elastic potential energy. It is given by

v =1

2 × Force × Change in length

=1

F L2

=1 F L

A L2 A L

=1

2 × Stress × Strain × Volume

P.E. stored per unit volume of wire is

u =12

stress × strain.


SOLVED EXAMPLES

BASED ON YOUNG’S MODULUSExample 1. A load of 2 kg produces an extension of 2 mm in a wire of length 3 m and 1 mm

in diameter. Calculate young’s modulus of elasticity of wire.


l = 2 mm = 2 × 10–3 m

L = 3 m

r =1

mm2

= 0.5 mm = 0.5 × 10–3 m

Now Y = 2

mg L

r l

= 3 2 3

2 9.8 3

(3.14) (0.5 10 ) 2 10

= 3.74 × 1010 Nm–2

Example 2. Find the maximum length of a steel wire that can hang with out breaking.

Breaking stress = 7.9 × 1012 dyne/cm2

density of steel = 7.9 g/cm3.

Solution. Breaking stress = 7.9 × 1012 dyne/cm2

density = 7.9 gram/cm3

We know Breaking Stress =F

A

=(A L)

A

g = Lg.

L =Breaking Stress

density g

=127.9 10

7.9 980

= 1.02 × 109 cm.

Example 3. Calculate the force required to stretch a steel wire 2 cm2 in cross section todouble its length. Y = 2 × 1011 Nm–2.

Solution. Here l = L , L = LA = 2 cm2 = 2 × 10–4 m2.

Y = 2 × 1011 Nm–2.

Y =F L

A L

F =YA L

L

=11 42 10 2 10 L

L

= 4 × 107 N.


Example 4. A steel wire of length 4.7 m and cross section 3.0 × 10–5 m2 stretches by the sameamount as a copper wire of length 3.5 m and cross section 4.0 × 10–5 m2, under a given load. Whatis the ratio of Young’s modulus of steel to that of copper.

Solution. Here for steel l1 = 4.7 m, a1 = 3.0 × 10–5 m2

l2 = 3.5 m, a2 = 4.0 × 10–5 m2

For the same value of load F and same change in length

Y1 = 1

1

F l

a l

Y2 = 2

2

F l

a l

1

2

Y

Y=

51 2

51 2

4.7 4 10

3.0 10 3.5

l a

a l

= 1.79.Example 5. Two long metallic strips are joined together by two rivets each of radius 2.0 mm.

Each rivet can stand a maximum shearing stress of 1.5 × 109 Pa. What is the maximum tensileforce that the strip can exert assuming that each rivet shares the stretching load equally?

Solution. Let the tensile force is F, which provides shearing force on the two rivets. Shearingforce in shared equally.

so, shearing force on each rivet =F

2

shearing stress on each rivet =F/ 2 F

A 2 A

Maximum shearing stress on each rivet is 1.5 × 109 Pa

maxF

2A= 1.5 × 109

or Fmax = 2A × 1.5 × 109

=22

27

(2 × 10–3)2 × 1.5 × 109

= 3.8 × 104 N.Example 6. A cable in replaced by another one of the same length and material but of twice

the diameter. How will this affect the elongation under a given load? How does this affect themaximum load it can support without exceeding the elastic limit?

Solution. We know

Y = 2 2

MgL 4 MgL

L Lr D

L = 2

4 L

D Ymg

or L 2

1

D.

It shows that if diameter is doubled then elongation will remain one fourthAnd mg D2.

If the diameter is doubled, the wire can support four times the original load.


Example 7. The length of a wire increases by 8 mm, when a weight of 5 kg is hung. If theconditions are the same but the radius of the wire is doubled what will be the increase in thelength?

Solution. Here l1 = 8 mm = 8 × 10–3 m

r1 = r, r2 = 2r

Y = 2

Lmg

r L

L 2

1

r

1

2

L

L

=

222

1

r

r

3

2

8 10

L

=2

2

(2 )

( )

r

r

L2 = 2 × 10–3 m.Example 8. A rubber string 10 m long is suspended from a rigid support at its one end.

Calculate the extension in the string due to its own weight. The density of rubber is 1.5 × 103

kg m–3 and Young’s modulus for the rubber is 5 × 106 Nm–2. (g = 10 ms–2).

Solution. Weight of the string W = mg

W = (Volume × density) g

= 10 × A × 1.5 × 103 × 10 (A is area of cross section of the string).

W

A= Longitudinal stress = 1.5 × 105 Nm–2 ... (i)

The weight of the string acts on its centre of gravity so it produces extension only in 5 mlength.

Let L is the extension produced then

Longitudinal strain =L

L 5

L ... (ii)

Young’s modulus =Stress

Strain

or 5 × 106 =51.5 10

L/ 5

[Using equation (i) and (ii)]

or L = 0.15 m.Example 9. Compute the elongation of the steel and brass wires

shown in figure. Unloaded length of steel wire = 1.5 m, unloaded length ofbrass wire = 1.0 m diameter of each wire = 0.25 cm. Young’s modulus ofsteel is 2.0 × 1011 Pa and that of brass is 0.91 × 1011 Pa.

Solution. For steel wire

Ls = 1.5 m, rs = 0.25

cm2

= 0.125 × 10–2 m

Fs = (6 + 4) × 9.8 N = 98 N

1.5 m stee l

4 .0 kg

1 .0 m Brass

6 .0 kg


Ys = 2.0 × 1011 Pa.

Ys = 2

F L

L

s s

s sr

Ls = 2

F L

Y

s s

ssr

= 2 2 11

98 1.5

3.14 (0.125 10 ) 2.0 10

= 1.5 × 10–4 m

Similarly for brass:Lb = 1.0 m Fb = 6 × 9.8 N

Lb = 2 2 11

6 9.8 1.0

3.14 (0.125 10 ) 0.91 10

= 1.3 × 10–4 mExample 10. A load of 4 kg produces an extension of 2 mm in a wire of 3 m in length.

Calculate Young’s modulus of elasticity of wire. Given radius of wire is 1 mm.

Solution. Here F = mg = 4 × 9.8 N

L = 2 mm = 2 × 10–3 m

L = 3 m

r = 1 mm = 10–3 m

Y = 2

mg L

L r

= 3 2 3

4 9.8 3

3.14 (10 ) 2 10

= 7.5 × 1010 Nm–2.Example 11. A wire increases by 10–4 of its length when it is subjected to a stress of 107

Nm–2. Calculate Young’s modulus of the material of wire.Solution. Here stress = 107 Nm–2

strain =L

L

= 10–4

Young’s modulus

Y =7

4

Stress 10

Strain 10

= 1011 Nm–2.Example 12. Calculate the percentage increase in the length of a wire of radius 1.25 mm

stretched by a force of 200 kg f. Young’s modulus of elasticity of the wire is 1.25 × 1011 Nm–2.

Solution. Here r = 1.25 × 10–3 m

F = 200 × 9.8 NY = 1.25 × 1011 Nm–2

We know

Y = 2

Mg L

L r


L

L

= 2 3 2 11

Mg 200 9.8

Y 3.14 (1.25 10 ) 1.25 10

r

= .3194 × 10–2

Percentage increase in length = 0.3194 × 10–2 × 100

= 0.32%.Example 13. The breaking stress of a material is 108 Nm–2. Find the greatest length of a wire

that could hang vertically without breaking. (density of the material = 3000 kg m–3).

Solution. � Here, breaking stress = 108 Nm–2

let the maximum length of the wire is L

Stress =Force

area

=AL g

A

= Lg.

L =Breaking Stress

g

=810

3000 9.8

= 3.4 × 103 m.Example 14. Four identical hollow cylindrical columns of steel support a big structure of

mass 50,000 kg. The inner and outer radius of each column are 30 cm and 40 cm respectively.Assuming the load distribution to be uniform calculate the compressive strain of each column.Young’s modulus of steel is 2.0 × 1011 Pa.

Solution. Here M = 50,000 kg

ri = 0.30 m, r0 = 0.40 m

Y = 2.0 × 1011 Nm–2

Area of cross section of 4 columns

= 4 2 20 ir r

= 4 (0.42 – 0.32)

= 4 × 0.07 m2

From formula Y = 2

Mg L

L r

L

L

= 2

Mg

Y r

= 11

50000 9.8

2.0 10 3.14(0.07) 4

= 2.78 × 10–6.


BASED ON BULK MODULUSExample 15. Compute the bulkmodulus of water from the following data: Initial volume =

100.0 lt, pressure increase = 100.0 atm final volume = 100.5 lt.

Solution. Here vi = 100 lt

vf = 100.5 lt P = 100 atm

= 100 × 1.01 × 105 Nm–2.

B =V P

V

=5100 100 1.01 10

(100.5 100)

= 2.026 × 109 Nm–2.Example 16. The compressibility of water is 10 × 10–10 m2/N. Find the decrease in volume

of 100 cm3 of water when subjected to a pressure of 15 × 106 Pa.

Solution. Here K = 10 × 10–10 m2/N V = 100 cm3

= 100 × 10–6 m3

P = 15 × 106 Pa.

We know B =P V

V

or1

K=

P V

V

V = K V P.

= 10 × 10–10 × 100 × 10–6 × 15 × 106

= 1.50 10–6 m3

Example 17. The bulk modulus of water is 2.3 × 109 Nm–2.

(i) Find the compressibility per atmosphere of pressure

(ii) How much pressure in atmosphere is needed to compress a sample of water by 0.1%?

Solution. Here B = 2.3 × 109 Nm–2

=

9

5

2.3 10

1.01 10

� 2.27 × 104 atm.

(i) Compressibility = 4

1 1

B 2.27 10

= 4.4 × 10–5 atm–1

(ii) Now B =V P

V

P = B

v

v


Butvv

= .1% = 0.1

100

P =0.1

100

(2.27 × 104)

= 22.7 atm.Example 18. A sphere contracts in volume by 0.04% when taken to the bottom of sea, one

km deep. Find bulk modulus of the material of sphere. Density of sea water is 1000 kg m–3.

Solution. Herevv

= 0.04%

P = 1000 × 103 × 9.8 Nm–2.

From formula B =V P

V

=31000 10 9.8

.04 /100

= 2.45 × 1010 Nm–2.Example 19. What is the density of ocean water at a depth where the pressure is 500 atm.

Given that density of water at the surface is 1.025 × 103 kg m–3?

B = 2.0 × 109 Nm–2? Take 1 atm = 105 Nm–2.

Solution. We know

B =P

v

v

or v =P

B

v

P = (500 – 1) = 499 atm

= 499 × 1.01 × 105 Nm–2.

B = 2.0 × 109 Nm–2.

v = 1 m3 m = 1.025 × 103 = 1025 kg.

v =5

9

1 499 1.01 10

2 10

= 0.025 m3

density of water at the given depth

=1025

1 0.025

=1025

9.975 = 1051 kg m–3.

Example 20. A sphere contracts in volume by 0.02% when taken to the bottom of sea oneKm deep. Calculate the bulk modulus of the material of sphere. Density of sea water is 1000 kgm–3.


Solution. Herevv

=0.02

100

P = 1000 m of water column

= 1000 × 103 × 9.8 Nm–2

B =P

v

v

or B =3P 1000 10 9.8

/ 0.02 /100

v v

= 4.9 × 1010 Nm–2.Example 21. A solid sphere of radius 10 cm is subjected to a uniform pressure = 5 × 108 Nm–2.

Determine the consequent change in volume.

Bulk modulus of the material of sphere is 3.14 × 1011 Nm–2.

Solution. Here r = 10 cm. = 0.1 m.

v =3 34 4 22

(0.1)3 3 7

r

P = 5 × 108 Nm–2

B = 3.14 × 1011 Nm–2.

B =P

V

v

v =3 8

11

P 4 3.14 (0.1) 5 10

B 3 3.14 10

v

= 6.67 × 10–6 m3.Example 22. Compare the densities of water at the surface and bottom of a lake 100 m deep,

given that the compressibility is 10–3/22 per atmosphere

(1 atm = 1.015 × 105 Pa)

Solution. B = 3

1 1

K 10 / 22 = 22 × 103 × 1.015 × 105 N/m2

= 22 × 1.015 × 108 Nm–2

Let V is the volume of 1 kg of water at the surface and v – v is its volume at the bottom ofthe lake, 100 m deep.

P = h dg = 100 × 103 × 9.8 Nm–2.

vv

=3

3 5

P 100 10 9.8

B 22 10 1.015 10

Density of Water at surface

Density of Water at the bottom=

1/

1/

v v v

v v v

= 1 vv


=3

3 5

100 10 9.81

22 10 1.015 10

= 0.99956.

BASED ON MODULUS OF RIGIDITYExample 23. A cube of aluminium of each side 4 cm is subjected to a tangential (Shearing)

force. The top face of the cube is sheared through 0.012 cm with respect to the bottom face. Find(i) Shearing Strain (ii) Shearing stress (iii) Shearing force. ( = 2.08 × 1011 dyne/cm2).

Solution. Here l = 4 cm

l = 0.012 cm

= 2.08 × 1011 dyne cm–2.

(i) Shearing Strain =0.012

4

x

l

= 0.003 rad.

(ii) Area of top face = l2 = (4 cm)2 = 16 cm2.

Modulus of rigidity =Shearing stress

Shearing strain

shearing stress = × shearing strain

= 2.08 × 1011 × 0.003

= 6.24 × 108 dyne cm–2.

Shearing force = Shearing Stress × area

= 6.24 × 108 × 16

= 9.984 × 109 dyne.Example 24. A square lead slab of side 50 cm and thickness 5.0 cm is subjected to a shearing

force (On its narrow face) of magnitude 9.0 × 104 N. The lower edge is riveted to the floor. Howmuch is the upper edge displaced, if the shear modulus of lead is 5.6 × 109 Pa?

Solution. Here l = 50 cm = 0.50 m

F = 9 × 104 N

= 5.6 × 109 Pa

Area of upper face on which shear force is applied

= (50 × 10–2) × (5 × 10–2)

= 250 × 10–4 m2

Now =F / A F L

L / L A x

or x =F L

A

=4

4 9

9 10 0.5

250 10 5.6 10

= 3.214 × 10–4 m.

Example 25. A metallic cube whose each side is 10 cm is subjected to a shearing force of1960N. The top face is displaced through 0.25 cm with respect to the bottom? Calculate theshearing stress, strain and shear modulus.


Solution. Here L = 10 cm = 0.1 m

F = 1960 N

x = 0.25 cm.

= 0.25 × 10–2 m.

Shearing stress =F 1960

A 0.1 0.1

= 1.96 × 105 Nm–2

Shearing Strain =L

x

=20.25 10

0.1

= 2.5 × 10–2.

Shearing modulus =Shearing Stress

Shear Strain

=5

2

1.96 10

2.5 10

= 0.784 × 107

= 7.84 × 106 Nm–2.Example 26. A metal cube of side 10 cm is subjected to a shearing stress of 106 Nm–2.

Calculate the modulus of rigidity, if the top of cube is displaced by 0.05 cm w.r.t. the bottom?Solution. Here Shearing Stress = 106 Nm–2

x = 0.05 cm

= 0.05 × 10–2 m

L = 10 cm = 0.1 m

From formula =F / A F L

/ L Ax x

=6

2

10 0.1

0.05 10

= 2 × 109 Nm–2.

Example 27. A cubical body 2.5 cm at a edge placed on the table is subjected to a shearingforce of 1.0 kg. The upper face of the cube is displaced by 0.5 cm. Calculate the shear modulus ofthe material of cube.

Solution. Here L = 2.5 cm = 2.5 × 10–2 m

A = (2.5 × 10–2)2 = 6.25 × 10–4 m2

F = 1.0 × 9.8 N

x = 0.5 × 10–2 m.

= F L

A X

=–2

4 2

1.0 9.8 2.5 10

(6.25 10 ) (0.5 10 )

= 7.84 × 104 Nm–2.Example 28. A rubber block 3.0 cm is clamped at one end with 10 cm side vertical. A

horizontal force of 30 N is applied on its free end. Calculate horizontal displacement of free endif modulus of rigidity is 2.0 × 105 Nm–2.


Solution.

=F/ A F L

X/ L A X

or X =F L

A

=2

4 5

30 3 10

(30 10 ) 2 10

= 1.5 × 10–3 m.

BASED ON POISSON’S RATIO AND ELASTIC POTENTIAL ENERGYExample 29. A material has Poisson’s ratio 0.5. If a uniform rod of it suffers a longitudinal

strain of 2 × 10–3, what is the percentage increase in volume?

Solution. Here = 0.5

L

L

= 2 × 10–3.

From formula =R / R

L/ L

orR

R

= –

L

L

R

R

= – 0.5 × 2 × 10–3 = – 1 × 10–3.

Volume of rod V = R2L.

% increase in volume is

100 vv

=R L

2 100R L

= [2 (– 1 × 10–3) + 2 × 10–3] × 100

= 0%.Example 30. A metallic wire is stretched by suspending weight from it. If is the

longitudinal strain and Y is the Young’s modulus. Show that elastic potential energy per unitvolume is given by 1/2 Y2.

Solution. Elastic potential energy u = 1

2 × Stress × Strain


We know Y =Stress

Strain

Stress = Y strain = Y

u =1

Y2

= .21Y

2

Example 31. A wire of length 2.0 m is stretched through 2.0 mm. The area of cross sectionof the wire is 2.0 mm2. Calculate (a) energy density of the wire and (b) elastic potential energystored in the wire. Given Y = 2 × 1011 Nm–2.

Solution. Here L = 2.0 m l = 2.0mm = 2.0 × 10–3 m

A = 2.0 × 10–6 m2

(a) Energy per unit volume = 21Y Strain

2

=

23111 2 10

2 102 2

= 105 J/m3.

(b) Elastic potential energy u = energy density × volume

= 105 × 2 × 10–6 × 2.0

= 0.4 J.Example 32. Calculate the increase in energy of a brass bar of length 0.4 m and cross

sectional area 2 cm2, when compressed with a load of 5 kg weight along its length(Y = 1.0 × 1011 Nm–2).

Solution. Increase in energy of brass bar

u =1

F L2

We know Y =F L

A l

l =F L

A Y

u =1 F L

F2 A Y

=2F L

2 AY

=

2

11 4

(5 9.8) 0.4

2 (1.0 10 ) (2 10 )

= 2.4 × 10–5 J.Example 33. The area of cross section of a wire is 1 mm2 and its length is 2m. How much

work will be done to increase its length by 0.1 mm. Young’s modulus of elasticity for the materialof the wire is 2 × 1011 Nm–2.


Solution. The work done to stretch a wire is given by

W =1

2 × Stress × Strain × Volume

=1

2 × Y × Strain2 × Volume

=1

2 × Y ×

2

A LL

l

=21 L

Y A2 L

=3 2

11 61 (0.1 10 )2 10 1 10

2 2

= 5 × 10–4 J.

EXERCISE

BASED ON YOUNG’S MODULUS1. A weight of 2 kg is suspended from the lower end of a metallic wire of cross section 5 mm2.

Calculate the stress produced. [Ans. 3.92 × 106 Nm–2]

2. The breaking stress of aluminium wire is 7.5 × 107 Nm–2. Find the greatest length ofaluminium wire, that can hang without breaking. = 2.7 × 103 kg m–3.

[Ans. 2.83 × 103 m.]3. The length of a metallic wire is 1 m. On applying a stress of 2.4 × 107 Nm–2, the length of

wire increases by 2 × 10–4 m. Calculate Young’s modulus of the material of wire.

[Ans. 1.2 × 1011 Nm–2]

4. A load of 4 kg produces an extension of 4 mm in a wire 3 mm length, and 1 mm in diameter.Calculate Young’s modulus of the material of wire. [Ans. 3.74 × 1010 Nm–2]

5. Calculate the percentage change in the length of a wire of diameter 2.5 mm stretched by aforce of 100 kg wt. Young’s modulus of elasticity of wire is 12.5 × 1011 dyne/cm2.

[Ans. 0.1597%]

6. A copper wire has a length of 2.2 m. Its diameter is 7 × 10–4 m. What will be the elongationin its length when 5 kg weight is suspended? (Y = 1.25 × 1011 Nm–2 and g = 10 ms–2.]

[Ans. 2.3 mm]

7. Calculate the weight to be suspended from a steel wire of length 2.0 m and radius 0.5 mm sothat length of the wire increases by 0.5 mm. Y = 2 × 1011 Nm–2, g = 9.8 ms–2

[Ans. 4.0 kg]

8. The length of a wire increases by 1% on loading a 2 kg weight on it. Calculate the linearstrain in the wire. [Ans. 0.01]

9. A wire elongates by 18 mm when a load of 10 kg is suspended from its lower end. Calculatethe elongation if radius of the wire is doubled, keeping all other factors same.

[Ans. 4.50 mm]

10. A uniform wire 6 m long weighing 40 g elongates by 0.8 mm when stretched by a load of 1kg. Given density of the material of wire is 8.9 g cm–3. Determine Young’s modulus of thematerial of wire. [Ans. 9.8 × 1010 Nm–2]


11. An aluminium wire of radius 1 mm and length 2.5 m stretches by 1 mm, when mass M issuspended from it. Calculate the value of M. Given Young’s modulus for aluminium is 7 ×1010 Nm–2 [Ans. 9 kg]

12. Calculate the percentage increase in the length of a wire of diameter 2.5 mm when it isstretched by a force of 50 kg wt. Young’s modulus of the wire is 12.5 × 1011 dyne cm–2.

[Ans. 0.08%]

13. A steel wire 4 mm in diameter is stretched between two clamps when its temperature is 40°C.Calculate the tension in the wire when its temperature falls to 30°C. Given coefficient oflinear expansion of steel = 11 × 10–6/°C and Y for steel is 21 × 1011 dyne/cm2.

[Ans. 2.9 × 102 N]

14. Two wires A and B are of the same material. Their length are in the ratio 1 : 2 and thediameters in the ratio 2 : 1. If they are pulled by the same force what will be the ratio of theirincrease in length? [Ans. 1 : 8]

15. The breaking stress of a material is 107 Nm–2. Calculate the greatest length of a wire thatcould hang vertical without breaking. ( = 3000 kg m–3.) [Ans. 340 m.]

16. A composite wire of uniform diameter 3.0 mm consisting of a copper wire of length 2.2 mand a steel wire of length 1.6 m stretches under a load by 0.7 mm. Calculate the load, giventhat the Young’s modulus for copper is 1.1 × 1011 Pa and for steel is 2.0 × 1011 Pa.

[Ans. 176.8 N]

17. For steel the breaking stress is 8.0 × 106 Nm–2 and the density is 8.0 × 103 kg m–3. Find themaximum length of steel wire which can be suspended without breaking under its ownweight. (g = 10 ms–2) [Ans. 100 m]

18. Calculate the percentage increase in the length of the wire by a stress of 1 kg wt mm–2

Young’s modulus of the wire is 1.0 × 1011 Nm–2. [Ans. 0.0098%]

Based On Bulk Modulus19. Calculate the pressure required to stop the increase in volume of a copper block when it is

heated from 50°C to 70°C. Coefficient of linear expansion of copper = 8.0 × 10–6 per °C andbulk modulus of elasticity = 1.3 × 1011 Nm–2. [Ans. 6.24 × 107 Nm–2]

20. What is the density of ocean water at a depth where the pressure is 80.0 atm. given that itsdensity at the surface is 1.03 × 103 kg m–3. Compressibility of water = 45.8 × 10–11 Nm–2.

[Ans. 1.034 × 103 kg m–3]

21. Compute the bulk modulus of water from the following data. Initial volume = 200.0 litre.Pressure increase = 100.0 atm. Final volume is 201.0 litre. [Ans. 2.02 × 109 Nm–2]

22. A sphere contracts by .02% when taken to the bottom of sea one km deep. Find the bulkmodulus of the material of sphere. = 1000 Kg m–3. [Ans. 4.9 × 1010 Nm–2]

23. A fluid of volume 1.5 × 10–3 m3 when subjected to a pressure change of 5.0 × 106 Nm–2 hasits volume reduced by 3.0 × 10–7 m3. Find the coefficient of bulk modulus of the fluid.

[Ans. 2.5 × 1010 Nm–2]

24. Calculate the change in volume which 2 m3 of water will undergo when taken from thesurface to the bottom of lake 100 m deep. Given Bulk modulus of water is 22.000 atm.

[Ans. 4.4 × 10–4 m3]25. What will be the decrease in the volume of 200 cm3 of water under pressure of 100

atmosphere. Given compressibility of water is 4 × 10–5 per unit atmosphere.

[Ans. 0.8 cm3]


26. 1 m3 of water is subjected to a pressure change of 2.1 × 104 Nm–2. Calculate decrease in itsvolume if bulk modulus of water is 0.21 × 1010 Nm–2. [Ans. 10 cm3]

27. A spherical ball (Bulk modulus of the material of ball is 1.03 × 1011 Nm–2) is subjected to apressure of 100 atm. Calculate fractional change in its volume. [Ans. 0.098%]

28. What will be the density of lead under a pressure of 20,000 N cm–2? (density of lead is 11.4g cm–3, and bulk modulus of lead is 0.80 × 10 Nm–2). [Ans. 11.7 g cm–3]

29. A solid sphere of radius 10 cm is subjected to a uniform pressure = 5 × 108 Nm–2. Determinethe consequent change in volume. Bulk modulus of the material of the sphere is equal to 3.14× 1011 Nm–2. [Ans 6.67 × 10–6 m3]

30. A solid copper cube 40 mm on each edge when subjected to a pressure of 2 × 107 Nm–2, thevolume of it decreases by 10.24 mm3. Calculate the bulk modulus of copper.

[Ans. 1.25 × 1011 Nm–2]

BASED ON MODULUS OF RIGIDITY31. A 5.0 cm. cube of gelatin placed on a table is subjected to a shearing force of 0.5 kg. The

upper surface of the cube is displaced by 1.0 cm. Calculate the shear modulus of gelatin.

[Ans. 3.92 × 104 Nm–2]

32. A square lead slab of side 50 cm and thickness 5.0 cm is subjected to a shearing force, sothat upper edge is displaced by 3.2 × 10–4 m. Calculate the shearing force. (Modulus ofrigidity for lead is 5.6 × 109 Nm–2). [Ans. 9 × 104 N]

33. A metallic cube whose each side is 10 cm is subjected to a shearing force of 1.96 × 103 N.The top face is displaced by 0.5 cm, with respect to the bottom? Calculate shear modulus.

[Ans. 3.92 × 106 Nm–2]

34. The upper face of a cube of edge 1 m moves through a distance of 1 mm relative to the lowerfixed face under the action of a tangential force of 1.5 × 108 N. Calculate the tangentialstress, shear strain, and modulus of rigidity.

[Ans. 1.5 × 108 Nm–2, 0.001 rad, 1.5 × 1011 Nm–2]

BASED ON POISSON’S RATIO AND ELASTIC POTENTIAL ENERGY35. A steel wire of 4.0 m is stretched through 2.0 mm. The cross sectional area of the wire is 2.0

mm2. If Young’s modulus of steel is 2.0 × 1011 Nm–2. Find

(a) Energy density of the wire

(b) The elastic potential energy stored in wire.

[Ans. 25 × 104 J/m3, 0.2 J]

36. Calculate the increase in energy of a brass bar of length 0.4 m and cross sectional area 2 cm2

when compressed with a load of 10 kg wt along its length.

Young’s modulus of brass is 1.0 × 1011 Nm–2. g = 9.8 ms–2 [Ans. 4.8 × 10–5 J]37. When the load on a wire is increased slowly from 2 kg to 4 kg the elongation increases from

0.6 mm to 1.0 mm. If g = 10 ms–2, then find the work done during the extension of the wire.[Ans. 14 × 10–3 J]

38. A metal rod, has a length L and area of cross section A. Show that work done in stretching

the rod by an amount L is 2YA ( L)

.2L

339

PRESSUREThe thrust exerted by a fluid per unit area of the surface in contact with it is called pressure.

Pressure =Thrust

Area

i.e. P =F

A.

S.I. Unit of pressure is Nm–2 (pascal Pa)

Dimensional formula [ML–1 T–2].

PASCAL’S LAWAccording to this law the pressure exerted at any point, on the liquid enclosed in a container,

is transmitted equally in all the directions.

HYDRAULIC LIFTIt is used to lift heavy loads and it is based on Pascal’s Law.

Suppose force f is applied on the piston of smaller area of cross section a.

Pressure exerted on liquid

P =f

a.

This pressure is transmitted to larger piston.

F = P × A

= Af

a

A > a So F > f.

Thus a heavy load can be lifted by applying smaller force.

�� 11UNIT


PRESSURE OF LIQUID COLUMNThe pressure exerted by a liquid column of height h and density d is given by

P = h d g

ATMOSPHERIC PRESSUREOur earth is surrounded by an envelope of air and gases upto a certain height. This envelope

of air or gases is called atmosphere. The pressure exerted by the atmosphere is called atmosphericpressure.

Numerically atmospheric pressure is equal to the pressure exerted by 76 cm of mercurycolumn.

1 atmospheric pressure = pressure exerted by 76 cm of Hg column.

= h d g

= 0.76 × 13.6 × 103 × 9.8

= 1.013 × 105 Nm–2.Other units of atmospheric pressure are

1 bar = 106 dyne cm–2 = 105 Nm–2.

1 torr = 1 mm of Hg

1 atm = 101.3 kPa = 1.013 bar = 760 torr.

IDEAL LIQUIDAn incompressible and non viscous liquid is called ideal liquid. In actual practice there is no

liquid which is ideal liquid.

STREAMLINED FLOWWhen liquid flows in such a way that each liquid particle, passing a certain point follows the

same path as was of preceding particle, which has passed through the same point, then the flow issaid to be stremlined and the path followed is called a streamline.

TURBULENT FLOWThe flow of liquid is streamlined flow only when the velocity of flow of liquid is below a

certain value called Critical value. When the velocity of flow exceeds the critical velocity, theflow is no longer streamlined and it becomes turbulent.

REYNOLD’S NUMBERIt is a combination of four physical quantities, given by

NR =VD .

where is density of liquid,

V is velocity of flow, D is diameter of the pipe and is viscosity of liquid.

If NR 2000 the flow of liquid is laminar.

NR 3000 the flow of liquid is turbulent.

2000 < NR < 3000 the flow of liquid is unsteady.

VISCOSITYThe property of liquid by virtue of which it opposes the relative motion between its different

layers is called viscosity.

Mechanics of Liquid 341

According to scientist Newton the viscous force between two adjacent layers of liquid isgiven by

F =A

x

v.

where is called coefficient of viscosity.

S.I. unit N sm–2. (decapoise).

C.G.S. unit dyne - sec cm–2 (Poise).

1 decapoise = 10 poise.

STOKES’ LAWSc. Stoke showed experimentally that the viscous force acting on a body moving slowing in

a liquid is given byF = 6 rvt

where = Viscosity of liquid

r = radius of the spherical body falling freely in liquid

vt = terminal velocity.

TERMINAL VELOCITYWhen a solid body falls through a liquid, it experiences two forces, force due to gravity in

downward direction and viscous force of liquid in upward direction. After falling through a certaindistance the two forces becomes equal and now it falls with a constant velocity called terminalvelocity. It is given by

vt =22 ( )

9

r g

where r = radius of spherical body = density of body = density of liquid = viscosity of liquid.

If >> then

vt =22

.9

r g

EQUATION OF CONTINUITYWhen an incompressible and nonviscous liquid flows in streamlined motion through a pipe

of nonuniform area of cross section, then the product of area of cross section and velocity of flowremains constant at every point. i.e., Av = constant

ENERGY OF A FLOWING LIQUIDThere are following three types of energies in a flowing liquid.

(i) Pressure EnergyPressure energy per unit volume of liquid = P.

(ii) Kinetic Energy

Kinetic energy per unit volume of liquid = 21

2v


(iii) Potential EnergyPotential energy per unit volume of liquid = gh.

Imp:Kinetic energy per unit weight of liquid is called velocity head.

Velocity head =2

.2g

v

Similarly pressure head = g

p

.

gravitational head = h.

BERNOULLI’S THEOREMAccording to Bernoulli’s theorem, when an incompressible and non viscous fluid (liquid or

gas) flows in streamlined motion through a pipe, then at every point of its path the total energy perunit volume remains constant. That is

21

2P g h v = Constant.

In case of horizontal pipe

21

2P v = Constant.

TORRICELLI’S THEOREMAccording to this theorem the velocity of efflux of a liquid from an orific is equal to the

velocity which the liquid acquires in falling freely from the free surface of the liquid to the orific

Velocity of efflux v = 2 .gh

VENTURIMETERIt is a device based on Bernoulli’s theorem which is

used to measure the rate of flow of a liquid in a pipe.

Volume of liquid per second is given by

Q = 1 2 2 21 2

2A A .

A A

gh

A1 and A2 are area of cross sections at X and Y respectively. h is the difference in heights of

liquid columns in two barometers.

POISEUILLE’S FORMULAThe volume of liquid flowing per second through a horizontal capillary tube of length l is

given by

Q =4

.8

P r

l

Where P = pressure difference between two ends of the tuber = radius of the tube and

= viscosity of liquid.

h

v1v2

A 2A 1

X

Y


SURFACE TENSIONThe property of liquid by virtue of which its free surface behaves like a stretched membrane

is called surface tension.

Numerically the surface tension of liquid is defined as the force per unit length in the planeof liquid surface, acting at right angles on either side of an imaginary line drawn in free surface ofliquid i.e.

T =F

.L

S.I. Unit Nm–1

Dimension [MT–2].

SURFACE TENSION DECREASES WITH RISE IN TEMPERATURE

SURFACE ENERGYThe potential energy per unit area of the surface film is called surface energy.

Let W is the work done to increase the surface area of the film by A then

T =W

.A

Where T = Surface tension of the liquid

From here another unit of surface tension comes out to be joule/m2.

ANGLE OF CONTACTWhen liquid is kept in a container, the free surface of liquid is always curved, may be convex

or concave. The angle formed inside the liquid, between the side of container and the tangent tothe liquid surface at point of contact is called angle of contact, for given solid liquid surface.

W ater in g lass con ta iner

(ang le of con tact, acu te)

M ercury in g lass con ta iner

(ang le of con tact ob tuse)

PRESSURE DIFFERENCE ACROSS THE CURVED SURFACE OF LIQUIDAs the free surface of liquid inside a container is curved, the pressure on cancave side always

exceeds the pressure on convex side.

Excess pressure inside a liquid drop = 2T

.R

Excess pressure inside a liquid bubble = 4 T

.R

Excess pressure inside an air bubble = 2T

.R

Total pressure inside a bubble formed at a depth d below the free surface of liquid

PTotal = Patm + Pliquid + Pexcess


= Patm + hdg + 2T

.R

Total pressure inside a liquid bubble = atm4 T

PR

CAPILLARITYWhen a Capillary tube open at the both ends is dipped vertically in water, the water rises up

in Capillary but if the capillary tube is dipped in mercury, then the mercury is depressed below theoutside level. The phenomenon of rise or fall of liquid in a capillary tube is called Capillarity.

The height to which the liquid rises or falls in a capillary is given by

h =2T cos

.rdg

Where T = Surface tension of liquid

= angle of contact

r = radius of Capillary

d = density of liquid.

Imp: If the Capillary is of insufficient height the liquid will not overflow.

The product of height and radius of curvature remains constant ie h R = Constant.

If the Capillary is of insufficient height, the radius of curvature will increase. In case the Capillary is not vertical in the liquid say it is inclinedat an angle to the vertical there length of liquid column incapillary is given by

l = h cos .

SOLVED EXAMPLES

BASED ON THRUST AND PRESSUREExample 1. A force of 100 N is applied on a nail whose tip has area of cross section 4 mm2.

Calculate the pressure on the tip.

Solution. Here F = 100 N

A = 4 mm2 = 4 × 10–6 m2.

P = 6

F 100

A 4 10

= 25 × 106 Nm–2.

Example 2. A vertical off-shore structure is built to withstand a maximum stress of 109 Pa.Is the structure suitable for putting up on the top of an oil well in Bombay high? Take the depth ofthe sea to be roughly 3 km and density of sea water 103 kg m–3.

Solution. Here h = 3 km = 3000 m.

Maximum stress = 109 Nm–2.

Pressure due to sea water = hdg

= 3000 × 103 × 9.8

= 2.94 × 107 Nm–2.

hl


This pressure is less than the maximum stress of 109 Nm–2 so structure is suitable putting onthe top of an oil well.

Example 3. How much pressure will a girl of weight 50 kg f exert on the ground if area ofher one foot is 50 cm2. (Take g = 10 ms–2).

Solution. Here F = 50 kg f = 50 × 10 N

A = 2 × 50 × 10–4 = 10–2 m2.

P = 2

F 500

A 10 = 5 × 104 Nm–2.

BASED ON PASCAL’S LAWExample 3. A hydraulic automobile lift is designed to lift cars with a maximum mass of 3000

kg. The area of cross section of the piston carrying the load is 425 cm2. What maximum pressurewould the smaller piston have to bear?

Solution. Here, Area of cross section of larger piston

= 425 cm2 = 425 × 10–4 m2.

Force F = 3000 × 9.8 N.

Pressure P = 4

F 3000 9.8

A 425 10

= 6.92 × 105 Nm–2.Example 4. The area of cross section of the smaller piston of a hydraulic press is 2 cm2. and

that of large piston is 40 cm2. How much weight can be raised by the larger piston by applying a200 kg f on the smaller piston?

Solution. Here a = 2 cm2 = 2 × 10–4 m2.

A = 40 cm2 = 40 × 10–4 m2

f = 200 kg f = 200 × 9.8 N.

Using formulaf

a=

F

A

4

200 9.8

2 10

= 4

F

40 10or F = 200 × 20 × 9.8 N

= 3.92 × 104 N.Example 5. Two pistons of hydraulic press have diameters in the ratio 1 : 10. Calculate the

force exerted by larger piston when 75 kg wt is placed on the smaller piston?

Solution. Here r : R = 1 : 10

f = 75 kg wt.

From formulaf

a=

F

A

or 2

75

r= 2

F

R

or F =2

R75

r

= (10)2 × 75

= 7500 kg wt.


Example 6. To lift an automobile of 2000 kg, a hydraulic pump with a large piston 900 cm2

in area is employed calculate the force that must be applied to small piston of area 10 cm2 to liftthe car.

Solution. Here M = 2000 kgA = 900 × 10–4 m2

a = 10 × 10–4 m2

f = ?

From formulaF Mg

A A

f

a

f =4

4

10 102000 9.8

900 10

� 218 N.

BASED ON LIQUID PRESSUREExample 7. Calculate the pressure exerted by a water column of height 10 m.

(g = 9.8 ms–2)

Solution. Here h = 10 m

g = 9.8 ms–2

P = hdg

= 10 × 103 × 9.8

= 9.8 × 104 Nm–2.Example 8. A tank with a square base of area 1.0 m2 is divided by a vertical partition in the

middle. The bottom of the partition has a small hinged door of area 20 cm2. The tank is filled withwater in one compartment and an acid (R.D. 1.7) in the other, both to a height of 4.0 m. Calculatethe force necessary to keep the door closed. (g = 9.8 ms–2)

Solution. Here ha = hw = 4 m

a = 1.7 × 103 kg m–3

w = 103 kg m–3

Area of door = 20 cm2

= 20 × 10–4 m2.

Pressure due to water Pw = hw dw g

= 4.0 × 103 × 9.8

= 3.92 × 104 Nm–2

Pressure due to acid = 4.0 × 1.7 × 103 × 9.8

= 6.66 × 104 Nm–2

P = Pa – Pw

= (6.66 – 3.92) × 104

= 2.74 × 104 Nm–2

Net force on the door = A P

= 20 × 10–4 × 2.74 × 104 = 54.88 N

� 55 N.Example 9. A U tube contains water and spirit, seperated by mercury. The mercury column

in the two arms is in level with 10.0 cm of water in one arm and 15 cm of spirit in the other.Calculate the relative density of spirit.

W ater Acid

H inged door


Solution. The mercury level in the two arms is in level therefore

Pressure due to water = Pressure due to spirit

hw dw g = hs ds g

ds = ww

s

hd

h

= 31010

12.5 = 0.8 × 103 kg m–3

R.D. of spirit = 0.8.Example 10. In the above example 15 cm of water and spirit each are further poured into the

respective arms of the U tube. What will be the difference in the levels of mercury in the two arms.The relative density of mercury is 13.6.

Solution. New height of water column = 10 + 15 = 25 cm

New height of spirit column = 12.5 + 15 = 27.5 cm

= 0.275 cm.

Here pressure difference at A and B is balanced bypressure due to mercury column of height h.

(hw dw g – hs ds g) = hm dm g

or hw dw – hs ds = hm dm.

.25 × 103 – 0.8 × 103 × 0.275 = hm × 13.6 × 103

or .25 – 0.8 × 0.275 = 13.6 hm.

or hm =.25 (0.8 0.275)

13.6

= 0.22 × 10–2 m.

= 0.22 cm.Example 11. A manometer reads the pressure of a gas in an enclosure as shown below in fig.

(a) When some of the gas is removed by a pump, the manometer reads as in fig (b). The liquid usedin the manometer is mercury and the atmospheric pressure is 76 cm of Hg.

A B

20 cm

A B

18 cm

To pum p

Fig. (a) Fig. (b)

(i) Give the absolute and gauge pressure of the gas in the enclosure for case (a) and (b) inunits of cm of mercury.

(ii) How would the levels change in case (b) if 13.6 cm of water (immiscible with mercury)is poured in to the right limb of the manometer.

Solution. Here atmospheric pressure P = 76 cm of Hg.

(i) Point A and B in figure (a) are at the same level.

Pressure at point A = Pressure at point B

A B

10 cmwater

12.5 cmSpirit

A

B

25.0 cmwater

27.5 cmSpirit

h


= Pat + 20 cm of Hg

= 96 cm of Hg.Gauge pressure at A is = 20 cm of Hg.

In figure (b) pressure at point A and B are equal

The absolute gas pressure at A is = P – 18 cm of Hg

= (76 – 18)

= 58 cm of Hg.

Gauge pressure at A = – 18 cm of Hg.

(ii) hw dw g = hm dm g

13.6 × 1 × g = hm × 13.6 × g

hm = 1 cm.

i.e. 13.6 cm of water column is equivalent to 1 cm of Hg column. Therefore 13.6 cm of water,poured in right limb will displace mercury level by 1 cm in the left limb. The difference in thelevels of mercury in two limbs is (20 – 1) = 19 cm.

Example 12. What is the absolute and gauge pressure of the gas above the liquid surface inthe tank shown. Density of oil is 820 kg m–3. density of mercury = 13.6 × 103 kg m–3. Given 1 atm.= 1.01 × 105 Pa.

Solution. Points A and B are at the same level inthe mercury column.

PA = PB

Now PA = P + (1.5 + 1.0) × 820 × 9.8

PB = P + (1.50 + 0.75) × 13.6 × 103 × 9.8

Here P is pressure of the gas in the tank and P isatmospheric pressure.

[P + (2.5 × 820 × 9.8)] = [P + 2.25 × 13.6 × 103 × 9.8]

Gauge pressure = P – P= (2.25 × 13.6 × 103 × 9.8 – 2.5 × 820 × 9.8)

= 2.8 × 105 Pa.Absolute pressure = Pat + PGauge

= 105 + 2.8 × 105

= 3.8 × 105 Pa.Example 13. Mercury is replaced by water is barometer, then what should be the length of

tube to show atmospheric pressure?

Solution. We know

1 atm pressure = 76 cm of Hg Column.

hw × dw g = hm dm g

hw =3

3

.76 13.6 10 9.8

10 9.8

= 0.76 × 13.6

= 10.3 m.Example 14. Calculate the length of mercury column in a barometer tube, when the

atmospheric pressure is 75 cm of mercury and the tube is inclined at an angle of 30° withhorizontal direction.

1.5 cm

Hg

0.75 m

A B


Solution. Here h = 75 cm. of mercury

= 30°

Let l is the length of mercury column in the barometer tube, then

h

l= sin 30°

or l = o

75

1/ 2sin 30

h

= 150 cm.

BASED ON REYNOLD’S NUMBER VISCOSITY AND POISEUILLE’S FORMULAExample 15. Calculate the velocity of water in a tube of radius 1.0 cm so that the flow is

streamlined. Given viscosity of water is 0.001 kg m–1 s–1 and NR = 1000.

Solution. Reynold number is given by

NR =D

v

v = RN

D

Here = 0.001 kg m–1s–1, NR = 1000.

= 103 kgm–3 and D = 2.0 cm = 2 × 10–2 m

v = 3 2

0.001 1000

10 2 10

= 0.05 ms–1.Example 16. A square metal plate of 10 cm side moves parallel to another plate with a

velocity of 10 cms–1, both plates immersed in water. If the viscous force is 200 dyne and viscosityof water is 0.01 poise, what is their distance apart?

Solution. Here A = 10 × 10 = 100 cm2.

= 0.01 poise.

F = 200 dyne dv = 10 cm s–1

= 0.1 ms–1

From formula F = Ad

dx v

dx =A

F

dv

=0.01 100 10

200

= 0.05 cm.Example 17. Show that Reynold number is dimensionless.Solution. We know Reynold number is given by

NR =Dv


Writing dimensions R.H.S =3 1

1 1

[ML ][LT ] [L]

[ML T ]

= [M° L° T°].i.e. Reynold number is dimensionless.Example 18. Two layers of liquid 0.2 cm apart have relative velocity 10 cms–1. Calculate

velocity gradient.

Solution. Here dv = 10 cm s–1

dx = 0.2 cm.

Velocity gradient =10

0.2

dv

dx

= 50 sec–1.Example 19. Show by method of dimensions, that in Poiseuille’s formula Q represents the

volume of liquid flowing per second.

Solution. Poiseuille’s formula is Q =4

8

pr

l

writing dimensions on R.H.S.

Q =1 2 4

–1 –1

[ML T ] L

[ ] LML T

= [L3 T–1]

L3 T–1 is dimensions of volume/time. So Q represents the volume of liquid flowing persecond.

Example 20. A flat plate of area 100 cm2 rests on a 2 mm thick layer of glycerine ( = 1.5decapoise). A force of 0.225 N is applied on the plate. Calculate the velocity of the plate by whichit slides.

Solution. Here = 1.5 decapoise (N s m–2)

A = 100 cm2 = 100 × 10–4 m2.

F = 0.225 N.

dx = 2 mm = 2 × 10–3 m.

From formula F = Ad

dx v

dv =–3

2

F 0.225 2 10

A 1.5 10

dx

= 3 × 10–2 ms–1.Example 21. Water at 20°C is escaping from cistern by way of a capillary tube 10 cm long

and 0.4 mm in diameter, at a distance of 50 cm below the free surface of the water in cistern.Calculate the rate of escaping of water. ( = 0.001 Nsm–2).


r =0.4

2 = 0.2 mm = 0.2 × 10–3 m.

p = 50 cm of water column.


p = 0.50 × 9.8 × 103 Nm–2

= 0.001 Nsm–2.

From Poiseuille’s formula,

Q =4

8

pr

l

.

=3 3 43.14 0.50 9.8 10 (0.2 10 )

8 0.001 0.1

= 3.08 × 10–8 m3.Example 22. Calculate the mass of water flowing in 20 minutes through a tube of diameter

1.0 mm and 0.4 m long. The pressure difference between the two ends of tube is equal to 0.2 m ofwater. ( = 8.9 × 10–4 Nsm–2)

Solution. Here t = 20 min = 20 × 60 sec.

r =1

mm2

= 0.5 mm = 0.5 × 10–3 m

l = 0.4 mP = 0.2 m of water

= 0.2 × 103 × 9.8 Nm–2

= 8.9 × 10–4 Nsm–2

Volume of water flowing in 1 sec

Q =4

8

pr

l

=3 3 4

4

3.14 0.2 10 9.8 (0.5 10 )

8 0.4 8.9 10

= 1.35 × 10–7 m3s–1

Mass of water flowing in 20 min.

= Q × t × = 1.35 × 10–7 × 20 × 60 × 103

= 0.162 kg.Example 23. What should be the average velocity of water in a tube of radius 0.005 m so that

the flow is just turbulent? Given viscosity of water is 0.001 Nsm–2.

Solution. Here r = 0.005 m

= 0.001 Nsm–2

For water flow to be just tubulent

NR = 3000

NR =Dv

v = R3

N 3000 0.001

D 10 (0.005 2)

= 0.3 ms–1.


Example 24. Glycerine flows steadily through a horizontal tube of length 1.5 m and radius1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10–3 kgs–1 what is thepressure difference between the two ends of the tube? Density of glycerine = 1.3 × 103 kg m–3 andvisocity of glycerine = 0.83 Nsm–2.

Solution. Here l = 1.5 m r = 10–2 m.

= 1.3 × 103 kg m–3.

= 0.83 Nsm–2

Mass collected per unit time = 4.0 × 10–3 kg s–1

Volume collected per unit time

Q =Mass collected per second

density

=3

3

4.0 10

1.3 10

m3s–1.

Q =4

8

pr

l

P = 4

8 Ql

r

=3

2 4 3

8 0.83 1.5 4.0 10

3.14 (1.0 10 ) 1.3 10

= 9.8 × 102 Nm–2.Example 25. A liquid flows through a pipe of 10 cm length at a rate 3.14 cm3/s and a

pressure of 104 dyne cm–2. Calculate the radius of the pipe. ( = 1.25 centipoise)


Q = 3.14 cm3/s

P = 104 dyne cm–2

= 1.25 centipoise = 1.25 × 10–2 poise.Using Poiseuille’s formula

Q =4P

8

r

l

r4 =8Q

P

l

=2

4

8 3.14 1.25 10 10

3.14 10

or r = 0.1 cm.Example 26. Water flows through a horizontal pipe of length 40 cm and radius 0.081 cm.

The pressure difference between the two ends of pipe is equal to 40 cm of water. Calculate thecoefficient of viscosity of water if 432 cm3 of water is collected in 6 minute, form the pipe.



r = 0.081 cm

p = 40 cm of water

= 40 × 1 × 981 dyne cm–2

Q =432

6 60 cm3 s–1.

Using Poiseuille’s formula

Q =4

8

pr

l

=4 43.14 40 1 981 (.081) 360

8Q 8 432 40

pr

l

= 0.0138 poise.Example 27. A flat square plate of side 10 cm moves over another similar plate with a thin

layer of 0.1 cm of liquid between them. The coefficient of viscosity of liquid is 0.98 PaS. Calculatethe force required to move one of the plates uniformly with a speed of 1 ms–1

Solution. Here A = 10 cm × 10 cm = 100 × 10–4 m2

dx = 0.1 cm = 0.1 × 10–2 m

= 0.98 Pas

v = 1 ms–1

From formula

F = Ad

dx v

=4

2

0.98 100 10 1

0.1 10

= 9.8 N.

Example 28. A square metal plate of 10 cm side moves parallel to another plate with avelocity of 10 cm s–1, both plates immersed in water. If the viscous force is 200 dyne and viscosityof water is 0.01 poise, what is their distance apart.

Solution. Here A = 10 cm × 10 cm = 100 cm2

= 0.01 poise

F = 200 dyne dv = 10 cms–1

We know F = Ad

dx v

dx =A

F

d v

=0.01 100 10

200

= 0.05 cm.


Example 29. A flat square plate of 20 cm side moves over another similar plate with a thinlayer 0.4 cm of liquid between them. If a force of 1 kg wt moves one of the plates uniformly witha velocity of 1 ms–1 calculate the coefficient of viscosity.

Solution. Here A = 20 cm × 20 cm

= 400 × 10–4 m2

F = 1 kg wt = 9.8 N

dv = 1 ms–1 dx = 0.4 cm = 0.4 × 10–2 m.

Using formula F = Ad

dx v

or =2

4

F 9.8 (0.4 10 )

A / 400 10 1dv dx

= 0.98 Nsm–2.Example 30. A metal plate 100 cm2 in area rests on a layer of castor oil ( = 15.5 poise) 0.2

cm thick. Calculate the horizontal force required to move the plate with a speed of 3 cm s–1.

Solution. Here A = 100 cm2 = 100 × 10–4 m2

= 15.5 Poise.

dx = 0.2 cm = 0.2 × 10–2 m

dv = 3 cm s–1 = 3 × 10–2 ms–1.

Using formula

F = Ad

dx v

=4 2

2

15.5 100 10 3 10

0.2 10

= 0.233 N.

BASED ON STOKE’S LAW AND TERMINAL VELOCITYExample 31. An iron ball of radius 0.3 cm falls through a column of oil of density 0.94 g cm–3.

It is found to attain a terminal velocity of 0.5 cms–1. Determine the viscosity of the oil. Given thatdensity of iron is 7.8 g cm–3.

Solution. Here r = 0.3 cm

v = 0.5 cms–1 = 7.8 g Cm–3 = 0.94 g cm–3.

v =22

( ) g9

r

or =22

( ) g9

r v

=22 (0.3) (7.8 0.94)

9809 0.5

=2 0.09 6.86 980

4.5

� 269.0 poise.


Example 32. An air bubble of 2 cm diameter rises steadily through water of density 103 kgm–

3 at the rate of 2 mms–1. Calculate the viscosity of water neglecting the density of air.

Solution. Here r = 1 cm = 10–2 m.

= 103 kg m–3

vt = 2 mm s–1 = 2 × 10–3 ms–1.

We know

vt =22 ( ) g

9

r

neglecting density of air i.e

vt =22

9

r g

or =22

9 t

r g

v

=2 2 3

3

2 (10 ) 10 9.8

9 2 10

= 1.09 × 103 Nsm–2.Example 33. In Millikan oil drop experiment what is the terminal speed of a drop of radius

2.0 × 10–5 m and density 1.0 × 103 kg m–3 ? Take the viscosity of air at the temperature of theexperiment to be 1.5 × 10–5 N sm–2. How much is the viscous force on the drop at that speed?

Solution. Given that r = 2 × 10–5 m

= 1.0 × 103 kg m–3

= 1.8 × 10–5 Nsm–2

Using formula

vt =22

9

r g (neglecting density of air)

=5 2 3

5

2 (2 10 ) 1 10 9.8

9 1.8 10

= 5.8 × 10–2 ms–1.Now viscous force in given by Stoke’s Law

F = 6 r vt

= 6 × 3.14 × 1.8 × 10–5 × 2.0 × 10–5 × 5.8 × 10–2

= 3.9 × 10–10 N.Example 34. Two exactly similar rain drops falling with terminal velocity of (2)1/3 ms–1

Coalesce to form a bigger drop. Find the new terminal velocity of the bigger drop.

Solution. Let r be the radius of each smaller drop and R that of bigger drop.

Volume of bigger drop = Volume of two smaller drop

34R

3 =

342

3r


orR

r= (2)1/3

Terminal velocity of smaller and bigger drops are

v1 =22 ( ) g

9

r

v2 =22 R ( ) g

9

or 1

2

vv

=2

2R

r

or v2 =2

1/ 2 2 / 31 2

R2 2

r

v

= 2 ms–1.Example 35. Calculate the terminal velocity of a steel ball of radius 1 mm falling through

glycerine. Relative density of steel is 8 and that of glycerine is 1.3. Viscosity of glycerine is 8.3poise.

Solution. Here r = 1 mm = 0.1 cm

density of steel = 8 g cm–3.

density of glycerine = 1.3 g cm–3

Viscosity = 8.3 poise

Terminal velocity is v =22 ( ) g

9

r

=22 (0.1) (8 1.3) 9.80

9 8.3

= 1.758 cm s–1.Example 36. A sphere is dropped under gravity through a fluid of viscosity . Taking the

average acceleration as half of the initial acceleration, show that the time taken to attain theterminal velocity is independent of the fluid density.

Solution. Suppose a sphere of radius r and density is falling in a liquid of density andviscosity . When the sphere just enters the fluid, the net downward force on it is

F =3 34 4

g g3 3

r r

= 34( ) g

3r

acceleration a = F/m =3

3

4 / 3 ( ) g

4 / 3

r

r

= g

.

As soon as the sphere attains terminal velocity acceleration becomes zero.


Average acceleration =0 ( )g

2 2

a .

Let ‘t’ is the time taken by the sphere to attain terminal velocity.

v =22

( )g9

r

Using first equation of motion

v = u + at

22( )g

9

r

= 0 g2

t

or t =24 �

9 �

r.

Example 37. A metallic sphere of radius 1.0 × 10–3 m and density 1.0 × 104 kg m–3 enters atank of water, after a free fall through distance h in the earth’s gravitational field. If its velocityremains unchanged after entering water, determine the value of h.

Given = 1.0 × 10–3 Nsm–2, g =10 ms–2, = 103 kg m–3]

Solution. Let velocity attained after through distance h is v then

v = 2 gh

This will become terminal velocity of sphere in water.

Hence we have

v =22 ( ) g

9

r

or v =3 2 4 3

3

2 (1.0 10 ) (1.0 10 1.0 10 )10

9 (1.0 10 )

= 20 ms–1.

h =2 20 20

2g 2 10

v

= 20 m.

Example 38. A small sphere falls from rest under gravity in a viscous medium producing heatdue to friction. Find how does rate of production of heat depend upon the radius of the sphere atterminal velocity.

Solution. We know terminal velocity is given by

v =22 ( ) g

9

r

and viscous force is given by F = 6 r v .

The rate of production of heat due to force F is


dQ

dt= F v

= 6 r v 2.

ordQ

dt=

222 ( )6 g

9

rr

dQ

dt r5.

BASED ON EQUATION OF CONTINUITY AND BERNOULLI’S THEOREMExample 39. Water flows through a pipe whose internal diameter is 2.0 cm at a speed of 1.0

ms–1. What should the diameter of the nozzle be, if the water is to emerge at a speed of 4.0 ms–1?Solution. Here r1 = 1.0 cm = 10–2 m r2 = ?

v1 = 1.0 ms–1, v2 = 4.0 ms–1

Using equation of continuity

a1v1 = a2v2

r12 v1 = r2

2 v2

or r22 =

2 42 21 1

2

1.0 10(10 )

4.0 4

r

vv

or r2 =210

2

d2 = 2 r2 = 10–2 m = 1 cm.Example 40. Calculate the speed at which the velocity head of water stream be equal to

80 cm of mercury column.

Solution. Here Velocity head = 80 cm of Hg

Velocity head =2

2g

v = 80 × 13.6 cm of water

80 × 13.6 =2

2 1000v

(g = 1000 cm s–2)

or v = 80 13.6 2 1000 = 1475.12 cm s–1

= 14.7512 ms–1

� 14.75 ms–1.Example 41. The cylindrical tube of a spray pump has a cross section of 8.0 cm2. One end

of which has 40 fine holes each of diameter 1 mm. If the liquid flow inside the tube is 1.5 m min–1

what is the speed of ejection of the liquid through the holes?

Solution. Here a1 = 8 cm2 = 8 × 10–4 m2

v 1 = 1.5 m min–1 = 1.5

m/s.60


radius of each hole =0.1

mm2

= 0.5 × 10–3 m.

a2 = r22 = (0.5 × 10–3)2 m2.

area of cross section of 40 holes

= 40 (0.5 × 10–3)2


a1v1 = a2 v 2

or v2 = 1 1

2

a

a

v

=4

3 2

(8 10 ) 1.5

(0.5 10 ) (40 60)

= 0.637 ms–1.Example 42. If the velocity of water in a 4 m diameter pipe is 5 ms–1. What is the velocity

in a 2 m diameter pipe which connects with it, both pipes flowing full?

Solution. Here r1 =4

2 = 2 m.

r2 =2

2 = 1 m.

v 1 = 5 ms–1


a1 v 1 = a2 v 2

v 2 = 1 1

2

a

a

v

v 2 =2 2

1 12 22

2 (2) 5

(1)

r

r

v

= 20 ms–1.Example 43. A pilot tube is mounted on an aeroplane wing to measure the speed of the plane.

The tube contains alcohol, and shows a level difference of 40 cm. What is the speed of the planerelative to air (sp. gravity of alcohol is 0.8 and density of air = 1 kg m–3).

Solution According to Bernoulli’s theorem

Kinetic energy per unit volume = Pressure energy per unit vol.

or 21

2v = hdg

or v =2 hdg

.


Here h = 40 cm of alcohol = 0.40 m of alcohol.

d = sp. gravity of alcohol = 0.8 = 0.8 × 103 kg m–3

g = 9.8 ms–2

= 1 kg m–3

v =32 0.4 0.8 10 9.8

1

= 79.2 ms–1.Example 44. Water emerges from a horizontal pipe of varying cross section at a rate of 30

litre per minute. Find the velocity of water at a point in the pipe when the radius of the pipe is2 cm.

Solution. Let a1 is area of cross section and v 1 is velocity of emerging of water then

a1 v 1 =30 litre

min

=330 10

60

= 0.5 × 10–3 m3s–1.

Let a2 is area of cross section of the pipe where the radius is 2 cm. v2 is the velocity of waterat that point.

a1 v 1 = a2 v 2

v 2 =3

1 12 2

2

0.5 10

3.14 (2 10 )

a

a

v

= 0.4 ms–1.Example 45. The velocity head of a stream of water is equal to 20 cm of mercury column.

What is the velocity of flow in the stream. The relative density of mercury is 13.6. (g = 9.8 ms–2)

Solution. Velocity head = 20 cm of mercury column.

or2

2g

v= 0.2 × 13.6 m (of water).

v = 2 9.8 0.2 13.6

= 7.3 ms–1.Example 46. Water flows out of a small hole in the wall of a large tank near its bottom. What

is the speed of efflux of water when the height of water level in the tank is 5.0 m.

Solution. Here h = 5.0 m

The velocity of efflux is given by

v = 2gh

= 2 9.8 5

= 9.9 ms–1.Example 47. Calculate the volume of water coming out per minute from a tank through an

opening 1 cm in diameter and 5.1 m below the level of water?

Solution. Here t = 1 min = 60 sec.

5 m

U


r =1

2 cm = 0.5 cm = 0.5 × 10–2 m.

h = 5.1 m.Volume of water flowing per second is given by

Q = A 2gh

= 2 2 ghr

= 3.14 (0.5 × 10–2)2 2 9.8 5.1

= 7.86 × 10–4 m3s–1

Volume of water flowing in 1 min

= 7.86 × 10–4 × 60

= 4.7 × 10–2 m3 min–1.Example 48. Calculate the total energy possessed by two kg of water at a point where the

pressure is 1.96 × 105 Nm–2. Velocity is 0.1 ms–1 and the height is 50 cm above the ground level.

Solution. Here p = 1.96 × 105 Nm–2

v = 0.1 ms–1

h = 50 cm = 0.5 m

= 103 kg m–3.

Pressure energy per kg =4

3

P 19.6 10

10

= 196 J.

Gravitational potential energy per kg = gh= 9.8 × 0.5 = 4.9 J

Kinetic energy per kg. = 2

2

v

=0.1 0.1

2

= 0.005 J.

Total energy per kg of water = 196 + 4.9 + 0.005= 200.9 J

Total energy of 2 kg of water = 200 9 × 2

= 401.8 J.Example 49. A horizontal pipe line carries water in a streamline flow. At a point along the

pipe where cross sectional area is 10 cm2, the water velocity is 1 ms–1 and the pressure is 2000 Pa.Find the pressure of water at another point where the cross sectional area is 5 cm2. ( = 103 kgm–3)

Solution. Here a1 = 10 cm2 = 10 × 10–4 m2

v1 = 1 ms–1

P1 = 2000 Pa, a2 = 5 cm2 = 5 × 10–4 m2. = 103 Kg m–3

Using equation of continuitya1 v 1 = a2 v 2


v 2 = 1 1

2

a v

a

=4

4

10 10 1

5 10

= 2 ms–1.

Using Bernoulli’s theorem

P1 + 1

2 v 1

2 = P2 + 22

1

2 v

or P2 = 2 21 1 2

1( )

2P v v

= 3 2 212000 10 (1 2 )

2

= 500 Pa.Example 50. The reading of a pressure meter attached with closed organ pipe is 2.5 × 105

Nm–2. On opening the valve of the pipe, the reading of the pressure meter reduces to 0.5 × 105

Nm–2. Calculate the speed of the water flowing in the pipe.

Solution. Here P1 = 2.5 × 105 Nm–2

P2 = 0.5 × 105 Nm–2

= 103 kg m–3

Using Bernoulli’s theorem for horizontal pipe

21 1

1P

2 v = 2

2 21

P2

v

or P1 = 22 2

1P

2 v

v 22 =

51 2

3

2(P P ) 2(2.5 0.5) 10

10

= 400

or v 2 = 20 ms–1.Example 51. Air is streaming past a horizontal aeroplane wing such that its speed is 120 ms–1

at the upper surface and 90 ms–1 at the lower surface. If the density of air 1.3 kg m–3, find thedifference in pressure between the two surfaces of the wing. If the wing is 10 m long and has anaverage width of 2m then calculate the gross lift on it.

Solution. Here v 1 = 120 ms–1

v 2 = 90 ms–1

= 1.3 kg m–3

and A = l × b = 10 × 2 = 20 m2.

Let P1 and P2 be the pressure of air at the upper and lower surface of the wing.

Applying Bernoulli’s theorem


21 1

1P

2 v = 2

2 21

P2

v

or (P2 – P1) = 2 21 2

1( )

2 v v

= 2 211.3 (120 80 )

2

=1

1.3 200 402

Nm–2.

Gross lift = (P2 – P1) A

=1

1.3 8000 202

= 8.2 × 104 N.Example 52. Calculate the rate of flow of glycerine of density 1.25 × 103 kg m–3 through the

conical section of a pipe if the radii of its ends are 0.1 m and 0.04 m and pressure drop across itslength is 10 Nm–2.

Solution. Here = 1.25 × 103 kg m–3

r1 = 0.1 m, r2 = 0.04 m

P = 10 Nm–2


21 1

1P

2 v = 2

2 21

P2

v

or 2 22 1( )v v =

1 22 (P P ) 2 P

2 22 1v v = 3

2 10

1.25 10

= 16 × 10–3 ... (i)

From equation of continuity a1v1 = a2v2

or1

2

vv =

22 2

21 1

a r

a r

=2

2

(0.04)

(0.1) = 16 × 10–2.

� v 1 = 16 × 10–2 v2 ... (ii)

Using equation (i) and (ii)

v 22 – (16 × 10–2 v 2)2 = 16 × 10–3

or v 22 (1 – 256 × 10–4) = 16 × 10–3

or v 22 =

3 3 4

4

16 10 16 10 10256 10000 256

110


or v 2 = 0.13 ms–1

Rate of flow = a2 v 2

= 3.14 (0.04)2 × 0.13

= 6.53 × 10–4 m3s–1.

Example 53. Water is flowing through two horizontal pipes connected together havingdiameters 3 cm and 6 cm. The speed of water in first pipe is 4 ms–1. Calculate the pressuredifference across the two pipes. ( = 103 kg m–3).

Solution. Here r1 =3

2 = 1.5 cm. = 1.5 × 10–2 m

r2 =6

2 = 3.0 cm. = 3.0 × 10–2 m.

v 1 = 4 ms–1

= 1 × 103 kg m–3.


a1 v 1 = a2 v 2

or v 2 =

21 1 1 1

22 2

a r

a r

v v

=2 2

2 2

(1.5 10 ) 4

(3.0 10 )

= 1 ms–1.

Now using Bernoulli’s theorem

21 1

1

2P v = 2

2 21

2P v

(P2 – P1) = 2 23 1 2

1( )

2 v v

= 3 2 2110 (4 1 )

2

= 7.5 × 103 Nm–2.Example 54. Calculate the least pressure required to force the blood from the heart to the top

of head (50 cm above heart). Given density of blood is 1.04 g cm–3.

Solution. Here h2 – h1 = 50 cm

= 1.04 g cm–3.


P1 + gh1 + 1

2 v 1

2 = P2 + gh2 + 22

1

2 v .

or (P1 – P2) = gh (h2 – h1) Considering v 1 = v 2.

= 1.04 × 981 × 50

= 5.1 × 104 dyne cm–2.


Example 55. In a test experiment on a model aeroplane in a wind tunnel, the flow speed onthe upper and lower surfaces of the wing are 70 ms–1 and 63 ms–1 respectively. What is the lift onthe wing if its area is 2.5 m2.

Take the density of air to be 1.3 kg m–3.

Solution. Let v1 and v2 be the speed of air on the upper and lower surfaces respectively andpressure are P1 and P2 respectively. Using Bernoulli’s theorem.

21 1

1P

2 v = 2

2 21

P2

v .

(P2 – P1) = 2 21 2

1

2 v v

=2 21

1.3 (70 63 )2

= 605.15 Nm–2

Force on the wing = (P2 – P1) A

= 605.15 × 2.5

= 1.5 × 103 N.

BASED ON SURFACE TENSION AND SURFACE ENERGYExample 56. A u shaped wire is dipped in a soap solution and removed. The thin soap film

formed between the wire and a light slider supports a weight of 1.5 × 10–2 N (which includes thesmall weight of the slider). The length of the slider is 30 cm. What is the surface tension of thefilm.

Solution. Here F = 1.5 × 10–2

L = 30 cm = 0.3 m

From formula T =F

2L

=21.5 10

2 0.3

= 2.5 × 10–2 Nm–1.Example 57. The surface tension of a soap solution is 0.04 Nm–1. Calculate the work done

to form a bubble of radius 1.0 cm, from this solution.

Solution. Here T = 0.04 Nm–1.

r = 1.0 × 10–2 m.

Total surface area of the bubble A = 2 × 4 r2

Where r is radius of the bubble = 1.0 × 10–2 m.

A = 8 × 3.14 (1.0 × 10–2)2

= 2.51 × 10–3 m2.

From formula T =W

A

W = T A


= 0.04 × 2.51 × 10–3

= .1004 × 10–3

= 10.04 × 10–5 J.Example 58. A mercury drop of radius 1.0 cm is sprayed into 106 droplets of equal size.

Calculate the energy expanded. (Surface tension of mercury is 32 × 10–2 Nm–1).

Solution. Let radius of the drop is R and that of each droplet is r.

34R

3 = 6 34

103

r

or r = 10–2 R = 10–2 (1.0) = 10–2 cm

= 10–4 m.

Surface area of drop = 4 R2

Surface area of 106 droplets= 106 × 4 r2

= 106 × 4 (10–2 R)2

= 4 R2 × 102.

A = 4 R2 × 102 – 4 R2

= 4 R2 × 99= 4 (10–2)2 × 99 m2.

From formula T =W

A.

W = A T= 4 × 3.14 (10–4) × 99 × 32 × 10–2

= 3.98 × 10–2 J.Example 59. The size of a rectangular soap film is increased from its size 3 cm × 4 cm to

3 cm × 5 cm. Calculate the work done if surface tension is 3.0 × 10–2 Nm–1.

Solution. Here Change in area A = (15 cm2 – 12 cm2) × 2

= 6 cm2 (A soap film has two free surface)

= 6 × 10–4 m2.

T = 3.0 × 10–2 Nm–1.

Work done W = T A

= 3.0 × 10–2 × 6 × 10–4

= 18.0 × 10–6 J.Example 60. A mercury drop of diameter D breaks up into 64 tiny drops. Find the change is

surface energy.

Solution. Let the surface tension of mercury is T.

Radius of mercury drop is (D/2).

Let radius of each tiny drop is r.

Volume of drop = Volume of 64 tiny drops

34 D

3 2

= 3464

3r

or3

D

2

= 64 r3


or r =D 1 D

2 4 8 .

Initial surface area of one drop = 4 (D/2)2 = D2.

Final surface area of 64 tiny drops =2

D4 64

8

= 4 D2

A = 4 D2 – D2

= 3 D2.

Change in surface energy = 3 D2 × T

= 3 TD2.Example 60. Two soap bubbles in vacuum having radii 3 cm and 4 cm respectively coalesce

under isothermal conditions to form a single bubble. What is the radius of new bubble.

Solution. Surface energy of smaller bubble = 2 × 4 21r T

= 8 21r T

Surface energy of second layer bubble = 8 22r T.

Let radius of new bubble is r.

By the conservation of energy

Surface energy of new bubble = 8 21r T + 8 2

2r T

or 8 r2 T = 8 21r T + 8 2

2r T

or r2 = 21r + 2

2r

= 32 + 42

= 52

r = 5 cm.Example 61. 2 × 10–5 J of work is done in blowing a soap bubble of radius r. Calculate

additional work to be done to blow it to a radius 2 r.

Solution. Work done in blowing the soap bubble of radius r

W = T A= T (4 r2)2

= 8 r2 T.

Additional work to be done to increase the radius from r to 2 r

W = T × 2 × 4 [(2r)2 – r2]

= T × 2 × 4 × 3r2.

= 3W

= 3 × 2 × 10–5

= 6 × 10–5 J.Example 62. If a number of little droplets of water, each of radius r, coalesce to form a single

drop of radius R, show that the rise in temperature will be given by

=3T 1 1

J Rr

.


Where T is the surface tension of water & J is the mechanical equivalent of heat.

Solution. Let n droplets coalesce to form single drop

Volume of n droplets = Volume of one drop

34

3n r = 34

R3

n r3 = R3

Decrease in surface area = n × 4 r2 – 4 R2

= 4 (n r2 – R2)

=3

24 Rn r

r

=3

2R4 R

r

= 3 1 14 R

Rr

Energy evolved = T × A

= T × 4 R3 1 1

Rr

Heat produced Q =3W 4 TR 1 1

J J Rr

But Q = ms

=34

R3 × 1 × 1 ×

34R

3 × 1 × =

34 T R 1 1

J Rr

or =3T 1 1

J Rr

.

BASED ON EXCESS PRESSURE IN DROPS AND BUBBLESExample 63. What would be the excess pressure inside an air bubble of 0.2 mm radius

situated just below the surface of water? (T = 0.07 Nm–1)

Solution. Here r = 0.2 mm = 0.2 × 10–3 m

T = 0.07 Nm–1

Excess pressure P = 3

2T 2 0.07

0.2 10r

= 700 Nm–2.


Example 64. What is the pressure inside a drop of mercury of radius 3.00 mm at roomtemperature? Surface tension of mercury at that temperature (20°C) is 4.65 × 10–1 Nm–1. Theatmospheric pressure is 1.01 × 105 Pa. Also give the excess pressure inside the drop.

Solution. Here r = 3.00 mm = 3.00 × 10–3 m.

T = 4.65 × 10–1 Nm–1.

Pat = 1.01 × 105 Nm–2.

Excess pressure inside the drop is

Pex =1

3

2T 2 4.65 10

3.00 10r

= 310 Nm–2.Total pressure inside the drop

P = Pat + Pex

= (1.01 × 105) + (310)

= 1.013 × 105 Pa.Example 65. A spherical shell with a hole in it is immersed into water to a depth of 10 cm

before any water enters it. The surface tension of water is 73 dyne cm–1.

Find the radius of the hole.

Solution. Here T = 73 dyne cm–1.

h = 10 cm

In equilibrium hdg =2T

r

r =2T 2 73

10 1 980hdg

= .014898

� 0.015 cm.Example 66. What should be the pressure inside a small air bubble of 0.1 mm radius situated

just below the surface? Surface tension of water = 7.2 × 10–2 Nm–1 and atmospheric pressure =1.013 × 105 Nm–2.

Solution. Here r = 0.1 mm = 0.1 × 10–3 mT = 7.2 × 10–2 Nm–1.

Excess pressure inside the bubble

Pex =2

3

2T 2 7.2 10

0.1 10r

= 1440 Nm–2 = .0144 × 105 Nm–2.

Pressure inside the bubble = Pat + Pex

= 1.013 × 105 + 0.0144 × 105

= 1.027 × 105 Nm–2.

Example 67. The pressure of air in a soap bubble of diameter 0.7 cm is 8 mm of water abovethe atmospheric pressure. Calculate the surface tension of soap solution.


Solution. Here r =0.7

2 = 0.35cm

P = 8 mm of water

= 0.8 × 1 × 981 dyne cm–2.

For a bubble

p =4T

r

T =R

4

p

T =0.8 1 981 .35

4

= 68.67 dyne/cm.

Example 68. Calculate the pressure inside a small air bubble of radius 0.01 mm situated at adepth of 20 cm below the free surface of water. (Take surface tension of water to be 75 dyne/cm).Atmospheric pressure is 76 cm of Hg.

Solution. Here r = 0.01 mm = 0.001 cm.

T = 75 dyne cm–1.

d = 1 g cm–3.

Atmospheric pressure P = h dg

= 76 × 13.6 × 981

= 1013961.6 dyne cm–2.

Pressure of liquid column of 20 cm = 20 × 1 × 981

= 19620 dyne cm–2

Excess pressure inside the air bubble =2T

r =

2 75

.001

= 150,000 dyne cm–2.

Total pressure inside air bubble = 1013961.6 + 19620 + 150,000

= 1183581.6 dyne-cm–2

� 1.18 × 106 dyne cm–2

Example 69. A u tube is made up of capillaries of bore 1 mm and 2 mm respectively. Thetube is held vertically and partially filled with a liquid of surface tension 49 dyne-cm–1 (angle ofcontact is zero). Calculate the density of liquid, if the meniscus level in the two capillaries differby 1.25 cm.

Solution. The excess pressure in the two limbs of the capillary u tube is given by

P =1 2

1 12T

r r

Let density of liquid is d then

h dg =1 2

1 12T

r r

.


or d =1 2

2T 1 1

h g r r

Here r1 =1

mm2

= 0.05 cm

r2 = 0.1 cm

T = 49 dyne cm–1

h = 1.25 cm

d =2 49 1 1

1.25 981 0.05 0.1

� 0.8 g cm–3.Example 70. The excess pressure inside a soap bubble of radius 3 mm is balanced by 2 mm

column of oil of specific gravity 0.8. Find the surface tension of soap solution.

Solution. Here r = 3 mm = 3 × 10–3 m

h = 2 mm = 2 × 10–3 m

d = 0.8 × 103 kg m–3.

Excess pressure inside soap bubble = Pressure exerted by 2 mm height of oil.

4T

r= h dg

or T = rhdg/4 = 1

4 × 3 × 10–3 × 2 × 10–3 × .8 × 103 × 9.8

= 1.175 × 10–2 Nm–1.

BASED ON CAPILLARITYExample 71. Mercury has an angle of contact equal to 140° with soda lime glass. A narrow

tube of radius 1.00 mm made of thin glass is dipped in a trough containing mercury. By whatamount does the mercury dip down in the tube relative to the liquid surface outside? Surfacetension of mercury at the temperature of experiment is 0.465 Nm–1, density of mercury = 13.6 ×103 kg m–3.

Solution. Here = 180°, r = 1.00 mm = 10–3 m.

T = 0.465 Nm–1

d = 13.6 × 103 kg m–3.

cos 140° = – 0.7660.

Using formula

h =2T cos

r dg

= 3 3

2 0.465 cos 140

10 13.6 10 9.8

=2 0.465 ( 0.7660)

13.6 9.8

= – 5.34 × 10–3 m.


Example 72. Two capillary tubes A and B are dipped in to a liquid which rises 8 cm and6 cm respectively above the outside level. Compare the diameters of tubes.

Solution. Here h1 = 8 cm, h2 = 6 cm.

Using formula h =2T cos

r dg

h1 =1

2T cos

r dg

h2 =2

2T cos

r dg

1

2

h

h= 2 2 2

1 1 1

2.

2

r r d

r r d

or 1

2

d

d= 2

1

6

8

h

h 3

4.

Example 73. A tube of 1 mm bore is dipped in to a liquid of density 0.8 g cm–3 and surfacetension 30 dyne cm–1. Calculate the height to which the liquid will rise in capilary. Also calculatethe length of liquid column in capillary when it is inclined at 30° to vertical. Take angle of contactto be zero.

Solution. Here r =1

mm2

= 0.5 mm = 0.05 cm

d = 0.8 g cm–3.

T = 30 dyne cm–1

Using formula h =2T cos

r dg

=2 30 1

0.05 0.8 9.8

= 1.51 cm.(b) If the tube is inclined at an angle 30° to the vertical then

h = o

1.51

cos 30 32

h

= 1.75 cm.Example 74. Two narrow bores of diameters 3.0 mm and 6.0 mm are joined together to form

a u tube open at both ends. If the u tube contains water, what is the difference in its level in twolimbs of the tube? Surface tension of water at the temperature of experiment is 7.3 × 10–2 Nm–1.Take the angle of contact to be zero density of water = 1.0 × 103 kg m–3. (g = 9.8 ms–2)

Solution. Here r1 =3.0

2 = 1.5 mm = 1.5 × 10–3 m


r2 =6.0

2 = 3.0 mm = 3.0 × 10–3 m.

T = 7.3 × 10–2 Nm–1.

d = 1.0 × 103 kg m–3

Now h1 =1

2T

r dg

h2 =2

2T

r dg

h1 – h2 =1 2

2T 1 1

dg r r

=2

3 3 3

2 7.3 10 1 1

10 9.8 1.5 10 3.0 10

= 3

2 7.3 10 1 1

1.5 3.010 9.8

=214.6 10 1

9.8 3

= 5.29 × 10–3 m= 5.29 mm.

Example 75. A capillary tube of radius 0.1 mm is dipped in a liquid of density 0.8 × 103 kgm–3 and surface tension 2.8 × 10–2 Nm–1. Calculate the height to which the liquid will rise incapillary. Angle of contact between the liquid and solid surface is zero. (g = 10 ms–2).

Solution. Here r = 0.1 mm = 0.1 × 10–3 m.

d = 0.8 × 103 kg m–3.

T = 2.8 × 10–2 Nm–1

= 0°

From formula h =2T cos

r dg

h =

2

3 3

2 2.8 10 cos 0

0.1 10 0 8 10 10.0

= 7 × 10–2 m

= 7 cm.Example 76. Water rises in a capillary tube to a height 9.0 cm, while mercury falls down by

3.4 cm in the same capillary. Assume angles of contact for water glass and mercury glass are 0°and 135° respectively. Determine the ratio of surface tensions of mercury and water. (cos 135° =– 0.71).

Solution. Here h1 = 9 cm = 0.09 m


d1 = 103 kg m–3 1 = 0°.

h2 = – 3.4 cm = – 0.034 m

d2 = 13.6 × 103 kg m–3 2 = 135°

h1 =1 1

1

2T cos,

rd g

h2 =

2 2

2

2T cos,

rd g

T1 =1 1

12cos

h d rg

T2 =2 2

22cos

h d rg

1

2

T

T =1 1 2

2 2 1

cos

cos

h d

h d

or2

1

T

T =2 2 1

1 1 2

cos

cos

h d

h d

=3

3

0.034 13.6 10 cos 0

0.09 10 (cos 135)

=0.034 13.6 1

0.09 ( 0.71)

= 7.28.

EXERCISE

BASED ON THRUST AND PRESSURE1. The two thigh bones each of cross sectional area 10 cm2 support the upper part of a human

body of mass 40 kg. Estimate the average pressure sustained by the femurs.

[Ans. 1.96 × 105 Nm–2]

2. A 50 kg girl wearing high heel shoes balances on a single heel of diameter 1.0 cm. Calculatethe pressure exerted by the heel on the floor. [Ans. 6.2 × 106 Nm–2]

3. A rectangular tank is 10 m, long 5 m broad and 3 m high. It is filled to the brim with waterof density 103 kg m–3. Calculate the thrust at the bottom and walls of the tank due tohydrostatic pressure. (g = 9.8 ms–2)

4. Atmospheric pressure is nearly 50 K Pa. How large the force does the air in a room exert onthe inside window pane that is 40 cm × 80 cm? [Ans. 16 kN]

5. Calculate the pressure of sea water at a depth of 400 m. Consider average density of seawater is 1.032 × 103 kg m–3. [Ans. 4.046 × 106 Nm–2]

6. Find the pressure exerted at the tip of a thumb pin if it is pushed against a board with a forceof 40 N. Assume the area of the tip to be 0.1 mm2. [Ans. 4 × 108 Nm–2]


BASED ON PASCAL’S LAW AND HYDRAULIC LIFT7. A hydraulic automobile lift is designed to lift cars with a maximum mass of 6000 kg. The

area of cross section of the piston carrying the load is 425 cm2. What maximum pressurehave to bear by the smaller piston. [Ans. 13.84 × 105 Nm–2]

8. The neck and bottom of a bottle are 2 cm and 20 cm. in diameter respectively. If the cork ispressed with a force of 1.2 kg f in the neck of the bottle, calculate the force exerted on thebottom of the bottle. [Ans. 120 kg f]

9. In a car lift compressed air exerts a force F1 on a small piston having a radius of 5 cm. Thepressure is transmitted to a second piston of radius 15 cm. The mass of the car to be lift is1350 kg. Calculate the pressure necessary to accomplish this work.

[Ans. 1.9 × 105 Nm–2]

10. In a hydraulic press used for compressing wool, the area of the piston is 0.2 m2 and the forceexerted along the piston rod is 200 N. If area of cross section of larger cylinder is 0.4 m2 findthe pressure produced in the cylinder and the total crushing force exerted.

[Ans. 1000 Nm–2, 400 N]

11. The diameter of the piston P2 is 50 cm and that of piston P1 is 10 cm. What is the forceexerted on P2 when a force of 1 N is applied on P1. [Ans. 25 N]

BASED ON PRESSURE EXERTED BY LIQUID COLUMN12. A column of water 50 cm high supports a 40 cm column of an unknown liquid. Determine

the density of the liquid. [Ans. 1.25 × 103 kg m–3]

13. The water pressure gauge shows the pressure at ground floor to be 250 k Pa. Calculate howhigh would water rise in the pipes of a building. (g = 10 ms–2) [Ans. 25 m]

14. A square tank is of side 10 m and 3 m high. It is filled to the brim with water of density 103

kg m–3. Calculate the gauge pressure and absolute pressure at the bottom of tank. Also findthe thrust on the bottom. [Ans. 1.3 × 105 Nm–2, 2.94 × 106 N]

15. The pressure due to a liquid column at a place is 2 × 105 Nm–2. If the density of the liquid is1.2 × 103 kg m–3, calculate the height of the liquid column. [Ans. 17.0 m]

16. What is the pressure on a swimmer 10 m below the surface of a lake? [Ans. 2 atm]

17. The pressure of water at the basement of a building is 1.96 × 105 Nm–2 and at the third flooris 9.8 × 104 Nm–2. Find the height of third floor. [Ans. 10 m]

18. Calculate the pressure at a depth of 200 m below the surface of a lake. Density of water is1000 kg m–3. (Neglect atmospheric pressure) [Ans. 19.6 × 105 Nm–2]

19. A barometer kept in an elevator accelerating upwards reads 76 cm of Hg. If the elevator isaccelerating upwards at 4.9 ms–2, What will be the air pressure in the elevator?

[Ans. 114 cm of Hg]

BASED ON SURFACE TENSION AND SURFACE ENERGY20. The surface tension of a soap solution is 0.07 Nm–1. How much work is required to form a

bubble of 1.0 cm radius from this solution. [Ans. 1.76 × 10–4 J]

21. A liquid drop of diameter D breaks up into 27 tiny drops. Find the resulting change inenergy. Surface tension of liquid is T. [Ans. 2 D2T]

22. Calculate the work done to form a bubble of radius 0.05 m. (Surface tension of liquid is 0.05Nm–1) [Ans. 3.14 × 10–3 J]

23. A rectangular film of soap solution is extended from (5 × 4) cm2 to (6 × 5) cm2. Calculatethe surface tension of soap solution if the work done is 3.0 × 10–4 J. [Ans. 0.15 Nm–1]


24. Calculate the work done in spraying a spherical drop of 1 mm radius into million droplets ofequal size. The surface tension of water is 0.072 Nm–1. [Ans. 8.96 × 10–5 J]

25. The surface tension of a soap solution 0.07 Nm–1. How much work is required to form abubble of 1.0 cm radius from this solution. [Ans. 1.76 × 10–4 J]

26. Calculate the energy released when 1000 small water drops each of same radius 10–7 mcoalesce to form one large drop. The surface tension of water is 7 × 10–2 Nm–1.

[Ans. 7.9 × 10–12 J]

27. If 400 erg of work is done in blowing a soap bubble to a radius r. Calculate the additionalwork to be done to blow it to a radius of 4 r. [Ans. 6000 erg]

28. Two soap bubble in vacuum having radii 1.5 cm and 2 cm respectively coalesce underisothermal conditions to form a single bubble. Calculate the radius of new bubble.

[Ans. 2.5 cm]

29. A film of water is formed between two straight parallel wires each 10 cm long and atseperation 0.5 cm. Calculate the work required to increase 1 mm distance between wires.(Surface tension of water is 72 × 10–3 Nm–1) [Ans. 1.44 × 10–5]

30. In increasing the area of a film of soap solution from 50 cm2 to 100 cm2, 3 × 10–4 J of workis done. Calculate the surface tension of soap solution. [Ans. 3.0 × 10–2 Nm–1]

31. A wire ring of 3 cm radius is rested flat on the surface of a liquid and is then raised. The pullrequired is 3.03 g more before the film breaks. Calculate the surface tension of liquid.

[Ans. 78.9 dyne cm–1]

32. Calculate the force required to take away a flat circular plate of radius 0.02 m from thesurface of water. The surface tension of water is 0.07 Nm–1. [Ans. 8.8 × 10–3 N]

33. A wire 0.1 m long is placed horizontally on the surface of water (surface tension 0.07Nm–1). Calculate the force needed to pull the wire from water surface. [Ans. 1.456 × 10–2 N].

34. The surface tension of soap solution is 0.03 Nm–1. 1.88 × 10–3 J of work is done to producea bubble of this soap solution. What will be the radius of it? [Ans. 0.05 m]

35. Calculate the work done in breaking a water drop of radius 1 mm into 1000 drops. Surfacetension of water is 72 × 10–3 Nm–1. [Ans. 8.14 × 10–6 J]

36. The length of a needle floating on water is 2.5 cm. How much minimum force in addition tothe weight of needle will be needed to lift the needle above the surface of water? Surfacetension of water is 7.2 × 10–4 N–1 cm–1. [Ans. 36 × 10–6 N]

BASED ON EXCESS PRESSURE IN DROPS AND BUBBLE37. Find the difference of air pressure between the inside and outside a soap bubble 5 mm in

diameter. Assume surface tension to be 1.6 Nm–1. [Ans. 2560 Nm–2]

38. Calculate the pressure inside a small air bubble of 0.1 mm radius situated just below thesurface of water? Surface tension of water = 7.2 × 10–2 Nm–1 and atmospheric pressure1.013 × 105 Nm–2. [Ans. 1.027 × 105 Nm–2]

39. If excess pressure inside a soap bubble is balanced by that due to a column of oil (specificgravity = 0.8) 3 mm high where R = 1.0 cm., find surface tension of the soap bubble.

[Ans. 5.88 × 10–2 Nm–1]40. Calculate the difference in air pressure between the inside and outside of a soap bubble of

radius 4 mm. Take surface tension of soap solution to be 1.4 Nm–1. [Ans. 1400 Nm–2]

41. Two soap bubbles have radii in the ratio 2 : 3. Compare the excess of pressure inside thesebubbles. Also compare the works done in blowing these bubbles. [Ans. 1.5, 4 : 9]


42. The air pressure inside a soap bubble of diameter 3.5 mm is 8 mm of water column abovethe atmospheric pressure. Calculate the surface tension of soap solution.

[Ans. 3.43 × 10–2 Nm–1]

43. The excess pressure inside a spherical drop of water of radius 10–3 m is 146 Nm–2. Calculatethe surface tension of water. [Ans. 7.3 × 10–2 Nm–1]

44. The pressure of air in a soap bubble is 8 × 10–3 m of water column above the atmosphericpressure. The surface tension of soap solution is 6.9 × 10–2 Nm–1. Calculate the radius of thebubble. [Ans. 3.5 mm]

45. A 0.02 cm liquid column balances the excess pressure inside a soap bubble of radius7.5 mm. Determine the density of the liquid. Surface tension of soap solution is 0.03 Nm–1.

[Ans. 8.2 × 103 kg m–3]

46. What is the pressure inside a drop of mercury of radius 3.00 mm at room temperature. Alsocalculate the excess pressure inside the drop. Given, S.T. of mercury is 4.65 × 10–1 Nm–1 andatmospheric pressure is 1.01 × 105 Nm–2. [Ans. 3.1 × 102 Nm–2, 1.01 × 105 Nm–2]

47. What would be the excess pressure above atmosphere inside an air bubble of 0.2 mm radiussituated just below the surface of water. The surface tension of water is 0.07 Nm–1.

[Ans. 700 Nm–2]

BASED ON CAPILLARITY48. Two capillary tube of radius 2.5 mm and 2.0 mm are held vertically inside water one by one.

How much high the water will rise in each tube. (Surface tension of water is 7 × 10–2 Nm–1,g = 10 ms–2) [Ans. 5.6 mm, 7 mm]

49. Water rises in a capillary tube to a height of 2.0 cm. In an other capillary whose radius is onethird of it how much water will rise? If the first capillary is inclined at an angle of 60° withthe vertical then what will be the position of the water in the tube?

[Ans. 6.0 cm, 4.0 cm]

50. Water rises to a height of 10 cm in a certain capillary tube. In the same tube the level ofmercury is depressed by 3.42 cm. Compare the surface tension of water and mercury. Takeangle of contact for water to be zero and for mercury 135°. [Ans. 1 : 6.57]

51. A capillary tube of radius 0.04 cm is dipped in a liquid of surface tension 9.8 × 10–2 Nm–1.Calculate the height to which the liquid will rise in capillary. [Ans. 5.0 cm]

52. Water rises in a capillary tube to a height of 2 cm. The surface tension of water at roomtemperature is 8 × 10–2 Nm–1. Calculate the radius of capillary tube. [Ans. 0.8 mm]

53. The tube of mercury barometer is 4 mm in diameter. Calculate the error in the reading due tosurface tension (Surface tension of mercury is 0.54 Nm–1, angle of contact 135°).

[Ans. 0.00288 m]

54. A capillary tube of radius 0.4 mm is dipped vertically in water. Find upto what height thewater will rise in the capillary, surface tension of water is 7.0 × 10–2 Nm–1.

[Ans. 3.57 cm]

55. A capillary tube of inner diameter 0.5 mm is dipped in mercury which is depressed down by2.1 cm in capillary with respect to outer liquid level. The specific gravity of mercury is 13.6calculate the surface tension of mercury. [Ans. .545 Nm–1]

56. Water rises in a Capillary tube to a height 2.0 cm. In an another capillary tube whose radiusis one third of it, how much the water will rise? If the first Capillary tube is inclined at anangle of 60° with the vertical then what will be the position of water in the tube.

[Ans. 6.0 cm, 4.0 cm]


BASED ON REYNOLD’S NUMBER VISCOSITY AND POISEUILLE’S FORMULA57. The flow rate from a tap of diameter 1.25 cm is 3 lt/min. The coefficient of viscosity of

water is 10–3 PaS. Characterise the flow. [Ans. turbulent]

58. A metal plate of area 0.10 m2 is connected to a 0.01 kgmass via a string that passes over an ideal pulley as shownin figure. A liquid with a film thickness of 0.3 mm is placedbetween the plate and the table. When released, the platemoves to the right with a constant speed of 0.085 ms–1.Find the coefficient of viscosity of the liquid.

59. Calculate the Reynold’s number for water flowing in a pipe of radius 2 cm, with an averagevelocity of 10 cm s–1. Given viscosity of water is 10–3 kg m–1 s–1 and density of water is 103

kg m–3. [Ans. 4000]

60. Check the dimensional consistency of the Poiseuilli’s formula

Q =4

1 2(P P )

8

r

l

Where symbols have their usual meaning.

61. A metal plate of area 0.02 m2 is lying on a liquid layer of thickness of 10–3 m and coefficientof viscosity of 120 poise. Calculate the horizontal force required to move the plate with aspeed of 0.025 ms–1. [Ans. 6 N]

62. What should be the maximum average velocity of water in a tube of diameter 2 cm so thatthe flow is luminar. The viscosity of water is 10–3 N m–2 S. [Ans. 0.1 ms–1]

63. A flat piece of 20 cm. square plate moves over another similar plate with a thin layer of 0.4cm of liquid between them. The coefficient of viscosity of liquid is 0.98 N S m–2. Calculatethe relative velocity between the two plates if a force of 1 kg wt is applied on it.

[Ans. 1 ms–1]

64. The difference of height between the two limbs of a manomometer connected at the ends of20 cm long tube of 1 mm radius is 28 cm. If 8 cm3 of water is collected in 3 minutecalculate coefficient of viscosity. [Ans. 1.212 Poise]

65. A metal plate of area 0.02 m2 is lying on a liquid layer of thickness 10–3 m and coefficient ofviscosity 120 poise. A force of 6 N is applied on the plate, calculate the speed of it.

[Ans. 0.025 ms–1]

66. A metal plate 0.04 m2 in area is lying on a liquid layer of thickness 10–3 m and coefficient ofviscosity of 140 poise. Calculate the horizontal force required to move the plate with a speedof 0.04 ms–1. [Ans. 22.4 N]

67. The relative velocity between two layers of water is 5.0 cm s–1. If the perpendicular distancebetween the layers is 0.1 cm find the velocity gradient. [Ans. 50 cm/s]

68. Show that if two capillaries of radii r1 and r2 having lengths l1 and l2 respectively are set inseries, the rate of flow Q is given by

Q =

1

1 24 4

1 2

P

8

l l

r r

Where p is pressure difference and is coefficient of viscosity of the liquid.

TM g

0.01 kg

F

Film


BASED ON STOKE’S LAW AND TERMINAL VELOCITY69. A drop of water of radius .021 × 10–3 cm. falls through air and covers 4.1 cm in 4 seconds.

Calculate the viscosity of air. The density of air is 0.001293 g cm–3.

[Ans 1.8 × 10–4 Poise]

70. A drop of water of radius 0.01 mm is falling through a medium whose density is 1.21 kgm–3. and coefficient of viscosity is 1.8 × 105 NS m–2. Find the terminal velocity of the drop.

[Ans. 1.46 × 10–5 ms–1]

71. Two exactly identical rain drops falling with terminal velocity of (2)1/3 ms–1, Coalesce toform a bigger drop. Find new terminal velocity of bigger drop. [Ans. 2 ms–1]

72. A metal sphere of radius 2 mm falls vertically in glycerine. Find the viscous force exerted bythe glycerine on the sphere where the speed of the sphere is 1 cm s–1 (Given viscosity ofglycerine is 8.0 poise) [Ans. 1.5 × 10–4 N]

73. An iron ball of radius 0.3 cm falls through a column of oil of density 0.94 g cm–3. It is foundto attain a terminal velocity of 0.5 cm s–1. Determine the viscosity of the oil. (density of iron= 7.8 g cm–3] [Ans. 268.9 poise]

74. Water is flowing through a pipe of nonuniform cross section with a velocity of 0.5 ms–1 andleaves the other end with a velocity of 0.7 ms–1. The pressure of water at one end is 103 N m–2.Calculate the pressure at the other end.

75. A gas bubble rises steadily at the rate of 2.5 × 10–3 ms–1 through a solution of density 2.25× 103 kg m–3. Calculate the radius of gas bubble if viscosity of liquid is 1960 poise.

[Ans. 2 cm]

76. Show that if n equal rain dropelets falling through air with equal steady velocity of 10 cm s–1

coalesce, the resultant drop attains a new terminal velocity of 10 n2/3 cm s–1.

[Ans. 10 n2/3 cm s–1]

77. The terminal velocity of a copper ball of radius 4.0 mm is falling through a tank of oil is 13.0cm s–1. Calculate the viscosity of oil. (Density of oil is 1.5 × 103 kg m–3, density of Copperis 8.9 × 103 kg m–3] [Ans 1.984 Nsm–2]

BASED ON CONTINUITY EQUATION AND BERNOULLI’S THEOREM78. Water is flowing through a cylindrical pipe of cross sectional area 0.09 m2 at a speed of 2.0

ms–1. If the diameter of the pipe is halved, then find the speed of flow of water through it.[Ans. 8.0 ms–1]

79. Water is flowing through a horizontal tube of non uniform cross section. At a place the radiusof the tube is 0.5 cm and velocity of water at that place is 20 cm s–1. What will be the radiusof the tube at one other place where speed of water is 5 cm s–1. [Ans. 1 cm]

80. Velocity of flow of water in a horizontal pipe is 20 ms–1. Find the velocity head of water.(g = 10 ms–2) [Ans. 20 m.]

81. Water flows into a horizontal tube whose one end is closed with a valve and the reading of apressure gauge attached to the pipe is 3 × 105 Nm–2. This reading of the pressure falls to 1× 105 Nm–2, when the valve is opened. Calculate the speed of water flowing into the pipe.

[Ans. 20 ms–1]

82. Water flows out of a small hole in the wall of a large tank near its bottom. What is the speedof efflux of water when the height of water level in the tank is 10 m. (g = 9.8 ms–1]

[Ans. 14 ms–1]

83. What volume of water will escape per minute from a tank through an opening 0.5 cm inradius and 5 m below the level of water. [Ans. 4.7 × 10–2 m3]


84. The reading of pressure meter attached with a closed pipe is 4.0 × 105 Nm–2. On opening thevalve of the pipe, the reading of the pressure meter is reduced to 3.0 × 105 Nm–2. Calculate

the speed with which water is flowing in pipe. [Ans. 10 2 ms–1]

85. A pilot tube is fixed in a main pipe of diameter 20 cm and difference of pressure indicatedby the gauge is 5 cm of water column. Find the volume of water flowing through main pipein one minute. [Ans. 1.87 m3]

86. A horizontal tube has different cross sectional area at points A and B. The diameter of A is4 cm and that of B is 2 cm. Two manometer limbs are attached at A and B. Where a liquidof density 8.0 g cm–3 flows through the tube, the pressure difference between the limbs of themanometer is 8 cm. Calculate the rate of flow of the liquid in the tube.

[Ans. 406 cm3 s–1]

87. Water at a pressure 4 × 104 Nm–2 flows at 2 ms–1 through a pipe of 0.02 m2 cross sectionalarea which reduces to 0.01 m2. What is the pressure in the smaller cross section of the pipe?

[Ans. 3.4 × 104 Nm–2]

88. Water is flowing through two horizontal pipes of different diameters are connected together.The speed of water in the first pipe is 4 ms–1 and the pressure is 2.0 × 104 Nm–2. The densityof water is 1 × 103 kg m–3. Calculate the speed and pressure of water in the second pipe. Thediameter of the pipes are 2 cm and 4 cm respectively. [Ans. 1 ms–1, 2.75 × 104 Nm–2]

89. A plane is in level flight at constant speed and each of its two wings has an area of 25 m2. Ifthe speed of air is 180 km h–1 over the lower wing and 234 km h–1 over the upper wingdetermine the plane mass. (density of air is 1 kg m–3) [Ans. 4400 kg]

90. The diameter of a pipe at two points where a venturimeter is connected is 8 cm and 5 cm anddifference of level in it is 4 cm. Calculate the volume of water flowing through the pipe persecond. [Ans. 1889 cm3 s–1]

91. Water stands at a depth H in a tank whose sides are vertical. A holeis made in one of the walls at a depth h, below the water surface.Calculate

(i) velocity of efflux

(ii) the range of water stream

(iii) the valve of h for which the range is maximum.

[Ans. v = 2 gh , R = 4 (H )h h ;H = 2h].

92. At what speed the velocity head of a stream of water be equal to 25 cm of mercury column?(g = 10 ms–2.) [Ans. 8.25 ms–1]

93. Water flows horizontally through a pipe line of non uniform area of cross section. If thepressure of water equals to 10 cm of Hg at a point where velocity of flow is 40 cm s–1. Whatis the pressure at another point where the velocity of flow is 50 cm s–1.

[Ans. 1.33 × 104 Nm–2]

94. Water flows in to a horizontal pipe whose one end is closed with a valve and the reading ofa pressure gauge attached to a pipe is 3 × 105 Nm–2. This reading of the pressure gauge fallsto 1 × 105 Nm–2 when the valve is opened. Calculate the speed of water flowing into thepipe. [Ans. 20 ms–1]

S

P

Qh

H

R

381

HeatHeat is a form of energy which flows from a body at high temperature to a body at low

temperature. S.I. Unit of heat is Joule. Dimensions –[ML2T–2]

TemperatureTemperature is that thermal property which has same value for two or more than two bodies

in thermal equilibrium.

ThermometerIt is an instrument which is used to measure the temperature of the body. Every

thermometer is based on a property called thermometric property e.g. Resistance of conductor,pressure, volume etc.

Thermometric PropertyIt is the property of a substance which changes regularly with temperature. For example,

resistance of a conductor increases with rise in temperature.

Temperature ScalesVarious temperature scales are —

(a) Celsius Scale: On this scale lower fixed point (melting point of ice) is at 0°C and upperfixed point (boiling point of water) is at 100°C. The distance between the two fixed points isdivided into 100 equal divisions.

(b) Fahrenheit Scale: On this scale lower fixed point is at 32°F and upper fixed point is at212°F. The distance between them is divided into 180 equal divisions.

(c) Kelvin Scale: The lower fixed point is at 273K and upper fixed point is at 373K. Thedistance between them is divided into 100 equal divisions.

Relation Between Different Temperature ScalesLet C, F and K are the temperatures of a body on Celsius, Fahrenheit and Kelvin scale,

respectively.

�� 12UNIT

100°C212°F

373K

K

273K32°F0°C

C F

Cels iusScale

Fahrenhe itScale

Ke lv inScale


0

100 0

C =

32 273

212 32 373 273

F K

100

C=

32 273.

180 100

F K

5

C=

32 273.

9 5

F K

We may also write above equation as

C =5

( 32)9

F or F = 9

325

C

C = K – 273 or K = 273 + C

F =9 9

( 273) 32 4595 5

K K

Important1. Let a physical quantity X has value Xt, X100 and X0 at t°C, 100°C and 0°C respectively.

Their unknown temperature t is given by

t =0

100 0

100.tX X

X X

2. Let Rt is the resistance of a conductor at t°C and R0 is the resistance at 0°C. Then

Rt = R0 (1 + t)

Where is temperature coefficient of resistance.

Linear ExpansionWhen a rod is heated to increase its temperature, then its length undergoes a change called

linear expansion.

l2 = l1 [1 + (t2 – t1)

where l2 is length at t2°C

l1 is length at t1°C and is called coefficient of linear expansion.

= 2 1

1 2 1

( )

( )

l l

l t t

It is defined as the increase in length per unit length per degree rise in temperature.

Areal or Superficial ExpansionWhen a solid sheet is heated, its area undergoes a change. It is called superficial expansion.

A2 = A1 [1 + (t2 – t1)]

where A1 = area of sheet at t1°C

A2 = area of sheet at t2°C

Thermal Properties of Matter 383

and = 2 1

1 2 1( )

A A

A t t

is called coefficient of superficial expansion,

which is defined as the increase in surface area per unit surface area per unit rise in temperature.

Cubical or Volume ExpansionWhen a solid body (in three dimensions) is heated, its volume will change. It is called

volume or cubical expansion.

V2 = V1 [1 + (t2 – t1)].

where V1 = Initial volume at t1°C

V2 = Final volume at t2°C

and =2 1

1 2 1

( )

( )

V V

V t t

is called coefficient of cubical expansion,

which is defined as increase in volume of the body per unit volume per degree increase intemperature.

, or has unit per degree Celsius or per Kelivn (°C–1 or K–1).Dim. [K–1] or [–1]

Relation between , and :

= .2 3

Joule’s Mechanical Equivalent of HeatSr. Joule performed an experiment to show that mechanical work is equal to heat. Let W

joule of mechanical work is converted into heat energy then

W Q

or W = JQ.

where J is called mechanical equivalent of heat. Mechanical equivalent of heat is numericallyequal to the work to be done to produce unit amount of heat.

J = 4.2 J Cal–1.

Specific HeatThe amount of heat required to raise the temperature of m mass of a substance by t is

given by

Q = m s t.

where S is specific heat.

s = .Q

m t

Specific heat of a substance is defined as the amount of heat required to raise thetemperature of unit mass of that substance by one Kelvin (degree Celsius)

S.I. unit J Kg–1 k–1

Dimension [M°L2T–2K–1].


Molar Specific HeatSometimes we have to express mass in moles. The molar specific heat is defined as the

amount of heat required to raise the temperature of one mole of that substance through oneKelvin.

S.I. Unit Joule/mole-K. (J mol–1 k–1]

Heat Capacity or Thermal CapacityWe know Q = ms t

Let t = 1°CQ = ms.

Here Q is called heat Q thermal capacity which may be defined as the amount of heatrequired to raise the temperature of a body through our degree Celsius or one Kelvin.

S.I. Unit Joule K–1 [ML2 T–2 K–1].

Water EquivalentThe water equivalent of a body is that mass of water which requires same amount of heat

as is required by that body for the same rise in temperature.Water equivalent of a body W = mass × specific heat

W = ms.

Principle of CalorimetryAccording to principle of calorimetry, when heat exchange takes place between two bodies,

then heat lost by hot body is equal to heat gained by cold body.

i.e. Heat gained = Heat lost.

Specific Heats of a GasThe temperature of gas may be increased by two different methods: (1) by keeping pressure

constant and (2) by keeping volume constant. Correspondingly in case of gases there are twodifferent types of specific heats:

(a) Specific heat at constant volume: It is defined as the amount of heat required to raisethe temperature of unit mass of gas through 1°C at constant volume. It is denoted by CV.

(b) Specific heat at constant pressure: It is defined as the amount of heat required to raisethe temperature of unit mass of gas through 1°C at constant pressure. It is denoted by Cp. Cp isalways greater than Cv.

Also Cp – Cv = R (Universal Gas Constant)

R = 8.3I Joule/mol – K.

Degree of FreedomThe total number of independent variables required to describe the state of a gas molecule

completely is called degree of freedom.

The number of degree of freedom is given by

n = 3 N – k.

n = no. of degree of freedoms

N = no. of independent relations between the particles.

For example — For a diatomic gas N = 2 and k = 1

n = 3 N – k

= 3 (2) – 1= 5


Law of Equipartition of EnergyAccording to this law, the energy of the system is equally distributed among the various

degree of freedom. Also the energy associated with each degree of freedom per molecule is 1

kT2

,

where k is Boltzmann’s constant.

SOLVED EXAMPLES

BASED ON TEMPERATURE AND MEASUREMENT OF TEMPERATUREExample 1. The triple point of neon and carbon dioxide are 24.57 k and 216.55 k

respectively. Express these temperatures on the Celsius and Fahrenheit scales.

Solution. Here

Triple point for neon = 24.57K.

Now C = K – 273.15

= 24.57 – 273.15

= – 248.58°C.

Also F =9

(C) 325

=9

5 (– 248.58) + 32 = – 415.44°F

For Carbon dioxide

Triple point = 216.55k.

C = K – 273.15

= 216.55 – 273.15

= – 56.6°C

and F =9

(C) 325

=9

5 (– 56.6) + 32

= – 69.88°C.Example 2. A platinum wire has resistance of 10 at 0°C and 20 at 273°C. Find the

value of coefficient of resistance.

Solution. Here t1 = 0°C R1 = 10 t2 = 273°C R2 = 20 .

Using formula =2 1

1 2 1

R R 20 10

R ( ) 10(273 0)t t

=1

273°C–1.


Example 3. Two absolute scales A and B have triple points of water defined to be 200A and350B. What is the relation between TA and TB?

Solution. A

B

T

T=

200 4

350 7

or TA = B4

T .7

Example 4. The electrical resistance in ohms of a certain thermometer varies withtemperature, According to approximate law

R = R0 [1 + 5 × 10–3 (T – T0)]

The resistance is 101.6 at the triple point of water and 165.5 at normal melting pointof lead (600.5K). What is the temperature when the resistance is 123.4?

Solution. Here T = 273 K R = 101.6 and at T = 600.5 K R = 165.5 .

101.6 = R0 [1 + 5 × 10–3 (273 – T0)] ... (i)

and 165.5 = R0 [1 + 5 × 10–3 (600.5 – T0)] ... (ii)

Divide Equation (ii) by (i)

165.5

101.6=

30

30

1 5 10 (600.5 )

1 5 10 (273 )

T

T

On solving T0 = – 49.3 k.

Using Equation (i)

R0 = 3

101.6.

1 5 10 [273 ( 49.3)] = 38.9.

For R = 123.4123.4 = 38.9 [1 + 5 × 10–3 (J + 49.3]

or T = 384.8K.Example 5. There is a certain temperature which has same reading on both Centigrate and

Fahrenheit thermometers. Find the temperature.

Solution. Let the temperature having same value on Centrigrade and Fahrenheit scale is x.

32

180

x = .

100

x

x – 32 =9

5x

or x = – 40

Temperataure is – 40°C or – 40°F.Example 6. A faulty thermometer has its fixed points marking at 5 and 95. What is the

correct temperature in Centigrade when this thermometer reads 59?

Solution. Let the reading on Centrigrade thermometer be C

C

100=

R L.F.P.

U.F.P. L.F.P.


C

100=

59 5

95 5

or C = 60°C.Example 7. Two ideal gas thermometers A and B use oxygen and hydrogen respectively.

The following observations are made:

Temperature Pressure PressureThermometer A Thermometer B

Triple point of water 1.250 × 105 Pa 0.200 × 105 Pa

Normal melting point

of sulphur 1.797 × 105 Pa 0.287 × 105 Pa

What is the absolute temperature of normal melting point of sulphur as read bythermometers A and B?

Solution. (i) For pressure thermometer A:

Ttr = 273 K

Ptr = 1.250 × 105 Pa.

P = 1.797 × 105 Pa

Normal freezing point of sulphur

T =5

tr 5tr

P 1.795 10 273T

T 1.250 10

= 392.46 k.(ii) Similarly, for pressure thermometer B:

T =5

5

0.287 10 273

0.200 10

= 391.75 k.Example 8. The resistance of the platinum wire of a platinum resistance thermometer at ice

point is 5 ohms and at the steam point is 5.93 ohms. In a hot bath the resistance is measured tobe 5.795 ohm. Calculate the temperature of the hot bath.

Solution. Here R0 = 5R100 = 5.93 and Rt = 5.795

t =0

100 0

R R100

R Rt

=5.795 5

1005.93 5

= 85.48°C.

BASED ON THERMAL EXPANSION (LINEAR EXPANSION SUPERFICIAL ANDCUBICAL EXPANSION

Example 9. A steel tape 1m long is correctly calibrated for a temperature of 270°C. Thelength of a steel rod measured by this tape is found to be 63.0 cm. on a hot day when thetemperature is 45.0°C. What is the actual length of the steel rod on that day? What is the length


of the same steel rod on a day when the temperature is 270°C. Coefficient of linear expansion ofsteel is 1.20 × 10–5/°C.

Solution. Here t1 = 27°C l1 = 63.0 cm.

t2 = 45°C l2 = ? = 1.20 × 10–5 / °C.

l2 = l1 [1 + (t2 – t1)]

= 63 [1 + 1.20 × 10–5 (45 – 27)]

= 63.0136 cm.The steel tape has been calibrated for at 27°C, so length of steel rod at 27°C = 63 cm.Example 10. A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of

same length and diameter. What is the change in the length of the combined rod at 250°C, if theoriginal lengths are at 40.0°C? Is there a thermal stress developed at the junction? The ends ofthe rod are free to expand. Coefficient of linear expansion of brass = 2.0 × 10–5 °C–1 and that ofsteel = 1.2 × 10–5 °C–1.

Solution. Here Lb = 50 cm t1 = 40°C t2 = 250°C b = 2.0 × 10–5 °C–1.

Change in length of brass rod

lb = Lb b t

= 50 × 2.0 × 10–5 × 210

= 0.21 cm.Now Ls = 50 cm. t1 = 40°C t2 = 250°C 5 = 1.2 × 10–5 °C–1

Ls = ls s t

= 50 × 1.2 × 10–5 × (250 – 40)

= 0.13 cm.

Total change in the length of the combined rod

= 0.21 + 0.13

= 0.34 cm.The ends of the rod are free to expand so no thermal stress is developed at the junction.Example 11. A brass wire 1.8m long at 27°C is held tant with little tension between two

rigid supports. If the wire is cooled to a temperature of – 39°C, what is the tension developed inthe wire if its diameter is 2.0 mm?

Coefficient of linear expansion of brass = 2.0 × 100–5 C–1.

Young’s modulus of brass is 0.91 × 1011 Pa.Solution. Here l = 1.8 m, t1 = 27°C t2 = – 39°C

r = 1.0 mm = 1.0 × 10–3 m.

= 2.0 × 10–5 °C–1, = 0.91 × 1011 Pa.

Using formula L = L t

Also =2

L

L

mg

r .

mg =2 L

L

r =

2 L

L

r t

= r2 t.

T = mg = 0.91 × 1011 × 3.14 × (1.0 × 10–3)2 × 2 × 10–5 × 66

= 0.91 × 3.14 × 2 × 66

= 3.77 × 102 N.


Examle 12. An iron ring which is 1 m in diameter is to be shrunk on a pulley which is1.005 m in diameter. If the temperature of the ring is 10°C, find the temperature to which it mustbe raised so that it will just slip on the circumference of the pulley. Coefficient of linearexpansion of iron is 0.000012 per °C.

Solution. Diameter of the ring at 10° = 1 m

Circumference of the ring = D = × 1 = m

Diameter of pulley = 1.005 m

Circumference of the pulley = × 1.005 m

Then iron ring is to be heated to such a temperature that its circumference becomes equalto that of the pulley.

Suppose the final temperature to which the ring must be heated

= t2

Original temperature = t1 = 10°C

Rise in temperature = t = (t2 – t1) = (t2 – 10)

Here, Lt = × 1.005 m; L0 = m,

= 0.000012 per °C

Lt = L0 (1 + t)

× 1.005 = × 1 [1 + 0.000012 (t2 – 10)]

1.005 = 1 + 0.000012 (t2 – 10)

t2 = 426.6 °CExample 13. The inner volume of a hollow brass sphere is 1000 cc at 0°C. What would be

its volume at 100°C? ( for brass = .000018 per C°). (D.H.S.)

Solution. Original volume = V0 = 1000 cc

t1 = 0°C, t2 = 100°C

Rise in temp. t = t2 – t1 = 100 – 0 = 100°C

= 0.000018 per °C = 3 = 3 × 0.00018 per °C

= 0.000054 per °C

Vt = V0 (1 + t)

= 1000 (1 + 0.000054 × 100)

Vt = 1000 + 5.4 = 1005.4 cc.Example 14. A brass disc at 20°C has a diameter of 30 cm and a hole cut in the centre is

10 cm in diameter. Calculate the diameter of the hole when the temperature of the disc is raisedto 50°C. ( for brass = 0.000018 per °C).

(P.U.)

Solution. Diameter of the hole = L = 10 cmLt = ?t1 = 20°C, t2 = 50°C,

Rise in temp. = t = (50 – 20) = 30°C = 0.000018 per °CLt = L (1 + t)

= 10 (1 + 0.000018 × 30)Lt = 10 + 0.0054 = 10.0054 cm

Hence the diameter of the hole at 50°C = 10.0054 cm


Example 15. A glass rod when measured with a zinc scale, both being at 20°C, appears tobe 1 metre long. If the scale is correct at 0°C, what is the true length of the glass rod at 0°C?

( for glass = 8 × 10–6 and for zinc = 26 × 10–6)

Solution. Suppose the original length of the zinc scale at 0°C = L0 = 100 cm

The original length of the glass rod at 0°C = L0 = ?

Final length of zinc scale at 20°C = LtFinal length of glass rod at 20°C = Lt

Rise in temperature = 20 – 0 = 20°C

Coefficient of linear expansion for zinc = = 26 × 10–6

Coefficient of linear expansion for glass = 8 × 10–6

Here, Lt = Ltor L0 (1 + t) = L0 (1 + t)

L0 = 0L (1 )

(1 )

t

t

= L0 (1 + t) (1 + t)–1

L0 = L0 (1 + t) (1 – t) = L0 [1 + ( – ) t]

L0 = 100 [1 + (26 × 10–6 – 8 × 10–6) × 20]

= 100 + 100 × 18 × 10–6 × 20

= 100 + 0.036

= 100.036 cm

Hence the true length of glass rod at 0°C = 100.036 cmExample 16. The density of a substance is 8.92 at 0°C and 8.80 at 80°C. Calculate its

coefficient of linear expansion.

Solution. Here, D0 = 8.92, Dt = 8.80, Rise in temp. = 80°C

Dt = 0D

(1 )t

8.80 =8.92 8.92

or 1 801 80 8.80

or =8.92 8.80 0.12

8.80 80 704

= 0.00017

=0.00017

3 3

= 0.0000567 per °C

Example 17. A rod of brass and that of iron differ by 5 cm in length at all temperatures.What are their lengths at 0°C if the coefficients of linear expansion of iron and brass 11 × 10–6

°C–1 and 18 × 10–6/°C respectively? (Bombay)

Solution. Length of the iron rod at 0°C = L0

Length of the brass rod at 0°C = L0L0 – L0 = 5 cm (given)

= 11 × 10–6/°C for iron

= 18 × 10–6/°C for brass

Suppose rise in temp. = t°C


Lt = L0 (1 + t)

Lt = L0 (1 + t)

Lt – Lt = L0 (1 + t) – L0 (1 – t)

Lt – Lt = (L0 – L0 ) + t (L0 – L0 )But Lt – Lt = 5 cm

L0 – L0 = 5 cm

5 = 5 + t (L0 – L0 )t (L0 – L0) = 0 ; L0 = L0

or L0 × 11 × 11–6 = L0 × 18 × 10–6

L0 = 011 L

18But L0 – L0 = 5

00

11LL

18 = 5

07 L

18= 5

L0 =90

7 = 12.85 cm

L0 = 12.85 – 5 = 7.85 cm.Example 18. A clock which keeps correct time at 0°C has a pendulum made of iron. How

many seconds will it gain or lose per day when the temperature rises to 30°C? ( for iron =0.000012). (Roorkee E.E.)

Solution. Here, the temperature rises and hence the clock will lose time.

At 0°C, T1 = 0L2

g ... (1)

At 30°C, T2 = 30L2

g ... (2)

Dividing (2) by (1)

2

1

T

T=

30 0

0 0

L L (1 )

L L

t

2

1

T

T=

121 (1 ) 1

2

tt t

Suppose T1 = 1 s

T2 =0.000012 30

1 12 2

t T2 = (1 + 0.00018) s

Number of vibrations made in one day


=2

86400

T

=86400

(1 0.00018) = 86400 (1 + 0.00018)–1

= 86400 (1 – 0.00018)

Loss per day = 86400 – 86400 (1 – 0.00018)

= 86400 × 0.00018 = 15.552 sExample 19. A steel rod 25 cm long has a cross sectional area of 0.8 sq cm. What force

would be required to stretch this rod by the same amount as the expansion produced by heatingit through 10°C? (I.I.T. Entrance Exam.)

Y for steel = 2 × 1012 dynes/cm2

for steel = 10–5 / °C

Solution. L = 25 cm a = 0.8 cm2

= 10–5 / °C, (t2 – t1) = 10°C

l = L (t2 – t1)

= 25 × 10–5 × 10 = 25 × 10–4 cm

Y =F × L

a l

F =12 42 10 0.8 25 10

25

Y al

L

F = 1.6 × 108 dynes = 1.6 × 102 newtonsExample 20. A clock which has a brass pendulum beats seconds correctly when the

temperature of the room is 30°C. How many seconds will it gain or lose per day when thetemperature of the room falls to 10°C? Take for brass = 0.000019 / °C.

(All India H.S. 1974)

1 day = 86400 s

Solution. As the temperature decreases, the clock will gain time

In the case of a pendulum,

T =L

2g

At 30°C, T1 =30L

2g

.. (i)

At 10° C, T2 =10L

2g

... (ii)


2

1

T

T =10

30

L2

L


=0

0

L (1 10)

L (1 30)

= (1 10 )(1 30 )

= 1 20

= (1 – 10 )

Since the clock keeps correct time at 30°C

T1 = 1 s

T2 = (1 – 10 )

= (1 – 10 .000019)

= (1 – 0.00019) s

Number of vibrations made in one day

=86400

(1 0.00019)

= 86400 (1 – 0.00019)–1

= 86400 (1 + 0.00019)

Gain per day = 86400 (1 + 0.00019) – 86400

= 16.42 s.Example 21. A piece of metal weighs 46g in air. When it is immersed in a liquid of specific

gravity 1.24 at 27°C it weighs 30 g. When the temperature of the liquid is raised to 42°C, themetal piece weighs 30.5 g. Specific gravity of the liquid at 42°C is 1.20. Calculate coefficient ofthe linear expansion of the metal. (IIT)

Solution. Volume of metal at 27°C = Volume of liquid displaced

=Loss in weight

Sp. gravity of the liquid

V1 =46.30 16

1.24 1.24

= 12.90 cm3.

Similarly at 42°C V2 =46 30.5

1.20

=15.5

1.20 = 12.92 cm3.

V2 = V1 [1 + (t2 – t1)]

or = 2 1

1 2 1

12.92 12.90

( ) 12.90(42 27)

V V

V t t

= 1.034 × 10–4 / °C.


or =41.034 10

z z

= 3.447 × 10–5 / °C.Example 22. The coefficient of volume expansion of glycerine is 49 × 10–5 °C–1. What is

the fractional change in its density for a 30°C rise in temperature?

Solution. Let M be the mass of glycerine, 0 its density at 0°C and t its density at t °C .Then

= 0

0

tV V

V t

density =Mass

Volume or V =

M.

=0

0

M M

t

Mt

=0

0

t

t

Fractional change in density is

0

0

t = t

= 49 × 10–5 × 30

= 0.0147Example 23. An iron sphere has a radius of 10 cm at a temperature of 0°C. Calculate the

change in the volume of the sphere if its final temperature is 80°C. Given coefficient of cubicalexpansion is 33 × 10–6 °C–1.

Solution. Here r = 10 cm

t = 80 – 0 = 80°C

= 33 × 10–6 °C–1.

Change in volume V = V t

=34

3r t

=3 64

3.14 (10) 33 10 803

= 11.04 Cu3.

Example 24. The density of a solid is defined as mass per unit volume is M

.V

Prove. =1 d

dt

Solution. Given =M

V


or = MV–1

Differentiating above eqn. w.r.t. temperature

d

dt

= – mV–2

dV

dt

=2

m dV

dtV

ord

dt

=

V

V V

m d

dt

ord

dt

=

V

m = 2 1

1

V V

V t

or = 1

.dV

V dt

or =1

.d

dt

BASED ON MECHANICAL EQUIVALENT OF HEATExample 25. The height of Niagra falls is 50 metres. Calculate the difference between the

temperature of water at the top and the bottom of the falls. (J = 4.2 × 107 ergs/cal)

(P.U. ; Gau. U.)

Solution. h = 50 metres = 5000 cm

Suppose difference of temperatures = °C = ?

Mass of water = m

Work done = W = mgh = m × 980 × 5000 ergs

Heat produced H = mS = m × 1 × calories

W = JH

m × 980 × 5000 = 4.2 × 107 × m ×t

t = 7

980 5000

4.2 10

= 0.117°C

Example 26. A piece of lead (Sp heat = 0.03) falls from a height of 20 metres. If onreaching the ground 50% of its energy is converted into heat, calculate the rise in temperature.

Solution. Suppose mass of lead = m, Sp. heat = 0.03

h = 20 metre = 2000 cm

Rise in temperature = = ?

Total work done = mgh = m × 980 × 2000 ergs

As 50% of the work is converted into heat, useful work done

W =980 2000

2

m

H = mS = m × 0.03 × W = JH


or980 2000

2

m = 4.2 × 107 × m × 0.03 × t

t = 7

980 2000

2 0.03 4.2 10

= 0.778°C

Example 27. A lead bullet strikes a target with a velocity of 480 m/s. If the bullet falls dead,calculate the rise in temperature assuming that all the heat developed is equally shared betweenit and the target. (Sp. heat of lead = 0.03)

Solution. Suppose mass of the bullet = m grams

Rise in temperature = t = ?

Velocity v = 480 m/s = 48000 cm/s

Work done = 2 21 1(48000)

2 2mv m ergs

H =2

7

(48000)cal

2 4.2 10

W m

J

Half of this heat energy is used to raise the temperature of the bullet.

Useful heat H =2

7

(48000)

2 2 4.2 10

m

cal

mst =2

7

(48000)

2 4.2 10

m

t = 7

48000 48000

0.03 4 4.2 10

= 457.14°C

Example 28. In one of his famous experiments, Joule churned water with a paddle wheeldriven by two loads each of mass 14 kilograms and each falling through a vertical height of 2.1metres. He had 6 kilograms of water (including water equivalent of calorimeter and paddle etc.)to be heated. After each churning he hauled the load up and let them fall again. Whattemperature rise would you expect him to find after 50 falls?

Solution. J = 4.2 J/cal, g = 9.80 m/s2

Here, n = 50, m = 14 kg, g = 9.80 m/s2, h = 2.1 m

Work done = 2n × mgh = 2 × 50 × 14 × 9.8 × 2.1 joules

Heat produced, H =2 50 14 9.8 2.1

4.2

W

J

calories

= 6860 calories

Mass of water = 7 kg = 7000 g

Suppose, rise in temp. = t°C

H = 7000 × t

7000 t = 6860

t = 0.98°C


Example 29. Calculate the increase in energy per atom of aluminium in ergs when thetemperature of a piece of aluminium increases by 1°C (27 g of aluminium contains 6 × 1023

atoms and sp heat of aluminium = 0.22).

Solution. Heat required to raise the temperature of 27 g of aluminium by 1°C

= 27 × 0.22 × 1 cal

Energy gained by 6 × 1023 atoms of aluminium

= 27 × 0.22 × 4.2 × 107 ergs

Increase in energy per atom of aluminium

=7

23

27 0.22 4.2 10ergs

6 10

= 4.158 × 10–16 ergsExample 30. A petrol engine consumes 25 kg of petrol per hour. The calorific value of the

fuel is 11.4 × 106 cal per kg. The power of the engine is 99.75 kilowatts. Calculate the efficiencyof the engine. (All India H.S. 1971)

Solution. Heat produced per hour = 25 × 11.4 × 106 cal

Heat produced per second =625 11.4 10

cal3600

Input energy per second =625 11.4 10 4.2

3600

watts

Output energy = 99.75 kilowatts

= 99750 watts

% Efficiency =Output

100Input

= 6

99750 100 3600

25 11.4 10 4.2

= 30%

Example 31. A steel drill making 180 revolutions per minute is used to drill a hole in ablock of steel. The mass of the steel block and drill is 180 grams. If the entire mechanical workis used up in producing heat and the rate of rise in temperature of the block is 0.5°C per second,find (a) the rate of working of the drill in watts and (b) the couple required to drive the drill. (Sp.heat of steel = 0.10 cal/g– °C. (I.I.T. Entrance. Exam. 1969)

Solution. Heat gained by the steel block and drill in one second

= mst

Here m = 180 g, S = 0.1,

t = 0.5°C per second

Heat gained in one second H = 180 × 0.1 × 0.5

= 9 cal/s

Work done per second = JH

W = 4.2 × 9 = 37.8 joules/sW = 37.8 watts


Let C be the couple required to drive the drill and the angle of rotation in one second, =3 × 2

= 6 radians

W = C

C =37.8

6

W

= 2 newton-metre.

Example 32. A heavy box of mass 400 kg is moved along the floor for a distance 20 m. Ifthe coefficient of sliding friction is 0.2, how many k cal of heat are produced? (g = 10 ms–2)

Solution. Here m = 400 kg, S = 20 m = 0.2.

Force of friction = mg

Work done against force of friction

W = ( mg) S

= 0.2 × 400 × 10 × 20

= 16,000 J.

Heat produced Q = 3

16,000

4.2 10

W

J

= 3.809 k cal.Example 33. Calculate the height to which a 60 kg man can climb, by eating a bread piece

which produces 8 × 104 calorie. The efficiency of human body is 30%.

Solution. Let height climbed by man = h

Increase in potential energy = mgh

Energy utilised by man, by a piece of bread = 4308 10

100

= 24 × 103 calorie

= 24 × 103 × 4.2 J.

mgh = 24 × 4.2 × 103

or h =324 4.2 10

60 10

= 168 m.

BASED ON SPECIFIC HEAT AND LATENT HEATExample 34. A geyser heats water flowing at the rate of 3.0 litres per minute from 27°C to

77°C. If the geyser operates on a gas burner, what is the rate of consumption of the fuel if theheat of combustion is 4.0 × 104 J/g?

Solution. Here,Volume of water flowing per second = 3.0 lt/min.

t = 77 – 27 = 50°C.

Latent heat of combustion is 4.0 × 104 J/g.

Heat absorbed by water Q = m s t.

= p × 4.2 × 103 × 50


= 63 × 104 J min–1.

Rate of combustion of fuel =4

4

63 10 J / min

J /4 10 g

= 15.75 g min–1.Example 35. In an experiment on specific heat of a metal a 0.20 kg block of the metal at

150°C is dropped in a copper calorimeter (of water equivalent 0.025 kg) containing 150 Cu3 ofwater at 27°C. The final temperature is 40°C. Compute the specific heat of the metal.

Solution. Here,

Mass of metal block m = 0.20 kg = 200 g.Fall in the temperature of block = (150 – 40) = 110°C

Heat lost by metal block = (200 × S × 110) Cal

where S is specific heat of metal in calorie.

Volume of water in calorimeter = 150 Cu3.

Mass of water = 150 × 1 = 150 g

density of water = 1 g Cu–3

Water equivalent of calorimeter W = 0.025 Kg = 25 g

According to principle of calorimetery,

Heat lost by metal block = Heat gained by water and calorimeter

200 × S × 110 = (150 + 25) × 1 × (40 – 27)

or S =175 13

200 110

= 0.1 cal g–1 °C–1

Example 36. A 10 kw drilling machine is used to drill a bore in a small aluminium blockof mass 8.0 kg. How much is the rise in temperature of the block in 2.5 minutes, assuming 50%of power is used up in heating the machine itself lost to the surroundings.

Sp heat of aluminium = 0.91 J g–1 °C–1.

Solution. Here P = 10 kW = 10 × 103 W S = 0.91 × 103 J Kg–1 °C–1

t = 2.5 min = 2.5 × 60 s, m = 8.0 kg

Heat energy absorbed by aluminium block

= (p t) × 50

100

= (10 × 103 × 2.5 × 60) × 1

2= 0.75 × 106 J

Let rise in temperature is t i.e.

= m S t

or t =6

3

0.75 10

8.0 0.91 10ms

� 103°C.


Example 37. A copper block of mass 2.5 kg is heated in a furnace to a temperature of500°C and then placed on a large ice block. What is the maximum amount of ice that can melt?

Sp. heat of copper = 0.39 J g–1 °C–1

Latent heat of fusion of water = 335 J g–1.

Solution. Here

Mass of copper block M = 2.5 kg = 2.5 × 103 g

Sp. heat of copper S = 0.39 J g–1 °C–1

t = 500 – 0 = 500°C

Let mass of ice melted = m

Latent heat of fusion = 335 J g–1

Heat lost by copper block = Heat gained by ice

(2.5 × 103) × 0.39 × 500 = m × 33

m =32.5 10 0.39 500

335

= 1455.2 g

= 1.455 kgExample 38. 5 grams of steam at 100°C is passed into a calorimeter containing a liquid and

the temperature of the liquid is found to rise from 32°C to 40°C. If the latent heat of steam is540 calories per gram, calculate the water equivalent of the calorimeter and its contents.

Solution. Mass of steam = 5 g

Temp. of steam = 100°C

Latent heat of steam = 540 cal per gram

Suppose water equivalent of calr. and its contents = w

Initial temp. of calorimeter and its contents = 32°C

Final temp. of the mixture = 40°C Heat lost by steam = 5 × 540 + 5 (100 – 40) = 3000 calories

Heat gained by calorimeter and its contents

= w × 8 cal

Heat gained = Heat lost

8w = 3000

or w = 375 gExampl 39. Calculate how much steam from water boiling at 100°C will just melt 200 g of

wax at 15°C. (Melting point of wax = 55°C, Sp. heat of wax = 0.7, Latent heat of fusion of wax= 35 cal/g.) (C.U.)

Solution. Suppose mass of steam = m

Mass of wax = 200 g

Sp. heat of wax = 0.7

Latent heat of wax = 35 cal per gram

Initial temperature of wax = 15°C

Melting point of wax = 55°C

Heat required to raise the temperature of wax

from 15°C to 55°C = 200 × 0.7 × (55 – 15)


= 5600 cal

Heat required to melt wax at 55°C = 200 × 35 = 700 cal

Total heat gained by wax = 5600 + 7000 = 12,600 cal

Heat lost by steam = m × 540 + m (100 – 55)

= 540 m + 45 m = 585 m

Heat lost = Heat gained

585 m = 12600

m = 21.54 gExample 40. 50 g of steam at 100°C is passed into a mixture of 100 g of ice and 200 g of

water at 0°C. Find the rise in temperature. Latent heat of ice = 80 cal per gram and that of steam= 537 cal per gram).

Solution. Mass of ice = 100 g

Mass of steam = 50 g

Mass of water = 200 g


Temp. of ice and water = 0°C

Latent heat of steam = 537 cal per gram

Latent heat of ice = 80 cal per gram

Let the final temp. be T°C

Heat gained by ice = 100 × 80 + 100 × T

Heat gained by water = 200 × T

Total heat gained by ice and water = 8000 + 300 T cal

Heat lost by steam = 50 × 537 + 50 (100 – T) calHeat gained = Heat lost

800 + 300T = 50 × 537 + 50 (100 – T)

T = 68.14°CExample 41. It takes fifteen minutes for an electric kettle to heat a certain quantity of water

from 0°C to the boiling point 100°C. It takesd 80 minutes to boil this water into steam.Determine the latent heat of steam.

Solution. Suppose mass of water = m

Heat required to raise its temperature

from 0°C to 100°C = 100 m cal.

Time taken = 15 minutes

Heat produced by electric kettle in 15 minutes = 100 m cal.

Heat produced by electric kettle in 1 minute =100

15

m cal.

Heat produced by electric kettle in 80 minutes =100 80

cal15

m

Suppose latent heat of steam = L cal per gram

Heat required by m g of water to boil off into steam

= mL cal


mL =100 80

15

m

or L = 533.33 cal.Example 42. How much ice at 0°C would a kilogram of steam at 100°C melt, if the

resulting water is at 0°C? Latent heat of ice = 80 cal/g and latent heat of steam = 536 cal/g.

(Bombay 1972)

Solution. Mass of ice = M g

Mass of steam = 1 kg = 1000 g


Final temp. = 0°C

Heat gained by ice = M × 80 cal

Heat lost by ice steam = 1000 × 536 + 1000 × 100

= 636000 cal


80 M = 636000

M = 7950 g

= 7.95 kgExample 43. A piece of metal weighing 60 g at 10°C was immersed in a current of steam

100°C. Calculate the amount of steam condensed. Specific heat of the metal = 0.1 cal/g –°C andlatent heat of steam = 540 cal/g. (D.H.S. 1974)

Solution. M = 60 g

s = 0.1 cal/g–°C

t1 = 10°C

t2 = 100°C

Heat taken by the metal = mS (t2 – t1)

= 60 × 0.1 (100 – 10)

= 540 calories.

Suppose the mass of steam condensed = m grams

Latent heat of steam, L = 540 cal/g Heat lot by steam, mL = m × 540


540 = 540 m

m = 1 g.Hence, the mass of steam condensed = 1 gram.Example 44. A refrigerator converts 50 g of water at 15°C into ice at –20°C in one hour.

Determine the quantity of heat removed per minute. (Specific heat of ice = 0.5).

(Indian School Certificate 1965)

Solution. Here m = 50 g

Heat removed in cooling water

from 15°C to 0°C = 50 × 1 × 15 = 750 cal.

Heat removed in converting water into ice at 0°C

= 50 × 80 = 4000 cal


Heat removed in cooling ice from 0°C to –20°C

= 50 × 0.5 × 20 = 500 cal

Total heat removed in one hour = 750 + 4000 + 500 = 5250 cals

Heat removed per minute =5250

60

= 87.5 cals/minute.Example 45. Heat is supplied at constant rate to 400 grams of ice at 0°C in a copper

container. All the ice is converted into water in 5 minutes. How much time will be required (i)to raise the temperature of water to its boiling point (100°C) and (ii) to change all the water at100°C into steam. (Indian School Certificate, 1969)

Solution. Heat required to melt 400 grams of ice

= 400 × 80 32000 cal

Time taken = 5 min = 300 s

Rate of supply of heat =32000 320

300 13 cal/second

(i) Heat required to raise the temperature of 400 g of water from 0°C to 100°C = 400 × 1 ×100 = 40000 cal.

Time taken =40000 3

320

= 375 s = 6 min 15 s.

(ii) Heat required to convert 400 grams of water at 100°C into steam at 100°C = 400 × 540= 216000 cal.

Time required =216000 3

320

= 2025 s

= 33 min 45 sTotal time required = 375 + 2025 = 2400 s = 40 min.Example 46. A mixture of 250 g of water and 200 g of ice at 0°C is kept in a calorimeter

which has a water equivalent of 50 g. If 200 g of steam at 100°C is passed through this mixture,calculate the final temperature and weight of the contents of the calorimeter.

(I.I.T. Entrance Exam. 1974)

Solution. Amount of heat required to raise the temperature of calorimeter, ice and waterfrom 0°C to 100°C

= (200 × 80) + (50 + 200 + 250) 100

= 16,000 + 50,000 = 66,000 calories

Mass of steam condensed =66000

540 = 122.22 g

The final temperature of the mixture = 100°C

The final weight of the contents = 200 + 250 + 122.22 g

= 572.22 gTherefore 572.22 g of water at 100°C will be the final contents of the calorimeter.


CALORIMETRY AND SPECIFIC HEAT OF GASESExample 47. What amount of heat must be supplied to 2.0 × 10–2 Kg of nitrogen at room

temperature to raise the temperature by 45°C at constant pressure? Given molecular weight of N2

is 28 and R = 8.3 J mol–1 k–1 and CV = 7

2 R.

Solution. Number of moles of N2 gas =2 32.0 10 10 20

28 28

Also Cp =7 7

R 8.32 2

J mol–1 k–1.

Heat supplied to the gas Q = n Cp t

=20 7

8.3 4528 2

= 933.75 J.Example 48. A liquid of sp heat 0.5 at 60°C is mixed with another liquid of Sp. heat 0.3 at

20°C. After mixing, the temperature of the mixture becomes 30°C. In what proportion by weightare the liquids mixed? (D.H.S. 1971)

Solution. Suppose,

Mass of liquid A = m1

Mass of liquid B = m2

Sp heat of A = s1 = 0.5

Sp heat of B = s2 = 0.3

Temp of A = t1 = 60°C

Temp of B = t2 = 20°C

Final temperature = t = 30°C

Fall in temp. of A = 60 – 30 = 30°C

Rise in temp of B = 30 – 20 = 10°C

Heat lost by A = m1s1 (t1 – t) = m1 × 0.5 × 30 calories

Heat gained by B = m2 s2 (t – t2) = m2 × 0.3 × 10 calories


m1 × 0.5 × 30 = m2 × 0.3 × 10

1

2

m

m=

0.3 10

0.5 30

15

.

Example 49. A copper ball weighing 3 kg after being heated in a furnace is taken out andplunged into 8 kg of water at 10°C. If the temperature of water rises to 25°C, find thetemperature of the furnace. (Sp. heat of copper = 0.1, water equivalent of the calorimeter = 50g).

Solution. Mass of copper ball = m1 = 3 kg = 3000 grams

Sp., heat of copper s1 = 0.1

Suppose the temperature of the furnace = t1°CFinal temperature = 25°C


Fall in temp. of the ball = (t1 – 25)

Heat lost by the ball = 3000 × 0.1 × (t1 – 25) ... (i)

Mass of water = 8 kg = 8000 grams

Water equivalent of the calorimeter = 50 grams

Temp. of water and calorimeter = 10°C

Rise in temperature of water and calorimeter

= 25 – 10 = 15

Heat gained by water and calorimeter

= 8000 × 1 × 15 + 50 × 15

= (8050) × 15 calories ... (ii)


3000 × 0.1 × (t1 – 25) = 8050 × 15

t1 = 427.5°C.Example 50. Equal volumes of mercury and glass have the same capacity of heat. Calculate

the sp. heat of a piece of glass of density 2.5 g/cc if the specific heat of mercury is 0.033 anddensity is 13.6 g/cc.

Solution. Suppose,

Mass of mercury = m1

Mass of glass = m2

Sp. heat of mercury = s1 = 0.033

Sp. heat of glass = s2 = ?

Volume of mercury = V

Density of mercury = d1 = 13.6 grams/ccDensity of glass = d2 = 2.5 grams/cc

Thermal capacity = Mass × sp. heat

Here, m1s1 = m2s2

But Mass = Volume × density

m1 = V × d1

and m2 = V × d2

V × d1 × s1 = V × d2 × s2 or d1 × s1 = d2 × s2

13.6 × 0.033 = 2.5 × s2

s2 = 0.18.Example 51. The sp. gravity of a certain liquid is 0.8 and that of another is 0.5. It is found

that the heat capacity of 3 litres of the first is the same as that of 2 litres of the second. Comparetheir specific heats.

Solution. Here, d1 = 0.8d2 = 0.5V1 = 3 litres = 3000 ccV2 = 2 litres = 2000 cc

Mass m1 = V1 × d1

m2 = V2 × d2

Thermal capacity of A = Thermal capacity of B

m1 × s1 = m2 × s2


V1 × d1 × s1 = V2 × d2 × s2

1

2

s

s= 2 2

1 1

2000 0.5

3000 0.8

V d

V d

=5

12 = 5 : 12.

Example 52. The densities of two substances are in the ratio 5 : 6 and their specific heatsare in the ratio 3 : 5 respectively. Compare their thermal capacities per unit volume.

Solution. Thermal capacity = Mass × specific heat

Thermal capacity per unit volume =mass sp. heat

volume

But,mass

volume= density

Thermal capacity per unit volume = density × sp. heat

Here 1

2

d

d=

5

6

and1

2

s

s=

3

5

Thermal capacity per unit volume of

Thermal capacity per unit volume of

A

B=

1 1

2 2

d s

d s

=1 1

2 2

d s

d s

=5 3 1

6 5 2

= 1 : 2.Example 53. Three liquids, A, B and C, are at temperatures of 30°C, 20°C and 10°C

respectively. When equal parts (by weight) of A and B are mixed, the temperature of the mixtureis 26°C and when equal parts (by weight) of A and C are mixed, the temperature is 25°C. Findthe resulting temperature when equal parts of B and C is mixed. (Roorkee, E.E.)

Solution. Suppose mass of each liquid = m

Sp heat of A = s1 Sphere of B = s2

Sp heat of C = s3 Temp. of A = 30°CTemp. of B = 20°C Temp. of C = 10°C

(1) When A and B are mixed, final temperature

= 26°C

Heat lost by A = Heat gained by B

m × s1 (30 – 26) = m × s2 (26 – 20)

4s1 = 6s2

or s2 = 12

3s


(2) When A and C are mixed final temperature

= 25°C

Heat lost by A = Heat gained by C

m × s1 (30 – 25) = m × s3 (25 – 10)

5 s1 = 15 s3

or s3 = 1

3

s

(3) When B and C are mixed

Suppose the final temp. = t

Heat lost by B = Heat gained by C

m × s2 (20 – t) = m × s3 (t – 10)

Substituting the values s2 = 12

3s

and s3 =1

3

s

12

(20 )3

m s t = 1 ( 10)3

sm t

or t = 16.67°C.Example 54. 50 g of water and an equal volume of alcohol of sp gravity 0.8 are poured one

after the other in the same calorimeter. They are found to cool from 60°C to 55°C in two minutesand 1 minute respectively. Find the sp heat of alcohol. The water equivalent of the calorimeter =2 g.

Solution. Suppose, sp heat of alcohol = s

Mass of water = m = 50 g

Fall in temperature = t = (60 – 55) = 5°C

Mass of alcohol = M = 50 × 0.8 = 40 g

Water equivalent of the calorimeter = w = 2 g

Time taken by water to cool from 60°C to 55°C

= t1 = 2 minutes = 120 s

Time taken by alcohol to cool from 60°C to 55°C

= t2 = 1 minute = 60 s

Applying the relation1

( )m w t

t

=

2

(M )s w t

t

(50 2)5

120

=

(40 2)5

60

s

or s = 0.6.Example 55. When a piece of metal weighing 48.3 grams at 10.7°C was immersed in a

current of steam at 100°C, 0.762 gram of steam was found to condense. Calculate the specificheat of the metal.


Solution. Let sp heat of metal is S

Heat required by the metal to raise its temperature from 10.7°C to 100°C = 48.3 × S × (100– 10.7) Cal.

Mass of steam condensed = 0.762 g

Latent heat of steam = 536 Cal/g

Heat lost by steam = 0.762 × 536 Cal.

Now Heat gained = Heat lost.

48.3 × S × 89.3 = 0.762 × 536

or s =0.762 536

48.3 89.3

= 0.095.

BASED ON DEGREE OF FREEDOM OF GASExample 56. Calculate the internal energy of 2g of oxygen at N.T.P.

Solution. Number of moles in 2g =2

32

Energy associated with 1 mole of oxygen=5

RT.2

Internal Energy of 2 g of oxygen =2 5

RT32 2

=2 5

8.31 27332 2

= 177.2 J.Example 57. A gas has molar heat capacity C = 37.35 J mol–1 k–1 in the process PT =

constant. Find the number of degree of freedom of molecules in the gas (R = 8.3 J mol–1 k–1).

Solution. The gas undergoes the process, given by

PT = Constant. ... (i)

for 1 mole of an ideal gas PV = RT

or P =RT

.V

where R is contant. Combining it with eq. (i) we have

2T

V= Constant

Differentiating2

2

2T T T dV

V V

d = 0.

V

V

d=

T2

T

d... (ii)


According to first law of thermodynamics

dQ = du + dw

where dQ = C dT dV = CV dT

and dw = P dV = V

RTV

d

= 2 R dT using Eq. (ii) C dT = Cv dT + 2R dT

C = Cv – 2RCv = C – 2R

C = 37.35 J mol–1 k–1 =37.35

R8.3

= 4.5 R (given)

CV = 4.5 R – 2R= 2.5 R

If f is degree of freedom then

CV = R2

f

R2

f= 2.5 R

or f = 5.Example 58. Calculate the total number of degree of freedom possessed by the molecules

in 2 Cu3 of H2 at N.T.P.

Solution. Molar volume = 22.4 m= 22400 Cu3

22400 Cu3 of H2 contains 6.02 × 1023 molecules

Number of molecules in 2 Cu3 of H2 = 236.023 10 2

22400

Number of degree of freedom associated with each H2 molecule = 5.

Total number of degree of freedom associated with 2 Cu3 of H2

=236.023 10 2

522400

= 2.68750 × 1020.

EXERCISE FOR PRACTICE

Based on Temperature and Measurement of Temperature1. Convert 98.8°F into degree celsius. [Ans. 37°C]

2. Convert 40°C in degree Fahrenheit. [Ans. 104°F]

3. The resistance of a resistance thermometer at 0°C is 3.70, at 100°C is 4.71 and atunknown temperature t the resistance is 5.29.

Calculate temperature coefficient of resistance[Ans. = 2.73 × 10–3 °C–1]

4. Find the value of absolute zero on Fahrenheit scale [Ans. 459.4°F]


5. Mercury in a faulty centigrade thermometer stands at 5° mark when placed in melting iceand at 95° when in steam at normal pressure. What will be the reading when it is placedin contact with a body at 40°C? [Ans. 41°C]

6. Calculate the temperature at which the centigrade reading is 1

2 of the Fahrenheit reading.

[Ans. 26.67°C]

7. Calculate the temperature on the centigrade scale whose value is half of the absolutereading. [Ans. 273°C]

8. There is a certain temperature which has the same reading on the centigrade and Reaumurscales. Find the temperature. [Ans. 0°C]

9. Calculate the temperature at which the reading in the centigrade thermometer is doublethat of the Reaumur thermometer. [Ans. 0°C]

10. In a faulty thermometer, the temperatures of ice and steam are marked as 2° and 99°. Whatwill be the correct temperature on the centigrade scale when this thermometer reads 50.5°?

[Ans. 50°C]

11. The column of mercury in a thermometer stands at 1° mark when placed in melting ice and101° mark when placed in steam at normal pressure. Calculate the correct temperaturewhen this thermometer reads 39°. [Ans. 38°C]

12. A faulty thermometer has its fixed points marked at 6 and 96. What is correct temperatureon the centigrade scale when this thermometer reads 87° ? [Punjab Pre-Univ. 1964]

[Ans. 90°C]13. The resistance of a platinum resistance thermometer is 2.0 ohm at 0°C and 2.5 ohm at

100°C. The temperature coefficient of resistance is 0.0025 per °C. Calculate thetemperature at which the resistance will be 2.3 ohm.

[Ans. 60°C]14. The resistance of a conductor is 20 ohm at 20°C and 60 ohm at 500°C. Calculate the

resistance of this conductor at 80°C. [Ans. 25]15. Calculate the temperature at which the resistance of a conductor will become twice of its

value at 0°C. (Given temperature coefficient of resistance is 1

273 °C–1)

16. The resistance of a platinum resistance thermometer are 2.5, 4.0 and 7.3, at 0°C,100°C and at an unknown temperature t°C. Calculate ‘t’. [Ans. 320°]

BASED ON THERMAL EXPANSION (LINEAR, SUPERFICIAL AND VOLUMEEXPANSION)

17. A bar of length 100 cm is heated from 0° to 100°C. Calculate the coefficient of linearexpansion of the material if the increase in length is 0.1 cm. [Ans. 10–5 °C]

18. A brass rod has a length of 100 cm at 25°C. Calculate its length at 75°C given that thecoefficient of linear expansion of brass is 0.000018 per °C. [Ans. 100.09 Cu]

19. An iron ring 3 metres in diameter is to be fixed over a wooden wheel of diameter 3.01metres. If the initial temperature of the ring is 20°C, find the temperature to which the ringmust be heated so that it can just be fitted on the wooden wheel. for iron = 0.000012 per °C.

[Ans. 297.78°C]20. A hollow brass sphere has an inner radius of 15 cm. Calculate the change in volume of the

hollow space when the sphere is heated from 25°C to 85°C ( = 0.000018 per °C).[Ans. 15.274 cc]

21. An iron disc has a hole of radius 10 cm cut in the middle. Calculate the radius of the holewhen the disc is heated from 10°C to 90°C ( = 0.000012 per °C). [Ans. 10.0096 cm]


22. The length of an iron rod when measured with a brass scale is 2 metres at 30°C. If thebrass scale is correct at 0°C, what is the true length of the iron rod at 0°C? ( for iron= 0.000012 and for brass = 0.000018). [Ans. 200.036 Cu]

23. The density of a substance at 100°C is 7.50 g/cc and at 0°C it is 7.25 g/cc. Calculate thecoefficient of linear expansion of the substance. [Ans. 1.15 × 10–4 / °C]

24. A zinc and an iron rod are each 50 cm at 0°C. Calculate the difference in the lengths at50°C. ( for zinc = 26 × 10–6, for iron = 12 × 10–6) [Ans. 0.035 Cu]

25. A clock which keeps correct time at 30°C has a pendulum rod made of brass. How manyseconds will it gain or lose per day when the temperature falls to 10°C?

( for brass = 18 × 10–6) [Ans. 15.552 s g/Cu]

26. A clock which keeps correct time in winter (mean temperature 15°C) has a pendulum madeof iron. How many seconds will it gain or lose per day when the temperature in summerincreases to a mean value of 35°C? ( for iron = 12 × 10–6) [Ans. 10.368s loss]

27. A copper disc at 0°C has a diameter of 100 cm and a hole cut in the centre is 10 cm indiameter. What will be the diameter of the hole at 100°C? ( for copper = 0.000016)

(Delhi H.S.) [Ans. 10.016 Cu]

28. The volume of a lead ball is 100 cc at 0°C and 100.85 cc at 100°C. Calculate the coefficientof linear expansion of lead. (Punjab Pre-Uni.)

[Ans. 28.33 × 10–6 / °C]29. A metal rod of diameter 1 cm measures 50 cm in length at 20°C. When it is heated to 95°C

its length becomes 50.06 cm. What is the coefficient of linear expansion of the rod? Whatwill be the length of the rod if it is cooled to 0°C? (Punjab Pre-Univ.)

[Ans. 16 × 10–6, L0 = 49.984 Cu]30. A solid occupies 1000 cc at 20°C. Its volume becomes 1016.2 cc at 320°C. What is its

coefficient of cubical expansion? Calculate also the coefficients of linear and superficialexpansion of the solid. (Punjab Pre-Univ.)

[Ans. 18 × 10–6 / °C, 36 × 10–6 / °C]31. A cubical block of platinum is raised in temperature from 0°C to 10°C. If the density of

platinum at 0°C is 21.5 g/cm3, calculate the change in density. [Coefficient of linearexpansion of platinum is 0.000009 / °C] (All India H.S.)

[Ans. 0.0058]32. The difference between the lengths of two rods A and B is 60 cm at all temperatures. Find

their original lengths ifA = 18 × 10–6 / °CB = 27 × 10–6 / °C (Bombay)

[Ans. 180 Cu, 120 Cu]33. A piece of metal weighs 40 g in air. When it is immersed in a liquid of specific gravity 1.24

at 27°C it weighs 30 g. When the temperature of liquid is raised to 42°C the metal pieceweights 30.5 g. Specific gravity of liquid at 42°C is 1.20. Calculate the coefficient ofcubical expansion of the metal. [Ans. 1.034 × 10–4 / °C]

34. An iron sphere has a radius of 10 Cu at a temperature of 0°C. Calculate the change involume of the sphere if it is heated to 100°C. Given = 1.1 × 10–6 °C–1.

[Ans. 1.383 Cu3]35. At 20°C the area of a metal plate is 100 Cu2. Calculate the area of the metal plate at 200°C.

Given = 0.000016 °C–1. [Ans. 100.576 Cu2]36. The density of mercury is 13.6 g Cu–3 at 0°C and its coefficient of cubical expansion is 1.82

× 10–4 °C–1. Calculate the density of mercury at 50°C. [Ans. 13.48 g Cu–3]


BASED ON JOULE’S MECHANICAL EQUIVALENT OF HEAT AND CALORIMETRY37. Calculate the power of a person in horse power who can chew 200 g of ice in one minute.

J = 4.2 joule per calorie. [Ans. 1.50 h.p.]

38. Calculate the height a 50 kg man can climb by using energy from a slice of bread whichproduces 100,000 cal. Consider efficiency of human body is 28%, g = 10 ms–2.

[Ans. 235.2 m]

39. Find the velocity with which a lead bullet having initial temperature 30°C should strike atarget so that it may just melt. Only half of the heat produced is absorbed by the bullet. Sp.heat of lead = 0.03 k cal Kg–1 °C–1. Latent heat of llead = 6 k cal kg–1. Melting point oflead = 330°C. (Given 1 K cal = 4.2 × 103 J] [Ans. 50.2 ms–1]

40. From what height should a piece of ice fall so that it melts completely? Only half of theheat produced is absorbed by the ice.

(Given Latent heat of ice = 3.4 × 105 J/kg and g = 10 ms–2] [Ans. 68 Kn]

41. If in stopping a car of mass 800 kg by applying brakes, 41.8 K cal of heat is produced, thenwhat was the speed of the car before applying the brakes. [Ans. 20.9 ms–1]

42. In a water fall, water falls from a height of 840 m on the ground. Determine the rise in thetemperature of water consider total heat remains with in water. [Ans. 2°C]

43. A 5 kg hammer falls with a velocity 1 ms–1 on a piece of lead of mass 300 of at 27°C. Howmany strokes will be required to melt the piece of lead. [Ans. 7524]

44. A lead bullet moving with a velocity of 300 ms–1 strikes a target and comes to rest.Calculate the rise in the temperature of the bullet if heat is equally shared by bullet andtarget.

Given specific heat of bullet (lead) = 0.032 Kcal / Kg °C. [Ans. 167.4°C]

45. A 2 kg spherical ball falls from a height of 3 m. If whole of its potential energy is convertedinto heat, then how much heat will be produced? [Ans. 14 Cal.]

46. 50 g of water at 80°C is poured into a calorimeter containing 40 g of water at 15°C. Findthe final temperature. Water equivalent of the calorimeter is 8 g. [Ans. 48.16°]

47. 40 g of water at 60°C is poured into a calorimeter whose temperature is 20°C. The finaltemperature of the two is 45°C. Find the water equivalent of the calorimeter. [Ans. 2.4 g]

48. A liquid of specific heat 0.6 at 75°C is mixed with another liquid of specific heat 0.4 at25°C. The final temperature of the mixture becomes 50°C. In what proportion by weightare the two liquids mixed? [Ans. 2 : 3]

49. 100 g of iron (S = 0.112) at 95°C is dropped into a calorimeter containing 150 g of an oilof specific heat 0.8. The final temperature of the mixture is 60°C and water equivalent ofthe calorimeter is 10 g. Calculate the initial temperature of the oil. [Ans. 56.985°C]

50. 4 g of steam at 100°C is mixed with 200 g of water at 10°C. Find the temperature of themixture. Latent heat of steam at 100°C = 540 calories per gram. [Ans. 22.36°C]

51. A copper calorimeter contains 120 g of water at 20°C. When 25 g of ice is added to it, theresultant temperature of the mixture is 5°C. Calculate the water equivalent of thecalorimeter. Latent heat of ice = 80 calories per gram. [Ans. 21.67 g]

52. A copper sphere weighing 5kg, after being heated in a furnace, is taken out and plungedinto 10 kg of water at 10°C. If the temperature of water rises to 20°C, find the temperatureof the furnace. Sp heat of copper is 0.1 and water equivalent of the calorimeter 0.05 kg.

[Ans. 221°C]53. Equal volumes of copper and mercury have the same thermal capacity. Calculate the sp

heat of copper of density 6.8 g/cc if the sp heat of mercury is 0.048 and density of mercuryis 13.6 g/cc. [Ans. 0.096]


54. The densities of two substances are in the ratio 5 : 6 and their specific heats are 0.16 and0.08 respectively. Compare their thermal capacities per unit volume. [Ans. 5 : 3]

55. The specific gravity of liquid A is 0.75 and that of B is 0.60. The thermal capacity of 5litres of A is equal to 3 litres of B. Compare their specific heats. [Ans. 12 : 25]

56. There are three liquids A, B and C. When 5 g of A at 60°C and 7 g of B at 30°C are mixed,the resultant temperature is 50°C. When 4 g of B at 30°C and 1 g of C at 40°C are mixedthe resultant temperature is 33°C. Calculate the resultant temperature if 6 g of A at 60°Cand 9 g of C at 44°C are mixed. [Ans. 50.43 °C]

57. The temperatures of equal masses of three different liquids A, B and C are 20°C, 30°C and40°C respectively. On mixing A and B, the temperature of the mixture is 26°C. On mixingA and C, the........................... [Ans. 0.614 Cal/gm °C]

59. 300 g of water and an equal volume of alcohol of mass 240 g are poured successively in thesame calorimeter and they are cooled from 55°C to 50°C in 2.5 minutes and 1.25 minutesrespectively. Find the sp. heat of alcohol. Water equivalent of the calorimeter is 20 g.

[Ans. 0.583 cal gm° C]

60. When a metal sphere weighing 50 g and at 10°C was immersed in a current of steam at100°C, 0.75 g of steam was found to condense. Calculate the sp heat of the metal.

[Ans. 0.09 Cal/gm °C]

61. 50 g of steam at 100°C is passed into a mixture of 100 g of ice and 200 g of water at 0°C.Find the rise in temperature. Latent heat of steam = 537 cal/g and latent heat of ice = 80cal/g. (Punjab Pre-Univ. 1964)

[Ans. 68.14°C]

62. A piece of lead weighing 500 g gives out 1200 calories of heat when it is cooled from 90°Cto 10°C. Find its specific heat, thermal capacity and water equivalent.

(Punjab Pre-Univ. 1965)

[Ans. (i) 0.03 (ii) 15 cal (iii) 15 g]

63. A hot ball of iron weighing 200 g is dropped into 500 g of water at 10°C. The resultingtemperature is 22.8°C. Calculate the temperature of the hot ball. Specific heat of iron = 0.08.

[Ans. 422.8°C]

64. The density of a liquid A = 0.5 g/cm3 and that of B = 0.6 g/cm3. It is found that the heatcapacity of 8 litres of A is equal to the heat capacity of 10 litres of B. Calculate the ratioof their specific heats. (Delhi H.S. 1967) [Ans. 1:5]

65. 100 c of water contained in a calorimeter weighing 200 g and of specific heat 0.1 takes 6minutes to cool from 40°C to 25°C. The same volume of alcohol using the samecalorimeter takes four minutes to cool through the same range of temperature. Calculatethe specific heat of alcohol. Density of water = 1 g/cm3 and density of alcohol = 0.8 g/cm3.

(All India H.S. 1967) [Ans. 0.75]

66. Two liquids A and B are at temperatures 60°C and 20°C respectively. Their masses are inthe ratio of 3 : 4 and their specific heats are in the ratio of 4 : 5. Calculate the resultanttemperature of the mixture if the liquids A and B are mixed. [Neglect the water equivalentof the calorimeters]. [(All India H.S. (Comtt.) 1967)]

[Ans. 25°C]

67. Two liquids A and B are at the temperatures of 75°C and 15°C respectively. Their massesare in the ratio of 2 : 3 and their specific heats are in the ratio of 3 : 4. Calculate theresultant temperature of the mixture when the two liquids A and B are mixed. Neglect thewater equivalent of the calorimeters. [Ans. 35°C]


BASED ON SPECIFIC HEAT AND LATENT HEAT68. Calculate the amount of heat required to convert 20 grams of ice at – 20°C to steam at

100°C. (Latent heat of ice = 80 cal per gram, latent heat of steam = 540 cal per gram andsp. heat of ice = 0.5). [Ans. 14600 Cal]

69. Calculate the mass of ether at 0°C to be evaporated, so that the heat absorbed by it canfreeze 6 grams of water at 0°C. Latent heat of vaporization of ether being 96 cal per gram.

[Ans. 5 g]

70. Two grams of steam at 100°C is passed into a calorimeter containing 70 grams of ice.Calculate the amount of ice melted. [Ans. 16 g]

71. What will be the result of adding 60 grams of ice to 80 grams of water at 60°C?

[Ans. 140 g of water at 0°C]


[Ans. 31.43 °C]


[Ans. Mixture of 68.75 g of water and 81.25 g of ice at 0°C]

74. 4 grams of steam at 100°C is passed into a calorimeter containing a liquid and thetemperature of the liquid rises from 40°C to 50°C. Calculate the total water equivalent ofthe calorimeter and its contents. (Latent heat of steam = 540 cal per gram) [Ans. 236 g]

75. Calculate the amount of steam that will melt 100 grams of wax at 20°C. (Melting point ofwax = 55°C. Sp. heat of wax = 0.7 and latent heat of fusion of wax = 35 cal per gram).

[Ans. 10.17 g]

76. A metal sphere weighing 20 grams at 120°C is dropped into the cavity of a block of ice andit melts 6 grams of ice. Calculate the sp. heat of the metal. [Ans. 0.2 cal/gm °C]

77. 40 grams of steam at 100°C is passed into a mixture of 80 grams of ice and 160 grams ofwater at 0°C. Find the final temperature of the mixture. (Latent heat of ice = 80 cal pergram. Latent heat of steam = 540 cal per gram). [Ans. 68.57 °C]

78. It takes 20 minutes for an electric kettle to heat a certain quantity of water from 0°C to theboiling point at 100°C. In how much time, the whole of water at 100°C will be convertedinto steam? (Latent heat of steam = 540 cal per gram). [Ans. 1 hr. 48 m]

79. 10 grams of a substance at 80°C, placed in the tube of a Bunsen’s ice calorimeter shows acontraction in volume of 0.45 cc. Calculate the sp heat of the substance if the contractionin volume when 1 g of ice melts is 0.09 cc. [Ans. 0.5 cal/gm °C]

80. When 1 kilogram of ice melts the contraction in volume is 90 cc. Calculate the mass of themetal of sp. heat 0.1 heated to a temperature of 400°C, which when dropped into an icecalorimeter causes a diminution in volume of 0.36 cc. [Ans. 8 g]

81. A substance weighing 25 grams at 80°C was dropped into a Bunsen’s ice calorimeter andthe liquid column in the capillary tube recedes through a distance of 20 cm. Calculate thesp. heat of the substance. Diameter of the tube is 1.4 mm (When 1 g of ice melts thechange in volume is 0.09 cc). [Ans. 0.137 cal/g°C]

82. A substance weighing 10 grams at 150°C was dropped into a Bunsen’s ice calorimeter andthe liquid column in the capillary tube (4.5 sq mm cross-section) contracts through 4.5 cm.Calculate the sp. heat of the substance. [Ans. 0.12 cal/gm°C]

83. Find the result of mixing 10 grams of ice at – 10°C with 10 grams of water at 10°C. Spheat of ice = 0.5 and latent of ice = 80 cal/gram. (Delhi H.S. 1964)

[Ans. Mixture of 9.375 of ice 2 10.625 g of water at 0°C]


84. 20 grams of sand at 50°C when dropped into a Bunsen’s ice calorimeter caused a decreasein volume of 0.27 cc. Calculate the specific heat of sand. (Delhi Pre-Medical 1964)

[Ans. 0.24 cal/gm°C]

85. What will be the result of mixing 400 grams of copper chips at 500°C with 500 grams ofmelting ice? (Punjab H.S. 1964)

[Ans. 250 g of ice melts and final temp. is 0°C]

86. The density of ice is 0.93 gram/cc at 0°C. A piece of metal weighing 150 grams is heatedto 100°C and is then placed in a Bunsen’s ice calorimeter. The decrease in volume is foundto be 1.876 cc. Calculate the specific heat of the metal. Latent heat of ice = 80 cal/gram.

[Ans. 0.133 cal/gm °C]

87. A solid of mass 20 grams is heated to a temperature of 80°C and quickly dropped in aBunsen’s ice calorimeter. The contraction in volume observed is 0.9 cc. The contracting involume when one gram of ice melts is 0.09 cc. Latent heat of ice = 80 cal/gram. Calculatethe specific heat of the solid. (All India H.S. 1966)

[Ans. 0.5 cal/gm °C]

88. A calorimeter of mass 150 grams and specific heat 0.1 contains 10 grams of ice and 90grams of water in thermal equilibrium at 0°C. A certain mass of steam at 100°C is passedinto the calorimeter and the final temperature of the mixture is 40°C. Calculate the mass ofsteam. Latent heat of ice = 80 cals/gram and latent heat of steam = 540 cal/gram.

(All India H.S. 1971) [Ans. 9 g]

89. An earthen pitcher loses 1 gram of water per minute due to evaporation. If the waterequivalent of the pitcher is 0.5 kg and the pitcher contains 9.5 kg of water, calculate thetime required for the water in the pitcher to cool to 28°C from its original temperature of30°C. Neglect radiation losses. Latent heat of vaporization of water in this range oftemperature is 580 cal/g. (I.I.T Entrance Exam. 1970)

[Ans. 34.5 min]

90. What is the result of mixing 10 grams of ice at 0°C and 100 grams of water at 100°C?

(D.H.S. 1971) [Ans. 83.63°C]

91. A copper calorimeter of mass 500 g contains 500 g of water at 30°C. A piece of ice of mass10 g at a temperature of – 5°C is dropped into the calorimeter. Find the final temperatureof water when the whole of ice has melted. (Bombay, 1971)

[Ans. 27.81 °C]

92. 540 g of ice at 0°C is mixed with 540 g of water at 80°C. Calculate the final temperatureof the mixture. (I.I.T. Entrance Exam., 1975)

[Ans. 0°C]

BASED ON SPECIFIC HEAT OF GASES AND DEGREE OF FREEDOMS93. A mole of oxygen at 0°C is heated at constant pressure to increase its volume by 10%.

(a) What is the required amount of heat? The specific heat of oxygen at constant pressureis 0.22 cal g–1 k–1.

(b) If the same amount of heat is given to the gas at constant volume, what will be the finaltemperature? Given R = 8.31 J mol–1 k–1. [Ans. 807.24 J, 38°C]

94. .5 mole of a polyatomic gas is heated from 25°C to 125°C at constant pressure. Calculatethe amount of heat given to gas if Cp = 8 cal mol–1 °C–1. [Ans. 400 cal]


95. The difference between two specific heats of a gas is 5000 J kg–1 k–1 and the ratio ofspecific heats is 1.6. Calculate the two specific heats.

[Ans. 8333.33 J kg–1 k–1, 13337.33 J kg–1k–1]

96. Calculate the specific heat at constant volume for a gas. Given specific heat at constantpressure is 28.77 Joule mol–1 k–1. R = 8.31 Joule mol–1 k–1. [Ans. 20.46 Joule/mol–k]

97. A cylinder of fixed capacity 44.8 litres contains helium gas NTP. Find the amount of heatrequired to raise the temperature of the gas by 15.0°C. (R = 8.31 J mol–1 k–1).

[Ans. 374 J]

98. Show that for monoatomic gas specific heat ratio is 1.66.

99. For diatomic gas specific heat ratio is 1.4. The specific heat at constant pressure is 6.93 cal/mol – k. Calculate its value at constant volume. [Ans. 4.95 Cal/mol - k].

100. Calculate total number of degree of freedom possessed by the molecules in 1 Cu3 of H2 gasat N.T.P. [Ans. 1.34375 × 1020]

101. Calculate total number of degree of freedom for a mole of diatomic gas at N.T.P.

[Ans. 30.1 × 1023].

102. Calculate the number of degree of freedom in 30 Cu3 of nitrogen at N.T.P.

[Ans. 4.03 × 1021]

417

Temperature:- Temperature is that thermal property which has same value for two or morethan two bodies in thermal equilibrium.

Thermal Equilibrium: When two or more bodies are kept in contact but there is noexchange of heat between them, then they are called in thermal equilibrium. Or two bodies are inthermal equilibrium with each other if they have same temperature.

Zeroth Law of Thermodynamics: It states that If two bodies A and B are in thermalequilibrium with a third body C, separately then bodies A and B will also be in thermal equilibriumwith each other.

Thermodynamic Variables: The thermodynamic state of a system can be described in termsof certain variables called thermodynamic variables. For example, in case of gases these variablesare volume, pressure, temperature, internal energy. entropy etc. All the thermodynamic variablescan be expressed in terms of P, V and T.

Thermodynamic Process: A thermodynamic process is said to occur if the thermodynamicvariables of the system change with time. Various types of thermodynamic process are:

(a) Isothermal Process:- The process that takes place at constant temperature is calledisothermal process.

(b) Adiabatic Process: The process in which heat exchange cannot take place betweensystem and surroundings is called adiabatic process.

(c) Isochoric Process: A thermodynamic process that takes place at constant volume is calledisochoric process.

(d) Isobaric Process: A thermodynamic process that takes place at constant pressure is calledisobaric process.

Reversible Process: It is the process which can be reversed in such a way that the system andsurroundings undergo same intermediate state as in the direct process.

For a process to be reversible following conditions must be satisfied:

(a) The system must be in mechanical equilibrium.

(b) The system must exist in thermal equilibrium.

(c) The system must be in chemical equilibrium.

Irreversible Process: A process which cannot be retraced back in opposite direction is calledirreversible process.

Work done by a Thermodynamic System during Expansion : Consider the volume of agas changes from V1 to V2, at constant pressure, then work done is

w = 2

1

V

V

PdvDuring expansion work is done by the gas and hence it is positive. In case of compression

work is done on gas and is taken as negative.

��

��13UNIT


The work is also equals to the area enclosed between the P-V curve and the volume axis.

Work done in an isothermal process: When a gas expands isothermally, the work done isgiven by

W = 2.3 n R T log 2

1

vv = 2.3 n R T log 1

2

PP

n = number of moles

R = Universal gas constant

T = absolute temperature

V1, V2 = Initial and final volume

P1, P2 = Initial and final pressure

Work done in an adiabatic process: When gas expands adiabatically so that its temperaturedecreases from T1 to T2, then work done is given by

W =R–1 (T1 – T2)

=1–1 (P1V1 – P2 V2)

Mayer’s Formula:- Let CP and CV are specific heats of a gas at constant pressure and volumethen

CP – CV = R

Internal Energy:- Internal energy of a system equals to the sum of kinetic energy andintermolecular potential energy. The internal energy of the system is a function of its absolutetemperature.

First Law of Thermodynamics: According to first law of thermodynamics, if heat dQ issupplied to a system, a part of it is used in increasing the internal energy by an amount dV and restenergy is used in doing external work dW is

i.e. dQ = dV + dW.

Second Law of Thermodynamics: There are two different statements of second law ofthermodynamics:

(i) Kelvin Planck’s Statement: It is impossible to construct an engine which, when operatesin a cycle, produces no effect other than extracting heat from a reservoir and performing an equalamount of work.

(ii) Clausius Statement: It is impossible for a self-acting machine unaided by any externalagency to transfer heat from a body at lower temperature to a body at a higher temperature.

Heat Engine: It is a device which converts heat energy into mechanical energy continuously.

Efficiency of Heat Engine: The efficiency of heat engine is defined as the ratio of outputwork to the input heat (heat energy extracted from source), i.e.,

=1

WQ

= 1 2

1

Q – QQ

= 1 – 2

1

QQ

Source

Sink

W.S.

Q2

Q1

W

T1

T2

Thermodynamics and Kinetic Theory of Gases 419

Carnot Engine: It is an ideal engine which is free from all the defects of an actual engine.The efficiency of a carnot engine is maximum between two temperature limits.

The efficiency of Carnot engine is given by

= 1 – 2

1

QQ

= 1 – 2

1

TT

T1 = Temperature of Source

T2 = Temperature of Sink

Refrigerator: The working of a refrigerator is reverse to that of heat engine. Here workingsubstance absorbs heat Q2 from Sink at temperature T2. External work W is supplied by an externalagency and a larger amount of heat Q1 is rejected to source at a higher temperature T2, i.e.,

Q1 = Q2 + W

The coefficient of performance of a refrigerator is given by

2QW

Heat Taken fromSinkWork done by external agency

= 2

1 2

QQ – Q = 2

1 2

TT – T

Methods of Transfer of heatConduction: The process of transmission of heat from one point to another through a

substance, without actual motion of the particles of medium, is called conduction. Generally inmetal heat is transmitted by conduction.

Consider a rod of area of cross section A and length ‘d’, whose two opposite faces aremaintained at temperature Q1 and Q2 (Q1 > Q2), then amount of heat that flows in time t is

Q = 1 2KA( ) t

d

where K is coefficient of thermal conductivity of the material of rod.

S.I. unit of K – J/s mK. or W m–1K–1.

Convection: It is the process of transmission of heat from one point to another due to actualmovement of medium particles. By this process heat transfer takes place in liquids and gases.

Radiation: In this process no intervening medium is required for heat transfer.

Stefan’s Law: This law states that-

“The total amount of energy radited per second per unit area of a perfectly black body isdirectly proportional to fourth power of its absolute temperature” i.e.,

E T4

or E = T4

where = 5.67 × 10–8 J/sm2k4 is called Stefan’s Constant.In case a black body (a body for which absorptive power is unity) maintained at temperature

T is placed in an enclosure at temperature T0 (T>T0) then the amount of heat energy radiated perunit area per second is given by

Q = (T4–To4)

This equation is called Stefan Boltzmann Law.If the body is not a black body, then

Q = e (T4–To4)

e is called emissivilty of the substance of body.

Source

Sink

W.S.

Q2

Q1 = (Q2 + W)

W

T1

T2


Wein’s Displacement Law: According to this law, the wavelength (m) corresponding towhich energy emitted by a black body is maximum is inversely proportional to its absolutetemperature Tv

min 1T

or min T = Constant

The constant is called Wein’s constant and its value is = 2.898 × 10–3 mK.

Newton’s Law of Cooling: This law states that the rate of cooling of a body is directlyproportional to the temperature difference between the temperature of liquid and surroundings.

Rate of Cooling (T –To)

Expression for Pressure interms of Kinetic Theory of Gases:

P =13

MV

2v = 13

2v

M = mass of gas

V = Volume of gas

2v = mean square velocity

= density of gas

Root Mean Square Velocity: We know

P = 13

MV

. 2v

or 2v = 3PVM =

3RTM as PV = RT

or 2v = Vrms = 3RTM

Average Kinetic Energy of a Gas:Average Kinetic energy of one mole of a gas is given by

E =32

RT.

and mean kinetic energy per molecule of a gas

E =32

RN

T

=32

kT.

where k = RN

is called Boltzmann’s Constant.

Boyle’s Law: It states that at constant temperature the volume of a given mass of gas isinversely proportional to its pressure.


i.e. V 1P

or PV = Constant

or P1V1= P2V2

Charle’s Law: It states that at constant pressure the volume of a given mass of a gas isdirectly proportional to its absolute temperature i.e., V T

VT

= constant, or 1

1

VT = 2

2

VT

Pressure Law: It states that at constant volume the pressure of a given mass of a gas isdirectly proportional to its absolute temperature.

P T

orPT

= Constant, or 1

1

PT = 2

2

PT

Perfect Gas equationPV = RT, (for 1 mole of a gas)

or 1 1

1

P VT = 2 2

2

P VT

SOLVED EXAMPLES

Based On Work done (To be Calculated from P-V Diagram)Example 1. A gas is taken from an original state A to state B and then intermediate state C,

and finally to initial position A as shown in Figure 13.1. Calculate the work done by the gas fromB to C to A.

Solution. Work done by the gas is equal to the areaABC

=12

× AC × AB

=12

× (7.0–3.0) (800–400)

=12

× 4.0 × 400

= 800 JExample 2. One mole of an ideal gas undergoes a

cyclic change ABCD. Calculate the work done.

Solution. Here the loop is traced in anticlock-wisedirection. The work done is negative.

Now work W = area ABCD

= AB × BC

= (5 – 2) (6 – 2)

= 3 × 4

= 12 J.

7

6

5

4

3

2

1

1 2 3 4 50

A B

CD

V (m )3

P (Nm )– 2

Fig. 13.2

Fig. 13.1

400

800

P(Nm )– 2

B

A C

3.0 7 .0V (m 3)


Based On first Law of ThermodynamicsExample 3. At STP the volume of one gram of water increases from 1cm3 to 1.091 cm3 on

freezing. Calculate the change in the internal energy. Given atmospheric pressure is 1.013 × 105

Nm-2, and latent heat of fusion is 80 Cal g-1.

Solution. Here m = 1g; V = 1.091 – 1.0 = 0.091cm3

Heat given by water during freezing

dQ = mL = 1 × 80 = 80 Cal.

Heat is given by water dQ= – 80 Cal.

Now dW = P × dV

= (1.013×103×0.091×10–6)

= 0.0092 J

=0.00924.18

= 0.00022 Cal.

Using first law of thermodynamics

dQ = dU + dw

dU = dQ – dW

= – 80 – 0.0022

= – 80.0022 Cal– ve sign shows that on freezing the internal energy of water decreases.

Example 4. 1.0 m3 of water is converted into 1671 m3 of steam at atmospheric pressure and100°C temperature. The latent heat of vaporisation of water is 2.3 × 106 J kg–1. If 2.0 kg of wateris converted into steam at its boiling point and atmospheric pressure, calculate the increase in theinternal energy of water.

Solution. Heat energy absorbed by water is

dQ = mL

= 2.0 × 2.3 × 106 = 4.6 × 106 J

Volume of given mass of water

V =mass

density = 3

2

10 = 2 × 10–3 m3.

Work done against external pressure is

dW = P × dV

= 1.01 × 105 × (1671 – 1) × 2 × 10–3

= 0.337 × 106 J

Using first law of thermodynamics.

dQ = dU + dW

dU = dQ – dW

= 4.6 × 106 – 0.337 × 106

= 4.263 × 106 JExample 5. Calculate the change in the internal energy of a block of copper of mass 500 g

when it is heated from 20°C to 70 °C. Given specific heat of copper is 0.1 Cal/g °C. Considerchange in volume of copper on heating is negligible.

Solution. m = 500 g, = 70 – 20 = 50°C.


s = 0.1 Cal/g°C.

dQ = ms = 500 × 0.1 × 50

= 2500 Cal.

As change in volume is negligible dV = 0

dW = PdV = 0.

According to first law of thermodynamics.

dQ = dU + dW

dU = dQ – dW

= 2500 – 0 = 2500 Cal

= 2500 × 4.2

= 1.05 × 104 J.

Based on Isothermal and Adiabatic ProcessExample 6. A cylinder with a movable piston contains 3 moles of hydrogen at STP. The walls

of the cylinder are made of a heat insulator and the piston is insulated by having a pile of sand onit. By what factor does the pressure of gas increase if the gas is compressed to half its originalvolume?

Solution. The process is adiabatic.

1 1P V = 2 2P V

or2

1

P

P =1

2

V

V

Here V2 = 1 .2

V

2

1

P

P =

1.41

1 / 2

V

V

= 2 1.4 = 2.64.

Example 7. Five moles of an ideal gas is compressed isothermally from an initial volume of5 liters to a final volume of 1 litre. Calculate the work done in this process, if the temperature ofgas is 27°C.

Solution. Here n = 5, V2 =1 litre, V1 = 5 litre.

T = 27 + 273 = 300 k.

W = nRT × 2.3026 log10 2

1

.V

V

= 5 × 8.31 × 300 × 2. 3026 101

log5

= 5 × 8.31 × 300 × 2.3026 (– 0.6990)

= – 2.006 × 104 J.Example 8. An oxygen cylinder of volume 30 litres has an initial gauge pressure of 15 atm.

and a temperature of 27°C. After some oxygen is withdrawn from the cylinder, the gauge pressuredrops to 11 atm. and its temperature drops to 17°C. Estimate the mass of oxygen taken out of thecylinder. Molecular mass M of O2 = 32.


Solution. Let initial pressure and temperature of oxygen cylinder is P1 and T1. When someoxygen is withdrawn the pressure becomes P2 and temperature becomes T2. n1 is initial number ofmoles present in cylinder and n2 is final number of moles left over.

P1 V = 1RT1

P2 V = 2 RT2.

Here P1 = 15 atm = 15 × 1.01 × 105 Nm–2

P2 = 11 × 1.01 × 105 Nm–2

T1 = 27 + 273 = 300 K

T2 = 17 + 273 = 290 K.

V = 30 litres = 30 × 10–3 m3

n1 =1

1

P V

RT = 5 315 1.01 10 30 10

18.258.3 300

Also n2 =2

2

P V

RT = 5 311 1.01 10 30 10

13.848.3 290

Number of moles of oxygen taken out

= 18.25 – 13.84

= 4.41

Mass of oxygen taken out = 4.41 × 32

= 141 gExample 9. A tyre pumped to a pressure of 2 atmosphere at 47°C suddenly bursts. Calculate

the temperature of escaping air? = 1.4 and for air (2)2/7 = 1.219.

Solution. For adiabatic expansion of gas

11

1

T

P

= 21

2

T

P

or 2T =1

21

1

PT

P

or T2 =

1

21

1

PT

P

.

Here T1 = 47 + 273 = 320 K

P1 = 2 atm, P2 = 1 atm = 1.4.

T2 =

0.4

1.41320

2

= 2/ 7

320 320

1.219(2)

T2 = 262.5 K

= 262.5 – 273

= – 10.5 °C


Example 10. Given mass of an ideal gas at NTP is allowed to expand (i) twice of its originalvolume slowly and (ii) four times of its original volume but suddenly. Calculate the final pressure.

Given = 1.4, (4)1.4 = 6.963

Solution. Let P1V1 and T1 be the initial pressure, volume and temperature of gas and finalvalues of these parameters be P2 V2 and T2 respectively.

(i) In first case expansion takes place slowly so it is isothermal process, i.e.,

P1 V1 = P2 V2.

or P2 =1 1

2

176

2

P V

V = 38 cm of Hg.

(ii) In second case expansion takes place suddenly so it is adiabatic process, i.e.,

1 1P V = 2 2 .P V

P2 = 11

2

VP

V

=1.4

176

4

= 1.4

76 7.6

6.963(4)

� 11cm of HgExample 11. A gas is compressed to one fourth of its initial volume adiabatically at a

pressure of 2 × 105 Nm–2. Calculate its new pressure. Also calculate the work done in this process

if initial volume is 2m3 ( = 1.4) and (4)1.4 = 6.96., 101

log (.6021)4 .

Solution. Here P1 = 2 × 105 Nm–2.

V1 = V

V2 =4

V.

Since compression is adiabatic, we have

1 1P V = 2 2P V .

or P2 =1

12

.V

PV

=1.4

52 10/ 4

V

V

= 6.96 × 2 × 105

= 1.39 × 106 Nm–2.Also work done

W = 1 1 2 21

( )1

P V P V


=5 61

(2 10 2 1.39 10 0.5)1.4 1

=5 61

(4 10 0.695 10 )0.4

or �51

(4 7) 100.4

�51

( 3 10 )0.4

= – 7.5 × 105 J

Example 12. A gas is suddenly compressed to 1

4th of its original volume. Calculate the rise

in the temperature, the original temperature is 27°C. and= 1.4, (4)0.4 = 1.741

Solution. Here1

2

V

V = .4 T1 = 27 + 273 = 300 K.

For adiabatic change 11 1T V = 1

2 2T V

T2=

11

12

VT

V

= 300 (4)1.4 – 1

T2 = 300 (4)0.4

= 300 × 1.741

= 522.3 K

= 249.3 °C.Example 13. Two moles of a gas at 227°C expand isothermally until its volume is doubled.

Calculate the amount of work done and heat absorbed.

Solution. Here n = 2, T = 227 + 273 = 500 K

V2 = 2 V1.

Work done during isothermal expansion is

W = 2.303 nRT 210

1

logV

V

= 2.303 × 2 × 8.3 × 500 × log10 2

= 2.303 × 2 × 8.3 × 500 × .3010= 5753.5 J.

= 5.75 × 103 J.

Heat absorbed =W

J =

35.75 10

4.2

� 1.369 × 103 calorieExample 14. Ten litres of air at 27°C and 76 cm of Hg pressure is compressed adiabatically

to a volume of 0.5 litre. Calculate its final temperature. ( = 1.41).


Solution. Here V1 = 10 litre, V2 = 0.5 litre.

T1 = 27 + 273 = 300 K.

Now 11 1T V = 1

2 2T V

or T2 = T1

11

2

V

V

=0.41

10300

0.5

= 300 × (20)0.41

= 300 × 3 .415

1024 K.

= 751.5 °C.Example 15. One mole of gas in a cylinder is compressed adiabatically until its temperature

rises from 17°C to 87 °C. Calculate the work done. Given = 1.4.

Solution. T1 = 17 + 273 = 290 K

T2 = 87 + 273 = 360 K.

= 1.4.

Work done in adiabatic compression of gas is given by

W = 2 1( )1

RT T

=8.3(360 290)

1 1.4

=8.3 70

0.4

= – 1452.5 J.

Example 16. Two samples of a gas initially at same temperature and pressure are compressed

from volume V to .2

V One sample is compressed isothermally and the other adiabatically. In which

sample is the pressure greater?

Solution. In isothermal compression

P1 V1 = P2 V2

or P2 =1

1 12

./ 2

V VP P

V V = 2P1.

In adiabatic compression

2P = 11 1

2(2) .

VP P

V

Since > 1

2P > P2.


Example 17. Three moles of hydrogen at NTP are allowed to expand adiabatically so thattemperature of the gas falls to 263 K. Find the work done by the gas. = 1.4.

Solution. Here n = 3, T1 = 273 K

T2 = 263 K.

= 1.4.

The work done by n moles of a gas in expanding adiabatically from temperature T1 to T2 isgiven by

W = 1 2( )1

n RT T

=3 8.3

(273 263)1.4 1

= 623 J.Example 18. 200 cc of a gas is compressed to 100 cc at atmospheric pressure (106 dyne cm–2).

Find the resultant pressure if the change is slow.

Solution. Here V1 = 200 cc, V2 = 100 cc

P1 = 1 atmospheric

When the change is slow it is isothermal change

P1 V1 = P2 V2

or P2 = 11

2

.V

PV

=200

1100

= 2 atm.

Based On Carnot EngineExample 19. The source temperature of a Carnot engine is 127°C. It takes 500 cal of heat

from the source and rejects 400 cal to the sink during each cycle. What is the temperature of sink?

Solution. Here T1 = 127 + 273 = 400 K.

Q1 = 500 cal. Q2 = 400 cal.

2

1

Q

Q=

2

1

T

T.

T2 =2

11

Q 400400

Q 500T

= 320 K

= 47° C.Example 20. A carnot engine takes in a thousand kilocalorie of heat from a reservoir at

627°C and exhausts it to a sink at 27°C. What is its efficiency?

Solution. Here Q = 1000 k.Cal = 106 Cal.

T1 = 627°C = 627 + 273 = 900 K

T2 = 27°C = 27 + 273 = 300 K.


Now efficiency of the engine is

= 2

1

1 100%T

T

=300

100%1900

= 66.6 %

Example 21. The efficiency of a carnot cycle is 1

.6

If on reducing the temperature of sink

by 65°C, the efficiency becomes 1

3, find the initial and final temperature between which the cycle

is working.

Solution. Let initial temperature of source is T1 and that of sink is T2.

= 2

1

1T

T

1

6= 2

1

1T

T ... (i)

Temperature of sink is reduced by 65, then efficiency is 1

3.

1

3= 2

1

651

T

T

... (ii)


T1 = 390 K = 117°C and T2 = 325 K = 52°C.

Example 22. A locomotive generates steam at a temperature of 190°C and ejects the wastesteam into the air (temperature 30° C). Calculate the thermal efficiency.

Solution. Here T1 = 190 + 273 = 463 K

T2 = 30 + 273 = 303 K.

The thermal efficiency of the locomotive is

= 2

1

1T

T

=303

1463

= 0.3456

= 34.56 %.Exmple 23. A Carnot engine operates between temperature 327 °C and 127 °C. It absorbs 200calorie per cycle from the source. Calculate the thermal efficiency and the heat rejected per cycleto the sink.

Solution. Here T1 = 327 °C = 327 + 273 = 600 K

T2 = 177 °C = 177 + 273 = 450 K

Q1 = 200 Cal cycle–1.


Thermal efficiency of a carnot engine is given by

= 2

1

1T

T

=450 1

1 25%600 4

.

Again = 2

1

Q1 .

Q

Q2 = heat ejected per cycle to the sink.

2

1

1T

T = 2

1

Q1 .

Q

or2

1

T

T =2

1

Q.

Q

or Q2 =2

1

T

T .Q1.

=450

200600

150 Cal cycle–1

Example 24. Calculate the efficiency of the Carnot’s engine working between the steam pointand ice point.

Solution. Here T1 = 100 + 273 = 373 K

T2 = 0 + 273 = 273 K

=2

1

1 100T

T

=273

1001373

= 26.81 %.Example 25. A Carnot engine whose source is at 400 K takes 200 calories of heat at this

temperature and rejects 150 calories of heat to the sink. What is the temperature of the sink?

Solution. Here Q = 200 Cal

Q2 = 150 Cal.

T1 = 400 K.

Using relation1

2

Q

Q =1

2

.T

T

200

150=

2

400

T

or T2 = 300 K.


Example 26. A Carnot engine whose sink is at 27 °C has an efficiency of 50 %. It is desiredto increase the efficiency to 70 %. By how many degrees should the temperature of source beincreased?

Solution. Here = 50 %, T2 = 27 + 273 = 300 K

= 2

1

1 100T

T

50 =1

3001 100

T

1

300

T =1

2

T1 = 600 KExample 27. A Carnot engine has the same efficiency (i) between 100 K and 500 K and (ii)

between TK and 900 K. Calculate the value of T.

Solution. Efficiency

= 2

1

1–T

T

100

1500

= 1900

T

or 1900

T =

400 4

500 5

or900

T=

4 11

5 5

T =900

5 = 180 K

Example 28. A Carnot cycle is performed by air initially at 327 °C. Each stage represents acompression or expansion in the ratio 1:6. Calculate (i) the lowest temperature and (ii) efficiencyof the cycle. Given = 1.4.

Solution. Here T1 = 327 + 273 = 600 K

1

2

V

V =1

6, = 1.4

(i) 12 2T V = 1

1 1T V

T2 =

11

12

VT

V

=1.4 1

1600

6


= 293 K

= 293 – 273 = 20°C

(ii) Efficiency = 2

1

1T

T

=293

1600

= 0.512

= 51.2 %Example 29. A Carnot engine whose sink is at 290 K has an efficiency of 30 %. By how

much the temperature of the source be increased to have its efficiency equal to 60 %, keeping thetemperature of sink constant.

Solution. Here T2 = 290 K, = 30 %

= 2

1

1 100T

T

30 =1

2901 100

T

30

100=

1

2901

T

On solving T1 =2900

7K

For = 60 %

60 =1

2001 100

T

1T =290 100

40

=2900

4 K

T = '1T T

=1 1

29004 7

= 310.7 KExample 30. A Carnot engine operates between 500 K and 400 K . If it absorbs 6 × 105 cal.

heat at higher temperature, how much work per cycle can the engine perform?

Solution. Here T1 = 500 K, Q1 = 6 × 105 Cal

T2 = 400 K


From formula = 2

1

1 100T

T

=400

1001500

= 20 %

Output work = 5 206 10

100

= 12 × 104 Cal

= 5.04 × 105 J

Based On RefrigeratorExample 31. A refrigerator has a coefficient of performance of 7. If the temperature in the

freezer is –23°C, what is the temperature at which it rejects heat?

Solution. Here T2 = – 23 + 273 = 250 K

= 7

From formula = 2

1 2

T

T T

7 =1

250

250T

T1 – 250 =250

7

T1 =250

2507

= 285.7 KExample 32. In a refrigerator, heat from inside at 270 K is transferred to a room at 300 K.

How many joules of heat will be delivered to the room for each joule of electrical energyconsumed?

Solution. Here T2 = 270 K

T1 = 300 K , W = 1J

From relation

= 2 2

1 2

Q T

W T T

Q2 = 2

1 2.

TW

T T

= 1 ×270

300 270= 1 × 9

= 9 J


Example 33. A refrigerator transfers heat from inside at 300 K to the surroundings at 350 K.It extracts 400 J of energy from the contents. Calculate

(i) Coefficient of performance

(ii) Work done by it

(iii) Heat energy supplied to the surroundings.

Solution. Here T1 = 350 K

T2 = 300 K

Q2 = 400 J

(i) Coefficient of Performance

=2

1 2

300

350 300

T

T T

= 6

(ii) Now =2Q

W.

or W =2Q

= 400

6

= 66.7 J(iii) Q = W + Q2

= 66.7 + 400= 466.7 J

= 111.1 CalExample 34. Ice in a cold storage melts at the rate of 4 kg per hour when the external

temperature is at 20° C. Find the minimum power output of the motor used to drive the refrigeratorwhich just prevents the ice from melting. (L = 80 cal/g and J = 4.2 J Cal–1)

Solution. Mass of ice melting per second

=4 1000 10

60 60 9g

To prevent melting of ice, heat to be removed per second

Q2 = m L

=10 800

809 9 Cal.

Also T2 = 273 K

T1 = 20 + 273 = 293 K

Coefficient of performance

=2 2

2 2

Q T

W T T

or W =1 2

22

T TQ

T


=800(293 273)

9 273

=800 20

9 273

cal

sec

= 6.51 cal

sec

= 6.51× 4.2 = 27.35 J/sExample 35. Assuming a domestic refrigerator as reversible engine working between melting

point of ice and the room temperature of 27°C, calculate the energy in joule that must be suppliedto freeze one kg of water. (L = 80 cal/gm. )

Solution. Here T1 = 27 + 273 = 300 K

T2 = 0 + 273 = 273 K

Heat to be removed

Q1 = mL

= 1000 g × 80 cal/g

= 8 × 104 cal.

Now1

2

Q

Q =1

2

T

T

Q1 = 12

2

.T

QT

=4300

8 10273

= 87912.1 cal.

Energy to be supplied = Q1 – Q2

= (87912.1 – 80,000)

= 7912.1 Cal

= 33231 J

Based On Thermal ConductivityExample 36. A thermocole cubical ice-box of side 30 cm has a thickness of 5.0 cm. If 40 Kg

of ice are put in the box, estimate the amount of ice remaining after 6 hrs. The outside temperatureis 45° and coefficient of thermal conductivity of thermocole is 0.01J/sm°C. Given heat of fusionof water = 335 × 103 J/kg.

Solution. Here A = 6 × (0.30)2

= .54 m2.

d = 5 × 10–2 m

t = 6 hr = 6 × 3600 s

(1 –2) = 45 – 0 = 45° C

K = 0.01 Js–1 m–1 °C–1 , L = 335 × 103 J kg–1.

From formula

Q = 1 2( )KA t

d


=0.01 .54 45 6 3600

0.05

= 104976 J

Now Q = mL

m = 3

104976

335 10

Q

L

= 0.313 Kg.

Mass of ice left in the box

= 4.0 – 0.313

= 3.687 kg

Example 37. Total surface area of a box to keep the water cold is 1.6 m2 and wall thicknessis 2.0 cm. The thermal conductivity of box is about 0.01 Js–1m–1 °C–1. It is filled with ice and waterat 0°C. What is the rate of heat flow into the box if the outside temperature is 30°C? Also calculatemass of ice that melts in one day.

Solution. Here K = 0.01Js–1 m–1 °C–1.

A = 1.6 m2, (1 – 2) = 30 °C

d = 2.0 × 10–2 m

Q =1 2( )KA t

d

orQ

t=

1 2( )KA

d

= 2

0.01 1.6 30

2.0 10

= 24 J/s.

Total heat flowing in one day

Q = 24 × 24 × 60 × 60

If m kg of ice melts in one day, then

Q = m L

m = 5

24 24 60 60

3.34 10

= 6.2 kgExample 38. A layer of ice 10 cm. thick in formed on a pond. The temperature of air is –10° C.

Calculate how long will it take for the thickness of ice to increase by 1 mm. Density of ice = 1,thermal conductivity if ice = 0.005 cal cm–1 s–1 °C–1, latent heat of ice = 80 cal g–1.

Solution. The quantity of heat flowing out of ice is given by

Q =1 2( )KA t

d

Also Q = mL

= (Volume × density) L

= (A × 0.1 × 1) × 80

= 8 A


8A = 1 2( )KA t

d

t =1 2

8

( )

d

K

t =8 10

16000.005 10

s

= 26.67 min.Example 39. The walls of an insulating box are made of insulating bricks 8 cm thick. On the

inside, the walls are lined with cork 4 cm thick. The inside temperature is 10°C and the outsidetemperature is 20°C. Determine the temperature at common surface.

given K (brick) = 0.000 4 units

K (cork) = 0.0001 unit.

Solution. Here

For bricks For cork

d1 = 8 cm. d2 = 4 cm

1 = 20 °C 2 = 10 °C

K1 = 0.0004 units K2 = 0.0001 units.

Let be the temperature of the common surface, then

Q1 =1 1

1

( )K A

d

For cork surface.

Q2 =2 2

2

( )K A

d

Rate of flow of heat will remain same

Q1 = Q2

or1 1

1

( )K A

d

=

2 2

2

( )K A

d

or.0004(20 )

8

=

0.0001 ( 10)

4

On solving = 16.6 °C.Example 40. The length of a rod of aluminium is 2.0 m and its area of cross section is 10.0

cm2. The temperature difference between its two ends is 200 °C. How much heat will flow in therod in 5.0 min? Given K = 2.0 × 10–1 kJ/s m°C.

Solution. Here l = 2.0 m, A = 10 cm2 = 10 × 10– 4 m2

= 1 – 2 = 200 °C

t = 5 × 60 sec. , K = 2.0 × 10–1 k J

sm C

= 200 J/s m °C


Q =1 2( )KA t

d

Q =4200 10 10 200 300

2.0

= 6000 J.Example 41. A 10 cm thick slab of surface area 0.36 m2 is placed on a hot surface at a

constant temperature of 100 °C. A block of ice at 0°C is placed on the upper surface of the slab.In half an hour 2.4 kg. of ice melts. Find the thermal conductivity of the material of the slab.(Latent heat of fusion of ice is 3.35 × 105 J/kg).

Solution. Here d = 10 cm = 0.1 m

A = 0.36 m2

= 100 – 0 = 100°C

m = 2.4 kg , L = 3.35 × 105 .J

kg

From relation Q =1 2( )KA t

d

Also Q = mL.

1 2( )KA t

d

= mL

or K =1 2

.

( )

m Ld

A t

=52.4 3.35 10 0.1

0.36 100 0.5 60 60

= 1.24 J/s m° C.Example 42. Steam at 100°C is passed into a copper cylinder 10 mm thick and of 100 cm2.

Water at 100°C collects at the rate of 150 g per minute. Find the temperature of the outer surfaceif the conductivity of copper is 0.8 cal s–1 °C–1 cm–1 and latent heat of steam 540 cal/gm.

Solution. Here 1 = 100°C, l = 10 mm = 1 cm

A = 100 cm2, t = 1 min = 60 sec,

K = 0.8 Cal/s cm°C

We know Q =1 2( )KA t

d

also Q = m L

m L =1 2( )KA t

d

150 × 540 = 20.8 100(100 ) 60

1

On solving 2 = 83.12°C


Example 43. Two vessels A and B made of different materials but having identical shape, sizeand wall thickness are filled with ice and kept at the same place. Ice melts at the rate of 100 g min–1

and 150 g min–1 in A and B respectively. Assuming heat enters both vessels through the wall onlycalculate the ratio of thermal conductivities of their materials.

Solution. Let the thermal conductivities of the two materials be K1 and K2, then we have

Q1 =1 1 2( )K A t

d

= m1 L

Q2 =2 1 2( )K A t

d

= m2 L

or1

2

K

K=

1

2

100 2

150 3

m

m

i.e. K1 : K2 = 2 : 3Example 44. Three bars of equal length and equal areas of cross section are connected in

series. Their thermal conductivities are in the ratio 2 : 4 : 3. The open ends of the first and last barsare at temperature 200°C and 18°C. Calculate the temperatures of both the junctions.

Solution.

1 2200 18

d d dA B C D

C C

Let the temperature of junction B and C be 1 and 2 respectively. In steady state, the rate offlow of heat will remain same

Q

t= 1 1 2(2 ) ( ) (200 ) 4 . ( )K A K A

d d

=23 . ( 18)K A

d

or 2 (200 – 1) = 4 (1 – 2) = 3 (2– 18 °C)

On solving

1 = 116°C

2 = 74°C

Example 45. A box 0.5 m2 in area and whose sides are 0.6 cm thick is filled with meltingice and kept in a room, the temperature of the room being 20°C. Calculate the ice that will melt in20 min. (K = 0.0004 cal cm–1 s–1 °C –1)

Solution. Here A = 0.5 m2 = 5000 cm2.

d = 0.6 cm.

= 1 – 2 = 20°C

t = 20 × 60 = 1200 sec

We know Q =1 2( )KA t

d

Also Q = m L


m L =1 2( )KA t

d

or m =1 2( )

.

KA t

d L

=.0004 5000 20 1200

0.6 80

= 1000 g

= 1 kg.

Based On Stefan’s Law, Newton’s Law of cooling and Wien’s Displacement LawExample 46. Assuming that the total surface area of the human body is 1.2 m2 and the surface

temperature is 30 °C. Find the rate at which energy is radiated from human body. (Stefan-Boltzmann constant = 5.67 × 10–8 W/m2 K4).

Solution. Here A = 1.2 m2, T = 30 + 273 = 303 K

e = 1, = 5.67 × 10–8 W/m2 K4.

From formula

E = A T 4

= 1.2 × 5.67 × 10–8 (3034)

= 573 W.Example 47. The surface temperature of a hot body is 1327°C. Find the wavelength at which

it radiates maximum energy. Given Wien’s constant = 0.2898 cm K.

Solution. Here T = 1327 + 273 = 1600 K.

b = 0.2898 cm K

By Wien’s Law m =b

T

=0.2898

1600

= 1.81125 × 10–4 mExample 48. A small hole is made in a hollow sphere whose walls are at 527°C. Find the

total energy radiated per second per cm2. Given Stefan’s constant = 5.7 × 10–5 erg cm–2 s–1 K–4.

Solution. Here T = 527 + 273

= 800 K

= 5.7 × 10– 5 erg cm–2 s–1 K–4.

Total energy radiated per second per cm2

E = T4

= 5.7 × 10–5 (900)4.

= 5.7 × 10–5 × 6561 × 108

= 5.7 × 6561 × 103

= 3.7397 × 104 ergExmple 49. A black body at 27°C surrounds another black body at –73°C. Calculate the net

heat transferred per second per square metre of the body at higher temperature. ( = 5.67 × 10–8

S.I. units)


Solution. T = 27 + 273 = 300 K

T0 = –73 + 273 = 200 K

= 5.67 × 10–8 S.I. Units.

Using Stefan Boltzmann Law

E = ( 4 40T T )

= 5.67 × 10–8 (3004 – 2004)

= 5.67 × 10–8 (34 –24) × 108

= 5.67 × (81 –16)

= 368.55 J/s m2

Example 50. A black body of surface area 200 cm2 is maintained at (227 °C). Calculate theenergy radiated in one minute. Given = 5.7 × 10–8 S.I. units

Solution. Here A = 200 cm2 = 200 × 10–4 m2

T = 227 + 273 = 500 K

= 5.7 × 10–8 S.I. units.

From formula E = A e T 4

= 200 × 10–4 × 1 × 5.7 × 10–8 (500)4

= 200 × 10–4 × 1 × 5.7 × 10–8 × 625 × 108

= 2 × 5.7 × 625 × 10–2

= 71.25 J/s

Energy radiated in one minute

= 71.25 × 60

= 4275 J

Example 51. The ratio of radiant energies radiated per unit surface area by two bodies is 16: 1.The temperature of hotter body is 1000 K. Calculate the temperature of colder body.

Solution. Here 1

2

E

E=

16

1 , T1 = 1000 K.

From formula E = T4

1

2

E

E=

414

2

T

T

16

1=

4

42

(1000)

T

or T2 = 500 KExample 52. How much faster does a cup of tea cool off from 80°C than from 27°C?

Temperature of the surrounding is 17°C?

Solution. Here T1 = 80 + 273 = 353 K

T2 = 27 + 273 = 300 K

T0 = 17 + 273 = 290 K

Using relation E 4 40( )T T


1

2

E

E=

4 41 04 4

2 0

T T

T T

=4 4

4 4

(353) (290)

(300) (290)

=84.546

10.272

= 8.23Example 53. Luminosity of Rigel star in Orion constellation is 17,000 times that of one sun.

If the surface temperature of sun is 6000 K, calculate the temperature of the star.

Solution. Here 1

2

E

E= 17,000 , T2 = 6000 K

According to Stefans’s Law

1

2

E

E=

4 41 14 4

2 2

T T

T T

T1 =

141

22

.E

TE

= 6000 (17000) 1/4

� 68511. 5 kExmple 54. A body cools from 80°C to 50°C in 5 minutes. Calculate the time it takes to cool

from 60°C to 30°C. The temperature of the surroundings is 20°C.

Solution. Here 1 = 80°C, 2 = 50°C, 0 = 20°C

= 1 2 80 5065 .

2 2C

t = 5 min

According to Newton’s Law of cooling

1 2

t

= K ( – 0)

80 50

5

= K (65 –20) ... (i)

In second case

1 = 60°C 2 = 30°C

=60 30

452

C

(60 30)

t

= K (45 – 20) ... (ii)


Dividing equation (i) by (ii)

5

t=

45

25

or t = 9 secExample 55. Two bodies A and B are kept in evacuated vessels maintained at a temperature

of 27°C. The temperature of A is 527°C and that of B is 127°C. Compare the ratio at which heat islost from A and B.

Solution. Here T0 = 27°C = 27 + 273 = 300 K

T1 = 527°C = 527 + 273 = 800 K

T2 = 127°C = 127 + 273 = 400 K.Using Stefan–Boltzmann Law

A

B

E

E=

4 41 04 4

2 0

T T

T T

=4 4

4 4

(800) (300)

(400) (300)

= 23.Example 56. Estimate the temperature of the surface of the sun from the following data:

average radius of earth orbit = 1.5 × 108 km; average radius of the sun = 7.0 × 105 km, solarradiants power on earth at noon-time is 1400 Wm–2. Assume the sun to be perfect black body.(Given = 5.67 × 10–8 S.I units )

Solution. Total energy radiated by sun

E = T4 × surface area of the sun

= T4 × 4R2 (T is temperature of surface of sun)

Energy falling per second per unit area of earth’s surface

=4 2

2

.4

4

T R

r

= 4 2

2

T R

r

Radiant Power on earth = 1400 Wm–2

1400 = 5.67 × 10–8 × T 4 ×

28

11

7.0 10 .1.5 10

or T4 =1614 2.25

105.67 49

or T =

1

4 414 2.2510

5.67 49

� 5803 K.

Example 57. Two stars radiate maximum energy at wavelength 3.6 × 10–7 m and 6.0 × 10–7

m respectively Calculate the ratio of their temperature.

Solution. Here m1 = 3.6 × 10 –7 m.

m2 = 6.0 × 10 –7 m.

m1 T1 = m2 T2.


2

1

T

T =7

17

2

3.6 10

6.0 10m

m

= 3 : 5Example 58. A metallic ball having surface area 200 cm2 and at a temperature of 527°C is

placed in an enclosure at 27°C. If the surface emissivity of the metal be 0.2, at what rate is heatbeing lost by radiation by the ball? (= 5.7 × 10–12 watt/ cm2°C4).

Solution. Here A = 200 cm2

T = 527°C = 527 + 273 = 800 K

T0 = 27°C = 27 + 273 = 300 K

e = 0.2

= 5.7 × 10–12 watt/ cm2 °C4

Amount of heat lost per second from surface area A

E = e A 4 44( )T T

= 0.2 × 5.7 × 10 –12 2 4

watt

cm C × 200 cm2 (8004 – 3004) × °C4

= 0.2 × 5.7 × 10–12 × 200 (8004 – 3004) × Watt

=12 4 4

7

0.2 5.7 10 200(800 300 ) cal

sec4.2 10

= 21.81 calsec

Example 59. Surface temperature of sun is 6000 K. If we consider it as a perfect black body,calculate the energy radiated away by the sun per second. Given radius of sun = 6.92 × 105 km and = 5.67 × 10–8 J m–2 s–1 K–4

Solution. The energy radiated per second by a black body of area A is given by

E = A e T4

Here A = 4 R2 = 4 × 3.14 × (6.92 × 108)2

T = 6000 K

e = 1.

E = 4 (3.14) (6.92 × 108)2 × 1 × 5.67 × 10–8 × (6000)4.

� 4.42 × 1026 W.

Based On Kinetic Theory Of GassesExample 60. Molar volume is the volume occupied by 1 mole of any (ideal) gas at standard

temperature and pressure. Show that it is 22.4 litres.

Solution. Here T = 0 + 273 = 273 K.

P = 1 atm = 1.01 × 105 Nm–2.

R = 8.31 J/mol–K

PV = RT

� V = 5

8.3 273

1.01 10

RT

P

= 22.4 × 10–3 m3.

= 22.4 litres.


Example 61. Estimate the total number of molecules inclusive of oxygen, nitrogen, watervapour and other constituents in a room of capacity 25.0 m3 at a temperature of 27°C and 1 atm.pressure (KB = 1.38 × 10–23 J/K).

Solution. Here V = 25.0 m3, T = 27 + 273 = 300 K.

We know PV = nRT (for n moles).

PV = n NKBT. BR

KN

nN =B

PV

K T

=5

23

1.01 10 25

1.38 10 300

� 6.12 × 1026

Example 62. An oxygen cylinder of volume 30 litre has an initial gauge pressure of 15 atmand a temperature of 27 °C. After some oxygen is withdrawn from the cylinder, the gauge pressuredrops to 11 atm and its temperature drops to 17°C. Estimate the mass of oxygen taken out of thecylinder, R = 8.31 J/mol –k, M = 32.

Solution. Here V1 = 30 litre = 30 × 10–3 m3 ; V2 = 30 litres = 30 ×10 –3 m3

P1 = 15 atm = 15 × 1.01 × 105 Nm–2

P2 = 11 atm = 11 × 1.01 × 105 Nm–2

T1 = 27 + 273 = 300 K

T2 = 17 + 273 = 290 K.

Using relation P1V1 = n1 RT1

n1 =5 3

1 1

1

15 1.01 10 30 10

8.3 300

P V

RT

= 18.3

Similarly n2 =5 311 1.01 10 30 10

8.3 290

= 13.9

Number of moles of oxygen taken out = 18.3 – 13.9

= 4.4

Mass of oxygen taken out of cylinder = 4.4 × 32

= 140.8 g.Example 63. A given mass of a gas at –73°C exerts a pressure of 40 cm of mercury. What

pressure will be exerted at 127 °C if the volume remains constant?

Solution. Here T1 = – 73 + 273 = 200 K

T2 = + 127 + 273 = 400 K

P1 = 40 cm of Hg

P2 = ?

Using relation 2

1

P

P= 2

1

T

T


P2 = 2

1

T

T × P1

=400

40200

= 80 cm of HgExample 64. Molecular weight of oxygen is 32. At STP volume of 1 g of oxygen is 700 cm3.

Find the value of gas constant R.

Solution. Here 1 atm = 1.01 × 105 Nm–2.

T = 273 K

M = 32.

Volume of 1 g of oxygen = 700 cm3

Molar Volume = 700 × 32 cm3.

= 700 × 32 × 10 –6 m3.

Using gas equation PV = RT

R =5 61.01 10 700 32 10

273

PV

T

= 8.31 J/mol –K.Example 65. The density of a gas is 1.25 kg m–3 at STP. Calculate its density at 27°C and 730

mm of Hg.

Solution. We know Density =mass

volume

Volume =mass

density

Let the mass of gas is 1 Kg.

� V1 =31

m1.25

, V2 =1

T1 = 273 K T2 = 27 + 273 = 300 K

P1 = 76 cm of Hg P2 = 730 mm of Hg

= 73 cm of Hg.

Now1 1

1

P V

T =2 2

2

P V

T.

76 1

1.25 273

=

73 1

300

or =73 1 1.25 273

76 300

= 1.09 Kg m–3


Example 66. In a certain region of space there are only 5 molecules per cm3 on an average.The temperature is 3 K. Calculate the pressure of gas. Given Boltzmann constant is K = 1.38 × 10 –23

J K –1.

Solution. Heren

V= 5 molecules/cm3.

= 5 × 106 /m3.

T = 3K.

PV = nkBT

P = Bn

k TV

= 5 × 106 × 1.38 × 10 –23 × 3= 20.7 ×10–17 Nm –2

Example 67. An air bubble of volume 1.0 cm3 rises from the bottom of a lake 40 m deep ata temperature of 12 °C. To what volume does it grow on reaching the surface where thetemperature in 35 °C? The atmospheric pressure is 1.0 × 105 Nm–2 and density of water is 103 kgm–3 (g = 9.8 N/kg)

Solution. Let P1 is pressure at the bottom of lake, then

P1 = P + hdg

= 1.00 × 105 + (40 × 103 × 9.8)

= 1.0 × 105 + 3.92 × 105

= 4.92 × 105 Nm–2.

At bottom volume V1 = 1.0 cm3. = 1.0 × 10–6 m3

T1 = 12° C = 12 + 273 = 285 K.

At the surface of the lake

P2 = 1.0 × 105 Nm–2, T2 = 35 + 273 = 308 K

Using gas law

1 1

1

P V

T = 2 2

2

P V

T

V2 =1 1 2

2 2

P V T

T P

=5 6

5

4.92 10 1.0 10 308

285 1.01 10

= 5.3 × 10–6 m3

Example 68. Five molecules of gas have speeds 2, 4, 6, 8 and 10 km s–1. Calculate theaverage speed and root mean square speed.

Solution. Average speed

vav = 1 2 3 4 5

5

v v v v v

=2 4 6 8 10

5

=

30

5 = 6 km/s


Root mean square speed

vrms =2 2 2 2 21 2 3 4 5

5

v v v v v

=2 2 2 2 22 4 6 8 10 220

5 5

= 44 = 6.63 kms–1

Example 69. The density of nitrogen at S.T.P. is 1.25 × 10–3 g/cm3. Determine the roots meansquare speed of nitrogen molecules.

Solution. Here = 1.25 × 10–3 g/cm–3 , P = 1 × 105 Nm–2

= 1.25 kg/m3

Using formula vrms =53 3 1 10

1.25

P

= 4.89 × 102 ms–1

Example 70. Calculate the root mean square velocity of nitrogen molecules at 27°C and76 cm of Hg pressure.

Solution. Here T = 27 + 273 = 300 K.

Root mean square velocity

vrms = 3RT

M

= 3

3 8.31 300

28 10

= 516.8 ms–1

Example 70. At what temperature is the root mean square speed of an atom in an argon gascylinder equal to the r.m.s. speed of a helium gas atom at – 20 °C? Atomic mass of argon = 39.9and that of helium is 4.0.

Solution. Root mean square speed of helium at temperature T is

vrms =3RT

M

But (vrms)argon = (vrms)Helium

3

39.9

RT=

3 253

4

R

or39.9

T=

253

4.0

or T =253 39.9

4

= 2523.7 K.


Example 72. Calculate the temperature at which r.m.s. velocity of gas molecules is double itsvalue at 27°C pressure is kept constant.

Solution. We know vrms =3RT

M

i.e vrms T

27

tv

v=

273

27 273

t

or3v

v=

273

300

t .

or 9 =273

300

t

t + 273 = 2700

t = 2700 –273 = 2427 K.Example 73. Calculate the temperature at which r.m.s. velocity of oxygen molecules will be

10 km s–1. Give molecular weight of oxygen is 32.

Solution. We know vrms =3RT

m

� T =2

3rmsMv

R

=3 232 10 (10 1000)

3 8.3

=532 10

3 8.3

= 1.29 × 105 KExample 77. Calculate the K.E. of hydrogen per g-mol at 0°C given that R = 8.3 joule/mol–k

Solution. K.E per g-mol of a gas is given by

K.E =3

2 RT

=3

2 × 8.3 × 273 = 3.4 × 103 J.

Example 75. Calculate the root mean square velocity of oxygen molecules at 15°C. (GivenR = 8.3 J/ mole –K).

Solution. Here T = 15 + 273 = 288 K.

R = 8.3 J/mol K.

M = 32 × 10–3 kg


vrms =3RT

m

= 3

3 8.3 288

32 10

= 473 ms–1

Example 76. Calculate the average energy of a helium atom at (i) room temperature 27°C (ii)the temperature on the surface of sun (6000 K) and (iii) the temperature of 107 K (kB = 1.38 × 10–23

J

mol K)

Solution. Here kB = 1.38 × 10–23 J/mol–k

Average K.E. per molecule/atom is E = 3

2 Bk T

(i) At 27 °C

E = 2331.38 10 (27 273)

2

= 6.2 × 10–21 J.(ii) At 6000 K

E = 2331.38 10 6000

2

= 1.242 × 10–19 J.

(iii) At 107 K E = 23 731.38 10 10

2

= 2.07 × 10–16 JExample 77. The kinetic energy of a molecule of oxygen at 0°C is 5.64 × 10–21J. Calculate

Avogadro’s number, using R = 8.31 J/mol–K.

Solution. The K.E. of a molecule of gas is given by

E =3

2 Bk T

=3

2

RT

N

N =3 .

2

R T

E

= 21

3 8.3 273

2 5.64 10

= 6.03 × 1023.

Example 78. Calculate the temperature at which the average kinetic energy of a molecule ofgas will be same as that of an electron accelerated through 1volt, Boltzmann constant kB = 1.4 ×10–23 J/mol


Solution. Here e = 1.6 × 10–19 C

V = 1 V , kB = 1.4 × 10–23 J/mol–K

Average K.E. of a molecule= energy of electron

3

2 Bk T = 1.6 × 10–19 × 1

T =19

23

2 1.6 10

3 1.4 10

= 7619 KExample 79. The temperature of an ideal gas is increased from 120 K to 480 K. If at 120 K

the root mean square velocity of gas molecules is V, than what will be the root mean squarevelocity at 480 K.

Solution. Here T1 = 120 K, T2 = 480 K

(vrms)1 = v (vrms)2 = ?

We know

vrms T

1

2

( )

( )rms

rms

v

v=

1

2

T

T

2( )rms

v

v=

120

480

or (vrms)2 = 2v.

Example 80. 0.014 Kg of nitrogen is enclosed in a vessel at a temperature of 27°C. Howmuch heat has to be transferred to the gas to double the rms velocity of its molecules? (kB = 1.38× 10–23 J/mol–K) and Avogadro’s number is 6 × 1023 mole–K.

Solution. Here T = 27 + 273 = 300 K.

We know vrms T .

To double the rms velocity temperature is to be made four times i.e. 4T.

Change in temperature = 4 T – T = 3T

= 3 × 300 = 900 K.

For diatomic gas CV =5

2R =

5

2 × 8.31

= 20.7 J/mole K.

Heat required = n. CV .dT

=0.014 1000

20.7 90028

= 9315 JExample 81. At what temperature will the average velocity of oxygen molecules be sufficient

so as to escape from earth? Escape velocity from the earth is 11.0 km s–1 and mass of one moleculeof oxygen is 5.34 × 10–26 kg. (kB = 1.38 × 10–23 J/K ).


Solution. Let Ve is escape velocity of oxygen molecule, then

3

2 BK T = 21

2emV

T =2

3e

B

mv

k

=26 3 2

23

(5.34 10 ) (11.0 10 )

3 1.38 10

= 1.56 × 105 KExample 82. From a certain apparatus the diffusion rate of hydrogen has an average value

of 28.7 cm3/s, while that of another gas is 7.2 cm3/s. Identify the gas.

Solution. The diffusion rate R of a gas is directly proportional to the rms to velocity of themolecule of gas, i.e .

R vrms

1

2

R

R =1

2

( )

( )rms

rms

v

v

=2

1

M

M �� Vrms = 3RT

M

M1 and M2 are molecular masses of hydrogen and unknown gas respectively.

28.7

7.2=

2

1

M

M

or2

1

M

M � 16

M2 � 16 M1 = 16 × 2 �� M1 = MH = 2

= 32

Unknown gas is oxygen

EXERCISE

Based On Work done (P-V diagram)1. In the diagram two paths ABC and ACB are shown

representing the thermodynamic changes in a gasfrom state A to state C. The internal energies of thegas at A and C are 20 J and 50 J respectively.Calculate

(i) How much heat will be necessary to besupplied to the gas to change the state from A to Calong the path AC?

(ii) If 50 J of heat is necessary to be supplied to thegas for the change of state from A to B. What willbe the internal energy of the gas at state B.

15

0 2 10

BA

C

P (

Nm

)–2

V (m )3

10

5


(iii) How much work will be done by gas on thepath BC? (Ans. 70 J, 40 J, 30 J)

2. A gas is taken from an original state P to anintermediate state R by the process shown. Itsvolume is then reduced to original volume fromR to P, by an isobasic process. Calculate the totalwork done by the gas from Q to R and R to P.

[Ans. 450 J]

Based On first Law of Thermodynamics3. 1 g of water at 100°C and atmospheric pressure is changed into steam at the same

temperature. The volume of 1 cm3 of water becomes 1671 cm3 on boiling. Calculate thechange in its internal energy (Latent heat of vaporisation = 540 cal/g)

[Ans. 499.8 calorie]

4. Calculate the change in the internal energy when 20 g of air in heated from 0°C to 5°C. Thespecific heat of air at constant volume is 0.172 cal/g°C [Ans. 17.2J]

5. In a certain process 20,000 calorie of heat is supplied to the system while the system does12000 J of work. Calculate the change in the internal energy. (Ans. 71.6 kJ)

6. 2 g of water at 100°C is converted into steam at the same temperature. The volume of 1cc ofwater becomes 1671 cc on boiling. Calculate the change in internal energy of the system, ifheat of vaporisation is 540 cal/g. [Ans. 999.6 cal]

7. The volume of a gas at the atmospheric pressure is 2.0 litre. On giving 300 J of heat to thegas its volume increases to 2.5 litre at the same pressure. Determine the change in theinternal energy of the gas. [Ans. 250 J]

8. Calculate the change in the internal energy of a block of copper of mass 200 g when it isheated from 25° to 75°C. Assume the change in volume as negligible. Sp heat of Copper is0.1 [Ans. 4200 J]

9. What amount of work will have to be done in converting 1g of water into steam at 100 °C annormal atmospheric pressure? The volume of 1 kg of water at 100°C is 10–3 m3 and that of 1kg of steam is 1.671 m3 and the normal pressure is 105 Nm–2. [Ans. 1.670 × 105 J]

10. Calculate the change in internal energy when 50 g of air is heated from 0°C to 2°C, thespecific heat of air at constant volume being 0.172 cal/g °C. [Ans. 72.2 J]

Based on Isothermal and Adiabatic Process

11. The air in a cylinder is compressed to 1

15 of its initial volume. If the initial pressure is 1.0

× 105 Pa and initial temperature is 300 K find the final pressure and temperature aftercompression (Given = 1.4) [Ans. 44.3 × 105 Pa, 886 K]

12. A volume of gas at atmospheric pressure is compressed adiabatically to half of its originalvolume. Calculate the resulting pressure. ( =1.4) [Ans. 200.5 cm of Hg]

13. A certain volume of gas at 27 °C and 105 Nm–2 pressure expands isothermally until itsvolume is doubled and then adiabatically until its volume is redoubled. Find the finalpressure of gas. [ =1.4, (2)1.4 =2.64]. [Ans. 1.89 × 104 Nm–2]

14. A gas whose initial temperature is 27 °C is compressed adiabatically to 8 times its initialpressure. Calculate the rise in the temperature ( = 1.5) [Ans. 300°C]

600

300

0 2 .0 5 .0

PR

Q

P (

Nm

)–2

V (m )3


15. The volume of a given mass of air is doubled in adiabatic expansion. If the initial pressureof the air is 80 cm of mercury, what will be the final pressure? [ = 1.41, (2)1.41 = 2.66]

[Ans. 30 cm of Hg]

16. A volume of gas at atmospheric pressure is compressed adiabatically to half of its originalvolume. Calculate the final pressure. ( = 1.4) [Ans. 200.5 cm of Hg ]

17. Air is compressed adiabatically to half of its initial volume. Calculate the change in itstemperature. [Ans. 0.319 times of initial temperature]

18. The initial pressure and volume of 1 mole of an ideal gas are 2 × 105 Nm–2 and 6 litrerespectively. The gas is compressed adiabatically so that its final pressure and volume

becomes 6.24 × 105 Nm–2 and 2 litre respectively. Calculate the work done 5

3 .

[Ans. –72 J]

19. Calculate the work done to compress an ideal gas of 30 litre at 1 atm isothermally to avolume of 3 litre. [Ans. 2.73 kJ]

20. One mole of a gas is isothermally expanded at 27°C till the volume is doubled. Calculate thework done. [Ans. 1726 J]

21. Calculate the work done to compress isothermally 1 g of hydrogen gas at N.T.P. to half itsinitial volume. Find the amount of heat evolved and change in internal energy.

[Ans. –786.2J, 187. 2 Cal, Zero ]

22. A quantity of perfect gas at 15°C is compressed adiabatically to one-fourth of its volume.Calculate the final temperature. ( = 1.4) and (4)0.4 = 1.74 [Ans. 228°C]

23. Five moles of hydrogen gas at STP compressed adiabatically so that its temperature becomes400°C. Calculate the amount of work done on the gas ( = 1.4) [Ans. – 4.2 × 104 J]

24. A certain mass of air at normal temperature is compressed (i) slowly (ii) suddenly to onethird of its volume. Find the rise in temperature, if any, in each case ( = 1.4)

[Ans. (i) 222 K (ii) Zero

Based On Carnot-Engine and Refrigerator25. A Carnot engine whose temperature of the source is 400 K takes 200 calories of heat at this

temperature and rejects 150 calories of heat to the sink. What is the temperature of the sink?Also calculate the efficiency of the engine. [Ans. 25 %]

26. A Carnot engine operates between temperature 327°C and 177°C. It absorbs 300 calories percycle from source. Calculate the thermal efficiency and heat rejected per cycle to the sink.

[Ans. 25%, 225 cal/cycle]

27. A Carnot engine works between the temperature 100°C and 10°C. Calculate its efficiency.

[Ans. 24 %]

28. The working substance in Carnot engine absorbs 200 calories of heat from source at 227°Cand rejects of 150 calories to sink. Calculate the temperature of sink. [Ans. 102°C]

29. A reversible heat engine takes heat from source at 527 °C and rejects heat to sink at 127°C.Calculate how many calories of heat is absorbed from source to produce mechanical work atthe rate of 750 watt. (1 Cal = 4.2 J) [Ans. 357.1 cal s–1]

30. A Carnot engine operates between 500 K and 400 K. If it absorbs 6 × 105 cal heat at highertemperature, how much work per cycle can the engine perform? [Ans. 5.04 × 105 J]

31. Calculate the efficiency of a Carnot engine operating between the temperature 100°C and400°C [Ans. 44.57 %


32. A Carnot engine has the same efficiency (i) between 100 K and 500 K and (ii) between TKand 900 K. Calculate the value of T. [Ans. 180 K]

33. The source temperature of Carnot engine is 227°C. It takes 500 cal of heat from source andrejects 400 cal of heat to sink. What is the temperature of sink? [Ans. 127 °C]

34. A refrigerator freezes water at 0°C into 10 kg ice at 0°C in a time interval of 30 min.Assuming the room temperature to be 22°C. Calculate minimum amount of power needed tomake 10 kg of ice [Latent heat of ice = 80 k cal/kg] [Ans. 150 W]

35. A Carnot engine working between 300 K and 600 K has output work 1000 J per cycle.Calculate the heat supplied to the engine from the source. [Ans. 2000 J]

36. A Carnot engine takes in 100 Cal of heat from the source at 400 K and gives up 70 cal to thesink. What is the temperature of the sink and thermal efficiency of the engine?

[Ans. 280 K, 30%]

37. The temperature inside a refrigerator is –3°C and the room temperature is 27°C. How manyjoules of heat will be delivered to the room for each joule of electrical energy consumed?

[Ans. 10 J]

38. A refrigerator is to maintain eatables kept inside at 9°C. If room temperature is 36°C.calculate coefficient of performance. [Ans. 10.4]

39. In a refrigerator heat from inside at 277 K is transferred to a room at 300 K. How manyjoules of heat shall be delivered to the room for each joule of electrical energy consumed.

[Ans. 13 J]

40. A Carnot engine working as a refrigerator between 250 K and 300 K receives 1000 Caloriesof heat from the reservoir at the lower temperature.

(i) Calculate the amount of heat rejected to the reservoir at high temperature.

(ii)Calculate the amount of work done in each cycle to operate the refrigerator.

[Ans. (i) 1200 cal (ii) 840 J ]

41. The efficiency of Carnot engine is 1

6. If on reducing the temperature of the sink by 65°C, the

efficiency becomes 1

3 find the initial and final temperature between which the cycle is

working. [Ans. 117°C, 52°C]

42. A refrigerator has to transfer an average of 263 J of heat per second from temperature 10°Cto 50°C. Calculate the average power consumed. [Ans. 35 W]

43. Refrigerator A works between – 10°C and 27°C while refrigerator B works between – 27°Cand 17°C, both removing heat equal to 2000 J from the freezer. Which of the two is betterrefrigerator. [Ans. Refrigerator A]

Based on Thermal Conductivity44. The opposite faces of square sheets of metal of thickness 2 mm and area 1000 sq. cm are

maintained at the temperature difference of 75°C. Calculate the amount of heat that flowsacross the sheet in 10 minutes. K = 0.25 [Ans. 5625000 cal]

45. The opposite faces of a metal cube of side 10 cm are maintained at the temperature ofboiling water (100°C) and melting ice (0°C). Calculate the amount of ice that melts in halfan hour, the conductivity of the metal being 0.30. [Ans. 6750 g]

46. The amount of heat that is conducted across a square sheet of silver per minute is 60 × 106

cal. The area of the sheet is 1 sq. metre and thickness of the sheet is 1 mm. Calculate thetemperature difference between the opposite faces. K = 0.99. [Ans. 1.01°C]


47. The difference of temperature between the outside and the inside of the room is 20°C.Calculate the quantity of heat that will be conducted in one minute across glass window1 metre square and thickness 2 mm. K for glass = 0.002. [Ans. 1200 cal]

48. A metal sheet 4 mm thick and 25 cm square has temperature difference of 40°C between itsfaces. If 7.2 × 106 cal of heat is conducted per hour, calculate the conductivity of the metal.

[Ans. 0.032]

49. Steam at 100°C is passed through an iron pipe 10 metre long and 1.5 mm thick. The innercircumference of the pipe is 20 cm. Water at 100° collects at the rate of 1.5 kg per minute.What is the temperature outside? K for iron = 0.2 and latent heat of steam = 540 cal pergram. [Ans. 99.494°C]

50. Calculate the amount of ice that will melt in one hour if kept in a wooden box of size 50 ×40 × 20 cm assuming that the atmospheric temperature is 25°C. K for the wood = 0.0004 andthickness of wood is 5 cm. [Ans. 846 g]

51. A box of area 5 × 104 sq cm is filled with ice and kept in a room whose temperature is 35°C.The thickness of wood is 4 cm. Calculate the conductivity of the material of the box if 10 kgof ice melts in one hour. [Ans. 0.00051]

52. There is a layer of ice 5 cm thick over the surface of a pond. The temperature of air is – 8°C.Calculate the rate of loss of heat of water in the pond per square metre. K for ice = 0.005.

[Ans. 80 cal/s]

53. A layer of ice 20 cm thick is formed on a pond. The temperature of air is – 6°C. Find howlong it will take for thickness of ice to increase by 2 mm (K for ice = 0.005 and density ofice= 0.9 g/cc). [Ans. 3 hrs 20 min]

54. A square iron plate of side 3 cm and thickness 8 mm has its opposite faces at 100°C and60°C respectively. How much heat flows through the plate per minute? K for iron = 0.16

[Ans. 4320 cal]

55. The opposite faces of a metallic plate of thickness 0.2 cm are at a difference of temperatureof 100°C and the area of the plate is 200 sq. cm. Find the quantity of heat that will flowthrough the plate in one minute. K = 0.2 [Ans. 12 × 105 cal]

56. A cubical vessel of 10 cm of side is filled with ice at 0°C and is immersed in a water bath at100°C. Find the time in which all the ice in the vessel will melt. Thickness of the wall of thevessel is 0.2 cm, coefficient of conductivity is 0.02 cal/s cm°C and density of ice is 0.916 g/cc. [Ans. 12.21 sec]

57. The temperature inside a room is 25°C and outside it is 10°C. How much heat will leaveroom in 10 minutes through a glass window 2 m long, 1 m wide and 0.4 cm thick. [K forglass = 0.002 ] [Ans. 9 × 105 cal]

58. A glass vessel with an insulating cover and with surface area of 100 sq. cm and 1.5 mm thickis filled with ice at 0 °C and placed in a second vessel maintained at a temperature of 100°C.Find the mass of ice that melts per minute when the flow of heat becomes steady. Latentheat of ice = 80 cal/ gram and K for glass = 0.00185 C.G.S. units. [Ans. 92.5 g/min]

59. Calculate the rate of loss of heat through a glass window of area 500 sq cm and thickness 5mm. The temperature inside the room is 20°C and outside is – 10°C. K for glass = 0.002C.G.S. units. [Ans. 60 cal/sec]

60. Two rods A and B are of equal length. Each rod has its ends at temperature T1 and T2. Whatis the condition that will ensure equal rates of flow of each through the rods A and B?

[Ans. K1 A1 = K2 A2]


Based on Stefan’s Law, Newton’s Law of Cooling and Wien’s Displacement Law61. Calculate the value of the surface temperature of the moon, using Wien’s displacement law.

[Ans. = 200K]

62. Calculate the radiant emittance of black body at a temperature of (i) 400 K (ii) 4000 K.

[Ans. 1452 watts/m2, 14520 watt/m2]

63. A furnace is at a temperature of 2000 K. At what wavelength will it radiate maximumintensity? [Ans. 14,400 A°].

64. Calculate the radiant emittance of a black body at a temperature of 200 K. ( = 5.672 × 10–8

W/m2 K4) [Ans. 90.75 W/m2]

65. Calculate the energy radiated per minute from the filament of an incandescent lamp at 2000K if the surface area is 5.0 × 10–5 m2 and its emissivity is 0.85. [Ans. 2313 J]

66. Calculate the energy radiated per second from the filament of a lamp at 3000 K, if thesurface area is 5.0 × 10–5 m2 and emissivity is 0.85. [Ans. 196.2 W]

67. Calculate the temperature at which a perfectly black body radiates energy at the rate of onewatt per cm2. [Ans. 648 K]

68. The temperature of a body is increased from 27°C to 127°C. By what factor would theradiation emitted by it increased? [Ans. 256/81]

69. A body which has surface area of 5.0 cm2 and a temperature of 727° C radiates 500 J ofenergy each minute. What is the emissivity? [ = 5.67 × 10–8 W/m2 K4] [Ans. 0.18]

70. How much energy is radiated per minute from the filament of an incandescent lamp at 3000K, if the surface area is 10–4 m2 and its emissivity is 0.4? [Ans. 11022.5 J]

71. Calculate the rate of loss of heat by radiation from a sphere of 10 cm radius, at a temperature200°C, when the surroundings are at a temperature of 20°C. ( = 5.7 × 10–5 erg/cm2 s deg4)

[Ans. 73.1 cal s–1]

72. Find at what temperature a black body loses thermal energy at the rate of 1 watt per cm2.

[Ans. 374°C]

73. Compare the ratio of loss of heat from a body at 1000°C and from the same body at 500°Cwhen placed in an enclosure at 15°C. [Ans. 7.48 : 1]

74. A body cools in 7 minutes from 60°C to 40°C. What will be its temperature after next 7minutes? Temperature of the surrounding is 10°C. [Ans. 28 °C]

75. Hot water filled in a pan cools from 94°C to 86°C in 2 minutes. The room temperature is20°C. How long will it take to cool from 71°C to 69°C? [Ans. 42s]

Based on Kinetic Theory of Gases76. An electric bulb of volume 250 cm3 was sealed off during manufacture at a pressure of 10–3

mm of mercury at 27°C. Calculate the number of air molecules contained in the bulb.

[Ans. 8×1013 molecules]

77. The pressure of a gas is changed at constant volume from 2000 Nm–2 to 3000 Nm–2. If theinitial temperature of the gas is 77°C what will be its final temperature? [Ans. 525 K]

78. The volume of a gas at pressure 1.2 × 107 Nm–2 and temperature 127°C is 2.0 litre. Find thenumber of molecules in the gas. [Ans. 4.35 ×1024]

79. The temperature of a gas is 27° C. The volume of this gas is increased 1.5 times at constantpressure. What will be its temperature? [Ans. 177 °C]


80. The temperature of gas is changed from 27°C to 127°C. If the initial pressure was 1500 Nm–2

calculate the new pressure given volume is constant. [Ans. 2000 Nm–2]

81. A given mass of a gas occupies 200 ml at 27°C. At constant pressure its temperature ischanged to 127°C. Calculate its new volume. [Ans. 267 ml].

82. Show that the number of molecules per unit volume of an ideal gas is given by APN

RT where

NA is Avogadro’s number.

83. At what temperature, pressure remaining constant, will the root mean square velocity of agas be half of its value at 0°C? [Ans. –204.75°C]

84. Calculate the temperature at which the average K.E. of the molecules of a gas is one-third ofthe average K.E. of its molecules at 180°C. [Ans. –122°C]

85. At what temperature, pressure remaining constant, will the root mean square velocity of gasbe half of its value at 0°C? [Ans. – 204.75 °C]

86. Calculate the temperature at which nitrogen molecules have root mean square velocity 5.06× 104 cm s–1. [Ans. 15°C]

87. Calculate the root mean square velocity of the molecules of oxygen at 0°C. The density ofoxygen at S.T.P. = 1.43 g litre–1 [Ans. 4.6 × 102 ms–1]

88. Find the molecular kinetic energy of 1 g of helium at S.T.P. What will be the energy at100°C? [Ans. 1.16 × 103J]

89. What is the root mean square speed of a hydrogen molecule at 300 K?

[Ans. 1.93 × 103 ms–1]

90. Calculate the ratio of K.E. of the molecules of gas at 27°C and 127°C. [Ans. 3 : 4]

91. The mean K.E. of 1 kg. mole of nitrogen at 27°C is 600 J. What will be its mean kineticenergy at 127°C. [Ans. 800 J]

92. At what temperature will oxygen molecules have the same root mean square velocity as thatof the hydrogen molecules at –100°C? [Ans. 2495 °C]

93. Estimate the fractional molecular volume to the actual volume occupied by oxygen gas atNTP. Take the radius of an oxygen molecule to be roughly 3 A°. [Ans. 3 × 10 –3 ]

94. At what temperature is the root mean square velocity of oxygen molecules twice their rootsmean square velocity at 27°C. [Ans. 927°C]

95. The temperature of a gas is –73° C . To what temperature should it be heated so that averageK.E. of the molecules is doubled? [Ans. 127°C]

459

Periodic Motion:- A motion which repeats itself again and again after equal intervals oftime is called periodic motion e.g. Motions of the arm of a watch.

Oscillatory Motion:- A motion in which a body moves to and fro (back and forth) about afixed point is called oscillatory or vibratory motion e.g. motion of Simple Pendulum.

Simple Harmonic Motion:- It is a special type of oscillatory motion, in which restore force(acceleration) is directly proportional to the displacement of body from mean position, and isalways directed towards the mean position. i.e.

Force – displacement

or acceleration – displacement

Here negative sign shows that force (acceleration) always points in the opposite directionsof displacement.

Equation of simple Harmonic Motion (SHM): The displacement of a particle from meanposition, performing SHM is given by

y = a sin t or y = a cos t

a = amplitude of SHM

= angular frequency.

Velocity ln SHM: - Rate of change of displacement of the particle at any instant of timeis called velocity i.e.

v =dy d

dt dt (a sin t)

= a cos t

=21 sina t

=2 21 /a y a

=2 2a y

(i) At mean position y = 0

v = aIt is the maximum value of velocity in SHM Called velocity amplitude i.e.

vmax = a(ii)At extreme positions y = a

v = 0Acceleration is SHM :- Rate of change of velocity of the particle at any instant is called

acceleration

��14UNIT


Therefore =dv d

dt dt (a cos t)

= – a sin t

= – 2 y.

(i) At mean position y = 0

= 0(ii)At extreme Position y = a

max = a2.

Time Period of SHM: - = 2 y.

or =y

Time Period T =2

2y

T =Inertia factor

2Spring factor

Simple Pendulum: - Time period of oscillation of a Simple pendulum is given by

T = 2l

g .

l = distance between the point of suspension and

centre of the bob.

Second’s Pendulum : - A Pendulum whose time period is two seconds is called second’sPendulum.

Its Length is 99.3 cm

Time period of a Body Suspended from a massless Spring: - Consider a body of mass mis attached to a massless spring and pulled downward. The body will execute SHM whose timeperiod is given by

T = 2l

g = 2

m

k .

l = extension produced in the spring

m = mass of body

k = spring constant of the spring

Important :- If a spring of spring constant k is divided into n equal parts and the body issuspended from one such part then

T = 2m

nk .

Oscillations 461

Combination Of Springs-(a) Series combination: - If two springs of spring constant k1 and k2 are connected in series

than the spring constant k of the combination is given by

1

k=

1 2

1 1

k k

or k = 1 2

1 2

k k

k k

� T = 2m

k =

1 2

1 2

( )2

m k k

k k.

(b) Parallel Combination:- If two springs of spring constant k1 and k2 are connected inparallel then the spring constant k is given by

k = k1 + k2

T =1 2

m2�

k + k

Motion of Body Floating Vertically in Liquid: -Consider a body (density ) of verticallength L floats in a liquid (density ) such that its length h is inside liquid then time period ofits oscillations is

T = 2L

g

= 2

h

g.

Motion Of a Body In a Tunnel Due Along The Diameter of Earth:- A body dropped in atunnel along the diameter of earth performs SHM whose time period is given by

T = 2R

g

R = Radius of earth.

Motion of Liquid in U tube: - If a liquid of density contained in a U tube upto height his depressed, it executes SHM and its time period is given by

T = 2h

g

or T = 22L

g

L = Total of liquid in U tube.

Simple Pendulum Of Infinite Length The time period of a simple pendulum of length l comparable with radius of earth is given

by.


T =1

2 .l R

gl R

(i) If l = R is radius of earth

T =. 1

2 .2

R R

R g

= 22

R

g .

(ii) If l =

T = 12 .

1

l R

R gll

=1

2 .1

R

R gl

T =1

2 .1

R

R g

= 2R

g .

Rotatory Simple Harmonic MotionIn case of rotatory SHM the time period is given by

T = 2I

I = moment of Inertia

= Torque acting on the rotationg body.

Energy of Particle Executing SHMThe K.E of a particle of mass m at any instant of time is

K.E = Ek = 2 2 21( )

2m a y

a = amplitude of SHM

y = displacement of Particle from mean

Oscillations 463

for y = 0

K.E. = 2 212

m a .

It is maximum value of kinetic energy .

The potential energy of the particle at any instant is given by

P.E. = 2 212

m y

P.E. is maximum when y = a

Maximum P.E =2 21

2m a .

Total energy of a particle executing SHM

= 2 212

m a .

SOLVED EXAMPLES

Based on displacement, velocity and AccelerationExample 1. A simple harmonic motion is given by

y = 10 sin (20 t + 0.5)

Write down its amplitude, angular frequency, time period and initial phase; if displacementis measured in metre and time in second.

Solution. Given y = 10 sin (20 t + 0.5)

Equation of SHM is y = a sin (t + ).

(i) a = 10 m(ii) = 20 rad/sec

(iii) frequency =20

.2 2

10Hz

(iv) Time period T =2 2

.20

sec10

(v) Initial Phase = 0.5 rad.

Example. 2 A particle in SHM is described by the displacement function

x (t ) = A cos (t + ) ; = 2

T

.

If the initial (t = 0) position of the particle is 1cm and its velocity is cms–1, What are itsamplitude and initial phase angle? Angular frequency of the particle is s–1.

Solution. Here t = 0, x = 1 cm, v = cm/s and = s–1.

x (t ) = A cos (t + )

v (t) = – A sin (t + ).dx

vdt

�

at t = 0, x (0) = 1 = A cos ... (i)


(0) = = – A sin or A sin = – 1 ... (ii)

squaring and adding

A2 cos2 + A2 sin2 = 2

or A = 2 cm.

Divide equation (ii) by (i).

tan = –1

or =34

.

Example. 3 A particle performs SHM with amplitude 0.5 cm and time period 1500

sec. If

initial phase is 4

write equation of SHM.

Solution. Here a = 0.5 cm

=1 1 1500

Hz./1500T

Equation of SHM is y = a sin (t + ).

y = 0.5 1500

sin 24

= 0.5sin 3000

4

cm

Example. 4 A body of mass 1 kg is executing SHM is given by

y = 6.0cos 1004

t

cm

What is the (i) amplitude of displacement (ii) frequency (iii) velocity

Solution. y = 6.0cos 1004

t

.

Standard equation of SHM is

y = a cos (t + ).


(i) a = 6.0 cm

(ii) = 100

or v =100 50

Hz.2 2

(iii) velocity v = .dy

dt

Oscillations 465

v = 6.0cos 1004

dt

dt

= 6.0 100sin 1004

t

v = 1600sin cms100

4t

Example. 5 The equation of SHM is given by

y = 3 sin 10 t + 4 cos 10 t

where y is in cm and time in sec. Calculate amplitude, period and initial phase.

Solution. Given y = 3 sin 10 t+ 4 cos 10 t ... (i)

Standard equation of SHM is

y = a sin (t + )

= a sin t cos + a cos t sin ... (ii)

Comparing equation (i) and (ii)

= 10 a cos = 3 ... (iii)

a sin = 4 ... (iv)

(i) Period T =2 2 1

10 5

= 0.2 sec.

Squaring and adding equation (iii) and (iv)

a2 = 32 + 42

or a = 5 cm

Also tan =4

1.3333

= 53.8°.

Example 6. A particle in SHM is described by the displacement function

x (t) = B sin (t + ) = 2

T

If the initial (t = 0) position of the particle is 1 cm and its initial velocity is cm–1, whatare its amplitude and initial phase angle? Given angular frequency of particle is s–1.

Solution. Here t = 0, x (0) = 1cm, v (0) = cm s–1.

= s–1

Now x (t) = B sin (t + )

v (t) = ( ) cos ( )d

x t B tdt

t = 0 x (0) = 1 = B sin or B sin = 1 ... (i)

v (0) = = B cos


or cos = 1 ... (ii)

Squaring and adding equation (i) and (ii)

B2 = 2

or B = 2 cm

Also divide equation (i) by (ii)

Tan = 1

=4

or 54

Example 7. A small body of mass 100 g is undergoing SHM of amplitude 100 cm andperiod 0.2 sec. What is the maximum value of the force acting on the body? What is the phasedifference between the acceleration of the body and its displacement?

Solution Here m = 100 g = 0.1 kg

a = 100 cm = 1 m, T = 0.20 sec.

Maximum force = m 0

= m (a2)

=2

20.1 1

0.2

= 98.6 NExample 8. A body is performing SHM of amplitude 1 m. Its velocity while passing through

mean position is 10 ms–2. Calculate its frequency.

Solution Here a = 1 m

v0 = 10 ms–1

When the body passes through mean position its velocity is maximum i.e.

v0 = a

10 =2

1T

v =1 10

2 3.14T

= 1.6 Hz

Example 9. Show that

y = a cos t + b sin t

represents SHM. Also determine its amplitude

Solution. General equation of SHM is

y = A sin (t + )

y = A sin t cos A cos t sin ... (i)

Given equation of SHM

y = a cos t + b sin t ... (ii)


b = A cos ... (iii)

Oscillations 467

a = A sin ... (iv)

Square and add

A = 2 2a bDifferentiating equation (ii)

v =dy

dt = – a sin t + b cos t

Again differentiating w.r.t. ‘t’

2

2

d y

dt= – a cos t – b sin t

= –2 (a cos t + b sin t)

= –2 y

i.e = – 2 yHence given equation represents SHM.

Example 10. A particle executes SHM of amplitude 25 cm and time period 3s. What is theminimum time required for the particle to move between the points 12.5 cm on either side of themean position?

Solution Here a = 25 cm, T = 3s

y = 12.5 cm.

y = a sin t

12.5 =2

25sin3

t

or1

2=

2sin

3

t

or2

sin3

t= sin

6

2

3

t=

6

or t =

1sec

4

Time taken by the particle to move between the two points 12.5 cm. on either side

=1

24

= 1

sec2

Example 11. A particle is moving with SHM in a straight line. When the distance of theparticle from the equilibrium position has values x1 and x2, the corresponding values of velocitiesare v1 and v2. Show that the time period of oscillation is given by

T =

1/ 22 22 12 21 2

2x x

v v

Solution. Here y = x1 v = v1


y = x2 v = v2

We know v =2 2a y

v1 = 2 21a x

v2 = 2 22a x

21v = 2 2 2

1( )a x ... (i)

22v = 2 2 2

2( )a x ... (ii)

Subtracting (i) from (ii)

22v – 2

1v =2 2 2

1 2( )x x

or 2 =

2 22 12 21 2

v v

x x

or =2 22 12 21 2

v v

x x

or2

T

=

2 22 12 21 2

v v

x x

or T =2 21 22 22 1

2x x

v v

2 22 12 21 2

-2�

-

x x

v v

Example 12. In what time after its motion begins with a particle oscillating according to

equation y = 5 sin 2

t

move from mean position to maximum displacement

Solution y = 5sin2

t

For maximum displacement

y = a = 5

5 = 5sin2

t

or sin2

t= 1 = sin

2

sin2

t= sin

2

or t = 1 sec

Oscillations 469

Example 13. In a simple pendulum oscillation, the amplitude is 2 cm and time period is 2sec. Calculate velocity and acceleration amplitude.

Solution a = 2 cm = 2 × 10–2 m

T = 2 sec

Velocity amplitude v0 = a = a × 2 = 2

aT

= 2 22 10

2

Velocity amplitude = 2 × 10–2 = 6.28 × 10–2 m/s.

Acceleration amplitude 0 =2

2a

T

=2

2 22 10

2

= 2 × 9.8 × 10–2

= 19.6 × 10–2 ms–2

Example 14. A particle executing linear SHM had a maximum velocity of 40 cm s–1 and amaximum acceleration of 50 cm s–2. Calculate its time period.

Solution. Here v0 = 40 cm s–1

0 = 50 cm s–2

v0 = a = 40 ... (i)

0 = a2 = 50 ... (ii)

Divide equation (ii) by (i)

=5

4

or2

T

=

5

4

or T =8

5

= 1.6 Sec.

Example 15. A Particle executing SHM has a maximum displacement of 4 cm and itsacceleration at a distance of 1 cm from its mean position is 3 cm s–2. What will be its velocitywhen it is at a distance of 2 cm from its mean position?

Solution. Acceleration = 2 y

Here acceleration = 3 cm s–2, y = 1 cm

3 = 2 × 1

or = 3 rad/sec

Now v =2 2a y


v = 2 23 (4) (2)

= 3 12

= 36 = 6 m/s

Example 16. A particle executing SHM along a straight line has a velocity of 8 ms–1 whenat a distance of 3 m from the mean position and 6 ms–1 when at a distance of 4 m from it.Calculate the time taken by the particle to travel 2.5 m from its mean position

Solution. Here y1 = 3 m v1 = 8 ms–1

y2 = 4 m v2 = 6 ms–1

We know v =2 2a y

8 =2 23a or 64 = 2 (a2 – 9)

6 =2 24a 36 = 2 (a2 – 16)

or64

36=

2

2

9

16

a

a

or16

9=

2

2

9

16

a

a

On Solving a = 5 m.

Now y = a sin t

2.5 = 5 sin t

or1

2= sin 2 t

or sin 6

= sin 2 t

or 2 t =6

or t = sec12

Example 17. For a particle performing SHM the acceleration is 24 cm/s2 at a distance of 6cm from its mean position. Calculate the time period of SHM.

Solution. Here = 24 cm/s2

y = 6 cm

= 2 y = 2

2y

T

� T2 =24 y

2 26 5 46 3

2 rad/s.

Oscillations 471

or T = 2y

= 6

2 3.1424

= 3.14 secExample 18. A horizontal platform vibrates up and down with SHM of amplitude 20 cm.

At what frequency will an object kept on the platform just loose contact with the platform.g = 9.8 ms–2.

Solution. The object will loose contact from the platform

mg = m = ma2

or =g

a

v =1

2 2

g

a

=1 9.8 7

2 3.14 0.2 2 3.14

� 1.1 Hz.

Example19. A body is vibrating with SHM amplitude 15 cm and frequency 4 Hz. Calculatethe maximum acceleration and the velocity when the displacement is 10 cm

Solution. Here a = 15 cm, v = 4 Hz

y = 10 cm

Maximum Acceleration is

0 = a2

=2

2a

T

= 15 (2 × 4)2

= 15 × 4 × 2 × 16

= 9408 cm/s2 = 94.1 m/s2

Also v =2 2a y

=2 22 a y v

=2 222

2 4 (0.15) (.10)7

� 2.81 m/sExample 20. A particle executes SHM of the period sec. The amplitude of SHM is 2 cm.

Find the maximum velocity. Also calculate its velocity at half its displacement

Solution. Here T = = 3.14 sec

a = 2 cm


Max velocity is

v0 = a = 2

2T

=2

2

= 4 cm/s

Now v = 2 2a y

= 2 22a y

T

= 2 22

2 1

= 2 3

= 2 1.732 = 3.46 cm/s

Based On SpringExample 21. A body of mass 12 kg is suspended by a coil spring of natural length 50 cm

and force constant 2.0 × 103 Nm–1. What is the stretched length of the spring? If the body ispulled down further by stretching the spring to a length of 59 cm and then released, what is thefrequency of oscillation of the suspended mass.

Solution. Here m = 12 kg l = 50 cm

k = 2.0 × 103 Nm–1

Extension produced in the spring is

y =mg

k = 3

12 9.8

2.0 10

= 0.0588 m

= 5.88 cm Stretched length of the spring = 50 + 5.88 = 55.88 cmAlso

v =31 1 2 10

2 2 3.14 12

k

m

= 2.06 Hz

Example 22. A Spring of force constant 1200 Nm–1is mounted horizontally on a horizontaltable. A mass of 3.0 kg is attached to the free end of the spring pulled sideways to a distance of2.0 cm and released .

(i) What is the frequency of oscillation of the mass

(ii)What is the maximum acceleration of mass?

(iii) What is the maximum speed of mass?

Solution. Here k = 1200 Nm–1, m = 3.0 kg

a = 2.0 cm = 2.0 × 10–2 m.

frequency of oscillation of the mass

v =1

2

k

m

Oscillations 473

=1 1200

2 3.14 3

= 3.18 Hz(ii)Maximum accretions is = a2

= a (2v)2

= (2.0 ×10–2) (20)2

= 8.0 ms–2

(iii) Maximum Speed = a= 2.0 × 10–2 × 20

= 0.40 ms–1

Example 23. Two bodies A and B of mass 1 kg and 2 kg respectively are rigidly connectedto a spring of force constant 400 Nm–1, the body B rests on horizontal table, from the restposition body A is compressed by a distance of 2 cm and released calculate (i) the frequency ofoscillation (ii) the maximum velocity of body A.

Solution. Given m = 1 kg, k = 400 Nm–1

a = 2 cm = 2 × 10–2 m

(a) v =1 1 400

2 2 3.14 1

k

m

� 3.18 Hz(b) Maximum velocity

v0 = a = a × 2 = 2 × 10–2 × 2 × 3.14 × 3.18

� 0.4 m/sExample 24. A body of mass 25 kg is oscillating on a spring of force constant 100 Nm–1.

Deduce the angular frequency of body.

Solution Here m = 25 kg

k = 100 Nm–1

Angular frequency =k

m

=100 10

25 5 = 2 rad/sec

Example 25. A body of mass 0.5 kg is suspended from a weight less spring. The length ofthe spring increases by 5.0 cm. When the body is slightly pulled down and let go, it begins tooscillate in a simple harmonic way. Find the force constant and periodic time of oscillations.

Solution. Here m = 0.5 kg , y = 5.0 × 10–2 m.

k = 2

0.5 9.8

5 10

mg

y

= 98 Nm–1

��

�


Also T = 2m

k =

0.52 3.14

98

= 0.448 secExample 26. A 5kg collar is attached to a spring of spring constant 500 Nm–1. It slides

without friction over a horizontal rod. The collar is displaced from its equilibrium position by10.0 cm and released. Calculate (a) the period of oscillations (b) maximum speed.

Solution. Here m = 5 kg , k = 500 Nm–1

a = 10.0 cm = 0.1 m.

(a) T = 2m

k

=5

2500

= 2

10

= 0.628 sec

(b) Also v0 = a = 2

aT

= 0.10 × 2

102

= 1 ms–1

Example 27. A spring balance has a scale that reads from 0 is 50 kg. The length of the scaleis 20 cm. A body suspended from this balance when displaced and released oscillates with periodof 0.6 sec. What is the weight of body?

Solution. Here y = 0.20 , m = 50 kg

k =mg

y = 50 9.8

0.2

= 50 × 49 Nm–1.

T = 2m

k

0.6 = 250 49

m

or m =2

0.36 50 49

4( )

kg

weight W = mg

= 2

0.36 50 499.8

4

= 220 NExample 28. A Spring is hung vertically and loaded with a mass of 100 g and allowed to

oscillate. Calculate the time period. When the spring is loaded with 200 gram it extends by 10 cm

Oscillations 475

Solution. Here m = 100 g = 0.1 kg

M = 200 g = 0.2 kg , y = 10 cm = 0.1 m

k =Mg

y

T = 2m

k = 2

my

Mg

=0.1 0.1

20.2 9.8

= 0.499 sec.

Example 29. The pan attached to a spring has a mass of 1 kg. A weight of 2 kg when placedon the pan, stretches the spring by 10 cm. What is the frequency with which the empty pan willoscillate?

Solution. Here m = 1kg , F = 2 kg wt = 2 × 9.8 N

y = 10 cm = 0.1 m

k =2 9.8

0.1

mg

y

= 196 Nm–1

v =1

2

k

m =

1 196 14

2 1 2

=7

= 2.23 Hz.

Example. 30 An impulsive force gives an initial velocity of –1.0 ms–1 to the mass in theunstretched position. What is the amplitude of the motion? Given x as a function of time t forthe oscillating mass. m = 3 kg and k = 1200 Nm–1

Solution. Velocity of mass in unstretched position of spring is –1 ms–1

vmax = 1 ms–1

Also =1200

3

k

m = 20 rad/s.

amplitude A = max 1

20

v

= 0.05 m

Now x = A sin t

= 5 sin 20 t

As initial impulse is negative the displacement is towards -ve x axis

x = – 5 sin 20 tExample 31. Two masses m1 = 1.0 kg and m2 = 0.5 kg are suspended together by a massless

spring of force constant k = 12.5 Nm–1. When they are in equilibrium position m1 is gentlyremoved. Calculate the angular frequency and the amplitude of oscillation of m2. g = 10 ms–2

Solution. x is the extension in the spring when both m1 and m2 are suspended then

F = (m1 + m2) g = ky


or y =1 2( )m m g

k

When m1 in removed extension decreases to y. m2 g = ky

or y =2m g

k

y – y =1 2 2( )m m g m g

k k

= 1m g

k

It is amplitude of oscillation of mass m2.

i.e., a =1 1.0 10

12.5

m g

k

= 0.8 m

Angular frequency =2

12.5

0.5

k

m

= 5 rad/s.

Example 32. Following figure shows four different spring arrangements. If the mass of eacharrangement is displaced from its equilibrium position and released, what is the resultingfrequency of vibration of each case?

K 1

K 2

(a )

K 1 K 2

(d )

m

(b) (C)

1m

2m

k1

k2

k1 k2

m

Oscillations 477

Solution. In above figure (a) and (b) springs are connected in parallel. Let the springconstant of two springs be k1 and k2. In parallel combination the extension in both the springswill be same say it is y

Restoring force in each spring is

F1 = k1 y , F2 = – k2 y

F = F1 + F2 = – k1 y – k2 y = – (k1 + k2) y.

If k is equivalent spring constant then

F = – ky

or – ky = – (k1 + k2) y

or k = (k1 + k2).

v =1

2

m

k = 1 2

12

m

k k

In figure (c) and, (d) , the springs are connected in series. In series combination, therestoring force will remain constant but extension in the two springs will be different say it is y1and y2 respectively.

F = – k1 y1 = – k2 y2

or y1 =1

F

k and y2 =

2

F

k

If y is the total extension in the series combination then

y = y1 + y2 = 1 2

1 1F

k k

y = 1 2

1 1

k kF

k k

.

Let equivalent spring constant in parallel combination is k then

y =F

k

1

k=

1 2

1 2

k k

k k

or k = 1 2

1 2

k k

k k

v =1

2

m

k =

1 2

1 2

( )12

m k k

k k

Example33. A tray of mass 12 kg is supported by two identical springsan shown. When the tray is pressed down slightly and released it executesSHM with a time period of 1.5 sec. What is the force constant of eachspring? When a block of mass M is placed on the tray the period of SHMchanges to 3.0 sec. What is the mass of the block?

Solution. Here both the springs are in parallel. Let the spring constantof each spring is k then

m m


Keq = k + k = 2k

T = 22

m

k

1.5 =12

22k

... (i)

or k =2

2

4 12

2 (1.5)

= 105.17 Nm–1

When block of mass M is placed on the pan the time period is 3 sec.

3 =12 M

22k

... (ii)

Divide (ii) by (i) 2 =12 M 2

2 12

k

k

4 =12 M

12

or M = 36 kgExample 34. A Trolley of mass 3.0 kg is connected to two identical springs each of force

constant 600 Nm–1. If the Trolley is displaced from its equilibrium position by 5.0 cm andreleased what is (i) the period of oscillations (ii) the maximumspeed of the trolley (iii) How much is the total energy dissipatedas heat by the time the trolley comes to rest due to dampingforces.

Solution. Here m = 3.0 kg

Both the springs are in parallel k = 600 + 600

= 1200 N/m.

a = 5.0 × 10–2 m

(i) T = 2m

k = 3

2 3.141200

= 0.314 sec

(ii) Maximum Speed = vmax = a

=k

am

= 2 12005.0 10

3 = 1 ms–1

(iii) Energy dissipated as heat = Initial Maximum K.E

=2max

1

2m v

=21

3(1)2

= 1.5 J

600 N/m 600 N/m3.0 Kg

Oscillations 479

Example 35. If the spring in the above problem are replaced by rubber bands. What is theanswer to (i) and (ii), above

Solution. A rubber band acts like a spring during stretching but provides no restoring forceduring compression

In case of rubber band k = 600 Nm–1.

(i) T = 2m

k =

32

600 =

3.14 2

10 2

= 0.44 sec(ii)Maximum speed of trolley

v0 = a

=2 a

T

=

2 3.14 0.05

0.44

= 0.71 ms–1

Example 36. Two identical springs each of spring constant k are connected in (i) series (ii)parallel and support a mass m. Calculate the ratio of their time periods

Solution. When two springs are connected in series then force constant will be k

2.

Ts = 2/ 2

m

k =

22

m

k

In parallel equivalent spring constant is 2k

Tp = 22

m

k

s

p

T

T=

2 2 /

2 / 2

m k

m k

=

2

1

Example 37. The time period of a body suspended from a spring of spring constant k is 4sec. The spring is divided and now the body is suspended from one part. What will be its newtime period.

Solution. T = 2m

k

or 4 = 2m

k ... (i)

The spring is divided in to four equal parts. The spring constant of each part will be 4k

T = 24

m

k ... (ii)


Divided (ii) by (i)

4

T =

/ 4

/

m k

m k =

4

m k

k m

T = 2 secExample 38. The period of oscillation of mass m suspended by an ideal springs is 2 sec. If

an additional mass of 2 kg be suspended the time period is increased by 1 sec. Find the value ofm.

Solution. Here m = m, T = 2 sec

m = (m + 2) , T = 3 sec

T = 2m

k

2 = 2m

k

3 =2

2m

k

or2

3=

2

m

m or 4m + 8 = 9m

or m = 1.6 kgExample 39. What will be force constant of the spring system shown in figure.

Solution. Springs of spring constant k1 and k2 are connected in parallel so their equivalent

spring constant is = 2 k1

This parallel combination is in series with spring of constant k2.

1

k=

1 2

1 1

2k k

1

k=

2 1

1 2

2

2

k k

k k

or k =

1 2

2 1

2k kk + 2k

Example. 40 Two particles A and B of equal masses are suspended from two masslesssprings of spring constant k1 and k2 respectively. If maximum velocities during oscillation areequal, then calculate the ratio of amplitudes of A and B

Solution We know =k

m

k

m

k1 k1

k2

Oscillations 481

or1

2

=

1

2

k

k

(vmax)1 = (vmax)2

a1 1 = a2 2

1

2

a

a= 2 2

1 1

k

k

Based on various examples of SHMExample 41. The acceleration due to gravity on the surface of moon is 1.7 ms–2. What is the

time period of simple pendulum on the moon if its time period on the earth is 3.5 sec. (g =9.8 ms–

2)

solution. Here gm = 1.7 ms–2 , Tm = ?

ge = 9.8 m/s2 , Te = 3.5 sec

Te = 2e

l

g

Tm = 2m

l

g

m

e

T

T =e

m

g

g

or Tm =9.8

3.51.7

� 8.4 sec.Example 42. The length of a simple pendulum is increased by 1cm, as a result the time

period in increased by 1

100sec. Find the original length of the pendulum.

Solution. We know T = 2l

g

Differentially both sides

dT = 1/ 22 1.

2l dl

g

= dll g

l =2 2

2

( ).( )

dl

g dT


or l =2 2

2

(3.14) (0.01)

19.8

100

= 1 m

Example 43. Calculate the percentage change in the time period of a simple pendulum ifthe length of the pendulum is increased by 2%.

Solution. We know

T = 2l

g

or T2 =24 l

g

Using error method

2 T

T

=

l

l

or 2

100T

T

= 100l

l

100T

T

=1

2%2

= 1%Example 44. Two pendulum begin to swing simultaneously. The first pendulum makes 9

oscillations when the other makes 7. Calculate the ratio of length of the two pendulum

Solution. Let the time taken be t

So time period of Ist pendulum is 9

t

and time period of 2nd pendulum in 7

t

9

t=

12l

g

7

t=

22l

g

or 1

2

l

l=

27

9

4981

Example 45. Two pendulum of length 121 cm and 100 cm start vibrating. At some instantthe two are in the same phase. After how many vibrations of the shorter pendulum the two willbe in phase in the mean position ?

Solution. Let T1 and T2 be the time periods of two pendulum

T1 =121

2g

. T2 = 100

2g

Oscillations 483

Let the shorter pendulum makes n vibrations, then the second pendulum will complete onevibration less i.e. (n – 1) to come in phase

n × 100

2g

= (n –1) 121

2g

n2 × 100 = (n –1)2 × 121

100 n2 = 121 n2 – 242 n + 121

or 21 n2 – 242 n + 121 = 0

On Solving n = 11Example 46. Simple pendulum oscillates with an amplitude 50 mm and time period 2s.

Calculate the maximum velocity of its bob.

Solution. Here a = 50 mm = 50 × 10–3 m

T = 2s

Maximum velocity

V0 = a = aT

=350 10 2

2

= 0.157 m/sExample 47. A pendulum which gives correct time beats second on ground at a place is

taken to the top of a tower 320 m high. Calculate the loss of time of the pendulum clock in oneday

Solution. The time period of simple pendulum is given by

T =l

g also g = 2

Gm

R

T =2lR

GM

T R

At height h

T1 (R + h)

1T

T= 1 .

R h h

R R

or 1T T

T

= .

h

R

or (T1 – T) = 6

2 320.

6.4 10

hT

R

= 10–4 sec.

Loss of time in one day = 10–4 × 43200

= 4.32 sec.


Example 48. A pendulum suspended from the roof of an elevator at rest has a time periodT1. When the elevator moves up with an acceleration a its time period becomes T2. When theelevator moves down with an acceleration a, its time period becomes T3. Show that

T1 = 2 3

2 22 3

2 T T

T T

Solution. When the lift is stationary

T1 = 2l

g or

2

21

4 lg

...(i)

When lift is going upward then

T2 = 2l

g a

or

2

22

4( )

lg a

T

...(ii)

When lift is going downward then

T3 = 2l

g a

or

2

23

4 l

T

= g – a ...(iii)

Adding equation (ii) and (iii)

2 2

2 22 3

4 4l l

T T

= 2g

or2 2

2 22 3

4 4l l

T T

=2

21

42

l

T

using (i)

On Solving

T1 =2 3

2 22 3

2 T T

T T

Example 49. A Second, pendulum is taken in a carriage. Find the time period of oscillationif the carriage moves with an acceleration of 5ms–2 vertically upwards. (g = 10 m/s2)

Solution. We know

T = 2l

g

For Second’s pendulum 2 = 2l

g

or l = 2 2

10gm

... (i)

When the carriage moves up with an acceleration of 5 ms–2

Oscillations 485

T = 2l

g a

= 2

10 12

(10 5)

Using Equation (i)

=2 10

15

=2

1.22 = 1.64

Example 50. The balance wheel of a watch has moment of inertia 2 × 10–8 kg m2 andtorsional constant of its hair spring is 9.8 × 10–6 Nm/rad. Calculate its frequency

Solution. Here = 9.8 × 10–6 Nm/rad

I = 2 × 10–8 kg m2

v =1

2 I

= 6

8

1 9.8 10

2 3.14 2 10

= 3.53 HzExample 51. A vertical U tube of uniform area of cross section upto a height 4.90 cm. Show

that if the water on one side is depressed and then released, its up and down motions in tube isSHM. Calculate its time period.

Solution. Here h = 4.9 cm , g = 9.80 ms–2

T = 2h

g

=24.9 10

29.80

= 1

214.14

= 0.444 sec.

Example 52. A Cubical body (side 0.1 m and mass 0.002 kg) floats vertically in water. It ispressed and then released so that it oscillates. The density of water is 1000 kg m–3 calculate thetime period of oscillations.

Solution. Here m = 0.002 kg

A = 0.1 × 0.1 = 0.01 m2.

= 1000 kg m–2 ]

Time period of oscillation

T = 2m

A g

T =3

0.0022

0.01 10 9.8

= 0.028 sec.


Example 53. In a U tube the level of water is at 60 cm. In one side level of water isdepressed and released so that it oscillates. Calculate its frequency of oscillation

Solution. Here h = 60 cm = 0.6 m

g = 9.8 m/s2

frequency of oscillation is

v =1

2

g

h =

1 9.8

2 0.6

v =1 49

2 3.14 3

=1

2 3.14 × 4.04

= 0.64 Hz.

Example 54. A test tube weighing 10 g and external diameter 2 cm is floated vertically inwater by placing 10 g of mercury at its bottom. The tube is depressed in water and released.Calculate the time period of its oscillations

Solution. Here m = 10 + 10 = 20 g = 0.02 kg

A = r2

= 3.14 (10–2)2 m2

= 103 kg m–3

Time Period T = 2m

A g

T = 4 3

0.022 3.14

3.14 10 10 10

= 0.5 secExample 55. A sphere is hung with a wire, 30° rotation of the sphere about the wire

genertates restoring torque of 4.6 Nm. If the moment of inertia of the sphere is 0.082 kg m2

deduce the frequency of angular oscillation

Solution. Here = 30° = 4.6 Nm

I = 0.082 kg m2

Frequency of oscillation

v =1

2 I

=7 4.6

2 22 0.082

= 1.65 Hz.

Example 56. The bottom of dip on a road has a radius of R. A sick show of mass M left alittle away from the bottom oscillates about this dip. Deduce an expression for the period ofoscillation.

Solution. Let the radius of dip is R and the line joining from O to rickshaw makes an anglewith vertical. Various forces acting on the rickshaw are

Oscillations 487

(a) Weight mg of rickshaw vertically downward

(b) Normal reaction perpendicular to ground i.e. T.

Resolving mg into two components we observe that mg cos is balanced by T. and mg sin acts as restoring force

F = – mg sin For small value of sin �

F = – mg

But angle =arc

radius

=AB

R

F =AB

MgR

or F =mg

ABR

Let is the acceleration in the rickshaw at the instant it is at A thenF = m

m =mg

ABR

or =g

ABR

Comparing it with

= –2 y

we say motion is SHM with

2 =g

R

=g

R

or T = 2 = 2

R

g

Based on energy in SHMExample 57. A body of mass 4.0 kg executes SHM of amplitude 0.5 m If the force constant

be 100 Nm–1 calculate its kinetic energy, potential energy and total energy, when it is half waybetween the equilibrium and extreme position.

Solution. Here m = 4.0 kg, a = 0.5 m

k = 100 Nm–1

angular frequency =k

m =

100

4 = 5 rad/sec

O

T

A

B

m g sin

m g cos m g


y =2

a = 0.25 m

K.E = 2 2 21( )

2m a y

= 2 2 214 5 (0.5 .25 )

2

= 2 × 52 × 0.75 × 0.25 = 9.375 J

P.E = 2 21

2m y

= 214 5 5(.25)

2

= 50 × .25 × .25

= 3.125 JT.E = 9.375 + 3.125

= 12.50 JExample 58. A Particle is executing SHM, of amplitude ‘a’. At what displacement from the

mean position, is the energy half kinetic and half potential?

Solution. According to problem

Kinetic energy = Potential energy

2 2 21( )

2m a y = 2 21

2m y

a2 – y2 = y2

2y2 = a2

y = 2

a

Example 59. At a time when the displacement is half the amplitude what fraction of thetotal energy is kinetic energy and what fraction is potential energy in SHM?

Solution. Displacement y =2

a

Total energy =2 21

2m a

Kinetic energy Ek =2 2 21( )

2m a y

=2

2 21

2 4

am a

=2

21 3.2 4

am

=2 21 3

2 4m a

Oscillations 489

=34

E

Potential energy = 2 21

2m y

=2

21

2 2

am

=2

21

2 2

am

=4E

Example 60. A harmonic oscillator of force constant 2 × 106 N/m and amplitude 0.01 m hastotal mechanical energy of 160 J. Calculate maximum K.E and maximum P.E

Solution. Here k = 2 × 106 Nm–1

a = 0.01 m

maximum k.E = 21

2ka

= 6 212 10 (0.01)

2

= 102 JAt mean position total energy = maximum K.E + minimum P.E

160 = 100 + min P.E

minimum P.E = 160 – 100

= 60 JExample 61. The displacement of a particle of mass 3g executing SHM is given by

y = 3 sin (0.2 t)

Calculate the K.E of the particle at a point which is at a distance equal to 1

3 of its

amplitude from its mean position.

Solution. Kinetic energy = 2 2 21( )

2m a y

=2

3 2 21 13 10 (0.2) (3) 32 3

= 0.48 × 10–3 J

Example 62. An object of mass 0.2 kg executes SHM along x axis with frequency of 25

Hz. At the position x = 0.04 m, the object has a kinetic energy of 0.5 J and potential energy 0.4J. Calculate the amplitude of oscillation.

Solution. Total energy in SHM is

E = 2 21

2m a


= 2 21(2 ) .

2m v a

a =1 2

2

E

mvHere E = 0.5 + 0.4 = 0.9 J

a =1 2 0.925 0.22

= 0.06 mExample 63. A spring is extended by 2 cm energy stored is 100 J. When extended by further

2 cm calculate the energy increased, stored in spring

Solution. v = 21

2kx

1

2

v

v=

21

2

x

x

2

100

v=

22

4

v2 = 400 J

Increase in energy = 400 – 100 = 300 J

EXERCISE

Based On Displacement velocity and Acceleration1. A simple harmonic motion is represented by

y = 5 sin (50 t + 0.5).

Give its amplitude, angular frequency, time period, initial phase. Displacement is measured

in metre and time in second. [Ans. 5m, 50 rad/s,25

Hz, 0.5rad]

2. A body of mass 25 kg is made to oscillate on a spring of force constant 225 Nkg–1. Deducethe angular frequency. [Ans. 3 rad s–1]

3. In a pendulum the amplitude is 0.05 m, and period of 2 sec. Compute its maximumvelocity.

[Ans. 0.1571 ms–1]

4. The maximum velocity of a particle executing SHM is 100 cm s–1 and the maximumacceleration is 157 cm–2. Determine the periodic time. [Ans. 4s]

5. The velocity amplitude of a particle executing SHM is 50 cm and acceleration amplitude is157 cm/s2. Determine periodic time. [Ans. 2.0s]

6. The acceleration of a particle in S.H.M is 2

2

cm s–2 when its displacements is 8 cm.

Calculate its time period. [Ans. 8s]

7. A point describes SHM along a line 4 cm long. Its velocity when passing through thecentre of line is 12 cms–1. Find its frequency [Ans. 0.955 Hz]

Oscillations 491

8. A body is vibrating with simple harmonic motion of amplitude 15 cm and frequency 4Hz.compute the maximum values of the acceleration and velocity[Ans. 94.8 ms–2, 3.77 ms–1]

9. An object executes SHM with an amplitude of 0.17 cm and a period of 0.84 s. Determine(a) frequency (b) the angular frequency of the motion. Write down equation ofdisplacement in SHM. [Ans. y = 0.17 sin 7.5 t ]

10. A particle executes SHM of amplitudes 25 cm and time period 3 sec. What is the minimumtime required for the particle to move between two points 12.5 cm on either side of themean position? [Ans. 0.5 sec]

11. The periodic time of a body executing SHM is 2 second. After how much time interval

from the instant t = 0 will the displacements be half of its amplitude? [Ans. 1

6s]

12. In what time after the motion begins, will a particle performing SHM given by

y = 7 sin 0.5 t

move from the mean position to maximum displacement . [Ans. 1 sec.]

13. A body is performing SHM with amplitude 15 cm and frequency 4 Hz. Calculate themaximum acceleration and the velocity when the displacement is 10 cm

[Ans. 94.7 ms–2 , 2.81 ms–1]

14. A particle executes SHM given by

y = 12 sin 2

10 4

t cm

Calculate (i) velocity at 2.5 sec (ii) Acceleration at t = 5 sec ][Ans. – 5.33 cm s–1, 3.35 cm s –2].

15. If sin cosy a t b t show that is represents SHM. Find the amplitude and the

length of the path [Ans. .2a b a b ]

16. A spring is mounted on a horizontal table. A mass attached to it is then pulled sideway toa distance of 3 cm and released. Calculate

(i) maximum acceleration of the mass

(ii)the maximum speed of the mass. [Ans. 12 ms–2, 0.6 ms–1]

17. A bob executes SHM of period 16 sec. Its velocity is 4 cms–1, 2 second after it has passedthe mean position. Find amplitude of the vibration of bob [Ans. 14.4 cm]

18. Two simple harmonic motions are respectively represented by

y1 = a sin (t – kx) and y2 = b cos (t – kx ) What is the phase difference between the two

SHM. [Ans. 2

]

19. The maximum velocity and acceleration of a particle executing SHM are equal inmagnitude. Calculate the time period. [Ans. 6.28 sec]

20. A particle vibrates in SHM along a straight line. Its maximum acceleration is 52 cm s–2.When its distance from the mean postion is 4 cm the velocity of the particle is 3 cm s–1.Calculate the amplitude and period of the vibrating particle [Ans. 5 cm, 2 sec]

21. The displacements of a particle from mean position performing SHM is given by x = 7 cos0.5 t. Calculate the time taken by particle to move from the postion of equilibrium tomaximum displacement position. [Ans. 1.0 sec]


Based On loaded spring

22. A mass of 2 kg is attached to the two springs of springs constant 100 n/m and 300n/m as shown. Calculate the time period in cash case

[Ans. (i) 1.02 sec (ii) 0.44 sec]

(i)

2 kg

k = 100 N /m1

k = 300 N /m2

(ii)

2 kg

k = 100 N /m1 k = 300 N /m2

23. The length of a spring increases by 0.25 m when a body of mass 0.60 kg is suspended fromit. Calculate the time period if a body of mass 0.24 kg is suspended and stretcheddownward and left. [Ans. 0.628 sec.]

24. A body of mass 0.5 kg suspended by an ideal spring oscillates up and down. the amplitudeof oscillation is 1 cm. and periodic time is 1.57 seconds. Determine (a) maximum speed ofbody (ii) maximum kinetic energy [Ans. 4 cm s–1 ; 4 × 10–4 J]

25. A body of mass 1 kg is made to oscillate in turns on springs of force constant 1, 4, 9 and16 N/m. Deduce the angular frequencies [Ans. 1 Hz, 2 Hz, 3 Hz, 4 Hz]

26. The length of a weight less spring increases by 2 cm when a weight of 1.0 kg is suspendedfrom it. The weight is pulled down by 10 cm and released. Determine the time period ofoscillation. [Ans. 0.28 s]

27. A 2.0 kg body is suspended by a spring. When in addition to it, another body of 300 g issuspended, the spring further stretches by 2.0 cm. If the second body is removed and thefirst body is made to oscillate then what will be the time period. [Ans. 0.73 sec.]

28. The length of a spring is changed by 0.1 m, the potential energy stored in it changes by 0.5J. Find the force constant of spring[Ans. 100 Nm–1]

29. Two bodies A and B of mass 1 kg and 2 kg respectively are rigidlyconnected to a spring (k = 400 Nm–1) The body B is at rest on a horizontaltable. The body A is compressed by 2 cm from its mean position andreleased. Calculate

(i) frequency of oscillation

(ii)maximum velocity of body

[Ans. (i) 3.18 Hz 0.4 m/s]

30. Two identical springs each of force constant k are connected in (i) series (ii) parallel andsupports a mass m. Calculate the ratio of their time periods. [Ans. 2 : 1]

31. The time period of a body suspended by a spring is T. What will be the new period if thespring is cut into two parts and (i) the body is suspended from one part (ii) suspended by

both parts in parallel. [Ans. ,22

T T]

��

�

�

Oscillations 493

32. A pan of mass 1 kg is attached to the lower end of the spring. A weight of 2 kg placed onthe pan stretches the spring by 10 cm. What is the frequency with which the empty panwill oscillate? [Ans. 2.23 Hz]

33. The frequency of oscillation of a mass m suspended by a spring is v1, if the spring is cut to

one half, the same mass oscillates with frequency v2, calculate the ratio 2

1

v

v [Ans. 2 ]

34. A mass M is suspended from a spring of neglible mass. The spring is pulled a little andthen released so that the mass executes SHM with a time period T. If the mass is increased

by m then time period becomes 5

4

T . Determine the ratio

m

M. [Ans.

9

16]

35. A weightless spring of length 60 cm and force constant 100 N/m is kept straight andunstretched on a smooth horizontal table with its ends rigidly fixed. A mass of 0.25 kg isattached at the middle of the spring and is slightly displaced along the length. Calculate

the time period of the oscillation of the mass. [Ans. 20

sec]

Based On Simple Pendulum36. An object is attached to the lower end of a light spring and set vibrating. The maximum

speed of the object is 15 cm/sec and the time period is 628 milli seconds. Calculate theamplitude of the motion. [Ans. 1.5 cm]

37. A simple pendulum performs SHM with a period of 2 seconds calculate the time taken byit to cover a displacement equal to half the amplitude from mean postion. [Ans. 0.5 sec]

38. A simple pendulum has time period T in vacuum. It is immersed in a liquid of density one

eight of density of material of bob. Calculate its new period. [Ans. 8

7T ]

39. Calculate the time period of second’s pendulum when its length in doubled.[Ans. 2.828s]

40. Calculate the length of second’s pendulum at moon. Acceleration due to gravity on thesurface of moon is one sixth of that on the earth. [Ans. 16.5 cm]

41. A pendulum clock gives accurate time. Calculate the error in the time per day if length ofthe pendulum is increased by 0.1% [Ans. 43.2 sec]

42. Calculate the percentage change in the time period of a simple pendulum if its length isincreased by 4%. [Ans. 2%]

43. A simple pendulum with a brass bob has a time period T. The bob is now immersed in a

non viscous liquid and oscillated, the density of liquid in 1

9 that of brass. Find the time

period of the same pendulum. [Ans. 3

8

T ]

44. The mass of the bob of a simple pendulum in 2.5 gm. The effective length of the string is1.0 m and its time period is 2.0 second. Find the value of g [Ans. 9.86 ms–2]

45. Calculate the percentage change in the time period of a simple pendulum if the pendulumis taken to a place where the value of g is 0.6% more. [Ans. 0.3%]

46. Two pendulums of length 100 cm and 110.25 cm start oscillating in phase. After how manyoscillations will they again be in the same phase? [Ans. 20]

Based On SHM47. If a tunnel is made inside the earth not passing through the centre and a ball is dropped at

one end of it then show that , the ball will execute SHM. Determine its time period.

(Radius of earth = 6.4 × 106 m) [Ans. 84.6 min]


48. An air chamber of volume V has a neck area of cross section ‘a’ into which a ball of massm can move up and down without any friction. Show that when the ball is pressed down alittle and released, it executes SHM. Obtain an expression for the time period ofoscillations assuming pressure volume variation of air to be isothermal .

[Ans. T = 22

mV

Ba ]

49. A cylindrical wooden block of cross sectional area 15.0 cm2 and mass 230 g is floated overwater with an extra weight 50 g attached to its bottom. The cylinder floats vertically. It isdisturbed from its mean postion and released. If the specific gravity of wood is 0.30 and g= 9.8 ms–2 deduce the frequency of oscillation of the block. [Ans. 1.15 Hz]

50. Derive an expression for the time period of a cylindrical wooden block of area of crosssection A floating over water of density when it is slightly depressed and released from

equilibrium position. [Ans. T = 2m

A g

]

51. Deduce an expression for the time period of liquid filled in a U tube. The total length of

liquid column is l. [Ans. T = 22

l

g ]

52. In a HCl molecule we may treat Cl to be of infinite mass and H alone oscillating. If theoscillation of HCl molecule shows frequency 9 × 1013 Hz, deduce the force constant.Avogadro’s number is 6 × 1026 per kg mole [Ans. 533.4 Nm–1]

53. A sphere is hung with a wire. 60° rotation of the sphere about the wire produces a restoringtorque of 4.1 Nm. If moment of inertia of the sphere is 0.082 kg m2, determine thefrequency of angular oscillation. [Ans. 1.1 Hz]

Based on energy in SHM54. The amplitude of vibration in SHM is A .For what value of displacement the K.E will be

equal to P.E. [Ans. 2A ]

55. Staring from the origin a body oscillates simple harmonically with a period of 2 s. After

what time will its K.E. be 75% of total energy? [Ans. 1

6 sec]

56. A particle is executing SHM of amplitude 4 cm. At what displacement from the mean

postion is the energy half kinetic and half potential? [Ans. 2 2 cm ]

57. A body of mass 10 kg executes SHM of amplitude 0.5 m. If the force constant be 100 Nm–1.Calculate its K.E. Potential energy and total energy, when it is half way between theequilibrium and extreme postion [Ans. 9.375 J, 3.125J, 12.50 J]

58. At a time when the displacement is half the amplitude, what fraction of total energy is

kinetic energy and what fraction is potential in SHM? [Ans. 3

,4 4

E E]

59. A body of mass 2kg executes SHM of amplitude 0.3m. If force constant is 50 Nm–1

Calculate (i) maximum K.E. (ii) maximum P.E. [Ans. 2.25 J, 2.25 J]

60. Two exactly identical simple pendulum are oscillating with amplitude 2 cm and 6 cm.Calculate the ratio of their energies of oscillations. [Ans. 1 : 9]

495

WaveWave is a kind of disturbance which travels from one place to another through vibratory

motion of the particles of medium.Waves are of two types:

(a) Electromagnetic waves

(b) Mechanical waves

Mechanical waves–are also of two types:(a) Transverse waves: Those waves in which the particles of the medium vibrate about

their mean position in a direction perpendicular to direction of propagation of wave arecalled transverse waves.

(b) Longitudinal waves: Those waves in which the particles of the medium vibrate abouttheir mean position in a direction parallel to direction of propagation of wave are calledlongitudinal waves.

Wave velocityThe distance covered by a wave per second is called wave velocity. It is denoted by v.

v = v = T

Where v = frequency

= wavelength of the wave

T = time period.

Velocity of Transverse wave(a) Velocity of transverse wave in a solid is given by

v =d

= modulus of rigidity.

d = density of solid.

(b) Velocity of transverse waves in a string is given by

v =T

m

T = tension in string

m = mass per unit length.

��15UNIT


Velocity of Longitudinal Waves(a) Solids: Velocity of longitudinal wave in a long rod is given by

v =Y

d

Y = Young’s modulus

d = density of solid

(b) Liquids: Velocity of longitudinal waves in a liquid is given by

v =B

.d

B = Bulk modulus

d = density of liquid

(c) Gases: Newton considered propagation of longitudinal waves through gases isisothermal process.

v =E

d

E = Isothermal bulk modulus

= Pressure (P)

v =P

d

According to Laplace propagation of longitudinal waves through gases is adiabatic process.

E = Adiabatic bulk modulus

= P

v =P

d

.

= Specific heat ratio = P

V

C

C

Factors Affecting Velocity of Sound Through Gases(i) Effect of pressure – No effect.

(ii) Effect of Temperature – Speed of longitudinal waves in gaseous medium is directlyproportional to square root of its absolute temperature i.e.

v T

or 1

2

vv

= 1

2

T

T.

Also vt = v0 + 0.61t

(iii) Effect of density – Velocity of sound in air is inversely proportional to the square root ofdensity i.e.

v 1.

d

Wave Motion 497

Wave EquationA simple harmonic plane progressive wave travelling along X axis is given by

y = a sin (t – kx)

= a sin 2 .T

t x k =

2.

= a sin 2

( ).vt x

In case wave is travelling along –ve X-axis then

y = a sin (t + kx)

= a sin 2T

t x

= a sin 2

( ).vt x

Phase and Phase DifferenceThe equation of simple harmonic progressive wave is

y = a sin 2 T

t x

The argument of sine gives phase of a particle at distance x from origin at time t.

i.e. = 2 .T

t x

Phase difference in terms of time

=2

.T

t

Phase difference in terms of distance

=2

X.

Reflection of WavesA simple harmonic progressive wave

y = a sin 2 T

t x

,

travelling along X-axis may be reflected from a rigid end or free end.

Equation of wave reflected from free end

y = a sin 2 T

t x

Equation of wave reflected from rigid end

y = – a sin 2 .T

t x


Stationary WavesWhen two identical, simple harmonic progressive waves travelling in opposite direction

superimpose, then the result wave produced by their superposition, does not travel in eitherdirection. It is called stationary wave or standing wave.

The stationary wave produced by the superposition of incident wave and wave reflectedfrom free end is given by

y =2 2

2 cos sin .T

x ta

The stationary wave produced by the superposition of incident wave and wave reflectedfrom rigid end is given by

y =2 2

2 cos sin .T

x ta

Stationary waves consist of alternate nodes and antinodes. The distance between two nearest

nodes or antinodes equals to .2

Laws of Vibration of a Stretched StringIf a string stretched between two points vibrates in p loops then frequency of its vibration

is given by

v = .2

p T

l m p = number of loops

(a) Law of length: Frequency of vibration of a stretched string is inversely proportional toits length

l

(b) Law of Tension: Frequency of vibration of a stretched string is proportional to squareroot of tension in the string i.e.

v T.

(c) Law of mass: Frequency of vibration of a stretched string is inversely proportional tosquare root of its mass per unit length i.e.

v 1

.m

Modes of Vibration of StringA stretched string can produce, both odd and even harmonics i.e.

1 : 2 : 3 ... = 1 : 2 : 3 : ...

Where 1 =2L

v

Wave Motion 499

Modes of Vibration of Organ Pipes(a) Open Organ Pipe: In case of open organ pipe both odd and even harmonics can be

produced i.e.

1 : 2 : 3 ... = 1 : 2 : 3 : 4 ...

Also the fundamental frequency is given by

= .2L

v

(b) Closed Organ Pipe: In case of closed organ pipe only odd harmonics can be producedi.e.

1 : 2 : 3 .. = 1 : 3 : 5 ...

Also fundamental frequency is given by

= .4 L

v

Resonance TubeIt is an example of closed organ pipe. Let L1 and L2 are the position of first and second

resonance respectively then speed of sound in air is given by

v = 2 (L2 – L1)

Also end correction is given by

e = 0.3D = 2 1L 3L

2

Where D is diameter of resonance tube.

BeatsWhen two sounding bodies of nearly equal frequencies are sounded together, then the

resultant sound produced by their superposition consists of periodic vibration in the intensity ofsound. The phenomenon is called beats.

One rise and one fall of intensity constitute one beat.

Number of beats heard per second is called beat frequency.

Beat frequency = 1 – 2.

Doppler’s Effect in SoundWhen there is a relative motion between source, observer and medium, the frequency heard

by the observer will be different from the natural frequency of source. The phenomenon of changein the apparent frequency of sound is called Doppler’s effect. Apparent frequency heard byobserver is given by

1 = .m o

m s

v v v

v v v

v = velocity of sound

vm = velocity of medium

vo = velocity of observer

vs = velocity of source.


v

S S O O

vovmvs

P P

Velocities are taken positive in the direction from source to observer and negative fromobserver to source.

Doppler’s Effect in LightIn case of light

= .sv

c

In case source is approaching observer we take positive sign and if source is receding awaythen we take –ve sign.

Characteristic of a Musical Sound(i) Pitch: is the characteristic of musical sound that helps the listner in distinguishing a

shrill note from a grave one. Pitch depends on frequency.

(ii) Quality: The characteristic of musical sound that distinguishes between two sounds ofsame pitch and loudness is called quality, which depends on intensity of overtones.

(iii) Loudness: The sensation of sound which enables us to distinguish between a loud anda faint sound is called loudness. It depends on the intensity of sound.

Units of LoudnessThe loudness of sound of intensity I is given by

L = 100

Ilog .

I

I0 is threshold of hearing.

If I = 10 I0

L = 010

0

10Ilog

I = 1 bel.

The loudness of sound is said to be 1 bel if its intensity is 10 times that of the threshold ofhearing.

1 decibel =1

bel .10

In decibel, the loudness of sound of intensity I is given by

L = 100

I10 log .

I

Wave Motion 501

Threshold of HearingThe lowest intensity of sound that can be perceived by the human ear is called threshold of

hearing.

SOLVED EXAMPLES

Based on Wave VelocityExample 1. A hospital uses an ultrasonic scanner to locate tumours in a tissue. What is the

wavelength of sound in a tissue in which the speed of sound is 1.7 km/s. The operating frequencyof scanner is 4.2 MHz.

Solution. Here v = 1.7 km/s = 1.7 × 103 m/s.

v = 4.2 MHz = 4.2 × 106 Hz.

=3

6

1.7 10

4.2 10

v

� 4.05 × 10–4 m.

Example 2. A balt emits ultrasonic sound of frequency 100 kHz in air. If their sound meetsa water surface. What is the wavelength of (i) the reflected wave (sound) (ii) the transmittedsound? Speed of sound in air = 340 ms–1 and in water is 1486 ms–1.

Solution. Here v = 100 × 103 Hz

va = 340 ms–1, vw = 1486 ms–1.

(a) a = 3

340

100 10a

v

v

= 3.4 × 10–3 m.

(b) w = 5

1486

10w

v

v = 1.49 × 10–2 m.

Example 3. The audible frequency range of a human’s ear is 20 Hz – 20 kHz. Convert thisinto the corresponding wavelength range. Take speed of sound in air at ordinary temperature tobe 340 ms–1.

Solution. Here v1 = 20 Hz, v2 = 20 × 103 Hz.

v = 340 m/s.

1 =1

340

20vv

= 17 m.

and 2 = 32

340

20 10v

v

= 17 × 10–3m

Example 4. A stone dropped from the top of a tower 300 m high splashes into water of pondnear the base of the tower. When is the splash heard at the top? Speed of sound is air = 340 ms–1.

Solution. Let time taken by the stone to reach water surface is t.

s =1

2ut gt2

300 = 210 9.8

2 t


or t2 =300

4.9

or t =300

4.9 = 7.82 sec.

Time taken by the splash sound to come to top is t.

t =distance 300

speed 340 = 0.88s

Total time taken = 7.82 + 0.88

= 8.7 sec.Example 5. A body vibrating with a certain frequency sends waves 15 cm long through a

medium A and 20 cm long through a medium B. The velocity of waves in A is 1200 cm/s. Findthe velocity in B.

Solution. Let frequency of wave is v.

In first case

1 = 15 cm, v 1 = 1200 cm/s

v2 = 1

1

1200

15

v

= 80 Hz.

In second case v2 = v 2

= 80 × 20

= 1600 cm/s = 16 m/s.

Example 6. A thunder clap was heard 5.5 sec. later than the accompanying light flash wasseen. How far away did the flash occur. Velocity of sound = 350 ms–1.

Solution. Here v = 350 m/s t = 5.5 sec.

s = v × t

= 350 × 5.5 = 1925 m.Example 7. Velocity of light in vacuum is 3 × 108 ms–1 and that in water is 2.3 × 108 ms–

1. If the wavelength of the wave in vacuum is 6 × 10–7 m. What will be its wavelength in water?What is the frequency of the wave?

Solution. Here c = 3 × 108 ms–1 a = 6 × 10–7 m

vw = 2.3 × 108 ms–1 w = ?

We know v = v.

or v = .v

C

a

= .

w

w

v

or w =8 7

8

2.3 10 6 10

3 10w a

c

v

= 4.6 × 10–7 m.

Wave Motion 503

Example 8. How far does sound travel in air when a tuning fork of frequency 250 Hzcompletes 100 vibrations. The speed of sound in air is 350 ms–1.

Solution. Here v = 250 Hz, v = 350 ms–1.

v = v

350

250v

v

=7

5m.

Distance covered in 100 vibrations = 7

1005

= 140 m.

Based on Velocity of Transverse WavesExample 9. A steel wire has a length of 12.0 m and a mass of 2.10 kg. What should be the

tension in the wire so that the speed of a transverse wave on the wire equals to speed of sound indry air at 20°C which is 343 ms–1?

Solution. Here L = 12.0 m.m = 2.10 kg

v = 343 ms–1.

We know v =T

.m

or v 2 = T/m

or T = m v2

= 22.10(343)

12.0

= 2.06 × 104 N.Example 10. A string of mass 2.50 kg is under a tension of 200 N. The length of the

stretched string is 20.0 m. If a transverse jerk is struck at one end of the string, how long doesthe disturbance take to reach the other end?

Solution. Here m =12.50 25

kg m20.0 200

, T = 200 N.

From formula v =T 200 200

25m

=200

5 = 40 ms–1.

Time taken =distance 20 1

speed 40 2 = 0.5 sec.


Example 11: The speed of transverse wave in a stretched string is 500 m/s. When thetension in the string is 2 kg wt. Calculate the speed of the transverse wave in the same stringwhen tension is changed to 8 kg wt.

Solution. Here v 1 = 500 m/s , T1 = 2 kg wt

T2 = 8 kg wt.

From formula v =T

.m

1

2

vv

= 1

2

T

T

or2

500

v=

2

8

v 2 = 1000 m/s.Example 12. A string having mass per unit length 2.25 g/m is under a tension of 10 N.

Calculate the velocity of the wave in string.

Solution. Here m = 2.25 g/m = 2.25 × 10–3 kgm–1

T = 10 N

v = 3

T 10

2.25 10

m

=410 100

2.25 1.5

=200

3

= 66.7 ms–1

Example 13. The mass of a stretched string is 0.5 g cm–1. Calculate the tension in the stringif speed of transverse wave on string is 36 ms–1.

Solution. Here m = 0.5 g cm–1

=3

2

0.5 10

10

kg m–1

= 0.5 × 10–1 kg m–1.

v = 36 ms–1.

v =T

.m

T = m v 2 = (36)2 (0.05)

= 64.8 N.

Wave Motion 505

Based on Speed of Longitudinal WavesExample 14. Calculate the speed of sound in a steel rod. Young’s modulus for steel is 2.1

× 1011 Nm–2. The density of steel is 7.8 × 103 kg m–3.

Solution. Here Y = 2.1 × 1011 Nm–2.

d = 7.8 × 103 kg m–3.

From formula v =11

3

Y 2.1 10

7.8 10

d

= 5.19 × 103 ms–1.Example 15. The density of air at STP is 1.29 kg m–3. Deduce the velocity of longitudinal

waves using

(i) Newton’s formula

(ii) Laplace formula.

Solution. Here d = 1.29 kg m–3

P = 1.01 × 105 Nm–2.

(i) According to Newton’s formula

v =P

d

=51.01 10

1.29

= 280.2 ms–1.(ii) According to Laplace formula

v =5P 1.41 1.01 10

1.29d

= 331.56 ms–1.

Example 16. At a pressure of 1.01 × 105 Nm–2, the volume strain of water is 5 × 10–5.Calculate speed of sound in water.

Solution. Here P = 1.01 × 105 Nm–2.

V

V

= 5 × 10–5

Bulk modulus of air =PV

V

=5

5

1.01 10

5 10

B = 2.02 × 109 Nm–2

v =9

3

B 2.02 10

10

d

= 1.414 × 103 ms–1.


Example 17. Velocity of sound at NTP in air is 332 ms–1. Find the velocity of sound at819°C.

Solution. Here v 1 = 332 ms–1 t1 = 0°C

t2 = 819°C

From formula 2

1

vv

= 2

1

273

273

t

t

2

332

v=

273 8194

273

v 2 = 2 × 332 = 664 ms–1.Example 18. The planet Jupiter has an atmosphere composed mainly of methane at a

temperature of – 130°C. Find the velocity of sound on the planet assuming gas = 1.30. Given R= 8.36 J/mol k.

Solution. Here t = – 130°C, = 1.30.

R = 8.36 J/mol-k, M = 16.

From formula v =RT

.M

=31.30 8.36 10 143

16

= 311.6 ms–1.Example 19. At what temperature will the velocity of sound in air be double than the

velocity in air at 0°C.

Solution. From formula 2

1

vv

=2

1

T

T

or2vv

=273

.0 273

t

4 =273

273

t

or t = 819°C.Example 20. At NTP, the speed of sound in air is 332 ms–1. What will be the speed of sound

in hydrogen (i) at NTP (ii) at 819°C and 4 atmospheric pressure. Air is 16 times heavier thanhydrogen.

Solution. At NTP v 1

d

H

a

vv

=H

16

1ad

d = 4.

Wave Motion 507

v H = 4 × 332 = 1328 ms–1.

(ii) v T

v 819 273 819

v 0 0 273

819

1328

v=

1092

273 = 2

v 819 = 2 × 1328

= 2656 ms–1.Example 21. A tunning fork of frequency 220 Hz produces sound waves of wavelength 1.5 m

in air at NTP. Calculate increase in wavelength when temperature of air is 27°C.

Solution. Here v = 220 Hz

1 = 1.5 m t1 = 0°C

t2 = 27°C

Velocity of sound in air at NTP v = v 1 = 220 × 1.5

= 330 ms–1.

Now 27

0

vv

=273 27

273

or v 27 =300

330273

27 = 27 300 330

273 220v v

= 1.57 m.

Increase in wavelength = 27 – 0 = 1.57 – 1.5 = 0.07 m.Example 22. A gas is a mixture of hydrogen and nitrogen having volumes in the ratio 2 :

1. If the velocity of sound in hydrogen at 0°C is 1300 ms–1, find the velocity of sound in gaseousmixture at 27°C.

Solution. Hydrogen and nitrogen are diatomic gases

H = N.

Density of mixture =Total mass

Total volume

dmix = H N2 V V

2 V + V

d d

= H N H2 V (14 ) 16

3 3

d V d d

V

Speed of sound in mixture at 0°C

v =mix H

P P 3

16d d


Speed of sound in hydrogen at 0°C

vH =H

P

d

H

v

v= H

H

P 3 3

16 P 4

d

d

v =3

13004 = 325 3 ms–1.

Speed of sound in mixture at 27°C.

vt = 0 1546

tv

=27

325 3 1546

= 591 ms–1.

Example 23. Show that the speed of sound in air increases by 61 cm s–1 for every 1°C riseof temperature.

Solution. Let at 0°C speed of sound in air is v0 and at 1°C speed is vt . Then

v0 0 273

and vt 73.t

0

tv

v=

273

0 273

t

vt =1/ 2

0273

273

tv

=1/2

0 1273

tv

= 01

1273 2

tv

But v0 = 332 ms–1.

vt = 332 1546

t

= 332 + 332

546t = (332 + 0.61 t) m/s.

Where t = 1°C

vt = 332 + 0.61 ms–1 = v0 + 0.61 ms–1.

or vt – v0 = 0.61 ms–1.

Wave Motion 509

Example 24. Show that speed of sound in hydrogen is four times the speed of sound inoxygen.

Solution. We know v 1

.d

Let speed of sound in hydrogen is vH and speed of sound in oxygen is v 0.

v H H

1

d

v 0 0

1

d

H

0

vv

=0

H

16

1

d

d 0 H:

16 :1

d d

v H = 4V0.Example 25. The speed of sound in air at 16°C is 340 ms–1. Find the wavelength in air of

a note of frequency 680 Hz at 160°C and 51°C.

Solution. At 16°C suppose the wavelength is 1.

v 1 = 340 ms–1. v = 680 Hz.

1 = 1V 340 1

680 2

v = 0.5 m.

(2) At 51°C consider the wavelength is 2.

Now 1

2

vv

= 1

2

273

273

t

t

2

340

v=

16 273 289 17

51 273 324 18

v 2 =340 17

17

= 360 ms–1.

2 = 2 360

380

vv

= 0.53 m.

Example 26. Find the ratio of velocity of sound in hydrogen gas ( = 7/5) to that in heliumgas ( = 5/3) at the same temperature. Given that molecular weight of hydrogen and helium are2 and 4 respectively.

Solution. v =RT

M

H

He

vv

=HeH

He H

M

M

=7 / 5 4 42

5 / 3 2 25 = 1.68.


Based on Progressive WaveExample 27. A transverse harmonic wave on a string is described by

y (x, t) = 3.0 sin (36 t + 0.018x + /4).

Where x, y are in cm and t in sec. The positive direction of x is from left to right.

(i) Is this a travelling or a stationary wave? If it is travelling, what are the speed anddirection of its propagation.

(ii) What are its amplitude and frequency?

(iii) What is initial phase at the origin?

(iv) What is the least distance between two successive crests in the wave?

Solution. Here y = 3.0 sin (36t + 0.018x + /4). (1)

Equation of S.H. Progressive wave is

y = a sin 02 2t x

T

(2)

Comparing equation (1) and (2)

a = 3.0 cm,2

T

= 36 or T =

2

36

= sec.18

2

= 0.018

or =2

cm.0.018

(i) The wave is travelling. It is travelling along negative x-axis.

(ii) amplitude = 3.0 cm.

frequency =18

Hz.

(iii) Initial phase = .4

(iv) Wave-length =2

0.018

= 349 cm

= 3.49 m.Example 28. For a travelling harmonic wave

y = 2.0 cos (10t – 0.008x + 0.35).

Where x and y are in cm and t is sec. What is the phase difference between oscillatry motion

at two points seperated by a distance of (i) 4 m (ii) 0.5 m (iii) 2

(iv)

3

4

Solution. Here y = 2.0 cos (10t – 0.008x + 0.35)

Wave Motion 511

Comparing it with equation y =2 2

A cosT

t x

2

= 0.008

or =2

cm.0.008

=

2 1m

0.008 100

= 2

m.0.8

From formula =2 × path difference.

(i) x = 4m

=2

0.8 42

= 3.2 rad.

(ii) x = 0.5 m

=2 0.8

0.52

= 0.4 rad.

(iii) x =2

=2

2

= rad.

(iv) x =3

4

=2 3

4

32

rad.

Example 29. A displacement wave is represented by

y = 0.5 × 10–3 sin (500t – 0.025x).

Here y and x are in cm and time in sec.

Calculate (i) amplitude (ii) the time period (iii) angular frequency (iv) wavelength.

Solution. Here y = 0.5 × 10–3 sin (500t – 0.025x).

Comparing it with y = a sin 2 2

.T

t x (i) a = 0.5 × 10–3 cm.

(ii)2

T

t= 500 t

T =2

500

=

sec250

.


(iii) Angular frequency =2

T

=2

250

= 500 rad/sec.

(iv)2

= 0.025

or =2

0.025

= 80 cm.Example 30. What is the amplitude, the wavelength and the velocity of the wave

represented by

y = 5 sin (6 t – 4x)

Where x, the distance and time are in S.I. Units.

Solution. Here y = 5 sin (6 t – 4x)

= 5 sin 2

2 3x

t

Comparing it with the standard equation

y = sin 2 .T

t xa

(i) a = 5 m

(ii) 1

T = 3 or v = 3Hz

(iii) =2

= 1.57 m.

(iv) Velocity v = v

= 32

= 32

ms–1.

Example 31. The speed of a wave in a stretched string is 30 ms–1. Calculate the phasedifference in radians between two points situated at a distance of 15.0 cm, if frequency of waveis 50 Hz.

Solution. Here v = 30 ms–1

v = 50 Hz.

x = 12.5 cm.We know v = v.

=30

50vv

= 0.6 m = 60 cm.

=2

x

=2

1560

= rad2

.

Wave Motion 513

Example 32. The equation of a plane progressive wave is

y = 10 sin 2 (t – 0.005x).

Where y and x are in cm and t in second. Calculate the amplitude frequency, wavelengthand velocity of wave.

Solution. Given y = 10 sin 2 (t – 0.005x) (1)

The standard equation for a harmonic wave

y = a sin 2 T

t x (2)

Comparing equation (1) and (2),

a = 10 cm.

T = 1 sec. or v = 1 Hz.

=1

0.005 = 200 cm.

Velocity v = = 1 × 200 = 200 cm s–1.

Example 33. A 5 watt source ends out wave in air at frequency 1000 Hz. Deduce theintensity at a 100 m distance, assuming spherical distribution. If c = 350 ms–1, and d = 1.3 kgm–3, deduce the displacement amplitude.

Solution. Here Power P = 5 watt.

v = 1000 Hz., r = 100 m

d = 1.3 kg m–3.

The intensity is given by I = 2

Power

4 r

= 2

5

4 3.14(100)

= 4 × 10–5 W m–2.

Also I = 222 a2 dc.

or a =1 I

2v dc

=51 4 10

3.14 1000 2 1.3 350

= 6.67 × 10–8 m.Example 34. A progressive wave of frequency 50 Hz is travelling with a velocity of

360 ms–1. How far apart are two points 60° out of phase?

Solution. Here v = 360 ms–1.

v = 50 Hz

= 60°

=2

x


3

=

2360

50

x

or x = 0.12 m.Example 35. A simple harmonic wave train is travelling in a gas in the positive direction

of x-axis. Its amplitude is 2 cm, velocity 45 ms–1 and frequency is 75 per second. Find out thedisplacement of the particle of medium at a distance 135 cm from the origin in the direction ofthe wave at t = 3 sec.

Solution. We know y = sin 2T

t xa

Here a = 2 cm,

v = 45 ms–1

v = 75 Hz.

=45

75vv

= 0.6m = 60 cm.

and T =1

sec.75

y = 12 sin 2

1 6075

x

for t = 3 sec, x = 135 cm

y =135

2 sin 2 3 7560

= 2 sin (– 4.5 )

= 2sin 42

= – 2 cm.

Example 36. Given below are the equations of two progressive waves travelling alongpositive direction of x-axis. Compare their intensities.

y1 = 2.0 sin (314 t – 1.57 x)

y2 = 5.0 sin (314 t – 1.57 x + 1.57).

Solution. The intensity of a wave is given by

I = 22 v 2a2c

I a2.

1

2

I

I =2 212 22

(2)

(5)

a

a = 4 : 25.

Based On Stationary WavesExample 37. The transverse displacement of a string (clamped at its two ends) is given by

Wave Motion 515

y (x, t) =2

0.06 sin cos 120 .3

xt

Where x, y are in m, and t in second. The length of the string is 1.5 m and its mass is 3.0× 10–2 kg. Answer the following :

(a) Does the function represent a travelling or a stationary wave ?

(b) Interpret the wave as a superposition of two waves travelling in opposite directions.What are the wavelength, frequency and speed of propagation of each wave.

(c) Determine the tension in the string.

Solution. Given y =2

0.06 sin3

x cos 120 t.

(a) It represents a stationary wave.

(b) Comparing it with standard equation of stationary wave

y =2 2

sin cos .T

x ta

2 x =

2

3

x = 3 m.

2

T

t= 120 t T =

1sec.

60

or v = 60 Hz.

Also v = v= 60 × 3

= 180 m/s.

(c) Tension T = m v 2T

vm

= 2 × 10–2 (180)2

= 648 N.Example 38. The transverse displacement of a string (clamped at its two ends) is given by

y (x, t) =2

0.06 sin cos 120 .3

xt

Where x, y are in m and t is in sec.

(i) Do all the points on the string oscillate with the same (a) frequency (b) phase(c) amplitude. Explain your answer.

(ii) What is the amplitude of a point 0.375 m away from one end.

Solution. Given y =2

0.06 sin cos 120 .3

xt

In this equation time dependent function cos 120 t represents frequency. This function isindependent of x so frequency of oscillation of all the particles is same. Also the phase will besame.

Amplitude is given by

=sin 2

0.063

x


Which depends on x so amplitude of all the particles will be different.

(b) For a particle at distance x = 0.375

A =2 0.375

0.06 sin3

= 0.06 sin 0.7854

= 0.06 × 0.707

= 0.042 m.Example 39. Stationary waves are set by the superposition of two waves given by

y1 = 0.05 sin (5 t – x)

y2 = 0.05 sin (5 t + x)

Where x and y are in m and t in sec. Calculate the displacement of a particle at x = 1m.

Solution. Here y1 = 0.05 sin (5 t – x)

y2 = 0.05 sin (5 t + x).

According to superposition principle.

y = y1 + y2

y = 0.05 sin (5 t – x) + 0.05 sin (5 t + x)

= 2 (0.05) sin 5 t cos x

Amplitude of stationary wave is

A = 0.1 cos x

When x = 1 m

A =o180

0.1 cos

= 0.1 cos 57.3°

= 0.1 × 0.5406

= 0.054 m.Example 40. The equation of a longitudinal stationary wave produced in a closed organ

pipe is

y =2

2 sin8

x cos 120 t

When x, y are in cm, t in second. Find

(a) the frequency, amplitude and wavelength of original progressive wave.

(b) separation between two consecutive antinodes.

Solution. Here y =2

2 sin cos 1208

xt

Comparing it with standard equation of stationary wave

y =2 2

2 sin cosT

x ta

2a = 2 a = 1 cm.

2 x =

2

8

x

Wave Motion 517

= 8 cm.

2

T

t= 120 t

or T =1

60 v = 60 Hz.

(b) Separation between two consecutive antinodes

=2

=

8

2 = 4 cm.

Based on Vibrations of Stretched StringsExample 41. A wire of specific gravity 7, one metre long and 1 mm diameter is stretched

by a weight of 11 kg. Calculate the pitch of the fundamental note.

Here, d = 7

l = 100 cm.

D = 1 mm = 1

cm10

T = 11 kg = 11 × 1000 × 980 dynes

v = ?

v =1 T

Dl d

=1 11 1000 980 7

1 22 710010

= 70 Hz.

Example 42. A string when attached by a weight of 4 kg gives a note of frequency 256.What weight will produce an octave of this note?

Solution. Here v1 = 256v2 = 2 × 256 = 512

T1 = 4 kg wt

T2 = ?

2

1

v

v= 2

1

T

T

512

256= 2T

4

or 2T

4= 2

Squaring T2 = 16 kg wt.


Example 43. A stretched wire under a tension of 1 kg wt is in unison with a fork offrequency 520. What change in tension would make the wire vibrate in unison with a fork offrequency 260?

Solution. Here, T1 = 1 kg

v 1 = 520

T2 = ?

v2 = 260

2

1

=

2

1

T

T

260

520= 2T

1

or 2T =1

2

Squaring T2 =1

4 kg wt

= 0.25 kg wt

Tension should be decreased by

= 1 – 0.25 = 0.75 kg wtExample 44. Calculate the frequency of the fundamental note of a string, 1 metre long and

weighing 2 g when stretched by a weight of 400 kg.

Solution. Here, l = 1 metre

Total mass = 2g = 0.002 kg

m =0.002

1 = 0.002 kg/metre

T = 400 kg = 400 × 9.8 newtons

v =1 T

2l m

v =1 400 9.8

2 1 0.002

v =1

400 49002

= 700

Example 45. A stretched wire of length 50 cm produces a note of frequency 320. Calculatethe frequency when the vibrating length is 32 cm.

Solution. Here, l1 = 50 cm

v1 = 320

l2 = 32 cm

v2 = ?

2

1

v

v =1

2

l

l

2

320

v=

50

32or v2 = 500.

Wave Motion 519

Example 46. A steel wire 50 cm long and of mass 5 g is stretched with a tension of 40 ×106 dynes. What is the frequency of its fundamental mode of vibration? What is the number ofthe highest overtone that could be heard by a person who can hear frequencies up to 10,000C.P.S.

Solution. Here, l = 50

Total mass of the wire = 5 grams

m =5 1

50 10 g/cm

T = 40 × 106 dynes

v =1 T

2l m

v =61 40 10

12 5010

v =81

4 10100

= 200

The frequency of the fundamental note = 200

Maximum frequency audible = 10000

Number of the highest overtone =10000

1200

= 50 – 1 = 49(Because, if frequency = 200

first overtone = 2 × 200

and so the 49th overtone

= 50 × 200 = 10000)Hz.

Example 47. A wire of length 50 cm when stretched by a load of 8 kg vibrates with afrequency of 280 vibrations per second. Find its mass.

Solution. Suppose, total mass of the wire = x kg

Here, length l = 50 cm = 0.50 metre

T = 8 kg = 8 × 9.80 newtons

Mass per unit length m =0.5

x kg/metre

= 280

v =1 T

2l m

280 =1 8 9.80

2 0.50.5

x

280 =1 8 0.50 9.80

1 x


280 =4 98

10x

=1

145x

20 =1

5x

Squaring 400 =1

5x

x =1

2000 = 0.005 kg

= 0.5 g.Example 48. A wire gives a fundamental note of 256 cycles per second when it has a

tension of 10 kg wt.

(i) At what tension will the string emit a frequency of 512 cycles per second?

(ii) How would you make the wire emit a note of 768 cycles per second keeping the tensionof 10 kg wt.

Solution. Here v1 = 256, T1 = 10 kg wt

v2 = 512, T2 = ?

2

1

v

v =2

1

T

T

512

256= 2T

10

Squaring, 4 = 2T

10

or T2 = 40 kg wt

(ii) Length in first case = l1

v1 = 256

In second case v2 = 768

and length is l2.

2

1

v

v =1

2

l

l1

vl

�

or768

256=

1

2

l

l

or l2 =1

3

l.

Therefore by changing the length to 1

3 of the length in first case, the wire will emit a note

of frequency 768.

Wave Motion 521

Example 49. The length of a wire between the two ends of a sonometer is 105 cm. Wherewould the two bridges be placed so that the fundamental frequencies of the three segments arein the ratio 1 : 3 : 15?

Solution. Here total length of string is 105 cm.

v1 : v2 : v3 = 1 : 3 : 15

v =1 T

21 m

i.e. v 1

l

l1 : l2 : l3 =1 1 1

: :1 3 15

= 15 : 5 : 1

or l1 =15

10521

= 75 cm

l2 =5

10521

= 25 cm.

l3 =1

10521

= 5 cm.

i.e. position of bridges is 75 cm and (75 + 25) = 100 cm.Example 50. A wire of linear mass density of 5.0 × 10–3 kg m–1 is stretched between two

rigid supports with a tension of 450 N. The wire resonates at a frequency of 420 Hz. The nexthigher frequency at which the same wire resonates is 490 Hz. Find the length of the wire.

Solution. Suppose the wire resonates at 420 Hz in its nth harmonic and at 490 Hz in its(n + 1) harmonic.

As vn =T

2

n

l m

420 =T

2

n

l m(i)

490 =1 T

2

n

l m

(ii)

Divide equation (i) by (ii)

490 1

420

n

n

or70

1420

=1

1n

.

n = 6.From equation (i)

420 = 3

6 450

2 5 10l or l = 2.14 m.


Example 51. In an experiment a wire stretched between two points vibrate in 4 loops undera tension of 10 N. Calculate the tension in the wire so that the same wire may vibrate in 8 loops.

Solution. Here n1 = 4 T1 = 10 N

n2 = 8

We know =T

2

n

l m.

As , l and m are constant.

T 2 = constant

or 10 × (4)2 = (8)2 T2

or T2 =16

1064

= 2.5 N.Example 52. A 14.4 g string of a sonometer is 72 cm long. What should be the tension in

the string in order that it may vibrate in 2 segments with a frequency of 256 Hz.

Solution. Here

mass per unit length m =3

2

14.4 10

72 10

= 0.02 kg m–1.

n = 2

= 256

If a string vibrates in n loops then

=T

2

n

l m

2 =2

2

T

4

n

ml

or T =2 2

2

4l m

n

=2 2 2

2

4(256) (0.72) (0.02)

(2)

= 679.47 N.Example 53. A stretched wire emits a fundamental note of 256 Hz. Keeping the stretching

force constant and reducing the length by 20 cm, the frequency becomes 320 Hz. Calculate theoriginal length of wire.

Solution. Here 1 = 256 Hz.

l1 = l

2 = 320 Hz.l2 = (l – 20)

Wave Motion 523

v1 =1

1 T.

2l m

v2 =2

1 T.

2l m

1

2

v

v= 2

1

20l l

l l

or256

320=

20l

l

or4

5=

20l

l

or 4l = 5l – 100

or l = 100 cm.Example 54. The fundamental frequency of one meter long steel wire is 300 Hz. Density

of steel is 8 × 10–3 kg/m3.

(i) Find the speed of transverse wave in wire

(ii) If tension of wire is increased by 4%. Calculate the percentage change in the frequencyof wire.

Solution. Here v = 300 Hz

l = 1 m, d = 8 × 103 kg m–3.

(a) From formula

v = where = 2l.

v = 2 = 2 × 300 × 1

= 600 ms–1

(b) We know v =1 T

2l m

vv

=1 T

2 T

l m

l m

.

Tension is increased by 4%.

100 v

=1

4 %2

= 2%.

Example 55. A wire having a linear mass density of 5.0 × 10–3 kg m–1 is stretched betweentwo rigid supports with a tension of 450 N. If the wire resonates at a frequency of 420 Hz thenext higher frequency at which the same wire resonates is 490 Hz. Find the length of the wire.

Solution. Let the wire vibrates at 420 Hz in its nth harmonic and 490 Hz in its (n + 1)harmonic.


420 =T

2

n

l m(i)

490 =( 1) T

2

n

l m

(ii)

or420

490=

1

n

n or n = 6

Using this value of n in equation (i)

420 = 3

6 450

2 5 10l

or l = 2.1 m.

Based on Organ Pipes and Rods Clamped at Its CentreExample 56. An organ pipe emits a sound of frequency 200 Hz. On blowing air harder into

it, a sound of frequency 3 × 200 Hz is heard. Is the pipe open or closed at the other end? Explain.

Solution. Fundamental frequency = 200 and overtone = 3 × 200

The overtone is three times the fundamental frequency. It means odd overtones are presentand even overtones are missing. Hence the pipe is closed at the other end.

Example 57. Velocity of sound in air is 32000 cm/s. A column of air produces resonancewith a tuning fork of frequency 320. Find the length of the air column.

Air column can be open as well as closed

Here v = 32000 cm/s, n = 320

(1) When the column is open at both ends

v =2l

v or l =

2

vv

=32000

2 320 = 50 cm.

(2) When the air column is closed at one end

v =4

v

l or l =

4

v

v

=32000

4 320 = 25 cm. or 0.25 m

Example 58. Two open organ pipes of length 50 cm and 50.5 cm produce 3 beats/s.Calculate the velocity of sound in air.

Solution. Here, l1 = 50 cm, l2 = 50.5 cm

Suppose the velocity of sound = v

v1 =12 100

v v

l

Wave Motion 525

v2 =22 2 50.5 101l

v v v

Beats/s = 3

or v 1 – v 2 = 3

100 101

v v= 3 or

10100

v = 3

or V = 30,300 cm/s. = 203 ms–1

Example 59. Two tuning forks A and B give 6 beats/second. A resounds with closed columnof air 15 cm long and B with an open column 30.5 cm long. Calculate their frequencies.

Solution. Suppose, frequency of A = 1

v1 =14 4 15 60l

v v v

Frequency of B = v 2

v2 =22 2 30.5 61l

v v v

Beats/s = 6

v 1 – 2 = 6

60 61

v v = 6 or 3660

v = 6

or v = 21960 cm/s.

v2 =21960

60 60v

= 366 Hz.

v2 =21960

61 61v

= 360 Hz.

Example 60. The frequency of the fundamental note of a tube closed at one end is 200 Hz.What will be the frequency of the fundamental note of a similar tube of the same length but openat both the ends.

Solution. For the closed tube v1 =4l

v... (i)

In the open tube v2 =2l

v... (ii)


2

1

v

v= 2

v2 = 2 v 1 = 2 × 200 = 400 Hz.Example 61. A fork of frequency 250 is held over a tube and maximum resonance is

obtained when the columns of air are 31 cm and 97 cm long. Find the end correction and velocityof sound in air.


Solution. Here, v = 250, l1 = 31 cm, l2 = 97 cm

End correction c = 2 13

2

l l

=97 93

2

= 2 cm.

v = 2v (l2 – l1)

= 2 × 250 × (97 – 31)

= 33000 cm/s = 330 m/s.

Example 62. The length of a cylindrical glass tube is 30 cm. The velocity of sound in air is330 metre per second. What is the fundamental frequency of vibration of the air column if

(i) both ends of the tube are open and

(ii) one end is closed and flat? (Neglect end correction)


v = 330 m/s = 33000 cm/s

(i) When both ends are open v =33000

2 2 30l

v

= 550 Hz

(ii) When one end is closed v =33000

4 4 30l

v

= 275 Hz

Example 63. An organ pipe filled with air has a fundamental frequency of 500 vibrations/s. The first harmonic of another organ pipe closed at one end and filled with carbon dioxide hasthe same frequency as that of the first harmonic of the open organ pipe. Calculate the length ofeach pipe (Velocity of sound in air = 330 m/s, velocity of sound in CO2 = 264 m/s.)

Solution. (1) Open end pipe containing air

v =2l

v

Here, v = 500, v = 330 m/s

500 =330

2l

l = 0.33 metre(2) Closed end pipe containing carbon dioxide

v =4l

v

Here v = 500, v = 264 m/s

500 =264

4l or l = 0.132 metre.

Example 64. A 25 cm long rod is clamped in the middle. It is set into longitudinalvibrations in the fundamental mode. What is the velocity of sound in the rod if the frequency is4500 hertz.

Wave Motion 527


v = 4500 hertz

= 2l = 2 × 25 = 50 cm

v = v= 4500 × 50

= 225000 cm/s

= 2250 m/s.Example 65. A resonance air column of length 16.6 cm resonates with a tunning fork of

frequency 512 Hz. Calculate the speed of sound in air.

Solution. Here l = 16.6 cm

= 512 Hz

For closed organ pipe v =4

v

l

v = v 4l

= 4 × 512 × 16.6

= 340 ms–1.Example 66. A resonance tube with a tunning fork of frequency 256 Hz. If the length of

resonated air column be 32 cm and 99 cm. Calculate the value of end correction and speed ofsound.

Solution. The end correction x = 2 13

2

l l

=99 96

2

= 1.5 cm.

Speed of sound v = v= 2v (l2 – l1)

= 2 × 256 (99 – 32)

= 34304 Cms–1

= 343.04 ms–1.

Example 67. The frequency of fundamental note of a tube closed at one end is 250 Hz.What will be the frequency of the fundamenal note of a similar tube of same length but open atboth ends.

Solution. Let length of each tube is l. The frequency of fundamental note is closed organpipe is

v1 =4l

v

For open organ pipe v2 =2l

v

1

2

v

v=

1

2v2 = 2 v 1

= 2 × 250

= 500 Hz.


Example 68. A steel rod 100 cm long is clamped at its middle. The fundamental frequencyof longitudinal vibration of the rod is given to be 2.53 kHz. What is the speed of sound in steel.

Solution. For steel rod clamped at middle

L =2

or = 2L

v = v= v × 2L

= 2.53 × 103 × 2 × 1

= 5060 ms–1.Example 69. A pipe 20 cm long is closed at one end which harmonic mode of the pipe is

resonantly excited by a 430 Hz source? Will this source be in resonance with pipe if both endsare open. Speed of sound is 340 ms–1.

Solution. Here length of pipe = 20 cm = 0.2 m.

speed of sound = v = 340 ms–1.Fundamental frequency of closed organ pipes

=4L

v

=340

4 0.2= 425 Hz.

The fundamental mode of closed organ pipe may be excited by a source of 430 Hz.

For open organ pipe v =340

2L 2 0.20

v

= 850 Hz.Given source will not be in resonance with an open organ pipe of same length.

Example 70. A metre long tube open at one end with a movable piston at the other end,shows resonance with a fixed frequency source (a tuning fork of frequency 340 Hz) when the tubelength is 25.5 cm or 79.3 cm. Estimate the speed of sound in air at the temperature ofexperiment. Ignore edge effect.

Solution. In case of closed organ pipe the frequency of nth overtone is given by

vn =(2 1)

4 L

n vn = 1, 2, 3 ....

L1 = 25.5 cm, L2 = 79.3 cm.

Then v = 2

1 2

(2 1)(2 1)

4L 4L

nn vv

or 340 = 1 2(2 1) (2 1).

4 25.5 4 79.3

n n

v vv

or 1(2 1)

25.5

n = 22 1

79.3

n

or 1

2

2 1

2 I

n

n

=25.5 1

.79.3 3

�

Wave Motion 529

It is possible if n1 = 1 and n2 = 2. i.e. 25.5 cm corresponds to fundamental frequency and79.3 cm corresponds to first overtone frequency.

Also v = 1

1

4L

2 1

v

n =

4 25.5 340

(2 1) – 1

= 34680 cm s–1

= 346.8 ms–1.

Based On BeatsExample 71. A note produces 4 beats/s with a tuning fork of frequency 512 and 6 beats/s

with a tuning fork of frequency 514. Find the frequency of the note.

Solution. In the first case:Frequency of the tuning fork = 512

Beats/s = 4

Possible frequencies of the note are 512 + 4 = 516

or 512 – 4 = 508

In the second case:Frequency of the tuning fork = 514

Beats/s = 6

Possible frequencies of the note are 514 + 6 = 520

or 514 – 6 = 508As the frequency of 508 is common in both, the only possible frequency of the note = 508.Example 72. A tuning fork and a siren produce 18 beats in 3 s. The siren has 16 holes and

is making 960 revolutions per minute. If the speed of the siren is reduced, the two notes will bein unison. Find the pitch of the tuning fork.

Solution. There are 18 beats in 3 seconds

Beats/s =18

63

Number of holes h = 16

Revolution per minute = 960

Revolution per second =960

60 = 16

Frequency of the sound produced by the siren

= n × h = 16 × 16 = 256

Frequency of the tuning fork is either

= 256 + 6 = 262

or 256 – 6 = 250

Since with the decrease in the speed of the siren, its frequency decreases and is in unisonwith the tuning fork, the frequency of the tuning fork = 250.

Example 73. Calculate the velocity of sound in a gas in which two waves of lengths 1 metreand 1.01 metres produce 10 beats in 3 s.

Solution. Number of beats in 3 seconds = 10


Beats/s =10

3

Here, 1 = 1 metre = 100 cm2 = 1.01 metres = 101 cm

Suppose velocity = v

Frequency of the first wave = v1

Frequency of the second wave = v2

v1 – v2 = beats/s = 10

3

v1 – v2 =10

3

But v1 =1 100

v v

v2 =2

V V

101

Substituting the values of v1 and v 2

100 101v v

=10

3

10100

v=

10

3

v =101000

3 = 33666.67 cm/s

= 336.67 m/s.Example 74. The prongs of a tuning fork A, originally in unison with a tuning fork B, are

filed. Now the two tuning forks on being sounded together produce 2 beats/s. What is thefrequency of A after filing, if the frequency of B is 250 cycles/s.

Solution. Frequency of A = 250

Frequency of B = 250

Beats after A is filed = 2

Since the frequency increases after filing, frequency of A after filing

= 250 + 2

= 252 cycles/s.Example 75. Two tuning forks A and B when sounded together give 4 beats per second. A

is then loaded with a little wax and the number of beats/s is found to decrease. If the frequencyof A is 256, find that of B.


Beats/s = 4

Frequency of B is either 256 + 4 = 260

or 256 – 4 = 252

Wave Motion 531

Suppose the frequency of B = 260. When A is loaded with wax, its frequency will be lessthan 256 and its difference with 260 will be more than 4. Therefore beats/second increase whichis not given. Therefore, the frequency of B cannot be equal to 260.

Consider the frequency of B to be 252. When A is loaded with wax, its frequency will beless than 256 and its difference with 252 will be less than 4. Therefore, the beats/second decreasewhich is given. Hence the frequency of B = 252.

Example 76. Two tuning forks produce 4 beats per second when sounded together. Thefrequency of one is 200 and when the other is loaded with a little wax, the beats stop. Find thefrequency of the second tuning fork.

Solution. Frequency of the tuning fork A = 200

Beats/s = 4

Frequency of B is either = 200 + 4 = 204

or 200 – 4 = 196

When B is loaded with wax, beats stop. It means the number of beats/second decreases.

Suppose the frequency of B = 204. After loading with wax its frequency can be equal to Abecause after loading the frequency decreases.

Hence the frequency of B = 204.Example 77. A fork of unknown frequency when sounded with one of frequency 288, gives

4 beats/second and when loaded with a little wax again gives 4 beats per second. How do youaccount for this and what was the unknown frequency?

Solution. Suppose A to be the tuning fork of known frequency and B of unknown frequency

Frequency of A = 288Beats/s = 4

Frequency of B befoe loading is either288 + 4 = 292

or 288 – 4 = 284After loading B, again 4 beats/s are produced Frequency of B after loading is either

288 + 4 = 292or 288 – 4 = 284

Suppose the frequency of B before loading is 292. Then after loading its frequency mustdecrease and it is possible that its frequency after loading = 284.

On the other hand if we consider the frequency of B before loading as 284, its frequencyafter loading cannot be equal to 292.

Hence, the frequency of B befoe loading = 292

and after loading = 284.

Example 78. Two tuning forks A and B give 5 beat/second. The frequency of A = 512.When B is filed, 5 beats/second are again produced. Find the frequency of B before and afterfiling.


Beats per second = 5

Frequency of B before filing is either

512 + 5 = 517

or 512 – 5 = 507

Since after filing 5 beats/second are again produced, frequency of B after filing is either


512 + 5 = 517

or 512 – 5 = 507

Let us consider the frequency of B before filing as 517. After filing the frequency increases.Therefore, after filing the frequency of B cannot be equal to either 517 or 507. Therefore,frequency of B cannot be equal to 517.

Consider the frequency of B before filing 507.

After filing its frequency can be equal to 517

Therefore, this is the possible value.

Hence, frequency of B before filing = 507Hence, frequency of B after filing = 517.Example 79. If two of the organ pipes when sounded together produce 3 beats per second.

Find their frequencies if one of the pipe is 33 cm long and the other 33.5 cm long.

Solution. Let frequency of one is then frequency of second is ( v + 3).

v = Constant.

( v + 3) 33 = × 33.5

or v = 198 Hz.

Other frequency is v + 3 = 198 + 3 = 201 Hz.Example 80. Two sitar strings A and B playing the note Ga are slightly out of tune and

produce beats of frequency 6 Hz. The tension in the string A is slightly reduced and the beatfrequency is found to reduce to 3 Hz. If the original frequency of A is 324 Hz. What is thefrequency of B?

Solution. Frequency of A = 324 Hz.

Beat frequency = 6 Hz.

Possible frequency of B = 324 ± 6

= 330 or 318 Hz.

On reducing the tension frequency of string A will decrease. If we take frequency of B tobe 330 number. of beats heard per second will increase. But beat frequency decreases from 6 to3 so frequency of B is 318 Hz.

Example 81. Two tuning forks P and Q give 6 beats per second. P resounds with a closedair column 15 cm long and Q with a similar, open column 30.5 cm long. Calculate thefrequencies of the forks.

Solution. Let the frequencies of forks P and Q be v1 and v2.

Then v1 – v2 = 6 ... (i)

Also for P v1 = 4L 4 15

v v

v2 =2L 2 30.5

v v

1

2

v

v=

30.5

30... (ii)


1 = 366 Hz2 = 360 Hz.

Wave Motion 533

Example 82. Two identical strings of same material produce 2 beat per second. The lengthof one is 50 cm and that of the other is 50.1 cm. Calculate the frequencies of the two wires.

Solution. We know v =1 T

2l m

v =1 T

2 50 m

For second string (v – 2) =1 T

2 50.1 m1

vl

�

2

v

v =

50.1

50

50 v = 50.1 v – 100.2

or 0.1 v = 100.2

or v = 1002 Hz. Frequency of second wire is 1002 – 2 = 1000 Hz.Example 83. A tuning fork of frequency 300 Hz is in unison with a sonometer wire. How

many beats per second will be heard if the tension of the wire is increased by two percent?

Solution. Here v1 = 300 Hz.

Tension of the wire is increased by two percent

If T1 = 100 N

T2 = 102 N.

We know v T

1

2

v

v=

100

102

or 2

1

v

v=

11/ 2

2102 21

100 100

2

1

v

v=

1 21

2 100

2

1

v

v=

11

100

or 2

1

v

v – 1 =

1

100

or (v2 – v1) = 1 300

100 100

v = 3.

Number of beats produced per second is 3.Example 84. In an experiment it was found that a tuning fork and sonometer wire gave 5

beats per second, both when the length of wire was 1 m and 1.05 m. Calculate the frequency offork.

Solution. Consider frequency of tuning fork is v

Possible frequencies of sonometer wire are (v + 5) and (v – 5)


Since v 1

l

(v + 5) 1

l... (i)

and (v – 5) 1

(1.05 )l... (ii)

Divide (i) by (ii)

5

5

v

v

= 1.05

or v = 205 Hz.Example 85. Calculate the speed of sound is a gas in which two sound waves of wavelength

1.00 m and 1.01 m produce 4 beats per second.

Solution. Let velocity of sound in the gas is v .

v1 =1 1.00

v v

v2 =2 1.01

v v

v1 – v2 =1.0 1.01

v v = 4.

or v = 404 ms–1.

BASED ON DOPPLER’S EFFECTExample 86. A train standing at the outer signal of a railway station blows a whistle of 400

Hz in still air.

(i) What is the frequency of the whistle for a platform observer when the train

(a) approaches the platform with a speed of 10 ms–1.

(b) recedes from the platform with a speed of 10 ms–1?

(c) What is the speed of sound in each case (speed of sound in air = 340 ms–1).

Solution. Here v = 400 Hz.

vs = 10 ms–1

v0 = 0 v = 340 ms–1.

(i) When train approaches the stationary observer

v 0 = 0 v s = + 10 ms–1

v =0

s

v vv

v v

=340 0

400340 10

= 412.12 Hz.

Wave Motion 535

(b) When train moves away from observer

v 0 = 0, v s = – 10 ms–1.

v =0

s

v

v vv v

=340 0

340340 ( 10)

= 388.57 Hz.

Example 87. A train stands at a platform blowing a whistle of frequency 400 Hz in still air.

(i) What is the frequency of the whistle heard by a man running

(a) towards the engine at 10 ms–1

(b) away from the engine at 10 ms–1?

(ii) What is speed of sound in each case?

(iii) What is the wavelength of sound received by the running man in each case?

( v = 340 ms–1)

Solution. Here v = 400 Hz, v = 340 ms–1

(i) (a) When man runs towards the engine

v 0 = – 10 ms–1, v s = 0.

v =0

s

v

v vv v

=340 10

400340 0

=350

400340

= 411.8 Hz

(b) When the man runs away from engine.

v 0 = + 10 ms–1, v s = 0.

v =0

s

v

v vv v

=340 10

400340 0

=330 400

340

= 388.2 Hz.

(ii) (a) In first case relative velocity of sound

v = v + v 0 = 340 + 10 = 350 ms–1

(b) In second case relative velocity of sound

v = v – v 0 = 340 – 10 = 330 ms–1.(iii) Wavelength of sound is not affected by the motion of the listener.

=340

400vv

= 0.85 m.


Example 88. A SONAR system fixed in a submarine operates at a frequency 40.0 kHz. Anenemy submarine moves towards the SONAR with a speed of 360 km h–1. What is the frequencyof sound reflected by the submarine. Take the speed of sound in water to be 1450 ms–1.

Solution. Here v = 40 kHz = 40 × 103 Hz.

v = 1450 ms–1

speed of enemy submarine = 360 km.

=5

36018

= 100 ms–1

First sound is heard by enemy submarine source i.e. enemy submarine is moving towardsSONAR i.e. source

v 0 = – 100 ms–1

v s = 0

Frequency of SONAR waves received by the enemy submarine.

v =0

s

v

v vv v

=1450 100

401450 0

= 42.76 × 103 Hz.

When sound is reflected, enemy submarine acts as a source of sound of frequency . Thissource moves with a speed of 100 ms–1, towards SONAR.

So v S = + 100 ms–1, v 0 = 0

v =0

S

v

v vv v

=31450 0

42.76 101450 100

= 45.93 × 103 Hz

= 45.93 KHz.

Example 89. Find the velocity of the source of sound when the frequency appears to be(a) double (b) half of the original frequency to a stationary observer velocity of sound = 340

ms–1.

Solution. (a) Here v = 2 v , v 0 = 0.

v =0

S

v vv

v v

2 v =S

340 0

340

vv

2 (340 – v S) = 340

2 v S = 340

or or v S = 170 ms–1

Wave Motion 537

(b) =2

2

=

340

340 Sv

On solving vS = – 340 ms–1.

–ve sign shows that source is moving away from observer.

Example 90. A police man detects a drop of 15% in the frequency of the horn of a car as itcrosses him. If the speed of sound is 340 ms–1, calculate speed of the car.

Solution. Initially source is moving towards observer

=S

v

v v

... (i)

After crossing the source is going away from observer

=S

v

v v

... (ii)

Dividing (i) by (ii)

=+

S

S

v v

v v

But

=85

100(given)

85

100= S

S

340

340

v

v

On solving vS = 27.6 ms–1.

Example 91. Two railway trains each moving with a velocity of 108 km h–1 cross eachother. One of the train gives a whistle whose frequency is 750 Hz. What is the apparent frequencyfor passangers sitting in the other train before crossing. Given velocity of sound = 340 ms–1.

Solution. Here vS = 108 km h–1

=5

10818

= 30 ms–1

v0 =5

10818

= – 30 ms–1.

As the observer and source are approaching each other

= 0

S

v v

v v

=340 30

750340 30

=370

750310

= 895 Hz.


Example 92. The Sirens of two fire engines have a frequency of 650 Hz each. A man hearsthe sirens from two engines. One approaching him with a speed of 36 km h–1 and the otherreceding from him at a speed of 54 km h–1. What is the difference in the frequency of the twosirens, heard by the man. Velocity of sound is 340 ms–1.

Solution. Let 1 is the frequency of the siren from the approaching engine.

vS = 36 km h–1 = 5

3618

= 10 ms–1.

v = 340 ms–1.

1 =s

v

v v

=340

650340 10

= 669.7 Hz.Let 2 be the frequency of the siren from the receding engine

2 =+ s

v

v v

=340

650340 10

= 622.5 Hz.Difference is frequency = 669.7 – 622.5

= 42.2 Hz.Example 93. An automobile moving at 35 ms–1 with its horn blowing ( = 1200 Hz) is

chasing a car moving at 20 ms–1. What is the apparent frequency of the horn as heard by thedriver being chased (v = 340 ms–1)

Solution. Here v0 = 20 ms–1 vS = 35 ms–1.

= 1200 Hz

v = 340 ms–1.

Source is moving towards observer and observer is moving away from source.

= 0

s

v v

v v

=340 20

1200340 35

=1200 320

305

= 1259 Hz.Example 94. A band playing music at a frequency is moving towards a wall at a speed of

vb. A motorist is following the band with a speed vm. If c is speed of sound obtain an expressionfor the beat frequency heard by the motorist.

Wave Motion 539

Solution. Frequency received by the motorist directly is

v = .m

b

cv

c

vv

Frequency received by him after reflection

v = .m

b

cv

c

vv

Number of beats heard per second

= v – v

=1 1

( )mb b

c vc c

vv v

= 2 2

2 ( )b m

b

cv

c

v v

v.

Example 95. A source of sound of frequency 256 Hz is moving rapidly towards a wall witha velocity of 5 ms–1. How many beats per second will be heard if sound travels at a speed of 330ms–1.

Solution. Given that v = 256 Hz

v s = 5 ms–1

v = 330 ms–1

v 0 = – 5 ms–1

v = 0

s

v

v vv v

=330 – ( 5)

256330 5

= 263.88 Hz.Number of beats heard per second = 263.88 – 256

= 7.88 Hz.

Example 96. The spectral line for a given element in light received from a distant star isshifted towards the longer wavelength by 0.032%. Deduce the velocity of the star in the line ofsight.

Solution. Since spectral line is shifted towards the longer wavelength hence frequencydecreases.

Apparent frequency v =s

cv

c

v

orv

v= 1s sc

c c

v v

/

/

c

c

= 1c

sv


or 1

= s

c

v

or v s = c

= 80.0323 10

100

= 9.6 × 104 m/s.

Example 97. A supersonic jet travels with twice the speed of sound in air. What is the angleof the conical wave front of the shock waves produced by jet?

Solution. Half angle of the conical wavefront of the shock waves produced by jet is

sin =1

2 2s

c c

c

v

or = 30°.

Required angle = 2 = 60°.Example 98. A radar wave has a frequency of 4.8 × 109 Hz. The reflected wave from an

aeroplane shows a frequency difference of 2.4 × 103 Hz on the higher side. Calculate the velocityof the aeroplane in the line of sight.

Solution. Here v = 4.8 × 109 Hz

v = 2.4 × 103 Hz

We know v =cv

v

or v =v

cv

=3

89

2.4 103 10

4.8 10

= 150 ms–1.

The velocity of aeroplane determined by radar wave is double of its actual velocity ofapproach

Actual velocity of aeroplane =150

2

= 75 ms–1.

Example 99. A police man on duty detects a drop of 10% in the pitch of the horn of a motorcar as it crosses him. If the velocity of sound is 330 ms–1. Calculate the speed of the car.

Solution. Given v = 330 ms–1

Frequency of horn v = 100 Hz

Apparent frequency v = 90 Hz

We know v =s

v

vv v

Wave Motion 541

90 =330

100330 s

v

On solving v s = – 36.67 ms–1

negative sign shows that source is moving away from observer.

Example 100. A spectral line of = 5000 Å in the light coming from a distant star isobserved at 5200 Å. Determine the recession velocity of the star.

Solution. Given = 5000 Å

= 5200 Å.

We know v s = c

=8200

3 105000

= 1.2 × 107 ms–1.

EXERCISE

1. A tuning fork produces waves of wavelength one metre in a medium A. Calculate thelength of the waves produced in a medium B, given velocity of waves in medium A = 300m/s and velocity of waves in medium B = 450 m/s. [Ans. 1.5 m]

2. A tuning fork having a frequency of 480 produces waves of length 2.5 metres. Calculate thevelocity of sound in the medium. [Ans. 1200 ms–1]

3. A tuning fork of frequency 512 produces waves of length 0.60 m in air. Calculate thevelocity of sound in air. Also calculate the wavelength of the sound waves produced by thesame tuning fork in a medium where the velocity of sound is 600 m/s.

[Ans. 307.2 ms–1, 1.17m]

4. A tuning fork makes one complete vibration in 1/256 second and the velocity of soundwaves is 345 m/s. Calculate the wavelength of sound waves produced by the tuning fork.

[Ans. 1.348 m]

5. On a rainy day, a thunder was heard 4 seconds after the flash of light was seen. How farwas the cloud from the observer? (Velocity of sound is 332 m/s.) [Ans. 1328 m]

6. The pitch of a siren is 500 vibrations/s. What is the wavelength of the sound wavesproduced ? Velocity of sound = 33500 cm/s. [Ans. 67 cm]

7. A body vibrating with a certain frequency produces waves of length 20 cm in medium Aand 30 cm in a medium B. If the velocity of sound in a medium B is 5400 cm/s. Calculatethe velocity of sound in the medium A. [Ans. 360 m/s]

8. If the frequency of a tuning fork is 400 Hz and velocity of sound in air is 320 ms–1. Findhow far sound travels while the fork completes 50 oscillations. [Ans. 40 m]

9. The frequency of a radio transmission centre is 30 MHz. Find out the wavelength of thewaves transmitted from the station. [Ans. 10 m]

10. The frequency of a tuning fork is 500 Hz and velocity of sound in air is 320 ms–1. Calculatethe distance travelled by the sound waves while the fork completes 20 oscillations.

[Ans. 12.8 m]11. A source of sound is placed at one end of an iron bar two kilometre long and two sounds

are heard at the other end at an interval of 5.6 seconds. If the velocity of sound in air is 330ms–1. Find the velocity of sound in iron. [Ans. 4348 ms–1]


12. A stone is dropped into a well and its splash is heard at the mouth of the well after aninterval of 1.45 sec. Find the depth of the well. Given that velocity of sound in air at roomtemperature is equal to 332 ms–1. [Ans. 9.9 m]

Based on Velocity of Transverse Waves13. A sitar wire is under tension 30N and the length between the bridges is 80 cm. A 10 m

sample of that wire has mass 2.2 g. Calculate the speed of the transverse wave on the wire.

[Ans. 369.27 ms–1]

14. A rope of mass 0.1 kg m–1 is held with a tension of 1000N. As a wave passes along the ropeevery point on the rope moves with SHM with a frequency of 10 Hz and an amplitude 0.1m. For this wave, what is

(i) the velocity (ii) the wavelength [Ans. 100 ms–1, 10 m]

15. The mass of a 8.0 m long wire is 0.02 kg. What is the speed of transverse wave in this wireat a tension of 1600N. [Ans. 800 ms–1]

16. The density of the material of a string is 7.5 × 103 kg m–3. If the string is subjected to astress of 3.0 × 108 Nm–2, find out the speed of the transverse waves in the string.

[Ans. 200 ms–1]

17. The speed of transverse wave in a stretched string is 250 ms–1 when the tension in thestring is 2 kg wt. Calculate the speed of transverse wave in the same string when tensionis changed to 8 kg wt. [Ans. 500 m/s).

18. A string of mass 2.25 g m–1 is under a tension of 40N. Calculate the velocity of wave in thestring. [Ans. 133.3 ms–1]

19. Steel wire has a length 10.0 m and a mass 2.5 kg. Calculate the tension in the wire so thatthe speed of the transverse wave on the wire equals to 320 ms–1. [Ans. 2.56 × 104 N]

20. Calculate the velocity of transverse wave in a copper wire 1 mm2 in cross section, under thetension produced by 1 kg wt. The density of copper = 8.93 kg m–3. [Ans. 33.12 m/s]

Based on Speed of Longitudinal Wave21. Calculate the velocity of sound in a steel rod. The Young’s modulus for steel is 2.1 × 1011

Nm–2 and density of steel is 7.8 × 103 kg m–3. [Ans. 5.19 × 103 ms–1]

22. The density of air at S.T.P is 1.29 kg m–3. Deduce the velocity of longitudinal waves using

(i) Newton’s formula (ii) Laplace correction.

[Ans. 280.2 ms–1, 331.56]

23. At a pressure of 1.01 × 105 Nm–2 the volume strain of water is 5 × 10–5. Calculate thevelocity of sound in water. [Ans. 1.414 × 103 ms–1]

24. The velocity of sound in air at 27°C is 360 ms–1. Calculate the velocity of sound in air at17°C. [Ans. 352.8 ms–1]

25. Calculate the speed of sound in oxygen if velocity of sound in hydrogen is 1248 ms–1.Given oxygen is 16 times heavier than hydrogen. [Ans. 312 ms–1]

26. Determine speed of sound in helium at 1.2 × 105 Pa and at temperature 400 K. Molecularweight of helium is 0.0040 kg/mol. and = 1.63. [Ans. 1163.8 ms–1]

27. Calculate the ratio of velocity of sound in hydrogen ( = 7/5) to that in helium gas ( = 5/

3) at the same temperature. [Ans. 42 / 5 ]

28. Calculate the temperature at which sound travels in hydrogen with the same speed as inhelium at N.T.P. Given density of helium is twice that of hydrogen. [Ans. – 110.5°C]

Wave Motion 543

29. At what temperature the speed of sound in oxygen will be equal to its speed in nitrogen at14°C. Given d0 : dH = 16 : 14. [Ans. 55°C]

30. At what temperature will the speed of sound be double of its value at 0°C. [Ans. 819°C]

31. Speed of sound in air is 332 ms–1. at N.T.P. What will be its value in hydrogen at N.T.P. if

density of hydrogen at N.T.P. is 1

16 that of air. [Ans. 1328 ms–1]

32. Calculate the velocity of sound in air at 30°C and 76 cm pressure of mercury if the densityof air at 0°C and pressure 75.5 cm of mercury is 0.00129 g cm–3. [Ans. 353.04 ms–1]

33. Calculate the speed of longitudinal waves in a metal rod if the material of rod has Young’smodulus of 1.2 × 1010 Nm–2; and density is 8.92 × 103 kg m–3. [Ans. 1160 ms–1]

34. A longitudinal wave sent by a ship to the bottom of the sea returns after a lapse of 2.64seconds. Calculate the depth of the sea. Given bulk modulus of water is 2.45 × 109 Nm–2.density of sea water is 1.2 × 103 kg m–3. [Ans. 1.89 km]

35. At what temperature velocity of sound in oxygen is same as that in nitrogen at 10°C?Given their densities at 0°C are in the ratio 16 : 14.[Ans. 50.43°C]

36. A stone is dropped in a well 78.4 metres deep and the impact of sound was heard after 4.2seconds. Find the velocity of sound. [Ans. 392 ms–1]

37. The frequency of a tuning fork is 384 and the velocity of sound in air is 352 m/s. Find howfar the sound has travelled while the fork completes 36 vibrations. [Ans. 33 m]

38. Find the temperature at which the velocity of sound in air is 1.25 times the velocity ofsound at 0°C. [Ans. 153.6°C]

39. The velocity of sound in air at 0°C and 76 cm pressure is 332 m/s. Calculate the velocityof sound at 136.5°C and 75 cm pressure. [Ans. 406.6 ms–1]

40. The velocity of sound in air at 51°C is 350 m/s. Find the wavelength in air of a note offrequency 288 at (i) 51°C and (ii) 127°C. [Ans. 1.215 m, 1.35 m]

41. Calculate the velocity of sound in oxygen if the velocity of sound in hydrogen is 1248 m/s.The density of oxygen is 16 times that of hydrogen. [Ans. 312 ms–1.]

42. The velocity of sound through hydrogen is 1248 metres/second. What will be the velocityof sound in a mixture of hydrogen and oxygen? The volumes of oxygen and hydrogen aremixed in the ratio 3 : 2. [Ans. 394.7 ms–1]

43. The velocity of sound at normal temperature and pressure is 332 m/s. Calculate thetemperature at which the velocity of sound is 415 m/s. [Ans. 153.6°C]

44. The velocity of sound in air at 51°C is 360 m/s. Calculate the velocity of sound in air at16°C. [Ans. 340 ms–1]

45. The velocity of sound in air at 16°C is 340 m/s. Calculate the velocity of sound in air at88°C. [Ans. 380 ms–1]

Based on Progressive (SHM) Waves46. The equation of transverse wave travelling in a rope is given by

y = 10 sin (0.01x – 2.00t)

Where y and x are in cm and t in seconds. Find the amplitude, frequency, velocity andwavelength of wave. [Ans. 10 cm, 1 Hz, 200 cm s–1, 200 cm]

47. A wave on a string is described by

y (x, t) = 0.005 sin (6.28x – 314t) in which all quantities are in S.I. Units.


Calculate (i) amplitude (ii) wavelength. [Ans. (i) 0.005 m (ii) 1 m]

48. Show that in aharmonic wave v (velocity) is ahead of phase by /2 relative to y andacceleration is further ahead of the phase by /2.

49. A plane progressive wave of amplitude 2 cm and frequency 100 Hz is travelling along + xaxis, with velocity 15 ms–1. Calculate the displacement y. The particle velocity and theparticle acceleration at x = 180 cm from the origin at t = 5 sec.

[Ans. y = 2 sin2

15

(7500 – x), 0, 400 cm s–1, 0]

50. A plane progressive wave is given by

y = 10 sin 2 0.02 25

t x Where y and x are expressed in cm and t in sec. Calculate

(i) amplitude (ii) velocity

(iii) frequency (iv) wavelength.

[Ans. (i) 10 cm, (ii) 1250 cm/s. (iii) 50 Hz, (iv) 25 cm.

51. Find the equation of a plane progressive wave travelling along the x axis in the positivedirection having amplitude 15 cm; speed 330 ms–1 and frequency 110 Hz.

[Ans. y = 0.15 sin 2 (110t – x/3) m]

52. A plane progressive wave on a string has a frequency of 50 Hz and wavelength of 100 cm.Its amplitude is 10 cm. Write the equation of the wave.

[Ans. y = 10 sin (314 t – 0.0628 x].

53. A plane progressive wave is represented by

y = 2 sin 4 cm.4 16

xt

Calculate the phase difference for two positions of the same particle which are occupied atthe time interval 0.8s apart and (ii) the phase difference at any given instant of time when

the two particles are 48 cm apart. [Ans. 4/5 rad, 3

4

rad.]

54. Spherical waves of frequency 1500 Hz are emitted from a 1.0W source in an isotropic nonabsorbing medium. What will be the intensity and amplitude of the wave 1.0 m from thesource if density of the medium be 1.3 kg m–3 and wave velocity is 300 ms–1.

[Ans. 7.96 × 10–2 wm–2, 2.14 × 10–6 m]

55. Calculate the speed of propagation of a wave given by

y = 0.5 sin (0.01 x – 3 t)

Where y and x are in metre and t in sec. [Ans. 300 m/s]

56. Given below the equation of a plane progressive wave

y = 2.0 sin 240 314

xt

.

Where y and x are in metres and t in second. Calculate the phase difference between twopoints 50 cm apart. [Ans. 21.89°]

Wave Motion 545

57. A 5W source sends out waves in air at frequency 1000 s–1. Deduce the intensity at a 100 mdistance assuming spherical distribution. If v = 350 ms–1 and = 1.3 kg m–3, deducedisplacement amplitude. [Ans. 6.67 × 10–8 m]

58. The equation of a travelling wave is

y = 0.07 sin (12 x – 500 t)

where the distances are measured in metre and time in second. Calculate the wavelengthand speed of the wave.

[Ans. 1

m,6

41.7 ms–1]

Based on Stationary Waves59. The component waves producing a stationary wave have amplitude, frequency and velocity

as 8 cm, 30 Hz and 180 cm s–1 respectively. Write the equation for stationary wave.

[Ans. u = 16 cos 3

x sin 60 t].

60. The equation of a longitudinal stationary wave produced in a closed organ pipe is

y = 6 sin 2

6

x cos 160 t.

Where x, y are expressed in cm and t in second find(i) the frequency, amplitude and wavelength of the original progressive wave.

(ii) separation between two consecutive antinodes.

[Ans. v = 80 Hz, a = 3 cm, = 6 cm (ii) 3 cm.

61. The equation of stationary wave is

y = 12 cos sin 205

xt

where x and y are in cm, t in second. Find out.

(i) frequency and wavelength.

(ii) velocity and amplitude of the wave.

[Ans. 10 Hz, 10 cm, (ii) 10 0 cm s–1, and 6 cm]

62. Stationary waves are produced by the superposition of two waves given by

y1 = 0.05 sin (5 t – x)

y2 = 0.05 sin (5 t + x).

Where x and y are in metres and t in sec. Find the displacement of a particle situated at adistance x = 1 m. [Ans. 0.054 m]

Based on Vibrations of String63. A stretched wire has a length of 60 cm and the mass per unit length is 0.04 g per cm.

Calculate the frequency of the fundamental note emitted if the wire is stretched by a forceof 9 million dynes. [Ans. 125 Hz]

64. A wire of specific gravity 7, length 25 cm and 0.5 mm in diameter is stretched by weightof 1.76 kg wt. Calculate the pitch of the fundamental note. [Ans. 224 Hz]


65. The distance between the two bridges of a sonometer wire when it is in unison with atuning fork is 40 cm. The diameter of the wire is 0.75 mm, the density is 7.0 g/cc and thestretching force is 4.4 kg wt. Calculate the frequency of the tuning fork. [Ans. � 148 Hz]

66. A steel wire is stretched with a force of 4.5 kg wt. The length of the wire is 0.25 metre.One metre length of the wire weighs 0.009 kg. Calculate the frequency of the fundamentalnote emitted by it. [Ans. 140 Hz]

67. A string when stretched with a load of 25 newtons emits a note of frequency 200 Hz.Calculate the stretching force required if the frequency is to be doubled. [Ans. 100 N]

68. A string when stretched by a weight of 9 kg gives a note of frequency 288 Hz. What weightwill produce an octave of this note? [Ans. 36 kg]

69. A stretched wire under a tension of 4 kg wt is in unison with a fork of frequency 480 Hz.What change in tension will make the wire vibrate with a frequency of 240 Hz?[Ans. 3 kg]

70. Calculate the fundamental frequency of a string 0.5 metre long and weighing 0.004 kg,when stretched by a load of 225 kg. [Ans. 525 Hz]

71. A piano wire of length 30 cm produces a note of frequency 426 Hz. Calculate the frequencywhen the vibrating length is reduced to 25 cm. [Ans. 511.2 Hz]

72. A tuning fork of frequency 288 Hz is sounded along with a sonometer wire of length 0.24metre and stretched by a tension of 39.2 newtons. Calculate the number of beats producedper second, if the mass of one metre length of the wire is 0.002 kg. [Ans. 3.67]

73. A wire gives a fundamental note of frequency 288 Hz when it has a tension of 36 newtons.(i) At what tension will the wire emit a note of frequency 432 Hz?

(ii) How would you make the wire emit a note of frequency 720 Hz, keeping the tension at36 newtons? [Ans. 81 N, l2 = 0.4 l]

74. A wire of uniform cross-section is stretched between two points 100 cm apart. The wire isfixed at one end and a weight is hung over a pulley at the other end. A weight of 9 kgproduces note of fundamental frequency 750 vibrations/second.(i) What is the velocity of sound in the wire?

(ii) If the weight is reduced to 4 kg what is the velocity of sound? What is the wavelengthof sound? What is the frequency of sound?

[Hint: = 2l, v = n ] [Ans. 1500 m/s 1000 m/s, 200 cm, 500 Hz]75. A sonometer is emitting sound of a certain frequency. How must the tension be changed to

raise the frequency in the ratio 5 : 3? [Ans. 25 : 4]76. A string 30 cm long gives a note of frequency 280 Hz when stretched by a load of 4 kg wt.

What alteration must be made(a) in length without altering the tension and(b) in tension without altering the length to make it produce a note of frequency 420 Hz.

[Ans. 10 cm decrease, 5 kg increase]77. Calculate the frequency of the fundamental note of a string 10 cm long and weighing 0.2

g when stretched by a weight of 4 kg. (g = 9.80 m/s2). [Ans. 700 Hz]78. A sonometer wire 100 cm in length has a fundamental frequency of 330 cycles/second.

Find the velocity of propagation of transverse waves along this wire and the wavelength ofresulting sound in air. Velocity of sound in air = 330 m/s. [Ans. (i) 660 ms–1 (ii) 1 m]

79. A string 40 cm long gives a note of frequency 256 Hz when stretched by a load of 5 kg wt.What alteration must be made

(i) in length without altering the tension and

(ii) in tension without altering the length to produce a note of frequency 384 Hz.

[Ans. l2 = 26.67 cm, T2 = 11.25 kg wt.]

Wave Motion 547

Based on Organ Pipes80. A ship sounds its siren and an echo from a rock is heard on the ship 3.2 seconds later. If

the velocity of sound in air is 340 m/s how far is the rock from the ship? [Ans. 544 m]

81. A man standing between two parallel hills fires a gun. He hears the first echo after 2.4 sand next after 4 s. Find the distance between the two hills. Velocity of sound in air is 350m/s. [Ans. 1120 m]

82. A man standing between two parallel cliffs fires a rifle. He hears the first, second and thethird echoes after 2, 3 and 5 seconds respectively. The distance between the two cliffs is850 metres. What is velocity of sound and what is the distance of the man from one of thecliffs? [Ans. 340 ms–1, 340 m, 510 m]

83. In a resonance air column apparatus, the first and the second resonance positions wereobserved at 18 cm and 56 cm with a tuning fork of frequency 480 Hz. Calculate thevelocity of sound and end correction. [Ans. 364.80 ms–1]

84. An organ pipe emits a sound of frequency 240 Hz. On blowing air at higher pressure,sound of frequencies 480 Hz, 720 Hz, 960 Hz etc. are heard. Is the pipe open or closed atthe other end? Explain. [Ans. Open organ pipe]

85. What is the frequency of the fundamental note emitted by an open organ pipe of length0.75 metre? The velocity of sound is 342 m/s. [Ans. 228 Hz]

86. Velocity of sound in air is 326 metres/second. A column of air produces resonance with atuning fork of frequency 326 Hz. Find the length of the air column, when (i) pipe is open;and (ii) pipe is closed at one end. [Ans. (i) 50 cm (ii) 25 cm]

87. With a tuning fork of frequency 480 Hz, the difference at the levels of water between thetwo successive positions of resonance is 0.35 metre. Calculate the velocity of sound in air.

[Ans. 336 ms–1]

88. Two open organ pipes of lengths 0.60 metre and 0.605 metre produce 5 beats in twoseconds. Calculate the velocity of sound in air. [Ans. 363 ms–1]

89. Two tuning forks A and B give 5 beats/s. A is in resonance with a closed organ pipe oflength 16 cm and B is in resonance with an open organ pipe of length 32.5 cm. Calculatetheir frequencies. [Ans. 325 Hz, 320 Hz]

90. A tuning fork of frequency 256 Hz is held over the tube of the resonance apparatus andmaximum resonance is obtained when the columns of air are 32 cm and 100 cm long. Findthe end correction and the velocity of sound in air. [Ans. 348.16 ms–1, 2 cm]

91. Find the length of a closed end organ pipe which produces a note of fundamental frequency480 cycles/s. Velocity of sound in air is 360 m/s. [Ans. 18.75 cm]

92. In a resonance air column apparatus the first and the second resonance positions wereobserved at 16 cm and 50 cm with a tuning fork of frequency 512. Calculate the velocity ofsound in air and the end correction. [Ans. 348.16 m/s]

93. A boy hears an echo of his own voice from a distant hill after one second. If the velocity ofsound is 340 m/s, what is the distance of the hill from the boy? [Ans. 170 m]

94. Calculate the frequency of the fundamental note emitted by an open end organ pipe oflength 0.60 m. The velocity of sound in air is 360 m/s. [Ans. 300 Hz]

95. In a resonance air column apparatus, the first and the second resonance positions wereobserved at 18 cm and 56 cm with a tuning fork of frequency 480 Hz. Calculate thevelocity of sound in air and the end correction. [Ans. 364.8 ms–1 1 cm]

96. A tuning fork of frequency 340 Hz is vibrated just above a cylindrical tube. The height oftube is 120 cm. Water is slowly poured into the tube. What is minimum height of waterrequired for resonance. Velocity of sound in air = 340 m/s. [Ans. 45 cm]


Based on Beats97. The disc of a siren has 60 holes. What must be the speed of rotation per minute of the siren

so that the note emitted by it is in unison with that of a tuning fork of frequency 512?

[Ans. 512 Hz]

98. A note produces 5 beats/s with two tuning forks of frequencies 500 Hz and 510 Hz whensounded separately. Find the frequency of the note. [Ans. 505 Hz]

99. A tuning fork and a siren produce 5 beats per second. The siren has 40 holes and rotates720 times per minute. If the speed of the siren is reduced, the two notes are in unison. Findthe frequency of the tuning fork. [Ans. 475 Hz]

100. Calculate the velocity of sound in a gas in which two waves of lengths 2 metre and 2.02metre produce 8 beats in 5 seconds. [Ans. 323.2 ms–1]

101. The prongs of a tuning fork A, originally in unison with a tuning fork B of frequency 320Hz, are filed and the number of beats produced/second is 5. Calculate the frequency of Aafter filing. [Ans. 325 Hz]

102. Two tuning forks A and B when sounded together produce 3 beats/s. When A is loaded withwax, the number of beats/s decreases. If the frequency of B is 480, find the frequency of A.

[Ans. 483 Hz]

103. Two tuning forks produce 4 beats/second when sounded together. The frequency of one is256 Hz and when the second is loaded with a little wax the beats stop. Find the frequencyof the second tuning fork. [Ans. 260 Hz]

104. A tuning fork of unknown frequency, when sounded with a tuning fork of frequency 384,produces 2 beats/s and when loaded with a little wax again produces 2 beats/s. What is thefrequency of the fork? Explain your answer. [Ans. 386 Hz and 382 Hz]

105. Two tuning forks A and B give 6 beats in two seconds. The frequency of A is 432. When Bis filed, again 6 beats are produced in two seconds. Calculate the frequency of B before andafter filing. [Ans. 429 Hz, 435 Hz]

106. In a diatonic musical scale, keynote C is 128. Calculate frequency of the notes in the scale.

[Ans. 128, 144, 160, 170.67, 192, 213.33, 240, 256 Hz.]

107. Two tuning forks A and B give 4 beats/second. The frequency of A is 480 cycles/second.When B is filed, again 4 beats/s are produced. Find the frequency of B before and afterfiling. [Ans. 476 Hz, 484 Hz]

108. Calculate the velocity of sound in a gas in which two waves of wavelengths 2 m and 2.02m produce 5 beats in 2 seconds. [Ans. 505 ms–1]

109. The disc of a siren has 50 holes. It rotates uniformly and makes 400 rotations in 1 minute40 seconds. Calculate the frequency of the note emitted. Also calculate the wavelength, ifthe velocity of sound in air is 340 m/s. [Ans. 200, 1.7 m]

110. Calculate the velocity of sound in a gas in which two waves of lengths 0.08 m and 0.81 mproduce 5 beats per second. [Ans. 324 ms–1]

111. A tuning fork of unknown frequency when sounded with another of frequency 256 cyclesper second gives 5 beats per second and when loaded with a certain amount of wax it gives4 beats/s. Find the unknown frequency. [Ans. 260 Hz]

112. Two open organ pipes of length 75 cm and 80 cm are sounded together. How many beatsper second will be produced by them? (Speed of sound in air = 332 ms–1] [Ans. 14]

113. The wavelength of two waves are 49 and 50 cm. If the room temperature be 30°C then howmany beats per second will be heard due to these two waves.(velocity of sound at 0°C = 332 ms–1) [Ans. 14 beats/sec.]

Wave Motion 549

114. A 70 cm long sonometer wire is in unison with a tuning fork. If the length of the wire isdecreased by 1.0 cm it produces 4 beats per second with the same tuning fork. Find thefrequency of the tuning fork. [Ans. 276 Hz]

115. In an experiment it was found that a tuning fork and a sonometer gave 5 beats/sec. bothwhen length of wire was 1 m and 1.05 m. Calculate the frequency of tuning fork.

[Ans. 205 Hz]

Based on Doppler’s Effect116. With what velocity should an observer move towards a stationary source so that the

apparent frequency is double of the actual frequency (velocity of sound is 350 ms–1]

[Ans. 350 ms–1]

117. Two cars are approaching each other on a straight road and moving with a velocity of 30km/hr. If the sound produced in a car is of frequency 500 Hz. What will be the frequencyof sound as heard by the person sitting in the other car, when the cars have crossed eachother and moving away from each other ? What will be the frequency of sound as heard bythe same person? (speed of sound is 330 ms–1) [Ans. 526 Hz]

118. A person is standing on a railway platform. An engine while approaching the platformblows a whistle of pitch 660 Hz. The speed of the engine is 36 km/hr. Velocity of sound is340 ms–1. Calculate the apparent pitch of the whistle as heard by the person.[Ans. 680 Hz]

119. In stationary observer, the apparent frequency of a moving source of sound appears 5

4 of its

real frequency. Find the speed and direction of motion of source (Vs = 330 ms–1).

[Ans. 66 ms–1 towards observer]

120. An observer is approaching a stationary source producing sound of frequency 90 Hz, witha velocity one tenth of the velocity of sound. What is apparent frequency heard by theobserver? [Ans. 99 Hz]

121. An automobile moving at 35 ms–1 with its horn blowing ( = 1220 Hz) is chasing a carmoving at 10 ms–1. What is the apparent frequency of the horn as heard by the driver beingchased? (speed of sound is 340 ms–1). [Ans. 1320 Hz.]

122. A source of sound of frequency 256 Hz is moving rapidly towards a wall with a velocity 5ms–1. How many beats per second will be heard; if sound travels at a speed of 330 ms–1.

[Ans. nearly 8 Hz.]

123. A source of sound is approaching a stationary listner with a speed of 166 ms–1. The listnerhears a sound whose apparent frequency is twice the real frequency. Find speed of soundwaves. [Ans. 332 ms–1]

124. Two persons A and B each sending a note a frequency 580 Hz are standing a few metresapart. Calculate the number of beats heard by each in one second when A moves towards Bwith a velocity of 4 ms–1. (speed of sound in air is 330 ms–1) [Ans. 7 beats/sec.]

125. A person is standing on a platform. A railway engine moving away from the person with aspeed of 20 ms–1 blows a whistle of pitch 1110 Hz. Calculate the apparent pitch of thewhistle as heard by the person. (velocity of sound = 350 ms–1) [Ans. 1050 Hz.]

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