Preview A Complete Resource Book in Physics for JEE Main 2019 by Sanjeev Kumar (2018)

Preview A Complete Resource Book in Physics for JEE Main 2019 by Sanjeev Kumar (2018) Preview A Complete Resource Book in Physics for JEE Main 2019 by Sanjeev Kumar (2018) Preview A Complete Resource Book in Physics for JEE Main 2019 by Sanjeev Kumar (2018) Preview A Complete Resource Book in Physics for JEE Main 2019 by Sanjeev Kumar (2018) Preview A Complete Resource Book in Physics for JEE Main 2019 by Sanjeev Kumar (2018)

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Sanjeev Kumar

PHYSICS

A Complete Resource Book in

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attempt to render any type of professional advice or analysis, nor is it to be treated as such While much care has been

taken to ensure the veracity and currency of the information presented within, neither the publisher nor its authors bear

any responsibility for any damage arising from inadvertent omissions, negligence or inaccuracies (typographical or

factual) that may have found their way into this book

This book is sold subject to the condition that it shall not, by way of trade or otherwise, be lent, resold, hired out,

or otherwise circulated without the publisher’s prior written consent in any form of binding or cover other than that

in which it is published and without a similar condition including this condition being imposed on the subsequent

purchaser and without limiting the rights under copyright reserved above, no part of this publication may be reproduced,

stored in or introduced into a retrieval system, or transmitted in any form or by any means (electronic, mechanical,

photocopying, recording or otherwise), without the prior written permission of both the copyright owner and the

publisher of this book

ISBN 978-93-530-6216-3

eISBN 9789353063429

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Preface vi

JEE Mains 2018 Paper viii

JEE Mains 2017 Paper xvi

Chapter 1 Unit and Dimension 1.1–1.56

Chapter 2 Kinematics 2.1–2.68

Chapter 3 Newton’s Law of Motion 3.1–3.70

Chapter 4 Work, Energy, and Power 4.1–4.48

Chapter 5 Impulse and Momentum 5.1–5.64

Chapter 6 Rigid Body Dynamics 6.1–6.52

Chapter 7 Gravitation 7.1–7.36

Chapter 8 Properties of Solids and Liquids 8.1–8.60

Chapter 9 Oscillations and Waves 9.1–9.102

Chapter 10 Heat and Thermal Expansion 10.1–10.40

Chapter 11 Heat Transfer 11.1–11.20

Chapter 12 Thermodynamics 12.1–12.34

Chapter 13 Electrostatics 13.1–13.108

Chapter 14 Current Electricity 14.1–14.56

Chapter 15 Magnetism and Magnetic Effect of Current 15.1–15.66

Chapter 16 Electromagnetic Induction 16.1–16.58

Chapter 17 Electromagnetic Waves 17.1–17.12

Chapter 18 Ray Optics and Wave Optics 18.1–18.86

Chapter 19 Modern Physics 19.1–19.64

Chapter 20 Semiconductor and Communication 20.1–20.54

Mock Test Paper 1 M1.1–M1.6

Contents

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About the Series

A Complete Resource Book for JEE Main series is a must-have resource for students preparing for JEE Main examination

There are three separate books on Physics, Chemistry, and Mathematics; the main objective of this series is to strengthen the fundamental concepts and prepare students for various engineering entrance examinations It provides class-tested course material and numerical applications that will supplement any ready material available as student resource

To ensure high level of accuracy and practicality, this series has been authored by highly qualifi ed and experienced faculties

About the Book

It gives me immense pleasure to present this book A Complete Resource Book in Physics For JEE Main 2019 This book

will help the students in building the analytical and quantitative skills necessary to face the examination with confi dence

This title is designed as per the latest JEE Main syllabus, spread across 20 chapters It has been structured in an user friendly approach such that each chapter begins with topic-wise theory, followed by suffi cient solved examples and then practice questions The brain-map section in every chapter will help the students to revise the important formulae The chapter end exercises are structured in line with JEE questions; where ample number of questions on single choice correct question (SCQ), multiple-type correct questions (MCQ), assertion and reasoning, column matching, passage based and integer type questions are included for extensive practice Previous 15 years’ questions of JEE Main and AIEEE are also

added in every chapter Hints and Solutions at the end of every chapter will help the students to evaluate their concepts and

numerical applications

Series Features

• Complete coverage of topics along with ample number of solved examples.

• Includes various types of practice problems with complete solutions.

• Chapter-wise Previous 15 years’ AIEEE/JEE Main questions.

• Fully solved JEE Main 2017 and 2018 questions are included in the book.

• 5 Mock Tests based on JEE Main pattern.

• 5 Free Online Mock Tests as per recent JEE Main pattern.

I dedicate this book to my family for their immense support and love Special thanks to my parents for their support and encouragement and to my wife Pallawi and my sons Haardik and Saarthak for sustaining me throughout this project

I would like to express my heartfelt gratitude to the Pearson team, without them I would not have been able to bring out this book

Any suggestions and comments from the readers would be highly appreciated Please communicate to us if there are any errors, misprints or other such concerns

Sanjeev Kumar

Preface

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S No Chapters 07 08 09 10 11 12 13 14 15 16 17 18

5 Centre of Mass, Impulse and Momentum 1 2 1 2 – – – 1 2 2 – 2 2

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1 The density of a material in the shape of a cube is

determined by measuring three sides of the cube and

its mass If the relative errors in measuring the mass

and length are respectively 1.5% and 1%, the

maxi-mum error in determining the density is

2 All the graphs below are intended to represent the same

motion One of them does it incorrectly Pick it up

Velocity

3 Two masses m1 = 5 kg and m2 = 10 kg, connected by an

inextensible string over a frictionless pulley, are

mov-ing as shown in the figure The coefficient of friction of

horizontal surface is 0.15 The minimum weight m that

should be put on top of m2 to stop the motion is

(A) 18.3 kg (B) 27.3 kg (C) 43.3 kg (D) 10.3 kg

4 A particle is moving in a circular path of radius a

under the action of an attractive potential U k

5 In a collinear collision, a particle with an initial speed

v0 strikes a stationary particle of the same mass If the final total kinetic energy is 50% greater than the origi-nal kinetic energy, the magnitude of the relative veloc-ity between the two particles, after collision, is (A) v0

(C) v0

2

6 Seven identical circular planar discs, each of mass M

and radius R are welded symmetrically as shown The

moment of inertia of the arrangement about the axis

normal to the plane and passing through the point P is

O

P

(A) 19

22

MR

(C) 73

22

MR

7 From a uniform circular disc of radius R and mass 9M,

a small disc of radius R

3 is removed as shown in the

figure The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is

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8 A particle is moving with a uniform speed in a circular

orbit of radius R in a central force inversely

propor-tional to the nth power of R If the period of rotation of

the particle is T, then

(A) T ∝ R3/2 for any n (B) T ∝R n/2 1 +

(C) T ∝ R (n+1)/2 (D) T ∝ R n/2

9 A solid sphere of radius r made of a soft material of

bulk modulus K is surrounded by a liquid in a

cylindri-cal container A massless piston of area a floats on the

surface of the liquid, covering entire cross section of

cylindrical container When a mass m is placed on the

surface of the piston to compress the liquid, the

frac-tional decrement in the radius of the sphere, dr

10 Two moles of an ideal monoatomic gas occupies a

volume V at 27°C The gas expands adiabatically to a

volume 2V Calculate (a) the final temperature of the

gas and (b) change in its internal energy

(A) (a) 189 K (b) 2.7 kJ

(B) (a) 195 K (b) -2.7 kJ

(C) (a) 189 K (b) -2.7 kJ

(D) (a) 195 K (b) 2.7 kJ

11 The mass of a hydrogen molecule is 3.32 × 10-27 kg

If 1023 hydrogen molecules strike, per second, a fixed

wall of area 2 cm2 at an angle of 45° to the normal, and

rebound elastically with a speed of 103 m/s, then the

pressure on the wall is nearly

(A) 2.35 × 103 N/m2 (B) 4.70 × 103 N/m2

(C) 2.35 × 102 N/m2 (D) 4.70 × 102 N/m2

12 A silver atom in a solid oscillates in simple harmonic

motion in some direction with a frequency of 1012/sec What is the force constant of the bonds connecting one atom with the other? (Mole wt of silver = 108 and Avogadro’s number = 6.02 × 1023 gm mole–1)

(A) 6.4 N/m (B) 7.1 N/m (C) 2.2 N/m (D) 5.5 N/m

13 A granite rod of 60 cm length is clamped at its middle

point and is set into longitudinal vibrations The sity of granite is 2.7 × 103 kg/m3 and its Young’s mod-ulus is 9.27 × 1010 Pa What will be the fundamental frequency of the longitudinal vibrations?

(A) 5 kHz (B) 2.5 kHz (C) 10 kHz (D) 7.5 kHz

14 Three concentric metal shells A, B and C of respective

radii a, b and c (a < b < c) have surface charge

densi-ties +σ, -σ and +σ respectively The potential of shell

b c

15 A parallel plate capacitor of capacitance 90 pF is

con-nected to a battery of emf 20 V If a dielectric material

of dielectric constant K=5

3 is inserted between the

plates, the magnitude of the induced charge will be (A) 1.2 nC (B) 0.3 nC

17 Two batteries with emf 12 V and 13 V are connected

in parallel across a load resistor of 10 Ω The internal resistances of the two batteries are 1 Ω and 2 Ω respec-tively The voltage across the load lies between (A) 11.6 V and 11.7 V (B) 11.5 V and 11.6 V (C) 11.4 V and 11.5 V (D) 11.7 V and 11.8 V

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18 An electron, a proton and an alpha particle having the

same kinetic energy are moving in circular orbits of

radii r e , r p , rα respectively in a uniform magnetic field

B The relation between r e , r p , rα is

(A) r e > r p = rα

(B) r e < r p = rα

(C) r e < r p < rα

(D) r e < rα < r p

19 The dipole moment of a circular loop carrying a

cur-rent I is m and the magnetic field at the centre of the

loop is B1 When the dipole moment is doubled by

keeping the current constant the magnetic field at the

centre of the loop is B2 The ratio B

B

1 2

LC the current exhibits

reso-nance The quality factor, Q is given by

(A) ω0L

ω0R L

21 An EM wave from air enters a medium The

elec-tric fields are

E2 =E02 xˆ cos[k(2z ct− )] in medium, where the

wave number k and frequency v refer to their values in

air The medium is non-magnetic If ∈r1 and ∈r2refer

to relative permittivities of air and medium

respec-tively, which of the following options is correct?

(A) ∈

r r

1 2

r r

1 22

(C) ∈

r r

1 2

12

22 Unpolarized light of intensity I passes through an ideal

polarizer A Another identical polarizer B is placed

behind A The intensity of light beyond B is found to

be 1

2. Now another identical polarizer C is placed

between A and B The intensity beyond B is now found

to be 1

8. The angle between polarizer A and C is

23 The angular width of the central maximum at a single

slit diffraction pattern is 60° The width of the slit is

1 μm The slit is illuminated by monochromatic plane waves If another slit of same width is made near it, Young’s fringes can be observed on a screen placed at

a distance 50 cm from the slits If the observed fringe width is 1 cm, what is slit separation distance? (i.e., distance between the centres of each slit.)

24 An electron from various excited states of hydrogen

atom emit radiation to come to the ground state Let

λn, λg be the de Broglie wavelength of the electron in

nth state and the ground state respectively Let Λn be the wavelength of the emitted photon in the transition

from the nth state to the ground state For large n, (A, B

26 It is found that if a neutron suffers an elastic collinear

collision with deuterium at rest, fractional loss of its

energy is P d; while for its similar collision with carbon

nucleus at rest, fractional loss of energy is P c The

val-ues of P d and P c are respectively (A) (0.89, 0.28) (B) (0.28, 0.89) (C) (0, 0) (D) (0, 1)

27 The reading of the ammeter for a silicon diode in the

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28 A telephonic communication service is working at

carrier frequency of 10 GHz Only 10% of it is utilized

for transmission How many telephonic channels can

be transmitted simultaneously if each channel requires

a bandwidth of 5 kHz?

(A) 2 × 103 (B) 2 × 104

(C) 2 × 105 (D) 2 × 106

29 In a potentiometer experiment, it is found that no current

passes through the galvanometer when the terminal of

the cell are connected across 52 cm of the

potentiom-eter wire If the cell is shunted by a resistance of 5 Ω,

a balance is found when the cell is connected across

40 cm of the wire Find the internal resistance of the cell

30 On interchanging the resistances, the balance point of

a meter bridge shifts to the left by 10 cm The tance of their series combination is 1 kΩ How much was the resistance on the left slot before interchanging the resistances?

(A) 990 Ω (B) 505 Ω

Answer keys

11 (A) 12 (B) 13 (A) 14 (B) 15 (A) 16 (B) 17 (B) 18 (B) 19 (C) 20 (A)

21 (C) 22 (C) 23 (A) 24 (A) 25 (D) 26 (A) 27 (C) 28 (C) 29 (B) 30 (C)

l l

∆ 3∆

p

m m

l l

100 100 3 100

= 1.5 + 3 × 1 = 4.5%

Hence, the correct option is (C).

2 Incorrect option is (A) because it represents two values of

velocity at the same instant, which is impossible.

Hence, the correct option is (A).

3 For stopping the motion:

Hence, the correct option is (B).

Hints and solutions

k r

2 2

2 2 2 2 0

5 Momentum conservation yields:

⇒v1 +v2 = 3v0

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Solving the two equations, we get

By parallel axis theorem:

Hence, the correct option is (D).

7 Mass per unit area = 9M2

2 2 2

9

9 9

k R

= ∆∆

V

P K

mg Ka

r

V V

mg Ka

1

∆

10 For adiabatic expansion

1 1 2π

13 For fundamental frequency:

4 = ⇒ 2l 4v = 2

f l

∴ =f v=

l

v l

1 2

2

10 3

.

kHz

14 Potential of shell B is given by:

b

q b

q c

1 4

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16 Wattless current = irms sinφ

= i o = ×

2

20 2

1 2 sinφ

1 2

mK qB

p p

α = 2 4× = 2

1

r

M M

r r

1 2 1 2 1 2 1 2

1 2

Also, r

r

B B

1 2 2 1

1 2

= =

B

1 2

r r

k k

1 2

1 2 2

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r r

K K

1

1 4

2

22

2 cos

2 2

I=I θ

I I

2 2

26 Velocity of neutron after the collision is:

Fractional loss in KE = −

1 2

1 2 1 2

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For collision with deuterium, m1 =m m; 2 = 2m

27 For a silicon diode the barrier potential is 0.7 volts

5

29 Let the voltage gradient be λ

+ = ⇒ =

5 5

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1 An observer is moving with half the speed of light

towards a stationary microwave source, emitting

waves at frequency 10 GHz What is the frequency of

the microwave measured by the observer?

(Speed of light, c = 3 × 108 m/s)

(A) 12.1 GHz (B) 17.3 GHz

(C) 15.3 GHz (D) 10.1 GHz

2 The following observations were taken for

determin-ing surface tension T of water by capillary method:

Diameter of capillary, D = 1.25 × 10-2 m; rise of water,

h = 1.45 × 10-2 m Using g = 9.80 m/s2 and the

3 Some energy levels of a molecule are shown in the

figure The ratio of the wavelengths, r=λ λ1/ ,2 is

4 A body of mass, m =10-2 kg is moving in a medium

and experiences a frictional force, F = -kv2 Its initial

speed is, v0= 10 m/s If, after 10 seconds, its energy is

1

8mv ,0 then the value of k will be

(A) 10-3 kg/s (B) 10-4 kg/m

(C) 10-1 kg/m/s (D) 10-3 kg/m

5 C p and C v are specific heats at constant pressure and

constant volume, respectively It is observed that

C p – C v = a, for hydrogen gas; C p – C v = b, for nitrogen gas The correct relation between a and b is

(A) a = b (B) a = 14b (C) a = 28b (D) a= 1b

14

6 The moment of inertia of a uniform cylinder of length

l and radius R about its perpendicular bisector is I What will be the ratio of l to R, such that the moment

7 A radioactive nucleus A with a half life T, decays into

a nucleus B At t = 0, there is no nucleus B At some value of t, the ratio of the number of B to that of A is 0.3 Then, t is given by

(A) t T= log

log

1 32

8 Which of the following statements is false?

(A) In a balanced wheatstone bridge if the cell and

the galvanometer are exchanged, the null point is disturbed

(B) A rheostat can be used as a potential divider

(C) Kirchhoff ’s second law represents energy

conservation

(D) Wheatstone bridge is the most sensitive when

all the four resistances are of the same order of magnitude

9 A capacitance of 2 μF is required in an electrical circuit across a potential difference of 1.0 kV A large number

of 1 μF capacitors are available which can withstand a potential difference of not more than 300 V The mini-mum number of capacitors required to achieve this, is

10 In the given circuit diagram, when the current reaches

steady state in the circuit, the charge on the capacitor

of capacitance C will be

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r2

r

C E

(A) CE r

r r

1 2

r

r r

2 2

(C) CE r

r r

1 1

12 In amplitude modulation, sinusoidal carrier frequency

used is denoted by ωc and the signal frequency is

denoted by ωm The bandwidth (∆ωm) of the signal is

such that, ∆ωm << ωc Which of the following

frequen-cies is not contained in the modulated wave?

(A) ωc (B) ωm + ωc

(C) ωc- ωm (D) ωm

13 In a common emitter amplifier circuit using an n-p-n

transistor, the phase difference between the input and

the output voltages will be:

14 A copper ball of mass 100 gm is at a temperature

T It is dropped in a copper calorimeter of mass 100

gm, filled with 170 gm of water at room temperature

Subsequently, the temperature of the system is found

15 In a Young’s double slit experiment, slits are separated

by 0.5 mm, and the screen is placed 150 cm away A

beam of light consisting of two wavelengths 650 nm

and 520 nm, is used to obtain interference fringes on the screen The least distance from the common cen-tral maximum to the point where the bright fringes due

to both the wavelengths coincide, is (A) 7.8 mm (B) 9.75 mm (C) 15.6 mm (D) 1.56 mm

16 An electric dipole has a fixed dipole moment p,

which makes an angle θ with respect to X-axis When

subjected to an electric field, E1 Ei

17 A slender uniform rod of mass M and length l is

piv-oted at one end, so that it can rotate in a vertical plane (see figure) There is negligible friction at the pivot The free end is held vertically above the pivot and then released The angular acceleration of the rod when it makes an angle θ with the vertical, is

18 An external pressure P is applied on a cube at 0°C,

so that it is equally compressed from all sides K is the bulk modulus of the material of the cube and α is its coefficient of linear expansion Suppose we want

to bring the cube to its original size by heating The temperature should be raised by

19 A diverging lens with magnitude of focal length 25 cm is

placed at a distance of 15 cm from a converging lens of magnitude of focal length 20 cm A beam of parallel light falls on the diverging lens The final image formed is

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(A) Virtual and at a distance of 40 cm from

20 An electron beam is accelerated by a potential

differ-ence V to hit a metallic target to produce X-rays It

produces continuous as well as characteristic X-rays

If λmin is the smallest possible wavelength of X-rays

in the spectrum, the variation of log λmin with log V is

21 The temperature of an open room of volume 30 m3

increases from 17°C to 27°C due to the sunshine The

atmospheric pressure in the room remains 1 × 105 Pa

If n i and n f are the number of molecules in the room

before and after heating, then n f – n i will be

(A) 1.38 × 1023 (B) 2.5 × 1025

(C) –2.5 × 1025 (D) –1.61 × 1023

22 In a coil of resistance 100 Ω, a current is induced by

changing the magnetic flux through it, as shown in the

figure The magnitude of change in flux through the coil is

10

0.5 sec

Current (amp)

Time

(A) 225 Wb (B) 250 Wb (C) 275 Wb (D) 200 Wb

23 When a current of 5 mA is passed through a

galva-nometer having a coil of resistance 15 Ω, it shows full scale deflection The value of the resistance to be put

in series with the galvanometer to convert it into a voltmeter of range 0–10 V is

(A) 2.045 × 103 Ω (B) 2.535 × 103Ω (C) 4.005 × 103 Ω (D) 1.985 × 103Ω

24 A time dependent force, F = 6t acts on a particle of

mass 1 kg If the particle starts from rest, then the work done by the force during the first 1 second will be

25 A magnetic needle of magnetic moment 6.7 × 10–2

Am2 and moment of inertia 7.5 × 10–6 kgm2 is ing simple harmonic oscillations in a magnetic field of

perform-0.01T Time taken for 10 complete oscillations is

(A) 8.89 s (B) 6.98 s (C) 8.76 s (D) 6.65 s

26 The variation of acceleration due to gravity g with

dis-tance d from centre of the earth, is best represented by (R = Earth’s radius)

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27 A body is thrown vertically upwards Which one of

the following graphs correctly represent the velocity

vs time?

(A)

t v

(B)

t v

(C)

t v

(D)

t v

28 A particle A of mass m and initial velocity of v collides

with a particle B of massm

2 which is at rest The

colli-sion is head on, and elastic The ratio of the de Broglie wavelengths λA to λB after the collision is

(A) λ

λ

A B

= 2

(B) λ

λ

A B

=23

(C) λ

λ

A B

=12

(D) λ

λ

A B

=13

29 A particle is executing simple harmonic motion with

a time period T At time t = 0, it is at its position of equilibrium The kinetic energy-time graph of the par-ticle will look like

30 A man grows into a giant such that his linear

dimen-sions increase by a factor of 9 Assuming that his sity remains same, the stress in the leg will change by

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.

0 10

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8 Conceptual.

9 To get capacitance of 2 μF, such that potential difference

across any one should not be more than 300 V, can be

∴ The minimum number of capacitors required = 4 × 8 = 32.

10 In steady state no current will pass through capacitor.

Current through cell, i E

r r

= + ( 2 )

Potential difference across capacitor = Potential difference

( ).

Taking voltage of point A as = 0

Then voltage at other points can be written as shown in figure

Hence voltage across all resistance is zero.

14 From principle of calorimetry, Heat lost by copper sphere = Heat gained by calorimeter

+ Heat gained by water

⇒ 100 × 0.1 × (T – 75) = 100 × 0.1 × 45 + 170 × 1 × 45

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⇒ T – 75 = 810

⇒ T = 885°C.

15 Let n1th order maxima of wavelength 520 nm coincides with

n2th order maxima of wavelength 650 nm.

That is, n1 × 520 = n2 × 650

n

1 2

5 4

So, 5th order maxima of wavelength 520 nm coincides with

4th order maxima of wavelength 650 nm Then, the least

dis-tance from central maxima is,

v = +40 cm (using lens formula)

⇒ Real image will be formed at a distance of 40 cm from L2 ,

i.e., converging lens (right of L2).

It is a straight line with -ve slope.

21 Using, PV = NRT (N = Number of moles), we get

5

R

Trang 24

v=6t = t

2 3

2 2

For 10 oscillations, time = 10T = 6.65 s.

A B B A

=λ =

λ 233

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A B

30 Stress,σ =F = =ρ = ρ

A

mg A

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Physics, technology and society, SI units, Fundamental and derived units Least count, Accuracy and

precision of measuring instruments, Errors in measurement, Signifi cant fi gures Dimensions of Physical

quantities, Dimensional analysis and its applications Basic Mathematics

PhYSiCAl QUAntitieS

The quantities which can be measured by an instrument

and by means of which we can describe the laws of physics

are called physical quantities Till class X we have studied

many physical quantities

For example, length, velocity, acceleration, force,

time, pressure, mass, density etc

Fundamental or

Basic Quantities

Derived Quantities

Supplementary Quantities Physical quantities are of three types

Fundamental (Basic) Quantities

These are the elementary quantities which covers the entire

span of physics

Any other quantities can be derived from these

All the basic quantities are chosen such that they

should be diff erent, that means independent of each other

(i.e., distance, time and velocity cannot be chosen as basic

quantities as v =d

t ) An International Organization named

CGPM: General Conference on weight and measures chose

seven physical quantities as basic or fundamental

(A)

Luminous Intensity

(Cd)

Amount of Substance (mol)

These are the elementary quantities (in our planet) that’s why chosen as basic quantities

In fact any set of independent quantities can be chosen as basic quantities by which all other physical quan-tities can be derived

(Cd)

Luminous

(R)

Resistance (mol)mole

Can be chosen as basic quantities (on some other planet, these might also be used as basic quantities)

But

(L)

Length Area(A) Velocity(v)

cannot be used as basic quantities asArea = (Length)2 so they are not independent

Chapter Highlights

Unit and Dimension

1

Trang 27

Derived Quantities

Physical quantities which can be expressed in terms of

basic quantities (M, L, T ) are called derived quantities.

Here [M1 L1 T –1] is called dimensional formula of

momentum, and we can say that momentum has

1 Dimension in M (mass)

1 Dimension in L (meter)

and –1 Dimension in T (time)

The representation of any quantity in terms of basic

quan-tities (M, L, T ) is called dimensional formula and in the

representation, the powers of the basic quantities are called

dimensions

Supplementary Quantities

Besides seven fundamental quantities two supplementary

quantities are also defined They are

• Plane angle (The angle between two lines)

• Solid angle

θ

DimenSionS

• Height, width, radius, displacement etc are a kind of

length So we can say that their dimension is [L]

here [Height] can be read as “Dimension of Height”

• Area = Length × Width

[Area] = [Length] × [Width]

Here p is not a kind of length or mass or time so p

shouldn’t affect the dimension of Area

Hence its dimension should be 1 (M0L0T 0) and we can say that it is dimensionless From similar logic we can say that all the numbers are dimensionless

[200]

[M0 L0 T0 ] = 1 Dimensionless

[–1]

[3]

1 2

• [Volume] = [Length] × [Height]

So dimension of volume will be always [L3] whether it

is volume of a cuboid or volume of sphere

Dimension of a physical quantity will be same, it doesn’t depend on which formula we are using for that quantity

• Work or Energy = force × displacement

[Work] = [force] [displacement]

= [M1L1T –2] [L]

= [M1L2T –2]

Trang 28

0 0 − 1

= [M0L0T –2]

• Torque = Force × Arm length

[Torque] = [force] × [arm length]

If two bodies of mass m1 and m2 are placed at r distance,

both feel gravitational attraction force, whose value is,

Gravitational force F g=Gm m

r

1 2 2

where G is a constant called Gravitational constant

[ ]

G m m r

2

Specific Heat Capacity

To increase the temperature of a body by DT, Heat required

If any spherical ball of radius r moves with velocity v in a

viscous Liquid, then viscous force acting on it is given by

Planck’s Constant

If light of frequency u is falling, energy of a photon is given by

E = hu Here h = Planck’s constant

Some Special Features of Dimensions

Suppose in any formula, (L + a) term is coming (where L

is length) As length can be added only with a length, so a should also be a kind of length

Similarly, consider a term (F – b ) where F is force A force

can be added or subtracted with a force only and give rise

to a third force So b should be a kind of force and its result

(F – b ) should also be a kind of force

Trang 29

Rule No 1: One quantity can be added / subtracted with a

similar quantity only and give rise to the similar quantity

SolveD exAmPleS

1 a

t2 = Fv + b

x2

Find dimension formula for [a] and [b ]

(here t = time, F = force, v = velocity, x = distance)

−

⎛

⎝⎜ 2⎞⎠⎟(V – b) = nRT Find the dimensions of a and b, where P is gas pressure,

v = volume of gas T = temperature of gas

Whatever comes in sin ( ) is dimensionless and entire [sin ( )] is also dimensionless

⇒

sin(- - -)

Dimensionless Dimensionless

Similarly:

⇒

cos(- - -)

⇒

tan(- - -)

⇒

2(- - -)

⇒

e(- - -)

⇒

loge(- - -)

3 a = F

V2 sin (bt)

(here v = velocity, F = force, t = time)

Find the dimension of a and b

Solution:

sin ( t)β

β β

Dimensionless

⇒ [ ] [t] = 1 [ ] = [T−1 ] Dimensionless

Trang 30

4 a =Fv22

b loge

22

(1) To Check the Correctness of the Formula

If the dimensions of the L.H.S and R.H.S are same, then we

can say that this equation is at least dimensionally correct

So this equation may be correct

But if dimensions of L.H.S and R.H.S is not same

then the equation is not even dimensionally correct

we have to check whether it is correct or not

Dimension of L.H.S is

[F] = [M1

L1T –2]Dimension of R.H.S is

[ ] [ ][ ]

m v r

2

= [ ]M LT

L

[ ][ ]

1 2

−

= [M1

L1T – 2]

So this equation is at least dimensional correct

⇒ We can say that this equation may be correct

6 A Boomerang has mass m surface Area A, radius of

curvature of lower surface = r and it is moving with velocity v in air of density r The resistive force on it should be

r

(A) 2rvA2

r log

rp

m Ar

A m

Ar m

⎛

⎝⎜ ⎞⎠⎟ (D)

2 2 2

rv A

r log

rp

Ar m

⎛

⎝⎜ ⎞⎠⎟

Solution:

Only C is dimensionally correct.

(2) We can Derive a New Formula Roughly

If a quantity depends on many parameters, we can mate, to what extent, the quantity depends on the given parameters!

esti-SolveD exAmPleS

7 Derive the formula of time period of simple pendulum

if

Mass pendulum

(m)

Length of the string ()

Gravitational acceleration

(g)

Time period of a simple pendulum

depends on

g m

Trang 31

The quantity “Some number” can be found

experimen-tally Measure the length of a pendulum and oscillate

it, find its time period by stopwatch

9 If velocity (v), force (F) and time (T) are chosen

as fundamental quantities, express (i) mass and

(ii) energy in terms of v, F and T

Trang 32

limitations of Dimensional Analysis

From Dimensional analysis

1 Dimensional analysis doesn’t give information about

the “Some Number” The dimensional constant

2 This method is useful only when a physical quantity

depends on other quantities by multiplication and

power relations

(i.e., f = x a y b z c)

3 It fails if a physical quantity depends on sum or

differ-ence of two quantities

(i.e f = x + y – z)

i.e., we cannot get the relation

S = ut + 1

2at

2 from dimensional analysis

4 This method will not work if a quantity depends on

another quantity as sine or cosine, logarithmic or

expo-nential relation The method works only if the

depen-dence is by power functions

We equate the powers of M, L and T hence we get only

three equations So we can have only three variables

(only three dependent quantities)

So dimensional analysis will work only if the quantity depends only on three parameters, not move than that

SolveD exAmPle

11 Can Pressure (P), density ( r) and velocity (v) be taken

as fundamental quantities?

Solution:

P, r and v are not independent, they can be related as

P = rv2, so they cannot be taken as fundamental variables

To check whether the P, r, and v are dependent

or not, we can also use the following mathematical method:

[P] = [M1L–1T–2] [r] = [M1L–3 T0]

[v] = [M0L1T–1] Check the determinate of their powers:

Measurement of any physical quantity is expressed in terms of

an internationally accepted certain basic standard called unit

Si units

In 1971, an international Organization “CGPM” (General Conference on weight and Measure) decided the standard units, which are internationally accepted These units are called SI units (International system of units)

SI Units of Basic Quantities

Base Quantity

SI Units

Length meter m The meter is the length of the path travelled by light in vacuum during

a time interval of 1/299, 792, 458 of a second (1983) Mass kilogram kg The kilogram is equal to the mass of the international prototype of

the kilogram (a platinum-iridium alloy cylinder) kept at International Bureau of Weights and Measures, at Sevres, near Paris,

France. (1889)

(Continued )

Trang 33

Base Quantity

SI Units

Time second s The second is the duration of 9, 192, 631, 770 periods of the radiation

corresponding to the transition between the two hyperfine levels of the ground state of the cesium-133 atom (1967)

Electric Current ampere A The ampere is that constant current which, if maintained in two

straight parallel conductors of infinite length, of negligible circular cross-section, and placed 1 metre apart in vacuum, would produce between these conductors a force equal to 2 × 10 –7 Newton per metre of length (1948)

Thermodynamic

Temperature

kelvin K The kelvin is the fraction 1/273.16 of the thermodynamic temperature

of the triple point of water (1967) Amount of

Substance

mole mol The mole is the amount of substance of a system, which contains

as many elementary entities as there are atoms in 0.012 kilogram of carbon-12 (1971)

Luminous Intensity candela Cd The candela is the luminous intensity, in a given direction, of a source

that emits monochromatic radiation of frequency 540 × 10 12 hertz and that has a radiant intensity in that direction of 1/1683 watt per steradian (1979).

Two Supplementary Units were also Defined

• Plane angle – Unit = radian (rad)

• Solid angle – Unit = Steradian (sr)

Other Classification

If a quality involves only length, then mass and time

(quan-tities in mechanics), and its unit can be written in MKS,

CGS or FPS system

MKS system meter kg sec

CGS system

cm gram sec

FPS system footpoundsec

• For MKS system: In this system Length, mass and

time are expressed in meter, kg and sec respectively

It comes under SI system

• For CGS system: In this system, length, mass and

time are expressed in cm, gram and sec respectively

• For FPS system: In this system, length, mass and time

are measured in foot, pound and sec respectively

SI Units of Derived Quantities

• Velocity =displacement

time

metersecond

→

So unit of velocity will be m/s

• Acceleration =change in velocity

Unit = J/s called watt (w)

Units of Some Physical Constants

• Unit of “Universal Gravitational Constant” (G)

F = G m m

r

( )(1 2)2

• Unit of specific heat capacity (S)

04

i i r

1 2 2

N

m= m01( )

( )( )

A m

(A)2

Unit of m0 =N m⋅

A2

Trang 34

SI Prefi x

Suppose distance between kota to Jaipur is 3000 m so

d = 3000 m = 3 × 1000 m

kilo(k)

= 3 km (here k is the prefi x used for 1000 (103))

Suppose thickness of a wire is 0.05 m

d = 0.05 m = 5 × 10 −2 m

centi(c)

= 5 cm (here c is the prefi x used for (10–2))

Similarly, the magnitude of physical quantities is over a

wide range So in order to express the very large magnitude

as well as very small magnitude more compactly, “CGPM”

recommended some standard prefi xes for certain power

= (5)(103g)(1002 cm)

= 5 × 10 5g cm

s3 (in CGS system) This unit g cm

3 2

Trang 35

SI Derived Units, Named After the Scientist

Expression in terms of base units

S, W A/V kg –1 m –2 s 3 A 2

11 Magnetic field Tesla T Wb/m 2 kg s –2 A –1

12 Magnetic flux Weber Wb V s or J/A kg m 2 s –2 A –1

13 Inductance Henry H Wb/A kg m 2 s –2 A –2

Trang 36

Some SI Units Expressed in Terms of the Special

Names and also in Terms of Base Units

Physical Quantity

SI Units

In terms of special names In terms of base units

we can say that if the unit is increased to 100 times

(cm → m), the numerical value became

we can also tell if in a formal way like the following

Magnitude of a physical quantity = (Its Numerical value) (unit)

unit.

19 Force acting on a particle is 5N If unit of length and

time is doubled and unit of mass is halved than the numerical value of the force in the new unit will be

4 timesHence the numerical value of the force will be 4 times

SiGniFiCAnt FiGUReS

Suppose you want to measure the length of a sheet of paper with an ordinary scale Place the zero mark of the scale exactly at one end of the sheet and read the mark at other end You may obtain a doubtful digit It means that exact reading is not possible It can be understood from the Fig. 1.1 shown below

Trang 37

Paper edge

27 28

Fig 1.1

The end of the sheet lies between 27.9 and 28.0 cm Then

you can estimate the distance between 27.9 cm and end of

the sheet in this way You mentally divide the 1 small

divi-sion into 10 equal parts and guess on which part the edge is

falling You may note down the reading as 27.96 cm In this

reading the digits 2, 7 and 9 are certain but 6 is doubtful All

these digits including doubtful digit are called significant

digits The rightmost or doubtful digit is called the least

sig-nificant digit and the leftmost digit is called the most

signif-icant digit We can define in this way the reliable digits plus

the 1st uncertain digit which are called significant digits or

significant figure So we can say, significant figures indicate

the precision of measurement and it depends on the least

count of the measuring instrument

the Rules for Determining the number of

Significant Figures are as follows

1 All the non-zero digits are significant.

Example: 156, 78 contains five significant figures.

2

All the zeros between two non-zero digits are signif-icant no matter where the decimal point is

Example: 108.006 contain six significant figures.

3 If the number is less than 1, the zeros on the right of

decimal point but to the left of 1st non-zero digit are

not significant

Example: In 0.002308 the under lined zeros are not

significant

4 All the zeros to the right of the last non-zero digit

(trailing zeros) in a number without a decimal point

are not significant, unless they come from experiment

Thus 123 m = 12300 cm = 123000 mm has three

significant figures The trailing zeros are not significant

But if these are obtained from a measurement, they are

1 the length 2.308 cm has four significant figures, but

in different units, the same value can be written as 0.02308 m or 23.08 mm All these number have the same number of significant figures It shows that loca-tion of decimal point does not matter in determining the number of significant figures

When there are zeros at the right end of the number, then there may be some confusion

2 If length is 500 mm and we don’t know least count of

the measuring instrument, then we can’t be sure that last digits (zeros) are significant or not

Scientific Notation

If the scale had marking only at each meter, then the digit 5 can be obtained by eye approximation So only 5 is signifi-cant figure, but if the markings are at each centimetre, then only 5,0 of the reading will be significant If the scale used have marking in millimetres, all three digits 5,00 are signif-icant To remove such ambiguities in determining the num-ber of significant figures, the best way is to report every measurement in scientific notation, every number should

express as a × 10b where a is between 1 and 10 and b is any +ve or –ve power of 10, and decimal is placed after the first digit

Now the confusion mention above can be removed

4.700 m = 4.700 × 102 cm = 4.700 × 103 mmHere power of 10 is irrelevant to the determination of sig-nificant figures

Significant Figure in Algebraic operation

To know the number of significant figures after an algebraic operation (Addition, subtraction, multiplication and divi-sion) certain rules can be followed which are as follows

1 In multiplication or division, the number of significant

digits in the final result should be equal to the number

of significant digits in the quantity, which has the imum number of significant digits

Example: If mass of an object measured is 4.237 gm

(four significant figures) and its volume is measured to

be 2.51 (three significant figure) cm3, then its density

= massvolume = 4 237

2 57

= 1.6486 gm/cm3 But it should be

up to three significant digits Density = 1.65 gm/cm3

2 In addition or subtraction the final result should retain

as many decimal places as are there in the number with the least decimal place

Trang 38

But the least precise measurement (227.2) gm is

cor-rect to only one decimal place So final should be

rounded off to one decimal place

So sum will be 663.8 gm

In subtraction also we follow the similar rule

Rounding off the Uncertain Digit

(least Significant Digit)

The least significant digit is rounded according to the rules

given below

1 If the digit next to the least significant (Uncertain) digit

is more than 5, the digit to be rounded is increased

by 1

2 If the digit next to the rounded one is less than 5, the

digit to be rounded is left unchanged

3 If the digit next to the rounded one is 5, then the digit

to be rounded is increased by 1 if it is odd and left

unchanged if it is even

The insignificant digits are dropped from the result

if they appear after the decimal point Zero replaces

them if they appear to the left of the decimal point

SolveD exAmPleS

21 Suppose we have to round off three significant digits to

15462

Solution:

In 15462, third significant digit is 4 This digit is to be

rounded The digit next to it is 6, which is greater than

5 The third digit should therefore be increased by 1

The digits to be dropped should be replaced by zeros,

because they appear to the left of decimal point thus

15462 becomes 15500 on rounding to three significant

figures

22 Round of the following numbers to three significant

digits (A) 14.745 (B) 14.750 (C) 14.650 × 1012

Solution:

(A) The third significant digit in 14.745 is 7 The

num-ber next to it is less than 5 So 14.745 becomes 14.7 on rounding to three significant digits

(B) 14.750 will become 14.8 because the digit to be

rounded is odd and the digit next to it is 5

(C) 14.650 × 1012 will become 14.6 × 1012 because the

digit to be rounded is even and the digit next to it

is 5

eRRoR AnAlYSiS in meASURementS

Measurement is an important aspect of physics Whenever

we want to know about a physical quantity, we take its measurement first of all

Instruments used in measurement are called ing instruments

various Cause of errors in measurement

Least Count Error

The least count error is the error associated with the tion of the instrument Least count may not be sufficiently small The maximum possible error is equal to the least count

resolu-Instrumental Error

This is due to faulty calibration or change in conditions (e.g thermal expansion of a measuring scale) An instrument may also have zero error A correction has to be applied

Trang 39

Systematic Error

The systematic errors are those errors that tend to be in

one direction, either positive or negative Errors due to air

buoyancy in weighing and radiation loss in calorimetry are

systematic errors They can be eliminated by manipulation

Some of the sources of systematic errors are:

possible value of the quantity under the given conditions of

measurement

Absolute Error

The magnitude of the difference between the true value of

the quantity and the individual measurement value is called

the absolute error of the measurement

(Arithmetic mean is taken as true value of number of

The absolute error may be positive or negative

Absolute Mean Error

The average of the mod of errors is called the average or

mean error

DXmean = (|DX1|+ |DX2|+ |DX3| … |DX n |)/n

Relative Error

The relative error is the ratio of the mean absolute error

DXmean to the mean value Xmean of the quantity measured

Relative error = DXmean /Xmean

Percentage Error

When the relative error is expressed in percent, it is called

as percentage errorThus, percentage error = (DXmean /Xmean) × 100%

Combination of errors

When we perform some experiment, different observations are used to get a result using algebraic operations like addi-tion, subtraction, multiplication, division etc

Now we need to calculate the errors in combination

of various mathematical operations

Errors in Sum or Difference

\ Elongation = (1.6 ± 0.8) cm

Errors in Product

Trang 40

Then (X ±DX)=(A±DA B) ( ±D B)

A A

B B

A A

B B

A A

B B

A A

B B

A A

B B

= ±⎛⎝⎜ + ⎞⎠⎟

Therefore maximum fractional error in product of two (or

more) quantities is equal to sum of fractional errors in the

individual quantities

SolveD exAmPle

24 The measures of the lengths of a rectangle are l = (3.00

± 0.01) cm and breadth b = (2.00 ± 0.02) cm What is

the area of the rectangle?

± 0.08 cm2 Area = (6.00 ± 0.08) cm2

Errors in Division

B

=

B

B as X

A B

A A

B B

A B

B B

∓

Here, D DA

A

B B

⋅ is small quantity, so can be neglected

Therefore, ±DX = ±D D

X

A A

B B

∓ Possible values of DX

X are

A

B B

A A

B B

A A

B B

−

⎛

⎝⎜ ⎞⎠⎟,⎛⎝⎜ + ⎞⎠⎟,⎛⎝⎜− − ⎞⎠⎟ and ⎛⎝⎜−D +D ⎞⎠⎟

A A

B B

Therefore, the maximum value of DX D D

X

A A

B B

= ±⎛ +

⎝⎜ ⎞⎠⎟

or, the maximum value of fractional error in division of two quantities is equal to the sum of fractional errors in the individual quantities

SolveD exAmPleS

25 The change in the velocity of a body is (12.5 ± 0.2) m/s

in a time (5.0 ± 0.3) s Find the average acceleration of the body within error limits

–2 Also

a

v v

t t

+

= Then ln( )X =nln( )A −mln( )B

Differentiating both sides, we get dX

dA

dB B

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